Lect23 - Physics 212 Lecture 23 A quote from one of you:...

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Unformatted text preview: Physics 212 Lecture 23 A quote from one of you: "Physicists define stress as force per unit area. The rest of humanity defines stress as physics." 40 30 20 10 Another quote: If waldo and carmen sandiego had a kid, would it be invisible? Confused Confident Avg = 3.1 Physics 212 Lecture 23, Slide 1 Physics 0 Plane Waves from Last Time E and B are perpendicular and in phase Oscillate in time and space Direction of propagation given by E X B E0 = cB0 Argument of sin/cos gives direction of propagation Physics 212 Lecture 23, Slide 2 Physics Preflight 2 BB No. Moving in the minus z direction No. Has Ey rather than Ex 40 30 20 10 0 Physics 212 Lecture 23, Slide 3 Physics Preflight 6 BB c=3.0 x 108 m/s Wavelength is equal to the speed of light divided by the frequency. λ= c 300, 000, 000 1 = = f 900, 000, 000 3 70 60 50 Check: Look at size of antenna on base unit 40 30 20 10 0 Physics 212 Lecture 23, Slide 4 Physics Doppler Shift Doppler Example Audio Doppler Example Visual The Big Idea As source approaches: Wavelength decreases Frequency Increases Physics 212 Lecture 23, Slide 5 Physics Doppler Shift for E&M Waves What’s Different from Sound or Water Waves ? Sound /Water Waves : You can calculate (no relativity needed) BUT Result is somewhat complicated: is source or observer moving wrt medium? Electromagnetic Waves : Electromagnetic You need relativity (time dilation) to calculate BUT Result is simple: only depends on relative motion of source & observer bserver ⎛ 1+ β ⎞ f′= f ⎜ ⎟ 1− β ⎠ ⎝ 1 2 β = v/c β > 0 if source & observer are approaching if β < 0 if source & observer are separating if Physics 212 Lecture 23, Slide 6 Physics Doppler Shift for E&M Waves f’ f v or f f’ v The Doppler Shift is the SAME for both cases ! f’/f ONLY DEPENDS ON THE RELATIVE VELOCITY ⎛ 1+ β ⎞ f′= f ⎜ ⎟ 1− β ⎠ ⎝ 1 2 Physics 212 Lecture 23, Slide 7 Physics Doppler Shift for E&M Waves A Note on Approximations ⎛ 1+ β ⎞ f′= f ⎜ ⎟ 1− β ⎠ ⎝ 1 2 β << 1 f ′ ≈ f (1 + β ) WHY ?? ⎛1 + β ⎞ ⎟ Taylor Series: Expand F ( β ) = ⎜ 1− β ⎠ ⎝ 1/ 2 around β = 0 F ( β ) = F ( 0) + Evaluate: F ′(0) F ′′(0) 2 β+ β + ... 1! 2! F ( 0) = 1 F ′(0) = 1 F (β ) ≈ 1 + β Physics 212 Lecture 23, Slide 8 Physics Doppler Shift for E&M Waves A Note on Approximations ⎛ 1+ β ⎞ f′= f ⎜ ⎟ 1− β ⎠ ⎝ 1 2 β << 1 f ′ ≈ f (1 + β ) 1/ 2 ⎛1 + β ⎞ ⎟ Taylor Series: Expand F ( β ) = ⎜ ⎝1− β ⎠ around β = 0 F ′(0) F ( β ) ≈ F (0) + β = 1+ β 1! F ′(0) = NOTE: F ( β ) = (1 + β )1 / 2 F (β ) ≈ 1 + 1 β 2 Physics 212 Lecture 23, Slide 9 Physics Red Shift Wavelengths shifted higher wavelength Frequencies shifted lower Star separating from us (Expanding Universe) Our Sun Star in a distant galaxy Physics 212 Lecture 23, Slide 10 Physics Example Police radars get twice the effect since the EM waves make a round trip: f ′ ≈ f (1 + 2β ) If f = 24,000,000,000 Hz (k-band radar gun) 24,000,000,000 c = 300,000,000 m/s v 30 m/s (67 mph) 31 m/s (69 mph) β 1.000 x 10-7 1.033 x 10-7 f’ 24,000,004,800 24,000,004,959 f’-f 4800 Hz 4959 Hz Physics 212 Lecture 23, Slide 11 Physics Preflight 7 A) B) C) ficlicker = 900 MHz BB Need to shift frequency UP Need to approach i>clicker 60 50 40 30 20 10 0 (β > 0) How fast would you need to run to see the i>clicker radiation? f ′ 1014 ⎛1 + β ⎞ = 9 = 105 = ⎜ ⎟ f 10 1− β ⎠ ⎝ ⎛1 + β ⎞ 10 = ⎜ ⎟ 1− β ⎠ ⎝ 10 1/ 2 1010 − 1 1 − 10−10 β = 10 = 10 + 1 1 + 10−10 Approximation Exercise: β ≈ 1 − (2 × 10−10 ) Physics 212 Lecture 23, Slide 12 Physics Waves Carry Energy Physics 212 Lecture 23, Slide 13 Physics Intensity Intensity = Average energy delivered per unit time, per unit area I≡ 1 dU A dt dU = u ⋅ volume = u Acdt Length = c dt Area = A I =c u Sunlight on Earth: I ~ 1000J/s/m2 ~ 1 kW/m2 Physics 212 Lecture 23, Slide 14 Physics Waves Carry Energy Physics 212 Lecture 23, Slide 15 Physics Comment on Poynting Vector Just another way to keep track of all this - Its magnitude is equal to I – Its direction is the direction of propagation of the wave Physics 212 Lecture 23, Slide 16 Physics Light has Momentum! If it has energy and its moving, then it also has momentum: Analogy from mechanics: p2 E= 2m dE 2 p dp mv dp = = dt 2m dt m dt = vF v→c For E&M waves: dE dU → = IA dt dt IA = cF I P= c Radiation pressure IF = pressure cA Physics 212 Lecture 23, Slide 17 Physics Preflight 4 60 50 40 30 20 10 0 BB But then again, what are we keeping constant here? WHAT ABOUT PHOTONS? Physics 212 Lecture 23, Slide 18 Physics PHOTONS We believe the energy in an E&M wave is carried by photons Question: What are Photons? Answer: Photons are “light particles”. Photons possess both wave and particle properties Particle: Energy and Momentum localized Wave: They have definite frequency & wavelength (fλ = c) Connections seen in equations: Energy = hf p = E/c = h/λ Planck’s constant Planck h = 6.63e-34 J-s Question: How can something be both a particle and a wave? Answer: It can’t (when we observe it) What we see depends on how we choose to measure it ! The mystery of quantum mechanics: More on this in PHYS 214 Physics 212 Lecture 23, Slide 19 Physics Exercise An electromagnetic wave is described by: where ˆ is the unit vector in the +y direction. j y x r E = ˆ 0 cos(kz − ωt ) jE z Which of the following graphs represents the z-dependence of Bx at t = 0? X (A) (B) (C) X (D) E and B are “in phase” (or 180o out of phase) Wave moves in +z direction y rr E E × B points in direction of propagation x r B r E = ˆ 0 cos(kz − ωt ) jE BB ˆ B = −iB0 cos(kz − ωt ) Lecture 23, Slide 20 zhysics 212 Physics P Exercise An electromagnetic wave is described by: What is the form of B for this wave? ˆj r i+ˆ E= E0 cos(kz + ωt ) 2 y x z (A) (B) ˆj r i+ˆ ( E0 / c) cos(kz + ωt ) B= 2 ˆj r i−ˆ ( E0 / c) cos(kz + ωt ) B= 2 (C) (D) ˆj r −i + ˆ ( E0 / c) cos(kz + ωt ) B= 2 ˆj r −i − ˆ ( E0 / c) cos(kz + ωt ) B= 2 BB ˆj r i+ˆ E= E0 cos(kz + ωt ) 2 Wave moves in –z direction +z points out of screen -z points into screen y E x rr E × B points in negative z-direction Physics 212 Lecture 23, Slide 21 Physics B Exercise An electromagnetic wave is described by: r E = ˆ 0 sin( kz + ωt ) jE Which of the following plots represents Bx(z) at time t = π/2ω ? BB (A) (B) (C) (D) Wave moves in negative z-direction y E B x r ˆ B = i ( E0 / c)sin(kz + ωt ) Bx = ( E0 / c)sin(kz + π / 2) at ωt = π/2: +z points out of screen -z points into screen rr E × B points in negative z-direction Bx = ( E0 / c){sin kz cos(π / 2) + cos kz sin(π / 2)} Bx = ( E0 / c) cos(kz ) Physics 212 Lecture 23, Slide 22 Physics Exercise A certain unnamed physics professor was arrested for running a stoplight. He said the light was green. A pedestian said it was red. The professor then said: “We are both being truthful; you just need to account for the Doppler effect !” Is it possible that the professor’s argument is correct? (λgreen = 500 nm, λred = 600 nm) BB (A) YES (B) NO • As professor approaches stoplight, the frequency of its emitted light will be shifted UP • The speed of light does not change • Therefore, the wavelength (c/f) would be shifted DOWN • If he goes fast enough, he could observe a green light ! Physics 212 Lecture 23, Slide 23 Physics Follow-Up A certain unnamed physics professor was arrested for running a stoplight. He said the light was green. A pedestian said it was red. The professor then said: “We are both being truthful; you just need to account for the Doppler effect !” How fast would the professor have to go to see the light as green? (λgreen = 500 nm, λred = 600 nm) BB (A) 540 m/s (B) 5.4 X104 m/s (C) 5.4 X 107 m/s 1+ β 1− β (D) 5.4 X 108 m/s Relativistic Doppler effect: f ′ = f f ′ 600 1+ β = = f 500 1− β 36(1 − β ) = 25(1 + β ) f ′ = f (1 + β ) β= 11 = 0.18 61 Note approximation for small β is not bad: c = 3 X 108 m/s fl v = 5.4 X 107 m/s β= 1 = 0.2 5 Change the charge to SPEEDING! Physics 212 Lecture 23, Slide 24 Physics ...
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