HW1Soln - 4 CHAPTER 2. GEOMETRY AS PHYSICS with constants c...

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Unformatted text preview: 4 CHAPTER 2. GEOMETRY AS PHYSICS with constants c and d depending on a to be determined. The special case considered in (a) with fl = *y = 7r/ 2 gives A=c(7r+oz)+d=cra2 holding for arbitrary 04. Thus, 0 = a2 and d = —a27r, giving Aza2(a+,6+'y—7r) or, what is the same thing: A a+fi+ry=7r+7 a 2-4. Draw examples of a triangle on the surface of a sphere for which: a) the sum of whose interior angles is just slightly greater than 7r. b) the sum of Whose angles is equal to 27r. c) What is the maximum the sum of angles of a triangle on a sphere can be according to (2.4)? Can you exhibit a triangle where the sum achieves this value.7 Solution: Consider the triangle contained within the equator and two lines of longitude differing by an angle a. That triangle has two right interior angles at the 4 PROBLEM 2.5 5 equator and the interior angle oz at the pole. The sum of the interior angles is 7r+a. By taking oz near zero, one has a triangle whose sum of angles is slightly bigger than 7r. By taking a 2: 7r, one has a triangle whose sum of angles is 27r. From (2.4), the maximum sum of interior angles occurs when A = 47rR2 — the area of the whole sphere —~ and is 57v. A triangle which nearly realizes this bound is the complement of a small equilateral triangle. The three interior angles are each (27r — 7r/ 3) and add up to 57r. 2-5. Calculate the area of a circle of radius 7“ (distance from center to circumference) in the two - dimensional geometry which is the surface of a sphere of radius a. Show that this reduces to 7rr2 when 7" << a. Solution: Refer to Fig.2.6 for the geometry. Consider an element of area at ((9, gb) spanned by coordinate intervals (old, do) The length of the edge of size d0 is add, the length of the edge of size do in asin ddqfi. Since the coordinate lines are orthogonal the area is (ad0)(a sin 0d¢) . The circle of radius 7“ lies at 0 : r/a. Integrating the element of area above r/a 27r A : / d6 do a2 sin 6d6d¢ 0 0 gives the result: A : 27ra2[1 — cos(r/a)] . For small r/a 2—6 [B] Consider a sphere of radius a and on it a segment of length 3 of a line of latitude that is a distance d from the north pole measured on the sphere. What is the angle between the lines of longitude that this segment spans? Is 5 8 CHAPTER 2. GEOMETRY AS PHYSICS This is, of course, the correct answer. The key step in the above evaluation is recognizing that the whole circle is covered by the coordinate range n = —\/ 27“ to a = V27". 2-8. [A] The surface of an egg is an axisymmetric geometry to a good approximation. In the line element for two—dimensional axisymmetric geometries (2.21), pick an f (6?) such that the resulting surface would resemble that of an egg. Calculate the ratio of the biggest circle around the axis to the distance from pole to pole. Solution: There are many solutions to this problem corresponding to the different choices of f (0) that make a surface like an egg which is smaller near one pole than the other. A simple linear choice is f(6) = 1 — (9/27r which varies between 1 at 0 = 0 (the larger end) and 1/2 at 6 = 7r (the smaller end). The circumference of a circle around the axis at 6 is 0(0) 2 27raf(0). The maximum circumference occurs at 0 = 0: 0(0) : 27r a. The distance d from pole to pole is d = fwdda 0 7T0: The ratio C/d is thus 2 for this example. 2-9. The surface of the Earth is not a perfect sphere. The polar radius of the Earth, 6357 km, is slightly less than the mean equatorial radius 6378 km. Suppose the surface of the Earth is modeled by an axisymmetric surface with a line element of the kind in (2.21) with f(()) = sin0(1+ esin2 6) for some small 6. What values of a and 6 would best fit reproduce the known polar and equatorial radii? 16 CHAPTER 3. NEWTONIAN PHYSICS Solution: The mass density in the sphere is - ———M—— — 3M - const (1) M _ (47TR3/3) -— 47rR3 — ' The spherical symmetry of the problem implies that (I) is a function only of the radius 7". Poisson’s equation (3.18) for the gravitational potential is 1 1 2 d<1> The general solution inside (7' < R) which does not diverge at r = 0 is @(7‘) 2 Ar? + B (3) Where A and B are constants. A is determined from (2) and (1) to be 27? 1 GM 1 Lara“? (—35“) fi' (4) B is determined by matching to the exterior solution <I>('r) : wGM/r at r 2 R to be B = ~3G'M/(2R). Thus, (13(7") : $22—4— [(%)2~3] , r<R (5) 2 —%¥- , 7“ > R. (6) 3-5. Consider the functional smog] = /0T [@2302 +22%) dt. Find the curve x(t) satisfying the conditions 56(0) 2 0, $(T) =1, which makes S an extremum. What is the extremum value 0fS[:c(t)] .7 Is it a maximum or minimum? 16 18 CHAPTER 3. NEWTONIAN PHYSICS which is greater than (3), as calculating a few values or a simple plot will show. 3-6. [B,E,C] Estimate the gravitational self—energy of the Moon as a fraction of the Moon’s rest mass energy. Is this ratio larger or smaller than the accuracy of the Lunar laser ranging test of the equality of gravitational and inertial mass? Solution: The gravitational self energy is of order GM2 1110011 Rmoon Eself N and the ratio to the rest energy is Eself GMmoon X X 1025) 3 X 10_11 Erest Rmonc2 (1.7 x 108)(3 x 1010)2 which is within the 10~13 accuracy of lunar laser ranging. 18 ...
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HW1Soln - 4 CHAPTER 2. GEOMETRY AS PHYSICS with constants c...

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