HW7Soln - PROBLEM 9.16 131 c) The first integral...

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Unformatted text preview: PROBLEM 9.16 131 c) The first integral (including the 2) is just the one in ( 9.54) and equals 27r. Show that the second integral gives (7r/2)(u1 + U2) and that this equals 7r G'M/(f2 to lowest order in 1/c2. d) Combine these results to derive (9.55). Solution: There is no need for a solution to this problem since the problem itself describes how to do it. 9-16. A beam of photons with a circular cross section of radius a is aimed towards a black hole of mass M from far away. The center of the beam is aimed at the center of the black hole. What is the largest radius a = amax of the beam such that all the photons in the beam are captured by the black . n 2 hole? The capture cross section is namax. Solution: From Figure 9.8 we learn that Wherever 1/1)2 is greater than the height of the barrier the photon will be captured. This is the condition that the impact parameter be less than x/27M photon will be captured. The cross section is a : 7r(\/§TM)2 = 277TM2 9-17. Calculate the deflection of light in Newtonian gravitational theory assuming that the photon is a “non-relativistic” particle which moves with speed c when far from all sources of gravitational attraction. Compare your answer to the general relativistic result. Solution: In Newtonian physics the energy integral for a particle of mass m is [of (9.31)] m—i E_1df2 £2 M dt with K E L/ m and G = 1 The angular momentum integral for an orbit in the 131 132 CHAPTER 9. THE GEOMETRY OUTSIDE A SPHERICAL STAR equatorial plane is 2 61¢ 6 = — . T dt An incoming “photon” with speed c = 1 and impact parameter b has Ezmc2/2, Lzmcb. Then for the shape of the orbit d¢_d¢/dt_ r [2 (g_f_+i_\r_>]‘% m 2T2 'r dr " dr/dt ‘ 53 Introducing u E b/r and following the discussion that led to (9.77) gives the following discussion for the total deflection angle Art is one pass. u 2M —* A¢=2/ ldu [1~UZ+TU] 2 0 c b where ul is the first turning point (zero of the denominator) in u. This is W 1 M = 2/0 [(11.1 ~ u) (u + u2)]_5 where U1U2 2 1 and U1 —- U2 2 ZM/czb. The result (6.9. using Mathematica) for small 2M / (:21) is M = 7r + (2M/c2b) This gives 6gb : ZM/CZb —— half the general relativistic result. 9-18. Suppose in another theory of gravity (not Einstein ’5 general relativity) the metric outside a spherical star was given by 2M 6182 2 (1 e T) [—61.52 + on? + New? + sin2 6d¢2)] Calculate the deflection of light by a spherical star in this theory assuming that photons move along null geodesics in this geometry and following the steps that led to (9.78). When you get the answer see if you can find a simpler way to do the problem. 132 140 CHAPTER 10. SOLAR SYSTEM TESTS Solution: Working through the derivation of the gravitational redshift in Section 9.2 for a PPN metric of the form (10.4) gives woo 1 GM 2 t:— A“*—*"§§+(fi*7’§)<a3)'b“- to quadratic order in (GM / 02R)2 For the Earth GMEB/C2R99 ~ 10—9. Corrections to the gravitational redshift from terms involving [3 and 7 are unimportant. The third order Doppler shift is a factor V/ c N 10‘5 from the second order effects. This is just below the 10—4 accuracy of Vessot & Levine. 10-3. Evaluate the maximum deflection of light by the Sun predicted by general relativity in seconds of arc. Solution: The maximum deflection occurs when b is approximately a solar radius and is 1.75”. 10—4. Derive (10.8) for the deflection of light as a function of the parametrized post-Newtonian parameters. Solution: The idea of this problem is to follow through the steps that led to (9.63) and (9.64) and then from (9.76) to (9.81) with the PPN metric (10.4) instead of the Schwarzschild metric. We present only the major steps. The conserved quantities e and E are e=Aut , E=r2u¢. (1) The condition 11 u u = 0 can be re—written as an effective equation for radial motion n2_i i_fi m dA '“.B .4 r2 The equation for the shape of the orbit ch/dr : u‘b/ur is @_B%«L_iy% @ d?“ r2 PROBLEM 10.5 141 where b2 = (6/6)? The angle Art swept out in the passage by the light ray is 1 Mafia—133 (—1———1~)_% (4) r2 b2A r2 where n is the turning point where the denominator vanishes. Insert (10.6) for A and B into (4), and transform to the dimensionless variable w E b/r. Then expand both B in the numerator and 1/A in the denominator in powers of M / b keeping only the leading correction in both places. The result is -1 2 1111 2 A¢=2 dw (1+Léww) (1+———24w—w2) (5) 0 where 101 = b/n. This is in a form where it can easily be looked up. The result expanded to leading order in M / b is M = 7r + $0 + 7) (M/b) (6) 10-5. Evaluate, in seconds of arc per century, the precession of the perihelion of Mercury, Venus and the Earth as predicted by general relativity. Semimajor axis 106 (km) Eccentricity Mass/Me; Period (yr) Mercury Venus Earth Solution: The precession for Mercury is 42.98”/century. So the precession for any other planet is given by amerc(1 " Gierc) (PMerc> 5¢ = (42,98”/century) a(1 “ 62) P where a, e and P are the semimajor axis, eccentricity, and period for the planet and aMerc, eMerC and PMerc are the same quantities for Mercury. 141 PROBLEM 12.5 161 12—5. An observer falls radially into a spherical black hole of mass M. The observer starts from rest relative to a stationary observer at a Schwarzschild coordinate radius of 10M. How much time elapses on the observer’s own clock before hitting the singularity? Solution: Geodesics in the Schwarzschild geometry are characterized by the two integrals e and 6. For radial geodesics, such as the one in this problem, Z = O. The value of e is determined by the condition that the particle starts from rest as seen by a stationary observer at r 2 10M. The energy measured by this observer is therefore just the rest energy m. Thus, szz—p-U (1) where p is the particle’s four momentum mu, and U is the four velocity of the stationary observer U=[(1——2—RA{>—%,0,0,0] . (2) Here, R : 10M. Evaluating (1) with (2) one finds for the component at of the four velocity of the particle at r 2 10M. m_(L3Myi_0_3MJt_i§ _ R ‘ 10M ‘2' We could obtain the same result by arguing that the radial velocity seen by the observer will be zero implying dr/dr = 0, so that the particle is stationary at the radius R 2 10M, and then using (2) for 11. Then for e [of (9.21)] 2M 2 6:0“td“= At other radii, using u - u = —1, n2_g_0_ug_31_1 dT— r_r 5 162 CHAPTER 12. BLACK HOLES which vanishes at r 2 10M as it must. The proper time to go from r 2 10M to r = 0 is thus 10M 2M 1 ’T 2/ dr (— ~ —) 0 7” 5 Writing 7' = 10M§, this is T210x/5M/01d£<—:——1>—§ NIH which works out to be T = fix/5 7rM 12—6. An observer decides to explore the geometry outside a Schwarzschild black hole of mass M by starting with an initial velocity at infinity and then falling freely on an orbit that will come close to the black hole and then move out to infinity again. What is the closest that the observer can come to the black hole on an orbit of this kind? How can the observer arrange to have a long time to study the geometry between crossing the radius r 2 3M and crossing it again .7 Solution: Orbits in the Schwarzschild geometry are characterized by values of e and t. For a given 6 the orbit which comes closest to the black hole is the one which has an energy just below the maximum of the potential. The radius of closest approach is just slightly larger than the radius of the maximum which is [cf (9-34)] Wam—{weir} As 6 / M becomes large, this radius becomes smaller approaching 7“ = 3M as rmaxz3M[1+6<—A;)2+~ ]. (1) In order not to fall into the black hole, (62 — 1) / 2 should be just below the maximum of the potential at this radius which is Miami . (2) 162 PROBLEM 12.7 163 The observer should therefore start with a very large t, with a large energy given by (2) — very nearly at the speed of light. He or she will then be able to explore down to the radius r 2 3M. Further, since the energy is close to the maximum of the potential the observer will spend a very long time in the vicinity of 7" = 3M, orbiting the black hole many times before escaping to infinity. Put difierently, the observer should come as close as possible to the circular light ray orbit at 'r = 3M. 12-7 A meter stick falls radially into a center of Newtonian gravitational attraction produced by one solar mass located at a point. Using Newtonian physics estimate the distance from the point at which the meter stick would break or be crushed. Solution: When the meter stick is a distance R from the center of attraction, the difference in gravitational force on its ends is approximately GMQm GMGm N R2 (R + d)? N R3 where m is the mass of the meter stick (N 100 g) and d is its length (1 m). The meter stick will break when this is comparable to the breaking force Fbreak at a radius R N (GMde>i I F break Guessing Fbreak ~ 1000 N N 200 lbs, one finds R N 200 km. 12-8. Can an observer who falls into a spherical black hole receive information about events which take place outside? Is there any region ofspacetirne outside the black hole which an interior observer cannot eventually see? Analyze these questions using diagram like the one in Figure 12.2. Solution: 163 ...
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HW7Soln - PROBLEM 9.16 131 c) The first integral...

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