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Unformatted text preview: Physics H7ABC Welcome to the archival Web page for U.C. Berkeley's Physics H7ABC, Honors Physics for Scientists and Engineers, Fall 1998, Spring 1999, and Fall 1999. Instructor: (Prof.) Mark Strovink. I have a research web page, a standardized U.C. Berkeley web page, and a statement of research interests. Physics H7A (Mechanics and Vibrations) Problem set solutions initially composed by E.A. ("Ted") Baltz Graduate Student Instructors: David Bacon and Elizabeth Wu Physics H7B (Electromagnetism and Thermal Physics): Most problem set solutions composed by Peter Pebler Graduate Student Instructor: Robin BlumeKohout Physics H7C (Physical Optics and Modern Physics): Problem set solutions composed by Graduate Student Instructor: Derek Kimball Most documents linked on this page are in PDF format. They are typeset except where indicated. The documents are intended to be displayed by Adobe Acrobat [Reader], version 4 or later (version 3 may also work). (You may optimize Acrobat's rendering of equations by unchecking "Use Greek Text Below:" on FilePreferencesGeneral.) You may rightclick to download a single .pdf file (19.5 MB) that includes every displayable image on this Web page. Physics H7A (Mechanics and Vibrations) Texts: Kleppner/Kolenkow, An Introduction to Mechanics; French, Vibrations and Waves. General Information including schedules and rooms. Course Outline including all reading assignments. Questionnaire that was filled out by the students. Typos in Kleppner & Kolenkow Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Set Set Set Set Set Set Set Set Set Set Set 1 2 3 4 5 6 7 8 9 10 11 Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Set Set Set Set Set Set Set Set Set Set Set 1 2 3 4 5 6 7 8 9 10 11 http://d0lbln.lbl.gov/h7abcweb.htm (1 of 3) [1/5/2000 9:50:55 AM] Physics H7ABC Practice Exam 1 Exam 1 Exam 2 Final Exam Solution Solution Solution Solution to to to to Practice Exam 1 Exam 1 Exam 2 Final Exam You may rightclick to download a single .tar.gz archive (1.0 MB) that includes the source files required to build every H7A file linked above. Physics H7B (Electromagnetism and Thermal Physics) Text: Purcell, Electricity and Magnetism. General Information including schedules and rooms. Course Outline including all reading assignments. Special Relativity Notes* also used in H7C. Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Set Set Set Set Set Set Set Set Set Set Set Set Set 1 2 3 4 5 6 7 8 9 10 11 12 13 Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Set Set Set Set Set Set Set Set Set Set Set Set Set 1 2 3 4 5 6 7 8 9 10 11 12 13 Midterm Final Exam Solution to Midterm Solution to Final Exam* * handwritten You may rightclick to download a single .tar.gz archive (20 MB) containing the source files required to build every typeset H7B file linked above. Physics H7C (Physical Optics and Modern Physics) Texts: Fowles, Introduction to Modern Optics; Rohlf, Modern Physics from alpha to Z0. General Information including schedules and rooms. Course Outline including all reading assignments. Note on H7C texts, required and supplementary. Errata for Fowles, Introduction to Modern Optics, second edition. Special Relativity Notes* also used in H7B. http://d0lbln.lbl.gov/h7abcweb.htm (2 of 3) [1/5/2000 9:50:55 AM] Physics H7ABC Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem Set Set Set Set Set Set Set Set Set Set Set Set 1 2 3 4 5 6 7 8 9 10 11 12 Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Set Set Set Set Set Set Set Set Set Set Set Set 1 2 3 4 5 6 7 8 9 10 11 12 Midterm 1 Midterm 2 Final Exam Solution to Midterm 1 Solution to Midterm 2 Solution to Final Exam * handwritten You may rightclick to download a single .tar.gz archive (0.5 MB) containing the source files required to build every typeset H7C file linked above. http://d0lbln.lbl.gov/h7abcweb.htm (3 of 3) [1/5/2000 9:50:55 AM] University of California, Berkeley Physics H7A, Fall 1998 (Strovink) General Information (1 Sep 98) Instructors: Prof. Mark Strovink, 437 LeConte; (LBL) 4867087; (home, before 10) 4868079; (UC) 6429685. Email: [email protected] . Web: http://d0lbln.lbl.gov/ . Office hours: M 3:154:15, Tu 1011, Th 1011. Mr. David Bacon, 214 LeConte; (UC) 6425430; (home, before 1 AM) 6669867. Email: [email protected] . Office hours: W 122. Ms. Elizabeth Wu, 214 LeConte; (UC) 6425430. Email: [email protected] . Office hours in 262 LeConte: M 1011. You may also get help in the 7A Course Center, 262 LeConte. Lectures: TuTh 11:1012:30, 2 LeConte. Lecture attendance is essential, since not all of the course content can be found in either of the course texts. Labs: In the second week, in 270 LeConte, please enroll in one of only 3 special H7A lab sections [(A) #134, M 46; (B) #241, Th 46; or (C) #312, F 810]. Section 134 is taught by Ms. Wu and Sections 241 and 312 are taught by Mr. Bacon. If you can make more than one of these lab (and section, see below) slots, please attempt to enroll in the earliest of these lab slots. Depending on crowding, you may be asked to move to a later lab. During "off" weeks not requiring lab apparatus, your lab section will still meet, in (#134) 336 LeConte, (#241) 395 LeConte, or (#312) 335 LeConte. Discussion Sections: Beginning in the second week, please enroll in the only one of the 1 hr H7A discussion sections corresponding to your H7A lab section: (A) #134, W 12, 343 LeConte; (B) #241, Tu 45, 395 LeConte; (C) #312, W 89, 385 LeConte. You are especially encouraged to attend discussion section regularly. There you will learn techniques of problem solving, with particular application to the assigned exercises. Texts (both required): Kleppner/Kolenkow, An Introduction to Mechanics, (McGrawHill, 1973). A.P. French, Vibrations and Waves, Paper Edition (Norton, 1971). Problem Sets: Twelve problem sets are assigned and graded, with solutions provided as part of the Syllabus. They are due on Wednesday at 5 PM on weeks (including Thanksgiving) in which there is no exam, beginning in week 2. Deposit problem sets in the box labeled "H7A" outside 201 LeConte. You are encouraged to attempt all the problems. Students who do not do them find it almost impossible to learn the material and to succeed on the examinations. Work independently. Credit for collective effort, which is easy to identify, will be divided among the collectivists. Late papers will not be graded. Your lowest problem set score will be dropped, in lieu of due date extensions for any reason. Syllabus: H7A has two syllabus cards. The first card is mandatory; it will be collected at the time of the first inclass examination. This card pays for the experiment descriptions and instructions that you will receive from your GSI at the beginning of each laboratory. The second syllabus card is optional. It entitles you to pick up printed solutions to the problem set assignments from Copy Central. These solutions will also be made available on the Web. Both cards will be available for purchase at Copy Central beginning in the second week of class. Exams: There will be two 80minute inclass examinations and one 3hour final examination. Before confirming your enrollment in this class, please check that its final Exam Group 9 does not conflict with the Exam Group for any other class in which you are enrolled. Please verify that you will be available for both inclass examinations (Th 24 Sep and Th 5 Nov, 11:1012:30), and for the final examination, F 11 Dec, 58 PM. Except for unforeseeable emergencies, it will not be possible for these inclass or final exams to be rescheduled. Passing H7A requires passing the final exam. Grading: 35% inclass examinations; 20% problem sets; 40% final exam; 5% lab. Grading is not "curved"  it does not depend on your performance relative to that of your H7A classmates. Rather it is based on comparing your work to that of a generation of earlier lower division Berkeley physics students, with due allowance for educational trends. COURSE OUTLINE Week No. Week of... Lecture chapter Topic (K&K = Kleppner/Kolenkow, Intro. to Mechanics ) (French = Vibrations & Waves) (Feynman = Lectures on Physics Vol. II) Introduction; vectors, kinematics. Problem Due Set No. 5 PM on... Lab 1 24Aug K&K 1.18 2 31Aug 1.92.3 Motion in polar coordinates; Newton's laws; units. LABOR DAY HOLIDAY Application of Newton's laws; forces. Momentum; center of mass. none (do experiment in lab=expt) (have discussion in lab=disc) expt 1 Wed 2 Sep disc 2 3 9Sep expt 16Sep disc 3 7Sep 2.42.5 4 5 14Sep 21Sep 24Sep 28Sep 3.1Note3.1 4.16 Work; kinetic energy. EXAM 1 (covers PS 13) Potential energy; nonconservative forces; energy conservation; power; collisions. Angular momentum; fixed axis rotation; rotation with translation. Vector angular momentum; conservation thereof. Noninertial systems; fictitious forces. Rotating coordinate systems; equivalence principle. Central forces. Planetary motion. 8 Damped forced harmonic oscillator. Transient response. EXAM 2 (covers PS 18) Coupled oscillator; beats. Fourier expansion in normal modes. Waves: travelling, sinusoidal, modulated; phase and group velocity. Longitudinal waves; sound. Boundary reflections of waves; Doppler effect. THANKSGIVING HOLIDAY 28Oct 6 4.714 disc 4 30Sep disc 5 7Oct expt 6 14Oct expt 7 21Oct disc 7 5Oct 6.17 8 12Oct 7.15 8.18.4 8.5Note8.2 9.19.5 9.69.7 French 1015,4345,7789 6270,9296 9 19Oct 10 26Oct 11 2Nov 5Nov 9Nov disc 12 1927,119129 161170,189196 201209,213215, 228234 4562,209212 170178,274279 disc 9 11Nov expt 10 18Nov disc 11 25Nov disc 12 Fri 4 Dec 13 16Nov 14 23Nov 15 16 26,27Nov 30Nov Feynman II.21,2,3,4,5;II.401,2,3 Fluid statics and nonviscous dynamics. LAST LECTURE (review) 7Dec FINAL EXAM (Group 9) (covers PS 112) 11Dec 58 PM Physics H7A Fall 1998 (Strovink) University of California, Berkeley Physics H7A STUDENT QUESTIONNAIRE Please complete this questionnaire if you are considering enrolling in Physics H7A. Among other purposes, it will be used to make up the initial class roll. Are you qualified for this course? Your interest in Honors Physics is already a good indication that you are. It is unlikely that the issue of your eligibility will come into question. If it does, you will be interviewed promptly by the instructor. Fall 1998 (Strovink) Last name__________________________First_______________________Initial_________ Registration No.___________________ SAT scores: SAT I Quantitative______ SAT I Verbal______ SAT II Math 1______ SAT II Math 1C______ SAT II Math 2______ SAT II Physics______ AP Calculus Exam: AP Physics Exam: Grade______ Grade______ AB B BC C SAT II Math 2C______ (circle one) (circle one) Year (Freshman, etc.)_____________ Year of last math course_______ Where taken?________________________________ Course title and no._______________________________________________________ Course text_______________________________________ Year of last physics course____ Grade received_______ Where taken?________________________________ Course title and no._______________________________________________________ Course text_______________________________________ Grade received_______ List majors that you are considering at Berkeley_______________________________ ___________________________________________________________________________ Why are you (might you be) interested in Honors in place of standard lowerdivision physics at Berkeley?__________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ University of California, Berkeley Physics H7A Fall 1998 (Strovink) TYPOS IN KLEPPNER & KOLENKOW This is a list of typos in Kleppner & Kolenkow that have been brought to the attention of H7A F’98 staﬀ by students in the course. Thanks to them for pointing these out. Page 205, Example 5.2: In the equation below “the diﬀerential of f is”, replace the second dy with dx. Page 276, Note 6.2: In the equation following E = K + U , the ﬁrst ˙ term 1 l2 φ2 should be multiplied by m. 2 Page 392, Equation 9.22: The ﬁrst instance of x should be x2 instead. University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 1 1. Specify the properties of two vectors a and b such that (a.) a + b = c and a + b = c. (b.) a + b = a − b. (c.) a + b = c and a2 + b2 = c2 . (d.) a + b = a − b. (e.) a + b = a = b. 2. K&K problem 1.2 “Find the cosine of the angle between A = (3ˆ + ˆ + k) and B = ijˆ ˆ − 3ˆ − k).” ˆ (−2i j 3. The relation between Cartesian (x, y, z ) and spherical polar (r, θ, φ) coordinates is: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ. Consider two points on a sphere of radius R: (R, θ1 , φ1 ) and (R, θ2 , φ2 ). Use the dot product to ﬁnd the cosine of the angle θ12 between the two vectors which point to the origin from these two points. You should obtain: cos θ12 = cos θ1 cos θ2 + sin θ1 sin θ2 cos (φ1 − φ2 ). 4. New York has North Latitude (= 90◦ − θ) = 41◦ and West Longitude (= 360◦ − φ) = 74◦ . Sydney has South Latitude (= θ − 90◦ ) = 34◦ and East Longitude (= φ) = 151◦ . Take the earth to be a sphere of radius 6370 km; use the result of Problem 3. (a.) Find the length in km of an imaginary straight tunnel bored between New York and Sydney. (b.) Find the distance of the shortest possible lowaltitude ﬂight between the two cities. (Hint: The “great circle” distance along the surface of a sphere is just Rθ12 , where θ12 is the angle between the two points, measured in radians.) 5. K&K problem 1.6 “Prove the law of sines using the cross product. It should only take a couple of lines. (Hint: Consider the area of a triangle formed by A, B, C, where A + B + C = 0.)” 6. K&K problem 1.11 “Let A be an arbitrary vector and let n be a unit vector in some ﬁxed diˆ rection. Show that A = (A · n) n +(n × A) × n.” ˆˆ ˆ ˆ 7. If the air velocity (velocity with respect to the air) of an airplane is u, and the wind velocity with respect to the ground is w, then the ground velocity v of the airplane is v = u + w. An airplane ﬁles a straight course (with respect to the ground) from P to Q and then back to P , with air speed u which is always equal to a constant U0 , regardless of the wind. Find the time required for one round trip, under the following conditions: (a.) No wind. (b.) Wind of speed W0 blowing from P to Q. (c.) Wind of speed W0 blowing perpendicular to a line connecting P and Q. (d.) Wind of speed W0 blowing at an angle θ from a line connecting P and Q. (e.) Show that the round trip ﬂying time is always least for part (a.). (f.) What happens to the answers to (b.)(d.) when W0 > U0 ? Interpret this limiting condition physically. 8. A particle moves along the curve y = Ax2 such that its x position is given by x = Bt (t = time). (a.) Express the vector position r(t) of the particle in the form ˆ ˆ r(t) = if (t) + jg (t) [or xf (t) + yg (t)] ˆ ˆ where i and j [or x and y] are unit vectors, and f (t) and g (t) are functions of t. (b.) Find the (vector) velocity v(t) as a function of t. (c.) Find the (vector) acceleration a(t) as a function of t. (d.) Find the (scalar) speed v(t) as a function of t. (e.) Find the (vector) average velocity v(t0 ) between t = 0 and t = t0 where t0 is any positive time. 9. Below are some measurements taken on a stroboscopic photograph of a particle undergoing accelerated motion. The distance s is measured from a ﬁxed point, but the zero of time is set to coincide with the ﬁrst strobe ﬂash: time (sec) 0 1 2 3 4 5 6 7 8 distance (m) 0.56 0.84 1.17 1.57 2.00 2.53 3.08 3.71 4.39 Plot a straightline graph, based on these data, to show that they are ﬁtted by the equation s = a(t − t0 )2 /2, where a and t0 are constants, and extrapolate the line to evaluate t0 . 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 1 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. You may remember the law of cosines from trigonometry. It will be useful for several parts of this problem, so we will state it here. If the lengths of the sides of a triangle are a, b, and c, and the angle opposite the side c is ψ , then c = a + b − 2ab cos ψ
2 2 2 (b.) This part is simple. Just subtract the vector a from both sides to see that b = −b. The only way that this can happen is if b = 0, the zero vector. (c.) This part can also be done by the law of cosines. Like part (a.), we have the following two equations c2 = a2 + b2 + 2ab cos θ This is just the law of cosines again, where θ is the angle between a and b. The problem states that c2 = a2 + b2 Comparing this with the equation above, we ﬁnd that cos θ = 0. This happens at θ = ±90◦ . This means that the vectors must be perpendicular to each other. (d.) Yet again, we can use the law of cosines. If the angle between a and b is θ, then the angle between a and −b is 180◦ − θ. The lengths of the sum and diﬀerence are a + b2 = a2 + b2 + 2ab cos θ a − b2 = a2 + b2 − 2ab cos θ For these to be equal, we need cos θ = 0, which happens when θ = ±90◦ . Again, this means that the vectors are perpendicular. (e.) Guess what? Yup, law of cosines. We know that a = b = a + b. Adding a to b is going to look like two vectors stuck together to form two sides of a triangle. If the angle between the vectors is θ, the law of cosines gives a + b2 = 2a2 + 2a2 cos θ where we have used the fact that a and b have the same length. We also know that a + b = a. Using this we get a2 = 2a2 + 2a2 cos θ Dividing by a2 , we ﬁnd a condition on the angle θ 1 ⇒ θ = 120◦ cos θ = − 2 (a.) When two vectors add up to a third vector, the three vectors form a triangle. If the angle between a and b is θ, then the angle opposite the side formed by c is 180◦ − θ. The law of cosines then tells us that c2 = a2 + b2 − 2ab cos (180◦ − θ) From trigonometry, remember that cos (180◦ − θ) = − cos θ which gives c2 = a2 + b2 + 2ab cos θ We know that a + b = c. equation, we get Squaring this c2 = a2 + b2 + 2ab If we compare this with the equation above, we can see that cos θ has to be equal to one. This only happens when θ = 0◦ . What this means is that the two vectors are parallel to each other, and they point in the same direction. If the angle between them were 180◦ , then they would be parallel but point in opposite directions. 2 2. K&K problem 1.2 We can use the dot product, also known as the inner product, of two vectors here. Remember that A · B = AB cos θ where θ is the angle between the vectors. We can use the formula for computing the dot product from the vector components A · B = Ax Bx + Ay By + Az Bz The vectors are given as follows: A = 3ˆ + ˆ + k ijˆ ˆ − 3ˆ − k. Multiplying, we ﬁnd that ˆ and B = −2i j A · B = −10. We need the lengths of A and B. Remember that A2 = A · A. This tells us that A2 = 11 and B2 = 14. Dividing, we ﬁnd that cos θ = √ −10 = −0.805 11 · 14 as we expected in the ﬁrst place. Now we calculate the dot product. Let x1 be the vector to the ﬁrst point and x2 be the vector to the second point. We ﬁnd that x1 ·x2 = R2 (sin θ1 sin θ2 cos φ1 cos φ2 + sin θ1 sin θ2 sin φ1 sin φ2 + cos θ1 cos θ2 ) This can be simpliﬁed if we remember the formula for the cosine of a sum of two angles. cos(θ ± φ) = cos θ cos φ ∓ sin θ sin φ Using this formula, we get the result x1 ·x2 = R2 (sin θ1 sin θ2 cos(φ1 − φ2 ) + cos θ1 cos θ2 ) To get the angle we just divide by the lengths of each vector, which are both R. This gives the ﬁnal result. cos θ12 = cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 − φ2 ) 4. This problem is an application of the results of problem 3. (a.) A straight tunnel between Sydney and New York can be represented by the diﬀerence of the vectors pointing to their locations. To say it another way, the distance between the ends of two vectors is the length of the diﬀerence of the vectors. Adjusting for the fact that latitude and longitude are not quite the same as the coordinates θ and φ, we ﬁnd the polar coordinates of the cities. XNY = (6370 km, 49◦ , 286◦ ) XSydney = (6370 km, 124◦ , 151◦ ) Converting to Cartesian coordinates (x, y, z ) using the formulas from problem 3 we get XNY = (1325 km, −4621 km, 4179 km) XSydney = (−4619 km, 2560 km, −3562 km) 3. Using the formulas on the problem set, we can convert the points on the surface of the sphere to Cartesian coordinates. x1 = R sin θ1 cos φ1 y1 z1 x2 y2 z2 = R sin θ1 sin φ1 = R cos θ1 = R sin θ2 cos φ2 = R sin θ2 sin φ2 = R cos θ2 As in problem 2, we need to know the length of these vectors in order to calculate the angle between them from the dot product. It is fairly obvious that the lengths of the vectors are just R, because that is the radius of the sphere; we will show this explicitly. (R, θ, φ)2 = R2 (sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ) Using the fact that sin2 φ + cos2 φ = 1, we get (R, θ, φ)2 = R2 sin2 θ + cos2 θ We can just repeat the previous step for θ now and get (R, θ, φ) = R 3 The distance between New York and Sydney through the earth is just XNY − XSydney . The result of the calculation is Distance = 12, 117 km (b.) Using the result from problem 3 to calculate the angle between Sydney and New York, we ﬁnd that cos θ12 = −0.809, thus θ12 = 144.0◦ . To calculate the distance along the earth’s surface we need to express this angle in radians. The conversion formula is θ(radians) = π θ(degrees) 180◦ Let A be an arbitrary vector and let n be a ˆ unit vector in some ﬁxed direction. Show that A = (A · n) n + (n × A) × n ˆˆ ˆ ˆ Form a triangle from the three vectors in this equation. Let B = (A · n) n and let C = ˆˆ (n × A) × n. Let the angle between A and ˆ ˆ n be θ. What this formula does is to break up ˆ the vector A into a piece parallel to n and a piece ˆ perpendicular to n. B gives the parallel piece. ˆ Its length is just B = A cos θ. The length of the perpendicular piece must then be A sin θ. Thus θ12 = 2.513 radians. Multiplying this by the radius of the earth, we get the “great circle” distance between New York and Sydney: Distance = 16, 010 km 5. K&K problem 1.6 This question asks you to prove the law of sines using the cross product. Let A, B, and C be the lengths of the vectors making the three sides of the triangle. Let a, b and c be the angle opposite each of those sides. The law of sines states that sin b sin c sin a = = A B C Remember that the length of the cross product of two vectors is equal to the area of the parallelogram deﬁned by them. Remember also that the the length of the cross product is equal to the product of the lengths times the sine of the angle between them: A × B = AB sin θ. We have three vectors to play with in this problem, and using the cross product we can compute the area of the triangle from any two of them. We ﬁnd that Area = AB sin c = BC sin a = AC sin b We just divide the whole thing by ABC and we recover the law of sines. 6. K&K problem 1.11 Examining the vector C, we see that inside the parentheses is a vector whose length is A sin θ and is perpendicular to n. This vector is then ˆ crossed into n. Since it is perpendicular to n, ˆ ˆ the length of the ﬁnal vector is A sin θ, which is what we want. Now we are just concerned with the direction. The ﬁrst cross product is perpendicular to the plane containing n and A. ˆ The second cross product is perpendicular to the ﬁrst, thus it is coplanar with n and A. It is ˆ also perpendicular to n. Thus it represents the ˆ component of A that is perpendicular to n. Be ˆ careful about the sign here. A useful vector identity that you will be seeing again is the socalled “BACCAB” rule. It is an identity for the triple cross product. A×(B × C) = B(A · C) − C(A · B) Its fairly obvious why this is called the BACCAB rule. Using this rule, we see that n×(n × A) = n(n · A) − A(n · n) ˆˆ ˆˆ ˆˆ This immediately gives A = (A · n) n + (n × A) ×n ˆˆ ˆ ˆ 4 Of course we haven’t derived the BACCAB rule here. It’s a mess. 7. The idea in all of the parts of this problem is that the plane must oppose any perpendicular wind speed to maintain its straight path. If the wind is blowing with a speed v perpendicular to the path, the plane’s airspeed must be −v perpendicular to the path. The airspeed is u, the wind speed relative to the ground is w, and the ground speed is v = u + w. u = U0 . Let the total distance traveled be D. (a.) No wind, w = 0 so u = v. T = D/U0 . (b.) Wind of speed W0 blowing parallel to the path. When the wind is going with the plane, v = W0 + U0 , when it opposes the plane, the ground speed is v = U0 − W0 . The time for the ﬁrst leg is T1 = D/2(U0 + W0 ). The time for the second leg is T2 = D/2(U0 − W0 ). The total time is the sum T= D 2 1 1 + U0 + W 0 U0 − W 0 (d.) Wind of speed W0 blowing at an angle θ to the direction of travel. The plane again needs to cancel the component of the wind blowing in the perpendicular direction. The perpendicular component of the wind speed is W0⊥ = W0 sin θ. As in part (c.) the airspeed in the parallel direction can be computed
2 2 U 0 = U⊥ + U 2 ⇒ U = 2 2 U0 − W0 sin2 θ In this case, the wind has a component along the direction of travel. This parallel component is W0 = W0 cos θ. On one leg of the trip, this adds to the ground velocity. On the other leg, it subtracts. This gives us the following formula: D 2 1
2 2 U0 − W0 sin2 θ + W0 cos θ T= + 1
2 2 U0 − W0 sin2 θ − W0 cos θ This can be simpliﬁed, and we get the ﬁnal answer, which agrees with part (a.) when W0 = 0. T= DU0 2 2 U0 − W 0 This can be simpliﬁed considerably: D
2 2 U0 − W0 sin2 θ 2 2 U0 − W 0 T= (c.) Wind of speed W0 blowing perpendicular to the path. This part is a little harder. The plane will not be pointed straight along the path because it has to oppose the wind trying to blow it oﬀ course. The airspeed of the plane in the perpendicular direction will be W0 , and we know what the total airspeed is, so we can calculate the airspeed along the path.
2 2 U 0 = U⊥ + U 2 ⇒ U = 2 2 U0 − W 0 If you look at this carefully, you will realize that it reduces to the correct answer for parts (a.), (b.), and (c.) with the proper values for W0 and θ. If θ = 90◦ , the wind blows perpendicular to the path and we get the result from part (c.). If θ = 0◦ , the wind blows parallel to the direction of travel and we recover the result from part (b.). (e.) (f.) This part requires some calculus. We need to do a minimization of a function. What this part asks is to study the travel time as a function of wind speed for an arbitrary angle θ. We need to consider the result from part (d.) as a function of the wind speed: D
2 2 U0 − W0 sin2 θ 2 2 U0 − W 0 The wind has no component along the path of motion, so the airspeed in the parallel direction is the same as the ground speed in the parallel direction. The ground speed is furthermore the same on both legs of the trip. The ﬁnal answer again agrees with part (a.) when W0 = 0 T= D
2 2 U0 − W 0 T (W0 ) = 5 For now we are going to ignore the fact that it also depends on U0 and θ. Remember that functions have maxima and minima at places where the derivative vanishes, so we need to take the derivative of T with respect to W0 :
2 2 2W0 U0 − W0 sin2 θ d T (W0 ) = D 2 2 dW0 (U0 − W0 )2 time, y = AB 2 t2 . In vector form, the position is then r(t) = xBt + yAB 2 t2 ˆ ˆ (b.) The vector velocity is obtained from the vector position by diﬀerentiating with respect to t v(t) = d r(t) = xB + y2AB 2 t ˆ ˆ dt − W0 sin2 θ
2 2 2 2 U0 − W0 sin2 θ(U0 − W0 ) The derivative is clearly zero when W0 = 0. In this case the travel time T = D/U0 as in part (a.). There is another case we have to worry about though. We divide out what we can to get an equation for another value where the derivative vanishes
2 2 2 2 U0 sin2 θ − W0 sin2 θ = 2U0 − 2W0 sin2 θ (c.) The vector acceleration is obtained from the vector velocity by again diﬀerentiating with respect to t a(t) = d v(t) = y2AB 2 ˆ dt (d.) The scalar speed is just the length of the √ velocity vector. Remember that A = A · A. v(t) = v(t)·v(t) = B 2 + 4A2 B 4 t2 This gives us the other point where the derivative is zero
2 2 2 2 − sin θ W 0 = U0 sin2 θ Notice that this point always occurs when the wind speed is greater than the air speed. No progress can be made against the wind if this is the case, so the trip cannot occur. The ﬁnal possibility to consider is the case where the wind speed is the same as the air speed. Looking at the formula, the time taken is inﬁnite. The only possibility is that the minimum is at W0 = 0. The ﬁnal piece of this problem is to observe what happens to the “time taken” when W0 > U0 . For one thing, it becomes negative. In some circumstances it can even become imaginary. There is really no interpretation of this other than “ask a stupid question, get a stupid answer”. The answer doesn’t make sense because the question didn’t make sense. The trip cannot occur when W0 > U0 , so it is meaningless to ask how long it would take. 8. A particle moves along the curve y = Ax2 and its x position is given by x = Bt. (a.) We can just plug the x equation into the y equation to get the y position as a function of (e.) The vector average velocity is the integral of the velocity vector over a time interval, divided by the time interval. In general, the (time) average of a quantity A is given by A= 1 (t2 − t1 )
t2 t1 A(t)dt Applying this formula, we see the integral that needs to be evaluated: v(t0 ) = 1 t0 1 = t0
t0 0 v(t)dt (xB + y2AB 2 t)dt ˆ ˆ t0 0 We could also use the fact that the integral of the velocity is the position to get a simpler looking formula for the average velocity v(t0 ) = 1 (r(t0 ) − r(0)) t0 Evaluating this integral, we get an answer that is not surprising v(t0 ) = xB + yAB 2 t0 ˆ ˆ 6 This is just (r(t0 ) − r(0))/t0 ! The average velocity is just the distance traveled divided by the time it took. 9. The idea behind this problem is to make a graph of position vs. time data and show that they ﬁt the equation s = a(t − t0 )2 /2. In addition you are supposed to ﬁnd t0 . The way to do this is to plot the square root of the distance vs. time, which will give a straight line graph: √ s = (a/2)(t − t0 ). The slope of this graph is approximately 0.168, so we can use that to extrapolate back to zero. We ﬁnd that the graph reaches zero at about t = −4.45, so this means that t0 = −4.45 to make the distance traveled equal zero at t = −4.45. University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 2 1. Calculate the following centripetal accelerations as fractions or multiples of g (= 9.8 m/sec2 ): (a.) The acceleration toward the earth’s axis of a person standing on the earth at 45◦ latitude. (b.) The acceleration of the moon toward the earth. (c.) The acceleration of an electron moving around a proton at a speed of 2×106 m/sec in a circular orbit of radius 0.5 Angstroms (1 Angstrom = 10−10 m). (d.) The acceleration of a point on the rim of a bicycle wheel of 26 in diameter, traveling at a constant speed of 25 mph. 2. K&K problem 1.17 “A particle moves in a plane...”. 3. K&K problem 1.20 “A particle moves outward along...”. 4. At t=0 an object is released from rest at the top of a tall building. At the time t0 a second object is dropped from the same point. (a.) Ignoring air resistance, show that the time at which the objects have a vertical separation l is given by t= t0 l +. gt0 2 8. K&K problem 2.5 “The Atwood’s machine shown in...”. 9. In the ﬁgure, the pulley axle has no friction and the pulley and cords have no mass. As the system is studied for various values of the external applied upward force F, it is found that there are regimes (ranges of F) for which (i) Neither block moves. (ii) Only the small block with mass m (m < M ) moves. (iii) Both blocks move. (a.) Find the values of F which deﬁne the transitions between regimes (i) and (ii), and between regimes (ii) and (iii). (b.) Find the accelerations of the masses within the regimes (b) and (c), expressed as functions of F, M , and m. 7. In the ﬁgure, two blocks are in contact on a table. The coeﬃcient of sliding friction between the blocks and the table is µ. A force F is applied to M as shown and the blocks begin to slide. Find the contact force between the two blocks. Is it the same if the force is applied to m instead of M ? Does the contact force depend on µ? How do you interpret this result for l < gt2 /2? 0 (b.) The above formula implies that there is an optimum value of t0 such that the separation l reaches some speciﬁed value l0 at the earliest possible value of t. Calculate this optimum value of t0 and interpret the result. 5. K&K problem 1.21 “A boy stands at the peak...”. 6. K&K problem 2.1 “A 5kg mass moves under the...”. 10. K&K problem 2.6 “In a concrete mixer...”. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 2 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. This problem is a simple application of the formula for centripetal acceleration. An object moving at speed v in a circular path of radius r has a centripetal acceleration directed inward: acentripetal = − v2 ˆ r r (c.) This time you are given the speed, not the angular velocity, so we use the ﬁrst formula. The acceleration is 8.0 × 1022 m/sec2 . This is 8.2 × 1021 g ! (d.) A point on the rim of a wheel is moving at the same speed that the wheel is rolling. The radius is 13 inches, and the velocity is 25 mph, so we need to convert these units. The oﬃcial deﬁnition of the inch is 1 inch = 2.54 cm. This gives 1610 meters per mile. The acceleration is then 379 m/sec2 , which is 38.6g . 2. K&K problem 1.17 In plane polar coordinates the velocity and acceleration are given by ˙ˆ v = rˆ + rθθ ˙r ¨ ˙r ˙˙ ˆ a = (¨ − rθ2 )ˆ + (rθ + 2rθ)θ r We know all of these things, so we can get a formula for the magnitudes of v and a, given that both the radial and angular velocities are constant (¨ = θ = 0). r¨ If we know instead the angular velocity ω of the object, we can use it instead. Remember that an object with angular velocity ω going on a circular path of radius r has speed v = ωr. Plugging into the above formula, we get r acentripetal = −ω 2 rˆ (a.) Standing at 45◦ latitude, the distance to the √ axis of the earth’s rotation is just rearth / 2. This is the radius to be used. The speed is just the angular velocity times this radius. The angular velocity is easy to guess, it’s 2π /day. Using rearth = 6370 km, the distance to the axis is 4504 km. Plugging in these numbers, the acceleration is 0.0238 m/sec2 . This is 2.4 × 10−3 g , a very small acceleration compared to the acceleration of gravity. (b.) This is the same sort of calculation, except that the angular velocity is now 2π /28 days. The mean radius of the moon’s orbit is 3.84 × 108 m. The centripetal acceleration is 0.0026 m/sec2 , which is 2.6 × 10−4 g . v= a= ˙ r2 + r2 θ2 ˙ ˙ r2 θ4 + 4r2 θ2 ˙˙ √ Plugging into these, we ﬁnd that v = and a = 20 m/sec2 . 3. K&K problem 1.20 52 m/sec The motion of a particle is given by r = Aθ, θ = αt2 /2 and A = (1/π ) m/rad. The sketch of this motion should look something like this: 2 4. Two objects are dropped from a building at times t = 0 and t = t0 . The distance that the ﬁrst has fallen as a function of time is just d1 = gt2 /2. The distance that the second has fallen is d2 = g (t − t0 )2 /2. When t > t0 , the separation between them is l = d1 − d2 . Thus we get l= (b.) We plug the expression for θ into r to get r = Aαt2 /2. From these we can get all of the ˙ necessary derivatives. r = Aαt, r = Aα, θ = αt, ˙ ¨ ¨ = α. and θ Using the equation for radial acceleration, we ﬁnd an expression for points where the radial acceleration is zero. ˙ ˙ ¨ ¨ ar = r − rθ2 = 0 ⇒ r = rθ2 Plugging in the expressions for the r and θ variables, we ﬁnd that Aα = Aα(αt2 )2 /2. Using the fact that θ = αt2 /2, we arrive at the result √ θ = 1/ 2. (c.) For this part, we set the magnitudes of the radial and tangential accelerations equal to each other and solve for the angle. Plugging into the formulas for the two accelerations, we get ˙ ¨ ar = r − rθ2 ¨ aθ = rθ + 2rθ ˙˙ Plugging into the various terms, we get 1 1 Aα − Aα3 t4 = Aα2 t2 + 2Aα2 t2 2 2 Plugging in the expression for θ where we can, we get 1 − 2θ2 = 5θ When the dust settles, there will be separate results depending on the value of θ. The ﬁnal results are √ 33 − 5 1 θ< √ : θ= 4 2 √ 1 33 + 5 θ> √ : θ= 4 2 1 12 1 2 gt − g (t − 2tt0 + t2 ) = gtt0 − gt2 0 2 2 20 Solving for t as a function of l, we get t= l t0 + gt0 2 When l < gt2 /2 this time is negative. This does 0 have a sensible interpretation, unlike the negative time in the airplane problem of the previous problem set. Think of the problem as if we wanted to throw both objects at the same time, but still have the initial conditions given. At t = 0, when we drop the ﬁrst object, from where do we have to throw the other object? The answer is we want to throw upwards from below in such a way that at time t = t0 , the ball has reached the peak of its path and is momentarily at rest at the point where the ﬁrst ball was dropped. During the time between t = 0 and t = t0 , the separation between the objects can be negative, meaning that the second one is below the ﬁrst. (b.) In this part we want to calculate the optimal value of t0 so that the separation reaches some value l0 at the earliest time possible. In other words, we want to minimize the function t(t0 , l0 ). First we take the derivative and set it to zero to ﬁnd local extrema. l0 1 dt = − 2 = 0 ⇒ t0 = dt0 2 gt0 2l0 g Notice that this is just the time that it takes the ﬁrst object to fall a distance l0 . The endpoints here are t0 = 0, where the separation remains at l = 0 forever and t0 = ∞, where the time to reach a separation of l0 is also inﬁnity. Thus this best time to drop it is at t0 = 2l0 /g . This means that the best thing to do is to drop the 3 second object when the ﬁrst object has already fallen a distance l0 . 5. K&K problem 1.21 This is another maximization problem. We want to know the optimal angle to throw a ball down a hill with slope angle φ. Splitting this into the x and y directions is the easiest way to do the problem. First put the origin at the top of the hill. If the ball is thrown up at an angle θ with speed v , the initial velocities are vx = v cos θ and vy = v sin θ. Treating this as a function of θ, we can maximize the range by diﬀerentiating with respect to θ. Note that the endpoints in this problem are not interesting. Throwing the ball straight up (θ = 90◦ ) and throwing it at an angle −φ both result in the ball traveling no distance in the x direction. 2v 2 dx = (cos (2θ) − 2 cos θ sin θ tan φ) dθ g 2v 2 (cos (2θ) − sin (2θ) tan φ) = 0 = g Solving for θ, we get cos(2θ) = sin(2θ) tan φ ⇒ cot(2θ) = tan φ Remembering that cot α = tan(π/2 − α), we see the ﬁnal result: Taking into account the acceleration of gravity, the positions are given by x = vt cos θ 1 y = vt sin θ − gt2 2 We need to know where the ground is in these coordinates. At a position x on the ground, the y coordinate is given by yground = −x tan φ. We can now ﬁnd the time at which the ball hits the ground. Plugging into the equation for distance traveled in y , we get 1 yground = −x tan φ = vt sin θ − gt2 2 1 ⇒ v sin θ − gt = −v cos θ tan φ 2 This gives the time at which the ball hits the ground: 2v (sin θ + cos θ tan φ) t= g We now plug this time into the equation for distance traveled in the x direction, giving the distance that the ball traveled: 2v 2 (cos θ sin θ + cos2 θ tan φ) x= g 2v 2 1 ( sin 2θ + cos2 θ tan φ) = g2 Note that on a level surface, when φ = 0, the optimal angle is 45◦ , as you might already know. 6. K&K problem 2.1 This is the ﬁrst problem where you are asked to consider the forces causing acceleration. The force on a 5 kg mass is given by F = 4t2 x − 3ty ˆ ˆ Newtons. Apply Newton’s second law of motion, namely F = ma, to get the acceleration, ˆ ˆ a = (4t2 /5)x − (3t/5)ym/sec. (a.) We can get velocity from acceleration by integrating v(t) − v(t0 ) =
t t0 tan π πφ − 2θ = tan φ ⇒ θ = − 2 4 2 a(t )dt Plugging the acceleration we just determined into this integral, and knowing that the velocity at t = 0 is zero, we get the velocity as a function of time: v(t) = 3 43 t x − t2 y m/sec ˆ ˆ 15 10 4 (b.) We get the position by integrating again: r(t) − r(t0 ) =
t t0 v(t )dt Applying this to the result of part (a.), and remembering that at t = 0 the mass is at the origin so r(0) = 0, we get the position as a function of time: 1 14 t x − t3 y m ˆ ˆ r(t) = 15 10 (c.) Now all that is left is to take the cross product of the position with the velocity. We ﬁnd that r×v = − 36 46 t6 t+ t ˆ= ˆ z z 150 150 150 7. This problem asks you to consider two blocks sliding on a table together. The larger block, with mass M , has ﬁve forces acting on it. They are F x, the applied force, a contact force that ˆ ˆ ˆ I will call −CM x, the force of gravity −g y, the normal force N y, and the force of friction. Beˆ cause there is no acceleration in the y direction, ˆ we can easily ﬁnd that N = M g , so that there is no net force in the y direction. From the norˆ mal force we can determine the force of friction. An object that is sliding with friction along a surface is acted upon by a force opposing the direction of motion with magnitude µN (= µM g ), where µ is the coeﬃcient of sliding friction. We now can write an expression for the acceleration of the large block in the x direction ˆ F CM Fx = − − µg ax = M M M There are two unknowns here, ax and CM . We need another equation. Luckily, there is another block that we can write equations about. The small block, having mass m, is aﬀected by four forces. They are the contact force, the force of gravity, the normal force, and the force of friction. The contact force is exactly opposite to the contact force on the ﬁrst block. This is due to Newton’s third law of motion, which states that every force has an equal and opposite force. ˆ Thus Cm = CM and the contact force is CM x. The force of gravity is just −mg y, the normal ˆ force exactly opposes gravity as before, +mg y, ˆ and the force of friction is again −µmg . We get the equation of motion for the second block, noticing that acceleration of the second block is the same as for the ﬁrst block because they are moving together: ax = CM − µg m We are only concerned with CM here, so we can simplify the solution of these equations. Subtract the second equation from the ﬁrst to get F − CM M 1 1 + M m =0 We can now solve for CM in terms of F : CM = F m M +m If we follow the same procedure when the force is acting on the second block, we get a very similar answer, but the mass in the numerator is the larger mass CM = F M M +m This is a factor of M/m larger, which is what we expect. The only force pushing on the second block in each case is the contact force, and the acceleration doesn’t depend on which side we push the combined system from. The 5 force of friction acts in proportion to the mass in this case, so it does not aﬀect this argument. It can be thought of as a force that acts on the combined system, not on the individual blocks, because it is proportional to the mass. Notice that neither of these expressions depend on the value of µ, which would indicate that the friction was aﬀecting the contact force. Solving for the acceleration, a=g M −m M +m The larger mass accelerates with magnitude a downward. The tension is found by plugging the acceleration into either of the starting equations T = Mg − Mg M −m M +m =g 2M m M +m 8. K&K problem 2.5 This is the ﬁrst pulley problem, and it won’t be the last. The pulley is massless and frictionless, and supports two masses M and m by a massless rope connecting them. The ﬁrst thing to notice about this problem is that the tension in the rope must be the same on both sides of the pulley. If the diﬀerent sides had diﬀerent tensions, there would be a tendency to cause an angular acceleration in the pulley. Since it is massless, this acceleration would be inﬁnite, which is unphysical, so the tensions must be equal. You will see this in more detail when you do rigid body motion later this term. 9. Nope, this isn’t the last pulley problem either. Again, the pulley and cords are massless and the pulley is frictionless. A force F is applied upward, and various things will happen depending on what F is. The ﬁrst thing to notice is that the pulley is massless. This means that the tensions on the two ropes must be equal, otherwise a ﬁnite angular force would be applied to a massless object, which again is unphysical. The second thing to notice is that the upward force must exactly balance the sum of the tensions. If this weren’t the case, there would be a net force applied to a massless object. This can’t happen, so to balance the forces we just need F = 2T . With these two points in mind, we can do the rest of the problem. (a.) The boundary between regimes (i) and (ii) is where the lighter block lifts oﬀ the ground. Consider the forces on this block. They are gravity −mg , the tension T , and the normal force N . The equation of motion is ma = T + N − mg . At the boundary between regimes (i) and (ii) when the block just barely can be lifted upward, the normal force N is zero, but so is the acceleration. This gives us T = mg . We know also that T = F/2, so the minimum force to lift the lighter block is F = 2mg . This is just twice the weight of the lighter block, which we expect because the force applied gets divided in half by the pulley. The boundary between regimes (ii) and (iii) is found in a similar way. The equation of motion is T + N − M g = M a. Again both a and N are zero at the transition point, so T = M g , which gives the force F = 2M g . The ﬁnal results are regime(i) {F < 2mg } regime(ii) {2mg < F < 2M g } regime(iii) {2M g < F } The second thing to notice is that the more massive block will fall and the less massive block will rise, and their accelerations will be the same, but in opposite directions. This just means that the rope isn’t stretching. For this problem I will set “down” to be the positive direction. The equation for the larger mass is M a = M g − T ⇒ T = M (g − a) Using the expression for tension derived in the above equation, the equation for the smaller mass is the following: ma = −mg + T ⇒ m(a + g ) = M (g − a) 6 (b.) Now we want to ﬁnd the accelerations for regimes (ii) and (iii). This is easy because we have already determined the equations of motion. For regime (ii), only the lighter block accelerates. The equation of motion is F/2 − mg = ma. This gives the result regime (ii) aM = 0 F am = −g 2m In regime (iii), the equation of motion of the ﬁrst block is the same, so we get the same result for the acceleration. The equation of motion of the larger block is F/2 − M g = M a. These give the results regime (iii) aM = am 10. K&K problem 2.6 The cement mixer’s drum has a radius R. We want to know how fast it can rotate so that the material will not stick to the walls all of the time. We just need to ﬁgure out at what speed the drum can oppose gravity all of the time. For a glob of material of mass m, the worst case is at the top. To remain in contact with the drum, at that point the glob must feel a downward force from the drum that is positive. In addition it feels the downward force mg due to gravity. So the total downward force on it must be at least mg . Now, what acceleration accompanies this force? We don’t want the glob to leave the drum, so its radial velocity must remain equal to zero. The only remaining possible downward acceleration is the centripetal acceleration due to the circular motion. This is ω 2 R. Equating the mass m times this acceleration to the total downward force, we conclude that mω 2 R ≥ mg , or ω ≥ g /R. For the material to not always stick, we need the ﬁnal result ω< g R F −g 2M F −g = 2m University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 3 1. (a.) A clothesline is tied between two poles, 10 m apart, in a way such that the sag is negligible. When a wet shirt with a mass of 0.5 kg is hung at the middle of the line, the midpoint is pulled down by 8 cm. What is the tension in the clothesline? (b.) A car is stranded in a ditch, but the driver has a length of rope. The driver knows that he is not strong enough to pull the car out directly. Instead, he ties the rope tightly between the car and a tree that happens to be 50 ft away; he then pushes transversely on the rope at its midpoint. If the midpoint of the rope is displaced transversely by 3 ft when he pushes with a force of 500 N (≈50 kg), what force does this exert on the car? If this were suﬃcient to begin to move the car, and the man pushed the rope another 2 ft, how far would the car be shifted, assuming that the rope does not stretch any further? Does this seem like a practical method of dealing with the situation? 2. A string in tension T is in contact with a circular rod (radius r) over an arc subtending a small angle ∆θ (see the ﬁgure). a package, it cuts into the package most deeply as it passes around corners, where r is least.) (c.) If the contact is not perfectly smooth, the values of the tension at the two ends of the arc can diﬀer by a certain amount ∆T before slipping occurs. The value of ∆T is equal to µ∆N , where µ is the coeﬃcient of friction between string and rod. Deduce from this the exponential relation T (θ) = T0 exp (µθ) where T0 is the tension applied at one end of an arbitrary arc θ of string and T (θ) is the tension at the other end. (d.) The above result expresses the possibility of withstanding a large tension T in a rope by wrapping the rope around a cylinder, a phenomenon that has been exploited by time immemorial by sailors. Suppose, for example, that the value of µ in the contact between a rope and a (cylindrical) bollard on a dock is 0.2. For T0 = 100 lb applied by the sailor, calulate the values of T corresponding to 1, 2, 3, and 4 complete turns of the rope around the bollard. (It is interesting to note that T is proportional to T0 . This allows sailors to produce a big pull or not, at will, by having a rope passing around a continuously rotating motordriven drum. This arrangement can be described as a force ampliﬁer). 3. A popular demonstration of inertia involves pulling the tablecloth out from beneath dishes with which the table is set. Suppose a tablecloth just covers the area s2 of a square table. A dish is in the exact center of the table. The coeﬃcient of sliding friction between the dish and the cloth is µ1 , and that between the dish and the table is µ2 . A dinner guest withdraws the cloth swiftly, but at a steady rate. Let the distance the dish moves while in contact with the mov (a.) Show that the force with which the string presses radially inward on the pulley (and hence the normal force ∆N with which the pulley pushes on the string) is equal to T ∆θ. (b.) Hence show that the normal force per unit length is equal to T /r. This is a sort of pressure which, for a given value of T , gets bigger as r decreases. (This helps to explain why, when a string is tied tightly around ing cloth be x1 and the distance it moves while in contact with the table be x2 . (a.) Solve for the maximum velocity v of the dish in terms of x1 , µ1 , and g . (b.) Do the same in terms of x2 , µ2 , and g . (c.) Show that in order for the dish just to remain on the table, x1 = (s/2) µ2 . µ1 + µ2 This situation serves as a simple model of the standard view of interactions (repulsive, in the present example) between elementary particles. 8. Find the center of mass of a semicircular hoop of radius R. (d.) Find the length of time during which the dish and tablecloth are in contact under conditions (c.). (e.) A pitfall for the dinner guest is that the dish may not slide at all, but instead merely move with the cloth. How does she avoid that? 4. A piece of string of length L, which can support a maximum tension T , is used to whirl a particle of mass m in a circular path. What is the maximum speed with which the particle may be whirled if the circle is (a.) horizontal; (b.) vertical? 5. K&K problem 2.31 “Find the frequency of oscillation...”. 6. K&K problem 2.35 “This problem involves... A block of mass m slides...”. 7. Two skaters (A and B ), both of mass 70 kg, are approaching one another, each with a speed of 1 m/sec. Skater A carries a bowling ball with a mass of 10 kg. Both skaters can toss the ball at 5 m/sec relative to themselves. To avoid collision they start tossing the ball back and forth when they are 10 m apart. Is one toss enough? How about two tosses, i.e. A gets the ball back? Plot the entire incident on a time vs. displacement graph, in which the positions of the skaters are marked along the x axis, and the advance of time is represented by the increasing value of the y axis. (Mark the initial positions of the skaters at x = + or − 5 m, and include the spacetime record of the ball’s motion in the diagram.) 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 3 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. (a.) This problem is a simple force balance. The component of the tension in the rope pointing up must balance the force of gravity pulling down on the shirt. We are going to ignore the eﬀect of gravity on the rope. The angle that the rope makes with the horizontal is just tan φ = 8 cm/5 m = 0.016. The sine of this angle is very close to this value. In fact sin φ = 0.015998, so we will use sin φ = 0.016. The equation for the force balance in the vertical direction is just 2T sin φ = mg This ensures that the force of tension balances gravity. There is a factor of two in front of the tension because the tension in each half of the rope acts on the mass. Solving this equation with the values given, the answer is T = 153 N will use sin φ = 0.12. We also need cos φ = 0.99. The force of tension is found by a force balance 2T sin φ = 500 N ⇒ T = 2083 N The force acting on the car is just the cosine of the angle times this tension, assuming that we want the force directed towards the tree. F = T cos φ = 2062 N Now we want to ﬁnd how far the car is shifted. The original right triangle had sides 3 ft, 25 √ ft, and 9 + 625 =25.18 ft. The rope will not stretch anymore, so the hypotenuse of this triangle will remain the same, 25.18 ft. The rope is pushed an additional 2 ft, so the short side now has length 5 ft. The long side must have √ length 634 − 25 = 24.68 ft. The car has moved twice this distance, since the are two triangles with total hypotenuse ≈ 50 ft; only 0.63 ft of car movement is produced by pushing the rope 2 ft. This isn’t a very practical way to get the car out of the ditch. 2. (a.) In this problem we will again use the fact that sin θ ≈ θ when θ 1. The angle that the rope makes with the normal force is ∆θ/2, so the tension in the normal direction is T sin(∆θ/2) ≈ T ∆θ/2. There are two tensions here, one from each side of the rope, and they must balance the normal force. The normal force on this section of string is then just ∆N = T ∆θ (b.) The length of rope that covers an angle ∆θ on a circular object is just r∆θ. θ of course is measured in radians. The normal force per length on the cylinder is ∆N/∆ . Plugging in the result for ∆N from part (a.), and using ∆ = r∆θ, we ﬁnd T ∆N = ∆ r (b.) This part is similar to the ﬁrst. In this case the force being applied to the rope is a man pushing on it, no gravity, but the method is the same. The force of 500 N must be balanced by tension in the rope. The angle tan φ = 3/25 = 0.12 is much larger, but the approximation sin φ ≈ tan φ ≈ φ is still very accurate, so we 2 (c.) When the tension on the rope is not constant, there can be slipping. The force countering this is friction. If the tension changes by an amount ∆T along a small section of rope, the frictional force must be equal to it. The frictional force is µ∆N when it is just about to slip. We thus get ∆T = µ∆N . Plugging in the results of (a.), we get ∆T = µT ∆θ. We are going to promote this relation between small quantities to a diﬀerential relation, so that we can get a diﬀerential equation to solve. dT = µT dθ ⇒ dT = µT dθ 3. A dish sits in the middle of a square table of side s. The coeﬃcient of friction between dish and tablecloth is µ1 . The coeﬃcient of friction between the dish and table is µ2 . The tablecloth is rapidly pulled out from under the dish. The dish moves a distance x1 while in contact with the moving tablecloth, and a distance x2 while in contact with the table. Let the mass of the dish be m, but we will see that this doesn’t matter. (a.) The tablecloth is being pulled out from under the dish, so the dish is sliding on the tablecloth. The frictional force tends to pull the dish to the edge of the table because this opposes the direction of the sliding. The normal force of the dish on the table is just mg , so the force of sliding friction is just µ1 mg . Newton’s second law then tells us that F = ma = mµ1 g ⇒ a = µ1 g This is a constant acceleration, so we can easily determine the amount of time that the dish is on the tablecloth and its maximum velocity. We know the total distance traveled is x1 , so we get the following equations for the time of contact t1 and the maximum velocity v : x1 = 1 µ1 gt2 1 2 v = µ1 gt1 This is a diﬀerential equation that we can solve by direct integration. dT dT = µT ⇒ = µdθ dθ T T (θ ) θ dT ⇒ = µdθ T T0 0 These integrals are ones that you should memorize if you haven’t yet. The result is ln T (θ) − ln T0 = µθ ⇒ T (θ) = T0 eµθ The tension increases exponentially, provided that the rope is about to slip. (d.) Here we are going to calculate some values for this ampliﬁcation of force. µ = 0.2 and T0 = 100 lbs. The values of T for 1, 2, 3, and 4 complete turns are as follows. One complete turn has angle 2π . Exp(2πµ) = 3.51, so the tension at the other end is 351 lbs. For two turns, the angle is 4π , so the ampliﬁcation factor is exp(4πµ) = 12.35 and the tension at the other end is 1235 lbs. For three complete turns, the tension is 4338 lbs. For four complete turns, the tension is 15240 lbs, almost 8 tons! These equations are easily solved for t and v : t1 = 2x1 µ1 g v= 2x1 µ1 g (b.) We do the same thing for the period when the dish slides on the table. This time, the frictional force tends to slow the dish down. The frictional force is −µ2 mg , so the acceleration is a = −µ2 g . The trip starts at the maximum velocity v , and ends with the dish at rest having moved a distance x2 . We can again solve for the travel time t2 and the maximum velocity v : 1 x2 = − µ2 gt2 + vt2 2 2 v = µ2 gt2 Again these equations are easily solved t2 = v µ2 g v= 2x2 µ2 g t2 = 2x2 µ2 g 3 (c.) In part (c.), we derived two expressions for the maximum velocity v . If we equate these we can get an expression relating the distances traveled x1 and x2 . 2x1 µ1 g = 2x2 µ2 g ⇒ µ1 x1 = µ2 x2 4. A string has length L and can support a tension T . A mass m is spun around on the end of the string. (a.) The string is spun horizontally. The centripetal acceleration is v 2 /L, so the force needed to supply this acceleration is mv 2 /L. In this case, the only force to consider is the tension. The maximum velocity is given by mv 2 =T ⇒ v= L LT m For the dish to remain on the table, we need the total distance traveled to be less than half the length of the table, x1 + x2 ≤ s/2. We can combine these equations to solve for the distance x1 that the dish spends on the tablecloth. We will consider the case where the dish stops at the edge of the table so the total distance traveled is s/2. From the previous equation we get a relation between x1 and x2 x2 = µ1 x1 µ2 Combining this with the previous equation, we get the ﬁnal answer x1 + x2 = µ1 s s x1 = ⇒ 1+ 2 µ2 2 s µ2 ⇒ x1 = 2 µ1 + µ2 The above would be correct if gravity could be ignored for part (a.). However, we know this cannot be the case, because then there would be no deﬁnition for the word “horizontal”. Gravity must be present. If so, the string makes an angle θ with the horizontal, and the radius of the spin is L cos θ. The centripetal acceleration is v 2 /L cos θ. This must be provided entirely by the tension in the rope in the radial direction, which is T cos θ. This gives a relationship between v and θ for a given T : TL mv 2 cos2 θ ⇒ cos2 θ = m TL The tension must also oppose the force of gravity downward, giving a second relation, which will be easier to handle when squared v2 = T sin θ = mg ⇒ sin2 θ = mg T
2 (d.) To ﬁnd the amount of time the dish is in contact with the tablecloth, we just use the result of part (a.), t1 = 2x1 /µ1 g . Plugging in our result, we get s g µ2 µ1 1 µ1 + µ2 Adding these two equations, and using cos2 θ + sin2 θ = 1, we get a condition on the velocity mg 2 L LT m 2 m2 g 2 v+ − = 1 ⇒ v2 = TL L2 m T The maximum velocity considering gravity is thus mg 2 L LT − v= m T t1 = (e.) It is possible that the dish won’t slide at all. This is the case of static friction. This force must be overcome but a suﬃciently hard tug. The tablecloth must accelerate at the beginning of the pull to get to its high, but constant velocity. If this acceleration is too low (a1 < µstatic g ), 1 the force of static friction will be suﬃcient to accelerate the dish at the same rate, and the host will not be pleased. If the initial acceleration is high enough, the force of static friction cannot impart the necessary acceleration to the dish, and it begins to slide. 4 (b.) The string is spun vertically. In this case we must also consider gravity. Gravity directly opposes the tension when the mass is at its low point. At any other point, the tension is less. At the low point, the two forces of tension and gravity oppose each other, so the tension must be higher to provide the centripetal acceleration. When the rope is at an angle θ to the vertical, the centripetal force is provided by two sources, gravity and tension mv = T (θ) − mg cos θ L T (θ) is largest at the bottom of the path, when θ = 0. Set T (0) = T , the largest allowed tension. This gives the ﬁnal answer mv 2 = T − mg ⇒ v = L 5. K&K problem 2.31 Find the eﬀective spring constants of these two spring systems. Remember that the frequency of oscillation is ω = k /m. (a.) Consider the point where the springs are attached to each other. There shouldn’t be any force acting there because the small point is massless. We can write equations for the displacements of the two springs x1 and x2 . This spring force acts like a tension in that it pulls from both ends. F1 = −k1 x1 = F2 = −k2 x2 ⇒ k1 x1 = k2 x2 The total displacement of the mass is x1 + x2 , and the total force is just the spring force of the bottom spring F = −k2 x2 . Applying the general relation F = −kx, where k is the spring constant of the combined spring system and x = x1 + x2 , we get F = −k2 x2 = −k (x1 + x2 ) ⇒ k = k2 x2 x1 + x2 (b.) In this case the thing to notice is that the displacements of the two springs must be equal, otherwise the support would be tilting. F1 = −k1 x F2 = −k2 x LT − Lg m
2 This relationship is usually written in a diﬀerent way that is easier to remember. You can check that it is correct: 1 1 1 = + k k1 k2 The ﬁnal result for the frequency is k1 k2 m(k1 + k2 ) ωa = The total force acting on the mass is just the sum of these two forces F = F1 + F2 . We can easily ﬁnd the eﬀective spring constant of the system: F = −kx = −k1 x − k2 x ⇒ k = k1 + k2 The ﬁnal result for the frequency is ωb = 6. K&K problem 2.35 (a.) In this problem we need to solve a diﬀerential equation. A block slides on a frictionless table inside a ﬁxed ring of radius l. The ring has a coeﬃcient of friction µ. We want to ﬁnd the velocity as a function of time. At time t = 0 the velocity is v0 . We will assume that the block moves in the circular path deﬁned by the ring. This makes it eﬀectively a one dimensional problem. k1 + k2 m Using the fact that x1 can be written in terms of x2 , we can remove the position dependence from the equation: k=
k2 k1 k1 k2 k2 ⇒ k= k1 + k2 +1 5 The ﬁnal result for the total angle traveled is µv0 1 ln 1 + t µ l θ(t) = There are two forces acting on the block in the plane of the table. They are the normal force exerted by the ring and the frictional force. The normal force merely makes the block move in the circular path that we assumed. We do need to know it though, because we want to calculate the frictional force. We ﬁnd it in the usual way, by requiring that it provide the centripetal acceleration. acentripetal = v l
2 Notice that the total distance traveled is inﬁnite if one waits for an inﬁnite time, even though the velocity approaches zero as time increases. We can of course write the x and y coordinates of the block as functions of time: x(t) = l cos(θ(t)) y (t) = l sin(θ(t)) ⇒ N= mv l 2 We can now write the equation of motion for the particle by Newton’s second law m µmv 2 dv µ dv = −µN = − ⇒ = − v2 dt l dt l 7. The two skaters each have mass 70 kg. Skater A carries a 10 kg bowling ball. Initially each skater is moving at 1 m/sec, and they are approaching each other. They are going to try to avoid collision by throwing the bowling ball back and forth. This problem uses conservation of momentum. Skater A starts out with pA = (70 + 10) × 1 = 80 kgm/sec of momentum. Notice that we must include the momentum of the bowling ball in the momentum of skater A when he is carrying it. This adds 10 kgm/sec to skater A’s 70 kgm/sec, since the bowling ball has a mass of m = 10 kg. Skater B is going in the opposite direction, so her momentum is negative, pB = −70 kgm/sec. Skater A throws the bowling ball to skater B in an attempt to stop the collision. Since there are no external forces on the system consisting of skater A and the bowling ball, the total momentum of these two objects is conserved. Throwing the bowling ball at 5 m/sec relative to the (initial) velocity of skater A gives it a momentum of 60 kgm/sec. This is because the bowling ball velocity is 6 m/sec when we add the initial velocity of 1 m/sec. Skater A is left with a momentum of pA = 20 kgm/sec, so he hasn’t reversed direction or stopped. Next we consider the system of skater B and the bowling ball. Again there are no external forces, so the momentum of skater B plus the bowling ball is conserved. The total momentum is 60 kgm/sec from the ball and −70 This equation can be solved by direct integration, as you saw in problem 2: µ dv =− ⇒ 2 v l
t dv µ dt =− 2 v l v0 0 µt 1 1 =− − ⇒ v0 v (t) l v (t) This result can be simpliﬁed to get K&K’s result v (t) = v0 1 + µv0 t l
−1 (b.) Now that we know the velocity of the block, ﬁnding the position is easy. It is easiest to describe the position in terms of the angle on the circle. We can easily determine the angular velocity as a function of time, because ω (t) = v (t)/l. We know the velocity, so to get the position we just integrate. Assume that θ = 0 at t = 0, which also means x = l, y = 0
t t θ(t) =
0 ω (t )dt =
0 t 1 v0 dt l 1 + µv0 t /l µv0 1 t = ln 1 + µ l 0 6 kgm/sec from skater B . When skater B catches the ball, she will then have all of this momentum, pB = −10 kgm/sec. After this exchange, skater B has the ball, and the two skaters are still approaching each other. One toss was not enough. To summarize the ﬁrst toss initial pA = 80 pB = −70 intermediate pA = 20 pB = −70 pball = 60 ﬁnal pA = 20 pB = −10 The second toss will be enough to stop the collision. We calculate the velocity of skater B (including the bowling ball): v = p/m = −0.125 m/sec. Now skater B throws the bowling ball to skater A. The bowling ball’s velocity will be −5.125 m/sec, so its momentum will be −51.25 kgm/sec. This leaves skater B with pB = 41.25 kgm/sec. This is in the opposite direction to her initial motion. Skater A gets all of the momentum of the ball again, so pA = −31.25 kgm/sec. This is also opposite to his initial direction. So, after two tosses, the skaters are moving away from each other and the collision is averted. Plotting position versus time for the two skaters, we get a graph like the following: It is fairly obvious that the center of mass lies on the line YCM = 0, or θ = 0. The center of mass is calculated by the following XCM = 1 M xλ dl where λ is the linear mass density and dl is a differential of length on the hoop. If the hoop has mass M , the linear mass density is λ = M/πR. We can use polar coordinates to integrate this. Remember that x = R cos θ for points on the hoop. The diﬀerential of length on the hoop is dl = Rdθ, so the integral we need to do is XCM = 1 M
π /2 −π/2 RM 2 cos θ dθ = R π π 8. This is an example of a center of mass calculation. Let’s put the hoop on a polar coordinate system so that it goes from θ = (−π/2, π/2). In cartesian coordinates this is on the right halfplane. University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 4 1. A very ﬂexible uniform chain of mass M and length L is suspended from one end so that it hangs vertically, the lower end just touching the surface of a table. The upper end is suddenly released so that the chain falls onto the table and coils up in a small heap, each link coming to rest the instant that it strikes the table. Find the force exerted by the table on the chain at any instant, in terms of the weight of the chain already on the table at that moment. 2. A supersonic plane of mass M has an airspeed v = 1000 m/sec. Its jet engine takes in 80 kg of air per sec, mixes it with 30 kg of fuel per sec, and compresses the mixture so that it ignites. The resulting hot gasses leave the engine with velocity 3000 m/sec relative to the plane. What thrust (force) does the engine deliver? 3. K&K problem 3.13 “A ski tow consists of...”. 4. This problem is designed to illustrate the advantage that can be obtained by the use of multiplestaged instead of singlestaged rockets as launching vehicles. Suppose that the payload (e.g. a space capsule) has mass m and is mounted on a twostage rocket (see ﬁgure). The total mass – both rockets fully fueled, plus the payload – is N m. The mass of the secondstage rocket plus the payload, after ﬁrststage burnout and separation, is nm. In each stage the ratio of burnout mass (casing) to initial mass (casing plus fuel) is r, and the exhaust speed is V , constant relative to the engine. noring gravity, is given by v = V ln N . rN + n(1 − r) (b.) Obtain a corresponding expression for the additional velocity u gained from the second stage burn. (c.) Adding v and u, you have the payload velocity w in terms of N , n, and r. Taking N and r as constants, ﬁnd the value of n for which w is a maximum. (d.) Show that the condition for w to be a maximum corresponds to having equal gains of velocity in the two stages. Find the maximum value of w, and verify that it makes sense for the limiting cases described by r = 0 and r = 1. (e.) Find an expression for the payload velocity of a singlestage rocket with the same values of N , r, and V . (f.) Suppose that it is desired to obtain a payload velocity of 10 km/sec, using rockets for which V = 2.5 km/sec and r = 0.1. Show that the job can be done with a twostage rocket but is impossible, however large the value of N , with a singlestage rocket. 5. A boat of mass M and length L is ﬂoating in the water, stationary; a man of mass m is sitting at the bow. The man stands up, walks to the stern of the boat, and sits down again. (a.) If the water is assumed to oﬀer no resistance at all to motion of the boat, how far does the boat move as a result of the man’s trip from bow to stern? (b.) More realistically, assume that the water offers a viscous resistive force given by −kv , where k is a constant and v is the velocity of the boat. Show that in this case one has the remarkable result that the boat should eventually return to its initial position! (a.) Show that the velocity v gained from the ﬁrststage burn, starting from rest and ig (c.) Consider the paradox presented by the fact that, according to (b.), any nonzero value of k , however, small, implies that the boat ends up at its starting point, but a strictly zero value of k implies that it ends up somewhere else. How do you explain this discontinuous jump in the ﬁnal position when the variation of k can be imagined as continuous, down to zero? For details, see D. Tilley, American Journal of Physics Vol. 35, p. 546 (1967). 6. The Great Pyramid of Gizeh when ﬁrst erected (it has since lost a certain amount of its outermost layer) was about 150 m high and had a square base of edge length 230 m. It is eﬀectively a solid block of stone of density about 2.5 g/cc. (a.) What is the minimum amount of work required to assemble the pyramid, if the stone is initially at ground level? (b.) Assume that a slave employed in the construction of the pyramid had a food intake of about 1500 Cal/day (1 Cal = 4182 joules). The Greek historian Herodotus reported that the job took 100,000 slaves 20 years. What was the minimum eﬃciency of a slave (deﬁned as work done divided by energy consumed)? 7. A particle of mass m, at rest at t=0, is subjected to a force f (t) whose magnitude at t=0 is F . This magnitude decreases linearly with time, becoming zero at time t = T . The direction of the force remains unchanged. What is the kinetic energy of the particle at time T ? 8. A wooden block of mass M , initially at rest on a table with coeﬃcient of sliding friction µ, is struck by a bullet of mass m and velocity v . The bullet lodges in the center of the block. How far does the block slide? 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 4 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. A chain of mass M and length L falls onto a table. Initially, the chain is hanging so that its lower end just touches the table. The chain is falling in gravity, so the velocity of a link that is falling is given by v = gt. The distance that the chain has fallen is given by x = gt2 /2. These two facts tell us how much of the chain is on the table at a given time. The density of the chain is M/L, so the mass of chain on the table is just M x/L: M (t) = 1M 2 gt 2L 2. The airspeed of a plane is v = 1000 m/sec. The engines take in 80 kg of air per second and mix it with 30 kg of fuel. The mixture is expelled after it ignites, and it is moving at a velocity of 3000 m/sec relative to the plane. We can calculate the thrust of this engine by calculating the rate of change of momentum. The fuel is ejected at a rate of 30 kg/sec, and it is given a velocity of 3000 m/sec relative to the plane. It started at rest with respect to the plane, so it need to be given the full velocity. The rate of change of momentum this corresponds to is dm dp = v = 30 × 3000 = 90, 000 kg − m/sec2 dt dt The air also contributes to the momentum. It is expelled at a rate of 80 kg/sec. Its velocity is already 1000 m/sec relative to the plane, so it only needs to gain 2000 m/sec of velocity in the engine. The rate of change of momentum that this corresponds to is dm dp = v = 80 × 2000 = 160, 000 kg − m/sec2 dt dt The total rate of momentum transferred to the exhaust by the plane’s engine is thus dp = 250, 000 kg − m/sec2 dt This rate of momentum transfer is equal to the thrust of the engine: Fthrust = 250, 000 N 3. K&K problem 3.13 This problem concerns the total force that a ski lift must exert to lift skiers to the top of a hill. There will be two parts to The rate at which the mass is falling on the table is just d M M M (t) = gt = v (t) dt L L At time t, the free elements of chain are moving with speed v (t). This is the velocity they have when they hit the table. The total rate at which momentum is being transferred is d M2 dp = v (t) M (t) = v (t) dt dt L Writing this in terms of M (t), the mass on the table at time t, we get the following: dp = 2M (t)g dt The rate of change of momentum should be familiar to you from Newton’s second law which states d F= p dt Thus the table must be exerting this force on the chain to slow it down. Remember also that the table exerts a normal force on the chain which is equal to the force of gravity FN = M (t)g Thus the total force that the table exerts on the chain is three times the weight of the chain on the table: F (t) = FN (t) + 2M (t)g = 3M (t)g 2 the force. The ﬁrst is just the force necessary to oppose the force of gravity on the skiers. The second is the force required to accelerate the skier at the bottom from rest to the speed of the lift. The rope is 100 meters long, and it is pulled at 1.5 meters per second. On average, one skier uses the tow rope every ﬁve seconds. This means the tow rope travels 5 × 1.5 = 7.5 meters between skiers, so 100/7.5 = 13 1 skiers are on the rope on aver3 age. Each skier weighs 70 kg, so the average total weight of the skiers who are on the rope is 933 kg. The component of the force of gravity that must be oﬀset by the rope is determined by the angle of the slope, which is 20◦ . The component of the acceleration of gravity that is directed down the slope is just g sin 20◦ = 0.342g . Therefore the force that the tow rope must exert to oﬀset that component of gravity is 933 × g × 0.342 = 3128 N. In addition, when a skier grabs the rope, he must be accelerated to the speed of the rope, 1.5 m/sec. The change in momentum for the skier is 1.5 × 70 = 105 kgm/sec. This change in momentum must be provided by the motor once every ﬁve seconds, which is how often skiers use the lift. On average, this force is 105/5=21 N. Therefore the total force that the lift must provide is, on average, 3128+21=3149 N. 4. A two stage rocket carries a payload of mass m. The total mass of the rocket is N m, and the mass of the second stage and payload is nm. In each stage, the mass of the fuel is a fraction (1r) of the total, so the mass of the casing is a fraction r of the total mass. The ﬁrst stage has a mass (N − n)m ,which is just the total minus the mass of the second stage. (a.) Since gravity can be ignored, the equation for rocket motion derived in class reduces to v (t) − v0 = V ln M0 . M (t) the ﬁrst burn. So the ﬁrst burn velocity gain is v = V ln N N = V ln . n + r(N − n) N r + n(1 − r) (b.) The method for this part is the same as for part (a.) because the ﬁrst equation guarantees that the velocity gain of a rocket is independent of its initial velocity. Here the initial mass is the full mass of the second stage, nm. The ﬁnal mass is nm minus the mass of fuel consumed in the second burn, which is (1 − r)(nm − m). This yields m(1 + r(n − 1)) for the ﬁnal mass, and a second burn velocity gain of u = V ln n n = V ln . 1 + r(n − 1) nr + (1 − r) (c.) Here we optimize n with all other parameters ﬁxed. We wish to maximize v + u. As V is ﬁxed, we choose equivalently to minimize Q = ln (V /(v + u)) in order to simplify the algebra. From (a.) and (b.) we have Q= N r + n(1 − r) nr + (1 − r) . N n Carrying out the division, Q = (r + Multiplying, Q = r2 + n 1 (1 − r)2 + r(1 − r)( + ) . N N n 1 n (1 − r))(r + (1 − r)) . N n Only the last term depends on n: √ 1 dn + = 0, n = N . dn N n (d.) For this value of n, the velocity gains from the ﬁrst and second burns are equal: √ N √ . v + u = 2u = 2V ln 1 + r( N − 1) (e.) A single stage rocket has the same values of N, r, and V. The initial mass is N m, as in part (a.), and the ﬁnal mass is m + r(N m − m), in To determine the velocity gain v from the ﬁrst burn, we need only to compute the mass of the rocket at the end of the burn. The initial mass is N m, while the mass of fuel burned by the ﬁrst stage is (N m − nm), the mass of the ﬁrst stage, multiplied by (1 − r). The diﬀerence m(n + r(N − n)) of these two masses is the residual mass after 3 analogy to part (b.) with N substituted for n. The ﬁnal velocity is v = V ln N . N r + (1 − r) this case zero. Centering the boat at x = 0, we can calculate the center of mass XCM = −(L/2)m M +m (f.) We want the ﬁnal velocity of the payload to be v = 10 km/sec, and we have a rocket with exhaust velocity V = 2.5 km/sec and r = 0.1. First let’s see if this can be done with a single stage rocket. Plugging into the result from part (e.), we see that 10 = 2.5 ln N 0.9 + 0.1N After the man is sitting at the bow, the center of the boat will be at some position x, which means that the man will be at a position x + (L/2). However, the center of mass will be in the same place. −(L/2)m M x + m(x + (L/2)) = M +m M +m mL . x=− M +m The boat has moved from its initial position. (b.) This time, the water exerts a viscous force F = −kv on the boat. We can show that the boat will always return to its original position. Newton’s second law gives the following equation. We want to use the total mass of the boat plus the man, because we don’t want the man accelerating relative to the boat (m + M )v = −kv ⇒ ˙ This gives v (t) = v0 exp − k (t − t0 ) M +m
v (t) v0 We try to solve for the necessary N N e4 = 0.9 + 0.1N 5.46N + 49.1 = N N = −11.0 . This answer doesn’t make any sense, which means that a single stage rocket can’t do the job. Let’s now look at the optimal two stage rocket, using the result from part (d.): N √ 10 = 5 ln 0.1N + 0.9 N Again we try to solve for N : N √ e= 0.1N + 0.9 N √ 0.739N + 6.65 N = N √ 0.261 N = 6.65
2 k dv =− v M +m t dt
t0 The distance traveled in this interval is just the integral of the velocity x(t) =
k (M + m)v0 1 − e− M +m (t−t0 ) k N = 650 . This rocket indeed can be built. 5. A boat of mass M and length L is ﬂoating at rest. A man of mass m is sitting at the stern. He stands up, walks to the bow and sits down again. (a.) There is no force from the water, therefore the net force on the system is zero. The momentum of the system is conserved, and the center of mass remains at the same velocity, in Now we just need to ﬁnd the initial velocity of the boat. When the man starts moving, say he applies an impulse ∆p. This is the same impulse that the boat must receive, but in the opposite direction. Thus, the velocity of the man is u = ∆p/m and the velocity of the boat is v = −∆p/M . This means that the velocity of the man relative to the boat is u − v = (m + M )∆p/M m. The man is now walking at constant speed relative to the boat. We plug in the initial velocity of the boat 4 v = −∆p/M to the solution of the diﬀerential equation and we ﬁnd the velocity of the boat v (t) = − k ∆p exp − (t − t0 ) M M +m must specify the time interval. We choose the interval from t = 0−, just before any motion starts, to t = ∞, at which time all motion must have stopped due to eﬀects of viscosity. At both of those times the total momentum of the boat+man system, whose rate of change is controlled by Fext , is zero. Therefore the impulse in question, which is equal to the diﬀerence P (∞) − P (0−) of the boat+man system, must vanish. The same impulse can also be written as
∞ At time τ = L/(u − v ) = LM m/(M + m)∆p, the man has reached the other end of the boat. The velocity of the boat is v (τ ) = − k ∆p exp − τ M M +m and it has traveled a distance (M + m)∆p x(τ ) = − kM k 1 − exp − τ M +m 0=
0− Fext dt
∞ 0− = −k dx dt = −k dt dx
0− He again applies an impulse, but this time it is −∆p. This gives the boat a change in velocity of +∆p/M . The total velocity of the boat is now v (τ ) = ∆p M 1 − exp − k τ M +m = −k (x(∞) − x(0−)) , where x is the position of the boat. This proves that the boat returns to its original position. (c.) The result of part (b.) says that any viscous force, no matter how small, results in the boat returning to its original location. The result of part (a.) says that when there is no viscous force, the boat moves some distance. Mathematically, the diﬀerence between the two results is due to the order in which the limits are taken. In part (a.), the ﬁrst thing done is to take the limit as k → 0, no viscous force. Then the limit as t → ∞ is taken. If we look at the result of part (b.), we ﬁrst take the limit as t → ∞, then we consider what happens when there is no viscous force. This is an instance in which we cannot reverse the order of taking limits. Denoting the results from part (a.) and (b.) by capital letters, we see that
t→∞ k→0 Using this as the initial velocity, we again solve the diﬀerential equation v (t) =
k k ∆p 1 − e− M +m τ e− M +m t M This is correct for all t > τ . We now calculate the total distance traveled in the second part of the trip. We take the ﬁnal time to be t = ∞. x(∞) = (M + m)∆p kM 1 − exp − k τ M +m This is exactly the opposite of the distance traveled in the ﬁrst part. Thus the boat will eventually return to its starting point. (b .) Here is a quick, elegant way to prove the result of part (b.). It deserves full grading credit. We do not mention only this method because, as seen above, the problem is amenable to solution by systematic calculation as well as brilliant insight. Consider the impulse applied by the force Fext of the water on the boat. To specify the impulse, which is the time integral of Fext , we lim lim A = lim lim B
k→0 t→∞ So much for the reason why, mathematically, the results of (a.) and (b.) are not the same. Physically, they are not in conﬂict. As the coeﬃcient k approaches zero in part (b.), the speed with which the boat ultimately migrates back to its original position approaches zero also. This cannot be distinguished by physical measurement from the limiting case (a). 5 6. The Great Pyramid at Gizeh is h=150 m high and has a square base of side s = 230 m. It has a density ρ = 2.5 g/cc. (a.) If all the stone is initially at ground level, it must be raised to its position in the pyramid. The work required to this is W = M ghcm = ρgz dV time. The direction remains the same. magnitude of the force is thus f (t) = F 1− t T The The force acts on a particle of mass m initially at rest. The kinetic energy at t = T is just the integral
T The volume element is the area of the square at a height z times dz , the diﬀerential of height. The square has side s at z = 0 and side 0 at z = h. The width of the square decreases linearly with height, so the width and area at height z is given by w(z ) = s 1 − z h A(z ) = s2 1 − z h
2 K=
0 F 1− t T T dx =
0 F 1− t T v dt We can ﬁnd v by applying Newton’s second law, but once we have it, we don’t need to do the integral because we know that K = mv 2 /2 f (t) = m dv =F dt 1− t T The volume element dV is given by dV = A(z ) dz . We can now perform the integral. Expanding the polynomial in z W = ρgs2
0 h We can just directly integrate both sides with respect to t, with limits t = 0 and t = T v (T ) = F T 2m z−2 z z +2 h h 2 3 dz We now have the answer K= This is a simple integral to perform: W = ρgs2 h2 121 −+ 234 = 1 ρgs2 h2 12 1 F 2T 2 8m Plugging in the values for these constants, we get the amount of work required to erect the pyramid W = 2.43 × 1012 Joules (b.) The slaves employed in building this pyramid consumed 1500 Calories per day, which is 6.3 × 106 joules per day. With 100,000 slaves working for 20 years, this is 730 million slavedays of work to build the pyramid. The total energy the slaves spent is thus 4.6 × 1015 joules. The eﬃciency thus implied is low, = 5.3 × 10−4 . This does not necessarily reﬂect a low intrinsic eﬃciency, since the slaves undoubtedly expended most of their energy on activities other than lifting the stone blocks to their ﬁnal position. 7. A force f(t) has magnitude F at t = 0, magnitude 0 at t = T , and it decreases linearly with 8. Instantaneously after the collision of the bullet and block, after the bullet has come to rest but before the frictional force on the block has had time to slow it down more than an inﬁntesimal amount, we can apply momentum conservation to the bulletblock collision. At that time the total momentum of the block+bullet system is (M + m)v0 , where v0 is the velocity of the block+bullet system immediately after the collision. Momentum conservation requires that momentum to be equal to the initial momentum mv of the bullet. Thus v0 = mv . M +m After the collision, the normal force on the block+bullet system from the table is (M + m)g , giving rise to a frictional force µN = µ(M + m)g 6 on the sliding block+bullet system. This causes a constant acceleration µg of that system opposite to its velocity. Take t = 0 at the time of collision. Afterward, the block+bullet system’s velocity in the horizontal direction will be v (t) = v0 − µgt. It will continue sliding until v (t) = 0, at which point the frictional force will disappear and it will remain at rest. Solving, the time at which the blockbullet system stops is t = v0 /(µg ) . The distance traveled in that time is 1 1 (v )2 x = v0 t − µgt2 = v0 t = 0 . 2 2 2µg Plugging in the already deduced value for v0 , this distance is x= m M +m
2 v2 . 2µg University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 5 1. Mass M rests on a table, and mass m is supported above it by a massless spring which connects the two masses. (a.) Find the minimum downward force that must be exerted on m such that the entire assembly will barely leave the table when this force is suddenly removed. (a.) Assuming a value of g independent of height, calculate how high the rocket would rise if ﬁred directly upward from rest at (P ). (b.) Let (Q) be a distance h vertically lower than (P ), and suppose that the rocket is ﬁred at (Q) after sliding down the frictionless chute. What is the velocity of the rocket at (Q) just before the spring is released? Just after the spring is released? (c.) To what height above (P ) will the rocket rise now? Is this higher than the earlier case? By how much? (d.) Remembering energy conservation, can you answer a skeptic who claims that someone has been cheated of some energy? 3. K&K problem 4.7 “A ring of mass M hangs...”. 2. It has been claimed that a rocket would rise to a greater height if, instead of being ignited at ground level (P ), it were ignited at a lower level (Q) after it had been allowed to slide from rest along a frictionless chute – see the ﬁgure. To analyze this claim, consider a simpliﬁed model in which the body of the rocket is represented by a mass M , the fuel is represented by a mass m, and the chemical energy released in the burning of the fuel is represented by a compressed spring between M and m which stores a deﬁnite amount of potential energy, U , suﬃcient to eject m. (This corresponds to instantaneous burning and ejection of all the fuel – i.e. an explosion.) Then proceed as follows: 4. Assume the moon to be a sphere of uniform density with radius 1740 km and mass 7.3×1022 kg. Imagine that a straight smooth tunnel is bored though the moon so as to connect any two points on its surface. The gravitational force on an object by a uniform sphere is equal to the force that would be exerted by the fraction of the sphere’s mass which lies at smaller radius than the object, as if that fraction were concentrated at the center of the sphere. (a.) Show that the motion of objects along this tunnel under the action of gravity would be simple harmonic (neglect friction with the walls of the tunnel). (b.) Consider this problem in the timereversed situation: Let the assembly be supported above the table by supports attached to m. Lower the system until M barely touches the table and then release the supports. How far will m drop before coming to a stop? Does knowledge of this distance help you solve the original problem? (c.) Now that you have the answer, check it against your intuition by (1) letting M be zero and (2) letting m be zero. Especially in the second case, does the theoretical result agree with your common sense? If not, discuss possible sources of error. (b.) Calculate the period of oscillation. (c.) Compare this with the period of a satellite travelling around the moon in a circular orbit at the moon’s surface. 5. K&K problem 4.23 “A small ball of mass m is placed on top...”. 6. In an historic piece of research, James Chadwick in 1932 obtained a value for the mass of the neutron by studying elastic collisions of fast neutrons with nuclei of hydrogen and nitrogen. He found that the maximum recoil velocity of hydrogen nuclei (initially stationary) was 3.3×107 m/sec, and that the maximum recoil velocity of nitrogen14 nuclei was 4.7×106 m/sec with an uncertainty on the latter of ±10%. What does this tell you about: (a.) The mass of the neutron (in amu)? (b.) The initial velocity of the neutrons used? (Take the uncertainty of the nitrogen measurement into account. Take the mass of a hydrogen nucleus as 1 amu and the mass of a nitrogen14 nucleus as 14 amu.) 7. K&K problem 4.13 “A commonly used potential energy function...”. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 5 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. A mass M rests on a table, and a mass m is supported on top of it by a massless spring connecting it to M . (a.) We want to ﬁnd the force F needed to push down on the spring so that the whole system will barely leave the table. We solve this by conservation of energy. The energy stored in a spring is given by kx2 /2 where x is the displacement from equilibrium of the spring. We can measure the gravitational potential of the small mass relative to the equilibrium point of the spring. Initially then, the spring is compressed with a force F + mg which is just the weight of the small mass plus the added force. Hooke’s law tells us that x = −(F + mg )/k . We can now write the initial energy of the system (F + mg )2 mg (F + mg ) + Ei = − k 2k The ﬁrst term is the gravitational energy relative to the equilibrium point of the spring, and the second term is the energy stored in the spring. We want the spring to be able to lift the mass M oﬀ the table. To do this it must apply a force equal to M g , its weight. When the spring is released, it will oscillate. At the top of the oscillation, there will be no kinetic energy. The displacement y of the spring must barely provide the force to lift the lower block: ky = M g . The energy here is the following Ef = M 2 g2 M mg 2 + k 2k The expression under the square root is just (M + m)g , so the expression simpliﬁes a lot: F = ±(M + m)g We obviously want the plus sign. (b.) This is a similar situation. The mass M is dangling and barely touching the table. The displacement of the spring just supports the weight of the block so kx = M g . At the other end, the small mass is momentarily at rest at some distance −y from equilibrium. The energies are Ei = M mg 2 M 2 g2 + k 2k ky 2 2 Ef = −mgy + We equate these energies and solve for y . −mgy + M mg 2 M 2 g2 ky 2 = + 2 k 2k Using the quadratic formula again ky = mg ± m2 g 2 + 2M mg 2 + M 2 g 2 Again the discriminant is a perfect square, and we want the positive value of y , so y= (2m + M )g k Conservation of energy tells us that these are the same, so now we can solve for F . − (F + mg )2 M mg 2 M 2 g2 mg (F + mg ) + = + k 2k k 2k The total distance that the mass falls is x+y = d. d=2 (M + m)g k Cancelling k and using the quadratic formula to solve for F + mg , F + mg = mg ± m2 g 2 + 2M mg 2 + M 2 g 2 This is just twice the displacement caused by the force in part (a.), which makes sense, because the displacement upwards should be the same as the displacement downwards. 2 (c.) (1) When M is zero, the necessary applied force is mg . This is just the weight of the small mass. The spring will bounce back with the same force, so this is what is needed to lift the whole assembly. The distance fallen makes sense also, because the spring starts at its equilibrium length. The mass wants to sit at mg/k below this, to just support the weight. Thus it will oscillate down to 2mg/k below this point. (2) When m is zero, the force needed to move the assembly is M g ; again this is the total weight. The distance traveled by the end of the spring in the second case is just 2M g/k . The end of the spring is M g/k away from equilibrium when it begins, so the total distance traveled by the end is 2M g/k . While this seems to work out, it does not necessarily agree with common sense: a massless spring would not seem to be able to pull a massive block oﬀ the table by virtue of its own motion. However, we realize that, as the spring mass approaches zero in this idealization, its maximum velocity approaches inﬁnity. This explains why the spring is still able to pull the block oﬀ the table, defying our intuition. 2. (a.) The rocket is ﬁred directly upward from the ground. The initial energy is just U , the energy in the fuel. After the fuel is spent, the fuel mass m is moving down at speed u and the remaining rocket mass M is moving upwards at speed v . Because momentum is conserved over this very short time, mu = M v . The energy of the system is given by conservation of energy, and at launch, all of the energy is kinetic: U= 1 M v2 1 M v 2 + mu2 = 2 2 2 1+ M m the answer to part (a.): d= U 1 M g 1 + M/m (b.) This is a little more involved. The rocket has gone around part of an oval track and is now a distance h below where it started. The gravitational energy (M + m)gh gets converted to kinetic energy, so we get the velocity v0 of the rocket before the fuel is used: (M + m)gh = 1 2 (M + m)v0 ⇒ v0 = 2 2gh Ignoring for the moment the gravitational energy, the energy of the rocket at this point is 1 2 Ei = U + (M + m)v0 = U + (M + m)gh 2 The spring (fuel) imparts a change in velocity ∆v to M and ∆u to m. As in part (a.), instantaneous conservation of momentum gives M ∆v = m∆u. After the spring is released, the energy corresponding to Ei is Ef = 1 1 M (v0 + ∆v )2 + m(v0 − ∆u)2 2 2 We know by conservation of energy that Ei = Ef , and we have M ∆v = m∆u, so we can ﬁnd ∆v : 2U ∆v = M (1 + M/m) The total velocity of M is now v = v0 + ∆v : v= 2gh + 2U M (1 + M/m) Now we want to consider the motion of the mass M alone. Its kinetic energy is M v 2 /2, which we can ﬁnd from the previous equation. KM = U 1 + M/m (c.) We can easily ﬁnd the kinetic energy of the remaining rocket, and, using energy conservation, the maximum height H to which it rises above its current position: K= Using v0 = √ 1 M (v0 + ∆v )2 = M gH 2 Energy for the mass M is now conserved, so we can just set KM = M gd, where d is the maximum height achieved by the rocket. This gives 2gh, we can solve for H : √ 2gh + ∆v 2gh + ∆v 2 H= 2g 3 Plugging in the result for ∆v , we arrive at the ﬁnal answer H =h+ U + M g (1 + M/m) hU M g (1 + M/m) force is outward, at the top the normal force will be negative. The radial equation of motion will be mac = N + mg cos θ ⇒ N (θ) = mg (2 − 3 cos θ) Now we can go back to the force equation for the ring and use this result. The total force on the ring will be zero, but the ring will just start to move upwards when the thread is slack, which is when T = 0. Using both of these facts, we get an equation for θ M g = 2mg cos θ(3 cos θ − 2) ∆H = hU M g (1 + M/m) This is just a quadratic equation in cos θ. Multiplying it out, we can apply the quadratic formula to get the answer cos θ = 11 ± 33 1− 3M 2m This is an interesting result. The ﬁrst term just gets the rocket back to the height where it started in the ﬁrst place. The second term gets it to the maximum height of the rocket in part (a.). The fact that the third term is positive means that the rocket actually ﬂies higher in this case. The gain in height is just the third term (d.) This result does not conﬂict with energy conservation, which says only that the total energy of the system is conserved. We have been neglecting what happens to the mass m, which will take away a smaller amount of energy in the second case. If we looked at the total energy of both pieces, it would be conserved. 3. K&K problem 4.7. This problem is one in which both force and energy need to be considered. The forces on the ring are gravity, the tension in the thread T , and the normal forces due to the beads. The forces on the ring in the vertical direction are Fring = T − M g − 2N (θ) cos θ where θ is the angle of the bead’s position from the top, and N (θ) is taken to be positive outward. The two beads will move symmetrically. We now need to ﬁnd the normal force N (θ). First we determine the velocity of the bead from conservation of energy. It yields the following: 1 E/2 = mgL(cos θ − 1) + mv 2 (θ) = 0 2 This gives the velocity, and thus the centripetal acceleration ac , as a function of θ: v 2 (θ) = 2gL(1 − cos θ) ⇒ ac = 2g (1 − cos θ) The centripetal acceleration is provided by gravity and the normal force. Since a positive normal There is a small problem here in that the discriminant can be negative, making the cosine of the angle complex. This of course is unphysical. The problem is that for suﬃciently small m, the motion of the small masses is never important enough to cause the tension in the rope to vanish, so our calculation is wrong from the start. Insisting that cos θ be real, we obtain the condition m> 3 M 2 Taking the positive root, the ﬁnal answer is θ = cos−1 11 + 33 1− 3M 2m 4. We assume that the moon is a uniform sphere of mass M = 7.3 × 1022 kg and radius R = 1740 km. A straight, frictionless tunnel connects two points on the surface. Given the mass and radius, the density is just ρ = 3M/4πR3 . We need to know the acceleration due to gravity at a distance r from the center of the moon. This is also straightforward. Recall that a spherical shell of mass exerts no force on objects inside it, so at a radius r, the only force we need to consider is 4 due to the mass in the moon interior to radius r. This is just the density times the volume interior to r, or M(r) = M r3 /R3 . The acceleration due to gravity is then just g (r) = −GM(r)/r2 = −GM r/R3 . Thus the acceleration due to gravity increases linearly as one moves away from the center of a uniform solid sphere. (a.) In a spherical polar coordinate system with its ˆ axis at the moon’s north pole, assume z that the tunnel lies in a straight line between (r, θ0 , φ) = (R, θ0 , 0) and (R, θ0 , π ), i.e. between two points at the same north latitude π/2 − θ0 having the largest possible diﬀerence in longitude. This means that the distance along a great circle between the ends of the tunnel is 2θ0 , while the distance from the center of the moon to the center of the tunnel is z0 = R cos θ0 . Now assume that the mass makes an angle ψ (− π < ψ < π ) 2 2 with a line connecting the center of the moon and the center of the tunnel, i.e. with the ˆ axis. z The distance of the mass from the center of the tunnel is then x = z0 tan ψ , while its distance from the center of the moon is r = z0 / cos ψ . We now need to know the component Fx of the gravitational force −GM mr/R3 which lies in the (x) ˆ direction of the tunnel, which makes an angle π/2 − ψ with the radial direction. This is Fx = − GM mr cos (π/2 − ψ ) R3 GM m z0 sin ψ =− R3 cos ψ GM m =− x R3 (c.) A satellite traveling in a circular orbit must have centripetal acceleration provided by gravity, which means that GM v2 = 2 = ω2 R R R From the last equality we see that the angular frequency ω of a circular orbit of radius R around the moon is the same as ω0 above. Of course the period is the same as well. 5. K&K problem 4.23. Two balls of masses M and m are dropped from height h and collide elastically. The small ball is on top of the larger ball. Conservation of energy for the system gives its speed v right before the balls hit the ground: (M + m)gh = 1 (M + m)v 2 ⇒ v = 2 2gh The ball M collides with the ground ﬁrst. In order to conserve energy, it must still have speed v instantaneously after it bounces from the ground. Now it immediately collides with the small ball. (Think of this problem as if there were a very small gap between the two balls so that the ﬁrst ball to hit the ground has a chance to bounce before the second one hits it.) We consider the elastic collision between the two balls, each moving at speed v towards the other. The easiest frame in which to study this collision is a comoving (inertial) frame that is instantaneously at rest with respect to the large ball M immediately after it has rebounded with √ velocity v = 2gh from its elastic collision with the ground. In this frame, M is instantaneously at rest, and m has (upward) velocity −2v . When the collision occurs, if m M as stated in the problem, M seems to m like a “brick wall” from which it bounces back elastically with the same speed. Thus, in the comoving frame immediately after the collision, m has velocity +2v . Finally, transforming back to the lab frame, m acquires an extra velocity increment v , for a total of 3v . Since the height that m reaches is proportional to the square of its velocity, this means that m reaches nine times the height from which it originally was dropped. This is like the force from a Hooke’s law spring with eﬀective spring constant keﬀ = GM m/R3 , yielding simple harmonic oscillation with resonant angular frequency ω0 = keﬀ = m GM R3 (b.) Plugging in values of M and R for the moon, and using T = 2π/ω0 , we get for the period of oscillation T = 6536 seconds = 109 minutes 5 A less elegant approach considers the collision between M and m in the lab frame. Here is it essential not to apply the approximation m M until near the end, since cancellations occur which may make nonleading terms more important than would initially be suspected. In the lab frame, conservation of momentum gives M v − mv = M VM + mVm We also have conservation of energy through this collision. This condition gives as before. 1 1 1 2 2 (M + m)v 2 = M VM + mVm 2 2 2 These are two equations in the two unknowns √ Vm and VM , since we already know v = 2gh. We are interested in Vm , which yields the de2 sired ﬁnal height Vm /2g of m, but we are not interested in VM . So we plan to eliminate VM by solving for it using the ﬁrst equation and then substituting for it in the second. Before proceeding with this algebra, it is convenient to substitute = m/M u = Vm /v U = VM /v so that all terms are dimensionless. The two equations above become 1− =U + u 1 + = U 2 + u2 Solving the ﬁrst equation for U , U =1− − u Substituting this value for U in the second equation, 1+ =1−2 +
2 Neglecting with respect to 3 or 1 in both quotients, the polynomial is u2 − 2u − 3 = (u − 3)(u + 1) with the physical solution u=3 Vm = 3v = h=
2 Vm 18gh 2g = 9h 6. This is a collision problem that has diﬀerent unknown quantities than those to which you are accustomed, but it is still solvable. We have two collisions to study, and the unknowns are the neutron mass and the initial and ﬁnal speeds of the neutrons. The initial speeds are the same, so there are four unknowns in total. We have two collisions, each of which yields two equations (one for momentum conservation, one for energy conservation since the collisions are elastic). Therefore the system can be solved uniquely. The directions of the scattered neutrons relative to the incident directions do not represent additional unknowns, since the maximum recoil velocities of the target nucleii will occur when the collisions take place headon, with the incoming neutrons bouncing straight back. Thus we can take this to be a one dimensional problem. The equations are the following (the energy equations have been multiplied by 2):
2 mn v = mn v + mH vH mn v 2 = mn v 2 + mH vH mn v = mn v + mN vN mn v 2 = mn v 2 2 + mN vN Solving these equations for mn and v requires careful algebra. We square the ﬁrst momentum equation to get a relation between v , mn , and v v2= (mn v − mH vH )2 m2 n − 2 u + 2 2u + 22 u + u2 0 = (1 + )u2 − 2 (1 − )u − (3 − ) 3− 1− u− 0 = u2 − 2 1+ 1+ Now we plug this into the ﬁrst energy equation mn v 2 = (mn v − mH vH )2 2 + mH vH mn 6 Expanding,
2 2 m2 vH − 2mn mH vvH + mn mH vH = 0 H MeV/c2 , well within his experimental range. The range of initial neutron velocity is given by v = 3.07 × 107 m/sec v+ = 2.82 × 107 m/sec v− = 4.13 × 107 m/sec . vH 7. K&K problem 4.13. potential is given by U= r0 r
12 Writing this as an equation for v , we get 1 v= 2 mH 1+ mn The LennardJones r0 r
6 This is fairly simple result. If we perform the same manipulations on the nitrogen equations, we will get an analogous result v= 1 2 1+ mN mn vN −2 We can now use these to solve for mn and v . Equating the right hand sides, we get a single equation for the mass. mn vH + mH vH = mn vN + mN vN mN vN − mH vH mn = . vH − v N We can now use this to ﬁnd the initial velocity of the neutrons: v= mH (vH − vN ) vH 1+ 2 mN vN − mH vH vH mN vN − mH vN = 2 mN vN − mH vH v H vN mN − mH = . 2 mN vN − mH vH (a.) We ﬁnd the minimum of this potential by diﬀerentiating it with respect to r and setting the results equal to zero: 12 dU =− dr r This is easy to solve: r0 r
12 r0 r 12 − r0 r 6 =0 = r0 r 6 ⇒ r = r0 The depth of the potential well is just U (r0 ) = − . Thus the potential well has a depth . (b.) We ﬁnd the frequency of small oscillations by making a Taylor expansion of the potential about r = r0 . Read section 4.10 in K&K for more information on this. We can write the potential as follows: U (r) = U (r0 ) + + 1 2 d2 U dr2 dU dr
r =r0 We want to know the mass of the neutron in amu, so we plug in mH = 1 and mN = 14 (greater accuracy is unnecessary, since the recoil velocities are measured only to 10%). We also look at both boundaries of the nitrogen velocity, calling these results m± and v± . Plugging in numbers, the values of mn are mn = 1.159 amu m+ = 1.415 amu m− = 0.911 amu . Chadwick’s experimental work is seen to be reliable; today’s accepted value for the neutron mass is 1.008665 amu, or 938.27231 ± 0.00028 r =r0 (r − r0 ) (r − r0 )2 + · · · We know that dU/dr = 0 at r = r0 , so we drop the middle term. U (r) ≈ − + 1 2 d2 U dr2 (r − r0 )2 r =r0 This is exactly the form of the potential of a mass on a spring. We only have to identify the spring constant. Remembering that Uspring = kx2 /2, we make the identiﬁcation k= d2 U dr2 r =r0 7 For the LennardJones potential, we already know the ﬁrst derivative, so we need to diﬀerentiate once more. 12 d2 U =2 dr2 r 13 r0 r
12 −7 r0 r 6 Plugging in r = r0 , we ﬁnd the eﬀective spring constant for this potential k= 72 2 r0 We now consider two identical masses m on the ends of this “spring”. Their (coupled) equations of motion are: ¨ mr1 = k (r − r0 ) mr2 = −k (r − r0 ) ¨ where r = r2 − r1 is the distance between the masses. Subtracting these two equations, we get mr = −2k (r − r0 ) ¨ The frequency of oscillation is then ω 2 = 2k/m. (Note that we could have obtained the same result by considering the twomass system to be a single mass of reduced mass mreduced = m1 m2 /(m1 + m2 )). Plugging in the above value for the eﬀective spring constant k , ω = 12 2 r0 m University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 6 1. K&K problem 6.1 “Show that if the total linear momentum...”. 2. K&K problem 6.3 “A ring of mass M and radius R lies...”. 3. Expansion of the previous problem: (a.) Let the azimuth of the bug on the ring be φ – that is, φ is zero when the bug starts walking, and 360◦ when the bug makes one revolution. Assume (for part (a.) only) that the ring is ﬁxed. Calculate the angular momentum l of the bug about the pivot in the previous problem, as a function of φ. Check that your result is consistent with what you used in the previous problem when φ was 180◦ . (b.) Now assume that the bug is ﬁxed at some azimuth φ on the ring, but that the ring itself is not ﬁxed, having angular velocity Ω about the pivot (opposite to the angular velocity of the bug when the bug was moving). Calculate the angular momentum l of the bug as a function of Ω and φ. (c.) Now assume that neither the bug nor the ring are ﬁxed. By requiring that the total angular momentum l + l of the bug about the pivot be balanced by the angular momentum of the ring, obtain an expression for the angular velocity Ω of the ring, as a function of φ. (d.) Get an integral expression for the angle θ through which the ring rotates, as a function of time, assuming that dφ/dt = ω = constant. You need not evaluate the integral. Note that the system is “bootstrapping” its way around the pivot! 4. K&K problem 6.5 “A 3,000lb car is parked on a...”. 5. A man begins to climb up a 12ft ladder (see ﬁgure). The man weighs 180 lb, and the ladder 20 lb. The wall against which the ladder rests is very smooth, which means that the tangential (vertical) component of force at the contact between ladder and wall is negligible. The foot of the ladder is placed 6 ft from the wall. The ladder, with the man’s weight on it, will slip if the tangential (horizontal) force at the contact between the ladder and ground exceeds 80 lb. How far up the ladder can the man safely climb? 6. K&K problem 6.8 “Find the moment of inertia of a uniform sphere...”. 7. K&K problem 6.14 “A uniform stick of mass M and length L is...”. 8. K&K problem 6.18 “Find the period of a pendulum...”. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 6 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. K&K problem 6.1 (a.) We know that the total linear momentum of the system is zero. (This would occur, for example, if we were in the center of mass frame.) P=
i (b.) The proof for this part is identical if angular momentum is replaced by torque and linear momentum is replaced by force. 2. K&K problem 6.3 This problem and the next concern the same system – that of a bug walking along a hoop that is free to pivot around a point on its edge. The hoop lies ﬂat on a frictionless surface. The ring has mass M and radius R, and the bug has mass m and walks on the ring with speed v . The key idea in this problem is conservation of angular momentum. About the pivot there is no net torque on the system, so the total angular momentum about that point is conserved. The ring starts at rest with the bug on the pivot, and the bug starts walking at speed v . Immediately after the bug starts walking, the total angular momentum measured about the pivot point continues to be zero. The ring is not yet moving, so it has no angular momentum; the bug has begun to move, but it is at r = 0, so it has no angular momentum yet. pi = 0 Examine the total angular momentum of the system. The vector from the origin to point i is denoted ri . The angular momentum in general depends on where the origin is: L=
i ri × pi We now want to ﬁnd the angular momentum about a new origin whose position vector is R in the current coordinate system. In this new system, the position vector of point i becomes ri − R. Each point has its position changed by the same amount. The new value of the angular momentum is Lnew =
i (ri − R) × pi Expanding, Lnew =
i ri × pi −
i R × pi Because R is the same for all points, we can pull it outside of the sum: Lnew =
i ri × pi − R ×
i pi We want to ﬁnd the angular velocity Ω of the ring when the bug is opposite to the pivot. The bug is moving at speed v on the ring, but the ring is also moving. The bug is at a distance 2R from the pivot, so the velocity of that portion of the ring which is under the feet of the =
i ri × pi − R × P We know that P = 0, so we are done. Lnew = L 2 bug is 2ΩR. The total velocity of the bug is thus v + 2ΩR. Next we need to know the moment of inertia of the hoop. A hoop has moment of inertia I = M R2 about its center of mass. We use the parallel axis theorem to ﬁnd the moment of inertia about a point on the edge. I = ICM + M d2 The distance d from the center of mass to the desired axis in this case is just R, so the moment of inertia of the hoop about a point on the edge is I = 2M R2 . We can now ﬁnd an expression for the total angular momentum of the system. For the hoop we use L = I Ω and for the bug we use L = mvr sin θ. The angle θ between the position vector and the velocity vector of the bug in this case is simply π/2, so sin θ is just 1. We now write the angular momenta of the two pieces Lbug = 2mR(v + 2ΩR) Lhoop = 2M R2 Ω extending from the center of the hoop to the bug and pivot, respectively. This is an isosceles triangle, with two equal sides of length R having an angle φ between them. If we deﬁne the azimuth of the bug on the hoop to be zero at the pivot, the angle φ is simply the azimuth of the bug on the hoop. The length r of the third side is found using the law of cosines: r2 = R2 + R2 − 2R2 cos φ ⇒ r2 = 2R2 (1 − cos φ) Using the trigonometric identity 1 − cos φ = 2 sin2 φ , we can get a simple result for r: 2 r = 2R sin φ 2 Since angular momentum about the pivot is conserved throughout the motion, We know that Lbug + Lhoop = 0. This gives the following expression: 2mvR + 4mΩR2 + 2M R2 Ω = 0 We solve this equation for Ω in terms of v and get Ω=− mv M R + 2mR Now we know v and r. The only thing left to determine is the angle between the position and velocity vectors. The ﬁrst step is to ﬁnd the angle between the position vector r of the bug and the line from the center of the circle to the bug. The isosceles triangle (like any triangle) has a total angle π , and its central angle is φ. The remaining two angles are equal, so they must be (π − φ)/2 each. Thus the three angles add up to π radians. Because the velocity vector v is tangent to the circle, the angle between r and v is π/2 minus this angle. Thus the angle between r and v is φ/2. We can now get the angular momentum of the bug. Assuming that the ring is ﬁxed, l = mvr sin θ from the bug alone, so l = 2mvR sin2 φ 2 Note that the minus sign means that the hoop rotates in a direction opposite to that in which the bug moves. This makes sense because the total angular momentum about the pivot point must vanish. 3. We now study the bug and hoop system in more detail. See the diagram in the previous problem. (a.) In this part we assume that the ring is ﬁxed. We want to calculate the angular momentum of the bug about the pivot point. The ﬁrst step is to ﬁnd the distance r between the bug and the pivot. We can do this using the law of cosines. Consider the triangle made by the line between the bug and the pivot, and the two radial lines (b.) In this part we assume that the ring is rotating with angular velocity Ω, but that the bug is ﬁxed on the ring. The velocity of the bug is just Ωr, where r is the same as was calculated in the previous part. The velocity in this case is always perpendicular to its position vector. This can be seen by remembering that the bug isn’t moving on the ring, so it must be in uniform circular motion about the pivot, with a velocity that is tangent to its present position. Therefore the angular momentum l of the bug is simply mvr, yielding l = mΩr2 = 4mΩR2 sin2 φ 2 3 (c.) We now allow both the bug and the ring to move. The total angular momentum of the bug is l + l from parts (a.) and (b.) respectively. To this we must add the angular momentum of the ring to get the total angular momentum of the system. From problem 2. we know that the total angular momentum must be zero. The angular momentum of the ring is I Ω, so we get the following equation. 4mΩR2 sin2 φ φ + 2mvR sin2 + 2M R2 Ω = 0 2 2 4. K&K problem 6.5 A car of mass m is parked on a slope of angle θ facing uphill. The center of mass is a distance d above the ground, and it is centered between the wheels, which are a distance l apart. We want to ﬁnd the normal force exerted by the road on the front and rear tires. It is easiest to do this problem choosing the origin as the point on the road directly below the center of mass. About this point there are three torques. The normal force on the front (Nf ) and rear (Nr ) set of wheels provides a torque, and also gravity provides a torque mgd sin θ because the car isn’t horizontal. However, the forces of friction on the tires don’t provide any torque because they are in line with the direction to the origin. The torque from the front wheels and the torque due to gravity tend to want to ﬂip the car over backwards, while the torque on the rear wheels opposes this tendency. We want the sum of the torques to vanish, because the (static) car is not undergoing any acceleration, angular or linear: l l + mgd sin θ − Nr 2 2 2mgd sin θ Nr − N f = l 0 = Nf We can get one more condition from the fact that the car is not undergoing linear acceleration perpendicular to the road. This means that the normal forces exactly cancel gravity: Nr + Nf = mg cos θ We can take the sum and diﬀerence of these two equations to get expressions for Nf and Nr . These are 1 cos θ + 2 1 Nf = mg cos θ − 2 Nr = mg d sin θ l d sin θ l We solve this for Ω in terms of φ. The result is Ω=− mv sin2
φ 2 M R + 2mR sin2 φ 2 This agrees with the result of problem 2. when the bug is at φ = π , opposite to the pivot. (d.) Finally, we want to ﬁnd an expression for the angle θ through which the ring rotates. We know that θ is related by a simple diﬀerential equation to the angular velocity Ω of the hoop Ω= dθ dt but we want to express Ω in terms of φ so that we can use the fact that Rdφ/dt, the speed v of the bug with respect to the rim of the hoop, is constant. We apply the chain rule to get Ω= dθ dφ dθ = dt dφ dt Substituting dφ/dt = v/R, where v is constant, we can write an integral for θ: θ=− R v
φ φo mv sin2 (φ /2) dφ M R + 2mR sin2 (φ /2) We can simplify this a little, but doing the integral is hard, which is why you weren’t asked to evaluate it. Setting the initial bug azimuth φ0 to zero and using the fact that dφ/dt = v/R is a constant so that φ = vt/R,
v t/R θ(t) = − 0 sin2 (φ /2) dφ (M/m) + 2 sin2 (φ /2) Plugging in θ = 30◦ , mg = 3000 lb, d = 2 ft, and l = 8 ft, we get Nr =1674 lb and Nf =924 lb. 4 5. We will solve this problem symbolically and wait until the end to plug in numbers. This is always good practice because it makes it a lot easier to check the units of the result and to explore whether the result is reasonable when the inputs have limiting values. We take M to be the mass of the man (M g = 180 lb) and m to be the mass of the ladder (mg = 20 lb). The length H of the ladder is 12 ft, and its point of contact with the wall is d = 6 ft from the wall. The angle that it makes with the wall is θ = arcsin (d/H ) = 30◦ . Finally, the force of friction on the ladder from max max = 80 lb. the ground is Ff ≤ Ff , where Ff There are ﬁve forces to consider in this problem. They are the two normal forces on the ladder, Ng from the ground and Nw from the wall; the force Ff of friction at the base of the ladder; and the two forces of gravity, M g on the man and mg on the ladder. This is a torque balance problem, so choosing a good origin makes it a lot easier. With this choice of the point of contact with ground, two of the ﬁve forces contribute no torque about that point. Not bad! As a sanity check we evaluate µ, the coeﬃcient of friction between the ladder and the ground. The normal force Nf from the ﬂoor is equal and opposite to (M + m)g , the sum of the weights of the ladder and the man. We are given max the maximum frictional force Ff , and we know max max = µN , so µ = Ff /((M + m)g ) = that Ff 80/200 = 0.4, a reasonable value. We now calculate the torques. To ﬁnd the maximum height h to which the man can climb without the ladder slipping, we assume that the ladder is about to slip. This means that the normal force Nw from the wall is equal and opposite max to Ff , exactly countering the maximum force of friction: since these two forces are the only forces in the horizontal direction they must sum to zero. The torque from the wall is then τw = max −Ff H cos θ, where the minus sign indicates that this torque pushes clockwise. The torque from the weight of the ladder is exerted at the midpoint of the ladder, its center of mass. The value of this torque is τm = mg H sin θ. Similarly, 2 the torque exerted by the weight of the man, who is a distance h up the ladder, is τM = M gh sin θ. Requiring these three torques to sum to zero, 0 = τM + τm + τw = M gh sin θ + mg Solving for h,
max Ff H cos θ − mg H sin θ 2 M g sin θ max Ff h m = cot θ − H Mg 2M m µ(M + m) cot θ − = M 2M µ(M + m) cot θ − (m/2) = √M 0.4(200) 3 − 10 = 180 = 0.7142 h = 8.571 ft H max sin θ − Ff H cos θ 2 h= 6. K&K problem 6.8 Because of the spherical symmetry, we work in spherical polar coordinates. To ﬁnd the moment of inertia we need to evaluate the integral I=
2 r⊥ ρdv where in these coordinates r⊥ = r sin θ is the perpendicular distance to the axis and dv = r2 dr d(cos θ) dφ is the element of volume. The integral to evaluate is thus I=M
R4 1 2π r dr −1 sin2 θ d(cos θ) 0 dφ 0 R2 1 2π r dr −1 d(cos θ) 0 dφ 0 where the denominator is the volume V of the sphere, needed to evaluate its density ρ = M/V . Substituting u ≡ r/R, I = M R2
1 0 u4 du
1 0 1 −1 sin2 θ d(cos θ)
1 −1 2π 0 dφ u2 du d(cos θ) 2π 0 dφ In both the numerator and the denominator, all three integrals have limits that do not depend on the other variables, so each integral can be 5 evaluated independently. The φ integrals cancel, and the u integrals have the ratio 3 1/5 = 1/3 5 The integrand in the cos θ integral in the numerator can be rewritten sin2 θ d(cos θ) = (1 − cos2 θ)d(cos θ) 1 cos3 θ = d(cos θ) − d 3 Therefore the ratio of the θ integrals is 2− 2 Putting it all together, I = M R2 × 7. K&K problem 6.14 When the stick is released, there are two forces acting on it, gravity at the midpoint, and the normal force at the point B . We use the point B as the origin, so the only torque about this point is provided by gravity. At the moment of release, the stick is still horizontal, so the torque is M gl τB = − 2 where the minus sign indicates that the torque pulls clockwise. We know that the moment of inertia of a thin stick about its endpoint is I = M l2 /3, so we can easily ﬁnd the angular acceleration α from τ = Iα. − 1 3g M gl = M l2 α ⇒ α = − 2 3 2l ω= 2 32 × = M R2 53 5
2 3 where the minus sign indicates that the acceleration is downward. Finally we use Newton’s second law to ﬁnd the normal force at B. We know the acceleration and we know the force of gravity, so this is a simple equation 3 1 N − Mg = − Mg ⇒ N = Mg 4 4 where the positive direction is up, opposite to the force of gravity. 8. K&K problem 6.18 We want to ﬁnd the equation of motion of the pendulum to determine the frequency. We will use the torque equation τ = Iα. If we choose the pivot point of the pendulum as the origin, only one force provides torque, the force of gravity. It acts on the center of mass of the pendulum, a distance lcm from the pivot point. The magnitude of this force is just (M + m)g . Thus the total torque is τ = −(M + m)glcm sin θ where θ is the angular position of the pendulum. Writing the torque equation and approximating sin θ ≈ θ, we get ¨ I θ = −(M + m)glcm θ We recognize that this is the equation for a simple harmonic oscillator. The angular frequency and period are thus (M + m)glcm I I (M + m)glcm = 2 3 The vertical acceleration of the center of mass is given by the simple formula a = αr, where r is the distance between the center of mass and point b, about which the stick is (instantaneously) executing circular motion. (This is analogous to the expression v = ωr.) Here this distance is r = l/2. This gives the acceleration of the center of mass: 3 a=− g 4 T = 2π All that is left is to evaluate lcm and I . The equation for the center of mass is easy to use. The center of mass of the rod is halfway along its length, and the disk is a distance l from the pivot, so lcm = ml/2 + M l M +m 6 This expression simpliﬁes the formula for the period I T = 2π (M + m/2)gl In the ﬁrst case, where the disk is tied to the rod, the moment of inertia is determined using the parallel axis theorem. The disk is ﬁxed to the rod, so, as the rod pivots, the disk must rotate at the same angular velocity. The moment of inertia of a stick about its end is ml2 /3. The moment of inertia of a disk about its center is M R2 /2. Because the center of mass is displaced a distance l from the origin, the parallel axis theorem tells us that the total moment of inertia of the disk is M R2 /2 + M l2 . Thus the total moment of inertia of the pendulum is I= 1 1 M R2 + M + m l 2 2 3 This gives the period of oscillation T = T = 2π M R2 /2 + (M + m/3)l2 (M + m/2)gl In the second case, where the disk is free to rotate on the rod, the moment of inertia is smaller. Because the disk is not ﬁxed, it has no tendency to rotate. Its eﬀective moment of inertia about the center of mass is thus zero. For our purpose it is the same as a point mass a distance l away from the origin. The new moment of inertia is I= 1 M + m l2 3 The period of oscillation in this case is also smaller: (M + m/3)l T = 2π (M + m/2)g This is the same as our answer for the ﬁrst case in the limit R → 0. University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 7 1. A trick cyclist rides his bike around a “wall of death” in the form of a vertical cylinder (see ﬁgure). The maximum frictional force parallel to the surface of the cylinder is equal to a fraction µ of the normal force exerted on the bike by the wall. Assume that the cyclist and his bike are small relative to the radius of the cylinder. energy occurs. (c.) At what radial distance from the axis of rotation do the men experience the greatest centrifugal force as they make their way to the center? 5. K&K problem 7.4 “In an oldfashioned rolling mill, grain...”. 6. K&K problem 7.5 “When an automobile rounds a curve...”. 7. K&K problem 8.2 “A truck...”. 8. K&K problem 8.4 “The center of mass...”. (a.) At what minimum speed must the cyclist go to avoid slipping down? (b.) At what angle φ to the horizontal must he be inclined at that minimum speed? (c.) If µ=0.6 (typical of rubber tires on dry roads) and the radius of the cylinder is 5 m, at what minimum speed must the cyclist ride, and what angle does he make with the horizontal? 2. K&K problem 6.24 “Drum A of mass M and radius R...”. 3. K&K problem 6.27 “A yoyo of mass M has an axle...”. 4. Two men, each of mass 100 kg, stand at opposite ends of the diameter of a rotating turntable of mass 200 kg and radius 3 m. Initially the turntable makes one revolution every 2 sec. The two men make their way to the middle of the turntable at equal rates. (a.) Calculate the ﬁnal rate of revolution and the factor by which the kinetic energy of rotation has been increased. (b.) Analyze, at least qualitatively, the means by which the increase of rotational kinetic 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 7 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. We will need to use ﬁctitious forces to solve this problem easily. Fictitious forces are never necessary, but they often simplify problems greatly. We choose the point of contact as the origin. There are two torques, τg caused by gravity and τc caused by the centrifugal force. Both act at the center of mass. (This is important to note: ﬁctitious forces always act at the center of mass!) We are assuming that the size of the cyclist l is very small in comparison with the radius R, so τg = +mgl cos φ τc = − mlv 2 sin φ R We want to know the angle where the cyclist is about to slip, so the normal force is mg/µ, equal to the centripetal force mv 2 /R. Substituting for mv 2 /R in the above equation, (a.) We want to know the minimum speed the cyclist needs not to slip down the side. The force of static friction must be mg to hold him up, so we require that µN = mg . The force of static friction can be less than µN , but we are setting it to the maximum to see what the limit is. Thus we need N > mg/µ. The only force acting horizontally in the system is the normal force, so it must entirely provide the centripetal acceleration, which is v 2 /R. We thus obtain mv 2 mg ≥ ⇒ v≥ R µ gR µ mgl cos φ = mgl 1 sin φ ⇒ tan φ = µ µ So when the cyclist is about to slip, he rides at an angle φ = tan−1 µ (c.) Taking µ = 0.6 and R = 5 meters, we ﬁnd that the cyclist must ride at a speed of least 9.0 meters per second, or 29.5 ft/sec, or 20.1 mph. On a road bike this is a mellow cruising speed. At this minimum speed, the angle the cyclist must make with the horizontal is 31 degrees. We caution you not to try this at home; it’s tough to get up to speed without crashing! 2. K&K problem 6.24 This problem is similar to many pulley problems that you have seen before. We need to apply both Newton’s second law and the torque equation to solve it. We denote the (positive downward) acceleration of the falling mass as a. There are two forces on it in the vertical direction, tension and gravity. Newton’s second law requires Mg − T = Ma We now need to apply the torque equation to both drums. For each drum we choose its own (b.) We now need to consider the ﬁctitious centrifugal force. The cyclist is in an accelerating frame of reference because he is moving in a circle. To correctly apply Newton’s second law in the cyclist’s frame, we must introduce the centrifugal force, which points outward with magnitude mv 2 /R. In the frame of the cyclist there are four forces: the normal force and the centrifugal force cancel each other, and the friction and gravity cancel each other. This is required to insure that, by deﬁnition, the cyclist isn’t accelerating in his own frame of reference. The problem is now reduced to a torque balance to ﬁnd the angle at which the cyclist is stable. 2 center as the origin. Each drum feels only one torque, the torque from the tension. This has a magnitude τ = T R in both cases. Notice that both of these torques have the same sign, thus the drums will tend to angularly accelerate in the same direction. Writing the torque equation for each drum, with angular accelerations α1 for the top drum and α2 for the bottom drum, T R = Iα1 T R = Iα2 assumption of a pure downward motion. Therefore the actual motion will be more complicated than this problem asks you to assume. 3. K&K problem 6.27 We need to apply both Newton’s law and the torque equation. The forces on the yoyo horizontally are the force F and the friction f . The vertical forces are the normal force and gravity, which immediately tell us that N = M g . We want to ﬁnd the maximum force we can apply with the yoyo not slipping. It is important to note that the force of friction, which stops the disk from slipping, is controlled by the coeﬃcient µs of static friction because the surface of a rolling wheel is at rest with respect to the ground. Since we are concerned with the maximum allowed force, we will consider the maximum allowed friction, which is µs N . Newton’s law gives us F − µs M g = M a The moment of inertia of the yoyo is I = M R2 /2. Because we want the yoyo to roll without slipping, we can use a = αR. The torque equation gives us µs M gR − F b = Iα = 1 M Ra 2 where the moments of inertia of both disks are the same, I = M R2 /2. From these equations it is easy to see that α1 = α2 ≡ α = T R/I , so the angular accelerations are both α= 2T MR We now need to ﬁnd a relation between a and α. The linear acceleration due to each disk is given simply by a = αR. There are two disks, both unwinding with the same angular acceleration α, so the linear acceleration of the bottom one is just a = 2αR. The previous equation becomes a= Ma 4T ⇒ T= M 4 Plugging this into the very ﬁrst equation that we got from Newton’s law, we ﬁnd the initial acceleration of the drum, assuming that it moves straight down, to be a= 4 g 5 We want to solve these two equations for F , the maximum allowed force. Eliminating a, we get F − µs M g = 2µs M g − 2F Solving for F , we get F = µs M g 3R R + 2b b R Will the drum in fact move straight down? For the moment assuming that the answer is “yes”, consider a (downward accelerating but nonrotating) frame with its origin at the (instantaneous) point of tangency between the lower drum and the tape. In this frame, the CM of the drum experiences a downward force mg and an upward ﬁctitious force 4 mg which does not quite com5 pensate it. Therefore it feels a net downward force 1 mg . About the chosen origin this force 5 causes a net (clockwise) torque, which causes the lower drum to swing to the left like a pendulum bob in this frame. This contradicts our Since R > b the applied force F is always larger than the frictional force, so the yoyo always accelerates to the right. 4. We will solve this problem symbolically and plug in numbers at the end. This is always a good practice because it makes it a lot easier to go back and check your work for correct dimensions and reasonable results for limiting cases. 3 (a.) Let the disk have mass M and radius R, and the two men each have mass m. If the men are momentarily at a radius r from the center of the disk, the total moment of inertia is given by I (r) = 1 M R2 + 2mr2 2 (b.) The extra kinetic energy comes from the work that the men must do against the (ﬁcitious) centrifugal force to make their way from the edge of the turntable to the center. This is the qualitative statement which the problem requests. Optionally, one can perform a quantitative analysis: Each man pushes against the centrifugal force to get to the center. The work they do is converted to rotational kinetic energy. The centrifugal force on each man is given by
2 Fc = m(ω (r))2 r = mω0 r The initial angular velocity of the disk is ω0 , so, when r = R, the initial angular momentum of the system is L = Iω0 = 1 M + 2m R2 ω0 2 There are no net torques acting on the system, so L is conserved. We can use L conservation, I (r)ω (r) = I (R)ω0 , to obtain the angular velocity of the system as a function of the radius of the men: ω (r) = ω0 M R2 + 4mR2 M R2 + 4mr2 1 + 4m/M 1 + 4mr2 /(M R2 ) 2 The work done is just Fc integrated from zero to R, doubled since each man does the same amount of work.
2 ∆W = 2mω0 1 + 4m M 2 0 R r dr 2 (1 + 4mr2 )2 MR The ﬁnal angular velocity ω is just the angular velocity when the men reach the center, ω (r = 0). ω = ω0 M + 4m = M 1+ 4m M ω0 You can look this up in a table, or notice that the top is proportional to the derivative of the bottom, so antidiﬀerentiating is not too hard: 4m 1+ M
2 Plugging in values, we ﬁnd that the ﬁnal angular velocity is 1.5 revolutions per second. The factor by which the kinetic energy has increased is 1 2 Kf 2 I (0)ω =1 2 Ki 2 I (R)ω0 Evaluating this, we ﬁnd Kf = Ki Simplifying, 4m Kf =1+ Ki M 4m Ki ∆K = M For the masses in the problem, Kf /Ki is equal to 3. The rotational kinetic energy is tripled.
22 1 2 1 + 4 m ω0 4MR M 1 22 22 4 M R ω0 − mR ω0 ∆W = 2 −2mω0 M R2 /8m 2 (1 + 4mr2 )2 MR R 0 Evaluating this, we get
2 M R 2 ω0 4 ∆W = 1+ 4m M 2 − 1+ 4m M Multiplying this out, ∆W = Simplifying,
2 ∆W = mR2 ω0 + 2 M R 2 ω0 4 4m 16m2 + M M2 4m2 2 2 R ω0 M
2 mR2 ω0 = 1+ 4m M 4 This is 4m/M times the initial kinetic energy, so, as expected, ∆W is equal to the kinetic energy gain ∆K that we already calculated. (c.) We want to ﬁnd where the maximum centrifugal force is felt. This is just a maximization problem. Diﬀerentiate Fc with respect to r and set it to zero, and also check the endpoints. d dFc = dr dr This gives 1− 1 16mr2 =0 M R2 1 + 4mr2 /M R2
2 mω0 r in magnitude. Since angular velocity is a vector, we can add these separate components to obtain the full angular velocity vector. In cylindrical coordinates, with ˆ pointing along the axis of z the orbit, R ω = − Ωˆ + Ωˆ r z b where the leading minus sign tells us that the radial component of ω is negative, i.e. the millstone is rotating clockwise about its horizontal axle. To calculate the angular momentum, we choose as an origin the intersection of the centerline of the vertical shaft and the centerline of the horizontal axle. Both the shaft and the axle are parallel to mirror symmetry axes of the millstone; thus we expect that the component of angular momentum due to Ω will be parallel to Ω , and the component of angular momentum due to ω will be parallel to ω . More quantiz tatively, the component Lz along ˆ is equal to (I + M R2 )Ω, where I = 1 M b2 is the moment 4 of inertia of a disk about a diameter and M R2 is added to I by use of the parallel axis theorem. Since Lz is constant, no torque is required to maintain it and we don’t need to consider it further. To calculate the radial component Lr of the angular momentum, we need I = M b2 /2, the moment of inertia of a disk about its center: 1 Lr = − M b2 ω 2 where the minus sign again reminds us that the millstone is rolling clockwise about its horizontal axle. Remember that, in cylindrical coordinates, the only unit vector which is constant is ˆ; the z radial and azimuthal unit vectors depend on θ. Even though the magnitude of Lr is constant, its direction is changing. In a time increment dt, the azimuth θ of the millstone axle with respect to the shaft changes by an angular increment dθ = Ωdt. This causes L to change by ˆ dL = θ Lr dθ ˆ = θ Lr Ωdt ˆ1 = −θ M b2 ω Ωdt 2 (1 + 4m/M )2 (1 + 4mr2 /M R2 )2 =0 We can solve this for r. Set x = 4mr2 /M R2 . Then 1 − (4x/(1 + x)) = 0 so x = 1/3, and 1 4mr2 = ⇒ r=R 2 MR 3 M 12m If we plug in the mass values for this problem, we √ obtain r = R/ 6. Since the centrifugal force is everywhere positive, and it is zero at the center, this extremum must in fact be the maximum. 5. K&K problem 7.4 Referring to the diagram, the stone orbits around the vertical shaft with orbiting angular velocity Ω. The velocity of the stone’s CM is thus v = ΩR. The stone is rolling without slipping on the ﬂat surface, so its rolling angular velocity is ω= R v =Ω b b 5 The torque is thus τ= dL ˆ1 = −θ M b2 ω Ω dt 2 the direction of its angular momentum would not change as the car turns left or right. So the ﬂywheel should be vertical with its spin axis pointing either sideways or forward. Deciding between these alternatives requires a more quantitative analysis. Finally we consider the forces on the system. The vertical shaft exerts a force on the horizontal axle, gravity pulls down on the millstone’s CM, and the normal force pushes up on the millstone. However, with respect to the chosen origin, the ﬁrst of these forces can exert no torque because it is applied at r = 0. The torque ˆ due to gravity is in the +θ direction, and the torque due to the normal force N of the ﬂat ˆ surface on the millstone is in the −θ direction. ˆ direction, we have Thus, in the +θ 1 −N R + M gR = τ = − M b2 ω Ω 2 1 b2 M g+ ωΩ = N 2R Substituting ω = RΩ/b, N =M 1 g + bΩ2 2 Of course, by Newton’s third law, the contact force exerted by the millstone upon the ﬂat surface is equal and opposite to N . As advertised, the eﬀective weight of the millstone for crushing grain is greater than M g ; this increment rises quadratically with the angular velocity. What keeps the millstone from accelerating upward, since the upward normal force on it is greater than the downward force of gravity? The force of the vertical shaft on the horizontal axle, which we ignored in the torque equation because it is applied at the origin, must push downward, in alignment with gravity, with the value M bΩ2 /2. Such millstones must have been in use before the time of Newton, so the beneﬁts of their increased eﬀective weight when rolling in a circle must have been discovered empirically rather than logically. 6. K&K problem 7.5 (a.) If the ﬂywheel were horizontal with its spin axis pointing up, it would have little eﬀect, since Let’s look at the car from the rear while it is in motion with speed v . First we’ll consider the car without any ﬂywheel. Suppose that the car is in the process of turning to the left, taking a turn of radius R0 . Adopt a reference frame attached to the car, with an origin halfway between the tires at the level of the road. It is easy to see why the act of turning causes the normal forces on the tires to become unbalanced. The sum of the torques on the car must remain zero if the car (assumed to have no suspension system, so it doesn’t lean) keeps all four tires on the road. With respect to the origin chosen, the forces of friction on both tires can exert no torque, because these forces act directly toward or away from the origin. Neither can the force of gravity exert a torque about this origin, for the same reason. That leaves Nl and Nr , the normal forces on the left and right sets of tires, and M v 2 /R0 , the ﬁctitious centrifugal force which pulls the CM to the right in this accelerating frame. Let the CM be a distance d above the road; let the rightleft separation of the wheels be 2D. Along −v, the sum of the torques is then ˆ −Nl D + Nr D − M v2 d=0 R0 Clearly Nr must exceed Nl if this equation is to be satisﬁed. This is the problem we are trying to solve with the ﬂywheel. 6 The torques that we just considered were ˆ along −v. If the ﬂywheel is to help, its angular momentum L should be directed so that, when the car turns left, the ﬂywheel produces a torque on the car equal to +M v 2 d/R0 along ˆ −v. By Newton’s third law, the torque of the car on the ﬂywheel should correspondingly be equal to +vM v 2 d/R0 . So, as the car turns left, ˆ the change in L of the ﬂywheel should be directed along +v. This will happen if the angular ˆ momentum vector of the (vertical) ﬂywheel is pointing to the right. This means that the ﬂywheel should rotate in the opposite direction as the tires. For simplicity, we’ll install it at height d from the road so as not to perturb the CM. It’s not necessary to reverse the ﬂywheel direction for right as opposed to left turns, because both the centrifugal force and the change in L will correspondingly reverse direction. (b.) Now that we have determined the ﬂywheel direction, we can calculate the desired magnitude L of the ﬂywheel’s angular momentum. We have dL = LΩ = M v 2 d/R0 dt where Ω = v/R0 is the angular velocity of the car around the turn. Solving, L = M vd For a diskshaped ﬂywheel of mass m and radius r, I = mr2 /2, and the ﬂywheel’s angular velocity should be ω= 2M vd mr2 (a.) The acceleration of the truck is A, and the mass and width of its rear door are M and w. The door starts fully open. The door can be thought of as a series of thin sticks, pivoted about their ends. The moment of inertia of the door is thus 1 I = M w2 3 The easiest way to ﬁnd the angular velocity of the door is to use work and energy. The rotational kinetic energy is given by K= 12 Iω 2 The work done by a torque on a system is given by W= τ · dθ In this system there is one torque of interest. We use the hinge of the door as the origin, so the only torque comes from the ﬁctitious force of acceleration. When the door has swung through an angle θ, this torque is given by τ= 1 M Aw cos θ 2 Note that w/2 is just the distance to the center of mass. From this we can easily calculate the work done from 0 to 90 degrees.
π /2 W=
0 1 1 M Aw cos θ dθ = M Aw 2 2 Substituting the expression for rotational kinetic energy, we ﬁnd the angular velocity of the door after it has swung through 90◦ : 1 12 Iω = M Aw ⇒ ω = 2 2 3A w This is independent of the turn radius R0 , which is very nice. We’ve achieved perfectly ﬂat cornering for a turn of any radius! Unfortunately, ω depends linearly on the velocity v of the car. So, unless we can come up with a quick easy way of varying the kinetic energy of a big ﬂywheel in concert with the square of the car’s speed, we’re not going to get rich installing these devices as highperformance vehicle options. 7. K&K problem 8.2 (b.) The force on the door needs to do two things. It needs to accelerate the door at a rate A, and it needs to provide the centripetal acceleration to make the door rotate. At θ=90 degrees, the torque is zero, so the angular velocity is not changing. Instantaneously, the door is in uniform circular motion. The force required to accelerate the door is just FA = M A 7 The force required to provide the centripetal acceleration is w Fc = M ω 2 2 The total force is the sum of these, and they act in the same direction. We substitute the value for ω from part (a.) to get 5 3 F = FA + Fc = M A + M A = M A 2 2 8. K&K problem 8.4 (a.) This is a torque balance problem. A car of mass m has front and rear wheels separated by a distance l, and its center of mass is midway between the wheels a distance d oﬀ the ground. If the car accelerates at a rate A, it feels a ﬁctitious force acting on the center of mass. This tends to lift the front wheels. When the front wheels are about to lift oﬀ the ground, the normal force on the front wheels, Nf is zero. This means that the normal force on the back wheels Nb must equal the weight of the car, Nb = mg . The simplest origin to use in this problem is the point on the road directly under the center of mass. Here there are only three torques, due to the two normal forces and ﬁctitious force. The torque from Nb exactly balances the torque from the ﬁctitious force when the wheels are about to lift, so we have l 1 1 Nb l = mAd = mgl ⇒ A = g 2 2 2d For the numbers given, A = 2g = 19.6 m/sec2 . (b.) For deceleration at a rate g , again we simply apply torque balance about the same origin. We also need the fact that Nf + Nb = mg Torque balance gives 1 1 Nf l − Nb l − mgd = 0 2 2 Substituting from the previous equation, we get Nf = Nb = 1d + 2 l 1d − 2 l mg mg Plugging in the numbers, we get Nf = 3mg/4 = 2400 lb, and Nb = mg/4 = 800 lb. University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 8 1. K&K problem 8.5 “Many applications...”. 2. K&K problem 8.11 “A high speed hydrofoil...”. 3. K&K problem 9.3 “A particle moves...”. 4. K&K problem 9.4 “For what values of n...”. 5. K&K problem 9.6 “A particle of mass m...”. 6. K&K problem 9.12 “A space vehicle is in circular orbit...”. 7. A satellite of mass m is travelling at speed V in a circular orbit of radius R under the gravitational force of a ﬁxed mass at the origin. a. Taking the potential energy to be zero at inﬁnite radius, show that the total mechanical energy of the satellite is −mV 2 /2. b. At a certain point B in the orbit (see ﬁgure), the direction of motion of the satellite is suddenly changed without any change in the magnitude of the velocity. As a result, the satellite goes into an elliptical orbit. Its closest approach to the origin is now R/5. What is the speed of the satellite at this distance, expressed as a multiple of V ? c. Through what angle α (see ﬁgure) was the velocity of the satellite turned at point B ? motion that its angular momentum with respect to the cloud is not changing. What attractive (central) force could account for such an orbit? 8. The commander of a spaceship that has shut down its engines and is coasting near a strangeappearing gas cloud notes that the ship is following a path that will take it directly into the cloud (see the ﬁgure). She also deduces from the ship’s 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 8 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. K&K problem 8.5 A gyroscope with mass M has angular velocity ωs and moment of inertia Is . It pivots at one end, and the center of mass is a distance l from the pivot. The angular momentum of the gyroscope is thus L = Is ωs The gyroscope undergoes an acceleration a perpendicular to the spin axis. The ﬁctitious force will create a torque of magnitude τ = M al The direction of this torque is perpendicular to both the acceleration and the gyroscope axis (down in the ﬁgure), causing the gyroscope axis to precess in the direction indicated by the angle θ. The magnitude of the angular momentum will not change, but the direction will. Thus the gyroscope axis will rotate around the direction of acceleration. The rate at which this happens is ω , and dL = Lω = τ dt This gives the following relation M la(t) = Is ωs ω (t) Both the acceleration and the angular velocity can depend on time. If we integrate both sides of this equation, we can get a relation between the ﬁnal velocity and the total angle of rotation. The integral of the acceleration is just the velocity and the integral of the angular velocity is just the angle: M lv = Is ωs θ ⇒ v = 2. K&K problem 8.11 Is ωs θ Ml A hydrofoil moves with respect to the earth’s surface at the equator with velocity v = 200 mi/hr directed along each of the four points of the compass. At rest with respect to the surface of the earth, the acceleration of gravity is g. We are asked to ﬁnd the eﬀective gravitational acceleration g that is felt by a passenger (of mass m) who is at rest with respect to the hydrofoil. The (ﬁctitious) Coriolus force on the passenger, which is proportional to the passenger’s (vanishing) velocity in the hydrofoil’s frame, must be zero if evaulated in this frame. The (ﬁctitious) centrifugal force on the passenger is Fcent = −mΩ × (Ω × R) = −m(Ω (Ω · R) − R(Ω · Ω )) = m(−Ω (Ω · R) + RΩ 2 ) where Ω (v) is the total angular velocity of the passenger, due both to the rotation of the earth ˆ and to the motion of the hydrofoil; R = Rx is a vector pointing from the earth’s center to the hydrofoil at the equator; and the “bac cab” rule is applied to the ﬁrst line. The hydrofoil’s velocity has one of the four directions (E,W,N,S) = (ˆ , −y, z, −z), yielding an angular velocity ω y ˆˆ ˆ due to hydrofoil motion relative to the earth’s surface: v z ˆ ˆˆ ω = (ˆ, −z, −y, y) R To this one must add the earth’s angular velocity Ω = Ωˆ z in order to get the total angular velocity Ω = ω + Ω of the hydrofoil. Evidently Ω is perpendicular to R, so ˆ Fcent (v) = xmR(Ω (v))2 Therefore g points along −x, i.e. toward the ˆ earth’s center, for all four directions (E,W,N,S) of v, as does g. Thus ∆g −R((Ω 2 − Ω2 ) g −g ≡ = g g g 2 For these four directions, Ω
2 − Ω2 = 2Ωω + ω 2 (E) = −2Ωω + ω 2 (W) = ω 2 (N and S) Since Ueﬀ is constant, d2 r/dt2 = 0, so if the particle acquires a nonzero radial velocity it will continue with the same radial velocity. If the particle moves with uniform radial velocity vr , the following equations are satisﬁed dr = vr dt L dθ = dt mr2 (t) Therefore R ∆g = (−2Ωω − ω 2 ) (E) g g R = (2Ωω − ω 2 (W) g R = (−ω 2 ) (N and S) g Using ω/Ω = 0.1931 and RΩ /g = 0.003432, we calculate ∆g/g  = 0.001325 from the 2Ωω term and ∆g/g  = 0.000128 from the ω 2 term. Thus ∆g/g = −0.001453 (E) = +0.001197 (W) = −0.000128 (N and S) 3. K&K problem 9.3 A particle moves in a circle under the inﬂuence of an inverse cube law force. This means that the potential is inverse squared and it is attractive. The eﬀective potential is given by Ueﬀ (r) = L2 A −2 2 2mr r
2 Solving the ﬁrst is easy: r(t) = r0 + vr t. Plugging this result into the second equation, we ﬁnd dθ L = dt m(r0 + vr t)2 We can solve this equation by direct integration, assuming that θ(0) = 0: 1 L 1 L dt = − 2 mvr r0 r(t) 0 m(r0 + vr t) √ We replace L with 2mA to get the ﬁnal answer θ(t) = θ(t) = 1 vr 2A m 1 1 − r0 r(t)
t 4. K&K problem 9.4 A particle moves in a circular orbit in the potential U = −A/rn . We want to know for which values of n the orbit is stable. The eﬀective potential is given by Ueﬀ = L2 A −n 2 2mr r The radial force is zero for a circular orbit, so we can ﬁnd the radius. 0= L2 2A dUeﬀ =− 3 + 3 dr mr r To ﬁnd the circular orbit radius we evaluate dUeﬀ /dr = 0: L2 nA dUeﬀ = − 3 + n+1 dr mr r This gives the radius of the circular orbit r0 when we set it to zero.
n r0 −2 = This shows that a circular orbit can have any radius, but there is only one possible magnitude of angular momentum, given by L2 = 2Am Plugging this value of the angular momentum into the eﬀective potential, we ﬁnd the peculiar result that Ueﬀ = 0 nAm L2 n Since r0 −2 must be a positive quantity for any value of n, and A > 0, this equation requires n>0 3 We now look at the second derivative of the eﬀective potential at r = r0 . If it is positive, then it is a potential minimum and the orbit is stable. 3L2 n(n + 1)A d2 Ueﬀ = − 2 4 dr mr rn+2 2 n(n + 1)A 1 3L − =4 >0 r m rn−2 Since r is always greater than zero we can divide it away. Substituting for rn−2 at r = r0 , (n + 1)L 3L − >0 ⇒ n<2 m m Putting both inequalities together, 0<n<2 Recall from the previous problem that when n = 2 the motion is barely unstable. When n = 0, U is constant, so there is no attractive force, therefore no circular orbit: this case is also unstable. 5. K&K problem 9.6 A particle moves in an attractive central force Kr4 with angular momentum l. If it moves in a circular orbit with radius r0 , the central force must provide exactly the necessary centripetal acceleration: l2 l2 mv 2 4 7 = Kr0 = ⇒ r0 = 3 r0 mr0 mK Relative to r = 0, the energy of the orbit is E= 1 1 5 mv 2 + Kr0 2 5 3l2 d2 U = + 4Kr3 dr2 mr4
7 Plugging in l2 = m2 v 2 r2 = mKr0 , we get 2 2 To ﬁnd the frequency of small radial oscillations, we must evaluate the second derivative of the eﬀective potential. Remember that for small oscillations the eﬀective spring constant k for radial motion is k= d2 Ueﬀ dr2 r0 The eﬀective potential is Ueﬀ = l2 1 + Kr5 2 2mr 5 The second derivative is easily found d2 U dr2 r0 3 = 7Kr0 Substituting the value of r0 , we ﬁnd the eﬀective spring constant k and the angular frequency ω of radial oscillation about the stable circular orbit: k = mω 2 = 7K ω= 7K l2 m mK l2 mK
3/14 3/ 7 6. K&K problem 9.12 A spacecraft of mass m orbits the earth at a radius r = 2Re . It will transfer to another circular orbit with radius r = 4Re . (a.) We know the radius of each orbit, so we can easily ﬁnd the energies of the two orbits. The energy of a bound orbit in a 1/r potential is given by GM m E=− A where A is the major axis of the elliptical (here the diameter of the circular) orbit. We can use this to ﬁnd the energies of the two orbits. The values of A are simply 4Re for the ﬁrst and 8Re where we have integrated the force to get the 5 potential. Plugging in v 2 = Kr0 /m, we get E= 1 7 1 5 5 Kr0 + Kr0 = Kr5 2 5 10 0 Substituting the above value for r0 , we get the ﬁnal result for the energy (relative to r = 0): E= 7 K 10 l2 mK
5/ 7 4 for the second. The energy input needed to go from one orbit to the other is at least GM m ∆E = − Re 11 − 84 GM m = 8Re Again this is due to the change in speed. ∆E = 1 1 2 2 mv3 − mv2 ⇒ v3 = 2 2 GM 4Re Finally we obtain the change in speed at point B ∆vB = 1 − 2 1 6 GM = 726 m/sec Re Plugging in the values given, ∆E = 2.34 × 1010 joules (b.) At point A, the rocket is ﬁred, putting the spacecraft in an elliptical orbit. The major axis of this orbit is A = 6Re . To ﬁnd the initial speed, we use the energy equation. The energy is partly gravitational potential energy and partly kinetic energy: E= GM m GM m 1 mv 2 − =− 20 2Re 4Re Since the two velocities at point A are both tangent to each other, and similarly for point B , the only changes in the velocities at either point are the changes in their magnitudes. 7. A satellite of mass m moves in a circular orbit of radius R at speed v . It is inﬂuenced by the gravity of a xed mass at the origin. (a.) The mechanical energy of the satellite is given by GM m 1 E = mV 2 − 2 R We know that gravity exactly provides the centripetal acceleration. GM mV 2 GM m ⇒ =V2 = 2 R R R The total energy is thus E= 1 1 mV 2 − mV 2 = − mV 2 2 2 Solving this equation for v0 , we ﬁnd the orbital speed GM v0 = 2Re The energy of the elliptical orbit is given by E=− GM m GM m ⇒ ∆E = 6Re 12Re The change in energy at point A is entirely due to a change in speed. ∆E = 1 1 2 2 mv1 − mv0 ⇒ v1 = 2 2 2GM 3Re The change in speed required at point A is thus ∆ vA = 2 − 3 1 2 GM = 865 m/sec Re (b.) At a certain point on the orbit, the direction of travel of the satellite changes. The magnitude of the velocity does not change, so the total energy of the orbit doesn’t change. We can now ﬁnd the kinetic energy at closest approach: 1 5GM m 1 E = − mV 2 = mv 2 − 2 2 R 1 2 = mv − 5mV 2 2 ⇒ v = 3V (c.) The circular orbit has angular momentum L1 = mRV , while the elliptical orbit’s angular momentum, evaluated at the perigee, is L2 = m 3 R 3V = L1 5 5 We repeat this analysis at point B . Conservation of angular momentum gives 2Re v1 = 4Re v2 ⇒ v2 = GM 6Re The energy of the new circular orbit is given by E=− GM m GM m ⇒ ∆E = 8Re 24Re 5 Therefore, just after the transition from circular to elliptical orbit, a fraction 3 of the original 5 velocity must remain tangential, while 4 of it 5 becomes radial (the squares of the two fractions must add to unity according to Pythagoras). Therefore the satellite turns through an angle 4/5 = 53.1◦ α = arctan 3/5 8. A spaceship is moving on a circular path that will take it directly through a gas cloud. The angular momentum with respect to the gas cloud is measured to be constant. We want to know what attractive central force causes this. Immediately we notice that as the ship passes through the center of the cloud, it velocity must become inﬁnite, because the angular momentum l = mvr is conserved. If l were zero, the ship could only fall straight into the cloud. We can express the circular trajectory of the ship as a function of θ by inspection: r(θ) = 2R cos θ (−π/2 < θ < π/2) Take φ to be the azimuth of the spaceship on the circle (−π < φ < π ), with φ ≡ 0 when θ = 0. Consider the isosceles triangle with sides r, r, and R. Requiring its angles to add up to π , it is easy to see that φ = 2θ. We are given two deﬁnite facts. One is that the ship’s angular momentum about the center of the cloud ˙ ˙ L = mr2 θ = 4mR2 θ cos2 θ is constant. (This expression conﬁrms our pre˙ vious observation that the ship’s velocity 2Rθ must be inﬁnite at the center of the cloud, where cos θ = cos π/2 = 0.) The second fact is that the spaceship moves in a circle of radius R. The ˙ centripetal force mRφ2 required to keep it in circular motion must be supplied by the component along R of the unknown attractive force F : ˙ ˙ F cos θ = mRφ2 = 4mRθ2 ˙ Using the previous equation for L to eliminate θ from this equation, F cos θ = 4mR L2 16m2 R4 cos4 θ L2 F= 4mR3 cos5 θ Finally, using the ﬁrst equation to eliminate cos θ, L2 32R5 F= 4mR3 r5 8L2 R2 F= mr5 Since L and R are constant, the unknown attractive force depends on the inverse ﬁfth power of the spaceship’s separation from the center of the cloud, for this particular spaceship trajectory. The last equation is the desired result. However, this is no simple force ﬁeld: its coupling to the spaceship is contrived to depend quadratically both upon the spaceship’s angular momentum about the cloud’s center and upon the radius of its circular orbit. As an alternative to considering the centripetal force that must be supplied by F , one can hypothesize that F is a conservative as well as a central force. (At least it is clear from the fact that the spaceship’s orbit is closed that there can be no monotonic decrease or increase in the total energy E ). From the above equation for L, one readily sees that the ship’s speed ˙ ˙ v = Rφ = 2Rθ is proportional to r−2 . Therefore the ship’s kinetic energy K is proportional to r−4 . If E is to be conserved, the potential energy U also must be proportional to r−4 so that it can cancel the r dependence of K ; its radial derivative −dU/dr = Fr must then be proportional to r−5 . The constant of proportionality is easily veriﬁed to be the same as is given above. University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 9 1. a. Expand in Taylor series: f (x) = ln (1 − x); f (x) = 1/(1 + x). b. Given two functions s(x) and c(x) such that ds/dx = c and dc/dx = s, prove that s(x) + c(x) = [s(0) + c(0)]ex . 2. a. French problem 14(b). b. French problem 19. c. Prove DeMoivre’s theorem, (cos θ + i sin θ)n = cos nθ + i sin nθ. 3. French problem 315. 4. At t = 0 a bullet of mass m and velocity v0 strikes a motionless block of mass M which is connected to a wall by a spring of constant k . The block moves with coordinate x(t) (along the direction of the bullet) on a frictionless table next to the wall. The bullet embeds itself in the block. If x(t) = [A exp (i(ωt + φ))], with A real, evaluate ω , A, and φ. 5. French problem 45. 6. French problem 48. 7. A piano has middle C = 256 Hz, and CabovemiddleC = 512 Hz. The white keys of its middle octave consist of middle C; D (1 step above middle C); E (2 steps); F (2.5); G (3.5); A (4.5); B (5.5); and CabovemiddleC (6 steps). Each step causes the frequency to be multiplied by a ﬁxed factor. a. Find the frequencies of D, E, F, G, A, and B. b. If G were tuned to a “perfect ﬁfth”, its second harmonic (2× the fundamental frequency) would be the same as middle C’s third harmonic. Find the beat frequency between the second harmonic of G and the third harmonic of middle C. (Pianos are tuned by listening for such beats.) Stringed instruments are tuned (for compositions in the key of C) so that this beat frequency is zero, producing a smoother tone. When a composition in a diﬀerent key is played, the stringed instrument can be retuned for that key, which would be impossible for the piano. 8. For an underdamped undriven harmonic oscillator with ω0 /γ ≡ Q = 100, ﬁnd the number of oscillations required to reduce the amplitude of oscillation by the factor eπ ≈ 23.1. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 9 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. (a.) The Taylor series for ln(1 − x) is found as follows: f (x) = ln(1 − x) = f (n) (0) n x n! n=0
∞ We recognize the sum as the Taylor series of ex . s(x) + c(x) = [s(0) + c(0)]ex 2. (a.) French problem 14(b). The magnitude of a complex number a + ib, √ where a and b are real, is just a2 + b2 . The phase angle θ is equal to tan−1 (b/a), where the quadrant is determined by the signs of both a √ √ and b. The ﬁrst vector (2 + i 3) has length 7 √ and phase θ = tan−1 ( 3/2) = 40.9◦ . The second √2 vector (2 − i 3) is merely the square of the complex conjugate of the ﬁrst vector. Therefore it has length 7 and −2× the phase, or −81.8◦ . (b.) French problem 19. The value of ii is a little odd, but here goes. We need to know how to ﬁnd the log of a complex number: ln z = ln z eiθ = ln z  + iθ There is an ambiguity here, because we can always add an integer multiple of 2π to θ. Here we will choose not to do so, but simply take the value of θ to be between −π and π . Thus we obtain ii = ei ln i = ei
2 where f (n) denotes the nth derivative of f . 1 1 ln(1 − x) = −x − x2 − x3 + · · · 2 3 ∞ n x ln(1 − x) = − n n=1 We do the same for f (x) = 1/(1 + x). 1 = 1 − x + x2 − x3 + · · · 1+x ∞ 1 = (−1)n xn 1 + x n=0 (b.) We have two functions c(x) and s(x) related as follows: dc ds =c =s dx dx An easy way to approach this problem is to solve these diﬀerential equations simultaneously. However, as is the case for most “easy” ways to do things, the mathematics leading up to the solution is somewhat advanced. Instead we will use the Taylor series to solve it. Expanding around x = 0, 1 s(x) = s(0) + c(0)x + s(0)x2 + · · · 2 1 c(x) = c(0) + s(0)x + c(0)x2 + · · · 2 Adding these two equations, we see that s(x) + c(x) = [s(0) + c(0)] xn n! n=0
∞ π/2 = e−π/2 = 0.2079 This means that paying 20 cents is bargain, but a very small one. (c.) Prove (cos θ + i sin θ)n = (cos nθ + i sin nθ). This is pretty trivial when we remember (cos θ + i sin θ)n = eiθ
n = einθ = cos nθ + i sin nθ 3. French problem 315. An oscillatory system loses energy according 2 to E = E0 e−γt . Q ≡ ω0 /γ . We deﬁne the Q value as two places in one oscillation. We want a place where it is zero and rising, because we know that x is increasing at the instant of contact. This is φ = −π/2 The solution is now x(t) = [A exp(i(ωt − π/2)]) We diﬀerentiate to get the velocity, and evaluate this at t = 0. v (0) = v = −Aω sin(−π/2) = Aω ⇒ A = This gives the ﬁnal result for the amplitude A= 5. French 45. (a.) A pendulum is forced by moving the point of support. The coordinate x gives the location of the pendulum bob, and ξ gives the location of the point of support. The forces on the pendulum are the damping force, which we must assume to be proportional to the absolute velocity of the pendulum, and the force of gravity. The force of gravity depends on the angle by which the pendulum is raised. This is proportional to the distance that the pendulum bob is displaced from the point of support, x − ξ . This gives the equation of motion m dx mg d2 x = −b − (x − ξ ) dt 2 dt l mv0 k (M + m) v ω (a.) Middle C on a piano is played, and the energy decreases to half of its initial value in one second. The frequency is 256 Hz. The angular frequency is this times 2π , so ω0 =1608.5/sec. We ﬁnd γ from 1 = (e−γ/2 )2 ⇒ γ = 0.693 2 Lastly , the Q of the oscillator is Q = 1608.5/0.693 = 2321 (b.) The note one octave above is struck (512 Hz). The decay time is the same, so the Q value is simply doubled: Q=4642. (c.) A damped harmonic oscillator has mass m = 0.1 kg, spring constant k = 0.9 N/m, and a damping constant b. The energy decays to 1/e in 4 seconds. This means that 1 = e−4γ ⇒ γ = 0.25 sec−1 e ⇒ b = mγ = 0.025 kg sec−1 The natural frequency ω0 = k /m = 3 Hz. Finally, the Q of the oscillator is Q = ω0 /γ = 12. 4. At t = 0, the bullet collides inelastically with the block, so only the momentum is conserved. The ﬁnal velocity of the block and bullet is given by mv0 mv0 = (M + m)v ⇒ v = M +m We now have the initial conditions for the oscillation. The initial position is x(0) = 0 and the initial velocity is v (0) = v . The frequency ω is given as usual by k /mass, but the mass in question is the total mass of the system: ω= k M +m 2 Using ω0 = g/l and γ = b/m, we put this in the standard form with ξ as a forcing term. dx d2 x 2 2 + ω0 x = ω 0 ξ +γ 2 dt dt (b.) The motion of the point of support is given by ξ (t) = ξ0 cos ωt. We use the formula for the amplitude of forced oscillation, but we note that in this case, the equation is dx d2 x 2 2 + ω0 x = ω0 ξ0 cos ωt +γ 2 dt dt The solution is given by x(t) = [A exp(i(ωt + φ))]. The initial position tells us that cos φ = 0. This is ambiguous because the cosine is zero in 3
2 The constant ω0 ξ0 takes the place of the F0 /m we normally see in this type of equation. The amplitude of the oscillation is thus given by 2 ω0 ξ0 2 (ω0 − ω 2 )2 + γ 2 ω 2 The solutions to this are ω2 = 1 2
2 2ω 0 − γ 2 ± 2 γ 4 + 12γ 2 ω0 A(ω ) = Plugging in the numbers, the two frequencies are ω = 3.147 sec−1 and ω = 3.113 sec−1 . 6. French 48. (a.) A mass is under the inﬂuence of a viscous force F = −bv . Let γ = b/m as usual. The equation of motion is dv + γv = 0 dt We can easily solve this equation by direct integration. v (t) = v0 e−γt We simply integrate this equation with respect to t to get the position. x(t) = C − v0 −γt e γ At exact resonance, ω = ω0 and the amplitude is ω0 = Qξ0 A(ω0 ) = ξ0 γ Now we want to ﬁnd Q. We are given that the forcing amplitude is ξ0 = 1 mm. The length of the pendulum is l = 1 m, so this gives ω0 = 3.13 sec−1 . We know that the amplitude falls oﬀ by a factor of e after 50 swings, or 50 periods. We know that A = A0 exp(−γt/2) so γt = 2. t is 50 periods, or 100π/Ω, where Ω is the frequency of free oscillation Ω= Then γt = 2 100π =2 γ Ω 50πγ =
2 ω0 + γ 2 /4 2 (50π )2 γ 2 = ω0 + γ 2 /4 2 ω0 + γ 2 /4 C is the integration constant that will allow us to ﬁt an initial condition. (b.) A driving force F = F0 cos ωt is turned on. We want to ﬁnd the steady state motion. We will use a complex exponential for the forcing term, with the understanding that we take the real part when we’re done. The new equation of motion is dx F0 iωt d2 x +γ = e 2 dt dt m Assume a solution of the form x(t) = Aeiωt where A is a complex number. Plugging this into the equation of motion, we see that −ω 2 A + iωγA = F0 /m F0 ⇒ A= m −ω 2 + iωγ Plugging in the numbers, we get γ = 0.0199. This gives us Q = 157, and A = 15.7 cm. (c.) We want to ﬁnd the frequencies where the amplitude is half of the resonant value. We merely solve
2 ω0 ξ0 2 (ω0 − ω 2 )2 + γ 2 ω 2 = ξ0 ω0 2γ This gives
2 2 4ω0 γ 2 = (ω0 − ω 2 )2 + γ 2 ω 2 Turning this into a quadratic equation for ω 2 , we get
2 4 2 0 = ω 4 + (γ 2 − 2ω0 )ω 2 + ω0 − 4ω0 γ 2 We write the denominator as the product of a magnitude and a phase. The magnitude is ω 4 + ω 2 γ 2 . The denominator has a negative real part and a positive imaginary part, so it is in the second quadrant with phase π − arctan (γ/ω ). Since the numerator is real, 4 the phase of A is minus the phase of its denominator, or arctan (γ/ω ) − π . According to the notation of the problem, the phase of the oscillation is −δ , so we ﬁnd δ = π − arctan (γ/ω ) The amplitude of the oscillation is just the magnitude of A, given by A = F0 /m ω4 + ω2 γ 2 − Plugging this value of v0 into the ﬁrst equation, C=− F0 /m F0 /m +2 =0 2 + γ2 ω ω + γ2 Collecting these results, the solution satisfying both boundary conditions is x(t) = F0 /m −γt e ω2 + γ 2 F0 /m ω4 + ω2 γ 2 cos(ωt − tan−1 (γ/ω )) The general solution to the problem is x(t) = C − − v0 −γt e γ F0 /m 7. Middle C is 256 Hz, and C above it is double that frequency, or 512 Hz. The scale is divided into 6 whole steps, or 12 half steps. The note after each half step is a constant multiple f of the frequency of the previous note. When we have gone up twelve half steps, the frequency will have doubled. The constant factor f is thus given by f 12 = 2 or f = 21/12 . (a.) The frequencies in the scale are thus C=256, D=287.4, E=322.5, F=341.7, G=383.6, A=430.5, B=483.3, C=512 Hz. These are zero, two, four, ﬁve, seven, nine, eleven, and twelve half steps above middle C, respectively. (b.) Middle C’s third harmonic is three times its fundamental frequency, or 768 Hz. The second harmonic of G is 767.133 Hz. The beat frequency is always just the diﬀerence between the two frequencies. In this case the beat frequency is 0.867 Hz, which is easily audible to the piano tuner. 8. An undriven oscillator that is underdamped has a Q of 100. We want to know how many oscillations it takes to damp by a factor of eπ . This just means that γt/2 = π . Now we want to write t in terms of the number of oscillations n. This takes n times the period, or t = 2πn/ω0 . This gives γn/ω0 = 1. Remember that Q = ω0 /γ , so n = Q = 100. ω4 + ω2 γ 2 cos(ωt + tan−1 (γ/ω )) At t = 0, we want x = 0. At t = 0, the last term B in the general solution is B (0) = − =− F0 /m ω4 ω4 + F0 /m ω2 γ 2 ω2 γ 2 cos(tan−1 (γ/ω )) ω ω2 + γ2 + F0 /m =− 2 ω + γ2 Thus the condition x(0) = 0 gives us one equation for C and v0 : C= F0 /m v0 +2 γ ω + γ2 The ﬁrst time derivative of B , evaluated at t = 0, is F0 /m ˙ B (0) = sin(tan−1 (γ/ω )) ω2 + γ 2 F0 γ/m =2 ω + γ2 Then, requiring the ﬁrst time derivative of the general solution to vanish at t = 0, the second equation for v0 and C is F0 γ/m ω2 + γ 2 F0 γ/m v0 = − 2 ω + γ2 0 = 0 + v0 + University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 10 1. French problem 56. 2. French problem 510. 3. French problem 514. 4. A transverse wave on a string has: • at x=0, a time variation of the form cos ωt + sin ωt with a period of 10−2 sec. • a wavespeed of 10 m/sec. • the appearance of propagating to the left (toward smaller x). • an amplitude of 0.01 m. Suppose that the string is described by y (x, t) = (A+ exp i(ωt − kx) + A− exp i(ωt + kx)) where A± are complex constants. a. Substitute numbers in place of all the variables (except x and t) in the above equation. b. Compute the maximum transverse speed and the maximum slope of the string. 5. If surface tension is included, the phase velocity of a surface wave on a liquid of mass density ρ and surface tension T is vph = gλ 2πT + λρ 2π and it is under tension S . a. What is its period of oscillation T ? b. At t = T /2, what is its shape? 7. A guitar string tuned to middle C (256 Hz) is “plucked” at exact center by giving it nonzero transverse velocity in a very small region. What tones (in Hz) will be heard in addition to 256 Hz? where λ is the wavelength and g is the acceleration due to earth’s gravity. a. Find the wavelength, frequency, and phase velocity of ripples on water which advance with minimum speed. b. Find the group velocity of such a wave. 6. A nondispersive string (ω/k = constant), initially at rest, has length L (0 ≤ x ≤ L), and mass per unit length µ. At t=0, its shape is y (x) = 3 sin 3πx πx + sin L L 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 10 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. French 56. (a.) All three springs are identical, with constant k . The equations of motion are d2 xA 2 2 + 2ω0 xA − ω0 xB = 0 dt2 d2 xB 2 2 + 2ω0 xB − ω0 xA = 0 dt2 We plug in a guessed solution, where the two masses oscillate at the same frequency, but with diﬀerent amplitudes A and B . This gives
2 2 −ω 2 A + 2ω0 A − ω0 B = 0 ⇒ B = A 2 2ω0 − ω 2 2 ω0 2 ω0 − ω2 position xA = 0. This happens when ω0 τ is in √ the second quadrant and 3ω0 τ is in the third: π − ω0 τ = 3ω0 τ − π 2π √ ω0 τ = 1+ 3 √ However, at t = τ , mass B will not have returned to its full original displacement since  cos τ  < 1. Thus, even though mass A will be back in place, mass B will not, and the system will not have returned to (plus or minus) its original state. In fact, because the ratio of the two normal frequencies is irrational, once both normal modes are excited the system can never return to its original state. 2. French 510. The equations of motion for this double spring system are as follows. The coordinate of the top mass is xA and the coordinate of the bottom mass is xB . d2 xA 2 2 + 2ω0 xA − ω0 xB = 0 dt2 d2 xB 2 2 + ω0 xB − ω0 xA = 0 dt2 Plugging in the standard guess that both masses oscillate at the same frequency, but at amplitudes A and B , we get the following equations.
2 2 (2ω0 − ω 2 )A = ω0 B 2 2 (ω0 − ω 2 )B = ω0 A 2 2 −ω 2 B + 2ω0 B − ω0 A = 0 ⇒ B = A 2 2ω0 Equating these we see that
2 4 (2ω0 − ω 2 )2 − ω0 = 0 Solving this quadratic equation, we ﬁnd that the 2 2 two frequencies are ω 2 = ω0 and ω 2 = 3ω0 . (b.) One mass is displaced by 5 cm. This excites each normal mode equally, with amplitude 2.5 cm. To see this, ﬁrst excite the ﬁrst normal mode with amplitude 2.5 cm. Now both masses are +2.5 cm from equilibrium. Now excite the second normal mode, also with amplitude 2.5 cm. This moves one mass forward 2.5 cm, and the other back 2.5 cm. One is now 5cm from equilibrium, and the other is at its equilibrium position. It is mass B that is displaced, so the masses obey xA = 2.5 cos ω0 t − 2.5 cos xB = 2.5 cos ω0 t + 2.5 cos √ 3ω0 t 3ω0 t Equating these, we get the quadratic equation
2 4 ω 4 − 3ω 0 ω 2 + ω 0 = 0 √ The solutions to this equation are
2 ω± (c.)√ After a time τ such that cos ω0 τ = cos 3ω0 τ , mass A returns to its equilibrium = 23 ω0 √ ±5 2 2 The amplitudes in these modes are easily found. For ω+ , we have √ 32 52 2 2 ω A+ = ω0 B+ 2ω0 − ω0 − 2 20 √ 1− 5 A+ B+ = 2 Likewise for ω− , √ 1+ 5 A− B− = 2 3. French 514. In the ﬁrst normal mode, the three particles have √ √ an amplitude ratio 2/2 : 1 : 2/2. The second normal mode has amplitude ratios 1 : 0 : −1. The third normal mode has amplitude ratios √ √ 2/2 : −1 : 2/2. 4. A wave is described by y (x, t) = A+ ei(ωt−kx) + A− ei(ωt+kx) Now we can ﬁnd the amplitude and frequency. The solution is y (x, t) = A cos ω t + kx − π 4 The amplitude is given as 0.01 m, and the period is 10−2 sec. The frequency ω = 2π/T = 200π sec−1 . The speed of waves on the string is c =10 m/sec, and we know that ω = ck , so this tells us that k = 20π m−1 . We now have the ﬁnal result y (x, t) = 0.01ei(200πt+20πx−π/4) = 0.01 cos(200πt + 20πx − π/4) (b.) We can now compute the maximum transverse speed and maximum slope. The transverse speed is dy = −2π sin(200πt + 20πx − π/4) dt The maximum value that the sine takes is 1, so the maximum transverse speed is 2π = 6.28 m/sec. We can likewise ﬁnd the maximum slope dy = −0.2π sin(200πt + 20πx − π/4) dx The maximum slope is thus π/5 = 0.628. 5. The phase velocity of surface waves is given by gλ 2πT + = λρ 2π g kT + ρ k (a.) We know that the wave is moving to the left. This corresponds to the second exponential. To see this, we note that a speciﬁc place on the wave train always has the same value of ωt ± kx. We want to see what happens when t increases. For the solution exp(i(ωt + kx)), we see that as t increases, x must decrease to stay on the same place in the wave. This is a left moving wave. We thus note that A+ = 0. We can now write the complex constant A− = Aeiδ , where A and δ are real. y (x, t) = Aei(ωt+kx+δ) vph = At x = 0 we know that the time dependence in proportional to cos ωt + sin ωt. This tells us that cos(ωt + δ ) ∝ cos ωt + sin ωt Using the formula for the cosine of a sum cos(ωt + δ ) = cos ωt cos δ − sin ωt sin δ For this to work we see that cos δ = − sin δ ⇒ δ = − π 4 (a.) Notice that at both zero and inﬁnite wavenumber, the phase velocity is inﬁnite. To ﬁnd the minimum phase velocity, we diﬀerentiate vph with respect to k and set to zero. T /ρ − g/k 2 dvph =0 = dk 2 k T /ρ + g/k k= ρg ⇒ λ = 2π T T ρg 3 The phase velocity at this wavenumber is vph = 4gT ρ
1/ 4 The frequency ω = vph k , which gives ω= 4gT ρ
1/ 4 term has frequency ω2 = 3cπ/L. The periods of these two oscillations are T1 = 2L/c and T2 = 2L/3c. The period of the total oscillation is the longer period, T1 . In one long period the fast oscillation has had exactly three periods. Thus, the period is T= 2L = 2L c µ S ρg = T 4ρg 3 T 1/ 4 (b.) The group velocity of this wave is given by vgr = dω dk √ k= ρg/T (b.) After a time T /2, the ﬁrst term has gone through a half period, and the second term has gone through one and a half periods. In both cases, this just means that there is a minus sign out front. y (x, T /2) = −3 sin 3πx πx − sin L L We know that ω = vph k , so ω= T k3 + gk ρ Taking the derivative with respect to k and evaluating at the wavenumber we found before dω dk √ = = 3T k 2 /ρ + g 2 T k 3 /ρ + gk 4gT 2g = k ρ
1/ 4 k= ρg/T k= √ ρg/T 7. A string of frequency 256 Hz is plucked in the exact center. This means that the even numbered modes are not exited at all. This is because the initial condition is symmetric around the middle of the string, and the even numbered modes are antisymmetric around the middle of the string. The odd numbered modes are also symmetric around the center of the string, so they survive. These frequencies are fn = (2n + 1)256 Hz, n = 0 . . . ∞ The slowest waves have the same phase and group velocities. This is a general result. Look at the equation for the frequency, and diﬀerentiate it to get the group velocity ω = vph k ⇒ dω dvph = vgr = vph + k dk dk We chose the phase velocity to be a minimum, so vgr = vph . 6. A string has tension S and linear mass density µ. This tells us the phase velocity of waves on it, c = S/µ. The string has length L. At t = 0, the string’s shape is y (x, 0) = 3 sin 3πx πx + sin L L (a.) The frequencies of each of these is ω = ck , so the ﬁrst term has ω1 = cπ/L and the second University of California, Berkeley Physics H7A Fall 1998 (Strovink) PROBLEM SET 11 1. French problem 69. 2. French problem 615(a). 3. French problem 89. Note that 1 m v 2 = 2 3 2 kT for the sodium atoms in the vapor, where T is the temperature in degrees Kelvin (◦ K), and k is Boltzmann’s constant, 1.38 × 10−23 J/◦ K. 4. French problem 812. 5. A tank is ﬁlled with water to a height H . A hole is punched in its wall a depth h below the surface of the water. a. Find the horizontal distance from the bottom of the tank that the stream of water hits the ground. b. Could a hole punched at a diﬀerent depth produce a stream with the same horizontal range? If so, at what depth? 6. Consider the stagnant air at the front edge of an airplane wing and the air rushing over a wing surface at speed v . Find the greatest possible value for v in streamline ﬂow, using Bernoulli’s equation and assuming that air is incompressible. Take the density of air to be 1.2×10−3 g/cm3 . Compare this numerical result with the speed of sound, 340 m/sec. 7. Verify by explicit computation in a Cartesian coordinate system that ∇ × (∇f (x, y, z )) = 0 8. (a.) Consider the function f (x, y, z ) ≡ x2 + y 2 − z 2 . At the point (x, y, z ) = (3, 4, 5), ﬁnd the direction of a vector ds (of small ﬁxed length) such that df /ds is a maximum. (b.) Consider the surface z (x, y ) = x2 + y 2 . At the point (x, y, z ) = (3, 4, 5), ﬁnd the direction of a vector du (of small ﬁxed length) which is normal to this surface. 9. A ﬂuid has a velocity ﬁeld v(x, y, z, t) = (yx − xy )ω (t) ˆ ˆ where ω is some function of time t. (a.) Prove that the ﬂuid density ρ(x, y, z, t) satisﬁes ∂ρ 1 ∂ρ ∂ρ −x = y ∂x ∂y ω ∂t (b.) Show that the angular velocity of the ﬂuid about the origin, evaluated at an arbitrary point, is half of ∇ × v evaluated at the same point. (c.) If ω (t) = ω0 = constant, prove that dv 2 = −rω0 dt where r = xx + yy , and dv/dt is the time rate ˆ ˆ of change of the velocity of an element of ﬂuid that is temporarily at (x, y, z ) at time t. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PROBLEM SET 11 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. French 69. (a.) The lowest resonant frequency of a room is 50 Hz. All integer multiples of this frequency are also resonant. The lowest two modes are excited. These are 50 Hz (the fundamental) and 100 Hz (the [ﬁrst] harmonic). The amplitude is maximum at t = 0. The time interval t = 1/200 sec is one fourth of a period for the fundamental and half a period for the harmonic. t = 1/100sec is one half of a period for the fundamental and a full period for the harmonic. These modes look like A1 = 10 µ. The last two equations become √ ξ (L/4) = 5 2 cos (100πt) + A2 cos (200πt) √ ξ (3L/4) = 5 2 cos (100πt) − A2 cos (200πt) As a trial solution, we assume that A2 is positive. Then the maximum displacement of 10 µ at √ 4 L/ is reached at t = 0; therefore A2 = (10 − 5 2) µ. The maximum displacement at 3L/4, a negative displacement in this case, is reached at t = 1/100 sec, when the harmonic has changed phase by a full 2π , but the fundamental has changed phase by only π and has therefore become negative. With the above values of A1 and A2 , the displacement at 3L/4 is equal to 10 µ, in agreement with the problem. 2. French 615(a). This is a bit messy, but bear with it. The string has length L. Its initial conditions are y (x, 0) = Ax(L − x) and (∂y/∂t)t=0 = 0. We write the solution as a Fourier series
∞ y (x, t) =
n=1 An sin nπx cos(ωn t − δn ) L (b.) We can write the total displacement as ξ (x) = A1 sin (πx/L) cos (100πt)+ + A2 sin (2πx/L) cos (200πt) where A1 and A2 are the unknown amplitudes for the fundamental and harmonic mode, respectively. In particular ξ (L/2) = A1 cos (100πt) + 0 A1 ξ (L/4) = √ cos (100πt) + A2 cos (200πt) 2 A1 ξ (3L/4) = √ cos (100πt) − A2 cos (200πt) 2 Since the amplitude at L/2 is due only to the fundamental, and is equal to 10 µ, we know that We know the solution at t = 0. Deﬁne Bn = An cos δn .
∞ y (x, 0) =
n=1 Bn sin nπx L We can now solve for the Bn using Fourier’s trick: Bn = 2 L
L 0 Ax(L − x) sin nπx dx L Integrating by parts once, notice that the surface term vanishes: Bn = 2A nπ
L 0 (L − 2x) cos nπx dx L 2 The L term integrates now. It always integrates to zero. Bn = − 4A nπ
L x cos
0 nπx dx L We integrate by parts again, and again the surface term vanishes. 4AL Bn = 2 2 nπ This integrates easily. 4AL2 nπx Bn = − 3 3 cos nπ L
L 0 L 0 The constant k = 1.38 × 10−23 J/Kelvin is Boltzmann’s constant, and the appropriate mass m ≈ 23mp is just the mass of the sodium atom. v 2 ≈ vmax , this gives T ≈ Approximating 900 Kelvin. This eﬀect can be used to measure the temperatures of objects in astronomy. 4. French 812. The Doppler eﬀect for a moving source and ﬁxed observer is given by ν0 1 − u cos θ v nπx dx sin L 8AL2 = 33 nπ ν (θ) =
n odd We see that Bn is zero for all even n. We expected this because the initial condition is symmetric around the center of the string. Now we tackle the velocities. ∂y ∂t
∞ =
t=0 n=1 An sin nπx ωn sin(δn ) = 0 L This equation is satisﬁed if we simply set all of the δn = 0, so that sin δn = 0 and cos δn = 1. This also means that An = Bn . We now have the full solution. We have rewritten the sum to only include odd n. y (x, t) = 8AL2 × (2m + 1)3 π 3 m=0 (2m + 1)πx cos(ω2m+1 t) L
∞ (a.) We want to ﬁnd the Doppler eﬀect for a ﬁxed source and a moving observer at velocity −u. We are going to do this diﬀerently than it was done in French, in order to provide an alternative and perhaps easier way to think about it. We are going to go into the rest frame of the wave crests, so the only thing moving will be the observer and the source. The velocities of the wave and the observer are Vw = (v, 0) Vo = (−u cos θ, u sin θ) Transforming into the rest frame of the wave by subtracting (v, 0), the velocity of the observer is Vo = (−v − u cos θ, u sin θ) The distance between wave crests is just L = v/ν0 , so the rate at which the observer crosses the wave crests is just the x velocity divided by this distance. This is the shifted frequency. u cos θ v × sin 3. French 89. This problem concerns a very important eﬀect called Doppler broadening. Sodium atoms emit light of 6000˚. The observed light varies in a A small frequency range of (6000 ± 0.02)˚. This A is caused by the thermal motion of the sodium atoms. The Doppler eﬀect tells us that v 0.02 ∆λ == = 3.33 × 10−6 λ c 6000 This gives the maximum velocity of the atoms vmax = 1000 m/sec. The thermal velocity is given by m v2 3 1 m v 2 = kT ⇒ T = 2 2 3k ν (θ) = ν0 1 + This trick doesn’t work for light because light doesn’t have a rest frame. In relativity, there is a diﬀerent Doppler formula that applies to both situations. It uses only the relative velocity of the source and the observer. 3 (b.) We want to know the approximate diﬀerence between the two formulas. We Taylor expand the ﬁrst formula, assuming that the speed is much less than the sound speed, u v. ν (θ) ≈ ν0 1 + u2 u cos θ + 2 cos2 θ + · · · v v The maximum velocity occurs when the pressure is zero. Plugging in P0 =105 N/m2 and ρ = 1.2 kg/m3 , we ﬁnd that v =408 m/sec. This is larger than the speed of sound, vs = 340 m/sec. 7. The gradient ∇f is equal to ˆ f = x ∂f ∂f ∂f ˆ ˆ +y +z ∂x ∂y ∂z This tells us the approximate diﬀerence between the two Doppler shifted frequencies ν (θ) − ν (θ) ≈ ν0 u2 cos2 θ v2 5. We can use Bernoulli’s equation to ﬁnd the velocities of these two ﬂows. (a.) If we look right as the ﬂow is leaving the tank, we see that it must be at atmospheric pressure. The stream is arbitrarily thin, so this is the only possibility. We then just use the gravitational potential to ﬁnd the velocities. If the hole were at the top of the tank, the velocity would be zero, so we use this as the constant. 1 P0 + ρg (H − h) + ρv 2 = P0 + ρgH 2 v (h) = 2gh The time it takes to hit the ground is given by the formula for constant acceleration. H −h= 12 gt ⇒ t = 2 2(H − h) g The x component of ∇ × (∇f ) is the diﬀerence between the y and z partial derivatives, respectively, of the z and y components of ∇f : (∇ × (∇f ))x = ∂ ∂f ∂ ∂f − ∂y ∂z ∂z ∂y Interchanging the order of diﬀerentiation in either of the terms, this expression is seen to vanish for wellbehaved f . By cyclic permutation, the y and z components of ∇ × (∇f ) vanish as well. 8. (a.) When the point of observation (x, y, z ) is displaced incrementally by ds, where ˆ ˆ ˆ ds = xdx + ydy + zdz points in an arbitrary direction, the change df in f (x, y, z ) is given by the chain rule: df = ∂f ∂f ∂f dx + dy + dz ∂x ∂y ∂z In this time, the water travels a horizontal distance vt, so the distance from the tank is d(h) = 4h(H − h) The righthand side can be rewritten as the dot product of ˆ ∇f = x and ds above: df = ∇f · ds For a ﬁxed length ds, this dot product is greatest when ds is parallel to ∇f . Therefore, when df /ds is a maximum, the direction of ds will be ∂f ∂f ∂f ˆ ˆ +y +z ∂x ∂y ∂z (b.) We notice that the previous formula says that water leaking from a depth h travels the same distance as water leaking from a depth H − h. 6. We again apply Bernoulli’s equation. The air at the leading edge is stagnant, and we will assume that it is at atmospheric pressure. Bernoulli’s equation gives 1 P0 = P + ρv 2 2 4 along ∇f . With f = x2 + y 2 − z 2 , this direction ˆ n is ˆ ∂x ˆ ∂y ˆ ∂z x ∂f + y ∂f + z ∂f ˆ n= ∇f  ˆ ˆ ˆ x2x + y2y − z2z = 2 x2 + y 2 + z 2 ˆ ˆ ˆ x6 + y8 − z10 =√ 2 9 + 16 + 25 ˆ ˆ ˆ x3 + y4 − z5 √ = 52 (b.) The surface z (x, y ) = described as x2 + y2 can be (c.) Suppose that the independent variables (x, y, z, t) upon which a vector A(x, y, z, t) depends change inﬁnitesimally, by (dx, dy, dz, dt). Then, by the chain rule, a component of A, e.g. Ax , changes by an amount dAx = dAz = dt = dAx = dt dA = ⇒ dt ∂Ax ∂Ax ∂Ax ∂Ax dt + dx + dy + dz ∂t ∂x ∂y ∂z ∂Ax ∂Ax ∂Ax ∂Ax + vx + vy + vz ∂t ∂x ∂y ∂z ∂ ∂ ∂ ∂Ax + vx + vy + vz Ax ∂t ∂x ∂y ∂z ∂ + v · ∇ Ax ∂t ∂ +v·∇ A ∂t xy plane and orthogonal to r⊥ . Thus the element is in circular motion about the z axis (to which r⊥ is perpendicular), with angular velocity ˆ Ω=z On the other hand ˆ ∇ × v = ω (t) z = 2ˆω (t) z ∂ x ∂y + ∂x ∂y v ˆ = zω (t) r⊥  0 = f (x, y, z ) = x2 + y 2 − z 2 This is the same f (x, y, z ) as in part (a.). Suppose the point of observation (x, y, z ) is displaced inﬁnitesimally by dv, where dv is on the surface f = 0. Then we would expect f not to change at all. However, according to the results of part (a.), df = ∇f · dv Therefore df can vanish only if dv is perpendicular to ∇f . Since dv can be any displacement which lies on the surface, this requires ∇f to be perpendicular to the surface. Therefore, the direction of the normal to the surface du in part ˆ (b.) is the same as the direction n of ds in part (a.), the direction of maximum change in f . 9. The ﬂuid velocity ﬁeld is ˆ v(x, y, z, t) = (ˆ x − xy )ω (t) y (a.) The equation of continuity (conservation of ﬂuid molecules) requires ∂ρ + ∇ · (ρv) = 0 ∂t Therefore ∂ρ ˆ = −ω ∇ · (ˆ xρ − xyρ) y ∂t ∂ρ ∂ρ 1 ∂ρ =y −x ω ∂t ∂x ∂y ˆ ˆ (b.) An element of ﬂuid at r⊥ = xx + yy has a ˆ velocity v = (ˆ x − xy )ω (t) that is always in the y (This is the convective derivative, yielding the total time rate of change of A.) In this problem A = v itself, so ∂ dv = +v·∇ v dt ∂t ∂ ∂ ˆ ˆ + vy (yx − xy ) = 0 + ω0 v x ∂x ∂y ∂ ∂ 2 ˆ ˆ +x (yx − xy ) = ω0 −y ∂x ∂y 2 ˆ ˆ = ω0 (−y y − xx)
2 = −ω0 r⊥ University of California, Berkeley Physics H7A Fall 1998 (Strovink) PRACTICE EXAMINATION 1 Directions. Do all three problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free 2 to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. You may use a calculator but you do not need one – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (25 points). A 4 kg mass moves in the xy plane under the inﬂuence of a constant force. At t=0, the particle’s position, velocity, and acceleration are shown in the diagram. d. What is the particle’s position at t=1 sec? e. What is the particle’s velocity when its absolute value (its speed) is at a minimum? 2. (40 points) A Millikan oil drop of mass m and charge q moves between two horizontal capacitor plates separated by a distance d. A battery of voltage V is applied to the plates, so that the electrical force on the drop is upward, of magnitude qV /d. When it is moving, the drop experiences an opposing drag force F = −k v, where v is its velocity and k is a constant. a. For t < 0 the drop is observed to be exactly stationary, despite the gravitational force that is exerted upon it. What is the voltage V in terms of the other constants? b. At t = 0 the plates are shorted out (V =0), and remain shorted thereafter. Calculate a, the downward acceleration of the drop immediately after the plates are shorted. c. As t → ∞ the acceleration of the drop becomes essentially zero. What is its downward velocity v then? d. For any time t > 0, write a diﬀerential equation containing v , its ﬁrst time derivative dv/dt, and constants. e. Find a solution for the downward acceleration a(t), valid for all t > 0. [Hint: Diﬀerentiate the answer to part (d.) with respect to time to get a simple diﬀerential equation for a(t). Solve it by integration. This problem will be graded on answers only – no part credit will be given. The answer to each question has three parts – value (3 points), unit (1 point), and direction (1 point). If for a particular question you believe the direction to be undeﬁned or irrelevant, please leave it blank. Organize your answers in columns as shown: part (a) (b) (c) (d) (e) value unit direction a. What is the particle’s momentum at t=0? b. What is the force acting on the particle? c. What is the particle’s momentum at t=0.375 sec? Use the result of part (b.) to determine the constant of integration.] 3. (35 points) A wooden block of mass M , initially at rest on a horizontal table with coeﬃcient of sliding friction µ, is struck by a bullet of mass m and velocity v . The bullet lodges in the center of the block. How far does the block slide? 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO PRACTICE EXAMINATION 1 Composed and formatted by E.A. Baltz and M. Strovink 1. For each of parts (a.) through (e.), specify magnitude, unit, and direction. Giving vector components is suﬃcient for magnitude and direction. The mass is 4 kg, the initial velocity is 3 meters per second, and the initial acceleration is 8 meters per second2 . The force acting on the particle is constant, so the acceleration is constant. (a.) The initial momentum is p = mv = 12 kg m/sec, in the +ˆ direction. y (b.) The force acting on the particle can be found by F = ma, which is 32 kg m/sec2 , or newtons, in the +ˆ direction. x (c.) At t = 0.375 = 3/8 sec, the momentum can be found easily because the acceleration is constant. The velocity in the +ˆ direction isn’t y changing because there is no force, but the +ˆ x velocity is given by vx = at = 3 m/sec. The momentum vector is thus (12,12) kgm/sec. Alternatively, we can write this as a magnitude and a unit vector. The magnitude of this vector √ is 12 2. The unit vector must have equal x and y components, and√ length must be one. This its ˆ gives p = (ˆ + y)/ 2. xˆ (d.) The position at t = 1 sec is given by the formulas for constant acceleration. In the y direction, there is no acceleration, so y is given simply by y = y0 + v0y t = 6 m. In the x direction there is a constant acceleration, so the position is given by x = x0 + v0x t + at2 /2 = 8 m. The position vector is thus (8,6) m. Alternatively, the length of this vector is 10 meters; its direction is given by the unit vector ˆ = (0.8ˆ + 0.6ˆ ). r x y (e.) The y component of the velocity is constant because there is no force in the y direction. The x component of the velocity vanishes at t = 0. 2 2 Therefore v = vx + vy can never be smaller ˆ than it is at t = 0, when it is 3 m/sec in the y direction. 2. An oil drop has mass m and charge q . It moves between plates a distance d apart, and voltage V is applied. The electrical force on the drop is qV /d upwards. When the drop is moving, it encounters a drag force F = −k v. (a.) At t < 0 the drop is stationary. The electrical force must balance the gravitational force, so mg = qV /d, giving the potential V V= mgd q (b.) The plates are shorted at t = 0, so there is no more electrical force. At this instant, the drop isn’t moving, so it feels only the gravitational force. Its acceleration is just the acceleration of gravity, a = g downwards. (c.) As t → ∞, the acceleration goes to zero. The velocity can be found by balancing the drag force with the gravitational force, kv = mg , so the terminal velocity is v (t → ∞) = mg k (d.) For t > 0 we can ﬁnd a diﬀerential equation for the velocity. Newton’s second law states that F = dp/dt which in this case can be written F = m dv/dt. There are two forces, gravity and the drag force. The equation we get is m dv = mg − kv dt (e.) Following the hint, we take d/dt of the answer for (d.): m dv d2 v = −k 2 dt dt Substituting a = dv/dt: m da = −ka dt 2 Rearranging: k da = − dt a m Integrating from 0 to t: ln a(t) − ln a(0) = − Exponentiating:
k a(t) = e− m t a(0) will remain at rest. Solving, the time at which the blockbullet system stops is t = v0 /(µg ) . k t m The distance traveled in that time is 1 1 (v )2 x = v0 t − µgt2 = v0 t = 0 . 2 2 2µg Plugging in the already deduced value for v0 , this distance is x= m M +m
2 From part (b.), a(0) = g , so a(t) = g e− m t The acceleration begins with value g and decreases exponentially with time constant equal to m/k . 3. Instantaneously after the collision of the bullet and block, after the bullet has come to rest but before the frictional force on the block has had time to slow it down more than an inﬁntesimal amount, we can apply momentum conservation to the bulletblock collision. At that time the total momentum of the block+bullet system is (M + m)v0 , where v0 is the velocity of the block+bullet system immediately after the collision. Momentum conservation requires that momentum to be equal to the initial momentum mv of the bullet. Thus v0 = mv . M +m
k v2 . 2µg After the collision, the normal force on the block+bullet system from the table is (M + m)g , giving rise to a frictional force µN = µ(M + m)g on the sliding block+bullet system. This causes a constant acceleration µg of that system opposite to its velocity. Take t = 0 at the time of collision. Afterward, the block+bullet system’s velocity in the horizontal direction will be v (t) = v0 − µgt. It will continue sliding until v (t) = 0, at which point the frictional force will disappear and it University of California, Berkeley Physics H7A Fall 1998 (Strovink) EXAMINATION 1 Directions. Do all four problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free to 2 approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Don’t use a calculator, which you don’t need – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (20 points) You wish to ﬂy (at low altitude) from Sapporo, Japan to Portland, OR. These cities are both at 45◦ north latitude (halfway between the equator and the north pole), and they are separated by 90◦ in longitude (azimuth). Consider the earth to be a sphere of radius R. a. (10 points) You ﬂy a course that keeps you at constant latitude, i.e. you ﬂy due east. Over what distance do you travel? b. (10 points) You ﬂy a “great circle” course that takes you from Sapporo to Portland in the shortest possible distance (without penetrating the earth, of course). What is that distance? 2. (30 points) A block of mass m slides on an inclined plane with a slope of 5/12 (i.e. the slope of the hypotenuse of a 51213 triangle). A massless rope, guided by a massless pulley, connects the block to a second block of mass m/13, which is hanging freely above a lower table. ity v0 . What is the coeﬃcient µ of sliding friction between block and plane? b. (15 points) After the hanging block hits the table, what is the distance s along the surface of the plane along which the top block continues to slide? (Assume that the plane is long enough that the block does not fall oﬀ. If you are unsure of your answer to part (a.), you may leave µ as an undetermined constant.) 3. (35 points) To determine the dependence of the force of air resistance upon the speed of a slowly moving object, one starts with three sheets of paper stacked together as they come out of the package. One then crumples them against one’s ﬁst as shown in the sketch. Next one separates a single sheet from the other two without changing their crumpled (pseudoconical) shapes. This yields two objects with the same shape, but with diﬀerent masses m and 2m, where m is the mass of a single sheet. Finally one releases these two objects simultaneously, and compares their motion in still air under the inﬂuence of gravity. a. (15 points) The twoblock system is observed to be moving with a constant veloca. (5 points) Instantaneously after the two objects are released, what is the ratio R of the (downward) acceleration of the heavier object to that of the lighter object? b. (15 points) Very soon after being released, the two objects reach terminal (constant) velocity due to the eﬀects of air resistance. After a long time, one observes √ that the object of mass 2m has dropped 2 times farther than the object of mass m. Assuming that the force of air resistance on these objects is proportional to v α , where v is the velocity and α is a constant exponent, what is the value of α? c. (15 points) Suppose that Mother Nature were to turn gravity oﬀ while these objects are falling. If one were willing to wait an arbitrarily long time, would they fall an arbitrarily long distance, or would that distance be bounded? Explain. 4. (15 points). An asymmetric barbell stands vertically at rest on frictionless horizontal ice. Mass M rests on the ice, and mass m is a distance h above it; the mass of the bar that rigidly connects these two masses is negligible. The dimensions of the masses can be neglected in comparison to their separation h. Take x = 0 to be the initial position of mass M . Mass m is given an inﬁnitesimal tap in the +x direction that produces a negligible momenˆ tum, but does eventually cause the barbell to topple over. You may assume that mass M remains in contact with the ice throughout the motion. When mass m hits the ice, at what horizontal coordinate xM will mass M be located? 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO EXAMINATION 1 1. In spherical polar coordinates, take Sapporo and Portland to be at (r, θ, φ) = (R, π/4, 0) = rS and (R, π/4, π/2) = rP respectively. (a.) Here the course is one quarter of a circle with its center on the earth’s axis of rotation at a point above the earth’s center. This circle has √ radius R sin θ = R 2/2. The distance traveled, s, is one quarter of its circumference: s= πR 2 1 R2 2π = . 4 2 4 √ √ the normal component of the gravitational force on block m, or N = mg cos θ . The frictional force Ff then will be Ff = µN = µmg cos θ opposite to the motion. (a.) Along the motion, the system consisting of mass m plus mass m/13 experiences a force due to gravity, consisting of the sum of the force mg sin θ on mass m and mg/13 on mass m/13. If the velocity of the system is constant, i.e. there is no acceleration, the gravitational and frictional forces must balance: µmg cos θ = mg (sin θ + 1/13) 5/13 + 1/13 = 1/2 . µ= 12/13 (b.) After the hanging block hits the table, the (massless) rope goes limp and has no further eﬀect on the sliding block. Opposite to the direction of motion, the net force on m is f = µmg cos θ − mg sin θ 5 mg 1 12 − = , = mg 2 13 13 13 producing a constant acceleration a = g/13 opposite to the motion. The block decelerates for a time t = v0 /a until it comes to rest. During that time, the distance traveled is
2 2 2 1 v0 1 v0 13 v0 1 v2 = = . s = v0 t − at2 = 0 − 2 a 2a 2a 2g (b.) The Cartesian coordinates of rS and rP are √ √ rS = R( 2/2, 0, 2/2) √ √ rP = R(0, 2/2, 2/2) . The angle between rS and rP is ψSP = arccos (rS · rP ) R2 = arccos (1/2) = π/3 . To calculate the minimum distance between Sapporo and Portland along the surface of the earth, we bisect the earth using a plane that contains these two cities as well as the earth’s center. The intersection of the earth with the bisecting plane is a circle with its origin at the center of the earth, having a circumference 2πR. Since both rS and rP lie in this plane, the course consists of the fraction ψSP /2π of this circumference. Therefore the distance traveled is s= π/3 πR ψSP 2πR = 2πR = . 2π 2π 3 (This is 0.333/0.354 ≈ 94% of the length of the “due east” course.) 2. Let θ = arctan (5/12) be the angle with which the plane is inclined. Since there is no acceleration (or motion) perpendicular to that plane, the normal force N on block m must be equal to 3. Two crumpled paper objects have the same force (proportional to v α ) due to air resistance, but diﬀerent masses 2m and m. (a.) Immediately after the two objects are released from rest, their velocity must still be negligible; otherwise they would have experienced inﬁnite acceleration. Likewise, the force of 2 air resistance, proportional to v α , is negligible at that time. So the only nonnegligible force acting on them is the force of gravity, 2mg and mg respectively, yielding an acceleration g in either case. So the ratio of accelerations is R = 1. (b.) After the objects reach terminal velocity (v2 and v1 respectively), and they no longer are accelerating, the forces due to air resistance and gravity must cancel:
α 2mg = Kv2 α mg = Kv1 , which is arbitrarily large when t is arbitrarily large. Therefore the distance fallen will be unbounded. [This result is similar to that obtained in Problem Set 3, Problem 6 (K&K 2.35).] 4. Since the ice is horizontal and frictionless, it cannot exert any force on the barbell in the x ˆ direction, either as the result of a contact force or a frictional force. Therefore the x coordinate xCM of the barbell’s center of mass, initially at rest at x = 0, must remain at x = 0. When mass m hits the ice and the barbell is horizontal, 0 = XCM = ⇒ xM M xM + m(xM + h) M +m m = −h . M +m where K is the unknown constant of proportionality. Taking the ratio of these two equations, v2 = 21/α v1 . The “long time” after the objects are dropped is very large compared to the “very soon” time at which they reach terminal velocity. So, to an excellent approximation, the distance they drop during the “long time” is proportional to the terminal velocity. Since mass 2m drops a factor √ 2 further, √ v2 = 2v1 . Comparing this to the previous equation, α=2. (c.) After gravity is turned oﬀ, the force of air resistance Kv 2 accelerates the falling object opposite to its direction of motion: m dv = −Kv 2 . dt Dividing through by v 2 , multiplying through by dt, and integrating, K 1 = t+C v m where C is a constant of integration that will be negligible with respect to Kt/m when t is suﬃciently large. Therefore, asymptotically as t → ∞, the downward velocity will be inversely proportional to t. Integrating this statement, the asymptotic distance traveled will increase as ln t, University of California, Berkeley Physics H7A Fall 1998 (Strovink) EXAMINATION 2 Directions. Do all three problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free to 2 approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Don’t use a calculator, which you don’t need – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (30 points) A cylinder of mass m and radius R initially is at rest on frictionless ice. About its C.M., the cylinder’s moment of inertia is mR2 /2. a. (15 points) The cylinder receives an instantaneous horizontal tap at a point on its circumference that is a vertical distance h above the ice. Immediately thereafter, it is observed to roll without slipping at a velocity v0 , even though the ice is frictionless. Calculate the value of h. b. (15 points) The cylinder continues to roll without slipping as long as the frictionless ice remains ﬂat. Eventually the ice slopes upward to form a hill as shown, all the while staying frictionless. To what maximum height H will the cylinder rise? a. (15 points) For this part, consider the bead and the rod to be glued together. The C.M. of the bead is observed to travel uniformly around a horizontal circle that is centered below the rod’s pivot. The circle’s radius is much smaller than H but much larger than the bead. With what angular velocity Ω does the bead move on this path? b. (15 points) For this part, the bead is no longer glued to the rod; instead it spins about the long axis of the rod with a constant, large angular velocity ω0 . About this axis, the moment of inertia of the bead is I . Again, the C.M. of the bead is observed to travel uniformly around a horizontal circle centered below the pivot. What restriction(s) can be placed on the radius r of this circle? 2. (30 points) A thin massless rod of length H hangs freely pivoted from the ceiling. At its end is a bead of mass m, negligible in size compared to H , acted upon by gravity. 3. (40 points) A tiny pebble moves on the frictionless inner surface of a vertical cone that has a halfangle of 45◦ . It is observed to be in uniform circular motion with constant velocity v0 . (You are not given the radius of this circle!) a. (15 points) what is the angular frequency Ω of this uniform circular motion? b. (15 points) The orbiting pebble is now nudged so that its new orbit diﬀers very slightly from the original circle. The nudge causes the orbit radius (the perpendicular distance from the pebble to the cone axis) to oscillate sinusoidally about its mean value. (All the while the pebble remains in contact with the inside surface of the cone.) Calculate the angular frequency ωr of this small radial oscillation. c. (10 points) Is the perturbed orbit “closed” – that is, does the perturbed orbit ever repeat itself? Explain. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO EXAMINATION 2 1.
2 we conclude that only Ktrans = 1 mv0 is avail2 able to be converted into potential energy mgH . Therefore the maximum height is 2 v0 2g H= . 2. (a.) The horizontal tap produces a (horizontal) linear impulse F dt ≡ J . With respect to the center of the cylinder, it also produces a (clockwise) angular impulse τ dt = Jb, where b = h − R is the impact parameter of the horizontal tap. Then, in terms of J , since the cylinder is initially at rest, mv0 = J Iω0 = J (h − R) Substituting I = 1 mR2 , and imposing the con2 dition v0 = Rω0 that the cylinder rolls without slipping, these equations become mv0 = J 1 v0 mR2 = J (h − R) 2 R These equations are mutually consistent only if h − R = R/2, or h= 3 R 2 (a.) Let β be the angle between the stick and the vertical. This part can easily be worked in the C.M. of the bead, where a centrifugal force mΩ2 r = mΩ2 H sin β balances the horizontal component mg tan β of the tension mg/ cos β in the stick. Or it can be worked in the lab, where the horizontal component of the tension supplies the necessary centripetal acceleration. Or, elegantly, the circular motion of the bead can be considered to be the superposition of an x pendulum and, delayed by onequarter of a period, a y pendulum with the same amplitude. When the approximation β 1, valid for part (a.), is applied, β cancels out, and any of these approaches yields the usual result Ω= g H (b.) Since the ice is frictionless, and the force of gravity acts on the C.M. of the cylinder, no torques about its axis can be exerted on the cylinder. Therefore its angular momentum and angular velocity ω = ω0 remain constant even on the hill. Considering that the kinetic energy of a rigid body decomposes into Ktrans + Krot , (b.) Take the origin to be the pivot point. If the bead is spinning with constant angular velocity ω0 about the stick’s axis, the vertical component 2 Iω0 cos β of the spin angular momentum L remains constant. But the horizontal component Lh = Iω0 sin β of L precesses with angular velocity Ω, as in a gyroscope. The torque ΩLh that is required to maintain this precession is the torque due to gravity, mgr = mgH sin β ; the stick can’t supply this torque because its end coincides with the origin. Then ΩIω0 sin β = mgH sin β The angular velocity of precession is Ω= mgH Iω0 where r is the perpendicular distance of the pebble to the cone axis. (In the second term, we are using the fact that, for a 45◦ cone with z = r, an increase ∆r causes an increase mg ∆z = mg ∆r in the true potential energy.) Then a circular orbit occurs when 0= 2l2 dUeﬀ =− + mg dr 2mr3 Substituting l2 = mv0 r × mΩr2 causes the r dependence to cancel: 0=− Ω= as before. (b.) Proceeding with the eﬀective potential method, we obtain the eﬀective spring constant keﬀ for radial motion by diﬀerentiating Ueﬀ again with respect to r: keﬀ = d l2 − 3 + mg dr mr 2 3l = mr4 3m2 r4 Ω2 = mr4 = 3mΩ2 √ √g keﬀ = 3Ω = 3 m v0 2m2 v0 Ω + mg 2m independent of β . Because sin β cancels out of the previous equation, there is no restriction on it and therefore no restriction on the orbit radius r = H sin β ; this circular motion can occur for any orbit radius r ≤ H provided that that ω0 is large enough to allow the spin angular momentum to dominate the other angular momenta in the problem. 3. g v0 Thus ωr = (a.) This part of the problem can be done by balancing forces; the normal force of the 45◦ cone on the pebble has equal horizontal and vertical components mg . When the horizontal component is equated to the centripetal acceleration mv0 Ω, we obtain immediately g Ω= v0 Anticipating what will be needed for part (b.), we can also solve part (a.) using the eﬀective potential l2 + mgr Ueﬀ = 2mr2 (c.) Since the ratio of ωr to Ω is irrational, an integer number of orbital cycles cannot occur in the same time interval as an integer number of radial cycles. Therefore the orbit never repeats itself and so is not closed. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) FINAL EXAMINATION Directions. Do all six problems (weights are indicated). This is a closedbook closednote exam except for three 8 1 × 11 inch sheets containing any information you wish on both sides. You are 2 free to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. Problem 1. (30 points) Northern Canada has two peculiar features: owing to lack of roads, most surface transportation occurs by train; and the principal fauna are tiny black ﬂies. Consider the elastic (kinetic energy conserving) headon collision of a locomotive of mass M and velocity V with a stationary black ﬂy of mass m. You may make any reasonable approximation concerning the relative magnitude of M and m. a. (15 points) With what velocity v does the ﬂy recoil from the locomotive? b. (15 points) Assuming that the (coasting) locomotive has ﬂat frontal area A, and there are N black ﬂies per cubic meter hovering over the track, apply the results of part (a.) to obtain a diﬀerential equation for V (neglect air resistance). Solve it to obtain V (t). Problem 2. (20 points) (see ﬁgure). Find the maximum value of b for which the rocket will hit the planet. Problem 3. (30 points) For decades, scientists have been designing a “space colony” in which thousands of people could exist while orbiting the sun. People would live on the inside curved surface of a large airﬁlled cylinder (length of order 10 km, radius R of order 1 km). The cylinder would rotate about its axis with an angular velocity ω such that earth’s gravitational acceleration g would be simulated by the centrifugal force acting near that surface. The curved surface would have dirt for farming, and also housing, factories, parks, hills, streams, a lake, etc. Sunlight would enter through one end; it would be controlled by mirrors and shutters to simulate day and night. There would be clouds and weather, etc. a. (5 points) Find the angular frequency ω of rotation. b. (10 points) Although many aspects of life in this colony would resemble life on earth, one peculiar feature would be the large Coriolis acceleration. When v (as seen by a colony inhabitant) is perpendicular to ω , the magnitude of the Coriolis acceleration aC can be expressed as aC = g A space probe is launched with initial velocity v0 and impact parameter b toward a very distant planet of radius R and very large mass Mp v vC where vC is a characteristic velocity. Find vC appropriate to the surface inhabited by the colonists. 2 c. (15 points) For residents in the colony, north is deﬁned to be in the direction of ω ; if a resident faces north her right hand points east. A baseball pitcher, new to the colony, ﬁres a ball toward the west with velocity v at his target a distance D away. If he were on the surface of the earth (where the Coriolis force is negligible), the ball would hit its target. In what direction (high, low, north, or south) does the ball miss its target? By what distance d does it miss (you may assume d D)? Problem 4. (30 points) A mass m is connected by a massless spring 2 of stiﬀness k = mω0 to a point of support xs . When the spring is relaxed, and xs = 0, the mass is at its equilibrium position x = 0. The mass moves only in the x direction, without friction. (along the same axis) is x. As usual, ξ satisﬁes the wave equation 1 ∂2ξ ∂2ξ + 2 2 =0 ∂x2 c ∂t where c is the (phase and group) velocity of sound waves in air. a. (3 points) Keeping in mind that the pipe is closed at both ends, write down the boundary conditions on ξ (0, t) and ξ (L, t). b. (12 points) The air inside the straight closed pipe is observed to carry a standing sinusoidal sound wave. What is the lowest angular frequency ωs with which this wave can vibrate? c. (3 points) The ends of the pipe are now opened and the pipe is bent into a hoop. The end at x = 0 is welded to the end at x = L, so that the pipe forms a continuously hollow circular torus (like a hula hoop) with a circumference equal to L. Suppose that the point of support is constrained by an external force to obey the following motion: xs = mA sin ωt, where A and ω are constants, and ω is not necessarily equal to ω0 . The external force does not act directly on the mass, but it nevertheless inﬂuences the mass because of the spring. a. (15 points) Find the particular solution xp (t) which would vanish if A were zero. b. (15 points) Find the solution x0 (t) which would be correct if the mass were ﬁxed at its equilibrium position and released at t = 0. Problem 5. (30 points) Consider a thin cylindrical pipe of length L, closed at both ends. The air inside the pipe can support longitudinal (sound) waves that propagate along the axis of the pipe. Let ξ (x, t) be the displacement (along the axis of the pipe) of an air molecule whose equilibrium coordinate Continue to consider sound waves that propagate along the (bent) axis of the pipe. As long as the circumference of the hoop is much larger than the pipe thickness, which is the case here, ξ (x, t) satisﬁes the same wave equation as before. However, since the pipe is now bent into a continuously hollow torus, x = 0 and x = L now describe the same coordinate along the pipe’s axis. More generally, ξ (x, t) and ξ (x + L, t) describe the displacement from equilibrium of the same molecule. In light of the above, write down the relationship between ξ (x, t) and ξ (x + L, t). 3 d. (12 points) The air inside the bent pipe is observed to carry a travelling sinusoidal sound wave. Keeping in mind the result of part (c.), what is the lowest angular frequency ωt that can characterize this wave? What is the ratio of ωt to the result ωs of part (b.)? Problem 6. (30 points) Nonviscous ﬂuid matter is in spherically symmetric, nonrelativistic ﬂow toward a black hole of mass M . Only the gravitational attraction of the black hole itself (as opposed to the gravitational attraction of other ﬂuid elements) is important to the ﬂuid motion. M is growing slowly enough to be taken as constant. a. (10 points) Consider Φ, the potential energy per unit mass of ﬂuid due to the gravitational attraction of the black hole. Starting from the standard formula for the gravitational force between two point objects, show that Φ(r) = − GM r Note that, in spherical polar coordinates, ∇·F= 1∂ 2 (r Fr ) + . . . r2 ∂r where r is the distance from the black hole. Note that in spherical polar coordinates ∇u = ˆ r ∂u + ... ∂r (You may use this result for Φ in subsequent parts of the problem.) b. (10 points) Because M is taken as constant, the ﬂuid ﬂow is steady: ∂ρ ∂v = =0 ∂t ∂t where ρ is the mass density. Also, the ﬂuid pressure p is known not to vary either with position or time. Away from the black hole, determine the dependence of ﬂuid velocity v upon r. c. (10 points) Away from the black hole, determine the dependence of ﬂuid mass density ρ upon r. 1 University of California, Berkeley Physics H7A Fall 1998 (Strovink) SOLUTION TO FINAL EXAMINATION Problem 1. a. We consider this headon collision in the center of mass. The center of mass velocity is M ≈V V =V M +m
∗ Problem 2. At the instant that the probe barely grazes the planet, it will have radius R and velocity vf directed tangentially to the planet. Angular momentum conservation requires mv0 b = mvf R b v f = v0 R Substituting for vf in the equation for energy conservation, we obtain 1 GMp m 1 2 2 mv0 = mvf − 2 2 R 12 1 2 b2 GMp v = v0 2 − 20 2R R b2 GMp −1 = R2 R b2 2GMp −1= 2 R2 v0 R b=R 1+ 2GMp 2 v0 R Using this approximation, in the C.M. the ﬂy approaches the locomotive with speed V . Since the collision is elastic, it bounces back with the same speed. Transforming back to the lab, the ﬂy has velocity v ≈ V + V = 2V b. In each collision, the momentum 2mV that is gained by the ﬂy is lost by the locomotive: ∆P = M ∆V = −2mV m ∆V = −2 V M In a time interval ∆t, the volume swept out by the front of the train is AV ∆t; this volume contains N AV ∆t ﬂies. So, for N AV ∆t collisions, ∆V V ∆V V2 dV V2 1 1 − V V0 m N AV ∆t M m = −2N A ∆t M m dt = −2N A M m = 2N A t M 1 V (t) = m 1 2N A M t + V0 = −2 V0 m 1 + 2N AV0 M t 12 v 20 Problem 3. a. mRω 2 = mg ω= b. 2ωv = aC = g vC = vC = vC = g 2ω g 2 1 2 R g gR v vC g R V (t) = where V0 is the velocity at t = 0. 2 c. FC = −2m(ω × v) ω is north, and −v is east; north × east is down. This is the direction in which the ball misses. aC = 2ωv = 2v 1 aC t2 2 1 g2 = 2v t 2 R D t= v g D2 d=v R v2 D2 g d= v R d= (We ignore the centrifugal force on the ball, because it is the same on the colony as on earth, and the pitcher already compensates for it.) As a sanity check, if D = 20 m and v = 40 m/sec (appropriate to baseball), and R = 1000 m, we obtain d ≈ 1 m. Indeed d is much smaller than D. Nevertheless, from the standpoint of the pitcher, the Coriolis force has a big eﬀect on his control. Problem 4. The equation of motion for x(t) is
2 mx = −k (x − xs ) = −mω0 (x − xs ) ¨ 2 2 x = −ω0 x + ω0 mA sin ωt ¨ 2 x + ω0 x = kA sin ωt ¨ be required otherwise, x0 (0) = 0 as well as ˙ x0 (0) = 0. The general solution to the homogeneous equation of motion (A = 0) is xh (t) = C cos ω0 t + D sin ω0 t The general solution to the full equation is obtained by adding xh to xp . Applying initial conditions, x0 (t) = kA sin ωt + C cos ω0 t + D sin ω0 t 2 ω0 − ω 2 g R x0 (0) = 0 ⇒ C = 0 x0 (0) = 0 ⇒ 0 = ˙ ωkA + ω0 D 2 ω0 − ω 2 kA ω D=− 2 − ω2 ω 0 ω0 ω0 sin ωt − ω sin ω0 t x0 (t) = kA 2 ω0 (ω0 − ω 2 ) Problem 5. a. ξ (x = 0, t) = ξ (x = L, t) = 0 b. ξ (x, t) = sin kx sin kL = 0 kL = nπ, n = 1, 2, . . . ω ≡ ck πc ωs = L c. ξ (x, t) = ξ (x + L, t) d. ξ0 exp (−iωt) a. try xp (t) = B sin ωt (−ω +
2 2 ω0 )B sin ωt = kA sin ωt kA B= 2 ω0 − ω 2 kA sin ωt xp (t) = 2 ω0 − ω 2 b. Because an inﬁnite force from the spring would 3 ξ0 exp (i(kx − ωt)) c. In steady ﬂow there can be no buildup of mass density ρ. Therefore the mass ﬂow ρv (kg/m sec) × 4πr2 (m2 ) through a spherical surface of radius r must be independent of r. So, using the result of part (a.), ρv ∝ r−2 ρ ∝ r−3/2 More formally, but equally acceptably, one can reach the same conclusion by applying the continuity equation ∂ρ + ∇ · (ρv) = 0 ∂t and using the fact that for steady ﬂow the ﬁrst term vanishes.
2 ξ (x, t) = exp(ikx) = exp(ik (x + L) 1 = exp(ikL) kL = 2nπ, n = 1, 2, . . . ω ≡ ck 2πc ωt = L ωt = 2ωs Problem 6. a. Per unit mass of ﬂuid, the force f is f = −ˆ r GM r2 We seek a function Φ(r) such that −∇Φ = f or equivalently, using spherical symmetry, Φ=− Clearly Φ(r) = − GM r fr dr satisﬁes either of these conditions. b. Since the ﬂow is steady, we can use Bernoulli’s equation (either along a streamline at constant (θ, φ), or, since the ﬂow is irrotational, anywhere outside the black hole): 12 ρv + p + ρΦ = constant 2 Only the ﬁrst and third terms are not constant, so they must have the same r dependence. Therefore v 2 and Φ have the same r dependence. So v ∝ r−1/2 University of California, Berkeley Physics H7B, Spring 1999 (Strovink) General Information Web site for this class: First link on http://d0lbln.lbl.gov/ . Instructors: Prof. Mark Strovink, 437 LeConte; (LBL) 4867087; (home, before 10) 4868079; (UC) 6429685. Email: [email protected] . Web: http://d0lbln.lbl.gov/ . Office hours: M 3:154:15, Th 1011. Mr. Robin BlumeKohout, 208 LeConte; (UC) 6425430. Email: [email protected] Office hours (to be held initially in 208 LeConte): M 12, W 34. You may also get help in the 7B Course Center, 206 LeConte. Lectures: TuTh 11:1012:30, 2 LeConte. Lecture attendance is essential, since not all of the course content can be found in the course text or handouts. Labs: In the second week, in 262 LeConte, please enroll in one of only 2 special H7B lab sections [(A) #401, F 24; (B) #402, F 46]. Both sections are taught by Mr. BlumeKohout. If you can make both of these lab (and section, see below) slots, please attempt to enroll in the earlier of these lab slots. Depending on crowding, you may be asked to move to the later lab. During "off" weeks not requiring lab apparatus, your lab section will still meet in the same room, 262 LeConte. Discussion Sections: Beginning in the second week, please enroll in the 1 hr H7B discussion section corresponding to your H7B lab section: (A) #401, M 23, 385 LeConte; (B) #402, W 45, 343 LeConte. You are especially encouraged to attend discussion section regularly. There you will learn techniques of problem solving, with particular application to the assigned exercises. Text (required): E.M. Purcell, Electricity and Magnetism (Berkeley Physics Course Volume 2), Second Edition (McGrawHill, 1985). At the beginning of the semester we will also use Chapters 22 through 26 (pp 493592) of Resnick, Halliday and Krane, Physics (Volume 1), Fourth Edition (Wiley, 1992). The publisher has granted permission to make it possible for students to purchase a Xerox copy of these pages from Copy Central. Problem Sets: Thirteen problem sets are assigned and graded, with solutions provided on the web and at Copy Central. They are due on Thursday at 5 PM on weeks in which there is no exam, beginning in week 2. Deposit problem sets in the box labeled "H7B" outside 208 LeConte. You are encouraged to attempt all the problems. Students who do not do them find it almost impossible to learn the material and to succeed on the examinations. Discuss these problems with your classmates as well as with the teaching staff; however, when the time comes to write up your solutions, work independently. Credit for collective writeups, which are easy to identify, will be divided among the collectivists. Late papers will not be graded. Your lowest problem set score will be dropped, in lieu of due date extensions for any reason. Syllabus: H7B has one syllabus card, which is mandatory. It will be collected at the time of the midterm examination. This card pays for the experiment descriptions and instructions that you will receive from your GSI at the beginning of each laboratory. Also, we expect you also to have the opportunity to purchase at Copy Central a copy of the above mentioned 100 pages of Resnick, Halliday and Krane. Finally, copies of solutions to the problem sets will also be available for purchase at Copy Central. These solutions will also be available on the Web. Exams: There will be one 80minute midterm examination and one 3hour final examination. Before confirming your enrollment in this class, please check that its final Exam Group 12 does not conflict with the Exam Group for any other class in which you are enrolled. Please verify that you will be available for the midterm examination (Th 4 Mar, 11:1012:30), and for the final examination, W 19 May, 811 AM. Except for unforeseeable emergencies, it will not be possible for the midterm or final exams to be rescheduled. Passing H7B requires passing the final exam. Grading: 25% midterm; 25% problem sets; 45% final exam; 5% lab. Grading is not "curved"  it does not depend on your performance relative to that of your H7B classmates. Rather it is based on comparing your work to that of a generation of earlier lower division Berkeley physics students, with due allowance for educational trends. COURSE OUTLINE
Week No. Week of... Lecture chapter Topic (RHK = Resnick/Halliday/Krane, Physics Vol. 1 ) (Purcell = Electricity and Magnetism ) Problem Due Set No. 5 PM on... Lab 1 18Jan MARTIN LUTHER KING HOLIDAY (18Jan) RHK (22), 25.7 Thermal expansion; heat transfer 23.123.4 Kinetic definition of temperature 23.523.6 24.124.4 25.425.6 26.126.4 Energetics of an ideal gas; equipartition Maxwellian distribution Heat capacities of an ideal gas; first law Second law of thermodynamics (do experiment in lab="expt") (have discussion in lab="disc") no lab 2 25Jan disc 1 Th 28Jan expt 2 4Feb disc 3 11Feb disc 4 18Feb expt 5 25Feb disc 3 1Feb 4 8Feb 5 15Feb RHK 26.526.9 Entropy Purcell 1.11.8 Electric charge PRESIDENTS' DAY HOLIDAY (15Feb) 1.91.15 Electric fields 2.12.6 Electric potential 2.72.13 3.13.4 3.53.8 11:10 AM  12:30 PM 4.14.11 Appendix A Appendix A 5.15.5 Gauss' law, Laplace's equation Electic fields around conductors Systems of conductors; capacitors MIDTERM EXAMINATION (covers PS 15) Electric currents Special relativity Special relativity Electric field in different frames of reference 6 22Feb 7 1Mar 4Mar 8Mar 8 expt 6 11Mar disc 7 18Mar disc 25Mar 9 15Mar 22Mar SPRING RECESS (2226 Mar) 10 29Mar 5.65.9 6.16.2, 6.46.5 6.3, 6.66.9 7.17.5 7.67.10 8.18.5 9.19.4; 9.6 10.110.6 10.710.12 11.111.6 11.711.11 Fields of moving charges Magnetic fields Vector potential; magnetic field transformation Faraday's law Inductance AC circuits Maxwell's equations Electric dipoles Electric fields in dielectric media Magnetic dipoles (but not monopoles) Magnetization LAST LECTURE (review) expt 8 1Apr disc 9 8Apr expt 10 15Apr disc 11 22Apr disc 12 29Apr disc 13 6May makeup 11 5Apr 12 12Apr 13 19Apr 14 26Apr 15 3May 10May INSTRUCTION ENDS (10May) FINAL EXAMS BEGIN (14May) 17May 19May 811 AM FINAL EXAM (Group 12) (covers PS 113) Physics H7B Spring 1999 (Strovink) University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 1 1. RHK problem 22.9 It is an everyday observation that hot and cold objects cool down or warm up to the temperature of their surroundings. If the temperature diﬀerence ∆T between an object and its surroundings (∆T = Tobj − Tsur ) is not too great, the rate of cooling or warming of the object is proportional, approximately, to this diﬀerence; that is, d ∆T = −A(∆T ), dt where A is a constant. Ths minus sign appears because ∆T decreases with time if ∆T is positive and increases if ∆T is negative. This is known as Newton’s law of cooling. (a) On what factors does A depend? What are its dimensions? (b) If at some instant t = 0 the temperature diﬀerence is ∆T0 , show that it is ∆T = ∆T0 exp (−At) at a time t later. 2. RHK problem 22.28 As a result of a temperature rise of 32 ◦ C, a bar with a crack at its center buckles upward, as shown in the ﬁgure. If the ﬁxed distance L0 = 3.77 m and the coeﬃcient of linear thermal expansion is 25 × 10−6 per ◦ C, ﬁnd x, the distance to which the center rises. side b is longer by ∆b. Show that if we neglect the small quantity ∆a ∆b / ab (see the ﬁgure), then ∆A = 2αA∆T . 4. RHK problem 25.47 The average rate at which heat ﬂows out through the surface of the Earth in North America is 54 mW/m2 , and the average thermal conductivity of the near surface rocks is 2.5 W/m·K. Assuming a surface temperature of 10 ◦ C, what should be the temperature at a depth of 33 km (near the base of the crust)? Ignore the heat generated by radioactive elements in the crust; the curvature of the Earth can also be ignored. 5. RHK problem 25.50 A cylindrical silver rod of length 1.17 m and crosssectional area 4.76 cm2 is insulated to prevent heat loss through its surface. The ends are maintained at temperature diﬀerence of 100 ◦ C by having one end in a waterice mixture and the other in boiling water and steam. (a) Find the rate (in W) at which heat is transferred along the rod. (b) Calculate the rate (in kg/sec) at which ice melts at the cold end. Hints: The thermal conductivity of silver is 428 W/m·K. The latent heat of fusion of water is 333 kJ/kg. 6. RHK problem 25.58 A container of water has been outdoors in cold weather until a 5.0cmthick slab of ice has formed on its surface (see the ﬁgure). The air above the ice is at −10 ◦ C. Calculate the rate of 3. RHK problem 22.30 The area A of a rectangular plate is ab. Its coefﬁcient of linear thermal expansion is α. After a temperature rise ∆T , side a is longer by ∆a and formation of ice (in centimeters per hour) on the bottom surface of the ice slab. Take the thermal conductivity and density of ice to be 1.7 W/m·K and 0.92 g/cm3 , respectively. Assume that no heat ﬂows through the walls of the tank. 7. RHK problem 23.16 A mercuryﬁlled manometer with two unequallength arms of the same crosssectional area is sealed oﬀ with the same pressure p0 of perfect gas in the two arms (see the ﬁgure). With the temperature constant, an additional 10.0 cm3 of mercury is admitted through the stopcock at the bottom. The level on the left increases 6.00 cm and that on the right increases 4.00 cm. Find the original pressure p0 . 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 1 1. RHK problem 22.9 It is an everyday observation that hot and cold objects cool down or warm up to the temperature of their surroundings. If the temperature diﬀerence ∆T between an object and its surroundings (∆T = Tobj − Tsur ) is not too great, the rate of cooling or warming of the object is proportional, approximately, to this diﬀerence; that is, equation, with dt substituted for dt, d ∆T = −A dt ∆T t t d ∆T =− A dt 0 ∆T 0 ln ∆T (t) − ln ∆T (0) = −At ∆T (t) ln = −At ∆T (0) ∆T (t) = exp (−At) ∆T (0) ∆T (t) = ∆T0 exp (−At) . 2. RHK problem 22.28 As a result of a temperature rise of 32 C◦ , a bar with a crack at its center buckles upward, as shown in the ﬁgure. If the ﬁxed distance L0 = 3.77 m and the coeﬃcient of linear thermal expansion is 25 × 10−6 per C◦ , ﬁnd x, the distance to which the center rises. d ∆T = −A(∆T ), dt where A is a constant. Ths minus sign appears because ∆T decreases with time if ∆T is positive and increases if ∆T is negative. This is known as Newton’s law of cooling. (a) On what factors does A depend? What are its dimensions? Solution: The LHS (and therefore the RHS) of the above equation have dimensions C◦ /sec, so A must have dimension sec−1 . Suppose that the heat ﬂowing between the object and its surroundings is conducted by a thermal barrier (i.e. a “skin” on the object that tends to insulate it from its surroundings). Then, from RHK Eq. 25.45, A should be proportional to the thermal conductivity of that barrier and to its area, and inversely proportional to the barrier’s thickness. (b) If at some instant t = 0 the temperature diﬀerence is ∆T0 , show that it is ∆T = ∆T0 exp (−At) at a time t later. Solution: Rearranging and solving the above Solution: In Physics H7B, all problems involving numbers should be solved completely in terms of algebraic symbols before any numbers are plugged in (otherwise it is much more diﬃcult to give part credit). Let L0 = ﬁxed distance = 3.77 m x = distance to which the center rises L = thermally expanded total length of the buckled bar (twice the hypotenuse of the right triangle whose legs are x and L0 /2) α = coeﬃcient of linear thermal expansion = 25 × 10−6 per C◦ ∆T = temperature rise = 32 C◦ 2 Then Then A + ∆A = (a + ∆a)(b + ∆b) = ab + a ∆b + b ∆a + ∆a ∆b A = ab A + ∆A − A = ab + a ∆b + b ∆a + ∆a ∆b − ab ∆ A = a ∆ b + b ∆ a + ∆a ∆ b ∆b ∆a ∆a ∆b + + = ab b a ab ∆b ∆a + ∆A ≈ ab b a ∆b ∆a = = α ∆T a b ∆A ≈ A(α ∆T + α ∆T ) ∆A ≈ 2α ∆T . A 4. RHK problem 25.47 The average rate at which heat ﬂows out through the surface of the Earth in North America is 54 mW/m2 , and the average thermal conductivity of the near surface rocks is 2.5 W/m·K. Assuming a surface temperature of 10 ◦ C, what should be the temperature at a depth of 33 km (near the base of the crust)? Ignore the heat generated by radioactive elements in the crust; the curvature of the Earth can also be ignored. Solution: Let H/A = heat ﬂow per unit area through Earth’s surface = 54 × 10−3 W/m2 k = thermal conductivity of near surface rock = 2.5 W/m·K T0 = temperature at earth’s surface = 10 ◦ C D = depth at which we wish to know the temperature = 33 × 103 m T = temperature at depth D Then, using RHK Eq. 25.45, ∆T H =k A ∆x T − T0 =k D HD = T − T0 Ak HD =T T0 + Ak 723 ◦ C = T . L = L0 + α ∆T x2 = (L/2)2 − (L0 /2)2 L0 x= (1 + α ∆T )2 − 1 2 = 0.0754 m . 3. RHK problem 22.30 The area A of a rectangular plate is ab. Its coefﬁcient of linear thermal expansion is α. After a temperature rise ∆T , side a is longer by ∆a and side b is longer by ∆b. Show that if we neglect the small quantity ∆a ∆b / ab (see the ﬁgure), then ∆A = 2αA∆T . Solution: Let A = original area of rectangular plate a = original width of plate b = original height of plate A + ∆A = thermally expanded area of plate a + ∆a = thermally expanded width of plate b + ∆b = thermally expanded height of plate α = coeﬃcient of linear thermal expansion 3 5. RHK problem 25.50 A cylindrical silver rod of length 1.17 m and crosssectional area 4.76 cm2 is insulated to prevent heat loss through its surface. The ends are maintained at temperature diﬀerence of 100 C◦ by having one end in a waterice mixture and the other in boiling water and steam. (a) Find the rate (in W) at which heat is transferred along the rod. Solution: Let L = length of cylindrical silver rod = 1.17 m A = area of rod = 4.76 × 10−4 m2 k = thermal conductivity of silver = 428 W/m·K ∆T = temperature diﬀerence between ends of rod = 100 C◦ . H = dQ/dt = rate at which heat is transferred along the rod. Then, using RHK Eq. 25.45 H = kA ∆T x ∆T = kA L = 17.4 W . 6. RHK problem 25.58 A container of water has been outdoors in cold weather until a 5.0cmthick slab of ice has formed on its surface (see the ﬁgure). The air above the ice is at −10 ◦ C. Calculate the rate of formation of ice (in centimeters per hour) on the bottom surface of the ice slab. Take the thermal conductivity and density of ice to be 1.7 W/m·K and 0.92 g/cm3 , respectively. Assume that no heat ﬂows through the walls of the tank. (b) Calculate the rate (in kg/sec) at which ice melts at the cold end. Solution: Let Lf = latent heat of fusion of water = 333 × 103 J/kg dm/dt = rate in kg/sec at which ice melts at the cold end Then, using RHK Eq. 25.7, Q = Lf m dm dQ = Lf dt dt dm H = Lf dt dm H = Lf dt dm . kg/sec = dt Solution: Let A = area of slab of ice on water’s surface h = present thickness of slab = 0.05 m T = temperature of air above ice = −10 ◦ C T0 = temperature at which water freezes = 0 ◦ C k = thermal conductivity of ice = 1.7 W/m·K ρ = density of ice = 0.92 × 103 kg/m3 Lf = latent heat of fusion of water = 333 × 103 J/kg H = dQ/dt = heat ﬂow (in W) through the ice dm/dt = rate of formation of ice (in kg/sec) on the bottom surface of the slab dh/dt = rate of change of ice thickness (in m/sec). 5.23 × 10−5 Hints: The thermal conductivity of silver is 428 W/m·K. The latent heat of fusion of water is 333 kJ/kg. 4 Then, using RHK Eqs. 25.45 and 25.7, ∆T ∆x T0 − T = kA h Q = Lf m dm dQ = Lf dt dt dm H = Lf dt dm T0 − T = Lf kA h dt ρhA = m 1 dm dh = dt ρA dt dm 1 Lf = ρALf dt kA T0 − T = ρALf h k T0 − T = ρLf h = 1.11 × 10−6 m/sec = 0.400 cm/hr . H = kA Note that the inverse dependence of dh/dt upon h requires h to increase only as the square root of the time t. Our numerical result for the rate of growth of the ice thickness is valid only when the ice has a particular thickness (5 cm). 7. RHK problem 23.16 A mercuryﬁlled manometer with two unequallength arms of the same crosssectional area is sealed oﬀ with the same pressure p0 of perfect gas in the two arms (see the ﬁgure). With the temperature held constant, an additional 10.0 cm3 of mercury is admitted through the stopcock at the bottom. The level on the left increases 6.00 cm and that on the right increases 4.00 cm. Find the original pressure p0 . Solution: Let ρ = density of Hg = 13.6 × 103 kg/m3 g = acceleration of gravity at earth’s surface = 9.81 m/sec2 L0 = initial height of gas in left arm of manometer = 0.50 m R0 = initial height of gas in right arm of manometer = 0.30 m L = ﬁnal height of gas in left arm of manometer = 0.44 m R = ﬁnal height of gas in right arm of manometer = 0.26 m A = crosssectional area of each manometer arm p0 = initial pressure in both arms of manometer pL = ﬁnal pressure in left arm of manometer pR = ﬁnal pressure in right arm of manometer NL = no. of gas molecules in left arm of manometer NR = no. of gas molecules in right arm of manometer kB = Boltzmann’s constant T = (constant) temperature 5 Applying the perfect gas law, p0 AL0 = NL kB T p0 AR0 = NR kB T pL AL = NL kB T pR AR = NR kB T p0 L0 p0 R0 L0 p0 L R0 p0 R = pL L = pR L = pL = pR R0 L0 − R L . (I) pR − pL = p0 Using Archimedes’ principle (ﬁrst equation on RHK page 387), the diﬀerence (L0 − L) − (R0 − R) in ﬁnal height of Hg between the two arms is proportional to the ﬁnal pressure diﬀerence: A(pR − pL ) = ρgA (L0 − L) − (R0 − R) (II) pR − pL = ρg (L0 − L) − (R0 − R) . Combining equations (I) and (II), p0 R0 L0 − R L = ρg (L0 − L) − (R0 − R) (L0 − L) − (R0 − R) R0 /R − L0 /L p0 = ρg = 1.526 × 105 Pa = 1.506 atm . University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 2 1. RHK problem 24.18 2. RHK problem 24.21 3. RHK problem 24.25 4. RHK problem 23.17 5. RHK problem 23.33 6. RHK problem 23.37 7. RHK problem 25.16 8. RHK problem 25.21 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 2 1. RHK problem 24.18 Solution: For ease of notation, here we denote the mean of any function f (v ) of the speed v of a gas molecule by
∞ 0 (c) v
2 = = f (v ) ≡ f (v )n(v )dv ∞ n(v )dv 0 vrms v0 2 v Cv 2 dv 0 v0 Cv 2 dv 0 15 5 v0 13 3 v0 32 v 50 ≡ v2 = 3 v0 . 5
v0 0 where n(v ) (called dN/dv in lecture) is the distribution of v . If this formula is used, n(v ) does not need to be normalized. With this notation, for example, v ≡ v . Proceeding with the problem, ¯ vrms ≡ (v − v ) ¯
2 vrms = (a) N≡ = Cv 2 dv v 2 (RHK Eq. 23.15)
2 2 2 2 0 ≤ (v − v )2 ¯ =v =v =v =v v≤v ¯
2 13 Cv 30 − 2v v + v ¯ ¯ − v 2v + v 2 ¯ ¯ − 2¯ + v v ¯
2 2 2 3N 3 =C. v0 3. RHK problem 24.25 Solution: n(E ) ∝ E 1/2 exp (−E/kT ) (RHK Eq. 24.27) Erms ≡ E2 = E2
∞ E 2 E 1/2 exp (−E /kT )dE 0 ∞ E 1/2 exp (−E /kT )dE 0 ∞ 0 ∞ 0 2 − v2 ¯ 0 ≤ v2 − v2 ¯
2 v≤ ¯ v2 v ≤ vrms . ¯ The equality occurs only when (v − v )2 = 0, ¯ i.e. all the molecules have the average speed v . ¯ 2. RHK problem 24.21 Solution: Using the notation introduced above, (b) v= = =
v0 v C v 2 dv 0 v0 Cv 2 dv 0 14 4 v0 13 3 v0 β ≡ 1/kT E
2 = = E 5/2 exp (−βE )dE E 1/2 exp (−βE )dE
∞ E 1/2 exp (−βE 0 ∞ E 1/2 exp (−βE )dE 0 d /dβ 2
∞ 0 )dE Z≡ E2 E 1/2 exp (−βE )dE d2 Z/dβ 2 . = Z 3 v0 . 4 The remaining deﬁnite integral Z has dimension (energy)3/2 . Since the limits of the integral are not ﬁnite, the only available quantity with which a dimensionful scale may be set is β , which has dimension 1/energy. Therefore the integral must 2 be equal to β −3/2 multiplied by some constant C : E2 = = = Erms = = d2 /dβ 2 C β −3/2 C β −3/2 − 3 − 5 C β −7/2 2 2 C β −3/2 15 −2 β 4 15 −1 β 4 15 kT . 4 vesc = escape velocity at earth’s surface m = generic molecular mass NA = Avogadro constant = 6.022 × 1023 molecules/mole MHyd = molar mass of H2 = 0.0020 kg/mole (RHK Table 23.1) MOxy = molar mass of O2 = 0.0320 kg/mole (RHK Table 23.1) kB = Boltzmann constant = 1.38 × 10−23 J/K Hyd Tesc (earth) = temperature (◦ K) at which rms H2 velocity is equal to escape velocity at earth’s surface Oxy Tesc (earth) = temperature (( K) at which rms O2 velocity is equal to escape velocity at earth’s surface Hyd Tesc (m) = temperature ((◦ K) at which rms H2 velocity is equal to escape velocity at moon’s surface Oxy Tesc (m) = temperature ((◦ K) at which rms O2 velocity is equal to escape velocity at moon’s surface Then 4. RHK problem 23.17 Solution: In Physics H7B, all problems involving numbers should be solved completely in terms of algebraic symbols before any numbers are plugged in (otherwise it is much more diﬃcult to give part credit). Let T = temperature of interstellar space = 2.7 ◦ K M = molar mass of H2 = 0.0020 kg/mole (RHK Table 23.1) NA = Avogadro constant = 6.022 × 1023 molecules/mole m = mass of H2 molecule = M/NA kB = Boltzmann constant = 1.38 × 10−23 J/K Then from RHK Eq. 23.20, 3 1 m v 2 = kB T 2 2 3kB T 2 v= m v2 vrms ≡ 3kB T m 3kB NA T = M = 183.5 m/sec . = 5. RHK problem 23.33 Solution: Let Re = radius of earth = 6.37 × 106 m Rm = radius of moon = 1.74 × 106 m 2 GMe /Re = g = gravitational acceleration at earth’s surface = 9.81 m/sec2 gm = gravitational acceleration at moon’s surface = 0.16g GMe m 1 2 mvesc = 2 Re 2GMe 2 vesc = Re = 2gRe 3 1 2 mvrms = kB Tesc 2 2 vrms = vesc (stated by problem) 3 1 2 mvesc = kB Tesc 2 2 3 1 m 2gRe = kB Tesc 2 2 2mgRe = Tesc . 3kB We use this general result to evaluate each of 3 the four cases posed: under the line. Similarly, for path 2, W =− =−
path 8 2 MHyd m= NA 2MHyd gRe Hyd Tesc (earth) = 3kB NA = 1.003 × 104 ◦ K 2MOxy gRe Oxy Tesc (earth) = 3kB NA = 1.604 × 105 ◦ K 2MHyd gm Rm Hyd Tesc (moon) = 3kB NA = 438 ◦ K 2MOxy gm Rm Oxy Tesc (moon) = 3kB NA = 7011 ◦ K . p dV p dV −
8 2 p dV −
2 2 p dV = −(12.5 kPa)(6 m3 ) + (5 kPa)(6 m3 ) − 0 = −45 kJ . 7. RHK problem 25.16 Solution: Let mv = (unknown) mass of vaporized material (ice), in kg mf = mass of fused material (ice) = 0.15 kg Lv = latent heat of vaporization of water = 2256 × 103 J/kg Lf = latent heat of fusion of water = 333 × 103 J/kg c = speciﬁc heat capacity of water = 4190 J/kg·C◦ Tv = temperature of steam = 100 ◦ C Tf = temperature of ice = 0 ◦ C T = ﬁnal temperature of steamice mixture = 50 ◦ C The fact that the container is thermally insulated means that the total heat Q transferred out of the steam molecules is transferred into the ice molecules: Q(lost by steam) = Q(gained by ice) mv (Lv + c(Tv − T )) = mf (Lf + c(T − Tf )) Lf + c(T − Tf ) mv = mf Lv + c(Tv − T ) = 0.033 kg . 8. RHK problem 25.21 Solution: Let Q = (unknown) heat transferred into sample Ti = initial temperature = 6.6 K Tf = ﬁnal temperature = 15 K m = mass of Al = 0.0012 kg C = heat capacity per mole of Al η = coeﬃcient of T 3 in expression for C = 3.16 × 10−5 J/mole·K4 MAl = molar mass of Al = 0.0270 kg/mole (RHK Appendix D) c = heat capacity per kg of Al = C/MAl At an altitude in the Earth’s atmosphere where the temperature is ≈ 1000 K, the preceding results imply that the rms velocity would be only √ a factor Tesc /T ≈ 10 below the escape velocity; because of leakage out of the tail of the velocity distribution, little hydrogen would be expected to remain. For √ oxygen, the rms velocity would be a factor ≈ 160 below the escape velocity, allowing that molecule to survive as an atmospheric component. 6. RHK problem 23.37 Solution: For path 1, the work W done on the gas is W =− =−
path 8 2 p dV (RHK 23.24) p dV −
8 8 p dV −
8 2 p dV = −(12.5 kPa)(6 m3 ) − 0 + (20 kPa)(6 m3 ) = 45 kJ , where we have evaluated each straightline segment by reading p oﬀ the graph, multiplying it by the diﬀerence in V to compute the area 4 With these deﬁnitions,
Tf Q=m
Ti c(T ) dT (RHK Eq. 25.4)
Tf = m MAl C (T ) dT
Ti C (T ) = ηT 3 Q=
Tf m η T 3 dT MAl Ti mη = T 4 − Ti4 4MAl f = 0.0171 J . University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 3 1. RHK problem 25.27 2. RHK problem 25.34 3. RHK problem 25.37 4. RHK problem 25.43 5. RHK problem 26.16 6. RHK problem 26.19 7. RHK problem 26.23 8. RHK problem 26.27 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 3 1. RHK problem 25.27 Solution: In Physics H7B, all problems involving numbers should be solved completely in terms of algebraic symbols before any numbers are plugged in (otherwise it is much more diﬃcult to give part credit). Let Q = (unknown) heat added to gas (J) n = no. of moles of gas = 4.34 Cp = molar speciﬁc heat of gas at constant pressure (J/mole·K) ∆T = change in temperature of gas = 62.4 K R = universal gas constant = 8.314 J/mole·K Eint = internal energy of gas (J) M = molecular weight of gas (kg/mole) v 2 = mean square velocity of gas molecules (m2 /sec2 ) (a) Q = nCp ∆T (RHK Eq. 25.17) 7 Cp = R (RHK Eq. 25.21) 2 7 Q = nR∆T 2 = 7880 J . (b) Eint = ∆Eint 5 nRT (RHK Eq. 23.36) 2 5 = nR∆T 2 = 5629 J . transferred between the gas and the ice water, but slowly enough to allow the pressure nevertheless to be deﬁned (as it is in RHK Fig. 25b); and that “slowly” means slowly enough that the gas and the ice water always have the same temperature. If so, the “quick” compression of the gas would occur along an adiabat, while the “slow” expansion would occur along an isotherm. Then W =−
V2 V1 p dV − V2 V2 p dV − V1 p dV .
V2 Further assuming that the gas is ideal, pV γ = p1 V1γ (adiabat) pV = p1 V1 (isotherm) W =− = −p1 V1γ =
V2 V1 p1 V1γ dV − 0 − Vγ V1 V2 p1 V1 dV V 1 1 V1 −1 − γ −1 − p1 V1 ln γ − 1 V2γ −1 V2 V1 (c) n n 1 M v2 2 3 nRT (RHK Eq. 23.31) 2 3 = nR∆T 2 = 3377 J . = 1 M ∆ v2 2 p1 V1 V1γ −1 V1 γ −1 − 1 − p1 V1 ln V . γ − 1 V2 2 The above is correct, given the assumptions, but it does not solve the problem; we are supplied neither the initial volume nor the number of moles of gas. Instead we are told that m = 0.122 kg of ice in the surrounding ice water are melted in one cycle. The heat −Q = Lf m required to melt this ice, where Lf = 333 kJ/kg is the latent heat of fusion of water, must be transferred from the gas (we call it −Q because +Q is deﬁned to be the heat transferred to the gas). Around one cycle, the ﬁnal temperature of gas is the same as the initial; its internal energy, which depends only on the temperature, can undergo no net change. Therefore, around the cycle, the work W done on the gas is given without any assumptions by ∆Eint = 0 ∆Eint = Q + W (1st Law) W = −Q = Lf m = 40626 J . 2. RHK problem 25.34 Solution: Plunging blindly ahead, we could start by assuming that “quickly” means quickly enough so that a negligible amount of heat is 2 3. RHK problem 25.37 Solution: (a) W =−
V2 (d) C≡ p dV Q (RHK Eq. 25.8) n ∆T 6nRT1 = n(4T1 − T1 ) = 2R . V1 p V = (from problem) V1 p1 W =−
V2 V1 p1 V dV V1 p1 V1 W V2 W W (b) 1 p1 2 =− V − V12 2 V1 2 V2 2 1 −1 = − p1 V1 2 V1 = nRT1 1 V2 2 = − nRT1 −1 2 V1 = 2V1 1 = − nRT1 (4 − 1) 2 3 = − nRT1 . 2 4. RHK problem 25.43 Solution: This problem is “overconstrained”: that is, too many pieces of information are provided. For example, TC need not have been supplied; it is uniquely determined by the facts that process BC is adiabatic; that VB = VA ; that pC = pA ; and that the gas is ideal monatomic. This is illustrated by the following calculation (not required as part of the solution):
γ γ pB VB = pC VC (adiabatic) γ γ pB V A = pA V C 3 nRT (ideal monatomic gas) 2 3 ∆Eint = nR(T2 − T1 ) 2 nRT2 = p2 V2 = (2p1 )(2V1 ) = 4p1 V1 = 4nRT1 T2 − T1 = 3T1 3 ∆Eint = nR(3T1 ) 2 9 = nRT1 . 2 Eint = (c) ∆Eint = Q + W (1st Law) Q = ∆Eint − W 3 9 = nRT1 − − nRT1 2 2 = 6nRT1 . pA TB TA pV = nRT TB p B = pA TA nRTA γ γ VA = pA nRTC γ γ VC = pA nRTA γ nRTC γ = pA pA pA TB γ γ T = TC TA A TB 1/γ T C = TA TA = 454.71497 K . If we needed to get exact answers, we would need to plug in this exact value of TC , rather than the approximate value of 455 K supplied in the problem. To proceed further, we choose not to use one known piece of information – not to use (as we did above) the speciﬁc relationship between p, V , T , and γ for an adiabatic transition. Because this choice is subjective and not unique, when our solutions are expressed in algebraic symbols we expect them also not to be unique. However, as long as the exact value of TC is plugged in, we expect any valid solution to yield the same numerical results. Let 3 TA = temperature at point A = 300 K TB = temperature at point B = 600 K TC = temperature at point C = 454.71497 K (see above discussion) n = no. of moles of monatomic ideal gas = 1.00 R = universal gas constant = 8.314 J/mole·K pA = 1.013 × 105 Pa. Then (a) Process AB: Process CA: ∆Eint = 3 nR(TA − TC ) 2 3 = − nR(TC − TA ) 2 = −1929.45 J .
VA W =− ∆Eint 3 = nR(TB − TA ) 2 = 3741 J .
VB p dV
VC W =− =− p dV
VA VA = −pVA + pVC (p = pA = pC ) pV = nRT W = −nRTA + nRTC = nR(TC − TA ) = 1286.30 J . Q = ∆Eint − W 3 = − nR(TC − TA ) − nR(TC − TA ) 2 5 = − nR(TC − TA ) (= Cp ∆T ) 2 = −3215.75 J . p dV
VA =0. Q = ∆Eint − W 3 = nR(TB − TA ) (= CV ∆T ) 2 = 3741 J . Complete cycle: Process BC: ∆Eint 3 = nR(TC − TB ) 2 3 = − nR(TB − TC ) 2 = −1812 J . Q = 0 (adiabatic) . W = ∆Eint − Q 3 = − nR(TB − TC ) 2 = −1812 J . ∆Eint ≡ 0 (state variable) . 3 W = − nR(TB − TC ) + nR(TC − TA ) 2 3 5 = −nRTA − nRTB + nRTC ) 2 2 = −525.55 J . 5 3 Q = nR(TB − TA ) − nR(TC − TA ) 2 2 3 5 = nRTA + nRTB − nRTC 2 2 = 525.55 J . 4 (b) pB TB = (V ﬁxed) pA TA TB pB = pA TA = 2.026 × 105 Pa . pC = pA = 1.013 × 105 Pa . nRTA VA = pA VB = VA nRTA = pA = 0.0246 m3 . VC TC = (p ﬁxed) VA TA TC VC = VA TA = 0.0373 m3 . 5. RHK problem 26.16 Solution: Consider a Carnot engine operating in reverse (as a refrigerator) between a cold reservoir at temperature TL = 276 K and a hot reservoir at TH = 308 K. Like all Carnot engines it is characterized by the equality TH QH  = (RHK Eq. 26.9) . QL  TL For operation as a refrigerator, the heat QH added to the gas by the hot reservoir is negative. Conversely, QL is positive. The net heat Q = QH + QL added to the gas over one complete cycle is negative. Since the internal energy Eint is a state function, over a complete cycle it must be conserved. Therefore, in one complete cycle, −Q must be balanced by the mechanical work W done on the gas. A ﬁgure of merit F for a heat pump, the ratio of −QH to W , is F= −QH W −QH = −QH − QL TH = TH − T L = 9.625 . The inventor claims to have achieved a ﬁgure of merit equal to F= −QH W 20 kW = 1.9 kW = 10.526 . This is slightly larger than the Carnot ﬁgure of merit. Any reversible heat pump will have the same ﬁgure of merit as a Carnot engine. The only other possibility would be that the inventor’s heat pump is irreversible. For example, friction in the refrigerator could convert a certain additional amount W of work directly to heat in each cycle. In the best case, all of the heat from W would be dumped into the hot rather than the cold reservoir. Then W would be added both to the numerator and to the denominator of F , reducing its physical value further below the value reported by the inventor. Therefore we are forced to reject the inventor’s claim. (Nevertheless, many patents indeed have been granted for processes that violate elementary physical laws.) 6. RHK problem 26.19 Solution: Again a Carnot engine is operated in reverse between a hot reservoir at TH and a cold reservoir at TL . Again QH is negative and QL is positive, and, since the refrigerator is reversible, TH QH  = . QL  TL Again ∆Eint must be zero over a complete cycle, so that W = −Q over the cycle. (a) W = −Q = −QH − QL −QH = QL −1 QL TH −1 = QL TL TH − T L . = QL TL 5 (b) Then −WE = Q1 + Q2 −Q2 Q1 T2 = Q1 1 − T1 WR = −Q3 − Q4 Q4 = −Q3 1 − −Q3 T4 = −Q3 1 − T3 −WR 1= WE = Q1 1 − = −Q3 1 − Q1 1 −
T2 T1 T4 T3 T2 T1 T4 T3 T4 T3 T2 T1 K≡ = QL W QL −QH − QL 1 = −QH QL − 1
TH TL = = 1 −1 TL . TH − T L (c) 1− −Q3 = Q1 1− 1− Q3  = Q1  1− TL = 260 K TH = 298 K TL K= TH − T L = 6.842 . . 7. RHK problem 26.23 Solution: Let Q1 = heat transferred to gas in engine from (hot) reservoir 1 (> 0) Q2 = heat transferred to gas in engine from (cold) reservoir 1 (< 0) Q1 = heat transferred to gas in refrigerator from (hot) reservoir 3 (< 0) Q1 = heat transferred to gas in engine from (cold) reservoir 4 (> 0) WE = mechanical work done on gas in engine (< 0) WR = mechanical work done on gas in refrigerator (> 0) 8. RHK problem 26.27 Solution: Let Wab = work done on gas in stroke ab, etc. W = work done on gas in cycle Wby eng = work done by engine in cycle p0 = smaller pressure = 1.01 × 105 Pa p1 = larger pressure = 2p0 V0 = smaller volume = 0.0225 m3 V1 = larger volume = 2V0 Qabc = heat added to gas during pair of expansion strokes e = eﬃciency of engine eCarnot = eﬃciency of Carnot engine operating between two temperatures with ratio p1 V1 /p0 V0 Then (a) W = Wab + Wbc + Wcd + Wda = 0 − p1 (V1 − V0 ) + 0 + p0 (V1 − V0 ) = −(p1 − p0 )(V1 − V0 ) Wby eng = −W = (p1 − p0 )(V1 − V0 ) = p0 V0 = 22725 J . 6 (b) Qabc = Qab + Qbc = Eint (c) − Eint (a) − Wab − Wbc 3 = nR(Tc − Ta ) − 0 + p1 (V1 − V0 ) 2 3 3 = p1 V1 − p0 V0 + p1 V1 − p 1 V0 2 2 5 3 = p1 V1 − p 1 V0 − p0 V0 2 2 13 p0 V0 = 2 = 147713 J . (c) e≡ Wby eng Qabc (p1 − p0 )(V1 − V0 ) =5 3 2 p1 V1 − p 1 V0 − 2 p0 V0 2 = 13 = 0.1538 . Tc − T a Tc p1 V1 − p 0 V0 = p1 V1 3 =. 4 (d) eCarnot = University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 4 1. RHK problem 26.36 2. RHK problem 26.40 3. RHK problem 26.43 4. Purcell problem 1.5 5. Purcell problem 1.8 6. Purcell problem 1.14 7. Purcell problem 1.26 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 4 1. RHK problem 26.36 Solution: Let n = no. of moles of ideal monatomic gas = 1.00 R = universal gas constant = 8.314 J/mole·K Then (a) Wabc = Wab + Wbc =−
b a (c) dEint ≡ 0 (state function) . dS ≡ 0 (state function) . p dV + 0 = −p0 (4V0 − V0 ) = −3p0 V0 . (b) ∆Eint (b → c) = 3 nR(Tc − Tb ) 2 3 = (pc Vc − pb Vb ) 2 3 = (8p0 V0 − 4p0 V0 ) 2 = 6p0 V0 . c δQ = T b c dEint − Wbc = T b c dEint = −0 T b 3 = nRT 2 c dT 3 nR = T b2 Tc 3 = nR ln 2 Tb 3 = nR ln 2 2 = 8.644 J/K . 2. RHK problem 26.40 Solution: Let n = no. of moles of ideal diatomic gas = 1.00 R = universal gas constant = 8.314 J/mole·K Then (a) ∆Sbc Eint ∆Sbc pV = constant (isotherm) V1 p2 = p1 V2 p1 . = 3 pV γ = constant (adiabat) V1 γ p3 = p1 V3 7 (diatomic) γ= 5 1 7/ 5 p3 = p1 3 = 0.215 p1 . T V γ −1 = constant (adiabat) V1 γ −1 T3 = T1 V3 1 2/ 5 = T1 3 = 0.644 T1 . Note that the last result does not depend on p0 or V0 , even though the problem asks us to express it in terms of p0 and V0 . 2 (b) Q12 = ∆Eint (1 → 2) − W12 = 0 + p1 V1 ln 3 Q23 = 1.099 p1 V1 . = ∆Eint (2 → 3) − W23 1 5 = − p1 V1 1 − 2 3 = −0.889 p1 V1 . ≡ 0 (adiabat) .
2/ 5 5 Eint = nRT (diatomic) 2 5 ∆Eint (1 → 2) = nR(T2 − T1 ) 2 =0. 5 ∆Eint (2 → 3) = nR(T3 − T2 ) 2 5 1 2/ 5 −1 = nRT1 2 3 5 1 2/ 5 = − p1 V1 1 − 2 3 = −0.889 p1 V1 . dEint ≡ 0 (state function) ∆Eint (3 → 1) = −∆Eint (1 → 2) − ∆Eint (2 → 3) 1 5 = −0 + p1 V1 1 − 2 3 = 0.889 p1 V1 .
2/ 5 −0 Q31 2 ∆S12 = 1 T = T1 Q12 ∆S12 = T1 p1 V1 = ln 3 T1 = nR ln 3 = 9.134 J/K . δQ T (isotherm) ∆S31 ≡ 0 (adiabat) . dS ≡ 0 (state function) ∆S23 = −∆S12 − ∆S31 = −nR ln 3 − 0 = −9.134 J/K . W12 = − =− 2 1 p dV
2 1 nRT1 dV V = −nRT1 ln V2 V1 = −p1 V1 ln 3 = −1.099 p1 V1 . W23 = −
3 2 p dV W31 =0. = ∆Eint (3 → 1) − Q31 1 5 p1 V1 1 − 2 3 = 0.889 p1 V1 . =
2/ 5 −0 3. RHK problem 26.43 Solution: In Physics H7B, all problems involving numbers should be solved completely in terms of algebraic symbols before any numbers are plugged in (otherwise it is much more diﬃcult to give part credit). Let m1 = initial amount of water = 1.780 kg (initial amount of ice is 0.262 kg) m2 = ﬁnal amount of water = (1.780 + 0.262)/2 = 1.021 kg Lf = latent heat of fusion of water = 333000 J/kg T0 = temperature of melting ice = 273 K Then 3 (a) Q(m1 → m2 ) = Lf (m2 − m1 ) ∆S (m1 → m2 ) =
2 1 δQ T T ≡ T0 Q(m1 → m2 ) ∆S (m1 → m2 ) = T0 Lf (m2 − m1 ) = T0 Lf (m1 − m2 ) =− T0 = −925.8 J/K . (b) dS ≡ 0 (state function) ∆S (m2 → m1 ) = −∆S (m1 → m2 ) Lf (m1 − m2 ) = T0 = 925.8 J/K . (c) Here the change of entropy of the environment in this cycle is calculated assuming that the heat to melt the ice is supplied at a temperature T>0 which is greater than T0 , for example by a Bunsen burner. Nevertheless, using the fact that the entropy of the environment is a state variable, we calculate its change by making use of a hypothetical reversible process, ∆S = Q/T : dSicewater ≡ 0 dSenviron = − Q(m1 → m2 ) Q(m2 → m1 ) − T0 T> 0 Lf (m1 − m2 ) =− − T0 Lf (m1 − m2 ) − T> 0 1 1 = Lf (m1 − m2 ) − T0 T> 0 >0 4. Purcell problem 1.5 Solution: Consider an element of charge dQ = λR dφ, where dφ is an element of azimuth around the semicircle (0 < φ < π ), and λ = Q/πR is the charge per unit length (in esu/cm) around the semicircle. Construct a Cartesian coordinate system with its origin at the center of the semicircle; choose x = R cos φ and y = R sin φ. Then the symmetry about x = 0 and z = 0 requires the electric ﬁeld at the origin from the full semicircle not to have any component in the x or z directions. So the net electric ﬁeld must be parallel to the y axis; it points toward −y if the charge Q is positive. At the origin, Coulomb’s law requires the above mentioned charge element dQ to create an element of electric ﬁeld dE which has a magnitude equal to dQ/R2 . However, only a fraction sin φ of that ﬁeld magnitude points in the −y direction. Therefore dEy = − dQ sin φ R2 λR dφ sin φ =− R2 Q R dφ = − πR 2 sin φ R Q = − 2 sin φ dφ πR π Q sin φ dφ Ey = − 2 πR 0 2Q =− π R2 2Q E = 0, − ,0 . π R2 5. Purcell problem 1.8 Solution: Let a be the ionic spacing of the onedimensional crystal. Place the ﬁrst positive ion at x = 0, two negative ions at x = ±a, two more positive ions at x = ±2a, etc. Consider Purcell’s Eq. 1.9: 1 U= 2
N j =1 k=j dSuniverse > 0 . qj qk . rjk 4 This is a double sum. As the number N of ions approaches ∞, the sum of the terms of the double sum which involve any particular ion will be the same as the sum of the terms involving any other particular ion (see the argument at the bottom of Purcell’s page 14). Thus the double sum reduces to a single sum: 1 U= N 2
N k=2 6. Purcell problem 1.14 Solution: This is similar to Purcell’s problem 1.5, discussed above, and we will use similar notation. Consider an element of charge dQ = λb dφ, where dφ is an element of azimuth around the circle (0 < φ < 2π ), and λ = Q/2πb is the charge per unit length (in esu/cm) around the circle. Construct a Cartesian coordinate system with its origin at the center of the circle; choose z as the coordinate along the axis normal to plane of the circle. Consider a line drawn from dQ to a point (0, 0, z ) on this axis. Deﬁne ψ to be the angle that this line makes with the plane of the circle. With these deﬁnitions, tan ψ = z/b and − π < ψ < π ; the distance from dQ to (0, 0, z ) is 2 2 b sec ψ . Because the conﬁguration is symmetric about x = 0 and y = 0, on the z axis the electric ﬁeld must point in the z direction, away from the plane of the ring if its charge Q is positive. At the point (0, 0, z ), Coulomb’s law requires the above mentioned charge element dQ to create an element of electric ﬁeld dE which has a magnitude equal to dQ/(b sec ψ )2 . However, only a fraction sin ψ of that ﬁeld magnitude points in the z direction. Therefore q1 qk , r 1k where we have chosen to sum only the terms involving ion 1. Furthermore, since the string of ions is symmetric about x = 0, we may consider in the single sum only the ions with x > 0, at the expense of multiplying the result by an extra factor of 2: U= 1 2N 2
N k=2; x>0 q1 qk . r 1k Here we evaluate r1k = a(k − 1), and we use the fact that the sign of q1 qk is equal to (−1)k−1 : 1 U = 2N 2 = Ne a
2 N k=2 N k=2 N −1 j =1 q1 qk a(k − 1) (−1)k−1 (k − 1) (−1)j . j N e2 = a dEz = Taking N → ∞ in the limit of the sum, U= N e2 a
∞ j =1 (−1)j j dQ sin ψ sec2 ψ λb dφ sin ψ =2 b sec2 ψ b2 = =
Q 2πb b dφ b2 sec2 ψ sin ψ N e2 ln (1 + 1) a e2 U = − ln 2 , N a =− where, following the hint, we have evaluated the sum by using the Taylor series expansion
∞ Q cos2 ψ sin ψ dφ 2πb2 Q cos2 ψ sin ψ 2π Ez = dφ 2πb2 0 Q cos2 ψ sin ψ . = b2 ln(1 + b) =
j =1 (−b)j −1 . j The problem thus reduces to ﬁnding the value of 5 ψ which maximizes the product cos2 ψ sin ψ : u ≡ sin ψ d u(1 − u2 ) 0= du = 1 − 3u2 u= 1 3 1 3 1 3 We have seen that dEA,y exactly cancels dEB,y for any choice of θ; therefore EC vanishes. ψ = arcsin z = b tan arcsin =b 1 . 2 7. Purcell problem 1.26 Solution: Place the origin of a Cartesian coordinate system at the center of the semicircle, with both parallel rods lying in the xy plane. Orient the y coordinate so that the rods extend to y = −∞. At point C , the origin of this coordinate system, any electric ﬁeld can point only in along the ±y direction, owing to the symmetry of the problem about x = 0 and z = 0. Purcell’s ﬁgure refers us to two elements of charge. The element at point A has a value dQ = λb dθ and generates an electric ﬁeld at the origin of magnitude λb dθ/b2 . Only a fraction sin θ of this ﬁeld points in the −y direction; thus λb dθ dEA,y = − 2 sin θ b λ = − sin θ dθ . b The ﬁeld from the element of charge at point B is slightly more complicated. This charge element has value dQ = λ dy , where dy  is an element of length along the straight rod, and y  = b tan θ. Therefore dQ = λb d tan θ = λb sec2 θ dθ. This element of charge lies a distance b sec θ away from the origin. Again, only a fraction sin θ of the ﬁeld generated by this charge element points in the +y direction. Putting it all together, dEB,y = + λb sec2 θ dθ sin θ b2 sec2 θ λ = + sin θ dθ . b University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 5 1. Purcell problem 1.16 2. Purcell problem 1.19 3. Purcell problem 1.29 4. Purcell problem 1.33 5. Purcell problem 2.1 6. Purcell problem 2.8 7. Purcell problem 2.19 8. Purcell problem 2.20 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 5 Solutions by P. Pebler 1 Purcell 1.16 A sphere of radius a was ﬁlled with positive charge of uniform density ρ. Then a smaller sphere of radius a/2 was carved out, as shown, and left empty. What are the direction and magnitude of the electric ﬁeld at points A and B ? The key is to consider the given distribution as a superposition of the two distributions at right. The electric ﬁeld will be the sum of the contributions from these two spheres, which are easy to evaluate. For points outside these spheres, we may treat them as point charges lying at their centers. The charges are 4 a Q1 = π 3 2 4 Q2 = πa3 ρ. 3 Consider the point A. The contribution from sphere 2 is zero since there is no ﬁeld at the center of a spherical distribution. The point A lies outside the sphere 1. EA = Q1 2π ˆ ˆ aρ y (−y) = (a/2)2 3
3 (−ρ) = − πa3 ρ , 6 Point B lies outside both spheres. EB = 2πa3 ρ 34 Q2 Q1 4πa3 ρ ˆ ˆ ˆ ˆ ˆ y+ y = − πaρ y (−y) + (−y) = − 2 2 2 2 a (3a/2) 3a 27a 27 2 Purcell 1.19 An inﬁnite plane has a uniform surface charge distribution σ on its surface. Adjacent to it is an inﬁnite parallel layer of charge of thickness d and uniform volume charge density ρ. All charges are ﬁxed. Find the electric ﬁeld everywhere. 1 The contribution due to the surface charge has magnitude 2π σ  and points away from or towards the surface depending on the sign of σ . To deal with the volume charge, we can treat it as a stack of very thin layers of charge and treat these layers as surface charges. We could add up all the contributions from these inﬁnitesimal layers by integrating. However, since the ﬁeld from an inﬁnite plane of charge does not depend on how far away you are, the contribution from each layer will be the same. So we will get the same answer by assuming the ﬁnite volume charge layer to be a surface with surface charge ρ t where t is the thickness of the layer in question. For x < 0, everything pushes to the left. ˆ ˆ ˆ E = 2πσ (−x) + 2πρ d(−x) = −2π (σ + ρ d) x Likewise, ˆ E = 2π (σ + ρ d) x x ≥ d. x<0 For the region 0 < x < d, the volume layer is split into two. We can thing of the right side as a single surface with surface charge ρ(d − x) pushing to the left, and the left side as a surface charge ρx pushing to the right. ˆ ˆ ˆ ˆ E = 2πρ x + 2πρ x x − 2πρ(d − x) x = (2πσ + 4πρ x − 2πρ d) x 0<x<d If we wished to consider the plane x = 0, we could say that the surface charge σ contributes nothing. ˆ E = −2πρ d x x=0 Notice that there is a discontinuity of 4πσ as we pass through zero. This is always the case for idealized surface charges. There is no discontinuity at x = d however. 3 Purcell 1.29 A spherical shell of charge of radius a and surface charge density σ is missing a small, approximately circular, piece of “radius” b a. What is the direction and magnitude of the ﬁeld at the midpoint of the aperture? The picture above assumes for simplicity that σ > 0. As a zero order approximation, we may consider the missing circle as inﬁnitesimal. The ﬁeld at left can be viewed as a superposition of the two distributions at right. We temporarily ignore points exactly at the surface. By considering the ﬁeld to the right or to the left we ﬁnd ˆ ˆ ˆ ˆ E = 4πσ x + 2π (−σ ) x = 0 + 2π (−σ )(−x) = 2πσ x, for the ﬁeld everywhere except at the surface. But for the distribution with the circle missing, there can be no discontinuity when passing through the hole, so the ﬁeld directly at the surface is also ˆ 2πσ x. This should be a good approximation when b a. But for a ﬁnite missing piece, this will not be the exact answer even at the center. To ﬁnd the contributions of higher order in b/a we can integrate. This is actually not too bad. 2 From symmetry considerations, we know the ﬁeld is radial in the center of the aperture. dEr = σda σR2 2π dθ cos(90 − θ/2) cos α = r2 2R2 (1 − cos θ)
θ sin θ sin 2 (1 + cos θ) θ (1 + cos θ) dθ = πσ cos dθ dθ = πσ 2 θ 2 sin θ 2 cos 2 = πσ
π Er = πσ θo cos θ θo dθ = 2πσ 1 − sin 2 2
3 We can take the initial angle θo to be b/R. Er = 2πσ 1 − b 1 − 2R 3! b 2R + ... 4 Purcell 1.33 Imagine a sphere of radius a ﬁlled with negative charge −2e of uniform density. Imbed in this jelly of negative charge two protons and assume that in spite of their presence the negative charge remains uniform. Where must the protons be located so that the force on each of them is zero? The forces on the protons from each other will be equal and opposite. Therefore, the forces on them from the negative charge distribution must be equal and opposite also. This requires that they lie on a line through the center and are equidistant from the center. The force on each proton at radius r from the negative charge will be proportional to the amount of negative charge lying inside a sphere of radius r. For purposes of ﬁnding the electric ﬁeld, we may treat all of this charge as if it were a point charge sitting in the center. We ignore all negative charge outside the radius of the proton positions. The negative charge inside the radius r is r3 2e. a3 The force on the right proton must be zero. q=− F= e(−2er3 /a3 ) e2 e2 ˆ ˆ x+ x= (2r)2 r2 (2r)2 The vector function Ey = 3x2 − 3y 2 Ez = 0 3 1−8 r3 a3 ˆ x=0 r= a 2 5 Purcell 2.1 Ex = 6xy represents a possible electrostatic ﬁeld. Calculate the line integral of E from the point (0, 0, 0) to the point (x1 , y1 , 0) along the path which runs straight from (0, 0, 0) to (x1 , 0, 0) and thence to (x1 , y1 , 0). Make a similar calculation for the path which runs along the other two sides of the rectangle, via the point (0, y1 , 0). Now you have the a potential function φ(x, y, z ). Take the gradient of this function and see that you get back the components of the given ﬁeld. ˆ We take the ﬁrst path in two parts. While moving along the x axis we have ds = dx x so that ˆ E · ds = Ex dx and while moving up parallel to the y axis we have ds = dy y and E · ds = Ey . E · ds =
x1 0 Ex dx + y1 0 Ey dy = x1 0 6xy dx + y1 0 (3x2 − 3y 2 ) dy When integrating along the x axis, y has the constant value y = 0 which we plug in to the ﬁrst integral. Along the second part of the path, x = x1 . E · ds = 0 +
y1 0 3 (3x2 − 3y 2 ) dy = 3x2 y1 − y1 1 1 We do the same thing along the second path. E · ds =
y1 0 (3(0)2 − 3y 2 ) dy + x1 0 3 6xy1 dx = −y1 + 3x2 y1 1 φ(x, y, z ) = 3x2 y − y 3 ∇φ(x, y, z ) = ∂ ∂ ∂ ˆ ˆ ˆ ˆ ˆ (3x2 y − y 3 ) x + (3x2 y − y 3 ) y + (3x2 y − y 3 ) z = 6xy x + (3x2 − 3y 2 ) y ∂x ∂y ∂z 6 Purcell 2.8 Consider an inﬁnitely long cylinder of radius a and uniform charge density ρ. Use Gauss’s law to ﬁnd the electric ﬁeld. Find the potential φ as a function of r, both inside and outside the cylinder, taking φ = 0 at r = 0. Our Gaussian surface both inside and outside the cylinder will be a cylinder of length L. Inside we have E · da = Er 2πrL = 4πQenc = 4ππr2 Lρ, E = 2πρ r ˆ r Outside, E · da = Er 2πrL = 4πQenc = 4ππa2 Lρ, 4 r < a. E= 2πρa2 ˆ r r r ≥ a. To ﬁnd the potential, we integrate radially outward from the center so that ds = dr ˆ. r φ(r) = φ(0) − E · ds = 0 −
r 0 2πρ r dr = −πr2 ρ r≤a For points outside the cylinder, φ(r) = φ(a) −
r a r 2πρa2 dr = −πa2 ρ − 2πρ a2 ln r a r > a. 7 Purcell 2.19 Two metal spheres of radii R1 and R2 are quite far apart compared with these radii. Given a total amount of charge Q, how should it be divided so as to make the potential energy of the resulting charge distribution as small as possible? Assume that any charge put on one of the spheres distributes itself uniformly over the sphere. Show that with that division the potential diﬀerence between the spheres is zero. Because the spheres are far apart, the energy will be essentially due to the energy of each sphere. We may assume that the charge on each sphere is uniformly distributed if the other sphere is very far away. To ﬁnd this energy we can use the standard formula adapted to surface charge, U= 1 2 σ φ da. The potential φ just outside a uniformly charged sphere is q/r and because the potential is continuous, this is also the potential at the surface. Then, U= 1 2 1 q2 q q2 r sin θ dφdθ = . 4πr2 r 2r Breaking up the charge into q and Q − q , U= q2 (Q − q )2 + . 2R1 2R2 If the minimum energy is obtained with q = qo , qo (Q − qo ) dU (qo ) = − = 0, dq R1 R2 Q − qo qo = . R1 R2 But these are just the potentials at both spheres. 8 Purcell 2.20 As a distribution of electric charge, the gold nucleus can be described as a sphere of radius 6 × 10−13 cm with a charge Q = 79e distributed fairly uniformly through its interior. What is the potential φo at the center of the nucleus, expressed in megavolts? For a uniformly charged sphere of radius a, E= Q ˆ r r2 r > a, 5 E= Qr ˆ r a3 r ≤ a. Since the potential is zero at inﬁnity, the potential at any point P is φ(P ) = −
P ∞ E · ds. We can make the path of integration come radial straight in. If the point P has r < a, φ(r) = − In SI units, φ(0) = 3 · 79(1.6 × 10−19 C ) 1 3Q = = 28.4 megavolts. 4π o 2a 4π (8.85 × 10−12 C 2 /N m2 )2(6 × 19−15 m)
r ∞ E r dr = − a ∞ Q dr − r2 r a Qr Q Qr2 Q − 3+ dr = 2 a a 2a 2a 6 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 6 1. Purcell problem 2.27 2. Purcell problem 2.29 3. Purcell problem 2.30 4. Purcell problem 3.1 5. Purcell problem 3.9 6. Purcell problem 3.17 7. Purcell problem 3.23 8. Purcell problem 3.24 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 6 Solutions by P. Pebler 1 Purcell 2.27 The electrostatic potential at a point on the edge of a disc of radius r and uniform charge density σ is φ = 4σr. Calculate the energy stored in the electric ﬁeld of a charged disk of radius a. We calculate the total energy by bringing in each inﬁnitesimal ring of charge from inﬁnity and adding up the energy for each ring. We assume that we have already built up the disc to radius r. We now bring in a ring of width dr and stick it on the edge. Recall that the energy necessary to bring in a test charge from inﬁnity to some point is just the potential at that point times the charge. (This is more or less the deﬁnition of the potential.) The potential just outside the disc where we are packing on the next ring is 4σr. The energy necessary is then dU = φ(r)dq = (4σr)(2πrdrσ ) = 8πσ 2 r2 dr. To add up all the rings integrate from 0 to a. U = 8πσ 2
a 0 8 8 Q r2 dr = πa3 σ 2 = πa3 3 3 πa2 2 = 8Q2 3πa 2 Purcell 2.29 Two nonconducting spherical shells of radius a carry charges of Q and −Q uniformly distributed over their surfaces. The spheres are brought together until they touch. What does the electric ﬁeld look like, both outside and inside the shells? How much work is needed to move them far apart? The ﬁeld of a uniformly charged shell is zero inside the shell and that of a point charge outside. Outside both shells, we have the ﬁeld of two point charges. Inside either shell, the ﬁeld is that of a single point charge at the center of the other shell. To ﬁnd the energy we use the following argument. Consider instead a uniform shell of charge −Q and a point charge Q a distance r from the center of the shell (but outside it). We know that outside the shell, the potential due to the shell is just −Q/r, so the energy needed to bring in the point charge is −Q2 /r and the energy needed to move it out is Q2 /r. However, this must be the same energy as that required to move out the shell while keeping the point charge ﬁxed. So we ﬁnd that the energy needed to move a shell out to inﬁnity in the ﬁeld of a point charge is Q2 /r. But since the other shell creates the ﬁeld of a point charge outside of it, this is also the energy needed 1 to separate our two shells. E= Q2 2a If you don’t like this argument, you can integrate a shell distribution times the potential of a point charge which isn’t too hard and ﬁnd the same answer. 3 Purcell 2.30 Consider a cube with sides of length b and constant charge density ρ. Denote by φo the potential at the center of the cube and φ1 the potential at a corner, with zero potential at iniﬁnity. Determine the ratio φo /φ1 . We imagine another cube with the same charge density but with twice the side length. Let the potential at the center of this cube be φo . The point at the center of this new cube lies at the corner of each of eight cubes of the original size. Because the potential is additive, we have φo = 8φ1 . We can also use dimensional arguments to ﬁnd φo . We can write φo = f (Q, s), where Q is the total charge, s is the side length and the functional form of f depends on the shape and nature of the distribution. We can now ask for what’s called a scaling law which tells us what happens if we multiply the variables Q and s by numerical factors while keeping all other details of the distribution the same. Whatever the functional form of f is, we know it has units of charge per length, the units of the potential. Fortunately, the only parameters carrying units which enter into f are Q and s. The only way then to get the right units is if f (Q, s) ∝ Q . s α f (Q, s). β The function f then satisﬁes the simple scaling f (αQ, βs) = In our case s = 2s and because we are keeping the charge density constant, Q = ρs 3 = ρ(2s)3 = 8Q. Then 8 φo = f (8Q, 2s) = f (Q, s) = 4φo , 2 4φo = 8φ1 , φo = 2. φ1 2 4 Purcell 3.1 A spherical conductor A contains two spherical cavities. The total charge on the conductor is zero. There are point charges qb and qc at the center of each cavity. A considerable distance r away is another charge qd . What force acts on each of the four objects A, qb , qc , qd ? Which answers, if any, are only approximate, and depend on r being relatively large? The force on qb and qc is zero. The ﬁeld inside the spherical cavity is quite independent of anything outside. A charge −qb is uniformly distributed over the conducting surface to cancel the ﬁeld from the point charge. The same happens with qc . This leave an excess charge of qc + qb on the outside surface of the conductor. If qd were absent, the ﬁeld outside A would be the symmetrical, radial ﬁeld E = qb + qc /r2 , the same as a point charge because the excess charge would uniformly distribute itself over the spherical outer surface. The inﬂuence of qd will slightly alter the distribution of the charge on A, but without aﬀecting the total amount. Hence for large r, the force on qd will be approximately Fd = qd (qb + qc ) ˆ. r r2 The force on A must be precisely equal and opposite to the force on qd . 5 Purcell 3.9 Two charges q and two charges −q lie at the corners of a square with like charges opposite one another. Show that there are two equipotential surfaces that are planes. Obtain and sketch qualitatively the ﬁeld of a single point charge located symmetrically in the inside corner formed by bending a metal sheet through a right angle. Which conﬁgurations of conducting planes and point charges can be solved this way and which can’t? The potential on each of the two lines A and B shown is zero because the contribution at each point on either line from any charge is cancelled by the opposite charge directly across from it. Therefore, the ﬁeld of a point charge in the corner of a bent conductor is the same as the ﬁeld from these four point charges. You should be able to see by looking at the ﬁrst few cases that this strategy will work any time we divide the space into an even number of wedges. This allows the contributions to the potential to cancel pairwise. For example, in the picture at right the potential is zero on lines A and B because all the charges come in equal and opposite pairs. The applicable angles are θn = 2π/(2n) = π/n, where n is an integer. This would not work for an angle of 120o . 3 6 Purcell 3.17 A spherical vacuum capacitor has radius a for the outer sphere. What radius b should be chosen for the inner spherical conductor to store the greatest amount of electrical energy subject to the constraint that the electric ﬁeld strength at the surface of the inner sphere may not exceed Eo ? How much energy can be stored? We ﬁrst need the capacitance of this capacitor. Assuming there is a charge Q on the inner shell and a charge −Q on the outer shell, the ﬁeld between the shells is E= Q ˆ. r r2
b a The potential diﬀerence is V =− Q a−b dr = Q , 2 r ab ab . a−b and the capacitance C = Q/V = The energy stored by this capacitor is U= 1 2 1a−b 2 Q= Q. 2C 2 ab The energy in the capacitor will depend on how much charge is on it. If we were allowed to put arbitrary amounts on, the energy would have no maximum. However, for a given b, the maximum ﬁeld near the inner sphere gives us the maximum allowed charge. This gives us the maximum stored energy for a given capacitor. Eo = Qmax b2 1 a − b 2 4 1 ab3 − b4 2 Eb = Eo 2 ab o 2 a Umax = Now we want to choose a b to make this as large as possible. 1 3ab2 − 4b3 2 ∂Umax (bmax ) = Eo = 0 ∂b 2 a 3a − 4bmax = 0 3 bmax = a 4 The energy is then Umax
3 1 a − 4a 2 3 = Eo a 2 a 3a 4 4 4 = 27 2 3 Ea. 512 o 7 Purcell 3.23 Find the capacitance of a capacitor that consists of two coaxial cylinder of radii a and b and length L. Assume L b − a so that end corrections may be neglected. Check your result in the limmit b − a a with the formula for the parallelplate capacitor. 4 A cylinder of 2.00 in outer diameter hangs with its axis vertical from one arm of a beam balance. The lower portion of the hanging cylinder is surrounded by a stationary cylinder with inner diameter 3.00 in. Calculate the magnitude of the force down when the potential diﬀerence between the two cylinders is 5 kV . The ﬁeld between charged cylinders is E= 2λ 2Q ˆ= ˆ, r r r rL
b assuming we have Q on the inside and −Q on the outside. The potential diﬀerence is V= 2Q dr 2Q b = ln . Lr L a a Just arrange your signs so that the capacitance comes out positive. C= L 2 ln(b/a) Let us now consider the general case where the potential diﬀerence is being held constant by a battery while the capacitance is changing. Initially we have charge and energy Q = CV 1 U = CV 2 . 2 1 U = (C + ∆C )V 2 . 2 After a change in capacitance ∆C , Q = (C + ∆C )V = Q + V ∆C The battery has done work on this system by moving this extra charge across the potential diﬀerence. Wb = (∆C )V 2 If the change in capacitance is caused by movement of the components, the electric ﬁeld does work on the plates or plate. W = F (∆L) From conservation of energy we have U + Wb = U + W, 1 (∆C )V 2 = (∆C )V 2 + W, 2 1 W = F (∆L) = (∆C )V 2 , 2 1 ∂C . F = V2 2 ∂L In our case we have 1 (16.7 statvolts)2 1 1 = = 172 dynes. F = V2 2 2 ln(b/a) 2 2 ln(3/2) 5 8 Purcell 3.24 Two parallel plates are connected by a wire. Let one plate coincide with the xz plane and the other with the plane y = s. The distance s is much smaller than the lateral dimensions of the plates. A point charge Q is located between the plates at y = b. What is the magnitude of the total surface charge on the inner surface of each plate? The total induced charge is −Q. We need to ﬁnd the fraction of induced charge on either conductor. For this we may notice that the fraction of induced charge on both planes will be the same for any distribution located at y = b because we may view it as the superposition of many little point charges. So we want to consider the simplest possible case which is a uniformly charged plane. (Once again, we are ignoring edge eﬀects.) Using a Gaussian pillbox with its left face inside the left plate and its right face at y , where 0 < y < b, the ﬁeld in the left region is ˆ El = 4πσ1 y. Similarly, the ﬁeld in the right region is ˆ Er = −4πσ2 y. Since the two conductors are connected by a wire, they are at the same potential so the line integral from the middle to the left and right should be the same. 4πσ1 (−b) = −4πσ2 (s − b) b σ2 = σ1 s−b Now switch back to the original problem. b Q2 = Q1 s−b Q1 = − s−b Q s Q1 + Q2 = −Q b Q2 = − Q s 6 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 7 1. Purcell problem 4.8 2. Purcell problem 4.20 3. Purcell problem 4.25 4. Purcell problem 4.26 5. Purcell problem 4.30 6. Purcell problem 4.31 7. Purcell problem 4.32 8. (Taylor & Wheeler problem 19) (a.) Two events P and Q have a spacelike separation. Show that an inertial frame can be found in which the two events occur at the same time. In this frame, ﬁnd the distance between the two events (this is called the proper distance). (Hint: one method of proof is to assume that such an inertial frame exists and then use the Lorentz transformation equations to show that the velocity βc of this inertial frame, relative to the frame in which the events were initially described, is such that β < 1, thus justifying the assumption made.) (b.) Two events P and R have a timelike separation. Show that an inertial frame can be found in which the two events occur at the same place. In this frame, ﬁnd the time interval between the two events (this is called the proper time). University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 7 Solutions to Purcell problems by P. Pebler 1 Purcell 4.8 A copper wire 1 km long is connected across a 6 volt battery. The resistivity of copper is 1.7 × 10−6 ohm cm, and the number of conduction electrons per cubic centimeter is 8 × 1022 . What is the drift velocity of the conduction electrons under these circumstances? How long does it take an electron to drift once around the circuit? The current will be VA V = I= R ρL and the current density J= V I = . A ρL The current density is the charge density times the drift speed and the charge density is ne where n is the number density of electrons. v= v= t= J V = ne ρLne (1.7 × 10−6 ohm cm)(105 6V = 2.8 × 10−3 cm/s cm)(8 × 1022 cm−3 )(1.6 × 10−19 C ) 1 year 105 cm = 3.6 × 107 s 2.8 × 10−3 cm/s 2 Purcell 4.20 A black box with three terminals a, b, and c contains nothing by three resistors and connecting wire. Measuring the resistance between pairs of terminals, we ﬁnd Rab = 30 ohm, Rac = 60 ohm, and Rbc = 70 ohm. Show that the contents cound be either of the following. Is there any other possibility? Are the two boxes completely equivalent, or is there an external measurement that would distinguish between them? For the ﬁrst box, the resistance between any two terminals involves two of the resistors in series with the third resistor extraneous. For example, Rab = 10 ohm + 20 ohm = 30 ohm. For the second box, the resistance between any two terminals involves one resistor in parallel with the other two in series. For example, Rab = 1 1 + 34 ohm 85 ohm + 170 ohm
−1 = 30 ohm. 1 The other two are easily veriﬁed. These are the only two ways to make these three resistances with only three resistors. For the two arrangements to be electrically identical, they must both draw the same currents given the same input voltages Va , Vb , and Vc . The details are somewhat messy. I’ll leave it to you to verify that given the input voltages Va , Vb , and Vc , both arrangements draw the currents Ia = Ib = Ic = 1 (7Va − 5Vb − 2Vc ), 170 1 (6Vb − Vc − 5Va ), 170 1 (3Vc − Vb − 2Va ). 170 Note that Ia + Ib + Ic = 0 as it must. 3 Purcell 4.25 A charged capacitor C discharges through a resistor R. Show that the total energy dissipated in the resistor agrees with the energy originally stored in the capacitor. Suppose someone objects that the capacitor is never really discharged because Q only becomes zero for t = ∞. How would you counter this objection? Assume that the capacitor initially has charge Q on it. The current as a function of time is i(t) = io e−t/τ where τ = RC and io = Vo /R = Q/CR. The power dissipated in a resistor is P = i2 R, so the total energy dissipated is
∞ ∞ E= 0 P dt = 0 R i2 exp (−2t/τ ) dt = R i2 o o Q2 RC 12 τ =R 2 2 = Q. 2 CR 2 2C This was the initial energy stored in the capacitor. We can ﬁnd the time it takes for the charge left on the capacitor to be one electron. The charge as a function of time is q = Q exp (−t/τ ). The time would be t = τ ln Q . e Because of the ln, even for macroscopic initial charges, the time wouldn’t be that large. 2 4 Purcell 4.26 Two graphite rods are of equal length. One is a cylinder of radius a. The other is conical, tapering linearly from radius a at one end to radius b at the other. Show that the endtoend resistance of the conical rod is a/b times that of the cylindrical rod. We consider the conical rod to be the series combination of little cylindrical rods of length dx. The radii of these little cylinders are b−a x. L We sum up the little resistances. r(x) = a +
L R= dR = 0 ρ dx =ρ A L 0 a L L dx = =ρ ρ2 2 π (a + (b − a)x/L) πab b πa ρL/πa2 is the resistance of the cylinder of radius a. 5 Purcell 4.30 Consider two electrodes 2 mm apart in vacuum connected by a short wire. An alpha particle of charge 2e is emmitted by the left plate and travels directly towards the right plate with constant speed 108 cm/s and stops in this plate. Make a quantitative graph of the current in the connecting wire, plotting current against time. Do the same for an alpha particle that crosses the gap moving with the same speed but at an angle of 45o . Suppose we had a cylindrical arrangement of electrodes. Would the current pulse have the same shape? The result of problem 3.24 tells us the induced charge on the electrodes for any given position of the alpha particle. The current is just the time derivative of this charge. q1 = −2e I= s−b s 2e db 2ev dq1 = = = 0.48 esu/s dt s dt s This pulse lasts for (0.2 cm)/(108 cm/s) = 2 ns. If the alpha particle travels at 45o , the normal √ speed and the current just decrease by 2. For concentric cylinders, the current pulse would have a diﬀerent shape. If you work out the details, 1 I∝ a + vt during the time that the alpha particle is in motion. 3 6 Purcell 4.31 Suppose a cube has a resistor of resistance Ro along each edge. At each corner the leads from three resistors are soldered together. Find the equivalent resistance between two nodes that represent diagonally opposite corners of the cube. Now ﬁnd the equivalent resistance between two nodes that correspond to diagonally opposite corners of one face of the cube. A total current I enters one node. It then has a choice of three directions to go. Because of the symmetry, each choice is identical to the others so the current must split up evenly so that I/3 goes through each resistor. Likewise, the current reaching the other node comes through three resistors each having current I/3. This leaves 6 resistors in the middle to share the current. Because each one is identical due to the symmetry, they must each have current I/6. To ﬁnd the voltage drop between the two nodes, follow a straight path from one to the other. V= I I 5 I Ro + Ro + Ro = RI 3 6 3 6 5 Req = Ro 6 In the second situation, because of the symmetry, we notice that all the resistors on the top square carry the same magnitude of current while all the resistors on the bottom square carry the same magnitude also. This tells us that there is no current through the two resistors indicated by arrows. We can therefore ignore them, because the circuit would behave the same without them. It is then easy to combine the remaining resistors. The top and bottom squares are parallel combinations of resistors 2Ro . R= 1 1 + 2Ro 2Ro
−1 = Ro This leaves us with one resistor Ro in parallel with three resistors Ro . Req = 1 1 + Ro 3Ro
−1 3 = Ro 4 4 7 Purcell 4.32 series. Find the input resistance (between terminals A and B ) of the following inﬁnite Show that, if voltage Vo is applied at the input to such a chain, the voltage at successive nodes decreases in a geometric series. What ratio is required for the resistors to make the ladder an attenuator that halves the voltage at every step? Can you suggest a way to terminate the ladder after a few sections without introducing any error in its attenuation? If we put another “link” on the left of this inﬁnite chain, we get exactly the same conﬁguration. If this inﬁnite chain has equivalent resistance Req , the new chain with the extra link can be described by the middle circuit. We can calculate the equivalent resistance of this circuit by considering Req and R2 in parallel, in series with R1 . But since this circuit is the same as the original, this equivalent resistance is again Req . Req = R1 + 1 1 + R2 Req
−1 This leads to the equation
2 Req − R1 Req − R1 R2 = 0, with positive solution Req = R1 +
2 R1 + 4R1 R2 2 . Now consider an arbitrary link with voltage diﬀerence Vi between top and bottom. To ﬁnd the voltage Vi+1 , we can replace the rest of the series with the equivalent resistor. We could get Vi+1 if we knew the current through R1 . For this purpose we can replace the ith link also with Req . The current through this equivalent circuit will also be the current through R1 . This current is just Vi /Req . Vi+1 = Vi − R1 Vi Req − R1 = Vi Req Req If we wish to halve the voltage each step, 1 R1 =, Req 2 4R1 = R1 +
2 R1 + 4R1 R2 , 5 R2 = 2R1 . If we wish to terminate the ladder without changing this property, we just replace the rest of the chain at any point with a resistor with resistance Req . 8 Taylor & Wheeler 19 (a.) Two events P and Q have a spacelike separation. Show that an inertial frame can be found in which the two events occur at the same time. In this frame, ﬁnd the distance between the two events (this is called the proper distance). (Hint: one method of proof is to assume that such an inertial frame exists and then use the Lorentz transformation equations to show that the velocity βc of this inertial frame, relative to the frame in which the events were initially described, is such that β < 1, thus justifying the assumption made.) (b.) Two events P and R have a timelike separation. Show that an inertial frame can be found in which the two events occur at the same place. In this frame, ﬁnd the time interval between the two events (this is called the proper time). Denote the frame in which P and Q were initially described by S , and the frame in which we wish them to be simultaneous by S . As usual, the origins of these frames coincide at t = t = 0. In S , if P and Q have a spacelike separation, their spatial separation must be nonzero. Orient the x and x axes along the direction of this separation, so that yQ = yP and zQ = zP but xQ = xP . Applying the Lorentz transformation between S and S , ctP = γctP − γβxP ctQ = γctQ − γβxQ ctQ − ctP = γ (ctQ − ctP ) − γβ (xQ − xP ) We wish the lefthand side to be zero. If it is, then γ (ctQ − ctP ) = γβ (xQ − xP ) (ctQ − ctP ) =β (xQ − xP ) Now, we are told that the separation between events Q and P is spacelike: c2 (tQ − tP )2 − (xQ − xP )2 < 0 This guarantees that β  < 1. It is straightforward to calculate the spatial separation of the two events in S by using the inverse Lorentz transformation: xP = γxP + γβctP xQ = γxQ + γβctQ xQ − xP = γ (xQ − xP ) + γβ (ctQ − ctP ) We have chosen β so that the two events are simultaneous in S ; this forces the last term to vanish. Substituting the value that we found for β , xQ − xP = γ (xQ − xP ) 6 xQ − xP = xQ − xP = 1 − β 2 (xQ − xP ) 1− ctQ − ctP xQ − xP
2 (xQ − xP ) xQ − xP = ± (xQ − xP )2 − (ctQ − ctP )2 where the sign is chosen so that xQ − xP has the same sign as xQ − xP . This proper distance is the smallest distance between the two events that can be reached in any reference frame. Similarly, when the separation between events P and R is timelike, xR − xP = γ (xR − xP ) − γβ (ctR − ctP ) γ (xR − xP ) = γβ (ctR − ctP ) (xR − xP ) =β (ctR − ctP ) ctR − ctP = γ (ctR − ctP ) + γβ (xR − xP ) ctR − ctP = ctR − ctP = 1 − β 2 (ctR − ctP ) 1− xR − xP ctR − ctP
2 (ctR − ctP ) ctR − ctP = ± (ctR − ctP )2 − (xR − xP )2 where the sign is chosen so that ctR − ctP has the same sign as ctR − ctP . This proper time is the smallest time interval between the two events that can be reached in any reference frame. 7 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 8 1. (Taylor and Wheeler problem 27) The clock paradox, version 1. On their twentyﬁrst birthday, Peter leaves his twin Paul behind on the earth and goes oﬀ in the x direction for seven years of his time at 24/25 the speed of light, then reverses direction and in another seven years of his time returns at the same speed. [In this most elementary version of the problem, we assume that the necessary periods of acceleration are inﬁnitesimal in duration, requiring Peter’s acceleration to be inﬁnite. Nonetheless, our plucky twin remains uninjured.] (a.) Make a spacetime diagram (ct vs. x) showing Peter’s motion. Indicate on it the x and ct coordinates of the turnaround point and the point of reunion. For simplicity idealize the earth as an inertial frame, adopt this inertial frame in the construction of the diagram, and take the origin to be the event of departure. (b.) How old is Paul at the moment of reunion? 2. Prove that tanh (η1 + η2 ) = tanh η1 + tanh η2 . 1 + tanh η1 tanh η2 Using the following masses in AMU, (proton) p (neutron) n (deuteron) H (helium 3) He (triton) H (alpha particle) 4 He
3 3 2 1.007825 1.008665 2.014102 3.016030 3.016050 4.002603 , calculate (to 5%) the kinetic energy released when one liter of heavy water (2 H)2 O undergoes deuteriumtritium fusion in an Hbomb. Express your answer in terms of tons of TNT (1 ton of TNT = 4.2×109 J of explosive energy). 4. The universe is ﬁlled with old cold photons that are remnants of the big bang. Typically their energy is ≈ 6.6 × 10−4 eV. A cosmonaut who is accelerated at 1 g for 10 years in her own rest frame attains a boost (= arctanh β ) of 10.34. As seen by her, what is the typical energy of these photons? 5. Prove that an isolated photon (zero mass) cannot split into two photons which do not both continue in the original direction. 6. The now retired Bevatron at Berkeley Lab is famous for having produced the ﬁrst observed antiprotons (you may have glimpsed whitemaned Nobelist Owen Chamberlain, one of the ﬁrst observers, being helped to his seat at Physics Department colloquia). An economical reaction for producing antiprotons is p+p→p+p+p+p, ¯ where the ﬁrst proton is part of a beam, the second is at rest in a target, and p is an antipro¯ ton. Because of the CP T theorem, both p and p must have the same mass (= 0.94 × 109 eV). ¯ At threshold, all four ﬁnal state particles have essentially zero velocity with respect to each other. What is the beam energy in that case? (The actual Bevatron beam energy was 6 × 109 eV). Using this relation, deduce Einstein’s law for the addition of velocities. 3. The thermonuclear “deuteriumtritium” reactions are:
2 2 2 H + 2 H → 3 He + n H + 2H → 3H + p H + 3 H → 4 He + n . These sum to 5(2 H) → 3 He + 4 He + p + 2n. 7. Using Eqs. 1.33 in the lecture notes, prove that E 2 − B 2 , where E (B ) is the magnitude of the electric (magnetic) ﬁeld, is a Lorentz invariant. 8. You shine a onewatt beam of photons on a crow, who absorbs them. Calculate the force (in N) on the crow. 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 8 1. (Taylor and Wheeler problem 27) The clock paradox, version 1. On their twentyﬁrst birthday, Peter leaves his twin Paul behind on the earth and goes oﬀ in the x direction for seven years of his time at 24/25 the speed of light, then reverses direction and in another seven years of his time returns at the same speed. [In this most elementary version of the problem, we assume that the necessary periods of acceleration are inﬁnitesimal in duration, requiring Peter’s acceleration to be inﬁnite. Nonetheless, our plucky twin remains uninjured.] (a.) Make a spacetime diagram (ct vs. x) showing Peter’s motion. Indicate on it the x and ct coordinates of the turnaround point and the point of reunion. For simplicity idealize the earth as an inertial frame, adopt this inertial frame in the construction of the diagram, and take the origin to be the event of departure. (b.) How old is Paul at the moment of reunion? Solution: On a spacetime (ct vs. x) diagram in Paul’s (unprimed) frame, Peter begins at (0,0) and proceeds with slope β −1 = 25 for a time interval 24 c∆t = γc∆t + γβ (∆x = 0) = γc∆t = 1
2 2. Prove that tanh (η1 + η2 ) = tanh η1 + tanh η2 . 1 + tanh η1 tanh η2 Using this relation, deduce Einstein’s law for the addition of velocities. Solution: tanh η = exp (2η ) − 1 = exp (2η ) + 1 tanh η1 + tanh η2 = exp (2η1 ) − 1 exp (2η2 ) − 1 = + exp (2η1 ) + 1 exp (2η2 ) + 1 2 exp (2η1 + 2η2 ) − 2 = exp (2η1 + 2η2 ) + 1 + exp (2η1 ) + exp (2η2 ) tanh η1 tanh η2 = exp (2η1 + 2η2 ) + 1 − exp (2η1 ) − exp (2η2 ) exp (2η1 + 2η2 ) + 1 + exp (2η1 ) + exp (2η2 ) 1 + tanh η1 tanh η2 = 2 exp (2η1 + 2η2 ) + 2 = exp (2η1 + 2η2 ) + 1 + exp (2η1 ) + exp (2η2 ) tanh η1 + tanh η2 2 exp (2η1 + 2η2 ) − 2 = 1 + tanh η1 tanh η2 2 exp (2η1 + 2η2 ) + 2 = tanh (η1 + η2 ) . = Suppose that all velocities are in the x direction. Take the velocity of frame S1 with respect to frame S to be β1 c; of frame S2 with respect to frame S1 to be β2 c; and of frame S2 with respect to frame S to be β3 c. β1,2,3 correspond to boost parameters (or rapidities) η1,2,3 according to the relation β1,2,3 = tanh η1,2,3 . The boost parameters have the unique property that they are additive, i.e. a boost of η1 followed by a boost of η2 is equivalent to a boost 1 − 24 25 25 c∆ t = 7 = 25 lightyr . c∆ t At Peter’s point of maximum excursion, (ct = 25, x = 24) lightyr. Peter then returns with slope β −1 = − 25 , reaching x = 0 at ct = 50 24 lightyr where he reunites with Paul. Peter has aged only 14 years, while Paul has aged 50 years (and has reached the age of 71). 2 of η1 + η2 . So, with the above deﬁnitions, η3 = η1 + η2 β3 = tanh η3 = tanh (η1 + η2 ) tanh η1 + tanh η2 = 1 + tanh η1 tanh η2 β1 + β2 = . 1 + β1 β2 This is Einstein’s law for the addition of velocities. 3. The thermonuclear “deuteriumtritium” reactions are:
2 2 2 contains 55.5 × NAvo = 55.5 × 6.023 × 1023 = 3.35 × 1025 molecules of heavy water. Each of the summed reactions requires ﬁve deuterium nuclei, or 2.5 molecules of heavy water, so 3.35 × 1025 /2.5 = 1.34 × 1025 summed reactions take place. The mass energy in one amu is equivalent to mc2 = 0.9315 × 109 eV, or, with 1 eV = 1.6×10−19 J, 1.49×10−10 J. Therefore the energy released is 0.026722 × 1.34 × 1025 × 1.49 × 10−10 = 5.33 × 1013 J. This is equivalent to the energy released by the explosion of 12.7 kilotons of TNT. A corollary is that about 80 liters – one Jeep gasoline tank – of (2 H)2 O are needed to make a one megaton Hbomb. This sets a lower limit on the degree to which an Hbomb can be miniaturized, irrespective of any espionage. 4. The universe is ﬁlled with old cold photons that are remnants of the big bang. Typically their energy is ≈ 6.6 × 10−4 eV. A cosmonaut who is accelerated at 1 g for 10 years in her own rest frame attains a boost (= arctanh β ) of 10.34. As seen by her, what is the typical energy of these photons? Solution: We know that the energymomentum fourvector (E/c, p) satisﬁes the same Lorentz transformation equations as the spacetime fourvector (ct, r): E /c = γE/c − γβpx px = −γβE/c + γpx py = py pz = pz , where the cosmonaut (in the primed frame) is assumed to be travelling with respect to the big bang’s (unprimed) frame with a velocity βc = tanh 10.34 in the x direction. [With respect to this large velocity, here we are neglecting the much smaller speed of 370 km/sec with which the solar system moves with respect to the big bang radiation; this was ﬁrst measured by a Berkeley group, including Profs. Smoot and Muller, in the 1970s.] H + 2 H → 3 He + n H + 2H → 3H + p H + 3 H → 4 He + n . These sum to 5(2 H) → 3 He + 4 He + p + 2n. Using the following masses in AMU, (proton) p (neutron) n (deuteron) 2 H (helium 3) 3 He (triton) H (alpha particle) He
4 3 1.007825 1.008665 2.014102 3.016030 3.016050 4.002603 , calculate (to 5%) the kinetic energy released when one liter of heavy water (2 H)2 O undergoes deuteriumtritium fusion in an Hbomb. Express your answer in terms of tons of TNT (1 ton of TNT = 4.2×109 J of explosive energy). Solution: The energy released in the summed reaction corresponds to a mass deﬁcit equal to ∆m = 5(2.014102) − 3.016030 − 4.002603 − 1.007825 − 2(1.008665) = 0.026722 amu. Heavy water has a density about 20 times that of ordinary 18 water, due to the extra two neutrons. Thus one liter of heavy water weighs 1.11 kg and .11 corresponds to 120 × 103 = 55.5 moles. It 3 On average, px = 0 for the big bang photons in the big bang’s frame; thus E /c = γ E /c = = 1 E /c 1 − β2 1 E /c 1 − tanh2 η where m is the particle’s rest mass. This is the fundamental equation for solving relativistic kinematics problems. The fundamental equation tells us that p · p = 0 and p = E/c for any massless particle like the photon. Returning to the problem, p · p = pa · pa + pb · pb + 2pa · pb 0 = 0 + 0 + 2Ea Eb /c2 − 2pa · pb = 2Ea Eb /c2 (1 − cos θab ) 1 = cos θab , where θab is the opening angle between the two photons. The last equation tells us that photons a and b must be travelling in the same direction (they are “collinear”); by conservation of momentum, that must be the direction of the initial photon. [For the case in which photons a and b do travel in the direction of the initial photon, which is allowed by the above kinematic calculation, the decay nevertheless is prevented by conservation of angular momentum. Angular momentum nonconservation in the collinear decay arises from the photon’s internal angular momentum (“spin”).] [Also, we note that the (electrically neutral) photon couples to electric charge, so, to lowest order, no electromagnetic interaction occurs when three photons meet at a common vertex. This is not the case for the strong force carriers (gluons), which also are massless; gluons both carry and couple to a diﬀerent kind of charge called “color”.] [How could we express the above solution in words? “If the two decay photons are not collinear, their combined invariant mass must be greater than zero. Since the initial state has invariant mass equal to zero, this violates energymomentum conservation.”] 6. The now retired Bevatron at Berkeley Lab is famous for having produced the ﬁrst observed antiprotons (you may have glimpsed whitemaned Nobelist Owen Chamberlain, one = E /c cosh η = 6.6 × 10−4 eV/c × cosh 10.34 E = 10.2 eV . Therefore, while the cosmic background radiation is in the far infrared as seen in the solar system, on average it is boosted to the ultraviolet as seen in the frame of the cosmonaut. 5. Prove that an isolated photon (zero mass) cannot split into two photons which do not both continue in the original direction. Solution: Assume that a photon decays into two other photons a and b. The photons have energymomentum fourvectors denoted by (E/c, p), (Ea /c, pa ), and (Eb /c, pb ), repectively. Both energy and momentum must be conserved in the decay. We can express this requirement in a single fourcomponent equation: (E/c, p) = (Ea /c, pa ) + (Eb /c, p)b ) . To save writing we will use the shorthand notation p ≡ (E/c, p); similarly for pa and pb . Rewriting the above equation in this shorthand notation, and taking the inner product of each side with itself, p = pa + p b p · p = (pa + pb ) · (pa + pb ) = pa · pa + pb · pb + 2pa · pb . In the above, the symbol “·” refers to the fourvector inner product, i.e. p · p ≡ E 2 /c2 − p · p . Since the inner product of any two fourvectors has the same value in any Lorentz frame, it is easiest to evaluate p · p in the rest frame of the particle; there one ﬁnds that p · p = E 2 /c2 − p2 = m2 c2 , 4 of the ﬁrst observers, being helped to his seat at Physics Department colloquia). An economical reaction for producing antiprotons is p+p→p+p+p+p, ¯ where the ﬁrst proton is part of a beam, the second is at rest in a target, and p is an antipro¯ ton. Because of the CP T theorem, both p and p must have the same mass (= 0.94 × 109 eV). ¯ At threshold, all four ﬁnal state particles have essentially zero velocity with respect to each other. What is the beam energy in that case? (The actual Bevatron beam energy was 6 × 109 eV). Solution: We shall use the notation of the previous solution. Denote by pa and pb the fourmomenta of the incident and target protons, each of which has mass m. At threshold, we are told that the four ﬁnalstate particles are at rest with respect to each other. Therefore, for kinematic purposes, they are equivalent to a single particle of mass 4m. Denote by pc the fourmomentum of this fourparticle state. Energymomentum conservation demands pa + pb = pc (pa + pb ) · (pa + pb ) = pc · pc pa · pa + pb · pb + 2pa · pb = pc · pc m c + m c + 2pa · pb = (4m) c
22 22 22 Heisenberg’s uncertainty principle. This is called “Fermi momentum”. When the target proton’s Fermi momentum is directed against the incoming beam proton, the energy available for the interaction can be augmented up to ≈ 20%. 7. Using Eqs. 1.33 in the lecture notes, prove that E 2 − B 2 , where E (B ) is the magnitude of the electric (magnetic) ﬁeld, is a Lorentz invariant. Solution: The equations for Lorentz transformation of the electric ﬁeld E and magnetic ﬁeld B may be derived from three facts: (φ, A) = a four vector 1 ∂A E = −∇φ − c ∂t B=∇×A, where φ is the scalar potential and A is the vector potential. The result of the derivation is Eq. 1.33 in the distributed relativity notes: E⊥ = γ (E⊥ + β × B⊥ ) B⊥ = γ (B⊥ − β × E⊥ ) E =E B =B , where β c is the velocity of frame S relative to S , the subscript ⊥ refers to the component perpendicular to β , and the subscript refers to the component parallel to β . Note that, in the ﬁrst two equations, the subscript ⊥ may be dropped from the last term, since taking the cross product with β automatically picks out the perpendicular part. Using the ﬁrst two equations,
2 2 γ −2 (E⊥ )2 = E⊥ + β 2 B⊥ + 2E⊥ · (β × B⊥ ) 2 2 γ −2 (B⊥ )2 = B⊥ + β 2 E⊥ − 2B⊥ · (β × E⊥ ) . 7m2 c2 = pa · pb = (Ea , pa ) · (m, 0) = Ea m Ea = 7mc2 = 7 × 0.94 × 109 eV = 6.58 × 109 eV . This is ∼ 10% more proton beam energy than the Bevatron (= 6 × 109 GeV) was able to supply! How then were Chamberlain, Segr`, Wiee gand, and Ypsilantis able to discover the antiproton at the Berkeley Bevatron in 1956? They took advantage of the fact that protons conﬁned inside the atomic nucleus have a signiﬁcant (∼ 200 MeV/c) rms momentum as a result of We rearrange the last line using the invariance under cyclic permutation of the triple product: a · (b × c) = b · (c × a) = c · (a × b) , a relation which may be found on the inside cover of Griﬃths (distributed in class), or, more 5 physically, may be understood from the fact that the triple product describes the (invariant) volume of a parallelopiped with sides a, b, and c. Cyclically permuting the triple product in the last line,
2 2 γ −2 (B⊥ )2 = B⊥ + β 2 E⊥ − 2E⊥ · (B⊥ × β ) 2 2 = B⊥ + β 2 E⊥ + 2E⊥ · (β × B⊥ ) 2 2 E⊥ − B⊥ 2 2 2 2 = E⊥ + β 2 B⊥ − B⊥ − β 2 E⊥ γ2 2 2 = (1 − β 2 )(E⊥ − B⊥ ) 2 2 2 2 E⊥ − B⊥ = E⊥ − B⊥ . 2 2 This demonstrates that E⊥ − B⊥ is conserved; E 2 − B 2 is conserved automatically since E and B are invariant under the transformation. 2 Finally, E 2 = E⊥ + E 2 , etc., because the dot product in the cross term vanishes. 8. You shine a onewatt beam of photons on a crow, who absorbs them. Calculate the force (in N) on the crow. Solution: Suppose that a photon in the ﬂashlight beam has an energy E . Then it must have momentum p = E/c. The beam power (= 1 W) is P = N E , where N is the number of photons emitted per second and E is their average energy. If the photons are totally absorbed by the crow, the momentum absorbed by the crow per second is F = N p = N E /c = P/c . Therefore the force F on the crow is F= 1W = 3.3 × 10−9 N . c University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 9 1. Purcell problem 5.3 2. Purcell problem 5.7 3. Purcell problem 5.10 4. Purcell problem 5.17 5. Purcell problem 6.4 6. Purcell problem 6.14 7. Purcell problem 6.18 8. Purcell problem 6.22 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 9 Solutions by P. Pebler 1 Purcell 5.3 A beam of 9.5 M eV electrons (γ = 20) amounting as current to 0.05 µA, is traveling through vacuum. The transverse dimensions of the beam are less than 1 mm, and there are no positive charges in or near it. In the lab frame, what is approximately the electric ﬁeld strength 1 cm away from the beam, and what is the average distance between the electrons, measured parallel to the beam? Answer the same questions for the electron rest frame. With γ = 20, the speed of the electrons will be essentially c. The number of electrons passing per second is νe = 0.05 × 10−6 A = 3.1 × 1011 1/s . 1.6 × 10−19 C The mean distance between electrons is c d= = 1 mm , νe and the charge per unit length is 4.8 × 10−10 esu = −4.8 × 10−9 esu/cm . 0.1 cm If we think of this as a continuous line charge, the ﬁeld strength a distance 1 cm away is λ=− 2λ = 9.6 × 10−9 statvolt/cm . r Since the electrons are moving, the distance between them will be contracted. Therefore, in the electron rest frame, they will be more spread out so that E=− d = γd = 2 cm . We might ﬁnd a new ﬁeld strength by E= E = 4.8 × 10−10 statvolt/cm , γ however, this will be only the average ﬁeld strength along a line parallel to the beam. Since the electrons are so far apart in this frame, there will be big variation in the ﬁeld. 2 Purcell 5.7 A moving proton has γ = 1010 . How far away from such a proton would the ﬁeld rise to 1 V /m as it passes? The electric ﬁeld strength of a moving point charge is (in SI units) E= 1 − β2 . 4π o r 2 (1 − β 2 sin2 θ )3/2 Q Q 4π o r
2 The maximum ﬁeld strength is directly beside the particle where θ = π/2. Then E= Q 1 =γ . 2 4π o r 2 1−β 1 We want the distance where the ﬁeld is 1 V /m. 1010 (1.6 × 10−19 C ) = 1 V /m 4π (8.85 × 10−12 C 2 /N m2 )r 2 r = 3.8 m 3 Purcell 5.10 In the rest frame of a particle with charge q1 another particle with charge q2 is approaching, moving with velocity v not small compared with c. If it continues to move in a straight line, it will pass a distance d from the position of the ﬁrst particle. It is so massive that its displacement from the straight path during the encounter is small compared with d. Likewise, the ﬁrst particle is so massive that its displacement from its initial position is small compared with d. Show that the increment in momentum acquired by each particle as a result of the encounter is perpendicular to v and has magnitude 2q1 q2 /vd. Expressed in terms of other quantities, how large must the masses of the particles be to justify our assumptions? In the ﬁrst approximation, we consider the trajectory to be a straight line. For this trajectory, the net impulse in the x direction will be zero. Using cylindrical coordinates, ∆ pr = Fr dt = 1 v Fr dx = q2 2πdv Er 2πd dx . The purpose of transforming this integral in this way is so that it becomes the electric ﬂux through an inﬁnite cylinder of radius d. We can evaluate it easily by using Gauss’s law. ∆ pr = q2 2πdv E · da = q2 2q 1 q 2 4πq1 = 2πdv dv In the rest frame of q2 , the situation is reversed, and q1 acquires the same y momentum in the opposite direction. This will be the same momentum in the rest frame of q1 because the perpendicular component of the momentum is unchanged by a Lorentz transformation. For this approximation to be good, the acquired y momentum must be much smaller than the x momentum, so that γmv 2q1 q2 , dv where m can represent either mass. 4 Purcell 5.17 Two protons are moving parallel to one another a distance r apart, with the same velocity βc in the lab frame. At the instantaneous position of one of the protons the electric ﬁeld strength caused by the other is γe/r2 . But the force on the proton measured in the lab frame is not γe2 /r2 . Verify that by ﬁnding the force in the proton rest frame and transforming that force back to the lab frame. Show that the discrepancy can be accounted for if there is a magnetic ﬁeld β times as strong as the electric ﬁeld, accompanying this proton as it travels through the lab frame. 2 In the rest frame of the protons, the force is e2 /r2 . In the lab frame this force is Fy = 1 e2 e2 ( = γ 2 + Fy b) , γ r2 r e2 r2 e v 1 e2 − 1 = −γβ 2 2 = − β γ 2 e . 2 γ r r c where the ﬁrst term is the electric force in the lab frame. The extra term is
( Fy b) = γ From the Lorentz force law e F = eE + v × B , c we can account for this extra force with a magnetic ﬁeld out of the page ˆ B = βE z . 5 Purcell 6.4 A long wire is bent into the hairpin like shape shown. ﬁnd an exact expression for the magnetic ﬁeld at the point P which lies at the center of the halfcircle. In the integral for the magnetic ﬁeld of an inﬁnite wire, each half of the wire contributes the same amount in the same direction, so for each of the half wires, we can take half of the formula for an inﬁnite wire. Also, the half circle will contribute half of the ring formula. All contributions point out of the page so Bz = 1 2I 1 2I (2 + π )I 1 2πI + + = . 2 cb 2 cb 2 cb bc 6 Purcell 6.14 A coil is wound evenly on a torus of rectangular cross section. There are N turns of wire in all. Assume that the current on the surface of the torus ﬂows exactly radially on the annular end faces, and exactly longitudinally on the inner and outer cylindrical surfaces. Show that the magnetic ﬁeld everywhere would be circumferential. Second, prove that the ﬁeld is zero at all points outside the torus, including the interior of the central hole. Third, ﬁnd the magnitude of the ﬁeld inside the torus as a function of radius. We set a coordinate system so the point P under consideration is in the z − x plane with coordinates (xo , 0, zo ). Consider a current loop at an angle φ from the x axis. One small section of the loop has coordinates (r cos φ, r sin φ, z ) and r = (xo − r cos φ, −r sin φ, zo − z ) . 3 The direction of this little current is arbitrary within the plane so the current direction vector can be written ˆ I = Ir ˆ + Iz z = (Ir cos φ, Ir sin φ, Iz ) . r Then the contribution to the magnetic ﬁeld will be in the direction ˆ ˆ ˆ I × r = [sin φ(Ir (zo − z ) + rIz )] x + [Iz (xo − r cos φ) − Ir cos φ(zo − z )] y − Ir xo sin φ z . However, for each section of current, there is a similar section that is identical except that φ → −φ. We see that the x and z components change sign. Therefore, these components cancel out and the net ﬁeld at the point P is in the y direction. For our coordinate system, this in the circumferential direction. Once we have this information, it is easy to use Ampere’s law. B · dl = 2πrB = B= 2N I cr 4π NI c a<r<b If we are outside the solenoid, the enclosed current will be zero and so the ﬁeld is zero. 7 Purcell 6.18 Two long coaxial aluminum cylinders are charged to a potential diﬀerence of 50 statvolts. The inner cylinder has an outer diameter of 6 cm, the outer cylinder an inner diameter of 8 cm. With the outer cylinder stationary the inner cylinder is rotated around its axis at a constant frequency of 30 Hz . Describe the magnetic ﬁeld this produces and determine its intensity in gauss. What if both cylinders are rotated in the same direction at 30 Hz ? The capacitance for two coaxial cylinders is C= L . 2 ln(b/a) Assuming we have equal amounts of positive and negative charge on the inside and outside respectively, we can ﬁnd the charge per unit length. CV V Q = = = 87 esu/cm L L 2 ln(b/a) The spinning charged cylinders are essentially perfect solenoids. The ﬁeld of a long solenoid is constant inside and practically zero outside. The solenoid formula is B= 4πIn , c where n is the number of turns per unit length. The quantity nI can be thought of as the charge per unit time per unit length passing through a line running parallel to the cylinder. You should be able to convince yourself that this quantity in our case is Qν/L, where ν is the frequency of revolution. With just the inner cylinder rotating with ν = 30 Hz , B= 4πQν 4π (87 esu/cm)(30 Hz ) = = 1.1 × 10−6 gauss cL 3 × 1010 cm/s r<a , and the ﬁeld elsewhere is zero. With a clockwise rotation as shown, the ﬁeld is into the page. 4 If the outer cylinder rotates in the same direction, it will produce a ﬁeld with the same magnitude but opposite direction. These two ﬁelds will cancel for r < a, but for a < r < b, the ﬁeld is 1.1 × 10−6 gauss out of the page. It is zero again for r > b. 8 Purcell 6.22 A constant B ﬁeld lies in the y − z plane. An arbitrary current loop lies in the x − y plane. Show, by calculating the torque about the x axis, that the torque on the current loop can be written N = m × B, where the magnetic moment m of the loop is deﬁned as a vector of magnitude IA/c where I is the current in esu/s and a is the area of the loop in cm2 , and the direction of the vector is normal to the loop with a righthand relation to the current. What about the net force on the loop? The force on a small section of loop is 1 dF = Idl × B . c Only the part of the force in the z direction will contribute to the torque about the x axis. We ˆ split the magnetic ﬁeld into two parts in the z and y directions. dl × (Bz z) is in the x − y plane, so we only need the y part of B. 1 dFz = IBy sin θ dl c From the ﬁgure we see that dl sin θ = dx so that dFz = 1 IBy dx and the torque is c Nx = 1 y dFz = IBy c
loop y (x) dx . This integral if we go forward along the top of the loop and back along the bottom is the area of the loop and 1 Nx = IBy a . c With m = Ia/c and pointing in the −z direction, 1 ˆ ˆ ˆ m × B = m × (By y) = mBy x = IBy a x . c For a constant ﬁeld there is no net force because F= and dF = 1 1 I dl × B = − I B × c c dl dl around the loop is zero. 5 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 10 1. Purcell problem 6.26 2. Purcell problem 6.28 3. Purcell problem 6.32 4. Purcell problem 7.4 5. Purcell problem 7.9 6. Purcell problem 7.11 7. Purcell problem 7.14 8. Purcell problem 7.16 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 10 Solutions by P. Pebler 1 Purcell 6.26 A round wire of radius ro carries a current I distributed uniformly over the cross ˆ section of the wire. Let the axis of the wire be the z axis, with z the direction of the current. Show 2 + y 2 )ˆ will correctly give the magnetic ﬁeld B of this that a vector potential of the form A = Ao (x z current at all points inside the wire. What is the value of the constant Ao ? The magnetic ﬁeld is the curl of the vector potential. ˆ ˆ B = ∇ × A = 2Ao y x − 2Ao x y If we use plane polar coordinates in the x − y plane, ˆ ˆ ˆ B = 2Ao r(sin φ x − cos φ y) = −2Ao r φ . We know that the magnetic ﬁeld circles in the counterclockwise direction for a current coming out of the page. We can ﬁnd the magnitude from Ampere’s law. 2πrB = B= 4π r2 I2 c ro 2I r 2 cro I . 2 cro The vector potential A therefore works with the constant Ao = − 2 Purcell 6.28 A proton with kinetic energy 1016 eV (γ = 107 ) is moving perpendicular to the interstellar magnetic ﬁeld which in that region of the galaxy has a strength 3 × 10−6 gauss. What is the radius of curvature of its path and how long does it take to complete one revolution? Magnetic forces do no work. They can only change the direction of the momentum. Because the force is perpendicular to the velocity, we can instantaneously think about the motion as being along a circle of some radius R. Because the ﬁeld and velocity are perpendicular, the magnitude of the force is evB F= . c If this were a nonrelativistic problem, we could ﬁnd the radius R by equating the force with mv 2 /R. In the relativistic case, this formula turns out to be correct with the replacement m → γm, but 1 this is something that must be proved. If we wait a time ∆t, the momentum will swing through some angle ∆θ, and ∆p = p∆θ. (Please note that ∆p is not the same thing as ∆p.) This angle ∆θ will also be the angle of the circle we go through in this time. Therefore, in the inﬁnitesimal limit ∆ → d, v = ωr. Consequently, pv dθ dp = pω = . =p dt dt R Equating this with the force, R= pc eB γ mc2 107 (1.5 × 10−3 ergs) = = 1 × 1019 cm . eB (4.8 × 10−10 esu)(3 × 10−6 gauss) The period is τ= 2πR 2π = ω v 2πR = 2.1 × 109 s . c 3 Purcell 6.32 Two electrons move along parallel paths, side by side, with the same speed v . The paths are a distance r apart. Find the force acting on one of them in two ways. First, ﬁnd the force in the rest frame of the electrons and tranform this force back to the lab frame. Second, calculate the force from the ﬁelds in the lab frame. What can be said about the force between them in the limit v → c? In the particle rest frame, the ﬁeld is just the Coulomb ﬁeld and the force magnitude is e2 /r2 . If we use the transformation formulas (14) in Purcell, the primed frame must be the particle rest frame. F= 1 e2 ˆ y γ r2 To ﬁnd the ﬁelds in the lab frame, it is easiest to transform them back from the rest frame where e ˆ B =0 . E =− 2y r Please note that most transformation formulas found in books assume that the primed frame is moving in the positive x direction of the unprimed frame. If you wish to use these formulas verbatim, you must choose your frames correctly. Here the particles are going to the right so the rest frame is the primed frame. To switch back to the lab frame, we need the inverse of the equations (6.60) in Purcell. We can accomplish this by simply switching the primes and the sign of β . Then E =E =0 , E⊥ = γ (E⊥ − β × B⊥ ) = γ E⊥ = −γ B =B =0 , 2 e ˆ y, r2 B⊥ = γ (B⊥ + β × E⊥ ) = −γβ The force is then F = (−e)E + e ˆ z. r2 e2 e2 e2 1 e2 (−e) ˆ ˆ ˆ ˆ y. v × B = γ 2 y − γβ 2 2 y = γ 2 (1 − β 2 ) y = c r r r γ r2 In the limit v → c, we see that F → 0. 4 Purcell 7.4 Calculate the electromotive force in the moving loop in the ﬁgure at the instant when it is in the position there shown. Assume the resistance of the loop is so great that the eﬀect of the current in the loop itself is negligible. Estimate very roughly how large a resistance would be safe, in this respect. Indicate the direction in which current would ﬂow in the loop, at the instant shown. We ﬁrst calculate the ﬂux. We will deﬁne the positive direction to be into the page. The ﬁeld is that of a wire. The current is given in SI, so we must use SI formulas. B · da = E =−
x+L x µo I µo Iw x + L w dr = ln 2πr 2π x (x + L)v v − x x2 = Lv µo Iw 2π x(x + L) µo Iw x d ΦB = − dt 2π x + L E = 2.1 × 10−5 V By choosing into the page as positive for ﬂux, we have also deﬁned clockwise as the positive way to go around the loop. Since E is positive, the induced current will be clockwise. 5 Purcell 7.9 Derive an approximate formula for the mutual inductance of two circular rings of the same radius a, arranged like wheels on the same axle with their centers a distance b apart. Use an approximation good for b a. From Purcell Eq. 6.41 (where a and b are interchanged relative to this problem) the ﬁeld along the axis of a ring is Bz = 2πa2 I . c(a2 + z 2 )3/2 a, and approximate the z component of the ﬁeld everywhere in We may use the information b the second loop as Bz = 2πa2 I c(a2 + b2 )3/2 2πa2 I , cb3 3 so the ﬂux is ΦB = 2πa2 I 2 2π 2 a4 πa = I. cb3 cb3 2π 2 a4 dI 1d ΦB = − 2 3 , c dt c b dt The induced emf is then E =− and the mutual induction is M= 2π 2 a4 , c2 b3 µo 2π 2 a4 . 4π b3 in cgs units. If you use SI formulas this becomes M= 6 Purcell 7.11 Two coils with selfinductances L1 and L2 and mutual inductance M are shown with the positive direction for current and electromotive force indicated. The equations relating currents and emf ’s are E1 = −L1 dI2 dI1 ±M dt dt E2 = −L2 dI1 dI2 ±M . dt dt Given that M is always to be taken as positive, how must the signs be chosen in these equations? What if we had chosen the other direction for positive current and emf in the lower coil? Now connect the two coils together as in b. What is the inductance L of this circuit? What is the inductance L of the circuit formed as shown in c? Which circuit has the greater selfinductance? Considering that the selfinductance of any circuit must be a positive quantity, see if you can deduce anything concerning the relative magnitudes of L1 , L2 , and M . Imagine ﬁrst that the current I2 is positive and increasing so that dI2 /dt > 0. In this case the magnetic ﬁeld due to coil 2 will point up through coil 1. As the current I2 increases, the ﬁeld it creates will increase and the ﬂux up through coil 1 will increase. By using Lenz’s law, we ﬁnd we need an induced current that will create a magnetic ﬁeld that will oppose this change in the ﬂux. In this case, the ﬁeld should point down through coil 1. To do this the induced current must ﬂow in the negative direction as it is deﬁned for coil 1. Thus, the induced emf must be negative and we need the negative sign. The same argument will tell you to choose the negative sign in the second equation also. (You should go through it yourself however.) 4 If the sign convention for coil 2 had been switched, the same argument would switch the sign in both equations. (Do it yourself though.) With the circuit in b, since both emf positive directions point in the same way, the total emf across the new circuit is E = E1 + E2 = −L1 dI2 dI1 dI1 dI2 −M − L2 −M . dt dt dt dt dI , dt We also have I = I1 = I2 so that E = −(L1 + L2 + 2M ) and the self inductance is L = L1 + L2 + 2M . With the circuit in c, the sign conventions “conﬂict” so that E = E1 − E2 = −L1 dI2 dI1 dI1 dI2 −M + L2 +M , dt dt dt dt dI , dt but with I = I1 = −I2 so that E = −(L1 + L2 − 2M ) and the self inductance is L = L1 + L2 − 2M . If the self inductance of a coil were negative, the circuit would be unstable – any change in current would result in more current the same direction which would built indeﬁnitely. Therefore we must have L > 0 and M≤ L1 + L2 . 2 5 7 Purcell 7.14 A metal crossbar of mass m slides without friction on two long parallel conducting rails a distance b apart. A resistor R is connected across the rails at one end; compared with R, the resistance of bar and rails is negligible. There is a uniform ﬁeld B perpendicular to the plane of the ﬁgure. At time t = 0 the crossbar is given a velocity vo toward the right. What happens then? Does the rod ever stop moving? If so, when? How far does it go? How about conservation of energy? Let us assume the magnetic ﬁeld is into the page, and let’s make that positive so that clockwise is positive for the loop. The ﬂux is then ΦB = bxB . The emf (in SI) is E =− d ΦB = −bvB = IR , dt so the current is counterclockwise. The bar will feel a force due to the magnetic charges moving through it. The force is ˆ F = I L × B = −I bB x . We can solve for the motion using F = ma. m b2 B 2 dv = −I bB = − v dt R
2 B 2 t/mR v (t) = vo e−b The bar never stops moving. It will approach the distance
∞ d= 0 vo e−b 2 B 2 t/mR dt = mRvo . b2 B 2 The lost kinetic energy is dissipated by the resistor. I= U= b Rv B
∞ o I 2 Rdt = ∞ o b2 B 2 2 −2b2 B 2 t/mR b2 B 2 2 mR 1 2 vo e vo 2 2 = mvo = R R 2b B 2 6 8 Purcell 7.16 The shaded region represents the pole of an electromagnet where there is a strong magnetic ﬁeld perpendicular to the plane of the paper. The rectangular frame is made of 5 mm diameter aluminum. suppose that a steady force of 1 N can pull the frame out in 1 s. If the force is doubled, how long does it take? If the frame is made of 5 mm brass, with about twice the resistivity, what force is needed to pull it out in 1 s? If the frame were 1 cm diameter aluminum, what force is needed to pull it out in 1 s? Neglect inertia of the frame and assume it moves with constant velocity. If we assume a constant velocity, the force necessary to pull out the loop will be equal in magnitude to the magnetic force on the loop. We will ignore signs here. The net force will be on the left wire of the frame. F = I hB The magnetic ﬂux will be something like Φ = LxB , where x is the length of loop in the ﬁeld. Then the emf is E = − d Φ = hvB . dt The current is I  = E/R and the resistance is R = ρL/A = ρL/πr2 , where L is the total length of the loop and r is the radius of the wire of which it is made. F = hBhBv vr2 A ∝ ρL ρ If the force is doubled, the speed doubles and it takes half the time or 0.5 s. If the resistivity doubles with the same speed, the force is halved so that F = 1 N . If the radius doubles with the same speed, the force is four times as great or 4 N . 7 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 11 1. Purcell problem 7.21. Assume that b2 and b2 b1 . a2 2. Purcell problem 7.22. Assume that the charge is distributed uniformly around the ring. 3. Purcell problem 7.23 4. Purcell problem 7.29 5. Purcell problem 8.5. For simplicity, if you wish, in parts (a) and (b) you may assume that the battery is connected only until just before the switch is closed. 6. Purcell problem 8.7 7. Purcell problem 8.11 8. Purcell problem 8.16. An inﬁnite LC ladder of the type studied in this problem is called a “lumpedelement delay line”. It is of great practical importance when one needs to delay an analog electrical pulse by a longer time than would be conveniently achieved by using the ﬁnite speed (≈ c) of propagation in a coaxial cable. University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 11 Solutions by P. Pebler 1 Purcell 7.21 A solenoid of radius a1 and length b1 is located inside a longer solenoid of radius a2 and length b2 . The total number of turns is N1 in the inner coil and N2 on the outer. Work out a formula for the mutual inductance M . The mutual inductances M12 and M21 are equal, so we are free to calculate the inductance with b1 , whatever coil is more convenient. We ﬁnd the ﬂux through the inner coil. If we assume b2 the ﬁeld through coil 1 will be fairly uniform and with the sign conventions shown, B2 = µo N2 I2 ˆ x, b2 along the common axis of both coils. Since there are N1 loops in this coil, the ﬂux through all of them is φ12 = πa2 N1 1 µo N2 I2 , b2 and the induced emf is E12 = − d µo πa2 N1 N2 dI2 1 Φ12 = − , dt b2 dt and the mutual inductance is M= µo πa2 N1 N2 1 . b2 2 Purcell 7.22 A thin ring of radius a carries a static charge q . This ring is in a magnetic ﬁeld of strength Bo , parallel to the ring’s axis, and is supported so that it is free to rotate about that axis. If the ﬁeld is switched oﬀ, how much angular momentum will be added to the ring? If the ring has mass m, show that it will acquire an angular velocity ω = qBo /2mc. We’ll assume the charge is uniformly distributed around the ring, with linear density λ = q/2πa. Then the torque about the z axis is τz = dτ z = aλf · dl = q 2π E · dl = q E. 2π 1 The emf is E =− so that τz = and Lzf − Lzi = − q 2πc(Φf − Φi ) = q πa2 Bo , 2πc q dΦ dLz =− , dt 2πc dt 1d Φ, c dt Lzf = qa2 Bo . 2c The moment of inertia of the ring is I = ma2 , and Lzf = Iω = ma2 ω = ω= qBo , 2mc qa2 Bo , 2c equal to half the cyclotron frequency of a particle with mass m and charge q in a ﬁeld B0 . 3 Purcell 7.23 There is evidence that a magnetic ﬁeld exists in most of the interstellar space with a strength between 10−6 and 10−5 gauss. Adopting 3 × 10−6 gauss as a typical value, ﬁnd the total energy stored in the magnetic ﬁeld of the galaxy. Assume the galaxy is a disk roughly 1023 cm in diameter and 1021 cm thick. Assuming stars radiate about 1044 ergs/s, how many years of starlight is the magnetic energy worth? The magnetic energy is U= 1 8π B 2 dV = 1 (3 × 10−6 gauss)2 (1021 cm)π (1023 /2 cm)2 = 3 × 1054 ergs , 8π and this is 3 × 1054 ergs = 3 × 1010 s = 900 yr 1044 ergs/s of starlight. 4 Purcell 7.29 Consider the arrangement shown. The force between capacitor plates is balanced against the force between parallel wires. An alternating voltage of frequency f is applied to the capacitors C1 and C2 . The charge ﬂowing through C2 constitutes the current through the rings. Suppose the timeaverage downward force on C2 exactly balances the time averaged force on the wire loop. Show that under these conditions the constant c is c = (2π )3/2 a Assume h b h
1/2 C2 f. C1 b and ignore the selfinductance of the wire loop. 2 The electric ﬁeld in capacitor C1 is 1 E = Eo cos ωt , s and the charge on it is Q = C1 V = C1 Eo cos ωt . The downward force will be the charge times the electric ﬁeld due to the bottom plate. This ﬁeld will be half of the total ﬁeld. E 2 C1 1 cos2 ωt F1 = EQ = o 2 2s Because the capacitance is C1 = a2 πa2 = , 4πs 4s and the time average of cos2 is 1/2, we may rewrite this as E 2C 2 ¯ F1 = o 2 1 . a If we have h B= 2I  , cr b, we can use the force of two long wires. The ﬁeld due to a wire is and the total force on the wire will be 1 2I 2 4πbI 2 1 =2 . F2 = I LB = 2πb c c ch ch The current is the time derivative of the charge on capacitor 2. I= dQ2 d = (C2 V ) = −C2 Eo 2πf sin ωt dt dt 4πb 2 2 C E (2πf )2 sin2 ωt c2 h 2 o 3 F2 = The time average of sin2 is also 1/2 so
2 2 8π 3 bC2 f 2 Eo C 2E 2 ¯ ¯ = F1 = 1 2 o , F2 = c2 h a c = (2π )3/2 a b h 1/2 C2 f. C1 5 Purcell 8.5 The coil in the circuit shown in the diagram is known to have an inductance of 0.01 henry . when the switch is closed, the oscilloscope sweep is triggered. Determine the capacitance C . Estimate the value of the resistance R of the coil. What is the magnitude of the voltage across the oscilloscope input a long time, say 1 second after the switch has been closed? Parts (a.) and (b.) of this problem may be approximated by assuming that the battery is disconnected when the switch is closed. However, the problem actually is not too bad with the battery connected, so we will solve the original problem. The answers are the same except for the ﬁnal voltage. If you work out the equation for the charge on the capacitor C , you will ﬁnd L d2 Q L + R2 + 2 dt R1 C dQ + dt R1 + R2 R1 V R2 Q = . C R1 If we assume that the resistance R2 of the inductor is much less that R1 , this becomes the LCR circuit equation. From the trace we see that ω= 2π · 4 = 8π × 103 Hz . 10−3 s 1 = 1.6 × 10−7 F . Lω 2 For low damping the capacitance is approximately C= Also from the trace, the amplitude falls oﬀ by a factor of e in about 0.5 × 10−3 s. e−Rt/2L = e−1 R= 2L = 40 ohms t 4 If we wait one second, a long time, things will settle so that a steady current passes through the inductor and the voltage across it will be due to the resistance. If the current is I , the voltage is V2 = IR2 and 20 V = I (105 ohm + R2 ) = V2 V = 8 mV . 6 Purcell 8.7 A resonant cavity of the form illustrated is an essential part of many microwave oscillators. It can be regarded as a simple LC circuit. The inductance is that of a toroid with one turn. Find an expression for the resonant frequency of this circuit and show by a sketch the conﬁguration of the magnetic and electric ﬁelds. 105 + 40 40 , The inner narrow circle will act as the capacitor while the outer ring is the solenoid. For a single turn toroid, the inductance is L= 2h b , ln c2 a πa2 a2 = , 4πs 4s and for the parallel plate capacitor, C= so that ω= c 1 = LC a 2s . h ln(b/a) The electric ﬁeld is concentrated in the circular gap, where its direction is vertical; the magnetic ﬁeld in the toroidal cavity is azimuthal in direction, with magnitude proportional to r−1 . 5 7 Purcell 8.11 An alternating voltage Vo cos ωt is applied to the terminals at A. The terminals at B are connected to an audio ampliﬁer of very high input impedance. Calculate the ratio V1 2 /Vo2 , where V1 is the complex voltage at terminals B . Choose values for R and C to make V1 2 /Vo2 = 0.1 for a 5000 hz signal. Show that for suﬃciently high frequencies, the signal power is reduced by a factor 1/4 for every doubling of the frequency. Since the impedance on the right is very large, the impedance of the circuit is approximately Z =R+ 1 , iωC and the magnitude is Z  = R2 + 1 ω2C 2 . This gives us the magnitude of the complex current. Io = Vo = Z  R2 Vo + 1/ω 2 C 2 The impedance of just the capacitor is ZC = 1 iωC ZC  = 1 . ωC This gives us the magnitude of the voltage across C . V1  = Io ZC  = √ Vo 2 R2 C 2 ω +1 1 V 1  2 =222 Vo2 ω R C +1 We would like an R and C such that 1 = 0.1 , [2π (5000 hz )]2 R2 C 2 + 1 which can be done with many values of R and C , for example R = 100 ohm C = 1 µF . For suﬃciently high frequencies we have P ∝ V1 2 ∝ ω −2 . A reduction of signal power by a factor 8 rather than 4 per octave can be achieved by substituting an inductor L for the resistor R. 6 8 Purcell 8.16 An impedance Zo is to be connected to the terminals on the right. For given frequency ω ﬁnd the value which Zo must have if the resulting impedance between the left terminals is Zo . The required Zo is a pure resistance Ro provided ω 2 < 2/LC . What is Zo in the special case ω = 2/LC ? We combine the impedances like resistances so that the total impedance is Z = ZL + 1
1 ZC + 1 ZL + Zo , with ZL = iωL and ZC = 1/iωC . We set this equal to Zo and simplify to obtain Zo = −ω 2 L2 + 2L/C . This will be pure real and thus a pure resistance if −ω 2 L2 + 2 ω2 < L >0 , C 2 . LC 2/LC , we have Zo = 0. In the special case ω = 7 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 12 1. Purcell problem 9.3 2. Purcell problem 9.5 3. Purcell problem 9.9 4. Purcell problem 9.10 5. Purcell problem 10.7 6. Purcell problem 10.17 7. Purcell problem 10.19 8. Purcell problem 10.21 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 12 Solutions by P. Pebler 1 Purcell 9.3 E= A free proton was at rest at the origin before the wave B= ˆ (−5 gauss) z 1 + [k (x + ct)]2 ˆ (5 statvolt/cm) y 2 1 + [k (x + ct)] came past with k = 1 cm−1 . Where would you expect to ﬁnd the proton after 1 µs? The proton mass is 1.6 × 10−24 g . To begin, we will neglect the magnetic force and see later if this is justiﬁed. In this case, the impulse due to the electric force will be in the y direction. The pulse only has an appreciable magnitude for a few nanoseconds, so we may extend the integral to inﬁnity. ∆p = Fe dt = e (5 statvolt/cm)ˆ y
∞ −∞ (5 statvolt/cm)πe dt ˆ y = 1 + (kct)2 kc ˆ ∆p = (2.5 × 10−19 g cm/s) y This corresponds to a speed of v = 1.6 × 105 cm/s . From the Lorentz force law e F = eE + v × B , c we see that because the electric and magnetic ﬁelds have the same strength, the magnetic force is smaller by FB v Fe c (5 × 10−4 )Fe , so our approximation is pretty good. The acceleration while the pulse is passing occurs for a very small time, so the position of the proton after one microsecond is essentially y = (1.6 × 105 cm/s)(1 × 10−6 s) = 0.16 cm . 2 Purcell 9.5 Ex = 0 Bx = 0 Consider the wave in free space Ey = Eo sin(kx − ωt) By = 0 Ez = 0 Bz = −Eo sin(kx − ωt) . Show that this ﬁeld can satisfy Maxwell’s equations if ω and k are related in a certain way. Suppose ω = 1010 hz and Eo = 0.05 statvolt/cm. What is the wavelength in cm? What is the energy density in ergs/cm3 , averaged over a large region? From this calculate the power density, the energy ﬂow in ergs/cm2 s. 1 We see immediately that ∇ · E = ∇ · B = 0. It’s also easy to calculate ˆ ∇ × E = kEo cos(kx + ωt) z , ˆ ∇ × B = kEo cos(kx + ωt) y , − ω 1 ∂B ˆ = Eo cos(kx + ωt) z , c ∂t c ω 1 ∂E ˆ = Eo cos(kx − ωt) y . c ∂t c The other two Maxwell’s equations will be satisﬁed if c= ω . k 1010 1/s 1 2π = = , λ 3 × 1010 cm/s 3 cm In this case, k= λ = 6π cm = 18.8 cm . The average energy density is
2 (0.05 statvolt/cm)2 Eo = = 9.95 × 10−5 erg/cm3 , 8π 8π and the average intensity (= power density) is
2 cEo = 3 × 106 erg/cm2 s . 8π 3 Purcell 9.9 The cosmic microwave background radiation apparently ﬁlls all space with an energy density of 4 × 10−13 erg/cm3 . Calculate the rms electric ﬁeld strength in statvolt/cm and in V /m. Roughly how far away from a 1 kW radio transmitter would you ﬁnd a comparable electromagnetic wave intensity? The average energy density is
2 Erms = 4 × 10−13 erg/cm3 , 4π and Erms = 2.2 × 10−6 statvolt/cm = 0.067 V /m . Assuming the transmitter projects in all directions, a distance r away, the intensity of the transmitter is
2 Erms (0.067 V /m)2 1 kW , = = 4πr2 376.73 ohms µo / o r = 2584 m . 2 4 Purcell 9.10 Find the magnetic ﬁeld at a point P midway between the plates of capacitor a distance r from the axis of symmetry. A current I is ﬂowing through the capacitor. We assume that the magnetic ﬁeld circles the capacitor axis. If the capacitor spacing is small, the electric ﬁeld will be fairly uniform and E= Q 4πsQ 4Q V = = =2, 2 s sC sπb b 4πr2 Q , b2 ΦE = πr2 E = and ignoring signs, 2πrB = B= 1 4πr2 dQ , c b2 dt 2rI . cb2 At the edge of the capacitor (r = b) this is the same as the magnetic ﬁeld around a long wire. 5 Purcell 10.7 A cell membrane typically has a capacitance around 1 µF/cm2 . It is believed the membrane consists of material having a dielectric constant of about 3. Find the thickness this implies. Other electrical measurements have indicated that the resistance of 1 cm2 of cell membrane is around 1000 ohms. Show that the time constant of such a leaky capacitor is independent of the area of the capacitor. How large is it in this case? What is the resistivity? The capacitance is given in Farads so we will use SI. The constant o appears in SI formulas. To deal with a dielectric material, we make the replacement o → . However, in SI, is not dimensionless. For example, if = 3 in cgs, the value in SI is 3 o . The capacitance of a parallel plate capacitor is C= so s = 2.66 × 10−9 m . We may view the leaky capacitor as a simple RC circuit, where the resistor and the capacitor are really the same element. The time constant is τ = RC = ρs A =ρ As , 1 cm2 A =3 = 1 × 10−6 F s s , which is independent of the area of the membrane. It is also independent of its thickness.. τ = (1000 ohms)(1 × 10−6 F ) = 1 × 10−3 s = ρ 3(8.85 × 10−12 C 2 /N m2 ) ρ = 3.8 × 107 ohm m τ = sec 3 6 Purcell 10.17 In the ﬁrst two cases, we assume the left dipole to be present and we bring in the right dipole from inﬁnity. We would like to do this in such a way that the work required is zero. This will be the case if the path of the dipole coming in is perpendicular to the force on it. We will bring in the second dipole on a straight line from the right. The force on it is F = (p · ∇)E , where E is the ﬁeld created by the other dipole. We wish to ﬁnd an orientation for the right dipole so that this force is perpendicular to the path. We can do this if the dipole is pointed to the right. In this case the force is F=p ∂E . ∂x p p ˆ ˆ z = − 3z . 3 r x The ﬁeld from the left dipole on the line of the path is E=− The force is then F=3 p2 ˆ z, x4 which is perpendicular to the path. Intuitively we can think of the dipole as two charges. The positive charge feels a force down and the negative charge feels a force up. But the positive charge is further away so the force on it is smaller. The net force is then up. But since this is perpendicular to the path, it still requires no work to bring in the dipole. The ﬁeld at the right dipole is E=− p ˆ z, d3 and the torque exerted on this is ˆ τ = p × E = pE sin θ y . The work we do in rotating the dipole is minus the work done by the ﬁeld. So
π W= π/2 pE sin θ dθ = pE = p2 . d3 This is good since the dipoles don’t like to point in the same direction in this orientation so it should take positive work to arrange it. In the second case, we rotate the dipole the same amount in the opposite direction and do the opposite work. W = −pE = − p2 d3 In the third situation, the ﬁeld from the left dipole points to the right along the path, so we can’t bring in the second dipole pointing to the right. However, if we bring it in pointing up, the 4 force on it will be up, perpendicular to the path. Taking the derivative is a little messy in this case, but we can ﬁnd the direction of the derivative intuitively. If you think of the dipole as two charges, the force on the positive charge will be up to the right, and the force on the negative charge will be up to the left. But since the ﬁeld is symmetric with respect to the x axis, the x components cancel out leaving a force up. The ﬁeld at the second dipole in this case has strength E = 2p/d3 . In analogy with the above results, the work to rotate it to the right is W = −2 p2 , d3 because the dipole wants to be in this orientation. The work for the ﬁnal situation is then W =2 p2 . d3 7 Purcell 10.19 If the ion is positive, the dipole will point away from it. The dipole ﬁeld at the ion location will then point towards the dipole and the force will be attractive. If the ion is negative, the dipole will point towards it and the dipole ﬁeld at the ion will be away from the dipole. The force again will be attractive. The polarization is p = αE = αq/r2 where q is the ion charge. The force on the ion is F =q 2q 2 α 2p =5. r3 r To ﬁnd the potential energy, we bring in the ion from inﬁnity. The work required is U =−
r ∞ αq 2 2αq 2 dr = − 4 . 5 r 2r For sodium, α = 27 × 10−24 cm3 . 4 × 10−14 erg = (4.8 × 10−10 esu)2 (27 × 10−24 cm3 ) 2r4 r = 9.4 × 10−8 cm 5 8 Purcell 10.21 The maximum ﬁeld strength is Em = 14 × 103 V = 5.5 × 108 V /m . 0.0000254 m The energy density is 1 12 E = (3.25)(8.85 × 10−12 C 2 /N m2 )(5.5 × 108 V /m)2 = 4.35 × 106 J/m3 . 2m2 Remember that in SI, we need to insert the 4.35 × 106 J/m3 = 3100 J/kg . 1400 kg/m3 This can raise the capacitor to a height mgh = (0.75)m(3107 J/kg ) , h = 238 m .
o. The energy per mass is then 6 University of California, Berkeley Physics H7B Spring 1999 (Strovink) PROBLEM SET 13 1. Purcell problem 10.13 2. Purcell problem 10.16 3. Purcell problem 11.2 4. Purcell problem 11.4 5. Purcell problem 11.7 6. Purcell problem 11.9 7. Purcell problem 11.12 8. Purcell problem 11.16 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 13 Solutions by P. Pebler 1 Purcell 10.13 Consider a parallel plate capacitor. The energy required to charge it to a potential diﬀerence V is E = CV 2 /2. The capacitance increases with a dielectric to C = Co = A/4πs. The potential diﬀerence is Es. Then AE 2 s2 1 = E 2 (As) , E = CV 2 = 2 8πs 8π and the energy density is E2 . 8π For a wave in a dielectric B = E2 B2 = . 8π 8π 2 Purcell 10.16 We use Gauss’s law inside the uniform spherical charge distribution. 4πr2 Er = 4πQenc = 4π E= 4π ρr 3 4π 3 rρ 3 √ E and the energy density in the magnetic ﬁeld is Let the sphere of density ρ be centered at the origin, and the sphere of density −ρ be centered at the location s. The total ﬁeld is E= 4π 4π 4π ρr + (−ρ)(r − s) = ρs . 3 3 3 In the middle of a long cylinder, we can ﬁnd the ﬁeld from Gauss’s law. 2πrLEr = 4π (πr2 Lρ) E = 2πρrˆ r We are using cylindrical coordinates here so ˆ points away from the axis. The total ﬁeld of two r cylinders with their axes displaced by s is E = 2πρrˆ + 2π (−ρ)(rˆ − s) = 2πρs . r r 1 3 Purcell 11.2 The magnetic ﬁeld of a current loop with its axis on the z axis has only a z component with Bz = 2πb2 I 2m =2 . c(b2 + z 2 )3/2 (b + z 2 )3/2 2m 2m =3 r3 z The dipole ﬁeld on this axis is all radial, which here is the z direction. Bz = Br = So Bz = z3 B (b2 + z 2 )3/2 z b. There is a 1% diﬀerence when and the loop ﬁeld approaches the dipole ﬁeld when z z3 = 0.99 , (b2 + z 2 )3/2 z = 12.2 b . 4 Purcell 11.4 The earth’s radius is about 6 × 108 cm so 0.62 gauss = 2m , (6 × 108 cm)3 m = 6.7 × 1025 erg/gauss = 6.7 × 1022 J/T . If we have a current loop of radius 3 × 108 cm, we need a current I where 0.62 gauss = 2π (3 × 108 cm)2 I , c[(3 × 108 cm)2 + (6 × 106 cm)2 ]3/2 I = 9.9 × 1018 esu/s = 3.3 × 109 A . 5 Purcell 11.7 We will use polar coordinates for the integration. We divide the surface into little strips subtended by the small change in polar angle dθ. The surface area of one of these strips is da = 2π (R sin θ)(Rdθ) . 2 The amount of charge on this strip is dq = σda = 1 Q 2πR2 sin θ dθ = Q sin θ dθ . 2 4πR 2 This charge revolves around with a frequency f = ω/2π , so it represents a little current dI = f dq = ωQ sin θ dθ . 4π
π /2 0 Each strip contributes a moment dm = A dI/c. m= 1 c A dI = 2 c π (R sin θ)2 ωQR2 ωQ sin θ dθ = 4π 2c
π /2 0 sin3 θ dθ = ωQR2 3c 6 Purcell 11.9 From Chapter 6, the ﬁeld from a ﬁnite solenoid is Bz = 2πIn (cos θ1 − cos θ2 ) . c For a semiinﬁnite solenoid, θ2 = π and with z measuring the distance of the point outside the top of the solenoid, Bz = 2πIn 1− c z2 z 2 + ro . We want to maximize Bz (dBz /dz ). dBz 2πIn =− dz c Bz z2 1 −2 2 2 (z + ro )3/2 z 2 + ro 1− z2 z 2 + ro =− 2πIn c
2 ro 2 (z 2 + ro )3/2 1 dBz ∝2 2 dz (z + ro )3/2 Taking a derivative and setting to zero yields the equation
2 2 3z 2 − ro = 3z z 2 + ro . 2 Squaring and solving the quadratic equation gives z 2 = ro /15. Only the negative root solves the original equation so z = −ro 1 . 15 This is slightly inside the solenoid. 3 7 Purcell 11.12 The potential of a single dipole in a magnetic ﬁeld can be chosen to be U = −m · B . This does not have the zero where we want our zero to be. However, for the purposes of ﬁnding work done in rotating the dipoles we may use W = Uf − U i . In the initial conﬁguration, the ﬁeld due to dipole 2 at m1 is as shown above with B2 = m2 . r3 m1 m2 sin θ1 . r3 The work required to rotate m1 is W1 = Uf − Ui = −m1 B2 cos(90 + θ1 ) − 0 = m1 B2 sin θ1 = To rotate m2 , we break up the ﬁeld from m1 into two parts with B1 = 2m1 cos θ1 r3 B1 = m1 sin θ1 . r3 The work to rotate m2 is then W2 = Uf − Ui = [−m2 B1 cos θ2 − m2 B1 cos(90 + θ2 )] − [−m2 B1 cos π ] = −m2 B1 cos θ2 + m2 B1 sin θ2 − m2 B1 2m1 m2 m1 m2 m1 m2 =− cos θ1 cos θ2 + sin θ1 sin θ2 − sin θ1 . 3 3 r r r3 the total work is W = W1 + W2 = m1 m2 (sin θ1 sin θ2 − 2 cos θ1 cos θ2 ) . r3 4 8 Purcell 11.16 The exterior ﬁeld of a uniformly magnetized sphere turns out to be that of a magnetic dipole with dipole moment m= 4π 3 rM . 3 This is something that needs to be proved, however. One can prove this by ﬁnding the ﬁeld from the bound current. The bound current density is Jb = c ∇ × M = 0 , and the bound surface current is ˆ ˆ Kb = c M × n = M sin θ φ . This is identical to the surface current of a rotating sphere with uniform surface charge. One can integrate to ﬁnd the vector potential which is that of a magnetic dipole at the center. We leave this to you as an exercise. The ﬁeld at the pole is B= 8π 2m (750 erg/gauss cm3 ) = 6280 gauss . = 3 r 3 m = 3140 gauss . r3 At the equator B= To ﬁnd the force, we need to know the force on a uniformly magnetized sphere in the ﬁeld of a dipole. Fortunately, this is simple due to the following argument. The force on the sphere on the right must be the same if we replace the sphere on the left with a dipole at its center. This force must be equal and opposite to the force on the imaginary dipole. But the ﬁeld from the sphere on the right at the dipole is that of a dipole, so the force between spheres is the same as the force between two dipoles. (This is not obvious without the argument just given.) F = m2 F= 3 8 db1z d = m2 dz dz 4π M 3
2 2m1 z3 =6 m1 m2 3 m1 m2 = z4 8 r4 r2 = 3.7 × 106 dynes = 37 N 5 University of California, Berkeley Physics H7B Spring 1999 (Strovink) EXAMINATION 1 Directions. Do all four problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free to 2 approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Don’t use a calculator, which you don’t need – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (25 points) A heat engine for which the working material is an ideal monatomic gas moves slowly enough that all parts of it are always in mutual equilibrium. It is described by a rectangular path on the T (absolute temperature) – S (entropy) plane, as in the ﬁgure. While on the path 1 → 2, the gas in the engine takes heat from a bath at high temperature T1 ; on the path 3 → 4, it returns heat to bath at lower temperature T3 . On the paths 2 → 3 and 4 → 1, the entropy has constant values S3 and S1 , respectively. c. (8 points) Deduce the value of the mechanical work
12341 p dV done by the gas on the rest of the universe over one complete cycle of the engine. d. (7 points) In one cycle, what fraction of the heat withdrawn from the hot reservoir is converted to mechanical work done by the gas on the rest of the universe? Hint: Keep in mind that the only parameters given in this problem are T1 , T3 , S1 , and S3 ; your answers, if nontrivial, must be expressed in terms of these parameters. 2. (25 points) In a hypothetical onedimensional system, thermal motion of atoms in the y and z directions is “frozen out”, so, eﬀectively, the atoms are able to move only in the x direction. In that direction, an atom has velocity v (−∞ < v < ∞). The fraction dF of atoms with velocity between v and v + dv is dF ≡ fv (v ) dv = exp − mv dv 2kT
2 2 a. (5 points) Write down the net change (∆U23 + ∆U41 ) in internal energy for the sum of the two paths 2 → 3 and 4 → 1. b. (5 points) Compute the net change T dS ∞ −∞ exp − mv dv 2kT , where fv (v ) is the probability density (RHK: “relative probability”) of the value v , m is the atomic mass, k is Boltzmann’s constant, and T is the absolute temperature. a. (10 points) Calculate the mean value of the square of v , i.e. v 2 . If you wish, you may 12341 over one complete cycle of the engine. leave your answer in the form of a ratio of deﬁnite integrals. Do not merely guess the answer. b. (15 points) Deﬁne E to be the kinetic energy 1 2 2 mv of an atom. The fraction dF of atoms with kinetic energy between E and E + dE is dF ≡ fE (E ) dE , where fE (E ) is the probability density of the value E . One might imagine fE (E ) to take the possible forms: fE (E ) ∝ E −1/2 exp − ∝ exp − E kT E kT ? E kT ? ? E kT ? 4. (25 points) The inﬁnite plane z = 0 carries a uniform surface charge density σ esu/cm2 . There are no other charges in the problem. ∝ E 1/2 exp − ∝ E exp − Which one form is correct, and why? 3. (25 points) A ﬁxed line charge of +λ esu/cm on the x axis extends from x = D to x = 2D, and a ﬁxed line charge of −λ esu/cm on the y axis extends from y = D to y = 2D. a. (5 points) Find the magnitude and direction of the electric ﬁeld E+ everywhere in the region z > 0. b. (5 points) Find the magnitude and direction of the electric ﬁeld E− everywhere in the region z < 0. c. (8 points) Consider a spherical surface of radius R centered at the origin. Find the electric ﬂux E · da through the top half (top hemisphere) of this surface. d. (7 points) Consider a second spherical surface, again of radius R, but now centered at the point (0, 0, 2R), so that it does not enclose any charge. Find the electric ﬂux E · da a. (10 points) Find the work required to bring a test point charge q from inﬁnity to the origin. Does your answer depend on the path you chose? If so, specify the path. b. (15 points) Calculate the mechanical force (magnitude and direction) that is required to keep the test charge at the origin. through the bottom half (bottom hemisphere) of this surface. 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO EXAMINATION 1 Directions. Do all four problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free to 2 approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Don’t use a calculator, which you don’t need – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (25 points) A heat engine for which the working material is an ideal monatomic gas moves slowly enough that all parts of it are always in mutual equilibrium. It is described by a rectangular path on the T (absolute temperature) – S (entropy) plane, as in the ﬁgure. While on the path 1 → 2, the gas in the engine takes heat from a bath at high temperature T1 ; on the path 3 → 4, it returns heat to bath at lower temperature T3 . On the paths 2 → 3 and 4 → 1, the entropy has constant values S3 and S1 , respectively. c. (8 points) Deduce the value of the mechanical work
12341 p dV done by the gas on the rest of the universe over one complete cycle of the engine. d. (7 points) In one cycle, what fraction of the heat withdrawn from the hot reservoir is converted to mechanical work done by the gas on the rest of the universe? Hint: Keep in mind that the only parameters given in this problem are T1 , T3 , S1 , and S3 ; your answers, if nontrivial, must be expressed in terms of these parameters. Solution: (a.) U of an ideal gas is a function only of T , so the isothermal segments 1 → 2 and 3 → 4 cause no change in U . Therefore a. (5 points) Write down the net change (∆U23 + ∆U41 ) in internal energy for the sum of the two paths 2 → 3 and 4 → 1. b. (5 points) Compute the net change T dS
12341 (∆U23 + ∆U41 ) = 12341 dU = 0 because U is a state function. (b.) T dS = (T1 − T3 )(S3 − S1 ) , 12341 over one complete cycle of the engine. the area of the rectangle in the ﬁgure. 2 (c.) p dV = − =− δW dU + T dS δQ where fE (E ) is the probability density of the value E . One might imagine fE (E ) to take the possible forms: fE (E ) ∝ E −1/2 exp − ∝ exp − E kT E kT ? E kT ? ? E kT ? 12341 12341 12341 12341 =0+
12341 = (T1 − T3 )(S3 − S1 ) . (d.)
12341 ∝ E 1/2 exp − ∝ E exp − p dV Q2 = (T1 − T3 )(S3 − S1 ) = T1 (S3 − S1 ) T3 . =1− T1 It is also acceptable to state that this is a Carnot engine and quote this standard result for its efﬁciency. 2. (25 points) In a hypothetical onedimensional system, thermal motion of atoms in the y and z directions is “frozen out”, so, eﬀectively, the atoms are able to move only in the x direction. In that direction, an atom has velocity v (−∞ < v < ∞). The fraction dF of atoms with velocity between v and v + dv is dF ≡ fv (v ) dv =
2 exp − mv dv 2kT 2 ∞ exp − mv dv 2kT −∞ 12341 2 T 1 p dV dS Which one form is correct, and why? Solution: (a.) From the deﬁnition of it that is given, this probability density is explicitly normalized:
∞ −∞ fv (v ) dv ≡ 1 . Using the standard method for taking the average, when fv is normalized, v2 = =
∞ , −∞ 2 ∞ v 2 exp − mv 2kT −∞ 2 ∞ exp − mv 2kT −∞ v 2 fv dv dv dv . where fv (v ) is the probability density (RHK: “relative probability”) of the value v , m is the atomic mass, k is Boltzmann’s constant, and T is the absolute temperature. a. (10 points) Calculate the mean value of the square of v , i.e. v 2 . If you wish, you may leave your answer in the form of a ratio of deﬁnite integrals. Do not merely guess the answer. b. (15 points) Deﬁne E to be the kinetic energy 1 2 2 mv of an atom. The fraction dF of atoms with kinetic energy between E and E + dE is dF ≡ fE (E ) dE , This answer is enough to earn full credit. For completeness, deﬁning β ≡ 1/kT and u ≡ 1 mv 2 , 2 we can rewrite this quotient as
2 2 m ∞ u1/2 exp (−βu) du −∞ ∞ u−1/2 exp (−βu) du −∞ ∞ 2∂ −1/2 v = ln u m ∂β −∞ 2∂ ln (Cβ −1/2 ) =− m ∂β 1 = mβ kT , = m =− exp (−βu) du 3 where, in the above, C is a constant whose value is immaterial here. (b.) dF ≡ fE (E ) dE dF fE = dE dF dv = dv dE dv ≡ fv (v ) dE fv (v ) =d1 2 dv ( 2 mv ) fv (v ) = mv E exp − kT . ∝ E 1/ 2 tain distance from the origin, there is a corresponding negative charge element located at the same distance from the origin (but in an orthogonal direction). Therefore, by symmetry, the electrostatic potential φ vanishes at the origin, as does the work W required to bring the charge in from inﬁnity: W = q φ(0) − φ(∞) = 0 . (b.) From the positive part of the charge distribution, the electric ﬁeld at the origin is ˆ E+ = −x λ dx x2 D λ λ − = −x ˆ D 2D λ . = −x ˆ 2D
2D 3. (25 points) A ﬁxed line charge of +λ esu/cm on the x axis extends from x = D to x = 2D, and a ﬁxed line charge of −λ esu/cm on the y axis extends from y = D to y = 2D. Likewise, from the negative part of the charge distribution, ˆ E− = +y The total electric ﬁeld is y−x λ ˆˆ √. E = E+ + E− = √ 2D2 The mechanical force F required to keep the test charge at the origin must oppose q E: λ . 2D a. (10 points) Find the work required to bring a test point charge q from inﬁnity to the origin. Does your answer depend on the path you chose? If so, specify the path. b. (15 points) Calculate the mechanical force (magnitude and direction) that is required to keep the test charge at the origin. Solution: (a.) For every positive charge element that is a cer x − y qλ ˆˆ √, F= √ 2D2 where the ﬁrst factor is its direction (at −45◦ to the x axis), and the second is its magnitude. 4. (25 points) The inﬁnite plane z = 0 carries a uniform surface charge density σ esu/cm2 . There are no other charges in the problem. 4 A parallel to the z = 0 plane, E · da = 4πQencl ((E+ )z − (E− )z )A = 4πσA ((E+ )z + (E+ )z )A = 4πσA (E+ )z = −(E− )z = 2πσ E+ = −E− = ˆ 2πσ . z It is acceptable simply to recall that the electric ﬁeld on either side of an inﬁnite sheet of charge has this value, in the absence of other charges. (c.) Again because the charge distribution is symmetric about the plane z = 0, substituting a sphere of radius R centered at the origin for the Gaussian pillbox used in the solution of part (a.),
top hemi top hemi a. (5 points) Find the magnitude and direction of the electric ﬁeld E+ everywhere in the region z > 0. b. (5 points) Find the magnitude and direction of the electric ﬁeld E− everywhere in the region z < 0. c. (8 points) Consider a spherical surface of radius R centered at the origin. Find the electric ﬂux E · da through the top half (top hemisphere) of this surface. d. (7 points) Consider a second spherical surface, again of radius R, but now centered at the point (0, 0, 2R), so that it does not enclose any charge. Find the electric ﬂux E · da through the bottom half (bottom hemisphere) of this surface. Solution: (a.) (b.) The charge distribution is symmetric about the plane z = 0, so E+ = −E− , and both ﬁelds are normal to the z = 0 plane. Using a Gaussian pillbox with ﬂat surface area E · da = E · da = b ot hemi E · da = 1 4πQencl 2 1 4ππR2 σ 2 = 2π 2 R2 σ . (d.) Because E is constant throughout the semiinﬁnite region z > 0, the ﬂux of E through the top of the hemisphere centered at (0, 0, 2R) is the same as the ﬂux in part (c.) through the top of the hemisphere centered at the origin. Since the hemisphere centered at (0, 0, 2R) contains no charge, the ﬂux of E through its bottom half must cancel the ﬂux through its top half. Therefore
b ot hemi E · da = −2π 2 R2 σ . 1 University of California, Berkeley Physics H7B Spring 1999 (Strovink) FINAL EXAMINATION Directions. Do all six problems (weights are indicated). This is a closedbook closednote exam except for three 8 1 × 11 inch sheets containing any information you wish on both sides. You are 2 free to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. Problem 1. (35 points) Four straight stainless steel wires of length x, crosssectional area A, and resistivity ρ are welded together so that they lie along four of the six edges of a regular tetrahedron, as shown. Is there a value of ω for which the voltmeter will measure Eout (t) = 0? Explain. Problem 2. (30 points) Experiments using heavy electrons (“muons”) have exploited the fact that, when a pi meson (“π + ”) decays at rest into a muon (“µ+ ”) plus a neutrino (“ν ”), the muon has a unique momentum with which its spin angular momentum is fully aligned. The reaction is π + → µ+ + ν . A beam of such muons is called a surface muon beam (because the pion is stopped near the surface of a solid target where it was produced by protons from a cyclotron). The muons in a surface beam are so well deﬁned that, if they were allowed to impinge normally on a book, nearly all would stop in the same page. Given that the muon mass is 3/4 of the pion mass, while the neutrino mass is negligibly small, compute the velocity of the muons in a surface beam, expressed as a fraction of the speed of light. The remaining two sides ab and cd are empty. Consider a and b to be electrical input terminals, and c and d to be electrical output terminals. a. (15 points) When c is shorted to d, what resistance is measured between a and b? b. (20 points) Suppose that wires ac and bd are replaced by two identical inductors L, and wires bc and ad are replaced by two identical capacitors C . Across terminals a and b is placed a source of input EMF Ein (t) = E0 cos ωt where E0 and ω are real constants. Across terminals c and d, an output EMF Eout (t) is measured by an ideal voltmeter which draws no current. 2 Problem 3. (35 points) A Panofsky quadrupole magnet consists of four long thin copper bars, pointing in the ˆ direction z (out of the page), arranged so that their inside surfaces form a square box of side 2b. The bars at y = ±b carry a uniform current density in the +ˆ direction while the bars at x = ±b carry the z same current in the −ˆ direction. Within the z box enclosed by the bars, the magnetic vector potential is A= α ˆ(y 2 − x2 ) , z 2 c. (15 points) Suppose that a diﬀerent region of space has ˆ ˆ B(x, y ) = A0 (xy − yx) , where A0 is a constant (a “bullseye” magnetic ﬁeld). Show that the current density along ˆ z must be nonzero everywhere in the region; give its magnitude and any dependence that it may have on x and y . (This magnetic ﬁeld acts as a converging lens in both the x and y projections. However, the beam particles are required to pass through the magnet’s currentcarrying element, which needs to be made as light as possible, e.g. of molten lithium.) Problem 4. (35 points) Consider a uniform region of space containing an insulating material with ﬁxed dielectric constant and magnetic permeability µ. a. (3 points) Write Faraday’s law in diﬀerential form. b. (3 points) For this material there are two diﬀerential versions of Ampere’s Law, as modiﬁed by Maxwell – one version uses the free current density Jfree , the other uses the total current density. Write down the version that uses Jfree (which is zero for this insulating material). c. (3 points) Expressing H in terms of B and µ, and D in terms of E and , taking advantage of the fact that and µ are constant in this material, rewrite equation (b.) in terms of B and E. d. (3 points) ∂ Take 1 ∂t of both sides of equation (c.). On the c lefthand side, interchange the order of diﬀeren∂ tiation, i.e. apply 1 ∂t to B before taking its curl. c e. (3 points) ∂ Use equation (a.) to substitute for 1 ∂t B. Now c you should have an equation in which E is the only vector ﬁeld that appears. f. (3 points) Use the identity ∇ × (∇ × E) = ∇(∇ · E) − ∇2 E where α is a constant. a. (10 points) Suppose that a particle of charge +e travels along ˆ at position (x, y ) = (0, y ). Show that z the particle is deﬂected toward (0,0) with a force that is proportional to y . (This means that, in the y projection, the Panofsky quadrupole acts as a converging lens. However, it acts as a diverging lens in the x projection. Fortunately, the combination of a converging and a diverging lens of equal strength remains slightly converging, if the two lenses are separated along their axis; this allows a pair of quadrupole magnets to focus a particle beam weakly in both the x and y projections. One of the ﬁrst experiments to use this fact discovered the antiproton at the Berkeley Bevatron in 1956.) b. (10 points) Prove that the current density J within the box enclosed by the bars (x < b and y  < b) must be zero. (This allows the box to be evacuated so that a particle beam can travel unimpeded within it.) 3 to eliminate ∇ × (∇ × E) from the lefthand side of equation (e.). g. (6 points) Give an argument, based on the absence of free charges in this insulator, and the strict proportionality of E to D, which allows you to ignore one of the terms on the lefthand side of equation (f.). h. (6 points) Your result should be a wave equation for E. Show that any function of (kx − ωt), where k and ω are constants, solves this equation. i. (5 points) Calculate ω/k , the phase velocity of the solution (h.). Evaluate it in terms of c, , and µ. Problem 5. (30 points) An (insulating) hollow spherical shell of dielectric with inner (outer) radius a (b) has “frozenin” polarization P=ˆ r q0 , 2π (a + b)r Problem 6. (35 points) Consider a hollow cubical box containing particles which make elastic collisions with its walls. a. (10 points) Suppose that the particles are molecules of a perfect gas. Using standard arguments about the number of molecules bouncing oﬀ an area of wall per unit time, and the momentum per collision that is imparted to the wall, prove that the pressure p (N/m2 ) and the kinetic energy density u (J/m3 ) of the gas are related by p= 2 u. 3 b. (15 points) Suppose that the box is ﬁlled not with molecules, but with electromagnetic radiation, which is quantized into photons. These photons can be considered to be massless particles which, like the perfect gas molecules, do not interact with each other and bounce elastically oﬀ the walls. Deduce the relationship between the pressure p and the energy density u of the electromagnetic radiation. b. (10 points) At suﬃciently high temperature, the electromagnetic radiation pressure inside the box would be suﬃcient to balance the ambient pressure (106 dynes/cm2 ) of the earth’s atmosphere at sea level. If this were to occur, what would be the root mean square magnetic ﬁeld (in gauss) inside the cavity? where r is the radius measured from its center and q0 is a constant. The dielectric encloses a conducting sphere of radius a which holds total free charge q0 (see the ﬁgure). At what values of r does E vanish? [Your answer may include particular values of r (including those which are not ﬁnite), and/or ranges of r.] Note that this (nonlinear) dielectric’s electric susceptibility is not deﬁned or supplied here, and it should not appear in your answer. University of California, Berkeley Physics H7C, Fall 1999 (Strovink) General Information (27 Aug 99) Web site for this class: First link on http://d0lbln.lbl.gov . Instructors: Prof. Mark Strovink, 437 LeConte; (LBL) 4867087; (home, before 10) 4868079; (UC) 6429685. Email: [email protected] . Web: http://d0lbln.lbl.gov . Office hours: M 3:154:15, 5:306:30. Mr. Derek Kimball, (UC research lab) 221 Birge, 6431829; (home, before 11) 5483115. Email: [email protected] . Office hours (to be held in 211 LeConte): Tu 24. You may also get help in the 7C Course Center, 262 LeConte. Lectures: TuTh 11:1012:30, 3 LeConte. Lecture attendance is essential, since not all of the course content can be found in the course text or handouts. On one or two occasions it is possible that the lecture normally held on Thursday will be given instead on Wednesday, at 4:305:50 PM, in 343 LeConte. Labs: Begin in the third week, in 278 LeConte. As soon as possible, please enroll in any one of the 7C lab sections that fits your schedule. Discussion Sections: Begin in the second week. Section 1: W 45, 343 LeConte; Section 2: W 5:306:30, also in 343 LeConte. You are welcome at either or both sections. You are especially encouraged to attend discussion section regularly. There you will learn techniques of problem solving, with particular application to the assigned exercises. Texts (required): Fowles, Introduction to Modern Optics, Second Edition (Dover paperback, 1989). Rohlf, Modern Physics from α to Z0 (Wiley, 1994). Supplementary text (recommended): Hecht, Schaum’s Outline of Theory and Problems of Optics (McGrawHill paperback, 1975). (Don’t confuse this with Hecht’s hardbound books on optics.) Problem Sets: Twelve problem sets are assigned and graded. Solutions will be available. Problem sets are due on Thursday at 5 PM on weeks in which there is no exam, beginning in week 2. Deposit problem sets in the box labeled "H7C" outside 211 LeConte. You are encouraged to attempt all the problems. Students who do not do them find it almost impossible to learn the material and to succeed on the examinations. Discuss these problems with your classmates as well as with the teaching staff; however, when the time comes to write up your solutions, work independently. Credit for collective writeups, which are easy to identify, will be divided among the collectivists. Late papers will not be graded. Your lowest problem set score will be dropped, in lieu of due date extensions for any reason. Syllabus: H7C has one mandatory syllabus card. It will be collected when the first midterm examination is handed back in lecture. This card pays for the 7C laboratory experiment descriptions and instructions. Copies of solutions to each problem set will also be available for separate purchase at Copy Central. Exams: There will be two 80minute midterm examinations and one 3hour final examination. Before confirming your enrollment in this class, please check that its final Exam Group 9 does not conflict with the Exam Group for any other class in which you are enrolled. Please verify that you will be available for the midterm examinations (Th 7 Oct and Th 11 Nov, 11:1012:30), and for the final examination, F 10 Dec, 58 PM. Except for unforeseeable emergencies, it will not be possible for the midterm or final exams to be rescheduled. Passing H7C requires passing the final exam. Grading: 30% midterms; 25% problem sets; 40% final exam; 5% lab. Grading is not "curved"  it does not depend on your performance relative to that of your H7C classmates. Rather it is based on comparing your work to that of a generation of earlier lower division Berkeley physics students, with due allowance for educational trends. Physics H7C Course Outline Fall 1999 (Strovink) Week No. 1 Week of... 23Aug Reading chapter Topic (Pu=Purcell, Fo=Fowles, Ro=Rohlf) Problem Due 5 PM Set No. on... 7C lab Ro 4, Pu A, Review of special relativity; Fo App. I relativistic transformation of EM fields Pu 9 Pu 10,11 Review of Maxwell's equations and EM waves… …in vacuum and in material; boundary conditions LABOR DAY Radiation by an accelerated charge Polarization Plane reflection/refraction Interference Coherence Multiple beams Multiple beams Diffraction Diffraction Optics of solids MIDTERM 1 (covers PS 15) Optics of solids, MaxwellBoltzmann distribution Planck's constant Planck's constant Wave properties of matter Uncertainty principle Probing the structure of matter Schroedinger equation Hydrogen atom Periodic table MIDTERM 2 (covers PS 19) Quantum statistics Quarks and leptons Unification of the forces THANKSGIVING Cosmology LAST LECTURE (review) NONE 2 30Aug NONE 1 2Sep 3 6Sep Pu B Fo 2 2 9Sep reflect/ refract geom optics 4 13Sep Fo 2 Fo 3 Fo 3 Fo 4 Fo 4 Fo 5 Fo 5 Fo 6 3 16Sep 5 20Sep 4 michelson 23Sep interferom diffract/ interfer NONE 6 27Sep 5 30Sep 7 4Oct 7Oct 8 11Oct Ro 2 Ro 3 Ro 3 Ro 5 Ro 5 Ro 6 Ro 7 Ro 8 Ro 9 6 14Oct polarIzation 9 18Oct 7 21Oct NONE photoelectric NONE 10 25Oct 8 28Oct 11 1Nov 9 4Nov atomic spectra radio halflife NONE 12 8Nov 11Nov 15Nov 13 Ro 12 Ro 17 Ro 18 10 11 18Nov 24Nov 14 22Nov 25Nov 29Nov 15 Ro 19 makeups 12 2Dec 16 6Dec 8Dec 10Dec 58 PM Final exams begin H7C FINAL EXAM (Group 9) (covers PS 112) updated 12/31/99 http://d0lbln.lbl.gov/h7cf99/h7cf99texts.txt NOTES ON H7C TEXTS: RELATIVITY TEXTS (to supplement distributed handwritten notes): Taylor and Wheeler, Spacetime Physics: Good for spacetime, Lorentz transformations as spacetime rotations, the boost parameter, and rocket problems. Rohlf, Modern Physics from alpha to Z0 (H7C required), chapter 4: Fair for relativistic kinematics  a Compton scattering derivation is given, though it is not the most elegant. Purcell, Electricity and Magnetism (H7B text), appendix A and chapter 9: Good for relativistic transformation of electric and magnetic fields. OPTICS TEXTS: Fowles, Introduction to Modern Optics (H7C required): Terse. Hecht, Schaum's Outline of Theory and Problems of Optics (H7C recommended): Students have found this to be a useful summary of formulae and a place to find simple practice problems. OPTICS TEXTS THAT PREVIOUSLY HAVE BEEN USED IN H7C: Pedrotti & Pedrotti, Introduction to Optics, 2nd ed. (PrenticeHall) Hecht, Optics, 3rd ed. (Addison Wesley) MODERN PHYSICS TEXT: Rohlf, Modern Physics from alpha to Z0 (H7C required) MODERN PHYSICS TEXTS THAT PREVIOUSLY HAVE BEEN USED IN H7C: These are listed in descending (subjective) order of usefulness: Blatt, Modern Physics (McGrawHill) Beiser, Concepts of Modern Physics, 5th ed. (McGrawHill) Ohanian, Modern Physics, 2nd ed. (PrenticeHall) Eisberg & Resnick, Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2nd ed. (Wiley) Serway, Modern Physics, 2nd ed. (Saunders) Tipler, Physics for Scientists and Engineers, Vol II, 3rd ed. (Worth) http://d0lbln.lbl.gov/h7cf99/h7cf99texts.txt [1/5/2000 9:52:25 AM] Errata Fowles, Introduction to Modern Optics, Second Edition M. Strovink This book was chosen for the optics portion of H7C because the selection and relative coverage of topics is excellent. As now reprinted by Dover, it is also bargain priced. For the most part Fowles tells you what you want to know, with brevity and insight. Perhaps more important for a general course like H7C, Fowles usually does not tell you what you do not want to know. This is in contrast to the pricey and encyclopædic Hecht, in which fundamental concept and arcane detail are mixed together without any obvious diﬀerentiation. The main problem with the book is that Fowles, while a good physicist, seems as an author not to be a suﬃciently concerned with correctness. The text is full of annoying lapses in accuracy of the drawings. Despite the existence of two editions, the book seems never to have been adequately proofread or revised. Therefore, to beneﬁt from this otherwise good book, you should correct the many errors in your copy, using this guide. Page 23, Eq. 2.16 should read H= Page 25, Eq. 2.23 should read I= µ0 1 n E0 H 0 = E0 2 2 2Z0 µ nE µ0 Z0 µ Page 41, Eqs. 2.49, 2.50, and 2.51: these equations assume that µ is the same in both media. Page 43 (an omission, not an error). To Eqs. 2.54 and 2.55 should be added: ts = tp = 2 cos θ cos θ + n cos φ 2 cos θ n cos θ + cos φ Page 45. The details in Fig. 2.12 are reliable only to a factor of ≈ 2, due to bad registration of the curves (printed in color in the hardcover version) with respect to the axes. Page 56, Problem 2.23: “Brewster window” should be “Brewster interface”, i.e. one interface between n = 1 and n = n. Page 64, last sentence, beginning “In this case the central fringe...”, and continuing onto page 65, should be ignored. This sentence would be correct only if plate A were not (even half) silvered. Page 75, Fig. 3.13: For your own increased comprehension, indicate the following on the ﬁgure: s is the distance between the sources Sa and Sb ; l is the distance between the receivers P1 and P2 ; r is the distance between the average of Sa and Sb , and the average of P1 and P2 . Page 76, sentence following Eq. 3.39 should read: “... small in comparison with τ0 .” Page 76, sentence preceding Eq. 3.40 should read: “We then have τb − τa = (r2b − r2a )/c, or approximately” Page 76, Eq. 3.40 should read: sl τb − τ a ≈ cr Page 77, Eq. 3.41 should read: ω (τb − τa ) = Page 77, Eq. 3.42 should read: lt = Page 77, Eq. 3.43 should read: lt = 1 λ 2θs rλ 2s ωslt =π cr Page 81, Eq. 3.46 should read: G(k ) = 1 2π ∞ −∞ W (x)e−ikx dx The factor 2π should appear to the power −1 rather than − 1 because there is no corresponding factor of 2 √ Conversely, in Eq. 3.28 on page 71 Fowles instead has adopted the neater convention of 1/ 2π in Eq. 3.45. √ inserting a factor 1/ 2π in the deﬁnitions of each of the Fourier transform pairs. Page 97, Fig. 4.7: For consistency with Eqs. 4.23, 4.27, and 4.32, in the [central] region with index n1 shown in the ﬁgure, k1 and k1 should be replaced by k and k . Page 98, last sentence: add at the end of this sentence “, with R + nT T /n0 = 1.” Page 137, Fig. 5.28; page 139, Fig. 5.29; and page 143, Fig. 5.32: The patterns should be centered on y = 0 or ν = 0 (again a regisration problem). Page 150, Problem 5.16: The last parenthesis should be “(Assume the trailing edge of the moon to be eﬀectively straight.)” Page 220, 4th sentence should read: “For left circularly polarized light, the direction of spin of the photon is parallel to the direction of propagation, whereas for right circularly polarized light, it is antiparallel to the direction of propagation. [This is a convention that depends on the charge of the particle (electron or positron) to which the photon couples; our (more usual) convention takes that charge to be positive.]” Page 297, Eq. 10.11 should read: 1 1 (n − 1)t 1 = (n − 1) − + f r1 r2 nr1 r2 Page 297, Eq. 10.12 should read: d2 = −f t 1−n r1 assuming that d1 and d2 are positive distances as drawn in Fig. 10.3. Page 298, Fig. 10.3. The rays are so badly drawn as to be meaningless. The top ray should emanate from the object in a direction parallel to the axis, and it should be bent only at principal plane H . The bottom ray should emanate from the image in a direction parallel to the axis, and it should be bent only at principal plane H . Page 307, 1st complete sentence should read: “The interference pattern can be made to shift by one fringe i.e. from a bright fringe to the adjacent dark fringe, or vice versa) by displacing either of the two mirrors M1 or M2 a distance of 1 wavelength.” 4 2 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 1 1. Two supernovæ are observed on earth in the direction of the north star, separated by 10 years. From the theory of supernovæ these are known to have identical (“standard candle”) light output, yet the ﬁrst is observed to have four times the light intensity of the second because it is closer. (a.) An astronomer theorizes that the two stars were at rest with respect to the earth, and that the ﬁrst supernova triggered the second. What is the maximum distance between the earth and the ﬁrst supernova under this hypothesis? (b.) A physicist theorizes that the two stars were traveling with the same (unspeciﬁed) velocity away from the earth, and that, in their common rest frame, the two supernovæ occurred at the same proper time. What is the minimum distance between the earth and the ﬁrst supernova under this hypothesis? 2. Inertial reference frames S and S coincide at t = t = 0. You may ignore the z dimension, so that a point in spacetime is determined by only three quantities r ≡ (ct, x, y ). The Lorentz transformation between S and S is given by ct ct x = L x , y y where L is a 3×3 matrix. (a.) Assume for this part that S moves with velocity V = βcx ˆ with respect to S . Using your knowledge of Lorentz transformations (no derivation necessary), write L for this case. (b.) Assume for this part that S moves with velocity x+y ˆˆ V = βc √ 2 with respect to S . Find L for this case. (Hint. Rotate to a system in which V is along the x ˆ axis, transform using your answer for part (a.), and then rotate back. Check that your result is symmetric under interchange of x and y , as is V, and that it reduces to the unit matrix as β → 0.) 3. Work out the Lorentz transformation matrix L for the general case in which β of frame S is directed along an arbitrary unit vector n = (nx , ny , nz ) as seen in frame S , e.g. ˆ r = Lr, L = ? 4. In a straight channel oriented along the ˆ z axis there are two opposing beams: • a beam of positrons (charge +e) with velocity +ˆβc. z • a beam of electrons (charge −e) with velocity −ˆβc. z Each beam is conﬁned to a small cylindrical volume of cross sectional area A centered on the ˆ axis. Within that volume, there is a uniz form number density = n positrons/m3 and n electrons/m3 . (a.) In terms of n, A, e, and β , calculate the total current I in the channel due to the sum of both beams (note I = 0). (b.) Use Amp`re’s Law to calculate the azimuthal e magnetic ﬁeld Bφ outside the channel a distance r from the ˆ axis. z Consider now a Lorentz frame S travelling in the ˆ direction with velocity βc relative to the z lab frame described above. (This β is the same β as above.) (c.) As seen in S , calculate the number density n+ of positrons within the cylindrical volume. (You may use elementary arguments involving 2 space contraction, or you may use the fact that (cρ, j) is a 4vector, where ρ is the charge density (Coul/m3 ) and j is the current density (amps/m2 ).) (d.) As seen in S , calculate the number density n− of electrons within the cylindrical volume. (e.) Calculate the radial electric ﬁeld Er seen in S . Do this both • by using the results of (c.) and (d.) plus Gauss’s law, and • by using the results of (b.) plus the rules for relativistic E and B ﬁeld transformations. 5. (Taylor and Wheeler problem 51) The clock paradox, version 3. Can one go to a point 7000 light years away – and return – without aging more than 40 years? “Yes” is the conclusion reached by an engineer on the staﬀ of a large aviation ﬁrm in a recent report. In his analysis the traveler experiences a constant “1g ” acceleration (or deceleration, depending on the stage reached in her journey). Assuming this limitation, is the engineer right in his conclusion? (For simplicity, limit attention to the ﬁrst phase of the motion, during which the astronaut accelerates for 10 years – then double the distance covered in that time to ﬁnd how far it is to the most remote point reached in the course of the journey.) (a.) The acceleration is not g = 9.8 meters per second per second relative to the laboratory frame. If it were, how many times faster than light would the spaceship be moving at the end of ten years (1 year = 31.6 × 106 seconds)? If the acceleration is not speciﬁed with respect to the laboratory, then with respect to what is it speciﬁed? Discussion: Look at the bathroom scales on which one is standing! The rocket jet is always turned up to the point where these scales read one’s correct weight. Under these conditions one is being accelerated at 9.8 meters per second per second with respect to a spaceship that (1) instantaneously happens to be riding alongside with identical velocity, but (2) is not being accelerated, and, therefore (3) provides the (momentary) inertial frame of reference relative to which the acceleration is g . (b.) How much velocity does the spaceship have after a given time? This is the moment to object to the question and to rephrase it. Velocity βc is not the simple quantity to analyze. The simple quantity is the boost parameter η . This parameter is simple because it is additive in this sense: Let the boost parameter of the spaceship with respect to the imaginary instantaneously comoving inertial frame change from 0 to dη in an astronaut time dτ . Then the boost parameter of the spaceship with respect to the laboratory frame changes in the same astronaut time from its initial value η to the subsequent value η + dη . Now relate dη to the acceleration g in the instantaneously comoving inertial frame. In this frame g dτ = c dβ = c d(tanh η ) = c tanh (dη ) ≈ c dη so that c dη = g dτ Each lapse of time dτ on the astronaut’s watch is accompanied by an additional increase dη = g dτ c in the boost parameter of the spaceship. In the laboratory frame the total boost parameter of the spaceship is simply the sum of these additional increases in the boost parameter. Assume that the spaceship starts from rest. Then its boost parameter will increase linearly with astronaut time according to the equation cη = gτ This expression gives the boost parameter η of the spaceship in the laboratory frame at any time τ in the astronaut’s frame. (c.) What laboratory distance x does the spaceship cover in a given astronaut time τ ? At any instant the velocity of the spaceship in the laboratory frame is related to its boost parameter by the equation dx/dt = c tanh η so that the distance dx covered in laboratory time dt is dx = c tanh η dt Remember that the time between ticks of the astronaut’s watch dτ appear to have the larger 3 value dt in the laboratory frame (time dilation) given by the expression dt = cosh η dτ Hence the laboratory distance dx covered in astronaut time dτ is dx = c tanh η cosh η dτ = c sinh η dτ Use the expression cη = gτ from part b to obtain dx = c sinh gτ dτ c 7. (Taylor and Wheeler problem 75) Doppler equations. A photon moves in the xy laboratory plane in a direction that makes an angle φ with the x axis, so that its components of momentum are px = p cos φ, py = p sin φ, and pz = 0. (a.) Use the Lorentz transformation equations for the momentumenergy 4vector and the relation E 2 /c2 − p2 = 0 for a photon to show that, in the rocket frame S (moving with velocity βr c along the x, x direction, and coinciding with the laboratory frame at t = t = 0), the photon has an energy E given by the equation E = E cosh ηr (1 − βr cos φ) and moves in a direction that makes an angle φ with the x axis given by the equation cos φ = cos φ − βr 1 − βr cos φ Sum (integrate) all these small displacements dx from zero astronaut time to a ﬁnal astronaut time to ﬁnd gτ c2 cosh x= g c −1 This expression gives the laboratory distance x covered by the spaceship at any time τ in the astronaut’s frame. (d.) Plugging in the appropriate numerical values, determine whether the engineer is correct in his conclusion reported at the beginning of this exercise. 6. Electrons (mc2 = 0.5 × 106 eV) are accelerated over a distance of 3.2 km from rest to a total energy of 5 × 1010 eV at SLAC (Stanford). (a.) To what boost η are the electrons ultimately brought? (b.) Assuming that the electrons are subjected to a uniform acceleration as observed in their comoving inertial frame, how many g ’s of acceleration do they feel? (c.) As observed in the lab, for what time interval is each electron in ﬂight? What is the corresponding proper time interval? Evaluate the ratio of the two intervals (a sort of average γ factor). (b.) Derive the inverse equations for E and cos φ as functions of E , cos φ , and βr . (c.) If the frequency of light in the laboratory is ν , what is the frequency ν of light in the rocket frame? This diﬀerence in frequency due to relative motion is called the relativistic Doppler shift. Do these equations enable one to tell in what frame the source of the photons is at rest? 4 8. Consider the following situation. A star is known, by means of external data, to be located instantaneously a distance D from an observer on earth. The external data do not tell us the rate of change of D with time. In her measurements, the observer corrects for aberration caused by the local velocity of the earth’s surface, due both to its daily rotation and its yearly orbit. Therefore we do not need to take into account these boring local phenomena in what follows. After making these corrections, the observer sees that the star is undergoing angular motion dψ/dt across the sky, such that D dψ/dt = c, where c is the speed of light. Finally, the observer measures the wavelength spectrum of light from this star, and ﬁnds its features not to be redshifted or blueshifted at all – they are exactly where they would be if the star were perfectly at rest with respect to the observer. Is it possible that this situation is physically reasonable? If so, what might be the true motion of the star with respect to the observer? If not, why not? Physics H7C Fall 1999 Solutions to Problem Set 1 Derek Kimball “Let’s get something straight here... e is real, 10 is just the number of ﬁngers we From Eq. (2), we see that if the distance between SN2 and SN1 is ∆x, the have.” distance between SN1 and earth is also ∆x. Then the elapsed time ∆tearth between detection of the two supernovae on earth, taking into account the propagation time  Prof. Nima ArkaniHamed, UC Berkeley of the light to the earth, is given by: If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! 2∆x c −c ∆x c = c∆t + ∆x. (4) If you liked problem 1 and you’re interested in astrophysics, general relativity, and cosmology, you should check out a paper by Saul Perlmutter, Michael S. Turner, and Martin White (Physical Review Letters, July 26, 1999, Volume 83, Issue 4, From Eq. (3), we know that c∆t ≥ ∆x, so we ﬁnd that: pp. 670673). This article and references therein describe an ongoing study of c∆tearth , (5) ∆x ≤ type Ia supernovae which have “standard candle” light output and have enabled 2 these scientists to measure largescale cosmological parameters. One of the most interesting results is that their data is consistent with a universe that is expanding or that ∆xmax = 5 light years. at an accelerating rate! Saul Perlmutter’s group is here at Berkeley and works at (b) LBL. A physicist theorizes that the two supernovae were traveling away from the earth Problem 1 at some velocity and occurred at the same proper time. In this case the two events have a spacelike or lightlike separation: First, let’s consider the implications of the diﬀerence in light intensity of the two supernovae (SN1 and SN2). These particular supernovae are known to have iden(6) c2 ∆t2 − ∆x2 ≤ 0. tical “standard candle” light output, i.e. the total light power P emitted is the same for SN1 and SN2. A small solid angle dΩ of the total light is detected on Thus, in the earth frame there is an observed time diﬀerence ∆tobs between SN1 and SN2, which from Eq. (6) must satisfy: earth, so the intensity of light I detected is given by: I= dΩ · P , 4πR2 (1) c∆tobs ≤ ∆xobs , (7) c∆tearth = c∆t + c where ∆xobs is the distance between SN1 and SN2 as observed in the earth frame. where R is the distance from a supernova to the earth at the time the light is As in part (a) we include the light propagation time, and ﬁnd that: emitted. Therefore the ratio of light intensities tells us the ratio of distances: c∆tearth = c∆tobs + ∆xobs ≤ 2∆xobs . (8) 2 I1 R2 = 2 = 4. (2) I2 R1 So in this case we ﬁnd that ∆xobs,min = 5 light years. (a) Problem 2 An astronomer theorizes that SN1 causes SN2, and that they are both at rest with (a) respect to the earth. Since the two events SN1 = (ct1 , x1 ) and SN2 = (ct2 , x2 ) This is just the traditional Lorentz matrix, only in 3D, so it is similar to the are causally related, there must be a timelike or lightlike separation between the expression (1.12) in Prof. Strovink’s notes on relativity, events: γ −γβ 0 ct ct 2 2 2 (3) c ∆t − ∆x ≥ 0, x = −γβ γ 0 · x . (9) y 0 0 1 y where ∆t = t − t and ∆x = x − x .
2 1 2 1 September 2, 1999 Physics H7C Fall 1999 (b) Solutions to Problem Set 1 Derek Kimball The general idea is to rotate to a system where we know the correct transform (from part (a)), and then rotate back. So we begin with: r = L · r. (10) Then we rotate the coordinate system with a rotation matrix R so that β is along x: ˆ Rr = RL · r = RLR−1 (Rr ). (11) In this frame we know the Lorentz transform Λ from part (a), so we ﬁnd that: RLR−1 = Λ. In other words, L=R
−1 (12) Figure 1: Relationship between n and the angles θ and φ employed in problem (3) ˆ in the ﬁrst approach. As you can see by inspection, this matrix is symmetric under interchange of x and y and reduces to the identity matrix as β → 0. Problem 3 Here are two common approaches to this problem. The ﬁrst method involves matrix multiplication in a manner similar to that employed in problem 2. The second involves determining a general vector formula for the Lorentz transform. ΛR. (13) The math can be made a little easier in these cases because rotations are described by orthogonal matrices which satisfy R−1 = RT where RT is the transpose of R. Now for the actual math. From (a) and Eq. (13) we ﬁnd: 1 1 0 0 γ −γβ 0 γ 0 · 0 L = 0 cosθ −sinθ · −γβ 0 0 sinθ cosθ 0 0 1 0 sinθ , cosθ (14) 0 cosθ −sinθ where we use straightforward 3D extensions of the usual rotation matrices. MulApproach 1 tiplying these matrices gives us: First, we’ll determine the rotation matrix R which will take us into the frame γ −γβcosθ −γβsinθ where β is along x. For me, it’s easier to think of this in terms of the angles θ and ˆ 2 2 −cosθsinθ + γcosθsinθ . (15) γcos θ + sin θ L = −γβcosθ φ as deﬁned in Fig. 1. 2 2 −γβsinθ −cosθsinθ + γcosθsinθ cos θ + γsin θ From Fig. 1, we notice that there is a natural correspondence between (nx , ny , nz ) We are given that: and θ, φ given by x+y ˆˆ cosφcosθ nx V = βc √ , (16) 2 ny = cosφsinθ . (18) nz sinφ so θ = π/4. Then L is given by: L= γ
−γβ √ 2 −γβ √ 2 −γβ √ 2 1+γ 2 1 2 (γ − 1+γ 2 −γβ √ 2 1) (17) Later, these relations will be used to express L in terms of nx , ny and nz . R is given by the multiplication of two rotation matrices A and B , R = B · A, September 2, 1999 (19) 1 2 (γ − 1) Physics H7C Fall 1999 where A rotates the axes about z by θ ˆ 1 0 0 cosθ A= 0 −sinθ 0 0 0 0 , 0 1 Solutions to Problem Set 1
0 sinθ cosθ 0 Derek Kimball (20) example, n is along x, Λ from Eq. (1.12) in Strovink’s notes is applicable and we ˆ ˆ ﬁnd: γ (ct − βx) ct x x + (γ − 1)x − γβct . (25) y = y z z Let r ≡ (x, y, z ), then by analogy with Eq. (25) we ﬁnd that: ct = γ (ct − βr · n) ˆ and r = r + n((γ − 1)r · n − γβct). ˆ ˆ . (28) (27) (26) and B rotates the axes about a new y (the yaxis after rotation by A) by φ so ˆ that x is along n: ˆ ˆ 1 0 0 0 0 cosφ 0 sinφ . B= (21) 0 0 1 0 0 −sinφ 0 cosφ Applying Eq. (13), we can now solve for L: L = A−1 · B −1 · Λ · B · A = AT · B T · Λ · B · A, where Λ is given by Eq. (1.12) in Prof. Strovink’s notes on relativity. We can then express these equations in terms of nx , ny and nz : γct − γβ (nx x + ny y + nz z ) ct x x + nx (γ − 1)(nx x + ny y + nz z ) − γβct (22) y = y + ny (γ − 1)(nx x + ny y + nz z ) − γβct z z + nz (γ − 1)(nx x + ny y + nz z ) − γβct The result of this rather tedious matrix multiplication, after some simpliﬁcation These expressions can be rewritten in matrix form, yielding L from Eq. (10) to using basic trigonometric identities, is given by L = be: γ −βγcosθcosφ −βγsinθcosφ −βγsinφ −βγcosθcosφ 1 + (γ − 1)cos2 θcos2 φ (γ − 1)sinθcosθcos2 φ (γ − 1)cosθsinφcosφ −βγsinθcosφ (γ − 1)sinθcosθcos2 φ 1 + (γ − 1)sin2 θcos2 φ (γ − 1)sinθsinφcosφ . −βγsinφ (γ − 1)cosθsinφcosφ (γ − 1)sinθsinφcosφ 1 + (γ − 1)sin2 φ (23) γ −βγnx −βγny −βγnz If we then use the relations given in Eq. (18) to reexpress Eq. (23) in terms of −βγnx 1 + (γ − 1)n2 (γ − 1)nx ny (γ − 1)nx nz x L= (29) nx , ny and nz we ﬁnd that: −βγny (γ − 1)ny nx 1 + (γ − 1)n2 (γ − 1)ny nz , y −βγnz (γ − 1)nz nx (γ − 1)nz ny 1 + (γ − 1)n2 z γ −βγnx −βγny −βγnz −βγnx 1 + (γ − 1)n2 (γ − 1)nx ny (γ − 1)nx nz which you will notice is the same result as the one obtained in approach 1. x (24) L= −βγny (γ − 1)ny nx 1 + (γ − 1)n2 (γ − 1)ny nz . y −βγnz (γ − 1)nz nx (γ − 1)nz ny 1 + (γ − 1)n2 z Problem 4 Approach 2 In this approach, we work out a vector formula for the Lorentz transformation using the fact that length contraction occurs only in the direction of β . If, for September 2, 1999 (a) The current I is the charge per second traveling through the channel, given by: I = nA(+e)(+βc) + nA(−e)(−βc) = 2nAeβc. (30) Physics H7C Fall 1999 (b) Solutions to Problem Set 1
as above. (d) Derek Kimball Ampere’s law (in SI units, feel free to use whatever units you like of course) is: B · d = µ0 Ienclosed . (31) Here we’ll just stick to the fourvector method. The relationship between the charge density of electrons seen in S and S is given by: cρ jz = γ −γβ −γβ γ · cρ jz , (39) So in our case, assuming an inﬁnitely long channel and using I from Eq. (30): Bφ · 2πr = µ0 2nAeβc, therefore Bφ = µ0 nAeβc . πr (33) (32) from which we ﬁnd the charge density: cρ = γcρ − βγjz . (40) (c) Let’s solve this both suggested ways... ﬁrst using length contraction. The density of positrons n+ is given by: n+ = N+ , A·d (34) The electron charge density in S satisﬁes cρ = n− (−e)c and the current density in S is given by jz = n− (−e)(−β )c. Plugging these into Eq. (40) allows us to solve for n− : n− = γ 1 + β 2 n− . (e) (41) First we solve the problem using Gauss’s law: where d is a unit length of the channel in the lab frame S and N+ is the total ρ number of positrons contained in this volume. This new frame S is the rest frame dV. (42) Er · dA = 0 of the positrons, so d = γd (sort of length uncontraction). Therefore the observed positron density in S is given by: Choosing a cylindrical Gaussian surface centered on the z axis with radius r and length d, we ﬁnd: n+ N+ n+ = = . (35) Aγd γ Er · 2πrd = Ad ρ+ + ρ− / 0 . (43) We can arrive at the same conclusion using the fact that cρ, j Considering only the zdirection, we have the relation: cρ jz = γ −γβ −γβ γ · cρ jz , is a fourvector. Using the relations for the charge density of positrons and electrons in S from Eqs. (38) and (41), we ﬁnd that: ρ+ + ρ− = (36) ne 1 − γ 2 (1 + β 2 ) = −2β 2 γne. γ (44) Combining these results, we ﬁnd for the radial electric ﬁeld Er seen in S : Er = −r ˆ β 2 γneA . π 0r (45) where the charge density in the lab frame S satisﬁes cρ = n+ ec and the current density in S is given by jz = n+ eβc. Thus from Eq. (36) we ﬁnd that: cρ = n+ ec = γ n+ ec − β 2 n+ ec . Consequently, n+ = n+ γ (38) September 2, 1999 (37) We can also solve this problem using the relativistic ﬁeld transformations for E and B given in Prof. Strovink’s notes (Eq. (1.33)), in particular: E⊥ = γ E⊥ + cβ × B . (46) Physics H7C Fall 1999 Solutions to Problem Set 1
Finally arriving at the solution: (47) xf = c2 (cosh(gτf /c) − 1). g ˆ ˆ Er = −cβγBφ r = −r β 2 γneA . π 0r Derek Kimball Employing Bφ from Eq. (33) and noting that Er = 0 in S , we ﬁnd that: (53) Where we use the fact that 0 µ0 = 1/c2 . This, of course, agrees with our result from Eq. (45) using Gauss’s law. (d) Problem 5 (a) Note the interesting fact that the number of seconds in a year is approximately Problem 6 π × 107 , a useful fact at cocktail parties and for backoftheenvelope calculations. (a) If you naively multiply the acceleration by the time, you ﬁnd: v = gt ≈ 10c. So, if you’re a believer in relativity, this can’t be right... (b) (48) The relativistic expression for energy E of particles with nonzero mass is given by Eq. (1.23) in Strovink’s notes: E = γmc2 , (54) If we plug in the numbers we ﬁnd that xf ≈ 1020 meters or 104 light years. This is just the ﬁrst leg of the journey, so the furthest distance the astronaut can reach is twice this, or 20,000 light years away! So the engineer was right... where m is the rest mass of the particles. Since γ = cosh(η ), the boost η is given This part is basically worked out in the text of the problem, so there’s nothing to by: say... E (55) η = cosh−1 mc2 (c) We start with dx = c · sinh(η )dτ = c · sinh where we use the expression cη = gτ. (50) (b) with from gτ dτ, c (49) Knowing from the problem that mc2 = 0.5 × 106 eV and Ef inal = 5 × 1010 eV, we can solve for η : η = 12.2 (56) Next we integrate the small displacements from 0 → τf where τf is the ﬁnal We can use the result obtained in problem 5, namely Eq. (53), replacing g some constant acceleration a. We also replace cosh(gτf /c) with γf , which “astronaut time.” part (a) we ﬁnd is γf ≈ 105 . This gives us: xf τf gτ dτ (51) dx = c · sinh c c2 0 0 xf ≈ γf a We can make a straightforward change of variable ξ = gτ /c: Solving for a and making the appropriate substitutions yields: c2 gτf /c xf = sinh(ξ )dξ. (52) a ≈ 3 × 1017 g g0 September 2, 1999 (57) (58) Physics H7C Fall 1999 (c) Solutions to Problem Set 1
(b) Derek Kimball We can use the relation between proper time dτ and time in the laboratory frame Now we use the inverse Lorentz transform: dt from problem 5: E/c cosh(η ) sinh(η ) 0 E /c (E/c)cosφ = sinh(η ) cosh(η ) 0 · (E /c)cosφ . dt = cosh(η )dτ. (59) (E /c)sinφ (E/c)sinφ 0 0 1 If we apply the relation cη = aτ , then integrating this expression yields: If we perform calculations similar to those in part (a), we ﬁnd: c c tlab = sinh(ηf ) = βf γf ≈ 10−5 s (60) a a E = E (cosh(η ) + sinh(η )cosφ ) = E cosh(η )(1 + βcosφ ). where tlab is the time interval in the lab frame. From cη = aτ we can quickly calculate the proper time interval: c τ = ηf ≈ 10−9 s a So, taking the ratio gives an “average” γ factor of 104 . Problem 7 (a) cosφ = (61) (c) and E cosφ + β (cosh(η )cosφ + sinh(η )) = . E 1 + βcosφ (66) (67) (68) Here, we can employ the relationship between energy and frequency of a photon, namely: E = hν, (69) We know that photons satisfy E 2 −p2 c2 = 0. Then, if we substitute the appropriate where h is Planck’s constant. Thus from Eq. (63) we solve for ν , ﬁnding the values from the problem into the equation describing the Lorentz transformation relativistic Doppler shift formula: for the fourmomentum (ignoring the zdirection), we ﬁnd: ν = νcosh(η )(1 − βcosφ). (70) cosh(η ) −sinh(η ) 0 E/c E /c (E /c)cosφ = −sinh(η ) cosh(η ) 0 · (E/c)cosφ . (62) If an observer knows only the frequency as observed in a given frame, one cannot (E /c)sinφ 0 0 1 (E/c)sinφ ﬁgure out what the frequency of light was in the rest frame of the source. Thus a measurement of light frequency in a particular frame does not directly tell us Solving for E gives us: about the velocity of the source. However, if we have prior knowledge of what the (63) frequency of light at rest should be (for example, wellknown atomic transitions in E = E (cosh(η ) − sinh(η )cosφ) = Ecosh(η )(1 − βcosφ). hydrogen or helium), we can tell something about the motion of the source. If we then ﬁnd the equation for px we can solve for cosφ : cosφ = E (cosh(η )cosφ − sinh(η )). E (64) Problem 8 This situation is physically reasonable, here is one example of how it could happen... Could the star be moving toward or away from us, even though the spectral features are not redshifted or blueshifted? The answer is yes, as can be seen from the relativistic Doppler shift given in Eq. (70). We demand that ν = ν , and then Substituting in the expression for E from Eq. (63) yields: cosφ = cosφ − β . 1 − βcosφ (65) September 2, 1999 Physics H7C Fall 1999 Solutions to Problem Set 1
γ−1 . βγ Derek Kimball ﬁnd a condition on the velocity of the source βc and the angle between β and the direction to earth φ: cosφ = (71) So long as this condition is satisﬁed, there is no restriction on the motion of the source (save that the source, if massive, cannot move at the speed of light!). Next we consider if some particular type of motion could increase the apparent velocity of the star across the sky. Once again the answer is yes. Consider the situation depicted in the ﬁgure to the right. Of course, the drawing is greatly exaggerated in dimensions since z D and t2 − t1 is diﬀerentially small, but hopefully it will give you the basic idea. Suppose the astronomer makes two measurements with which she determines the motion of the star across the sky. The star is moving toward the earth in this case, so it takes the light detected in the ﬁrst measurement longer to get to the earth. Suppose that the star gets closer to the earth by z between the times it emits the detected light. Then the time between the two light measurements on earth is: t2 − t1 = ∆t − z/c. (72) where ∆t is the time it takes the star to move to the new location in the earth frame. Then the apparent angular motion is given by: D D ∆θ dθ = . dt ∆t − z/c (73) So in fact (which is clear if you try some reasonable numbers), this apparent velocity can exceed the real velocity of the source by quite a bit, enough to make the star look like it’s going c or faster. There are real cases of this in astronomy... for example at the center of the galaxy there are stars whose apparent velocity greatly exceeds c (of course they’re redshifted and blueshifted all over the place, but you get the idea...)! September 2, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 2 1. A closely spaced circular parallel plate capacitor with long axial leads has a small (temporarily constant) current I passing through it (because the voltage across it is changing very slowly). The plates are perfectly conducting and have radius b. (a.) Use Gauss’s law to ﬁnd the time rate of change dE/dt of the electric ﬁeld within the plates. You may assume that the charge densities on the inside surfaces of the plates do not vary appreciably across those surfaces. (b.) Use the Amp`reMaxwell equation to ﬁnd the e magnitude and direction of the magnetic ﬁeld halfway between the plates, at a radius r < b from the axis. (c.) Use Amp`re’s law to evaluate the magnetic ﬁeld e in the vicinity of one of the long axial leads, far from the capacitor. Compare it to the answer for (b.). (d.) Suppose instead that I varies slowly. Far from the fringe of the capacitor, would you expect the electric ﬁeld to vary slightly with r? Explain. 2. (based on Purcell 10.14.) Consider three closely spaced parallel plate capacitors of the same square area and plate separation. The ﬁrst capacitor C1 consists only of those plates and vacuum. Both C2 and C3 are halfﬁlled with an insulating material having dielectric constant , but the dielectric is arranged in diﬀerent ways: C2 ’s dielectric extends over the full plate area, but ﬁlls only the half gap closest to one of the plates; C3 ’s dielectric extends from one plate to the other, but covers only half of the gap area. (The dielectric boundaries are always either parallel or perpendicular to the plates.) Calculate the capacitances C2 and C3 , expressed as a ratio to C1 . 3. (based on Purcell 10.23.) Consider an oscillating electric ﬁeld, E0 cos ωt, inside a dielectric medium that is not a perfect insulator. The medium has dielectric constant and conductivity σ . This could be the electric ﬁeld of some leaky capacitor which is part of a resonant circuit, or it could be the electric ﬁeld at a particular location in an electromagnetic wave. Work in SI units. Show that the Q factor, deﬁned by Q=ω energy stored , average power dissipated is ω/σ for this system, and evaluate it for seawater at a frequency of 1000 MHz. The conductivity is 4 (ohmm)−1 , and the dielectric constant may be assumed to be the same as that of pure water at the same frequency, ≈ 78 . 0 What does your result suggest about the propagation of decimeter waves through seawater? 4. (based on Purcell 10.24.) / 0 , ﬁlls A block of glass, refractive index n = the space y > 0, its surface being the xz plane. A plane wave traveling in the positive y direction through the empty space y < 0 is incident upon this surface. The electric ﬁeld in this wave is ˆ Ei sin (ky − ωt). There is a wave inside the z glass block, described exactly by E = ˆ E0 sin (ky − ωt) z B = x B0 sin (ky − ωt) . ˆ There is also a reﬂected wave in the space y < 0, traveling away from the glass in the negative y direction. Its electric ﬁeld is ˆ Er sin (ky + ωt). z Of course, each wave has its magnetic ﬁeld of 2 amplitude, respectively, Bi , B0 , and Br . The total magnetic ﬁeld must be continuous at y = 0, and the total electric ﬁeld, being parallel to the surface, must be continuous also. Show that this requirement, and the relation of B0 to E0 given in the equation B0 = √ µ0 E0 , suﬃce to determine the ratio of Er to Ei . When a light wave is incident normally on a vacuumglass interface, what fraction of the energy is reﬂected if the index n is 1.6? 5. (based on Purcell 11.11.) Write out Maxwell’s equations as they would appear if we had magnetic charge and magnetic charge currents as well as electric charge and electric currents. Invent any new symbols you need and deﬁne carefully what they stand for. Be particularly careful about + and − signs. Work in SI units. 6. (based on Purcell 11.17.) An iron plate 0.2 m thick is magnetized to saturation in a direction parallel to the surface of the plate. A “10 GeV/c” muon having momentum p, with pc = 1010 eV, moving perpendicular to the plate’s surface, enters the plate and passes through it with relatively little loss of energy. (This is possible because the muon, of mass m with mc2 ≈ 108 eV, is ≈ 200 times heavier than an electron, so it radiates 2002 times fewer photons.) Calculate approximately the angular deﬂection of the muon’s trajectory. Take the saturation magnetization of iron to be equivalent to 1.5 × 1029 electron magnetic moments per m3 (the electron magnetic moment is µB ≈ 6 × 10−5 eV per Tesla). 7. Fowles 1.4. 8. Fowles 1.6. Physics H7C Fall 1999 Solutions to Problem Set 2 Derek Kimball “The purpose of physics is to understand the universe... the purpose of mathemat (b) ics is, well, obscure to me...” The AmpereMaxwell equation, since there is no real (conduction) current J be Prof. Seamus Davis, UC Berkeley tween the plates of the capacitor, reduces to: If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! ∇ × B = µ0
0 ∂E ∂t (5) If you’re interested in the possibility of magnetic monopoles, you might want to Then using our result from part (a) and integrating (we choose an Amperian loop look up a paper by Blas Cabrera (Physical Review Letters, vol. 48, no. 20, centered on the z axis of radius r), we ﬁnd: 1982 pp. 137881), where the possible detection of a single magnetic monopole µ0 Ir2 is discussed. There have been no further monopoles detected since that time, so B · d = 2πrBφ = (6) this report remains unconﬁrmed. There is also an excellent discussion of magnetic b2 monopoles in J.D. Jackson’s Classical Electrodynamics. ˆ Thus we ﬁnd the magnetic ﬁeld in the φ direction to be: A discussion of the additional problem presented in discussion section this week can be found in a paper by Robert Romer (American Journal of Physics vol. 50, µ0 Ir . (7) Bφ = no. 12, 1982 pp. 108993). 2πb2 Problem 1 (a) (c) Far from the capacitor, there is no changing electric ﬁeld and therefore only conWe use Gauss’s law and choose a cylindrical surface of radius r centered on the axis duction current, so this is the familiar Ampere’s law: (we’ll call it z ) of the parallel plate capacitor, far from the edges of the capacitor ˆ B · d = µ0 Iencl , (8) (r b). Then: E · dA = Qencl
0 = κπr2 / 0 , (1) from which we ﬁnd a magnetic ﬁeld in the φ direction: Bφ = µ0 I . 2πr (9) where κ is the surface charge density of the capacitor. We ﬁnd directly from Eq. (1) that: E = (κ/ 0 )ˆ. z (2) Since there is a current I , the surface charge density changes with time by an amount: dκ I = 2, dt πb (3) which you will note is equivalent to Eq. (7) when r → b. Also note that inside the capacitor, the magnetic ﬁeld grows with r while far from the capacitor the ﬁeld falls as 1/r. (d) Let’s consider the electric ﬁeld in two diﬀerent regions. First, we’ll consider E far from the capacitor in the vicinity of one of the long axial leads (as in part (c)). where we assume the current is ﬂowing in the z direction. So from Eqs. (2) and The changing current produces a changing magnetic ﬁeld, and from Maxwell’s ˆ (3), we ﬁnd that: equations we know this creates an electric ﬁeld: dE = dt I z. ˆ 2 0 πb (4) September 2, 1999 E·d =− ∂B · dA. ∂t (10) Physics H7C Fall 1999 From Eq. (9), we see that
∂B ∂t Solutions to Problem Set 2
is given by: ∂B µ0 ∂I ˆ = φ. ∂t 2πr ∂t (11) Problem 2 Derek Kimball We choose an Amperian loop as indicated in Fig. 1. There is no electric ﬁeld perpendicular to the wire (along r). This can be deduced from symmetry considˆ A C= erations. Suppose there was an electric ﬁeld in the r direction. How does it know ˆ d whether to point in the +ˆ or −r direction? That has to be decided by either the r ˆ direction of the current or the change in current. If we reverse these quantities, where A is the area of the plates and d is the plate separation. So for C1 : the electric ﬁeld in the r direction should reverse. But on the opposite sides of the ˆ 0A wire, these quantities have opposite signs! The only way this can be true is if the C1 = d electric ﬁeld in the r direction is zero. ˆ Furthermore, we know that the electric ﬁeld must go to zero as r → ∞. But since E · d = 0, it must be the case that we have an electric ﬁeld in the z direction which ˆ varies with r. In other words, it is apparent that the electric ﬁeld is larger closer to the wire (z axis). This can be done explicitly, of course, from Eqs. (10) and (11): E (r2 ) − E (r1 ) = We can simplify the problem by thinking of C2 and C3 as two capacitors in series or in parallel, respectively (Fig. 1). The capacitance C of a parallel plate capacitor is given by: (15) (16) C2a
Axial Lead r1 ˆ φ r ˆ r2 Amperian Loop Figure 1 C2
z ˆ C2b C3 C3a C3b µ0 ∂I ln r1 /r2 . 2π ∂t (12) Let’s now consider the electric ﬁeld inside the capacitor, far from the fringe (as in part (b)). Once again we apply Eq. (10), but in this case: µ0 r ∂I ˆ ∂B = φ, ∂t 2πb2 ∂t (13) Figure 2
For C2 we break up the problem into two parts, solving for C2a and C2b (shown in Fig. 1), then determining C2 using: C2 = 1 1 + C2a C2b
−1 We see that there is also a component of the electric ﬁeld in the z direction which ˆ varies with r by utilizing similar arguments as those presented above: E (r ) = − µ0 r2 ∂I . 4πb2 ∂t (14) . (17) From Eq. (15) we can ﬁnd C2a and C2b , where: C2a =
0A = 2C1 d/2 (18) September 2, 1999 Physics H7C Fall 1999 and C2b = A =2 d/2
0 Solutions to Problem Set 2
where the current density is given by Ohm’s law: C1 . (19) J = σE. Once again taking the time average, we ﬁnd: (20) P=
2 σE0 . 2 Derek Kimball (28) So with a wee bit of algebra, we ﬁnd that: C2 = 2C1 . 0/ + 1 (29) Similarly for C3 , we break up the capacitor into two parts C3a and C3b , and then The Qfactor is the ratio of these two quantities, U and P , multiplied by the solve for C3 using: frequency: C3 = C3a + C3b . We use Eq. (15) to solve for C3a and C3b , ﬁnding: C3a = and C3b = A/2 = C1 . d 20 (23)
0 A/2 (21) Q= ω . σ (30) d = 1 C1 2 (22) If we plug in the numbers for seawater, we ﬁnd that Q ≈ 1.1. This suggests that decimeter waves cannot propagate very far in seawater, since the energy in the wave falls to 1/e its initial value in about one decimeter! Problem 4 First, we can write down the the electric and magnetic ﬁelds of the incident, transmitted and reﬂected waves: So here the overall capacitance is given by: C3 = C1 (/ 2
0 + 1). (24) z Ei sin (ky − ωt) ˆ xBi sin (ky − ωt) ˆ z E0 sin (k0 y − ωt) ˆ xB0 sin (k0 y − ωt) ˆ Problem 3 The energy per unit volume U stored in an electromagnetic wave is given by: U= 1 2 E2 + 12 B µ = E2. (25) If we then time average the energy, we ﬁnd that the average energy stored is:
2 U = E0 z Er sin (ky + ωt) ˆ xBr sin (ky + ωt) ˆ (31) cos2 (ωt)dt = 2 E0 . 2 (26) We note that k0 = nk since the transmitted wave is in glass. Then we can impose the condition √ B  =  µE  (32) The average power P dissipated per unit volume is given by the relation: P= J2 , σ (27) 1 on each of the waves, and demand that the Poynting vector, S = µ0 E × B , is along the direction of propagation of the waves. This ﬁxes the amplitudes and signs of September 2, 1999 Physics H7C Fall 1999 the magnetic ﬁelds with respect to the electric ﬁelds: z Ei sin (ky − ωt) ˆ √ x 0 µ0 Ei sin (ky − ωt) ˆ z E0 sin (k0 y − ωt) ˆ √ x µ0 E0 sin (k0 y − ωt) ˆ z Er sin (ky + ωt) ˆ √ −x 0 µ0 Er sin (ky + ωt) ˆ Solutions to Problem Set 2
Problem 5 Derek Kimball If there were magnetic charges, a magnetic charge density ρm and a magnetic current density Jm would appear in Maxwell’s equations. To avoid confusion, let’s denote the traditional electric charge density ρe and electric current density Je . We can place both of these, with some constants c1 and c2 which will be deﬁned later, in Maxwell’s equations to make them nice and symmetric: ∇ · E = ρe / (33)
0 ∇ · B = c1 ρm ∂B + c2 Jm ∂t ∂E ∇ × B = µ0 0 + µ0 Je ∂t ∇×E =− Now we consider the ﬁelds at y = 0, the interface between the block of glass and vacuum. We require that the electric and magnetic ﬁelds parallel to the surface of the glass satisfy: E =E B B = . µ µ After substitution, this leaves us with two equations: −Ei + Er = −E0 Ei + E r =
0 (38) (34) We can go further and work out a relationship between magnetic charge density and current density. We begin by demanding that magnetic charges and currents satisfy the continuity equation, namely: ∇ · Jm + ∂ρm = 0. ∂t (39) E0 . (35) Then if we take the divergence of the new third Maxwell’s equation, we get: ∇ · ∇ × E = −∇ · ∂B + c2 ∇ · Jm . ∂t (40) We can then eliminate E0 from these equations yielding the ratio of Er to Ei : Er = Ei / /
0−1 0 There is a vector derivative rule that states for any vector ﬁeld A, ∇ · ∇ × A = 0. The energy is proportional to E 2 (as can be readily seen by considering the Poynt So the lefthand side of (40) is 0. The derivatives on the right hand side, ∇ and ∂ , can be swapped and we get: ing vector S ), and in this case the index of refraction n = / 0 . Thus the ratio ∂t of reﬂected to incident energy Ur /Ui is given by: ∂ − ∇ · B + c2 ∇ · Jm = 0. (41) 2 2 ∂t Er n−1 Ur = = . (37) Ui Ei n+1 From the second Maxwell equation we know that ∇ · B = c1 ρm , so we ﬁnd: For n = 1.6, 5% of the energy is reﬂected. ∂ρm + c2 ∇ · Jm = 0. (42) −c1 ∂t If we then apply the continuity equation, Eq. (39), we ﬁnd that c1 = −c2 ≡ c. September 2, 1999 +1 . (36) Physics H7C Fall 1999 Thus the ﬁnal form of Maxwell’s equations is: ∇ · E = ρe /
0 Solutions to Problem Set 2 Derek Kimball ∇ · B = cρm ∇×E =− ∂B − cJm Anyhow, in this case it’s no problem, we ﬁnd that: ∂t ∂E (46) B = µ0 M . ∇ × B = µ0 0 + µ0 Je , (43) ∂t where c is a constant of proportionality between the magnetic charge unit and the inside the iron plate and zero outside the plate. magnetic ﬁeld it produces (the equivalent of 1/ 0 for electric ﬁelds). This problem now reduces to the traditional problem of solving for the Problem 6 cyclotron orbit of a moving charged particle in a magnetic ﬁeld. An imIron Plate The ﬁrst part of this problem is to portant diﬀerence, as pointed out by M calculate the magnetic ﬁeld B inside Paul Wright in section (thanks!), is Iron Plate the magnetized iron. We can use the that in this case we need to be careauxiliary ﬁeld H to make our job a ful about relativistic corrections to little easier. We know that: the radius of the cyclotron orbit. B To ﬁnd the radius of the cyclotron H= − M. (44) θ µ0 ←d→ orbit R, we balance the Lorentz force qvB with the relativistic centrifugal Also, we have the relation: force γmv 2 /R. This tells us: H · d = If ree , (45) γmv . (47) R= qB Amperian Loop where in our problem If ree = 0 evθ where R is the radius of the circular erywhere. We choose an Amperian orbit. If you take a look at Fig. 4, loop as pictured in Fig. 3 (M is in hopefully the simple geometric arguthe z direction), taking advantage of ˆ ments suggested convince you that the planar symmetry of the problem θ in fact: (we can assume the iron plate is inﬁR z ˆ nite). Since the component of H perdqBc dqBc R sin θ = d/R = = , pendicular to the surface of the iron γmvc pc plate must be zero based on symme(48) y ˆ try, and outside the iron plate H → 0 where θ is the deﬂection angle and as y → ±∞, we conclude that in fact Figure 4 d is the thickness of the plate. The x ˆ H = 0 everywhere. rest of the problem is working out I would like to pause here and point the correct units... out that this conclusion is not enFigure 3 tirely trivial. If there is no free curFirst let’s get B in SI units. B = µ0 M = 4π × 10−7 N/A2 ·1.5 × 1029 electron rent If ree , that does not necessarily magnetic moments per m3 ·9 × 10−24 J/T, or about 1.7 T. Then dqcB = 108 eV, September 2, 1999 mean that H = 0 everywhere. The fundamental reason for this is that in order to completely determine a vector ﬁeld you must know both its curl and divergence. Only in cases where we have planar, cylindrical, toroidal or solenoidal symmetry can we conclude that ∇ · H = 0, and get H quickly. This is diﬀerent from Ampere’s law with B where we always know ∇ · B = 0. So, be careful when using H ! Physics H7C Fall 1999 so sin θ ≈ θ = a deﬂection... Problem 7 Fowles 1.4 The 3D wave equation is: ∇2 f = 1 ∂2f u2 ∂t2
dqBc pc Solutions to Problem Set 2
−2 Derek Kimball = 10 rad. That’s only about half a degree, so not too big of where k = 2π/λ is the wave vector. We can also express ω in terms of u and λ: 2πu . λ If you take the derivative of ω with respect to λ: ω = ku = dω 2πu 2π du =− 2 + . dλ λ λ dλ Now we calculate (49)
dλ dk : (55) (56) λ2 dλ =− . (57) dk 2π We employ spherical coordinates, and since our wavefunction is a function only of If we then substitute the expressions in Eqs. (56) and (57) into Eq. (54), we arrive at our result: r, ∇2 is also a function only of r: ∂u (58) ug = u − λ . ∂f 1∂ ∂λ ∇2 f = 2 r2 . (50) r ∂r ∂r (b) Plugging in the spherical harmonic wavefunction f = 1 eı(kr−ωt) , we get: r We use similar tricks to derive the result: 2 1 1 λ0 dn k =− . (59) (51) ∇2 f = − eı(kr−ωt) = −k 2 f. ug u c dλ0 r If we evaluate the righthand side of Eq. (49), and use the fact that k = ω/u, we ﬁnd that: 1 ∂2f = −k 2 f. u2 ∂t2 which veriﬁes that f is a solution to the 3D wave equation. Problem 8 Fowles 1.6 (a) Let’s begin by deriving ug = u − λ ∂u . ∂λ (53) and dλ0 λ2 =− 0 . (63) dω 2πc Substituting these results back into Eq. (60) gives us the answer we were looking for: 1 1 λ0 dn =− . (64) ug u c dλ0 That’s all folks! (52) We begin by noting that dk dλ0 dk 1 = = · ug dω dλ0 dω Let’s write the wave vector in terms of λ0 and n: 2πn k= λ0 We can then take some derivatives, and ﬁnd that: 2πn 2π dn dk =− 2 + dλ0 λ0 λ0 dλ0 (62) (60) (61) We can begin by using Fowles (1.33), the deﬁnition of the group velocity: ug = dω dλ dω = · , dk dλ dk (54) September 2, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 3 1. (based on Purcell B.1.) An electron of rest mass me and charge e, moving initially at a constant velocity v , is brought to rest with a uniform deceleration a that lasts for a time t = v/a. Compare the electromagnetic energy radiated during the deceleration with the electron’s initial kinetic energy. Express this ratio in terms of two lengths: the distance that light travels in time t, and the classical electron radius r0 , deﬁned as r0 ≡ e . 4π 0 me c2
2 the dimensions of area and is called a scattering total cross section. The energy radiated, or scattered, by the electron, and thus lost from the plane wave, is equivalent to the energy falling on an area σ . (The case considered here, involving a free electron moving nonrelativistically, is often called Thomson scattering, after J.J. Thomson, the discoverer of the electron, who ﬁrst calculated it.) 3. (based on Purcell B.4.) The master formula Prad = 1 2 e2 a2 . 4π 0 3 c3 To carry out this calculation, you need a formula like Purcell Eq. (B.6) that relates the instantaneous radiated power Prad to the instantaneous acceleration a. In SI units, this formula is Prad = 1 2 e2 a2 . 4π 0 3 c3 (Note that, as far as one has been able to tell experimentally, the electron actually is consistent with having zero radius, and it must have a radius at least several orders of magnitude smaller than the “classical radius” r0 .) 2. (based on Purcell B.3.) A plane electromagnetic wave with angular frequency ω and electric ﬁeld amplitude E0 is incident on an isolated electron. In the resulting sinusoidal oscillation of the electron, the maximum acceleration is eE0 /m, where e is the electron’s charge. Averaged over many cycles, how much power is radiated by this oscillating charge? (Note that, when the maximum acceleration of the electron rather than its maximum amplitude is held ﬁxed, the power radiated is independent of the frequency ω .)
2 Divide this average radiated power by 0 E0 c/2, the average power density (per unit area of wavefront) in the incident wave. The quotient σ has is useful for particles moving relativistically, even though v c was assumed in Purcell’s derivation of it. To apply it to a relativistic situation, all we have to do is (i) transform to a comoving inertial frame F in which the particle in question is, at least temporarily, moving slowly; (ii) apply the master formula in that frame; and (iii) transform back to any frame we choose. Consider a highly relativistic electron (γ 1) moving perpendicular to a magnetic ﬁeld B. It is continually accelerated (in a direction perpendicular both to its velocity and to the ﬁeld), so it must radiate. At what rate does it lose energy? To answer this, transform to a frame F moving momentarily along with the electron, ﬁnd E in that frame, and thereby ﬁnd Prad . Now show that, because power is energy/time, Prad must be equal to Prad . This radiation generally is called synchrotron radiation. It is both a blessing and a curse. The blessing is that intense beams of UV and Xray photons are created at synchrotrons designed for that purpose, such as Berkeley Lab’s Advanced Light Source. These beams are essential for 2 many studies and uses such as semiconductor lithography. The curse is that synchrotron radiation prevents circular electron accelerators of practical size (up to tens of km in circumference) from exceeding about 1011 eV in energy, much weaker than the 1012 eV proton beams that have been available at Fermilab for a decade. 4. Fowles 2.4. The solution to this problem was sketched in lecture on 7 Sep. In Fowles’ notation, E0 and B0 are the same as the E1 and B1 discussed in class. 5. Fowles 2.7. 6. Fowles 2.10. 7. (a.) For an ideal linear polarizer with its transmission axis at an arbitrary angle φ with respect to the x axis, calculate the Jones matrix. (As usual, the beam direction is z , φ is an angle in the xy plane, and φ is positive as one rotates from x toward y .) (b.) For a linear polarizer, show that its Jones matrix M is not unitary, i.e. M∗ = M−1 j i . This ij means that the action of the wave plate violates timereversal invariance. This makes sense because, for general polarization, the irradiance of a light beam is reduced after passing through the plate. 8. A wave plate is made out of a birefringent crystal whose lattice constants are diﬀerent in the “fast” and “slow” directions of polarization. This leads to diﬀerent indices of refraction for the two polarizations. If the x axis is along the “slow” direction of the plate, x polarized light accumulates a phase shift δ with respect to light polarized in the “fast” or y direction, with δ= ωD (nx − ny ) . c Here nx > ny if the x direction is “slow”, ω is the (ﬁxed) angular frequency of the light, and D is the thickness of the plate. Because the absolute phase of the light is of no experimental interest, the eﬀect of the wave plate is equivalent to multiplying the x component of (complex) E1 by exp (iδ/2) and the y component by exp (−iδ/2). (a.) Write the Jones matrix for the ideal wave plate just described. (b.) Calculate the Jones matrix for the general case in which the “slow” plate axis lies at an angle φ from the x axis, where, as in Problem 7, φ is an angle in the xy plane. (c.) For the wave plate in (b), show that its Jones matrix M is unitary, i.e. M∗ = M−1 j i . This ij means that the action of the wave plate is timereversal invariant. For any polarization, the irradiance of a light beam is unaﬀected by traversing the plate (though its polarization may change dramatically). Physics H7C Fall 1999 Solutions to Problem Set 3 Derek Kimball “Niels Bohr (not for the ﬁrst time) was ready to abandon the law of conservation of energy. It is interesting to note that Bohr was an outspoken critic of Einstein’s light quantum (prior to 1924), that he discouraged Dirac’s work on the relativistic electron theory (telling him, incorrectly, that Klein and Gordon had already succeeded), that he opposed Pauli’s introduction of the neutrino, that he ridiculed Yukawa’s theory of the meson, and that he disparaged Feynman’s approach to quantum electrodynamics.”  Prof. David Griﬃths, Reed College, excerpted from Introduction to Elementary Particles I believe that the answer is frame dependent (as is the case for most relativity paradoxes). If you’re freefalling with the particle, you don’t see any radiation. If the charge is accelerating with respect to you, then you see radiation. This can be shown with the general relativistic ﬁeld transformations. What about energy conservation? Well, I’m no Niels Bohr, so I think that if you change your acceleration into the frame of the particle, everything works out... but to prove this seems a bit complicated... anyhow, good stuﬀ to think about, keep up the great work! Thanks! Problem 1 If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! First we’ll calculate the electromagnetic energy radiated Prad ∆t during the decelAn interesting (to me) point concerning the physical meaning of orthogonal po eration lasting for ∆t = v/a, which is given by: larization was raised after discussion section the other day. Fowles says that two waves E1 and E2 whose complex electric ﬁeld amplitudes satisfy: E1 · E∗ = 0 2 are orthogonally polarized. (1) Prad ∆t = 1 2 e2 av . 4π 0 3 c3 (2) The electron’s initial kinetic energy K is just: 1 K = mv 2 , (3) For linear polarization, there is a simple geometric analogy. Linearly polarized 2 light (e.g., in the x direction) is orthogonal to light with a perpendicular linear ˆ polarization (e.g., in the y direction). So no light will get through two linear and the ratio Prad ∆t/K is given by: ˆ polarizers which are “orthogonal” in the Euclidean geometry sense. However, a Prad ∆t 4 1 e2 this picture breaks down for more complicated polarization states, e.g. circular = . (4) 2 K 3 4π 0 mc cv polarizations. For example, two circular polarizations whose electric ﬁelds are always at right angles to each other are not orthogonal! The distance d traveled by light in ∆t is cv/a, and the classical radius of the elecOrthogonality for polarization states can be understood using notions from lin tron r0 is given by the formula in the problem set: r0 ≡ e2 / 4π 0 mc2 . Therefore: ear algebra. In this sense the complex vector space of polarization states can be Prad ∆t 4 r0 spanned by two linearly independent, complex vectors  any two linearly indepen= . (5) K 3d dent complex vectors are said to be orthogonal. This is what Fowles means when he says two polarization states are orthogonal. Here’s an interesting fact that might help you remember some important lengths Another interesting question raised after discussion section, although it is a bit in physics: beyond the scope of the course, was whether or not a freefalling (in a gravitational ﬁeld) charged particle radiates. You would expect that it might not based on the One of the most important constants in physics is the ﬁnestructure constant α, equivalence principle, which basically states that a freefalling frame is equivalent which sets the strength scale for the electromagnetic force: to an inertial frame. However, if you observe the charge from the surface of some 1 e2 planet, you would see a charged particle undergoing acceleration. This is a little ≈ , (6) α= c 137 funny, since you would expect that the charge either loses energy or it doesn’t... After thinking about it a little and consulting some wise general relativity texts where e is in CGS units. You may recognize the upper division physics course, (e.g., Wald’s General Relativity or Misner, Thorne and Wheeler’s Gravitation), Physics α−1 . September 16, 1999 Physics H7C Fall 1999 Solutions to Problem Set 3 Derek Kimball You can guess the the classical radius of the electron by setting the rest energy of the electron equal to the potential energy stored in a spherical shell of radius r0 Problem 3 with charge e on the surface. We transform to a comoving inertial frame F in which the electron is temporarily e2 mc2 = . (7) at rest. The electric and magnetic ﬁelds in F are given by: r0 In a few weeks, you’ll learn about Compton scattering (photonelectron scattering). The Compton wavelength of the electron is given by r0 /α. The Bohr radius, the radius of an electron’s orbit in the hydrogen atom, is given by r0 /α2 . Problem 2 The electron oscillates sinusoidally with the acceleration given by: a= The power Prad radiated is given by: Prad =
2 1 2 e4 E0 sin2 ωt, 4π 0 3 c3 m2 E⊥ = γ E⊥ + cβ × B 1 B⊥ = γ B⊥ − β × E c E =E B =B . (12) eE0 sin ωt. m (8) The force on the electron is the Lorentz force given by: F =e E+v×B , (13) (9) but in the comoving frame F the electron’s velocity is zero. Since in the lab frame F the electric ﬁeld is zero, the force acting on the electron in F is: F = eE⊥ = eγcβB. (14) and of course if we average over many cycles... Prad =
2 1 1 e4 E0 . 4π 0 3 c3 m2 (10) Thus, the acceleration a is: a= eγcβB m (15) Next if we divide this result by the average power density U in the incident wave we get the scattering cross section σ : This acceleration can be used in our old pal which describes the power radiated: σ= e4 1 2 m2 c4 . 6π 0 (11) Prad = 1 2 e2 a2 , 4π 0 3 c3 (16) 2 You might notice that σ = (8/3)πr0 , r0 being the classical electron radius from which gives us: problem 1... Prad = 1 2 e4 γ 2 β 2 B 2 . 4π 0 3 m2 c (17) When we transform back to frame F , the energy transforms as ∆E → γ ∆E and the time transforms as ∆t → γ ∆t. Therefore, since power is just ∆E/∆t, Prad = Prad . September 16, 1999 Physics H7C Fall 1999 Problem 4 Fowles 2.4 Solutions to Problem Set 3
This is pretty straightforward. Here’s the prescription: Given the electric ﬁeld of the wave, E , in the form: Derek Kimball For this problem we employ the complex exponential form of the wave functions for E and H : E = Re E0 e H = Re H0 e The Poynting vector is given by: S = E × H. Using the expressions from Eq. (18), we ﬁnd the Poynting vector is: S = Re E0 ei(k·r−ωt) × Re H0 ei(k·r−ωt) (20) (19)
i(k·r−ωt) i jbe E = E0 ˆ + ˆ iθ ei(k·r−ωt) , the Jones vector is given by: Ex Ey = E0 1 beiθ . (24) i(k·r−ωt) (25) . (18) You can normalize the Jones vector if you want, but if you didn’t feel like doing that in this problem that’s okay too. So pretty much we can just write down the answers, here they are: (a) √
1 √ 2 1 √ 2 We can expand the exponentials in terms of sines and cosines and ﬁnd the real parts: S = Re(E0 ) cos (k · r − ωt) − Im(E0 ) sin (k · r − ωt) × Re(H0 ) cos (k · r − ωt) − Im(E0 ) sin (k · r − ωt) .
2 Ex Ey = 2E0 . (26) (b) (21) Ex Ey = √ 5E0
1 √ 5 2 √ 5 If we expand this expression and time average (i.e. we set sin (k · r − ωt) and cos2 (k · r − ωt) equal to 1 and sin (k · r − ωt) cos (k · r − ωt) equal to 0), then we 2 get: S= 1 Re(E0 ) × Re(H0 ) + Im(E0 ) × Im(H0 ) . 2 (c) (22) . (27) This expression, by inspection, is equivalent to: 1 ∗ S = Re E0 × H0 , 2 which veriﬁes the claim. Problem 5 Fowles 2.7 September 16, 1999 (23) (d) Ex Ey = √ 2E0 1 √ 2 − √i 2 . (28) Ex Ey = 2E0 2 2 1+i 2 √ . (29) Physics H7C Fall 1999 Problem 6 Fowles 2.10 We’ll start with an arbitrary polarization: Ex Ey = a beiθ . Solutions to Problem Set 3
(b) Derek Kimball
∗ A quick way to check if the matrix is unitary is to multiply M by MT if it equals the identity matrix: M · MT (30)
∗ and see = cos2 φ sin φ cos φ = sin φ cos φ sin2 φ cos2 φ sin φ cos φ . sin φ cos φ sin2 φ (36) cos2 φ sin φ cos φ sin φ cos φ sin2 φ Now we’ll send it through a linear polarizer. Let’s orient the linear polarizer at Well that’s not the identity matrix, so the Jones matrix of a linear polarizer is not 45o , so our resultant polarization is given by: unitary. Ex Ey = 1 2 11 11 a beiθ = 1 2 a + beiθ a + beiθ = a + beiθ 2 1 1 (31) Problem 8 We can ignore the amplitude out front. Now we’ll sent it through a quarterwave (a) plate with the fast axis horizontal. Ex Ey = 10 0i 1 1 = 1 i The Jones matrix for the ideal wave plate is: , (32) M= (33) eiδ/2 0 0 e−iδ/2 . (37) which is indeed circular polarization! What happens if we change the order? Ex Ey = 1 2 11 11 10 0i a beiθ = a + beiθ 2 1 1 , (b) which is linear polarization. So circular polarized light is created only by placing For the general case we just do the matrix multiplication as in problem 7 (a): the optical elements in the proper order. cos φ − sin φ eiδ/2 0 cos φ sin φ M= sin φ cos φ − sin φ cos φ 0 e−iδ/2 Problem 7 1 + eiδ sin φ cos φ eiδ cos2 φ − sin2 φ = e−iδ/2 . (38) iδ − 1 + e sin φ cos φ cos2 φ − eiδ sin2 φ (a) For this problem, we can use the technique for changing the basis of a matrix (c) operator that we employed in the ﬁrst problem set to ﬁnd Lorentz transforms in Now we check for unitarity just as in problem 7 (b): rotated frames (see PS1 solutions, problem 2). Namely, M = R−1 MR. (34) M · (M ) eiδ/2
T∗ = e−iδ/2 where M is the Jones matrix and R is the appropriate rotation matrix. For a linear polarizer with transmission axis at an arbitrary angle φ: M= cos φ − sin φ sin φ cos φ 10 00 cos φ sin φ − sin φ cos φ = eiδ cos2 φ − sin2 φ − 1 + eiδ sin φ cos φ 1 + eiδ sin φ cos φ cos2 φ − eiδ sin2 φ = 10 01 · (39) e−iδ cos2 φ − sin2 φ − 1 + e−iδ sin φ cos φ 1 + e−iδ sin φ cos φ cos2 φ − e−iδ sin2 φ cos2 φ sin φ cos φ . So the Jones vector for the waveplate is a unitary operator! sin φ cos φ sin2 φ (35) Bye for now! September 16, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 4 1. A plane wave polarized in the x direction is norˆ mally incident upon an ideal quarterwave plate which has its slow axis oriented at +45◦ with respect to the x axis. Next it is reﬂected at norˆ mal incidence by a perfectly conducting mirror. Finally the wave passes back through the same quarterwave plate. After completing this journey... (a.) What is the ﬁnal wave’s irradiance, relative to the incident wave? (b.) What is the ﬁnal wave’s state of polarization? (If linearly polarized, specify the polarization direction; if circularly polarized, state whether rightor lefthanded.) Both of your answers should be justiﬁed. 2. A ﬁberoptic cable consists of a cylindrical core with refractive index n1 surrounded by a sheath with refractive index n2 , where 1 < n2 < n1 . The ends of the cable are cut perpendicular to the cable axis and polished. A point source of light is placed on the cable axis a negligible distance away from one end. What fraction of the total light emitted by the point source is transmitted by the core to the (distant) end of the cable? (You may ignore losses due to reﬂection at the cable ends.) 3. A vector ﬁeld F(r) is equal to the curl of a vector potential A which is a continuous function of r: F(r) = ∇ × A(r). Prove that F⊥ is continuous across any surface, where “⊥” refers to the component of F which is perpendicular to the surface. 4. A transverse electromagnetic wave in vacuum is normally incident on a semiinﬁnite slab of electrically insulating material which has = 0 but µ = κm µ0 . That is, the material has trivial dielectric but nontrivial magnetic properties. Consider the ferromagnetic limit κm → ∞. In that limit, what is the ratio of peak electric ﬁeld E max in the transmitted wave to the same quantity Emax in the incident wave? 5. Plane waves propagating in the ±z directions bounce between two semiinﬁnite nonconducting materials. The ﬁrst material occupies the region z < 0, and the second occupies z > L, where L is a positive ﬁxed distance. These (hypothetical) materials have no unusual magnetic properties (µ = µ0 ), but they have inﬁnite dielectric constant, / 0 = ∞. (a.) What components of E and H vanish, and where do they vanish? Justify each answer that you give. (b.) What angular frequencies for the light are possible? 6. A nonrelativistic particle of mass m in one (x) dimension can be represented by a (complex) wavefunction Ψ(x, t) ∝ exp (i(kx − ωt)) where h ¯ 2 k 2 (x) = 2m(E − V (x)), E = hω is the total and V (x) is the potential ¯ energy, and 2π ¯ is Planck’s constant. If V (x) h is piecewise ﬂat, at the discontinuities in V the Schr¨dinger equation demands that both Ψ and o ∂ Ψ/∂x remain continuous. Consider the reﬂection of a particle of initial kinetic energy T by a potential barrier of height ∆V < T . Show that the ratio of reﬂected to incident amplitudes is given by the same formula 2 as for the normal reﬂection of an electromagnetic wave at a dielectric interface with µ = µ0 everywhere, provided that the refractive index in the quantum mechanical case is taken to be proportional to k . 7. Fowles 3.1. 8. Fowles 3.6. Physics H7C Fall 1999 “Physicists are like 3% of rats.” Max Zolotorev, Lawrence Berkeley National Laboratory Solutions to Problem Set 4 Derek Kimball This result can also be arrived at using Fowles’s reﬂection matrix (page 52). The beam now passes back through the quarter waveplate, but now the wave sees the fastaxis oriented at +45o . So we ﬁnd that: Ex Ey 1 =√ 2 1i i1 E0 ·√ 2 1 i = −iE0 0 1 (4) If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! Problem 1 (a) As demonstrated in last week’s problem set (problem 8), an ideal quarterwave plate is described by a unitary Jones matrix. This means that the irradiance of the light beam is unaﬀected by traversing the plate. Also, the mirror is a perfect conductor, so 100% of the light is reﬂected. So the ﬁnal wave’s irradiance must be the same as the incident wave’s. In other words the resultant light is linearly polarized in the y direction. ˆ Problem 2 First let’s derive Fowles’ result regarding the acceptance angle α for a ﬁberoptic cable (pages 4647). At the ﬁrst phase transition as the light enters the ﬁberoptic n2 cable, we have from (b) Snell’s law: θ α The wave, initially polarized along the x direction, ﬁrst passes through the quarter ˆ β sin α = n1 sin β. (5) n1 wave plate, whose fast axis is oriented at 45o with respect to the initial light polarization: We want θ = π/2 − β to be greater than or equal to the critn2 ical angle, sin−1 n Then the light bounces oﬀ a perfectly conducting mirror. This reverses the sign of for total internal reFigure 1 the Poynting vector, which in turn changes the sign of the Bﬁeld relative to the ﬂectance, where n = n2 /n1 . With a little Eﬁeld, since (as can be shown from Maxwell’s equations): trigonometry it can be shown that for these conditions, i(k·r −ωt) ˆ E = E0 η e n2 − n2 1 2 1ˆ sin β = . (2) B = k × E, n1 v Ex Ey 1 =√ 2 1 −i −i 1 E0 0 E0 =√ 2 1 −i (1) where v = c/n is the phase velocity of light in a medium and η is the direction of ˆ the light’s electric ﬁeld. The light electric ﬁeld undergoes a phase shift of π upon relection (as can be deduced from the boundary conditions), but for circularly polarized light the direction of rotation (clockwise or counterclockwise) of the electric ﬁeld with respect to a ﬁxed coordinate system is preserved. However, we are now viewing it from the opposite direction (since k changed sign). Therefore the handedness of polarization has changed upon reﬂection: Ex Ey E0 =√ 2 1 i (3) September 23, 1999 Combining Eqs. (5) and (6) yields: sin α = n2 − n2 , 1 2 (7) (6) which proves Fowles’s assertion. The next step is to compute the solid angle of light accepted by the ﬁberoptic cable, given by:
2π α 0 ∆Ω =
0 sin θdθdφ = 2π (1 − cos α). (8) Physics H7C Fall 1999 Solutions to Problem Set 4
(1 − cos α) . 2 Derek Kimball Divided by the total solid angle (4π ), this yields the fraction T of the total light where Er , Ei , Et are the reﬂected, incident and transmitted electric ﬁeld amplitransmitted by the core to the end of the cable: tudes, respectively, and T= Problem 3 A vector ﬁeld F(r) is equal to the curl of a vector potential A(r), so we know that the divergence of F(r) is zero: ∇ · F(r) = ∇ · (∇ × A(r)) = 0. Then we know that: ∇ · F(r)dV = F(r) · dA = 0 (11) (10) Problem 5 (a) Plane waves propagating in the ±z directions must satisfy Eqs. (15) at the interfaces (z = 0 and z = L). At either interface we have, since Z2 → 0, 2Z2 Et = → 0. Ei Z1 + Z2 (18) (9) Z1 , 2 = µ1,2
1, 2 . (16) In particular, for the ferromagnetic material described, Z2 → ∞ while Z1 is ﬁnite, so: Et 2Z2 = → 2. Ei Z1 + Z2 (17) Choose a volume of vanishing thickness δ about a surface of area A (where δ is always normal to the surface), then from Eq. (11) we have that: F⊥ r + δ A − F⊥ (r)A + O(δ ) = 0 (12) where O(δ ) indicates a term of order δ , which arises from some ﬁnite amount of Therefore E vanishes in the material, i.e. E=0 for z < 0 and z > L. So at the “ﬂux” of F(r) out the sides of the volume. interfaces, E=0. The condition for continuity of F⊥ (r) is that for each > 0, there exists a δ > 0 The magnetic ﬁelds of the plane waves propagating in the ±z directions between such that F⊥ r + δ − F⊥ (r) < . From Eq. (12) we know that: the materials must satisfy E+ ×H+ = E− ×H− (i.e. the Poynting vectors of right and lefttraveling waves must be oriented in opposite directions). So while O(δ ) . (13) the electric ﬁelds cancel (E+ = E− ) at the interfaces the magnetic ﬁelds must F⊥ r + δ − F⊥ (r) = A add (H+ = H− )! Also we have the boundary condition H = H , so that if H is Given any > 0, clearly we can choose δ to make: ﬁnite on one side of the interface, it must also exist on the other side. So there can be components of H everywhere. O (δ ) <. (14) A (b) Therefore F⊥ (r) is continuous. Our requirements from part (a) set up a standing wave, where the components of E and H are π out of phase. The wave is time independent in order to assure that Problem 4 E=0 at the interfaces for all times t, so we can postulate: From Strovink’s treatment of reﬂection/refraction at a plane interface between (19) E = E0 sin kz, insulators, we have for normal incidence: which works so long as k = N π/L where N is an integer. So we get the condition Er Z2 − Z1 = on angular frequency from k = ω/c, which implies E Z +Z
i 1 2 Et 2Z2 = , Ei Z1 + Z2 (15) September 23, 1999 ω= N πc . L (20) Physics H7C Fall 1999 Problem 6 Solutions to Problem Set 4 Derek Kimball P r We have two regions as shown in Figure 2, with k1 and k2 in each (deﬁned as in the problem, where they are dependent on the potential V and the particle’s total energy U, which is conserved, U = T + V). In region 1 we have the wavefunctions: Aei(k1 x−ωt) + Be−i(k1 x+ωt) , and in region 2 we have a transmitted wavefunction: Cei(k2 x−ωt) . Continuity of the wavefunctions across the boundary (x=0) demands: A + B = C. Since
∂ψ ∂x (21) a r+asin θ a r+2asin θ θ (22) V (x) (23) U=T+V Figure 3 The sum of the amplitudes of the waves A coming from the slits at point P (see Figure 3) are given by the proportionality: Region 2 x or, (25) Figure 2 A ∝ eikr 1 + eika sin θ + e2ika sin θ
8 is also continuous: (24) Region 1 ∆V k1 (A − B ) = k2 C. Substituting Eq. (24), we get: (23) into A ∝ eikr + eik(r+a sin θ) + eik(r+2a sin θ) (27) k1 − k2 B = A k1 + k2 (28) If we assume n ∝ k , then we get the formula for normal reﬂection of an EM wave at a dielectric interface with µ = µ0 : n1 − n2 B = A n1 + n2 (26) 6 4 2 Problem 7
0.001 0.0005 0.0005 0.001 Fowles 3.1 September 23, 1999 Figure 4 Physics H7C Fall 1999 Solutions to Problem Set 4
I (θ) ∝ A2 ∝ (1 + cos(ka sin θ))
2 Derek Kimball The interference pattern is given by the norm square of the amplitude: (29) The pattern that you get when you plot this function depends on what value you choose for ka. Let’s take a = 1 mm and k = 12,000 mm−1 , then our pattern is shown in Figure 4 as a function of θ. Problem 8 Fowles 3.6 Light passes through the gas cell twice, so the optical path diﬀerence dop is given by: dop = c∆t = 2l 2l − = 2l(n − 1) c/n c (30) n changes as the gas ﬁlls the cell, and since I ∝ 1 + cos (2πdop /λ), a new fringe appears every time dop = 1/2. Thus the total number of fringes N is given by: N =2 4l(n − 1) 2l(n − 1) = . λ λ (31) Plugging in the suggested values gives us N = 203 fringes. September 23, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 5 1. It is known that, in Region 1 (y > 0, 1 = 4 0 , µ1 = µ0 ), there exists a plane wave propagating in the x direction. ˆ a. What is µ2 in Region 2 (y < 0, 2 = 0 )? b. What state(s) of polarization (E along y, ˆ, or ˆz both) can the plane wave be in? Why? 2. Fowles 2.23. Substitute “interface” for “window” – that is, consider just one glassair interface, not two. The “degree of polarization” is deﬁned by Fowles’ Eq. (2.26). 3. A beam of light travelling in the plane y = 0 is refracted by the interface z = 0 between two insulators: vacuum (z < 0) and a material (z > 0) with µ = µ0 and with dielectric constant , where > 0 . In the semiinﬁnite region z < 0, the angles of incidence and reﬂection, with respect to the z axis, are 60◦ . In the semiinﬁnite region z > 0, the angle of refraction, with respect to the same axis, is 30◦ . (a.) Taking the reﬂected and refracted angles to be as given, calculate / 0 for the material. (b.) The incident beam is righthand circularly polarized. What is the state of polarization of the reﬂected beam? Explain your answer. (c.) Calculate the ratio R of irradiances R= Ireﬂected . Iincident 6. Fowles 4.4. 7. A camera lens is purplish because it is optically coated to minimize reﬂection at the center of the visible spectrum. (The coating parameters therefore are not optimized for red or blue light.) Consider a plane EM wave in vacuum with wavelength λ normally incident on a semiinﬁnite piece of glass with refractive index n > 1 and unit permeability µ/µ0 = 1. Choose the thickness and the refractive index of a coating on the glass in order to force the reﬂected wave to vanish. 8. Show that the matrix equation (Fowles 4.24) for a singlelayer ﬁlm in fact is an equation that merely transforms the total (complex) ET and HT just to the right of the right hand interface to the total (complex) E0 and H0 just to the left of the left hand interface. Show, therefore, that, for a multilayer ﬁlm, the overall transfer matrix is equal to the product of the transfer matrices for the individual ﬁlms, as (Fowles 4.28) asserts without proof. 4. Fowles 3.11. 5. Fowles 3.13. Physics H7C Fall 1999 Solutions to Problem Set 5 Derek Kimball “Guys, don’t worry about midterms. They’re not the best measure of your worth We have the following boundary conditions at the interface: as a physicist. In fact, I did rather poorly on my ﬁrst physics midterm, I got (1) (2) something like 3 out of 40. Of course, everyone else got 1 out of 40, but that’s not 1 E⊥ = 2 E ⊥ really the point...” (1) (2) µ1 H⊥ = µ2 H⊥ (1) (2)  Prof. Nima ArkaniHamed, UC Berkeley E =E If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! Problem 1 H
(1) =H (2) (5) First let’s see if E can be along y . In this case, from Eqs. (5), we have that: ˆ
(1) 1 Ey = (2) 2 Ey ⇒ (6) (1) 4Ey = (2) Ey (a) Boundary conditions at the interface (y = 0) between Region 1 and Region 2 imply there will be EM waves propagating in the xdirection in both regions. Without ˆ knowing their polarizations just yet, let the electric ﬁelds of the waves be given by: E1 e
i(k1 x−ωt) and we also know that for the wave in Region 1:
(1) Hz = (1) Hz = 2 1 (1) Ey ⇒ 0 (1) Ey , µ1 µ0 (7) E2 ei(k2 x−ωt) . (1) and for the wave in Region 2:
(2) Hz = (2) Hz = 2 (2) Ey ⇒ 0 (1) Ey At the interface between Region 1 and Region 2, since the electric and magnetic ﬁeld amplitudes in both regions are independent of x and t, the only way to satisfy boundary conditions at all positions and times is if: ei(k1 x−ωt) = ei(k2 x−ωt) For example, at t = 0, this requires that: k1 = k2 (3) (2) µ2 1 2 0 µ0 (2) Ey = 2 (1) µ0
(2) (8) We see that these relations verify that Hz = Hz as demanded by the boundary conditions in Eqs. (5). The relations are consistent, so therefore such a polarization is allowed. Can E be along z ? We have from Eqs. (5) that ˆ
(1) (2) Ez = Ez which means the indices of refraction in the two regions must be the same. Thus we have,
1 µ1 0 µ0 (9) = 2 µ2 0 µ0 For the wave in Region 1: (4)
(1) Hy = (1) Hy = 2 1 (1) Ey ⇒ 0 (1) Ez . which gives us µ2 = 4µ0 . (b) September 30, 1999 µ1 µ0 (10) Physics H7C Fall 1999 In Region 2:
(2) Hy = (2) Hy 2 (2) Ey ⇒ 0 (1) Ey . Solutions to Problem Set 5
E E E ts = E E rp = E E tp = E rs = Derek Kimball transmission of TM and TE waves, using the deﬁnitions on page 43 of Fowles: µ2 TE 1 = 2 µ0 (11) TE We see that this veriﬁes the boundary condition on Hy from Eqs. (5):
(1) µ1 Hy √ (1) = 2 0 µ0 Ey = √ (2) (1) µ2 Hy = 2 0 µ0 Ez . TM (15)
TM (12) So in fact the wave can be in either polarization state. Problem 2 From the boundary conditions for E and H at the interface (Eqs. (5)), we ﬁnd some basic relations between the quantities in Eqs. (15): ts = rs + 1 ntp + rp = 1 (16) At the Brewster angle rp = 0, so tp = 1/n. Employing the fact that I ∝ nE 2 Reﬂection at the Brewster angle transmits all TM light and reﬂects part of the TE (thanks to Paul Wright!), we have for the partial polarization: light. The degree of polarization P (for linear polarized light) is given by Fowles 2 Eq. (2.27): 1 rs 1 (ts − 1)2 P= = (17) n t2 + t2 n t2 + 1/n2 s p s Imax − Imin P= (13) Imax + Imin Fowles works out the general formulae (Fresnel’s equations) for reﬂection/refraction at a plane interface, and in particular for ts we have from Fowles The light is initially unpolar(2.56): ized, so it is 50% TE and TM+TE TE 50% TM. The light transmit2 cos θ sin φ . (18) ts = ted has reduced TE intensity sin (θ + φ) θ = θB (since some is reﬂected at the interface), but the TM intenIf we combine Fowles Eq. (2.64) sity remains the same. Consequently, the transmitted light tan θB = n (19) is partially polarized: with Snell’s Law IT E IT M − IT E = . P= sin θ IT M + IT E IT M + IT E n= , (20) sin φ (14) φ Now we can apply the Fowles formalism for reﬂection and Figure 1 TM+TE September 30, 1999 we have the Brewster condition: θ + φ = π/2, (21) Physics H7C Fall 1999 Solutions to Problem Set 5 Derek Kimball which follows from the fact that sin φ = cos θ. Knowing also then from Eq. (21) From which we have the condition: that sin (θ + φ) = 1, we ﬁnd that: (27) −Ei + Er = −3Et . (22) ts = 2 cos2 θ Using the trigonometric identity sec2 = tan2 +1 and Eq. (19) we ﬁnd that cos2 θ = Adding Eq. (25) to Eq. (27) gives us (Er /Ei = 1/2)T E , or (Ir /Ii = 1/4)T E . The intensity in the TE component is half the initial intensity, so in total Ir /Ii = 1/8. 1/(n2 + 1) so ts = 2 n2 + 1
2 2 (23) Problem 4 Here we treat the tungsten ﬁlament as a relatively long straight wire of thickness s = 0.1 mm. The distance between the ﬁlament and an aperture is r. We want a transverse coherence width lt ≥ 1 mm. Then from Fowles Section 3.7, and in particular Eq. (3.42), we ﬁnd that: lt = rλ ≥ 1mm s (28) Plugging Eq. (23) into Eq. (17), with a little algebra, gives us: P= n[(n − 1)] . 1 + 6n2 + n4 (24) For glass, where n = 1.5, we have that P ≈ 12%. Problem 3 (a) Note that the Brewster condition (Eq. (21)) is met. Thus from Eq. (19) we have √ that n = tan 60o = 3. Since µ = µ0 , we ﬁnd that = 3 0 . (b) Since the Brewster condition is met, the reﬂected light is 100% TE. Therefore reﬂected light is linearly polarized along y . ˆ (c) If we assume that the tungsten lamp has a central wavelength of 5000 ˚, then Eq. A (28) demands that r ≥ 200 mm. If a doubleslit aperture is used, the slits should be oriented parallel to the lamp ﬁlament, otherwise the thickness of the wire s would have to be replaced with the length of the wire, which is naturally much greater than s. This would force r to be much greater. Problem 5 The power spectrum of the Gaussian pulse f (t) is given by G(ω ) = g (ω )2 , where The light is initially circularly polarized, so it is an equal superposition of linear in our case g (ω ) is: polarizations (TE and TM). Since the TM component has 100% transmission, it ∞ A suﬃces to consider the transverse electric case (E along y ) where all reﬂection ˆ g (ω ) = √ dt exp −at2 + i(ω − ω0 )t (29) 2π −∞ occurs. We have the boundary condition that E is continuous, so It is ﬁrst useful to derive a result about Gaussian integrals. It turns out that ∞ −ax2 dx converges, so let’s set it equal to some constant c. Now consider the Also H is continuous, and we have for the incident, reﬂected and transmitted −∞ e integral over the entire plane: waves the following components of H in the x direction: ˆ
( Hxi) = −Ei ( Hxr) = Er ( Hxt) = −Et 0 Ei + E r = E t . (25) µ0
0 cos 60o cos 60o (26) ∞ −∞ e−ax dx 2 ∞ −∞ e−ay dy = c2 . 2 (30) µ0 Next we convert the integral into polar coordinates (r2 = x2 + y 2 ).
∞ 0 30 cos 30o µ0 e−ar rdr
0 2 2π dφ = c2 . (31) September 30, 1999 Physics H7C Fall 1999 Solutions to Problem Set 5
2 Which is relatively straightforward with the substitution u = ar , du = 2ardr. From this integral we ﬁnd that c2 = π/a. Derek Kimball √ So θ ∝ N . Since the radius of the fringes r is proportional to θ, it follows √ immediately that r ∝ N . Thus all we need to do is to convert the integral in Eq. (29) to a form resembling Problem 7 a Gaussian. This can be done by completing the square in the exponent: −at2 + i(ω − ω0 )t = −a t2 − 2i (ω − ω0 )2 ω − ω0 t− 2a 4a2
2 + ( − ω0 )2 4a2 (32) This solution follows directly from the discussion of antireﬂecting ﬁlms in Fowles (page 99). We want to choose the thickness of the ﬁlm to be λ . Then the reﬂectance 4 is zero if the index of refraction of the coating n =
0 ω − ω0 = −a t − i 2a (ω − ω0 )2 − 4a . Problem 8 Now the integral in Eq. (29) is just a Gaussian integral, which is no longer a problem... Fowles Eq. (4.24) states: Working through the constants gives us: (ω − ω0 ) A g (ω ) = √ exp − 4a 2a
2 1 n0 . (33) which is equivalent to: 1 n0 + 1 −n0 r=M 1 nT t (38) G(ω ) = g (ω )2 is clearly of the same form, and so G(ω ) is a Gaussian function centered at ω0 . Problem 6 The condition for a fringe maximum to occur is given by Fowles Eq. (4.10): 2N π = Use the small angle approximation: 2N π = θ2 4π nd 1 − λ 2 + δr . (35) 4π nd cos θ + δr . λ (34) E0 + 1 −n0 E0 = M 1 nT ET (39) The total E and H just to the right of the righthand interface are: ERH = ET HRH = nT ET . The total E and H just to the left of the lefthand interface are: ELH = E0 + E0 HLH = n0 E0 − n0 E0 . (41) (40) When the relations in Eqs. (40) and (41) are substituted into Eq. (39), we ﬁnd that: ELH HLH =M ERH HRH (42) We are told that the zeroth order fringe (N = 0) has zero radius (∴ θ = 0), so we have: δr = − and subsequently: 2N π = 4π θ2 nd . λ 2 (37) September 30, 1999 4π nd λ (36) Therefore the overall transfer matrix Mtot is merely the product of transfer matrices for the individual ﬁlms Mi . This follows from induction. Suppose there are n ﬁlms with transfer matrices M1 , M2 , ..., Mn . Let the ﬁelds to the right of the last ﬁlm be En and Hn . The ﬁelds just to the left of the nth ﬁlm are: En−1 Hn−1 = Mn En Hn (43) Physics H7C Fall 1999 the ﬁelds just to the left of the (n1) En−2 Hn−2 and so on... This argument leads to Fowles Eq. (4.28) as stated. = Mn−1
th Solutions to Problem Set 5
ﬁlm are: = Mn−1 Mn En Hn (44) En−1 Hn−1 Derek Kimball September 30, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 6 Problems 14 are suitable review problems for Midterm 1. Problems 58 involve new material beyond the range covered by Midterm 1. 1. Model an animal’s eye as a sphere composed of vitreous humor, backed by a retina. If the eye is focused at inﬁnity, what is the refractive index of the humor? 2. A point source of isotropic light is located at the center of a small hemispherical hole in the plane end face of a cylindrical light guide with refractive index n = 2, permeability µ = µ0 . What fraction of the light emitted can be transmitted a long distance (relative to its radius) by the light guide? (A number is required.) 3. Using a combination of optical devices (polarizers, wave plates...), design an optical system that will pass righthand circularly polarized light without changing its polarization, but will completely block lefthand circularly polarized light. This system is called a “righthand circular analyzer”. Use Jones matrices to prove that your design will work. 4. A Michelson interferometer produces fringes on its screen (which is not quite perfectly aligned). It is fed by laser light polarized out of the interferometer plane along ˆ. With equal path lengths z the fringe visibility V ≡ (Imax − Imin )/(Imax + Imin ) is unity. (a.) An ideal linear polarizer, with transmission axis z oriented at 90◦ to ˆ, is placed in one leg of the interferometer. What is V ? Explain. (b.) Same as (a.) except that a second linear polarizer, with transmission axis oriented at 45◦ to ˆ, z is added in the same leg upstream of the ﬁrst (i.e. closer to the halfsilvered mirror). your calculation for V . Show 5. Two identical horizontal thin slits in a black plate are centered at y = ± h , where y is the 2 vertical coordinate. A screen with vertical coordinate Y is located a distance D downstream. If an analyzer is present, it is located just upstream of the screen. Fraunhofer conditions apply, i.e. D, and smallangle approximations can kh2 be made, i.e. y  D, Y  D. Plane wave A is normally incident on the top slit and plane wave B is normally incident on the bottom slit, with ˆ ˆ EA ∝ [(x + iy)ei(kz−ωt) ] and ˆ ˆ EB ∝ [(x − iy)ei(kz−ωt) ] , with denoting the real part. When either slit is blocked and no analyzer is in place, the intensity I (Y = 0) ≡ I0 . When neither slit is blocked, ﬁnd I (Y )/I0 , where I0 is deﬁned above, for the following cases: (a.) No analyzer is in place. (b.) The analyzer accepts only y polarized light. ˆ (c.) The analyzer accepts only righthand circularly polarized light E ∝ [(x − iy)ei(kz−ωt) ]. ˆ ˆ 6. Prove that
N exp (iφn ) =
n=1 sin N ∆φ/2 ¯ exp (iφ), sin ∆φ/2 where ∆φ ≡ φn+1 − φn , ¯ and φ is the average of the φn . 7. A plane wave of wavelength λ is incident on (A) 2 no screen, i.e. all the light passes through; (B) a black disk of radius R; (C) a black screen with a circular hole of the same radius R. The relative intensities seen by an observer on the axis at distance D downstream are in the ratio IA : IB : IC = 1 : 1 : 0. Fraunhofer conditions do not apply to this geometry, although the obliquity and 1/r factors do not vary appreciably across the screen. (a.) Find the smallest possible value of R that is consistent with the above conditions, expressed in terms of D and λ. (b.) In this problem the screen aperture functions gB for case (B) and gC for case (C) sum together to give the aperture function gA for case (A). For the particular R that you obtained for part (a), the intensities IB and IC also sum together to give IA . For what other choices of R would that be true? Explain. 8. A plane electromagnetic wave propagates in the ˆ direction within a good conductor (λEM wave z δ (= skin depth) λplasma ). Evaluate the total power lost per square meter due to Joule (ohmic) heating in the region 0 < z < ∞. Show that this is equal to the average value of S at z = 0. You may take µ = µ0 for this conductor. Physics H7C Fall 1999 Solutions to Problem Set 6 Derek Kimball At the question period after a Dirac lecture at the University of Toronto, somebody where n = 2 is the refractive index. The percent of light transmitted (intensity) is in the audience remarked: “Professor Dirac, I do not understand how you derived T = t2 = 4/9 in our case. the formula on the top left side of the blackboard.” n1 “This is not a question,” snapped Dirac, “it is a statement. Next question, please.”  George Gamow, excerpted from Thirty Years that Shook Physics, a very fun book on the people involved in the early development of quantum mechanics. If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! Problem 1 The eye is focused at inﬁnity, so we assume the rays incident on the eye are all parallel (Fig. 1). From the geometry of the diagram in Fig. 1 it is clear that θ = 2φ. Snell’s law demands that sin θ = n sin φ where n is the refractive index Figure 2 of the humor. We can make the small angle approximation (making the realistic assumption that light passes only through a small iris in the center of the front of the eye) and just say that θ = nφ, which gives us n = 2. Next consider the diagram in Fig. 2. We require that ϕ ≥ sin−1 (1/n) for total internal reﬂection. From geometry this demands that α ≤ cos−1 (1/n). We can now integrate to ﬁnd the total solid angle ∆Ω of light accepted into the light guide: eye 2π α θ sin θdθdφ = 2π (1 − cos α) = 2π (1 − 1/n) (3) ∆Ω = φ θ = 2φ φ 0 0 The percent of light accepted is then ∆Ω/4π , or 1/4. So then the fraction of light that travels an appreciable distance is given by the fraction of light transmitted through the interface in the correct direction which is 1/9. Problem 3 α n2 θ Figure 1 Problem 2 We start out with righthand circularly polarized light and send it through a All light coming from the point source is normally incident on the surface of the quarterwave plate with the fast axis vertical: hemispherical hole in the end of the light guide. Thus the amplitude of the electric ﬁeld of transmitted light Et is given by (from Strovink and/or Fowles): 10 1 1 · = . (4) 0 −i −i −1 Et 2Z2 = , (1) E0 Z2 + Z1 Next, we send the light through a linear polarizer with the transmission axis at Et where Z1,2 = µ1,2 / 1,2 . Thus the transmission coeﬃcient t = E0 is given by: 45o : t= 2 , 1+n (2) October 14, 1999 1 2 1 −1 −1 1 · 1 −1 = 1 −1 . (5) Physics H7C Fall 1999 Solutions to Problem Set 6 Derek Kimball So we have 100% transmission for rightcircularly polarized light. For left circularly polarized light, no light is transmitted: 1 2 1 −1 −1 1 · 1 0 0 −i · 1 i = 0 0 . (6) So now at the output we have two waves from the diﬀerent arms of the interferometer (supposing light is propagating in the x direction): ˆ E1 = E0 E0 z+ ˆ y ˆ 4 4 E2 = E0 eiφ z , ˆ (9) So the right hand circular analyzer works as claimed. Problem 4 where φ is the phase diﬀerence induced by the diﬀering path lengths for the arms of the interferometer. Thus the intensity of light at the output is given by: I=2 E0 4
2 2 + E0 + 2 2 E0 cos φ 4 (a) (10) ˆ If a linear polarizer at 90o to z (the direction of light polarization) is placed in a leg so we see that for Imax and Imin : of the Michelson interferometer, no light travels in one leg of the interferometer. 9 2 E2 Imax,min = E0 ± 0 . Then there will be no fringes and no interference, so V ≡ (Imax − Imin )/(Imax + 8 2 Imin ) = 0. This is clear since Imin = Imax if there are no fringes. (b) (11) Using these results in our equation for fringe visibility we ﬁnd that V ≡ (Imax − Imin )/(Imax + Imin ) = 4/9 Now, with a linear polarizer at 45o to z upstream of the ﬁrst linear polarizer, ˆ there is light transmitted in both legs of the interferometer. Interference will not Problem 5 occur for light of orthogonal polarizations, so only light polarized in the z direction ˆ contributes to the fringes. (a) The linear polarizer at 45o transmits: We have two light beams, whose electric ﬁeld amplitudes are given by: 1 11 E0 1 1 = · E0 . (7) EA ∝ Re[(ˆ + iy )ei(kz−ωt) ] x ˆ 0 1 2 11 2 and The next polarizer only transmits light in the orthogonal direction, so the transx ˆ EB ∝ Re[(ˆ − iy )ei(kz−ωt) ] , mitted light is given by with Re denoting the real part. We are not given the slit widths, so let’s assume E0 0 . that they are small enough to be ignored in our analysis... 1 2 The light bounces oﬀ the mirror, which preserves polarization, passes through With no analyzer in place, the two beams are orthogonally polarized so there is the second polarizer with no loss of amplitude, then passes through the polarizer no interference. Thus the intensity at the screen is simply the sum of the two at 45o (which now appears to be at 45o with respect to the direction of light individual intensities, which with small angle approximations is roughly I ≈ 2I0 . propagation): (b) 1 E0 0 E0 −1 1 −1 ˆ · = . (8) With an analyzer that accepts only y polarized light, the two light beams have the 1 1 2 −1 1 2 4 same polarization after the analyzer and then can interfere. The interference is October 14, 1999 Physics H7C Fall 1999 Solutions to Problem Set 6 Derek Kimball that of a typical double slit experiment (Young’s experiment), as solved in Fowles as the diﬀerence of two inﬁnite geometric series: pp. 5961. Of course, half the amplitude of each wave has been removed by the N N analyzer, so the resulting interference pattern is given by: exp (iφn ) → αn exp (iφ1 ) exp (in∆φ) I (Y ) = I0 1 + cos 4 π hY λD . (12) =
n=0 ∞ n=1 n=0 ∞ αn+1 exp (iφ1 ) exp (in∆φ) −
n=0 αN +n+1 exp (iφ1 + N ∆φ) exp (in∆φ) = (c) (1 − x)(N +1) exp (iφ1 + N ∆φ) (1 − x) exp (iφ1 ) − 1 − (1 − x) exp (i∆φ) 1 − (1 − x) exp (i∆φ) (13) The analyzer blocks out lefthand circularly polarized light so there is contribution only from EB . Therefore the intensity at the screen is I ≈ I0 . We now let x go to 0, and we then have:
N Problem 6 exp (iφn ) =
n=1 exp (iφ1 + N ∆φ) exp (iφ1 ) − 1 − exp (i∆φ) 1 − exp (i∆φ) exp (−iN ∆φ/2) − exp (+iN ∆φ/2) exp (−i∆φ/2) − exp (+i∆φ/2) (14) We want to prove that sin N ∆φ/2 ¯ exp (iφ), exp (iφn ) = sin ∆φ/2 n=1 where ∆φ ≡ φn+1 − φn , ¯ and φ is the average of the φn .
a Recall that for a geometric series 0 arn = 1−r , where r < 1. Let us rewrite the sum above as the sum of two inﬁnite series. In order to use the geometric series formula, let’s multiply each term in the series by an amplitude αn = (1 − x)n which ensures that αn exp (iφn ) < 1. We can require that the αn ’s are so close to unity that they are very well approximated by 1 for the ﬁrst N terms. We are only worried about the convergence of the tail of the series, which is taken care of with this postulate. We could even take the limit as x → 0 to make this argument more mathematically sound. ∞ N = exp φ1 + N −1 ∆φ 2 from which we deduce that:
N exp (iφn ) =
n=1 sin N ∆φ/2 ¯ exp (iφ) sin ∆φ/2 (15) proving the original conjecture. Problem 7 As discussed in Fowles pp. 126128, this problem can be solved using Fresnel zones. The radius of the N th Fresnel zone in this case is given by: √ (16) R = N λD since the factor L = (1/h + 1/h )−1 = D in our case (see Fowles Eq. 5.36). (a) (C) The smallest radius for which the light from Fresnel zones cancel is that which includes the ﬁrst two. In this case the optical disturbance is given by Now let’s write the series
N exp (iφn )
n=1 Up = U1  − U2  ≈ 0. October 14, 1999 Physics H7C Fall 1999 Solutions to Problem Set 6 Derek Kimball (A) With no screen, the optical disturbance is half that due to the ﬁrst Fresnel The power lost per square meter due to ohmic heating is given by the relation: zone, 1 P =σ E 2 dV Up = U1 . 2 σ µσω 2 σ ˜ (B) If we block out the ﬁrst two Fresnel zones, the optical disturbance is approxiE0 2 exp (− z) = E2 = (21) 2 2 2 µσω 0 mately half that due to the third zone, or Up = So as you can see, the choice of R= satisﬁes all the required conditions. (b) This will hold whenever a similar situation occurs, i.e. an even number of Fresnel zones are blocked by the black disk. The optical disturbances UA and UB will always sum to equal the optical disturbance without the screen U0 because they are complementary apertures. However, only when either UA or UB is zero can the squares (∝ I ) be equal. √ So whenever R = 2nλD where n is an integer, this is the case. Problem 8 From Maxwell’s equations in conducting media, we get the wave equations: ˜ ∇2 E (z, t) = E0 ei(κz−ωt) ˜ ∇2 B (z, t) = B0 ei(κz−ωt) where κ2 = µ ω 2 + iµσω. For a good conductor, κ= µσω (1 + i). 2 (19) (18) 1 1 U3  ≈ U1 . 2 2 √ The average value of S  is given by ﬁnd
1 µE × B . Plugging in values from above, we 2 E2 µσω 0 2λD S  = σ 2 (22) which conﬁrms the conjecture stated in the problem. (17) Also from Maxwell’s equations for a conducting medium, we ﬁnd that: B0 = κ . ω (20) October 14, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 7 1. Fowles 6.1. 2. Fowles 6.2. 3. Fowles 6.6. 4. Rohlf 2.13. Note that this problem is more diﬃcult than it ﬁrst appears to be. You may not assume that the average number of unexpected events found, per 106 interactions, is equal to unity! 5. Rohlf 2.38. 6. Assume that the sun is a blackbody of temperature 5800 K and radius 7 × 108 m, located 1.5 × 1011 m from the earth. Assume further that the earth is a gray body which absorbs part of the radiation incident upon it from the sun, and then reradiates it isotropically. Neglect any other eﬀects which could heat the earth. Calculate the surface temperature of the earth under these assumptions. 7. Greenhouse eﬀect. Take the sun’s blackbody spectrum to have its peak in the yellow ( = 0.58 µm). Take Tsun = 5800 K and Tearth = 300 K. (a.) Use Wien’s Law (λmax ∝ T −1 ) to estimate the wavelength at the peak of the earth’s blackbody spectrum. (b.) Imagine that the (visible) sun’s rays pass through the clear glass of a greenhouse. On the ﬂoor of the greenhouse is black dirt with ≈ 1, which absorbs these rays. In contrast, the (infrared) blackbody radiation from the dirt is totally absorbed by the glass and reradiated (half in and half out). Assuming that the temperature outside the greenhouse is 300 K, estimate the temperature inside it. 8. Nuclear winter (inverse Greenhouse eﬀect). According to some experts (though this is controversial), after nuclear war a thin layer of dust would remain in the upper atmosphere of the earth. To a ﬁrst approximation, the dust absorbs all light from the sun, which is near visible in wavelength. The dust then reradiates that energy in the infrared, to which it is transparent: half in toward the earth, half out to space. The dust is nearly transparent to the earth’s outward radiation, also in the infrared. If the peacetime surface temperature of the earth is 300 K, what would that temperature become after nuclear war? To relate to it physiologically, express this latter temperature in ◦ F. Physics H7C Fall 1999 Solutions to Problem Set 7 Derek Kimball Above the front door of Niels Bohr’s cottage was nailed a horseshoe. A visitor who saw it exclaimed: “Being as great a scientist as you are, do you really believe that a horseshoe above the entrance to a home brings good luck?” “No,” answered Bohr, “I certainly do not believe in this superstition. But you know,” he added with a smile, “they say that it does bring luck even if you don’t believe in it!”  George Gamow, excerpted from Thirty Years that Shook Physics. equations break down in the vicinity of the resonance, which is where we must work to solve this problem. So here we bring back the γ 2 ω 2 terms, but continue to assume there are few electrons. In this case our formulas for κ and n are given by: n≈1+ and
2 ω0 − ω 2 N e2 2 2m 0 (ω0 − ω 2 )2 + γ 2 ω 2 (4) If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! Problem 1 κ≈ γω N e2 . 2 2m 0 (ω0 − ω 2 )2 + γ 2 ω 2 (5) If we take the derivative of n with respect to ω and set it equal to zero, we ﬁnd two positive roots yielding the values for the max and min of the the function n, This problem makes sense only if you make some rather poorly motivated ap namely proximations. In particular, we must assume that we are far away from resonance ω = ω0 1 ± γ/ω0 . 2 (namely that ω0 − ω 2 γ ω ). We also can assume that there are very few electrons 2 (namely that N e0 1, which is wellmotivated by the fact that κ is much less than It is safe to assume, since damping is small, that this value can be approximated m n, i.e. few absorbers). If we make these approximations, the results follow almost by the ﬁrst order Taylor expansion: immediately. If you don’t make these assumptions, then the results are clearly ω = ω0 ± γ/2. incorrect (see Figure 6.1 in Fowles, which is nothing like the equations Fowles asks us to derive). Thus, we’ll make these assumptions! If we plug these values into our expression for κ (Eq. 5), we see that these values are those where κ attains half its maximum value. Then we can apply these approximations to equations 6.34 and 6.35 in Fowles. We ﬁnd that: Problem 3 1 N e2 (1) n2 − κ2 ≈ n2 ≈ 1 + 2 We are given that σ = 6.8 × 107 mho/m and that Ne = 1.5 × 1028 electrons/m3 . m 0 ω0 − ω 2 Using these values in the appropriate Fowles formulas gives us the desired anUsing a ﬁrst order Taylor expansion, we then ﬁnd that: swers... n≈1+ N e2 1 2 2m 0 ω0 − ω 2 (2) (a) Plasma frequency ωp = (b) Problem 2 Once again Fowles attempts to confuse us by implying the above results can be applied in the solution to this problem when they can’t. This is because the above October 21, 1999 Relaxation time τ= µ0 σc2 = 1.6 × 10−13 s 2 ωp N e2 = 6.9 × 1015 s−1 m0 Since n is approximately 1, κ is given by: N e2 γω κ≈ . 2 2m 0 (ω0 − ω 2 )2 (3) Physics H7C Fall 1999 (c) Our frequency with a wavelength of 10−6 m is given by ω= 2πc = 1.9 × 1015 s−1 . λ Solutions to Problem Set 7 Derek Kimball our conﬁdence in the value of a we have measured and then convolve it with the probability for seeing a second event. The probability density function, in this case that for Poisson statistics, is given by: fp (x) = e−a ax x! (6) Real and imaginary parts of the index of refraction can be derived from from Fowles Eqs. 6.55 and 6.56: n2 − κ2 = 1 − 2nκ = Clearly, ωp , ω τ −1 , so we get n2 − κ2 ≈ 1 − 2nκ ≈ ωp ω
2 2 ωp ω 2 + τ −2 2 ωp 1 ωτ ω 2 + τ −2 where x is a nonnegative integer and a is the average value of x. The probability density function (with known parameter a) allows us to predict the frequency with which random data x will take on some particular value. We ﬁrst want to calculate a value for a = n · p (where n is the number of interactions and p is the probability for an event), a distribution function based on the most likely values for a and our conﬁdence in those values. The most likely value of p from the data is 10−6 . For a conservative upper limit (without assuming very much about the prior probability distribution), we can estimate that the lower limit of p (at a 95% conﬁdence level) must be that for which the probability of seeing one event in 106 interactions is at least 5%: fp (1) = e−np np ≥ 0.05, (7) = −12.2 2 1 ωp = 0.044. ωτ ω 2 We can then solve these equations for n and κ, and with a little algebra we get: n = 0.006 κ = 3.5. (d) The reﬂectance is given by the HagenRubens formula, R=1− 8ω 0 ≈ 1. σ which tells us that np ≥ 0.05, or p ≥ 5 × 10−8 . Furthermore, we know that the upper limit at 95% conﬁdence level on p can be found from: fp (1) = e−np np ≤ 0.95, which tells us that p ≤ 5.14 × 10−6 . We want to be 90% sure we’ll see a second event. We could guess that if we’re 95% sure p is bigger than pmin = 5 × 10−8 and 95% sure that we’ll see at least one more event after n2 interactions using this value for p, we’ll be 90% sure to see a second event. We’re 95% sure we’ll see at least one more event if the probability to see zero events is less than 0.05: fp (0) = e−n2 pmin ≤ 0.05, (8) Problem 4 This problem, as Prof. Strovink pointed out, is a little bit tricky. Our experimenter ﬁnds 1 event in 106 interactions. Now we want to be 90% sure we ﬁnd a second which gives us event. How many interactions do we need? The basic problem is that we don’t n2 ≥ 6 × 107 interactions. really know the average number of events we should see per 106 interactions, which is needed to calculate how many more interactions are necessary to be 90% sure This is a conservative upper limit on the number of interactions we need before we’ll see a second event. So we’ll have to try to ﬁgure out some function describing we’ll see another event. October 21, 1999 Physics H7C Fall 1999 Solutions to Problem Set 7 Derek Kimball ˜ The above analysis gives us some idea of what our “likelihood distribution” L We can take the derivative of N with respect to time to obtain: for p looks like... it is peaked at 1 × 10−6 and nears zero at both 5 × 10−8 and P dV dN 5.14 × 10−6 . It is well described by the function = . (10) dt kT dt p −p/p0 ˜ L(p) = e We know from the statement of the problem that the number of CO2 molecules is p0 proportional to the surface area of the beer bubbles, where p0 = 10−6 is the most likely value for p. It is not an accident that this dN looks exactly like fp (1). This, as Prof. Strovink explained to me, is just Bayes’ = Cr2 . dt assumption of a uniform prior probability distribution – meaning that we assume we found the most likely value of p0 in our experiment and the distribution of Also, assuming a spherical bubble, we know that probabilities is that from, in our case, Poisson statistics. dV dr A little more mathematical rigor can be applied if we take our “likelihood distri= 4πr2 . dt dt ˜ bution” L for p and convolve it with the restriction that the probability for zero events must be less than 10%. This approach yields more or less the same result Plugging these into Eq. (10), we ﬁnd that: as Rohlf’s answer, which is reasonable since in the Bayesian approach we assume dr CkT a prior probability distribution with p0 = 10−6 as the central value. Rohlf just = , (11) dt 4πP assumed that a priori we knew the probability for an event to occur would be p0 = 10−6 . If we try this approach, we ﬁnd that: which indicates the radius of the bubble increases linearly with time.
∞ 0 ∞ 0 ˜ L(p)en2 p dp ≤ 0.1 ˜ L(p)dp Problem 6 Plugging in our assumed prior probability distribution or “likelihood distribution” The power given oﬀ by the sun is S (power radiated per unit area) times the and using the substitution u = (1/p0 + n2 )p in the numerator’s integral, we ﬁnd: surface area of the sun, which is: 1 p0
∞ 0 The portion of this power received by the earth is scaled down by the emissivity From which we can calculate n2 = 2.16 × 106 for 90% CL that we will see a second factor (earth is treated as a gray body) and the crosssectional area of the earth event. This is far smaller than our original rough estimate, but assumes a prior over the surface area of a sphere with a radius equal to the distance between the probability distribution. This is probably the more correct approach. earth and the sun: Well, as you can probably tell, this problem was quite diﬃcult for me, so don’t 2 πRE (in) 4 2 PE = (σTS ) · (4πRS ) (13) feel too bad if you had some trouble as well... 2 4πRES Problem 5 As evidenced by the sampling of problems from Rohlf, we can guess his two main interests are particle physics and beer. The expression for the number of particles from the ideal gas law is: N= PV . kT (9) October 21, 1999 The power reradiated by the earth is given by: PE In equilibrium, PE
(in) (out) 4 2 = ( σTS )(4πRE ). p −p/p0 −n2 p 1 e e dp = 2 2 ≤ 0.1. p0 p0 · (1/p0 + n2 ) 4 2 PS = (σTS ) · (4πRS ) (12) (14) = PE (out) . If we equate these expressions, we ﬁnd that:
4 2 1 RS TE 4 = 4 R2 , TS ES (15) Physics H7C Fall 1999 from which we deduce that TE = 290 K. Problem 7 Solutions to Problem Set 7 Derek Kimball Since we only get half the light power from the sun that we used to down on the earth, the new temperature on earth TE is given simply by: TE = Brrrr...... 1 2
1/ 4 TE = 252 K = −5.3 o F. (a) We apply Wien’s law to get the peak of the earth’s blackbody spectrum. First, the constant can be derived from plugging in the known parameters for the sun: λmax = C T C = λmax,sun TS = (0.58 µm)(5800 K) = 3364µm K. Applying Wien’s law to the earth, we get: λmax,earth = C = 11.2 µm. TE (b) If half the power reradiated from the dirt is radiated back into the greenhouse at every interface with the walls, in equilibrium the power radiated by the dirt should be twice that incident from the sun. The power from the sun hitting the dirt of the greenhouse is: PS
(dirt) 4 2 = (σTS )(4πRS ) Area of dirt , 2 4πRES (16) The power reradiated by the dirt is: Pout Setting 2Pout
(dirt) (dirt) 4 = σTdirt · (Area of dirt) (17) = PS (dirt) , we ﬁnd that:
4 Tdirt = 4 2 TS R S 2 2 RES Tdirt = 333 K. Problem 8 October 21, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 8 1. (a.) Consider a distant star with the same luminosity and surface temperature as the sun. A person (who is as “eﬃcient” as one of Rutherford’s graduate students) can see the star if 250 visible photons per second pass through her pupil, which has a radius of 2 mm when nearly fully dilated. What is the maximum distance at which the star is visible to the naked eye? (b.) How many cosmic photons per second per square cm were incident on the NobelPrizewinning microwave antenna of Penzias and Wilson? 2. (a.) The maximum energy of photoelectrons from aluminum is 2.3 eV for incident radiation of 0.2 µm and 0.9 eV for radiation of 0.313 µm. Use these data to calculate Planck’s constant and the work function of aluminum. (b.) An aluminum photocathode receives incident radiation of 0.313 µm. When the intensity of this radiation is 1 mW, a current of 1 µA is observed in a circuit that detects the photoelectrons that are liberated. Estimate the quantum eﬃciency of the photocathode. 3. Rohlf 3.20. 4. (a.) The power radiated by an accelerated charge e is given in classical physics by the formula P= 1 2e2 2 a (SI units), 4π 0 3c3 to the correspondence principle, when n is very large this should agree with a proper quantum mechanical calculation.) (b.) The decay rate for an electron in an orbit may be deﬁned to be the power radiated, P , divided by the energy emitted in the decay. (The decay rate is the inverse of the lifetime). Use the Bohr theory expression for the energy radiated, and the expression for P from part (a.) to calculate the “correspondence” value of the decay rate when the electron makes a transition from orbit n to orbit n − 1. What is the value of this decay rate when n = 2? (This will not agree exactly with the true quantum theory, since the correspondence principle will not hold when n is not 1.) What is the decay rate when the transition is from an orbit n to an orbit n − m? (c.) Use the value of the “lifetime” of an electron in an n = 2 Bohr orbit, calculated in part (b.), to estimate the uncertainty in the energy of the n = 2 energy level. How does it compare with the energy of that level? 5. Rohlf 3.42. Do the ﬁrst part of the problem. Then answer the ﬁnal question posed in this problem for two extreme cases: (a.) The muon capture probability is of the same order of magnitude as the decay probability. (b.) The muon capture probability is many orders of magnitude smaller than the decay probability. 6. There exists a fundamental constant of nature O whose value is 25,813 ohms. Calculate the ratio between the numerical value of O and the numerical value of Z0 , the characteristic impedance of free space. Using any hint that you can obtain from this ratio, determine the algebraic value where a is the acceleration. Using this formula, calculate the power radiated by an electron in a Bohr orbit characterized by the quantum number n. No numbers are required. (According 2 of O, expressed in terms of other fundamental constants. 7. Rohlf 3.56. 8. Rohlf 4.54. Physics H7C Fall 1999 Stop talking and write down the Hamiltonian! Solutions to Problem Set 8
(b) Derek Kimball  I. B. Khriplovich, Novosibirsk State University (Russia), during a seminar at The cosmic background radiation ﬁlls the universe roughly isotropically (there is no solid angle suppression), and the temperature of the radiation is T = 2.74 K. Berkeley. The incident number of photons per square cm on Penzias and Wilson’s antenna If you have any questions, suggestions or corrections to the solutions, don’t hesitate is given by an equation similar to Eq. (3): to email me at [email protected]! ∞ 2πc dλ 4 hc/λkT (4) = 2.6 × 1012 photons s−1 cm−2 . n= λ (e − 1) 0 Problem 1 Problem 2 (a) First we determine how many visible photons are emitted from the surface of the sunlike star. The radiated power per unit area per unit wavelength dR is given (a) dλ by the Planck distribution: The maximum energy of a electron ejected by the photoelectric eﬀect is given by: dR 2πhc2 = 5 hc/λkT . dλ λ (e − 1) (1) hc − Φ, λ (5) We want to convert this quantity into the number of photons per second per unit where Φ is the work function. We have two data points to use in this relation, area per unit wavelength dn , which can be done by dividing dR by the energy per with which we can determine h: dλ dλ photon: hc − Φ = 2.3 eV 0.2µm hc . (2) Ephoton = ω = λ hc − Φ = 0.9 eV 0.313µm We integrate over the visible spectrum to get photons per second per unit area, using T = 5800 K for the temperature of the sunlike star:
700 nm Subtracting these equations and solving for h gives us (3) h ≈ 2.6 × 10−21 MeV · s. n= 400 nm dλ 2πc λ4 (ehc/λkT − 1) = 6.3 × 1025 photons s−1 m−2 . Substituting this value of h back into one of the photoelectric eﬀect equations gives The total number of photons given oﬀ by the sun per second N is n times the us the work function 2 surface area of the star, which is roughly 4πRS = 6 × 1018 m2 . This gives us Φ = 1.58 eV. 44 3.8×10 photons per second from the star! We scale this by the solid angle subtended by the observer’s eye, (b) 2 πreye Quantum eﬃciency of the photocathode is just the ratio of emitted electrons to 2, 4πRSE incident photons. The number of photons is the power of light over the average and solve for RSE such that the eye receives at least 250 photons. This gives us
max RSE ≈ 1019 m ≈ 1300 light years. energy per photon at this wavelength: Nγ = Pλ = 1.6 × 1015 photons/sec. hc October 28, 1999 Physics H7C Fall 1999 Solutions to Problem Set 8
which gives us:
∞ Derek Kimball The number of electrons is just the current over the charge per electron: Ne = I = 6.2 × 1012 electrons/sec. qe n= 0 dλ λ 8πhc = 8π hc λ5 (ehc/λkT − 1) ∞ 0 dλ . λ4 (ehc/λkT − 1)
dλ λ2 , (11) Taking the ratio gives us the quantum eﬃciency: QE = 0.0039 = 0.39%. This was using the correct value for h. If you used the value of h you obtained from part (a) of this problem, you would ﬁnd QE = 0.0025 = 0.25%. hc Making the appropriate substitution x ≡ hc/λkT and dx = − kT desired result: we get the n = 8π kT hc 3 0 ∞ dx ex x2 ≈ (3.17 × 1019 eV−3 · m−3 )(kT)3 . −1 (12) (c) The average energy per photon is u/n. u = 4R/c can be found from the StefanBoltzmann law (the factor of 4 comes from averaging over all angles, Rohlf Eqs. (3.17) and (3.18)), u = 4R/c = 4σ (kT )4 . c (13) Problem 3 (a) For thermal radiation, the average energy per photon is given by: 1 E= n
∞ 0 dn dE · E dE (6) Therefore, the average energy per photon is found from the ratio of Eqs. (13) and (12) to be: E= 4σ (kT ) u = . n c · (3.17 × 1019 eV−3 · m−3 ) (14) The energy per unit volume u is given by:
∞ u=n E = 0 dE · E
du dE dn dE (7) Plugging in the constants gives us: E ≈ 2.7 kT. (15) We can employ the fact that from Eq. (7), dn = E dE , which gives us du dE dn dE =E dE dλ dE dλ This reduces to our desired result, simply that: dn hc dn du =E = . dλ dλ λ dλ (b) The total photon density n is given by:
∞ ∞ 0 (8) (d) For the number density of photons, we obtain from Eq. (12): n = (3.17 × 1019 eV−3 · m−3 )[(8.62 × 10−5 eV/K)(2.74 K)]3 ≈ 4 × 108 m−3 . (9) For the energy density, we from Eq. (15) we ﬁnd that E ≈ 0.64meV. n= 0 dλ dn = dλ dλ λ du , hc dλ (10) October 28, 1999 Physics H7C Fall 1999 Problem 4 (a) Acceleration a for a circular orbit is given by: v2 a= . r Solutions to Problem Set 8 Derek Kimball The decay rate γ is then the ratio of the radiated power (we assume that the electron is in the nth orbit until the moment it decays) to the energy diﬀerence between the levels: γ= P . ∆E (24) If we plug in our result from part (a) and use n = 2, we get (16) γ = 108 s−1 . Angular momentum L, applying the Bohr quantization condition, is given by: L = mvr = n . Solving for v from Eq. (17) and substituting into Eq. (16), we obtain: a= The Bohr radius r is given by: r = (4π 0 ) n2 2 . me2 L n = 2 3. m2 r 3 mr
2 22 If the decay is from the nth level to the (n − m)th level, we merely adjust the (17) energy diﬀerence in Eq. (23): ∆E = −13.6 eV 1 1 − . n2 (n − m)2 (25) (18) (19) Using Eq. (19) in conjunction with Eq. (18), we ﬁnd an expression for a in terms of fundamental constants: a= n = m2 r 3
22 and employ this equation in Eq. (24). Qualitatively, we see that if the energy difference is greater and the power radiated is the same, the decay rate will decrease. This is an example of the limitations of the Bohr model, since although it correctly predicts the order of magnitude of the transition rates it does not correctly predict the dependence of transition rates on the energy diﬀerence between levels, which actually scales as ω 3 . (c) There is an energy time uncertainty principle, which can be derived from ∆x∆p ≥ /2 in the following handwaving fashion: ∆E ∆t = p ∆p m m ∆x p = ∆x∆p. 1 4π 0 3 me . n4 4 6 (20) Using Eq. (20) in the classical expression for radiated power gives us: 1 2e2 2 P= a= 4π 0 3c3 (b) The energy En in the nth level of the Bohr atom is given by: En = − α2 mc2 −13.6 eV =− , 2 2n n2 (22) 1 4π 0
7 2 m2 e14 . 3 c3 n8 8 (21) We use the lifetime (1/γ ) as the uncertainty in time, and then ﬁnd for the uncertainty in energy: ∆E = γ 2 (26) so the energy radiated in a n → n − 1 transition is: ∆E = −13.6 eV 1 1 − . n2 (n − 1)2 (23) The value of ∆E is (108 s−1 ) · (197.3 MeV fm·(3 × 1023 fm/s)−1 ), or 6 × 10−8 eV. The energy diﬀerence between the ﬁrst and second levels in hydrogen is 10 eV, so the linewidth is smaller than the energy diﬀerence by nine orders of magnitude! October 28, 1999 Physics H7C Fall 1999 Problem 5 Solutions to Problem Set 8
Problem 7 Derek Kimball In calculations involving the Bohr model, the electron mass is replaced by the (a) reduced mass of the muonproton system, which is near the mass of the muon: The typical electron velocity in the Bohr model is v = αc, for the deuteron we m µ mp = 95 MeV. m= replace α by αs . So we have v = αs c = 3 × 107 m/s for both the proton and mµ + m p neutron. Energy levels in the Bohr model are linear with respect to the electron mass, and (b) are given by The reduced mass in the deuteron is roughly mp /2, so the nuclear “Bohr radius” 13.6 eV 95 MeV En = − (27) r is given by: · 2c 2 · 197.3 MeV · fm n2 0.5 MeV = 4 fm. r= ≈ mp c2 αs 0.1 · 938 MeV for the muonproton system. (a) (c)
12 2 A free muon decays with a characteristic lifetime 2.2 × 10−6 s, primarily in the The binding energy of the deuteron is roughly 2 αs mc = 2 MeV. mode: Problem 8 ¯ µ → e− + νe + νµ . If the capture probability was large, it would make the lifetime shorter compared First, we equate the relativistic centripetal force to the electrostatic force acting to the lifetime of a muon (at rest), since the two rates would add. on the electron: ke2 γmv 2 , = r2 r If the capture probability was small enough to be neglected, then the eﬀect of and proceed to solve for the radius: time dilation would lengthen the lifetime of the muon (since in the muonproton ke2 system, the muon has a characteristic velocity αc), compared to a muon at rest. r= . γmv 2 (b) Problem 6 We can also use the Bohr quantization condition (28) (29) If O is divided by the characteristic impedance of free space Z0 , we get (2α)−1 . pr = n So in conjunction with the relativistic expression for the momentum Z0 , O= 2α p = γmv probably. to ﬁnd the radius: r= γ mv . (30) Setting Eqs. (29) and (30) equal, we can solve for the velocity, and we obtain the desired result v = αc. October 28, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 9 1. Rohlf 5.31. 2. Rohlf 6.12. 3. Rohlf 6.29. 4. Rohlf 6.32. 5. Rohlf 7.4. 6. Rohlf 7.5. 7. Show that the “conservation of probability” law ∂ ∂ j = 0, where ψ∗ ψ + ∂t ∂x ∗ h ¯ ∂ψ ∂ψ ψ∗ − ψ ≡ j, 2mi ∂x ∂x and j is the probability current in one dimension x, holds if ψ (x, t) is a solution of the Schr¨dinger o equation with a potential V (x), provided that V (x) is real. Therefore absorptive potentials are imaginary. 8. A particle of mass m is bound in an inﬁnitely deep onedimensional potential well extending from x = 0 to x = L. At t = 0 it is described by a wavefunction of the form u(x) ∝ sin (πx/L) + sin (2πx/L). (a.) Normalize u(x). (b.) When the particle’s energy E is measured for the ﬁrst time, what value(s) could be obtained? (c.) What is the expectation value H of the particle’s kinetic energy? (Cogent arguments can substitute for some algebra here, and are encouraged.) (d.) At what time t0 is the probability for the particle to be located on the righthand side of the well (L/2 < x < L) a maximum? Reasoning rather than detailed calculation is needed. Please supply it. Physics H7C Fall 1999 Solutions to Problem Set 9 Derek Kimball “When the rules of quantum mechanics were formulated in the 1920’s they rep The strength of the electric force is given by resented a revolutionary break with the past, and an enormous extrapolation from ke2 1.44 MeV · fm experience. Since they were something very new, they could not be derived from = 0.36 MeV/fm. Fe = 2 = x0 (2 fm)2 something old and incorrect, that is, classical physics. Instead they had to be formulated by guessing, intuition, and inspiration. Their ultimate justiﬁcation was, and is, logical consistency and agreement with experiment.” (c)  Prof. Eugene D. Commins, U.C. Berkeley. If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! Problem 1 Problem 2
dσ It is convenient to write the diﬀerential cross section as d cos θ instead of dσ bedθ cause it makes integration over solid angles a little easier, since the integral always involves cos θ and the diﬀerential solid angle contains the term sin θdθ = −d(cos θ). The maximum acceleration of the proton is given by: amax = kx0 c2 F = = 1.3 × 1029 m/s2 ∼ 1028 g. m mc2 (4) (a) The uncertainty principle yields an estimate for the minimum momentum of a It is relatively straightforward to show that either method of solving for the total proton trapped in the nucleus: cross section gives the same result, since ∆p ≈ 2∆r . (1) dσ d cos θ dσ dσ = · = − sin θ . dθ d cos θ dθ d cos θ If we integrate over all angles, we obtain for the total cross section:
+1 π For the kinetic energy of the proton, we obtain: Ek =
2 22 c (∆p)2 197.3 MeV · fm = = = , 2 2 (∆r )2 2m 8m(∆r) 8mc 8(938 MeV)(2 fm)2 (2) σ=
−1 d(cos θ) dσ = d cos θ dθ sin θ
0 1 dσ . sin θ dθ from which we ﬁnd Ek = 1.4 MeV (b) Problem 3 (a) We’ll suppose, for the sake of this “back of the envelope” calculation, that the ki The maximum kinetic energy that can be transferred to a gold nucleus in a collision netic energy found in part (a) can be set equal to the potential energy at maximum with a 6 MeV αparticle would be when the collision is headon and the αparticle displacement in a classical harmonic oscillator: bounces straight back. Because the gold nucleus is very massive compared to the αparticle, the amount of kinetic energy transferred to the gold nucleus should be 1 (3) small, so roughly vi = −vf where vi and vf are the initial and ﬁnal velocities of the Ek = kx2 20 αparticle. Thus, Mgold V = 2mα vi . Using this result in the equation for kinetic The magnitude of the restoring force at maximum displacement is given by kx0 . energy, we ﬁnd: So we ﬁnd: 2 4mα 2mα vi 1 1 4·4 1 2Ek 2 Mgold V 2 = Mgold · 6 MeV ≈ 0.49 MeV = ( mα v i ) = = 1.4 MeV/fm. F= 2 2 Mgold Mgold 2 197 x0 November 4, 1999 Physics H7C Fall 1999 (b) Solutions to Problem Set 9
σ = 2π σ = 2π k q1 q 2 mv 2 k q1 q2 mv 2
2 cos θ1 Derek Kimball d(cos θ)
cos θ2 2 If we go into the rest frame of the αparticle (S ), we ﬁnd that because the αparticle is very massive compared to the electron, the energy transferred to the αparticle (to the electron in the lab frame) is small. Therefore we can approximate that in S , for a headon collision mα V = 2me vi , where vi is the speed of the αparticle in the lab frame and V is the recoil speed of the αparticle in S . Transforming into the lab frame, we ﬁnd that the recoil speed of the electron Problem 5 ve = 2vi . So the kinetic energy transferred to the electron is Ek = 1 4me me (2vi )2 = 2 mα
2 mα v i 2 1 (1 − cos θ)2 cos θ1 − cos θ2 (1 − cos θ1 )(1 − cos θ2 ) = 4 · (0.511 MeV) 6 MeV ≈ 3.3 keV. 3730 MeV Problem 4 A particle is conﬁned to the region −L/2 < x < L/2. As discussed in section, this means that any state (wavefunction) of the particle can be described as a superposition of eigenfunctions of the energy operator (the Hamiltonian). These eigenfunctions “span” the Hilbert space corresponding to our system (a Hilbert space is an inﬁnite dimensional vector space which is a subspace of the vector space of all continuous complex functions). Since inside the inﬁnite potential well the particle is free, our Hamiltonian H is given by: H=− Eigenfunctions of H are ψn = ψm = nπx 2 cos L L ∂2 . 2m ∂x2
2 (a) The relationship between diﬀerential scattering cross section dσ and the impact parameter b is given by Rohlf (6.18): dσ = 2πbdb The total scattering cross section is derived from this expression:
b1 (5) σ = 2π
b2 bdb = π (b2 − b2 ). 1 2 (6) mπx 2 sin L L where n = 1, 3, 5... and m = 2, 4, 6.... They have the eigenvalues En = n2 π 2 2 2mL2 Using Rohlf (6.40) k q1 q2 mv 2
2 1 + cos θ 1 − cos θ = b2 where n = 1, 2, 3.... These eigenfunctions are orthonormal, meaning that
L/2 −L/2 ∗ ψi (x)ψj (x)dx = 0 we ﬁnd that the total scattering cross section is given by: σ = 2π k q1 q2 mv 2
2 cos θ1 − cos θ2 . (1 − cos θ1 )(1 − cos θ2 ) (7) if i = j , and
L/2 −L/2 ∗ ψi (x)ψi (x)dx = 1. (b) Integrating explicitly gives us the same result:
cos θ1 (a) dσ d cos θ Assume the particle is in an eigenstate of energy. The probability that the particle is found in the region 0 < x < L/2 is 1/2 by symmetry. This is because November 4, 1999 σ=
cos θ2 d(cos θ) Physics H7C Fall 1999 Solutions to Problem Set 9
x2 = u2 sin (2u) sin (2u) 4 cos (2u) 2L2 u3 − + − 3 π3 m 6 4 8 4 x2 = L2 L2 − 2 2. 12 2n π Derek Kimball
mπ/2 the potential is symmetric about x = 0, so every eigenfunction is symmetric or antisymmetic about x = 0. The square of any eigenfunction is symmetric about x = 0. It is clear that probability does not depend on n because all of the eigenfunctions are symmetric or antisymmetric. (b) .
−mπ/2 For odd n, the procedure is pretty much the same... you even end up with the same result. 2 L/2 2 x cos2 (nπx/L)dx. x2 = The probability Pc that a particle in the ground state is in the central half of the L −L/2 box is given by the integral: Making the usubstitution u = nπx/L, we obtain: L/4 L/4 2 mπ/2 Pc dxψ1 2 = dx cos2 (πx/L). (8) 2L2 L −L/4 x2 = 3 3 u2 cos2 udu −L/4 m π −mπ/2 From which we ﬁnd: mπ/2 u2 sin (2u) sin (2u) 4 cos (2u) 2L2 u3 x=L/4 x2 = 3 3 + − − . 2 x sin (2πx/L) mπ 6 4 8 4 −mπ/2 + Pc = = 0.82. L2 (4π/L) x=−L/4 L2 L2 − 2 2. x2 = 12 2n π The probability decreases with n, and at large n approaches the classical limit √ Pc = 0.5. Taking the limit as n → ∞, we see that the rms value of x approaches L/ 12. Problem 6 Problem 7 We intend to prove the conservation of probability law: ∂ ∂ ψ∗ ψ + j = 0, where ∂t ∂x (9) ψ∗ (a) The average value (expectation value) of x2 as a function of n is given by: x2 =
L/2 ∗ dxψn · x2 · ψn . ∂ψ ∂ψ ∗ − ψ ≡ j, 2mi ∂x ∂x −L/2 and j is the probability current in one dimension x, where ψ (x, t) is a solution of We use the eigenfunctions discussed in problem (5), solving ﬁrst for even n. In the Schr¨dinger equation with a real potential V (x). o this case x2 is given by: We can start with the timedependent Schr¨dinger equation: o x2 = 2 L
L/2 −L/2 x2 sin2 (nπx/L)dx. − ∂2 ∂ ψ (x, t). + V (x) ψ (x, t) = i 2 2m ∂x ∂t 2 (10) Making the usubstitution u = nπx/L, we obtain: x2 = 2L2 m3 π 3
mπ/2 −mπ/2 Then take the complex conjugate of (10): − November 4, 1999 ∂2 ∂∗ ψ (x, t). + V (x) ψ ∗ (x, t) = −i 2m ∂x2 ∂t
2 u2 sin2 udu (11) Physics H7C Fall 1999
∗ Solutions to Problem Set 9 Derek Kimball Now multiply (10) by ψ (x, t) and (11) by ψ (x, t), then subtract the equations. which correspond to the energy eigenvalues: We obtain: n2 π 2 2 2 2 . En = ∂2 ∂2 ∂∗ ∂ 2mL2 ψ∗ − ψ−ψ − ψ∗ = i ψ∗ ψ + ψ ψ . (12) 2m ∂x2 2m ∂x2 ∂t ∂t From which we can deduce: − Now consider
∂ ∂x j : (18) 2mi ψ∗ ∂2 ∂2 ψ − ψ 2 ψ∗ ∂x2 ∂x = ∂ (ψψ ∗ ) . ∂t (13) From the postulates of quantum mechanics (discussed in section), we know that any state of an isolated system corresponds to a function in the corresponding Hilbert space. This Hilbert space is spanned by the eigenfunctions of a Hermitian operator (which corresponds to an observable, in this case energy). Therefore, if u (x) were a state in our system, it could be represented as a superposition of diﬀerent eigenfunctions of energy:
∞ ∂ ∂ j= ∂x ∂x 2mi ψ∗ ∂ψ ∗ ∂ψ −ψ ∂x ∂x (14) u (x) =
n=1 cn ψn (x). (19) ∂2 ∂2 ∂ j= ψ∗ 2 ψ − ψ 2 ψ∗ ∂x 2mi ∂x ∂x If we use Eq. (15) in Eq. (13), we obtain the conservation of probability law: ∂ ∂ j = 0. ψ∗ ψ + ∂t ∂x (15) (16) From equation (17), it is clear that all of the ψn ’s vanish at zero, whereas u (x) does not. Thus u (x) is a wavefunction that extends beyond our Hilbert space, or in other words is a particle not conﬁned in the inﬁnite square well, which creates a dilemma... one which is easily solved by use of u(x) as the initial state of the particle. u(x), by the way, is the wavefunction that would be obtained if the potential suddenly (which can be quantitatively deﬁned) sprung up from nowhere and captured a particle formerly in u (x). (a) Problem 8 There was a correction to this problem, speciﬁcally that the initial wavefunction of the particle in the box is supposed to be The normalization condition is:
L 0 dxu(x)2 = 1. (20) u(x) ∝ sin (πx/L) + sin (2πx/L) instead of u (x) ∝ exp (iπx/L) + exp (i2πx/L). It is useful to consider the problem with u (x). If we solve for the energy eigenfunctions of the Hamiltonian for this problem H=− we obtain: ψn (x) = nπx 2 sin , L L (17) ∂2 , 2m ∂x2
2 Let u(x) = C (sin (πx/L) + sin (2πx/L)), then normalization implies: C2
0 L dx sin2 (πx/L) + 2 sin (πx/L) sin (2πx/L) + sin2 (2πx/L) = 1 Note that u(x) is a superposition of the ﬁrst two energy eigenfunctions (given by Eq. (17)). Since eigenfunctions are orthonormal (discussed in problem (5)), the normalization condition reduces to: C2
0 L dx L ψ1 2 + ψ2 2 2 November 4, 1999 Physics H7C Fall 1999 from which we conclude 2 C= . L The same result can be obtained through explicit integration. (b) Solutions to Problem Set 9 Derek Kimball If we include the time evolution of u(x) as described by the timedependent Schr¨dinger equation, the constants in front of ψ1 and ψ2 acquire a timeo dependence: u(x, t) = c1 (t)ψ1 (x) + c2 (t)ψ2 (x). (23) The phase between the two wavefunctions ψ1 and ψ2 oscillates at a frequency: Since u(x) is a superposition of the ﬁrst two energy eigenfunctions, a measurement ∆E of the energy of the particle will yield either E1 or E2 (given in Eq. (18)), each , (24) ω= with a 50% probability. A measurement of an observable will always yield an eigenvalue of the corresponding Hermitian operator. Prof. Strovink mentions that where ∆E = E2 − E1 . Since u(x, 0) has maximum probability to be found on the this is the ﬁrst measurement. This is important, since from another postulate of lefthand side, when quantum theory we know that after measuring the energy, the wavefunction of the ∆E · t0 = nπ particle is subsequently described by the energy eigenfunction corresponding to the eigenvalue of energy obtained in the measurement. (n = 1, 3, 5...) the probability to be found on the righthand side is a maximum. So (c) nπ . t0 = ∆E The expectation value of the energy is E= E1 + E2 . 2 This follows from the fact that u(x) is a superposition of the ﬁrst two energy eigenfunctions with equal probability to be found in either state. Thus repeated measurements on identical systems will yield E1 half the time and E2 the other half. (d) There are some subtle and important points in this part of the problem. As you saw in problem (5), a particle in an energy eigenstate of a symmetric potential always has an equal probability to be found on either the left or righthand side of the potential. This is not true for a superposition of energy eigenstates. This is readily seen by evaluating the expectation value of x for u(x):
L x=
0 xu(x)2 dx (21) x= 2 L L 0 x sin2 πx + x sin2 L x= 2πx L + 2x sin πx sin L 2πx L dx (22) L 16L − . 2 9π 2 November 4, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 10 1. A particle of mass m in a harmonic oscil2 lator potential V (x) = 1 mω0 x2 has possible 2 deﬁniteenergy wavefunctions un (x) with ener¯ gies En = hω0 (n + 1 ), where n is zero or a pos2 itive integer. The particle is in thermal equilibrium with a bath at temperature T . Its probability of having total energy E (relative to the bottom of the well) is proportional to the Boltzmann factor e−E/kT where k is Boltzmann’s constant. ¯ Calculate the average energy E of the particle, true for any temperature. Then take the high ¯ temperature limit kT >> ¯ ω0 and show that E h reduces to kT , the classical result. 2. Consider the potential 1 mω 2 x2 of the harmonic 2 oscillator, where ω is a constant. Deﬁne the operator A ≡ y + iq , where y ≡ x mω/2 and q ≡ √ p/ 2mω . Use the fact that x and p are physical observables so that x† = x and p† = p, where † denotes the Hermitian conjugate. Remember that c† = c∗ when c is a constant. H = H† is the Hamiltonian. For this potential, prove that A = y − iq ¯ [A, A† ] = h H = ω (q 2 + y 2 ) ¯ H = ω (A† A + h/2) h H = ω (AA† − ¯ /2) ¯ [H, A† ] = hωA† [H, A] = −¯ ωA h (This is the formal basis for the assertion, proved in lecture, that a harmonic oscillator has energy ¯ levels En = hω (n+ 1 ), where n is an integer ≥ 0.) 2 3. Consider the angular momentum operator L ≡ r × p = (¯ /i)r × ∇. h For example, Lx ≡ (¯ /i) y (∂/∂z )−z (∂/∂y ) . Deﬁne L± ≡ Lx ±iLy . h
† Prove that [Lx , Ly ] = i¯ Lz h [Ly , Lz ] = i¯ Lx h [Lz , Lx ] = i¯ Ly h [L+ , L− ] = 2¯ Lz h [L− , Lz ] = hL− ¯ [L+ , Lz ] = −¯ L+ h [L2 , Lz ] = 0 [L2 , L± ] = 0 ¯ L2 = L− L+ + L2 + hLz z L2 = L+ L− + L2 − ¯ Lz h z This is the formal basis for the assertion, proved in lecture, that angular momentum is quantized in integral or (for intrinsic [spin] angular momentum) halfintegral units of h. ¯ 4. The spherical harmonic Ylm (θ, φ) is an eigenfunc¯ tion of L2 with eigenvalue h2 l(l + 1) and also of ¯ Lz with eigenvalue hm. It is normalized so that
∗ dΩ Ylm (θ, φ)Ylm (θ, φ) = 1. (a.) Show formally that L+ Yll = 0. (Hint: evaluate dΩ(L+ Yll )∗ (L+ Yll ). ) (b.) In lecture it was proved that L− is a lowering operator: L− Ylm = C− (l, m)Yl,m−1 , where C− (l, m) is a constant depending on l and m. Using the normality of the Ylm ’s, derive the value of C− (l, m)2 . 5. Consider the problem of an electron bound to an inﬁnitely heavy nucleus, here using our modern understanding of orbital and spin angular momenta. Suppose that the nucleus is spinless, and 2 that the atom is in a state of deﬁnite orbital angular momentum l = 2. Moreover, the atom is in a state of deﬁnite projection mj = 5 of its to2 tal (spin + orbital) angular momentum on the z axis. (a.) What value(s) of total angular momentum quantum number j is (are) possible? Why? (b.) Deﬁne the expectation value of the cosine of the angle between the atom’s orbital and spin angular momenta as follows: cos θ ≡ L·S L2 S 2 . Evaluate cos θ for this problem. (Hint. Consider J 2 , where J = L + S.) 6. Rohlf 7.27. 7. Rohlf 8.21. 8. Rohlf 8.25. Physics H7C Fall 1999 Solutions to Problem Set 10
Next we use the fact that for a geometric series
∞ n=0 Derek Kimball “It’s really quite straightforward, there’s nothing mystical about it.”  Prof. Eugene D. Commins, U.C. Berkeley, on the subject of the EinsteinPodolskyRosen paradox. arn = a . 1−r If you have any questions, suggestions or corrections to the solutions, don’t hesitate Thus for the average energy, we have: to email me at [email protected]! Problem 1 ¯ E=
∂ − ∂β 1 1−e−β ω0 1 1−e−β ω0 + 1 ω0 . 2 (6) ¯ The average energy of a particle E in a harmonic oscillator potential V (x) = From which we ﬁnd: 1 22 2 mω0 x is given by: ω0 e−β ω0 1 ω0 1 ¯ E= ω0 = ω /kT ω0 . + (7) + −β ω0 U 0 1−e 2 e −1 2 ¯ E= , (1) N This is an expression for the average energy of the particle at any temperature. If where U is the total energy of N particles in identical potentials (this is the usual we take the high temperature limit (kT ω0 ), eβ ω0 ≈ 1 + β ω0 , so imaginary ensemble of identical quantum mechanical systems used to calculate expectation values, etc.). The total energy is given by: 1 ¯ E ≈ = kT. β ∞ ∞ 1 −nβ ω0 U= N0 En e−En β = N0 ω 0 n + , (2) e 2 n=0 n=0 where β = 1/(kT ) and the total number of particles is:
∞ Problem 2 Here, we consider the potential 1 mω 2 x2 of the harmonic oscillator, where ω is 2 a constant. We deﬁne the operator A ≡ y + iq , where y ≡ x mω/2 and q ≡ √ p/ 2mω . Since x and p are physical observables, x† = x and p† = p and subsequently y † = y and q † = q . (a) (4) (b) A, A† = (y + iq )(y − iq ) − (y − iq )(y + iq ) = [y, y ] + [y, −iq ] + [iq, y ] + [iq, −iq ] , We know that y and q commute with themselves, and that [y, q ] = −[q, y ]. A† = (y + iq )† = y † − iq † = y − iq N=
n=0 N0 e −nβ ω0 . (3) The average energy is then given by the expression: U = N
∞ −nβ ω0 n=0 N0 ω0 ne ∞ −nβ ω0 n=0 N0 e 1 ω0 . + 2 Cancelling out common factors of N0 and noting that (n ω0 )e−βn we can simplify Eq. (4) to:
∂ − ∂β U = N ∞ −nβ ω0 n=0 e ∞ −nβ ω0 n=0 e ω0 =− ∂ −βn e ∂β ω0 + 1 ω0 . 2 Employing these results, we ﬁnd that (5) November 18, 1999 A, A† = −2i[y, q ]. Physics H7C Fall 1999 From the deﬁnition of y and q , [y, q ] = x From which we conclude: A, A† = −2i[y, q ] = . 1 p2 + mω 2 H= 2m 2 Since p2 = 2mωq 2 and x2 = 2y 2 /(mω ), we ﬁnd H = ω (q 2 + y 2 ). (d),(e) Note that from the deﬁnitions of A and A† , we have y= and q= Thus we ﬁnd that y2 = and q2 = A + A† 2 A − A† . 2i (c) p mω 1 ,√ = [x, p] = i /2. 2 2 2mω Solutions to Problem Set 10
(f) Since constants commute with anything /2, A
† Derek Kimball = 0. Thus we get: H, A† = ω AA† , A† = ω AA† A† − A† AA† = ω A, A† A† = ω A† . (g) Similarly, [H, A] = ω AA† , A = ω AA† A − AAA† = ωA A† , A = − ω A. Problem 3 Consider the angular momentum operator L ≡ r × p = ( /i)r × ∇. For example, Lx ≡ ( /i) y (∂/∂z ) − z (∂/∂y ) . Deﬁne L± ≡ Lx ± iLy . (a) [Lx , Ly ] = [ypz − zpy , zpx − xpz ] = [ypz , zpx ] + [zpy , xpz ] − [zpy , zpx ] − [ypz , xpz ] (8) Recall that in ﬁguring out these commutation relations, it often helps to think of the commutators as operators acting on functions. This is especially helpful in dealing with commutators involving derivatives. We’ll look at each of the terms in the above expression individually: [ypz , zpx ] = − 2 y [ypz , zpx ] = − 2 yz ∂2 − ∂z∂x ∂∂ z + ∂z ∂x
2 2 z 12 A + A†2 + AA† + A† A 4 ∂∂ y ∂x ∂z ∂2 ∂ = − 2y ∂x∂z ∂x y ∂ + ∂x 2 yz 1 −A2 − A†2 + AA† + A† A 4 From (c) and the above considerations we have that H = ω (q 2 + y 2 ) = ω AA† + A† A . 2 [ypz , zpx ] = − 2 y [zpy , xpz ] = − 2 z [zpy , xpz ] = − 2 zx + A† A ∂ ∂x
2 ∂∂ x + ∂y ∂z
2 x ∂∂ z ∂z ∂y
2 By adding and subtracting AA† or A† A where appropriate, AA† = AA† − A† A + A† A = A, A† + A† A = and also A† A = A† A − AA† + AA† = A† , A + AA† = − + AA† . These expressions can be used in our above expression for H, and from them we ﬁnd ω AA† + A† A = ω A† A + /2 = ω AA† − /2 . H= 2 November 18, 1999 ∂2 + ∂y∂z x ∂ + ∂y xz ∂2 ∂z∂y [zpy , xpz ] = [zpy , zpx ] = − 2 z 2 x ∂ ∂y
2 ∂∂ z + ∂y ∂x ∂2 + ∂y∂x z z ∂∂ z ∂x ∂y ∂2 ∂x∂y [zpy , zpx ] = − 2 z 2 22 Physics H7C Fall 1999 [zpy , zpx ] = 0 [ypz , xpz ] = − 2 y ∂∂ x + ∂z ∂z ∂2 [ypz , xpz ] = − 2 yx 2 + ∂z [ypz , xpz ] = 0 ∂∂ y ∂z ∂z ∂2 2 xy 2 ∂z
2 Solutions to Problem Set 10
[L+ , L− ] = 2 Lz x (e) [L− , Lz ] = [Lx − iLy , Lz ] = [Lx , Lz ] − i[Ly , Lz ] [L− , Lz ] = −i Ly − i(i Lx ) = (Lx − iLy ) [L− , Lz ] = L− ∂ i ∂y ∂ i ∂x (f) [L+ , Lz ] = [Lx + iLy , Lz ] = [Lx , Lz ] + i[Ly , Lz ] [L+ , Lz ] = −i Ly + i(i Lx ) = − (Lx + iLy ) [L+ , Lz ] = − L+ Derek Kimball Now we can put these simpliﬁed expressions into Eq. (8), and we ﬁnd: [Lx , Ly ] =
2 x ∂ ∂ −y ∂y ∂x =i x −y [Lx , Ly ] = i (xpy − ypx ) [Lx , Ly ] = i Lz (b),(c) The arguments used in (a) can be basically repeated, just changing the (g) identities of some of the variables. Or you can argue that since space is rotationally L2 , Lz = L2 , Lz + L2 , Lz + L2 , Lz = L2 , Lz + L2 , Lz x y z x y invariant, if we rotate our coordinate system in such a way that x → y , y → z 2 and z → x, the same commutation relation holds with the appropriate change of L , Lz = Lx Lx Lz − Lz Lx Lx + Ly Ly Lz − Lz Ly Ly coordinate names. The basic principle is that for any (i, j, k) which are a cyclic Now we employ a common trick in calculating commutation relations. We add permutation of (x, y, z ), we have: and subtract terms which allow us to substitute commutators we know into our [Li , Lj ] = i Lk . expression. We know that in general if we interchange two operators in a commutator, the result of the commutator acquires a negative sign. In other words, for any two operators A, B : [A, B ] = −[B, A]. Thus if (i, j, k) are an anticyclic permutation of (x, y, z ) (e.g., (y, x, z )), we have: [Li , Lj ] = −i Lk . (d) [L+ , L− ] = [Lx + iLy , Lx − iLy ] = [Lx , Lx ] + i[Ly , Lx ] − i[Lx , Ly ] + [Ly , Ly ] An operator always commutes with itself, so [Lx , Lx ] = 0, Thus we have [L+ , L− ] = −2i[Lx , Ly ] = −2i(i Lz ) November 18, 1999 L2 , Lz = 0 [Ly , Ly ] = 0. L2 , Lz = Lx Lx Lz −Lx Lz Lx +Lx Lz Lx −Lz Lx Lx +Ly Ly Lz −Ly Lz Ly +Ly Lz Ly −Lz Ly Ly L2 , Lz = Lx [Lx , Lz ] + [Lx , Lz ]Lx + Ly [Ly , Lz ] + [Ly , Lz ]Ly Maybe, if you are like me, when you ﬁrst see this trick it seems quite clever. This trick is used very often because of the potential noncommutativity of operators. Suppose we have two operators A, B which don’t commute. If we have AB and want to get BA for some reason, we can use AB = AB − BA + BA = [A, B ] + BA. This is a very useful operator identity! Anyhow, continuing on with the math by substituting in results from (a),(b) and (c) of this problem: L2 , Lz = −i Lx Ly − i Ly Lx + i Ly Lx + i Lx Ly Physics H7C Fall 1999 Solutions to Problem Set 10 Derek Kimball This relation proves it is possible to ﬁnd simultaneous eigenfunctions of both operators. This is an important result, so it probably won’t hurt to see why this is Problem 4 the case once more. Suppose we have operators A, B where [A, B ] = 0. Consider an eigenfunction of B , ψb , with eigenvalue λb . Is Aψb an eigenfunction of B ? It The spherical harmonic Ylm (θ, φ) is an eigenfunction of L2 with eigenvalue and also of Lz with eigenvalue m. It is normalized so that is, since BAψb = ABψb = A(λb ψb ) = λb (Aψb ). ∗ dΩ Ylm (θ, φ)Ylm (θ, φ) = 1. As you can see, this result relies on the fact that A and B commute. Thus measurement of one observable does not aﬀect measurement of the other observable. This shows that there is no fundamental quantum uncertainty in measurement of (a) observables which correspond to commuting operators. So we can ﬁnd a function Consider the integral: ψa b which is is an eigenfunction of both A and B with eigenvalues λa and λb , as ∗ we do for L2 and Lz . dΩ(L+ Yll )∗ (L+ Yll ) = dΩYll (L− L+ )Yll (h) We can use arguments analagous to those above to show that: L , Lx = 0,
2 2 l(l +1) where we make use of the fact that:
∗ ∗ (L+ Yll ) = Yll L† = Yll L− + ∗ L , Ly = 0. 2 If you want to avoid the math, you can just use the isotropy of space to claim From problem 3 parts (i),(j) we have: that L2 should not preferentially commute with a particular direction in space. It L− L+ = L2 − L2 − Lz . z immediately follows that 2 Conveniently, Yll is an eigenfunction of L and Lz with eigenvalues L2 , L± = 0 l respectively. Thus we obtain (i),(j) First, let’s consider L+ L− and L− L+ : L+ L− = (Lx + iLy )(Lx − iLy ) = L2 + iLy Lx − iLx Ly + L2 x y L+ L− = L2 + L2 − i[Lx , Ly ] = L2 + L2 + Lz x y x y To go from L+ L− to L− L+ we use the identity L+ L− = [L+ , L− ] + L− L+ = L− L+ + 2 Lz . Since and we have: L2 = L− L+ + L2 + Lz z L2 = L+ L− + L2 − Lz z L2 = L2 + L2 + L2 x y z L+ L− = L2 + L2 + Lz , x y L− L+ = L2 + L2 − Lz x y (b) dΩ(L+ Yll )∗ (L+ Yll ) = − Therefore, it must be that L+ Yll = 0
∗ dΩYll 2 2 l(l + 1) and l(l + 1) − l2 − l Yll = 0. We can use quite similar methods to ﬁnd C− (l, m). Consider the integral: dΩ(L− Ylm )∗ (L− Ylm ) = From problem 3 parts (i),(j) we have: L+ L− = L2 − L2 + Lz . z Ylm is an eigenfunction of L2 and Lz with eigenvalues Thus we have the integral dΩ(L− Ylm )∗ (L− Ylm ) =
∗ dΩYlm 2 2 ∗ dΩYlm (L+ L− )Ylm l(l +1) and m respectively. l(l + 1) − m2 + m Ylm Because the Ylm ’s are orthonormal, we have that November 18, 1999 Physics H7C Fall 1999 C− (l, m)2 =
2 Solutions to Problem Set 10
l(l + 1) − m2 + m n1 1 1 2 2 1 3 2 3 n2 1 2 1 2 3 1 3 2 n3 1 1 1 1 1 1 1 1 Energy 6 9 9 12 14 14 17 17 Derek Kimball Problem 5 (a) Since the atom is in a state of deﬁnite projection mj = 5/2 of its total (spin + orbital) angular momentum on the z axis, j ≥ 5/2. This follows from the fact that possible values for mj range between +j and −j . What values of j are possible for a one electron atom with orbital angular momentum l = 2? The values of j range between l + s and l − s where s is the electron spin. Therefore, in general Problem 7 j = 5/2, 3/2 are possible, but since the atom is in the mj = 5/2 state we know j = 5/2. (a) (b) Consider J 2 = (L + S) = L2 + S 2 + 2L · S.
2 2 Rohlf 8.21 Since the atom is in an eigenstate of J 2 , L2 and S 2 , we have that: L·S= 2 (j (j + 1) − l(l + 1) − s(s + 1)).
2 Of course, ﬁrst we specify that we know l and s. The possible values for j are j = l + s, l + s − 1, ..., l − s. We choose one particular value of j . Then there are 2j + 1 states with total angular momentum j . There are many ways to show this result. You could start with a “stretched” state (mj = ±j ) and use the raising or lowering operator, for example. But if you simply note that mj can take on the possible values mj = j, j − 1, ..., −j + 1, −j and count these states, we see immediately that the number of states is 2j + 1. (b) The quantum numbers of the n = 2 states of hydrogen in terms n, l, ml and ms are shown in the following table: n 2 2 2 2 2 2 2 2
π2 2mL2 )
2 With j = 5/2, l = 2, and s = 1/2, we obtain L · S = L2 S 2 = So we ﬁnd, √ cos θ = 2 3
2 . Also,
2 l(l + 1) · s(s + 1) = 3 √. 2 Problem 6 Rohlf 7.27 The energies for the particle in the 3D box are given by: En1 ,n2 ,n3 = π n2 + n2 + 4n2 . 2 3 2mL2 1
22 l 1 1 1 1 1 1 0 0 ml 1 1 0 0 1 1 0 0 ms 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 The following table shows the ﬁrst 5 unique energies (in units of quantum numbers of the states that possess them. Note there are 8 states in total in the table. and the The quantum numbers of the n = 2 states of hydrogen in terms n, l, j and mj are shown in the following table: November 18, 1999 Physics H7C Fall 1999 n 2 2 2 2 2 2 2 2 l 1 1 1 1 1 1 0 0 j 3/2 3/2 3/2 3/2 1/2 1/2 1/2 1/2 mj 3/2 1/2 1/2 3/2 1/2 1/2 1/2 1/2 Solutions to Problem Set 10 Derek Kimball So we get energy shifts proportional to ml + 2ms . For hydrogen in the n=3 state with a strong B ﬁeld, we have the following possible values for the angular momentum quantum numbers and energy shifts in units of e B /(2m): ml 2 2 1 1 0 0 1 1 2 2 ms 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 ∆E 3 1 2 0 1 1 0 2 1 3 Here there are also 8 states. The system we are considering is described by an 8D Hilbert space, so any complete, orthonormal set of eigenfunctions which span the space must consist of 8 states. (c) Since Jz = Lz +Sz , for the state mj = 3/2, l = 1, ml = 1 and ms = 1/2. This is the “stretched” state, and we can readily convert from the n, l, j and mj basis to the n, l, ml and ms basis by making the correspondence between the stretched states and employing the raising and lowering operators. Note also that J± = L± + S± . (b) Such transformations are used quite often and are tabulated (these are the famed In the absence of a magnetic ﬁeld the energy separation ∆E0 of the 3p and 1s ClebschGordan coeﬃcients). states is 13.6 eV = 12.1 eV. ∆E0 = 13.6 eV − (d) 9 The electric dipole transition selection rules demand that the diﬀerence in the If mj = 1/2, then we can have projection of the orbital angular momentum on the z axis bewteen the initial and (l, ml , ms ) = (0, 0, 1/2) ﬁnal states of an atomic transition must obey (l, ml , ms ) = (1, 0, 1/2) (l, ml , ms ) = (1, 1, −1/2). Problem 8 Rohlf 8.25 ∆ml = 1, 0, −1. So the energies of the photons Eγ can be Eγ = 12.1 eV ± eB , 12.1 eV. 2m (a) The magnetic dipole moment µ of a hydrogen atom, in the limit of a strong B ﬁeld, is given by: e µ=− (L + 2S) 2m and the energy shift ∆E due to the external ﬁeld is given by ∆E = −µ · B. November 18, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 11 1. Rohlf 8.41. 2. Rohlf 9.3. 3. Rohlf 9.10. 4. Rohlf 9.25. 5. Rohlf 9.31. For those of you itching to do more, here is a preview of the ﬁrst three problems in Problem Set 12: 1. Rohlf 12.5. 2. N electrons each of mass m are conﬁned within a (formerly) cubic inﬁnite potential well that has been “squashed” almost ﬂat: V = 0 for (0 < x < L and 0 < y < L and 0 < z < L), V = ∞ otherwise. Here 1 (cube is “squashed” in the z direction) and N 1. The electrons do not interact with each other and are at very low temperature so that they ﬁll up the available states in order of increasing energy. Take N 1, so that the z part of each electron’s wavefunction may be assumed to be the same (lowest possible kz ). Thus the problem is reduced to two dimensions. Calculate the diﬀerence ∆ between the energy of the most energetic electron (Fermi energy) and the energy of a ground state electron, using the approximation N 1. ∆ should depend on m, N , and L, but not . 3. Write an integral equation for the fraction F of nonrelativistic fermions in a gas at ﬁnite temperature T which have energy above the Fermi energy EF . The density of states is proportional to E 1/2 and the probability that a state is occupied is 1 exp β (E − EF ) + 1 where β = (kT )−1 . You don’t need to perform the integration, but you should set up the integral so that doing it would yield the correct answer without any additional physical reasoning. Physics H7C Fall 1999 Solutions to Problem Set 11
Problem 3 Rohlf 9.10 Derek Kimball “Civilization as we know it is based on ten ideas. These are Newton’s three laws, the three laws of thermodynamics and Maxwell’s equations. Everything that you see around you which diﬀerentiates modern times from the past is based on these concepts. Soon, Schrodinger’s equation will join these ten ideas. People will tell you that civilization is about art, or literature, or architecture... but these components of civilization have more or less been the same for thousands of years. The only real diﬀerence between today and thousands of years ago is physics. If civilization collapsed tomorrow, we could rebuild it in the same fashion armed with these ideas.”  Prof. Seamus C. Davis, U.C. Berkeley. An atom has two electrons in the dsubshell. What are the possible values of total z angular momentum? Well, we know that the total orbital angular momentum ltot can range from l1 + l2 to l1 − l2 . Thus ltot = 4, 3, 2, 1, 0. If ltot is even, then the spatial part of the wavefunction Ψspatial is symmetric, and if ltot is odd, then Ψspatial is antisymmetric. With two electrons, the total spin can be stot = s1 + s2 = 1, 0. If stot = 1 then the spin function Ξspin is symmetric. The total wavefunction Φtotal = Ψspatial · Ξspin If you have any questions, suggestions or corrections to the solutions, don’t hesitate must be antisymmetric since we are dealing with identical fermions. Therefore if to email me at [email protected]! stot = 1 then Ψspatial must be antisymmetric, meaning that ltot is odd (1 or 3). In this case the total angular momentum j can take on the values j = 4, 3, 2, 1, 0. Problem 1 Rohlf 8.41 If stot = 0 then the spin function Ξspin is antisymmetric, and Ψspatial must be symmetric. In this case ltot is even (0, 2 or 4). The possible values of j are 0, 2 or Here we wish to show that the average value of 1/r is independent of the orbital angular momentum for the hydrogen atom. This result is straightforward if we 4. introduce the virial theorem (see, e.g., B.H. Bransden and C.J. Joachain, Intro The largest j value possible is 4, so the possible mj values are: duction to Quantum Mechanics, pgs. 227228), which states (in one particular mj = −4, −3, −2, −1, 0, 1, 2, 3, 4 form) that for a spherically symmetric potential V (r) ∝ rn one has for a stationary state: 2 T = n V (r) , Rohlf 9.25 where T is the kinetic energy. This is analagous to a classical result of the same Problem 4 name. The average potential energy for hydrogen is proportional to 1/r , as is the kinetic energy and hence the total energy from the virial theorem. We know (a) that the total energy for the hydrogen atom (from the Bohr model) is independent of l and depends only on the principal quantum number n. Consequently, 1/r is Sodium atoms are placed in a magnetic ﬁeld of 1.5 T. The Zeeman splitting of the also independent of the orbital angular momentum. ground state (n = 0, l = 0) is given by the shift of energy due to the diﬀerent spin states the single valence electron can have. Energy shifts are Problem 2 Rohlf 9.3 E± = ±µB B We seek a totally antisymmetric wavefunction Ψ for 3 electrons in terms of ψa (r1 ), where µB is the Bohr magneton. The numerical value for the splitting is given by ψb (r2 ) and ψc (r3 ). The wavefunction must be totally antisymmetric because we ∆E = 2µB B = 2(6 × 10−5 ev/T)(1.5 T) = 1.8 × 10−4 eV. have three identical fermions. Such a wavefunction is given below. It is easily veriﬁed that under particle interchange it ﬂips sign. (b) 1 Ψ = √ [ψa (r1 )ψb (r2 )ψc (r3 ) − ψa (r1 )ψc (r2 )ψb (r3 ) + ψb (r1 )ψc (r2 )ψa (r3 ) 3! −ψb (r1 )ψa (r2 )ψc (r3 ) + ψc (r1 )ψa (r2 )ψb (r3 ) − ψc (r1 )ψb (r2 )ψa (r3 )] If 1/3 of the sodium atoms are in the higher energy state, then the Boltzmann factor 1/3 = 1/2. e−∆E/(kT ) = (1) 2/3 November 23, 1999 Physics H7C Fall 1999 From which we calculate kT = Hence, T ≈ 2.9 K (c) Same as above: e−∆E/(kT ) = 49/51. From which we calculate and T ≈ 51 K Problem 5 Rohlf 9.31 kT = 4.4 × 10−3 eV ∆E = 2.5 × 10−4 eV. ln 2 Solutions to Problem Set 11 Derek Kimball A sample of Na atoms are placed in a 1.0 T magnetic ﬁeld. We calculate the energy shifts for the 3s1/2 , 3p1/2 and 3p3/2 states. Note that in this case µB B = 6 × 10−5 eV/T. For the 3s1/2 state (s = 1/2, l = 0, and j = 1/2), the Lande factor is given by: gL = 1 + j (j + 1) + s(s + 1) − l(l + 1) . 2j (j + 1) = 2 Therefore the energy shifts, given by ∆E = µz Bz = gL µB Bmj are ∆E = ±µB B. Similarly, for the 3p1/2 state, gL = 2/3 so 1 ∆E = ± µB B. 3 For the 3p3/2 states gL = 4/3 so 2 ∆E = ± µB B, 3 ± 2µB B. November 23, 1999 1 University of California, Berkeley Physics H7C Fall 1999 (Strovink) PROBLEM SET 12 1. Rohlf 12.5. 2. N electrons each of mass m are conﬁned within a (formerly) cubic inﬁnite potential well that has been “squashed” almost ﬂat: V = 0 for (0 < x < L and 0 < y < L and 0 < z < L), V = ∞ otherwise. Here 1 (cube is “squashed” in the z direction) and N 1. The electrons do not interact with each other and are at very low temperature so that they ﬁll up the available states in order of increasing energy. Take N 1, so that the z part of each electron’s wavefunction may be assumed to be the same (lowest possible kz ). Thus the problem is reduced to two dimensions. Calculate the diﬀerence ∆ between the energy of the most energetic electron (Fermi energy) and the energy of a ground state electron, using the approximation N 1. ∆ should depend on m, N , and L, but not . 3. Write an integral equation for the fraction F of nonrelativistic fermions in a gas at ﬁnite temperature T which have energy above the Fermi energy EF . The density of states is proportional to E 1/2 and the probability that a state is occupied is 1 exp β (E − EF ) + 1 where β = (kT )−1 . You don’t need to perform the integration, but you should set up the integral so that doing it would yield the correct answer without any additional physical reasoning. 4. Rohlf 12.16. 5. Rohlf 17.27. 6. Rohlf 18.11. 7. Rohlf 19.18. 8. Rohlf 19.29. Physics H7C Fall 1999 Solutions to Problem Set 12
f (0) = 1 f (1) = 0.79 f (2) = 0.62 f (3) = 0.44 f (4) = 0.32 f (5) = 0.12 f (6) = 0.6 f (E > 6) = 0 Derek Kimball “Entropy in the universe is always increasing. At some point, the universe will reach its maximum state of entropy and then no work can be done. The universe will become a cold, lifeless place. This is known as the heat death of the universe. Get ready, it’s coming...”  Prof. Seamus C. Davis, U.C. Berkeley. If you have any questions, suggestions or corrections to the solutions, don’t hesitate to email me at [email protected]! Problem 1 Rohlf 12.5 Consider a system of 6 spin1/2 fermions having a total energy of 10 units. The fermions are in a quantum mechanical system where the ground state has 0 energy units, the ﬁrst excited state has 1 energy unit, the second excited state has 2 energy units, etc. Determine the energy distribution function df /dE . Make an estimate of the Fermi energy. We make a table of the possible distributions and the spin degeneracy of each state (i.e., if there is an isolated electron, it can be either spin up or spin down). We then total up number of times an electron is found in a state, and from this total we arrive at our distribution function f , which describes the probability to ﬁnd an electron in a particular energy level. The total number of times we ﬁnd an electron in an energy level for a distinct distribution (including spin degeneracy) is shown on the bottom line. Degen. 4 4 1 16 4 1 4 4 TOTAL Energy 0 2 2 2 2 2 2 1 1 68 The Fermi energy EF is where this distribution has f (EF ) ≈ 1/2. This is seen from the FermiDirac distribution: fF D (E ) = 1 e(E −EF )/(kT ) + 1 fF D (EF ) = 1/2. This is around E = 3 for this system. It doesn’t work out exactly because this is a discrete system with a small number of possible distributions. If the number of particles was greatly increased, the distribution would become increasingly welldescribed by fF D , which is derived in the large N limit. Problem 2 1 2 2 2 1 1 0 2 2 54 2 1 0 0 1 2 2 1 2 42 3 0 1 0 1 0 2 2 0 30 4 0 0 2 1 0 0 0 1 22 5 0 1 0 0 1 0 0 0 8 6 1 0 0 0 0 0 0 0 4 The energy of the 3D inﬁnite square well in this case is given by: En = π n2 z n2 + n2 + 2 . x y 2 2mL
22 is small, so since the number of particles is chosen to be small ( N 1), the temperature must be large in order to excite states with nz = 1. We consider the case of low temperature, where always nz = 1. In this case, the problem is eﬀectively 2D, with the energies:
22 π 1 n2 + n2 + 2 . x y f , which in this case is discrete, is just the ratio of the various total number of 2mL2 times a particle is found in an energy level in one of the distributions to the ground state (E = 0) number. So f (E ) is a discrete function described by the following We seek the diﬀerence ∆ between the ground state energy and the Fermi energy, both of which have a term relations: 22 π 2mL2 2 December 3, 1999 En = Physics H7C Fall 1999 Solutions to Problem Set 12
Problem 3 Derek Kimball which will cancel out. First, let’s calculate the Fermi energy. The FermiDirac distribution at T = 0 is given by fF D = 1 E < EF fF D = 0 E > EF . We start with the knowledge that the density of states is proportional to E 1/2 and the probability of occupation is P (E ) = 1 . eβ (E −EF ) + 1 The electrons will try to achieve the lowest energy possible, ﬁlling up states in accordance with the Pauli exclusion principle. The total number of particles is given by: ∞ Let dN fF D (E )dE 2 N= dE 0 dN = cE 1/2 . dE where the 2 is for spin degeneracy of the electrons. At T = 0, this integral becomes: where c is a constant. The fraction of nonrelativistic fermions in a gas of ﬁnite temperature T above the Fermi energy is given by the integral: EF dN dE. N= 2 ∞ dE P (E ) dN dE 0 dE EF F= ∞ . P (E ) dN dE dE 0 Now we must determine the density of states for 2D. The energy of a state (ne22 π glecting the common factor 2mL2 2 ) is ∝ N 2 where N 2 = n2 + n2 and is the square The denominator of the above equation is just the total number of particles, which x y of the total number of available states. The derivative of N 2 with respect to energy is easiest to evaluate at T = 0 where, since P (E ) is just the FermiDirac distribuis tion function, we have: dN 1 d2 N = 2πN P (E ) = 1 E < EF dE 4 dE P (E ) = 0 E > EF . where the factor of 1/4 arises because we consider only positive nx , ny and the 2π is from the integration about a ring of thickness dN in nspace. From this we can So the denominator is simply explicitly solve for dN : dE √ 4mL2 1 2 2mL −1/2 dN = 23 = E . dE πN π2
EF 0 dN dE = dE EF 0 cE 1/2 dE = 2 3/ 2 cE . 3F Now we employ this expression for dN in our integral for the total number of dE particles: √ √ EF 2mL 1/2 2 2mL −1/2 N= 2 E dE = 8 EF . π2 π2 0 Solving for EF yields EF = π N2 128mL2
24 So employing all of our information, the simplest expression we can get for F is: F= 3 −3/2 E 2F
∞ EF eβ (E −EF ) E 1/2 dE +1 Problem 4 Rohlf 12.16 We want to deduce the expression for density of states of a relativistic electron ˆ gas. We ﬁrst get the density of states with respect to k (where k is the electron wave vector) in the usual manner for a 3D particle in a box problem. The volume of a shell of thickness dk in k space (considering only positive kx , ky , and kz ) is (4πk 2 dk )/8. From the boundary conditions ki = (π/L)ni (where i = x, y, z and December 3, 1999 Physics H7C Fall 1999 Solutions to Problem Set 12 Derek Kimball the box is length L on a side), we know the number of states per unit volume of k space is (L/π )3 . So dN k2 V = dk 2π 2 where V is the volume of the box. Dividing by volume, to get the density of states per unit volume, and converting to momentum p using the deBroglie relation p = k , we have: dn 4πp2 = 3. dp h To convert this expression into density of states per unit energy ρ(E ), we use: ρ(E ) = dn dn dp = . dE dp dE where s1 = 1/2 and s2 = 1/2 are the spins of the quarks. The orbital angular momentum l of charmonium can be l = 1, 0 in the n = 2 state. The total (spin + orbital) angular momentum is given by J = L + S = l + s, ... , l − s. So we have the following possible states: j = 2, 1, 0 j=1 j=1 j=0 which total six. (b) The unobserved state has s = 0, l = 1 and j = 1 by inspection. (c) By analogy to similar states in the chart of charmonium, the energy diﬀerence between states with aligned vs. antialigned spins of the quarks is ∼ 100 MeV (the energy diﬀerence of the ψ (2s) and ηc (2s) states and ψ (1s) and ηc (1s) states, see Rohlf pg. 494). One could imagine that the energy splitting arises because of some spinspin interaction between the quarks, so the s = 0, l = 1, j = 1 state of charmonium (called the or smiley particle) should be split in energy from the χc1 (2p) state by ∼ 100 MeV. So the mass of should be roughly 3400 MeV. (d) The particle is not observed because the method by which all the states of charmonium were observed involved creation of a ψ (2s) particle and subsequent electromagnetic decay. The energy of emitted photons was measured and the spectrum of charmonium particles was established. Note that decays of ψ (2s) → change both s and l by 1. Such a decay would involve interaction with both the electric (to change l) and magnetic (to change s) components of the photon and is highly suppressed. for s = 1, l = 1 for s = 0, l = 1 for s = 1, l = 0 for s = 0, l = 0 The relativistic expression for momentum in terms of energy is: p= Therefore 1 c E 2 − m2 c4 . E dp =2. dE cp ρ(E ) = 2 4πpE c2 h3 where the factor of 2 is for spin degeneracy. The relativistic momentum is p = γmv and v ≈ c. The relativistic energy is E = γmc2 . Substituting these expressions in, ρ(E ) = 2 4πm2 c 2 γ h3 We now solve for ρ(E ): Problem 5 Rohlf 17.27 (a) There are six states of charmonium with n = 2. Charmonium is a bound state of a charm and anticharm quark. We don’t have to worry about symmetrization of the wavefunction because these are not identical fermions. The total spin s of the system can be: s = s1 + s2 = 1, 0 December 3, 1999 Physics H7C Fall 1999 Problem 6 Rohlf 18.11 Solutions to Problem Set 12
(b) Derek Kimball The tau particle has ﬁve times as many decay channels as the muon, so the phase space is increased by a factor of 5. So the tau lifetime is given, from the above arguments, by: 5 (a) 1 105.7 MeV = 0.3 ps. τ τ = τµ 5 1777 MeV Consider this problem in the context of the theory of weak interactions as it existed before Glashow, Weinberg, Salam discovered how to unify it with the theory of electromagnetic interactions. In this context (Rohlf p. 509), the coupling constant Rohlf 19.18 for weak interactions is the Fermi constant GF . According to Rohlf’s Eq. (18.28), Problem 7 3 GF has dimensions GeV fm . In a system of “natural” units in which = c = 1, we can transform a length (fm) into an inverse energy (inverse GeV) using the fact We have redshift parameter z = 2. We can employ the formula (19.15) on pg. 539 of Rohlf: that c ≈ 0.2 GeV fm. In natural units, GF therefore has dimensions GeV−2 . 1+β (1 + z )2 = Since GF is a coupling constant, like the ﬁne structure constant α, it describes 1−β the strength of a quantum mechanical amplitude. The rate is proportional to the square of the modulus of this amplitude. Thus the decay rate W (which is inversely where β = v/c as usual. We can solve for β : proportional to the muon lifetime τµ ) is proportional to the square of GF . (1 + z )2 − 1 = 0.8 β= 2 Now we have a dilemma. If GF  were the only dimensionful component of W , (1 + z )2 + 1 W would have units of GeV−4 . However, using the fact that = 6.6 × 10−25 GeV sec, in natural units we know that W must have units of sec−1 or GeV. So far we and then use β in Hubble’s law to determine the distance to the galaxy. Hubble’s law is are oﬀ by ﬁve powers of energy! βc d= The solution is to bring in the only other relevant dimensionful quantity around, H0 the muon mass mµ . Remembering that mc2 is the same as m in natural units, where H is the Hubble constant. We ﬁnd: 0 we ﬁnd that we need ﬁve powers of mµ in the numerator of W to make its units correct. Therefore its inverse, the muon lifetime, must have ﬁve powers of mµ in βc 0.8 × (3 × 108 m/s) d= = = 3400 Mpc. its denominator. H0 7 × 104 m/s · Mpc− 1 Alternatively, we can approach this problem a bit more formally. Fermi’s Golden rule says that the decay rate W is given by: W= 2π M2 × (phase space) Problem 8 Rohlf 19.29 where M is a transition amplitude obtained from perturbation theory (you’ll learn 2 all about this in 137B). In this case, all we need to know is that M ∝ 1/MW . By dimensionality, we need to cancel the mass of the W boson with something, the best guess is the mass of the muon. The phase space available to the decay products in this case is proportional to the available energy, in other words the muon mass again. So M2 × (phase space) ∝ m5 µ which again implies τµ ∝ m−5 . µ As a rough estimate, we simply set the thermal energy of particles in the early universe ∼ kT equal to the mass of 2 bottom quarks (actually a bottom and antibottom, which have the same mass). The bottom quarks must be produced in pairs so that “beauty” is conserved, since the b and ¯ have equal and opposite beauties. b You begin to wonder where this stuﬀ comes from. Anyhow, the mass of 2 bottom quarks is 10 GeV, which implies T = 1014 K. The characteristic expansion time texp comes from texp = 1 = H (t) 2.7 K T
2 2.7 K 3c2 = 5 × 1019 s 8πρG T 2 . December 3, 1999 Physics H7C Fall 1999 Solutions to Problem Set 12 Derek Kimball You should probably check out the discussion on pp. 558559 of Rohlf. So we can estimate: 4 × 1020 s · K2 texp ≈ = 4 × 10−8 s. T2 And that’s all folks! Good luck on your ﬁnals! Merry winter break! December 3, 1999 University of California, Berkeley Physics H7C Fall 1999 (Strovink) EXAMINATION 1 Directions. Do both problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free to 2 approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Don’t use a calculator, which you don’t need – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (58 points) In a freeelectron laser, a beam of relativistic electrons is subjected to a transverse magnetic ﬁeld that varies sinusoidally with lab coordinate z , the (average) beam direction: B = xB0 cos ˆ 2πz λ0 where t is the time as observed at the origin of S , compute ω0 in terms of the constants previously given. c. (8 points) Consider an electron of charge −e and mass m whose average position is x , y , z = (0, 0, 0) as observed in S . In this frame, its velocity is so small that you may ignore v × B with respect to E . In frame S , making this approximation, compute the electron’s motion y (t ). (In case you didn’t get part b. exactly right, leave your answer in terms of ω0 .) d. (8 points) The electric dipole moment p of a distribution of N point charges qi at positions ri is deﬁned as
N where B0 and λ0 are constants. In the lab, the z component of the electrons’ velocity is vz = β0 c where β0 is a constant. a. (8 points) Consider a Lorentz frame S moving with velocity z β0 c = ˆβ0 c with respect to the lab. The Lorentz transformation for electromagnetic ﬁelds is p= E =E B =B E⊥ = γ0 (E⊥ + β0 × cB⊥ ) cB⊥ = γ0 (cB⊥ − β0 × E⊥ ) ,
2 where γ0 ≡ (1 − β0 )−1/2 . Calculate the electric ﬁeld E seen in S ; continue to express it in terms of 2πz/λ0 . i=1 ri qi . The power P (t) radiated by a charge distribution with timevarying dipole moment p(t) is P= 1 2 (d2 p/dt2 )2 . 4π 0 3 c3 b. (8 points) Deﬁning 2πz/λ0 ≡ ω0 t , As seen in S , calculate P , the timeaveraged power radiated by a single electron in the freeelectron laser. e. (8 points) Energy and time both transform as the 0th component of a fourvector. Calculate P , the timeaveraged power radiated by a single electron as observed in the lab. You may leave your answer in terms of P. f. (10 points) For light, the relativistic Doppler shift is ω= ω . γ0 (1 − β0 cos θ) e. (8 points) Can a linear combination of a righthand and a lefthand circularly polarized plane wave in the region −L < y < L propagate in the z direction? If not, why not? If so, what combination(s) would be possible? Explain fully. Calculate the ratio λ0 /λ, where λ0 , as before, is the characteristic length describing the spatial variation in the lab of the freeelectron laser’s magnetic ﬁeld, and λ is the wavelength of the light that its electrons radiate in the forward direction, as observed in the lab. Express this ratio in terms of γ0 , in the limit β0 → 1. g. (8 points) What is the state of polarization of the freeelectron laser’s light? Explain. 2. (42 points) Semiinﬁnite regions y > L and y < −L are ﬁlled by perfect conductor, while the intervening slab −L < y < L is ﬁlled by dielectric with constant dielectric constant and permeability µ. a. (8 points) In SI units, write Maxwell’s equations for E and H inside the dielectric. Do not write any terms involving free charges or free currents, which both vanish there. b. (8 points) Prove that Ex and Ez both must vanish at y = ±L. For parts c. and d. only, assume, for −L < y < L, that the ﬁelds are given by Ephysical = Hphysical = yE2 exp i(kz − ωt) ˆ (xH1 + ˆH3 ) exp i(kz − ωt) ˆ z , where E2 , H1 , and H3 are unknown complex constants, and k and ω are unknown real constants. c. (8 points) Prove that H3 = 0. d. (10 points) Calculate the ratio H1 /E2 in terms of known quantities. University of California, Berkeley Physics H7C Fall 1999 (Strovink) SOLUTION TO EXAMINATION 1 Directions. Do both problems (weights are indicated). This is a closedbook closednote exam except for one 8 1 × 11 inch sheet containing any information you wish on both sides. You are free to 2 approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Don’t use a calculator, which you don’t need – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (58 points) In a freeelectron laser, a beam of relativistic electrons is subjected to a transverse magnetic ﬁeld that varies sinusoidally with lab coordinate z , the (average) beam direction: B = xB0 cos ˆ 2πz λ0 gives us the total electric ﬁeld seen in S : E = β0 × cB⊥ = γ0 β0 cB0 cos 2πz (z × x) ˆˆ λ0 2πz = γ0 β0 cB0 cos y. ˆ λ0 b. (8 points) Deﬁning 2πz/λ0 ≡ ω0 t , where t is the time as observed at the origin of S , compute ω0 in terms of the constants previously given. Solution. We need a Lorentz transformation to relate (ct, z ) to (ct , z ). Taking advantage of the fact that z = 0, we minimize algebra by choosing the inverse transformation: z = γ0 z + γ0 β0 ct = 0 + γ0 β0 ct = γ0 β0 ct 2π 2π z= γ0 β0 ct λ0 λ0 ≡ ω0 t 2πγ0 β0 c ω0 = . λ0 c. (8 points) Consider an electron of charge −e and mass m whose average position is x , y , z = (0, 0, 0) as observed in S . In this frame, its velocity is so small that you may ignore v × B where B0 and λ0 are constants. In the lab, the z component of the electrons’ velocity is vz = β0 c where β0 is a constant. a. (8 points) Consider a Lorentz frame S moving with velocity z β0 c = ˆβ0 c with respect to the lab. The Lorentz transformation for electromagnetic ﬁelds is E =E B =B E⊥ = γ0 (E⊥ + β0 × cB⊥ ) cB⊥ = γ0 (cB⊥ − β0 × E⊥ ) ,
2 where γ0 ≡ (1 − β0 )−1/2 . Calculate the electric ﬁeld E seen in S ; continue to express it in terms of 2πz/λ0 . Solution. There are no electric or parallel magnetic ﬁelds in the lab frame so the third equation with respect to E . In frame S , making this approximation, compute the electron’s motion y (t ). (In case you didn’t get part b. exactly right, leave your answer in terms of ω0 .) Solution. The Lorentz force on the electron in S is given by: F = −eE − ev × B , but we can ignore the v × B term since v is always small. So then we have an equation for the acceleration: me a = −eE . Plugging in E from (a.) and integrating twice with respect to t , y (t ) = e cos ω0 t γ0 β0 cB0 . m (ω0 )2 Plugging this result into the formula given above for radiated power P , P=
22 2 1 2 e4 γ0 β0 c2 B0 cos2 ω0 t . 4π 0 3 m2 c3 Since cos2 ω0 t = 1/2, the time average of P is given by: P =
222 1 1 e4 γ0 β0 B0 . 4π 0 3 m2 c e. (8 points) Energy and time both transform as the 0th component of a fourvector. Calculate P , the timeaveraged power radiated by a single electron as observed in the lab. You may leave your answer in terms of P. Solution. Since both energy and time transform as the 0th component of a fourvector, if we have measured a change in energy of the electron ∆E and a change in time ∆t in the electron’s rest frame S , then the same quantities in the lab frame are given by ∆E = γ0 ∆E and ∆t = γ0 ∆t . So P = ∆E /∆t = ∆E/∆t =P. d. (8 points) The electric dipole moment p of a distribution of N point charges qi at positions ri is deﬁned as
N p=
i=1 ri qi . The power P (t) radiated by a charge distribution with timevarying dipole moment p(t) is P= 1 2 (d2 p/dt2 )2 . 4π 0 3 c3 This was also solved in problem set 3, problem 3! f. (10 points) For light, the relativistic Doppler shift is ω= ω . γ0 (1 − β0 cos θ) As seen in S , calculate P , the timeaveraged power radiated by a single electron in the freeelectron laser. Solution. The second derivative of the timevarying electric dipole moment for a single electron in the free electron laser is given by: d2 p = −ea , dt 2 where a is the acceleration found in (c.): a =− e γ0 β0 cB0 cos ω0 t . m Calculate the ratio λ0 /λ, where λ0 , as before, is the characteristic length describing the spatial variation in the lab of the freeelectron laser’s magnetic ﬁeld, and λ is the wavelength of the light that its electrons radiate in the forward direction, as observed in the lab. Express this ratio in terms of γ0 , in the limit β0 → 1. Solution. In S , we know from (c.) that the electron is oscillating with angular frequency ω0 . Then, in S , the EM radiation produced by that electron has the same angular frequency. A forward observer in S , upon whom the beam impinges with θ = 0, sees this radiation with a Doppler shifted angular frequency ω0 that, by the above Doppler formula, is equal to ω0 = ω0 . γ0 (1 − β0 ) not write any terms involving free charges or free currents, which both vanish there. Solution. Within the dielectric, D = E and B = µH, where and µ are constants. The sourcefree equations are ∇·B=0 B = µH ⇒∇·H=0 and ∂B ∂t B = µH ∂H ⇒ ∇ × E = −µ ∂t ∇×E=− The useful sourcedependent equations are the variety that depend on free rather than total charges and currents, because free charges and currents are zero in the dielectric. These are ∇ · D = ρfree = 0 D= E ⇒∇·E=0 and ∇ × H = Jfree + D= E ∂E ⇒∇×H= ∂t b. (8 points) Prove that Ex and Ez both must vanish at y = ±L. Solution. Electric ﬁelds vanish in perfect conductors, because the inﬁnitely mobile free charges instantaneously rearrange themselves to shield out any externally applied electric ﬁeld. Consider a rectangular loop with long side S and short side s. One long side lies in the conductor, parallel to the plane y = L; the other long side lies in the dielectric. In the limit s → 0, the righthand side of Faraday’s law, E · dl = − d dt B · da , ∂D ∂D = ∂t ∂t Substituting ω0 = 2πc/λ, and plugging in the value of ω0 from (b.), 2πγ0 β0 c 1 2πc = λ λ0 γ0 (1 − β0 ) λ0 β0 = λ 1 − β0 1 + β0 β0 = 1 + β0 1 − β0 2 = γ0 β0 (1 + β0 ) .
2 This reduces to 2γ0 in the relativistic limit, a famous (and simple) result. If, say, λ0 is 0.1 m and the electron energy is 500 MeV (γ 2 ≈ 106 ), the FEL or wiggler can be made to radiate in the far UV (λ ≈ 0.05 µm), where no conventional laser is available. g. (8 points) What is the state of polarization of the freeelectron laser’s light? Explain. Solution. From (c.) we know that the electron oscillates in the y direction. The onaxis ˆ radiation from the dipole is propagating in the z direction, so light is polarized orthogonal to ˆ z . From the formula derived in class describing ˆ dipole radiation, we know the radiation is polarˆ ized in the θ direction where the vertical axis is deﬁned by the direction the dipole oscillates in. Hence we can conclude that the light is linearly polarized in the y direction. ˆ 2. (42 points) Semiinﬁnite regions y > L and y < −L are ﬁlled by perfect conductor, while the intervening slab −L < y < L is ﬁlled by dielectric with constant dielectric constant and permeability µ. a. (8 points) In SI units, write Maxwell’s equations for E and H inside the dielectric. Do vanishes because the area vanishes, and the contributions of the short parts to the rectangular loop on the lefthand side also vanish. The only nonvanishing contribution to the lefthand side is E · S in the dielectric. This proves that E in the dielectric must vanish at y  = L. This inˆ ˆ cludes E in either the x or z directions, which both are parallel to the interface. For parts c. and d. only, assume, for −L < y < L, that the ﬁelds are given by Ephysical = Hphysical = yE2 exp i(kz − ωt) ˆ (xH1 + ˆH3 ) exp i(kz − ωt) ˆ z , Setting the two values of H1 /E2 equal, − ω k =− µω k k √ µ. = ω Plugging this value for k/ω into either of the equations for H1 /E2 , H1 =− E2 . µ where E2 , H1 , and H3 are unknown complex constants, and k and ω are unknown real constants. c. (8 points) Prove that H3 = 0. Solution. As usual we require the complex electromagnetic ﬁelds to satisfy Maxwell’s equations (not just their (physical) real part). When their dependence on r and t is of the form exp i(k · r − ωt) , the operators ∇· and ∇× reduce to ik· and ik×, respectively, while the operator ∂/∂t reduces to −iω . Using the ﬁrst Maxwell equation in (a.), ∇·H=0 ˆ ik · (ˆH1 + z H3 ) = 0 x k = zk ˆ ⇒ ikH3 = 0 . d. (10 points) Calculate the ratio H1 /E2 in terms of known quantities. Solution. Using the methods of (c.), the second Maxwell equation in (a.) requires ˆ ik (ˆ × y )E2 = +iωµxH1 zˆ −ik xE2 = +iωµxH1 ˆ ˆ H1 k = − µω E2 while the fourth Maxwell equation in (a.) requires ˆ ik (ˆ × x)H1 = −iω y E2 zˆ ik y H1 = −iω y E2 ˆ ˆ ω H1 . =− E2 k e. (8 points) Can a linear combination of a righthand and a lefthand circularly polarized plane wave in the region −L < y < L propagate in the z direction? If not, why not? If so, what combination(s) would be possible? Explain fully. Solution. From (b.) we know that Ex = 0 at y  = L. In this part (only!) we are asked to assume that a plane wave is propagating in the z direction within the dielectric. This means ˆ that E must be ⊥ to z and it cannot depend on ˆ x and y . So, if Ex vanishes at the boundaries y  = L, it must vanish throughout the dielectric. Thus the only nonzero component of E lies in the y direction. We then ask, what comˆ bination of RH polarization (∝ x − iy ) and LH ˆ ˆ polarization (∝ x + iy ) add to pure y polarizaˆ ˆ ˆ tion? Evidently, the RH and LH waves must have equal amplitude and opposite sign. University of California, Berkeley Physics H7C Fall 1999 (Strovink) EXAMINATION 2 Directions. Do all four problems (weights are indicated). This is a closedbook closednote exam except for two 8 1 × 11 inch sheets containing any information you wish on both sides. You are free 2 to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Calculators are allowed but not essential – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (25 points) The basis of scalar diﬀraction theory is the FresnelKirchoﬀ integral formula. In Fowles notation (Eq. 5.11), this formula states Up = − × ikU0 exp (−iωt) × 4π exp ik (r + r ) n · ˆ − n · ˆ dA ˆr ˆr rr a. (10 points) Let δ be the maximum value of x2 + y 2 on the aperture plane for which the aperture is not opaque. Thus, for this part of the problem, there are three characteristic lengths: λ, δ , and D. By moving around in the plane z = D, restricting her own coordinates X, Y such √ D, the observer ﬁnds that X 2 + Y 2 that the optical disturbance there is proportional to the Fourier transform of g (x, y ). As someone who understands the physics of diﬀraction, you realize that this information implies that a single strong condition must be satisﬁed which relates λ, δ , and D. Write down this condition. (You needn’t prove it, and you may omit factors of order unity.) b. (15 points) For this part of the problem, take the aperture function to be g (x, y ) = 0, x < 0 g (x, y ) = 1, x > 0 . This describes a “knife edge” at x = 0 extending from y = −∞ to y = ∞. Therefore, in this part of the problem, δ = ∞: the strong condition of part a. cannot be satisﬁed. In this part of the problem, the observer is ﬁxed at (0, 0, D), i.e. at X = Y = 0. With this aperture in place, the observer records an irradiance Ia . With the aperture completely removed (g ≡ 1), the observer records an irradiance I0 . Give the ratio Ia /I0 . To receive credit you must explain why this ratio is correct. where r is a vector from the (point) source to a point on the aperture, r is a vector from the observer to the same point on the aperture, U0 exp i(kr − ωt) r is the optical disturbance at a point on the aperture, Up is the optical disturbance at the observer, ω = ck = 2πc/λ is the angular frequency of the light, dA is an element of aperture area, and n is the normal to dA. [Note that, ˆ in a typical geometry (source on the left, aperture in the middle, observer on the right, and n ˆ pointing to the left), n · ˆ is positive while n · ˆ ˆr ˆr is negative, so that both terms in the square bracket are positive.] Consider this simple geometry: Let z be the axis pointing from left to right. Place the source at (x, y, z ) = (0, 0, −D), the observer at (X, Y, D), and the aperture in the plane z = 0. The aperture is characterized by an aperture function g (x, y ) such that g = 1 where the aperture is open, and g = 0 where the aperture is opaque. 2. (25 points) James Rainwater was awarded the Nobel Prize in the 1980’s for experiments done at the Nevis (Columbia) cyclotron in the 1950’s. He measured the sizes of nuclei using their interactions with muons (heavy electrons) which were in orbit about them. In the following, use the Bohr picture to describe the muon orbit. For ease of numerical computation, you may take the natural length unit h/me c to be 400 fm; the ratio mµ /me of ¯ muon to electron masses to be 200; and the ﬁne structure constant α to be 1/150. You may neglect the diﬀerence between the muon’s actual and reduced mass. A muon in n = 1 Bohr orbit reacts with (is “captured” by) a Z = 50 nucleus before it decays: µ− + (A, Z ) → (A, Z − 1) + νµ , where the neutrino νµ has negligible rest mass. Assuming that the initial and ﬁnal nuclei have the same inﬁnitely large rest mass and therefore a negligible kinetic energy, what is the neutrino energy expressed in units of me c2 ? (1% accuracy is suﬃcient.) 3. (25 points) Consider the elastic scattering of a photon from an inﬁnitely massive, perfectly reﬂective, spherical target of ﬁnite radius R (like a bowling ball polished to a mirror ﬁnish). The bowling ball is centered on the origin. The photon is incident along the ˆ direction and scatters (reﬂects) into z the direction (θ, φ), where θ and φ are the usual spherical polar angles. Note that θ = 0 means that the photon remains undeﬂected. For this problem, ignore diﬀraction and any other eﬀects which arise from the wavelike properties of the photon. a. (10 points) What is the total scattering cross section σT , corresponding to any deﬂection of the photon? (You don’t need a calculation here, just a correct answer and a convincing explanation for it.) b. (15 points) Calculate the diﬀerential cross dσ , dΩ where dΩ = sin θ dθ dφ is an element of solid angle. (When you integrate your result over the full solid angle, do you conﬁrm your answer to a.?) 4. (25 points) A nonrelativistic particle of mass m is conﬁned to a onedimensional box extending from x = 0 to x = L. Here a “box” is a square potential well with inﬁnite sides. a. (10 points) In terms of n and other constants, write down the energies En , 1 ≤ n < ∞, measured with respect to the bottom of the potential well, that the particle is allowed by Schr¨dinger’s equation to have. o b. (15 points) Deﬁne N (E ) to be the total number of allowed states with energy ≤ E . Taking n 1, so that the distribution of E is approximately continuous, calculate the density of states ρ(E ) ≡ dN . dE section University of California, Berkeley Physics H7C Fall 1999 (Strovink) SOLUTION TO EXAMINATION 2 Directions. Do all four problems (weights are indicated). This is a closedbook closednote exam except for two 8 1 × 11 inch sheets containing any information you wish on both sides. You are free 2 to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Calculators are allowed but not essential – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (25 points) The basis of scalar diﬀraction theory is the FresnelKirchoﬀ integral formula. In Fowles’ notation (Eq. 5.11), this formula states Up = − × ikU0 exp (−iωt) × 4π exp ik (r + r ) n · ˆ − n · ˆ dA ˆr ˆr rr a. (10 points) Let δ be the maximum value of x2 + y 2 on the aperture plane for which the aperture is not opaque. Thus, for this part of the problem, there are three characteristic lengths: λ, δ , and D. By moving around in the plane z = D, restricting her own coordinates X, Y such √ D, the observer ﬁnds that X 2 + Y 2 that the optical disturbance there is proportional to the Fourier transform of g (x, y ). As someone who understands the physics of diﬀraction, you realize that this information implies that a single strong condition must be satisﬁed which relates λ, δ , and D. Write down this condition. (You needn’t prove it, and you may omit factors of order unity.) Solution. In order for the optical disturbance Up (X, Y ) to be the Fourier transform of g (x, y ), our system must satisfy the Fraunhofer condition (see discussion in Fowles Section 5.6). The basic idea of this condition is that the spherical curvature of the wavefront at the aperture must be small compared to the wavelength of the light, allowing us to treat the light at the aperture as a plane wave. Omitting factors of order unity, the Fraunhofer condition is δ2 Dλ . This condition implies that the obliquity factor n · ˆ − n · ˆ is constant over the aperture, ˆr ˆr the quantity eikr /r is nearly constant, and the quantity eikr /r ≈ eikr . With these approximations, the optical disturbance Up (X, Y ) ∝ eikr dA. where r is a vector from the (point) source to a point on the aperture, r is a vector from the observer to the same point on the aperture, U0 exp i(kr − ωt) r is the optical disturbance at a point on the aperture, Up is the optical disturbance at the observer, ω = ck = 2πc/λ is the angular frequency of the light, dA is an element of aperture area, and n is the normal to dA. [Note that, ˆ in a typical geometry (source on the left, aperture in the middle, observer on the right, and n ˆ pointing to the left), n · ˆ is positive while n · ˆ ˆr ˆr is negative, so that both terms in the square bracket are positive.] Consider this simple geometry: Let z be the axis pointing from left to right. Place the source at (x, y, z ) = (0, 0, −D), the observer at (X, Y, D), and the aperture in the plane z = 0. The aperture is characterized by an aperture function g (x, y ) such that g = 1 where the aperture is open, and g = 0 where the aperture is opaque. b. (15 points) For this part of the problem, take the aperture function to be g (x, y ) = 0, x < 0 g (x, y ) = 1, x > 0 . This describes a “knife edge” at x = 0 extending from y = −∞ to y = ∞. Therefore, in this part of the problem, δ = ∞: the strong condition of part a. cannot be satisﬁed. In this part of the problem, the observer is ﬁxed at (0, 0, D), i.e. at X = Y = 0. With this aperture in place, the observer records an irradiance Ia . With the aperture completely removed (g ≡ 1), the observer records an irradiance I0 . Give the ratio Ia /I0 . To receive credit you must explain why this ratio is correct. Solution. According to the FresnelKirchoﬀ integral, the optical disturbance Up arises from a superposition of secondary waves which originate at the z = 0 plane. When the semiinﬁnite screen is in place, due to the symmetry of the system exactly half the secondary waves are blocked. Thus Up = 1 U0 , 2 A muon in n = 1 Bohr orbit reacts with (is “captured” by) a Z = 50 nucleus before it decays: µ− + (A, Z ) → (A, Z − 1) + νµ , where the neutrino νµ has negligible rest mass. Assuming that the initial and ﬁnal nuclei have the same inﬁnitely large rest mass and therefore a negligible kinetic energy, what is the neutrino energy expressed in units of me c2 ? (1% accuracy is suﬃcient.) Solution. The binding energy of the muon in the Bohr model is given by: BE = 1 mµ c2 (Zα)2 2 and in our case Zα ≈ 1/3 and mµ c2 ≈ 200me c2 . So BE ≈ 11me c2 . Since the rest energies of the initial and ﬁnal nuclei are taken to be the same, the kinetic energy of the neutrino must be equal to the rest mass of the muon minus the binding energy, or KE (νµ ) ≈ 200me c2 − 11me c2 ≈ 189me c2 . or, in terms of intensity I ∝ U 2 , 1 I =. I0 4 2. (25 points) James Rainwater was awarded the Nobel Prize in the 1980’s for experiments done at the Nevis (Columbia) cyclotron in the 1950’s. He measured the sizes of nuclei using their interactions with muons (heavy electrons) which were in orbit about them. In the following, use the Bohr picture to describe the muon orbit. For ease of numerical computation, you may take the natural length unit h/me c to be 400 fm; the ratio mµ /me of ¯ muon to electron masses to be 200; and the ﬁne structure constant α to be 1/150. You may neglect the diﬀerence between the muon’s actual and reduced mass. 3. (25 points) Consider the elastic scattering of a photon from an inﬁnitely massive, perfectly reﬂective, spherical target of ﬁnite radius R (like a bowling ball polished to a mirror ﬁnish). The bowling ball is centered on the origin. The photon is incident along the ˆ direction and scatters (reﬂects) into z the direction (θ, φ), where θ and φ are the usual spherical polar angles. Note that θ = 0 means that the photon remains undeﬂected. For this problem, ignore diﬀraction and any other eﬀects which arise from the wavelike properties of the photon. a. (10 points) What is the total scattering cross section σT , corresponding to any deﬂection of the photon? (You don’t need a calculation here, just a correct answer and a convincing explanation for it.) Solution. The total cross section σT is the cross sectional area of the photon beam that suﬀers any deﬂection as a result of interaction with the target. Neglecting diﬀractive eﬀects, the only photons scattered are those which intercept the area of a hemisphere of radius R, projected into the z = 0 plane. This is a circle of area πR2 . Thus σT = πR2 . b. (15 points) Calculate the diﬀerential cross section dσ , dΩ where dΩ = sin θ dθ dφ is an element of solid angle. (When you integrate your result over the full solid angle, do you conﬁrm your answer to a.?) Solution. A photon with impact parameter b = x2 + y 2 intercepts the sphere at a point on the sphere described by b ≡π−ψ . R Just before it hits the sphere, it is travelling in the direction θ0 = 0 . θs = π − arcsin Before impact, the angle that the photon makes with the normal to the sphere is ψ . Since the angle of incidence is equal to the angle of reﬂection, its direction changes by ∆θ = π − 2ψ . Therefore the ﬁnal angle θ of the photon is θ = θ 0 + ∆θ = 0 + π − 2ψ b . R Rearranging and diﬀerentiating, = π − 2 arcsin arcsin πθ b =− R 2 2 b = R sin = R cos db = − πθ − 2 2 θ 2 An element dσ of beam cross section is equal to b db dφ. Substituting from above, dσ = b db dφ R2 2 R2 = 4 R2 = 4 R2 dσ = dΩ 4 = cos θ θ sin dθ dφ 2 2 sin θ dθ dφ dΩ . This is an isotropic (constant) diﬀerential cross section. Integrated over ∆Ω = 4π , it yields σT = πR2 as in (a.). Note that the isotropy of the diﬀerential cross section doesn’t follow automatically from the spherical symmetry of the potential (an inﬁnite wall at r = R). A diﬀerent spherically symmetric potential, for example the Coulomb potential, yields the dramatically diﬀerent Rutherford result 1 dσ ∝ dΩ sin4 . θ 2 4. (25 points) A nonrelativistic particle of mass m is conﬁned to a onedimensional box extending from x = 0 to x = L. Here a “box” is a square potential well with inﬁnite sides. a. (10 points) In terms of n and other constants, write down the energies En , 1 ≤ n < ∞, measured with respect to the bottom of the potential well, that the particle is allowed by Schr¨dinger’s equation to have. o Solution. We measure the energy E of the particle with respect to the bottom of the well, where V ≡ 0. We seek solutions of the timeindependent Schr¨dinger equation o − h ¯ 2 ∂2 + V (x) uE (x) = EuE (x) , 2m ∂x2 R θ sin dθ . 2 2 with the boundary condition (because of the inﬁnite potential wall) uE (0) = uE (L) = 0 . The solutions are of the form uE (x) ∝ sin kn x with kn = nπ/L. Therefore, from the timeindependent Schr¨dinger equation, o n2 π 2 ¯ 2 h2 ¯ 2 kn h = , En = 2m 2mL2 with 1 ≤ n ≤ ∞. [As posed, the problem doesn’t require a proof like the above; you just need to write down the correct values of En .] b. (15 points) Deﬁne N (E ) to be the total number of allowed states with energy ≤ E . Taking n 1, so that the distribution of E is approximately continuous, calculate the density of states ρ(E ) ≡ dN . dE Expressing ρ(E ) in terms of E and other constants, we substitute n2 = 2mL2 E π2 ¯ 2 h π2 ¯ 2 h mL2 ρ(E ) = 2 2 2mL2 E π¯ h m L . = π ¯ 2E h Solution. The diﬀerence in energy between two adjacent states is ∆E ≡ En − En−1 = n2 − (n − 1)2 = (2n − 1) π2 ¯ 2 h 2mL2 π2 ¯ 2 h . 2mL2 So when we increase the number of states by ∆N = 1 we increase the maximum energy by ∆E . The density of states is just the ratio: ρ(E ) ≡ dN dE ∆N ≈ as n → ∞ ∆E 2mL2 = (2n − 1)π 2 ¯ 2 h 2 mL ≈ as n → ∞ . nπ 2 ¯ 2 h University of California, Berkeley Physics H7C Fall 1999 (Strovink) FINAL EXAMINATION Directions. Do all six problems (weights are indicated). This is a closedbook closednote exam except for three 8 1 × 11 inch sheets containing any information you wish on both sides. You are free 2 to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Calculators are allowed but not essential – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (35 points) One circularly polarized photon is trapped between two parallel perfectly conducting plates, separated by a distance L, which are parallel to the photon’s electric and magnetic ﬁelds. a. (10 points) If the photon has the smallest deﬁnite energy possible under these circumstances, at a point halfway between the plates describe a possible motion of its electric ﬁeld vector. b. (10 points) Same for its magnetic ﬁeld vector. c. (15 points) For this plate conﬁguration, making no restriction on the photon energy, evaluate the density of photon states d2 N dL dλ where N is the number of states and λ is the photon wavelength L. Take into account the possible states of circular polarization. 2. (35 points) A linearly (x) polarized plane EM wave travˆ elling along ˆ is incident on an opaque baﬄe z located in the plane z = 0. The baﬄe has two slits cut in it, which are of inﬁnite extent in the y direction. In the x direction, the slit widths ˆ ˆ are each a and their centertocenter distance is d. (Obviously d > a, but you may not assume that d a.) The top and bottom slits are each an equal distance from x = 0. The diﬀracted image is viewed on a screen located in the plane z = L, where L d; also 2 λL d , where λ is the EM wavelength. Quarterwave plates are placed in each slit. They are identical, except that the top√ plate’s “slow” (highindex) axis is along (x + y)/ 2 (+45◦ with ˆˆ respect to the x axis), while the bottom plate’s ˆ √ slow axis is along (x − y)/ 2 (−45◦ with respect ˆˆ to the x axis). ˆ a. (15 points) What is the state of polarization of the diﬀracted light that hits the center of the screen, at x = y = 0? Explain. b. (20 points) At what diﬀracted angle θx does the ﬁrst minimum of the irradiance occur? 3. (35 points) A lens has an f number (ratio of focal length to diameter) equal to F . The lens is used to concentrate sunlight on a ball whose diameter is equal to the diameter of the sun’s image. The ball is convectively and conductively insulated, but it freely radiates energy outward so that its temperature can approach an equilibrium value Tb . a. (10 points) The sun subtends a halfangle of ≈ 0.005 radians. Is the size of its image “diﬀractionlimited”, i.e. determined largely by the eﬀects of diﬀraction? Make an orderofmagnitude argument assuming that the lens is a typical camera lens, with a radius of order 10−2 m. b. (25 points) Assuming the sun to be a blackbody of temperature T , calculate the ball’s temperature Tb . Neglect reﬂection by the lens. (Hint: your answer should depend only on T and F .) 4. (30 points) You are given a Hamiltonian H= 1 (LR + RL) , 2 b. (10 points) The N particles are identical bosons. c. (15 points) The N particles are identical spin 1 fermions. 2 6. (30 points) In the rest frame S of a star, ignoring the gravitational redshift, some of the photons emitted by the star arise from a particular atomic transition with an unshifted wavelength λ . When these photons are observed on earth, they are shifted to longer wavelength λ = λ + ∆λ because the star is receding from the earth with velocity β0 c due to the Hubble expansion of the universe. Astronomers measure this redshift by means of the parameter z , deﬁned by z≡ ∆λ . λ where R and L are two operators such that [L, R] = E0 with E0 a constant. You are also given an eigenfunction uE (x) of H, such that HuE = EuE , where E , another constant, is the energy eigenvalue. Prove that H(RuE ) = (E + E0 )(RuE ) , i.e. R is a raising operator. 5. (35 points). Consider a harmonic oscillator potential 1 2 V (x) = mω0 x2 2 in one dimension. An even number N of particles of mass m are placed in this potential. There are no special interactions between the particles – no signiﬁcant mutual electrostatic repulsion, gravitational attraction, etc., compared to the strength of their interaction with the harmonic potential itself. You may use what you already know about the levels of a harmonic oscillator. The system is in its ground state, i.e. T = 0 Kelvin. Calculate the total energy E of the N particle system, relative to the bottom of the well, for the cases a. (10 points) The N particles are distinguishable. For light, the relativistic Doppler shift is ω= ω . γ0 (1 − β0 cos θ) a. (10 points) In the “neighboring star” limit 1, show that β0 is approximately equal β0 to the measured z . b. (10 points) In the “distant star” limit γ0 1, derive an expression for γ0 in terms of the measured z . c. (10 points) The observation of Supernova 1987A marked the dawn of a new astronomy, in which humans are able to detect fermions (neutrinos) as well as bosons (photons) from (spatially or temporally) resolved sources outside the solar system. About a dozen such neutrinos were detected in each of two huge underground water tanks. The photons from Supernova 1987A were redshifted by z ≈ 10−5 . Taking the Hubble constant to be H0 ≈ 0.7 × 10−10 yr−1 , for how many years did the neutrinos from SN1987A travel before humans observed them? University of California, Berkeley Physics H7C Fall 1999 (Strovink) SOLUTION TO FINAL EXAMINATION Directions. Do all six problems (weights are indicated). This is a closedbook closednote exam except for three 8 1 × 11 inch sheets containing any information you wish on both sides. You are free 2 to approach the proctor to ask questions – but he or she will not give hints and will be obliged to write your question and its answer on the board. Calculators are allowed but not essential – roots, circular functions, etc., may be left unevaluated if you do not know them. Use a bluebook. Do not use scratch paper – otherwise you risk losing part credit. Cross out rather than erase any work that you wish the grader to ignore. Justify what you do. Box or circle your answer. 1. (35 points) One circularly polarized photon is trapped between two parallel perfectly conducting plates, separated by a distance L, which are parallel to the photon’s electric and magnetic ﬁelds. a. (10 points) If the photon has the smallest deﬁnite energy possible under these circumstances, at a point halfway between the plates describe a possible motion of its electric ﬁeld vector. Solution. The electric ﬁeld of the light E is parallel to the conducting plates, so E must vanish at the plates. Therefore we have E (0, t) = E (L, t) = 0 . Of course, E (z, t) is a solution to the wave equation: 1 ∂2 ∂2 − 2 2 E (z, t) = 0 . ∂z 2 c ∂t In order to satisfy both the wave equation and the boundary conditions, we choose ˆ ˆ E (z, t) = E0 sin (kz ) exp (−iωt) x ± iy , where k= nπ L Halfway between the plates (z = πct L E ( , t) = E0 exp −i 2 L
L 2 ), we ﬁnd that x ± iy . ˆ ˆ So E moves in a circle with angular frequency ω = πc/L. b. (10 points) Same for its magnetic ﬁeld vector. Solution. From Maxwell’s equations, ∇×E =− Therefore we have ∂By ∂Ex =− ∂t ∂z ∂Ey ∂Bx = . ∂t ∂z From the above relations, it follows that the spatial dependence of B (z, t) is given by cos (πz/L). Halfway between the plates B = 0. c. (15 points) For this plate conﬁguration, making no restriction on the photon energy, evaluate the density of photon states d2 N dL dλ where N is the number of states and λ is the photon wavelength L. Take into account the possible states of circular polarization. ∂B . ∂t where n = 1, 2, ... The smallest deﬁnite energy is achieved when k = π/L (consequently ω = πc/L), in which case we have for the electric ﬁeld: E (z, t) = E0 sin πct πz exp −i L L x ± iy . ˆ ˆ Solution. We have from part (a.) that k = nπ/L, so in terms of wavelength λ (recalling k = 2π/λ): 2L . n= λ Since we have 2 possible polarization states, the total number of states N as a function of wavelength is 4L N= . λ The derivative with respect to wavelength is 4L dN =2 dλ λ and so we obtain for the density of photon states 4 d2 N = 2. dλdL λ or lefthand circular polarization; light that exits the bottom slit (slow axis at −45◦ ) is in a state of polarization 1 √ 2 1 −i −i 1 1 0 1 =√ 2 1 −i , or righthand circular polarization. At the center of the screen, light from each slit contributes equally; the state of polarization is proportional to 1 2 1 i 1 +√ 2 1 −i = √ 2 1 0 ; it is x polarized like the incident beam. (See ˆ Fowles page 34 and Table 2.1 for the Jones vectors and matrices.) b. (20 points) At what diﬀracted angle θx does the ﬁrst minimum of the irradiance occur? Solution. Right and lefthand polarized states are orthogonal; they do not interfere. To see this formally (though this is not required as part of the solution), consult Fowles Eq. 3.11; the interference term there is proportional to E∗ · E1 ∝ ( 1 −i∗ ) 2 1 i = 1 + i2 = 0 . 2. (35 points) A linearly (x) polarized plane EM wave travˆ elling along ˆ is incident on an opaque baﬄe z located in the plane z = 0. The baﬄe has two slits cut in it, which are of inﬁnite extent in the y direction. In the x direction, the slit widths ˆ ˆ are each a and their centertocenter distance is d. (Obviously d > a, but you may not assume that d a.) The top and bottom slits are each an equal distance from x = 0. The diﬀracted image is viewed on a screen located in the plane z = L, where L d; also 2 λL d , where λ is the EM wavelength. Quarterwave plates are placed in each slit. They are identical, except that the top√ plate’s “slow” (highindex) axis is along (x + y)/ 2 (+45◦ with ˆˆ respect to the x axis), while the bottom plate’s ˆ √ slow axis is along (x − y)/ 2 (−45◦ with respect ˆˆ to the x axis). ˆ a. (15 points) What is the state of polarization of the diﬀracted light that hits the center of the screen, at x = y = 0? Explain. Solution. Light that exits the top slit (slow axis at +45◦ ) is in a state 1 √ 2 1 i i 1 1 0 1 =√ 2 1 i , Since there is no interference between the light from the top and bottom slit, the resulting irradiance is just twice that expected from a single slit of width a. According to Fowles Eq. 5.18, this pattern is proportional to sin β β
2 , where in this problem’s notation β= and k = 2π/λ. β = π , or 1 ka sin θx 2 The ﬁrst minimum occurs at sin θx = λ . a 3. (35 points) A lens has an f number (ratio of focal length to diameter) equal to F . The lens is used to concentrate sunlight on a ball whose diameter is equal to the diameter of the sun’s image. The ball is convectively and conductively insulated, but it freely radiates energy outward so that its temperature can approach an equilibrium value Tb . a. (10 points) The sun subtends a halfangle of ≈ 0.005 radians. Is the size of its image “diﬀractionlimited”, i.e. determined largely by the eﬀects of diﬀraction? Make an orderofmagnitude argument assuming that the lens is a typical camera lens, with a radius of order 10−2 m. Solution. The image of a distant point source formed at the focal plane of a lens is actually a Fraunhofer diﬀraction pattern where the aperture is the lens opening. The image becomes diﬀraction limited when the size of the image is near the size of an Airy disk. From this condition, we have the Rayleigh criterion: 2θ > 1.22λ , D Solution. The sun radiates total power
4 2 PS = σTS · 4πRS , where TS is the sun’s surface temperature and RS is the sun’s radius. The lens collects a fraction of this light power given by π (D/2)2 ∆Ω = , 2 Ω 4πRES where RES is the distance from the earth to the sun. The entirety of this light is focused on the ball. The ball reradiates power
4 2 Pb = σTb · 4πRb . In equilibrium we have the light power absorbed by the ball equal to the light power radiated by the ball. Setting the two equal, we obtain
4 TS 2 D 2 RS 22 = Tb r b . 2 4 RES where θ is the halfangle subtended by the sun (≈ 0.005 rad), λ is the wavelength of the light, and D is the diameter of the lens. We can assume λ ≈ 600 nm for the sun, D = 2 × 10−2 m, so the image is not diﬀraction limited if θ > 2.5 × 10−5 rad , which is clearly satisﬁed in this problem. Thus the image is not diﬀraction limited. (Note that this question requires only an orderofmagnitude analysis. So you don’t need to know anything about the details of Rayleigh’s criterion, or Airy disks, to get full credit; all that you need to say is that the diﬀraction angle is of order λ/D, which here is much smaller than the sun’s angular width.) b. (25 points) Assuming the sun to be a blackbody of temperature T , calculate the ball’s temperature Tb . Neglect reﬂection by the lens. (Hint: your answer should depend only on T and F .) To relate this result to F , we note that 2rb /f = 1/F where f is the focal length. Also we have rb /f ≈ RS /RES ≈ θ, where θ is the halfangle subtended by the sun. Employing the above relations in the equation relating the sun’s temperature to the ball’s temperature:
4 4 T b = TS 1 d2 θ 2 4 2 = 16F 2 TS . 4rb So we ﬁnd that Tb = TS 2 1 . F 4. (30 points) You are given a Hamiltonian H= 1 (LR + RL) , 2 where R and L are two operators such that [L, R] = E0 with E0 a constant. You are also given an eigenfunction uE (x) of H, such that HuE = EuE , where E , another constant, is the energy eigenvalue. Prove that H(RuE ) = (E + E0 )(RuE ) , i.e. R is a raising operator. Solution. 1 H (Ru) = LRR + RLR u 2 1 = LRR − RLR + 2RLR u 2 1 = [L, R]R + 2RLR u 2 1 = E0 R + 2RLR u 2 1 = E0 R + RLR − RRL + RRL + RLR u 2 1 = E0 R + R[L, R] + RRL + RLR u 2 1 = 2E0 R + R(RL + LR) u 2 1 = 2E0 R + R(2H ) u 2 = (E + E0 )Ru . The system is in its ground state, i.e. T = 0 Kelvin. Calculate the total energy E of the N particle system, relative to the bottom of the well, for the cases a. (10 points) The N particles are distinguishable. Solution. The energy levels of a harmonic oscillator are En = n + 1 h ¯ ω0 , 2 where n = 0, 1, 2 . . . . Nothing prevents mutually noninteracting distinguishable particles from occupying the same spatial wavefunction. At T = 0 this will be the ground state n = 0. Then E= N h ¯ ω0 . 2 b. (10 points) The N particles are identical bosons. Solution. Same as (a.). All the identical bosons are in the ground state at T = 0. c. (15 points) The N particles are identical spin 1 fermions. 2 Solution. Each spatial wavefunction can accommodate two identical spin 1 fermions, one 2 with spin up, one with spin down. At T = 0 the lowest occupied state is the ground state, with energy 1 h E 0 = ¯ ω0 , 2 while the highestenergy occupied state has (Fermi) energy equal to EF = E0 + N − 1 ¯ ω0 . h 2 5. (35 points). Consider a harmonic oscillator potential V (x) = 1 2 mω0 x2 2 in one dimension. An even number N of particles of mass m are placed in this potential. There are no special interactions between the particles – no signiﬁcant mutual electrostatic repulsion, gravitational attraction, etc., compared to the strength of their interaction with the harmonic potential itself. You may use what you already know about the levels of a harmonic oscillator. The total energy is N times the average energy, which is the mean of E0 and EF : E=N = E0 + EF 2 N N 2E0 + − 1 ¯ ω0 h 2 2 N2 h ¯ ω0 . = 4 6. (30 points) In the rest frame S of a star, ignoring the gravitational redshift, some of the photons emitted by the star arise from a particular atomic transition with an unshifted wavelength λ . When these photons are observed on earth, they are shifted to longer wavelength λ = λ + ∆λ because the star is receding from the earth with velocity β0 c due to the Hubble expansion of the universe. Astronomers measure this redshift by means of the parameter z , deﬁned by z≡ ∆λ . λ Solution. When γ0 the solution to (a.), 1, β0 ≈ 1. Then, from z ≈ 2γ 0 − 1 1+z . γ0 ≈ 2 Full credit is given with or without the “1” term. c. (10 points) The observation of Supernova 1987A marked the dawn of a new astronomy, in which humans are able to detect fermions (neutrinos) as well as bosons (photons) from (spatially or temporally) resolved sources outside the solar system. About a dozen such neutrinos were detected in each of two huge underground water tanks. The photons from Supernova 1987A were redshifted by z ≈ 10−5 . Taking the Hubble constant to be H0 ≈ 0.7 × 10−10 yr−1 , for how many years did the neutrinos from SN1987A travel before humans observed them? Solution. From Rohlf Eq. (19.17) (necessary for solving assigned problem 19.18), the velocity v with which SN1987A is receding from Earth is v = H0 d , where d is its present distance from Earth. Neutrinos are nearly massless and travel essentially at the speed of light c. Using the result of part (a.), the travel time T of the neutrinos from SN1987A was T= = d c For light, the relativistic Doppler shift is ω= ω . γ0 (1 − β0 cos θ) a. (10 points) In the “neighboring star” limit 1, show that β0 is approximately equal β0 to the measured z . Solution. ω γ0 (1 − β0 cos θ) 2πc = λ 1/λ 1/λ = γ0 (1 − β0 cos θ) λ = γ0 (1 − β0 cos θ) λ ∆λ = γ0 (1 − β0 cos θ) − 1 λ z = γ0 (1 − β0 cos θ) − 1 . ω= For a receding star, cos θ = −1, and so z = γ0 (1 + β0 ) − 1 . When β0 Then 1, γ0 ≈ 1 to second order in β0 . z ≈ 1 + β 0 − 1 = β0 . b. (10 points) In the “distant star” limit γ0 1, derive an expression for γ0 in terms of the measured z . v H0 c β0 = H0 z ≈ H0 ≈ 1 × 10−5 0.7 × 10−10 yr−1 ≈ 1.4 × 105 yr . ...
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This note was uploaded on 04/06/2009 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Lanzara
 Physics, The Land

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