HW6Soln - 120 CHAPTER 9. THE GEOMETRY OUTSIDE A SPHERICAL...

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Unformatted text preview: 120 CHAPTER 9. THE GEOMETRY OUTSIDE A SPHERICAL STAR for any 7‘ through which the observer falls. The largest value occurs at the radius of the shell B. As desoribed in the box (with c and G restored) 2GM 02R N 1. (2) Using this to eliminate M from (1) gives R>(C2h>%. (3) gmax Putting in the numbers gives from (3) and (2) R 2, 20, 000mm , M ,2 13, 000 MG (4) Not on this planet!! 9-5. Sketch the qualitative behavior of a particle orbit that comes in from infinity with a value of 8 exactly equal to the maximum of the effective potential Veg. How does the picture change is the value of 6 is a little bit larger than the maximum or a little bit smaller. Solution: Were 8 exactly equal to the maximum value of the potential, the orbit would come in from infinity, spiral around the black hole, approaching ever closer to the circular orbit at the potential maximum. The figure below shows what happens if the value of 5 is slightly below the maximum. J This figure was made with the Mathematica program, Shape of Orbits in the Schwarzschild Geometry, on the book website. The value for If is 4.3, the 120 PROBLEM 9.6 121 maximum of the effective potential is VmaLX = 00401391, and E = 0.0401300. The orbit spirals the center and then returns to infinity. Were 8 slightly greater than the maximum of the potential, the orbit would spiral around the center, and then plunges into it. 9—6. [S] An observer falls radially inward towards a black hole of mass M (exterior geometry the Schwarzschild geometry) starting with zero kinetic energy at infinity. How much time does it take, as measured on the observer’s clock, to pass between the radii 6M and 2M? Solution: From (9.37) we have for the inward a radial orbit (E = 0) starting from zero initial velocity (6 = 1) (tr—ca Therefore the proper time to pass between 6M and 2M is T = Meiji dr (3—1)?th dr (H (6% — 2%) : 5.59M. n i 1 9-7. Two particles fall radially in from infinity in the Schwarzschild geometry. One starts With e = 1, the other With e = 2. A stationary observer at r = 6M measures the speed of each when they pass by. How much faster is the second particle moving at that point? Solution: The energy E seen by a stationary observer at radius r is E:_p.uobs:__’r£.._ W where V is the speed of the particle determined by the observer and uobs is the observer’s four-velocity. The last equality holds because in an orthonormal 121 122 CHAPTER 9. THE GEOMETRY OUTSIDE A SPHERICAL STAR basis where ea -efi 2 nag, the usual kinematic relationships of special relativity holds. For a stationary observer at radius r, ugbs 2 U,ng 2 ufbs 2 0 and 1 2M ‘5 = (1“ T) (2) so that uobS - uobs 2 —1. From (1) and (9.9) E2 —p-uobS 2 —m (u Hobs) 2 mgaguo‘ugbs 1 2M — = m (1 — —) 2 u‘ 7" From this we can solve for V to find V(e)2—1’(e2—l+g-A£)% 8 1' which is less than 1 as it must be. In particular for r 2 6M, mam». m—T—lbéé. 9—8. A spaceship is moving without power in a circular orbit about a black hole of mass M. (The exterior geometry is the Schwarzschild geometry.) The Schwarzschild radius of the orbit is 7M. a) What is the period of the orbit as measured by an observer at infinity? b) What is the period of the orbit as measured by a clock in the Spaceship? Solution: 122 PROBLEM 9.9 123 a) From (7.48) (2 = dgb/dt : (M/r3)1/2. A clock at infinity measures t, so that Q is the angular velocity as measured from infinity. The period measured at infinity is T3 % 3 P00 2 —— = 27r 2 27r75M =116.M forr=7M. b) To get the period as measured on the spaceship we need To find dt/dr note that for a circular orbit in the equatorial plane _ 2M dt 2 2 05¢ 2 “"1 * ‘0‘?) +7" 2M dt 2 2 2 dt 2 1‘7) “9 (a?) < 2 —1 z (3E) ( —1 Thus dt/dT = (7/4)”2 and and the period Pspaceship : (4/7)1/2P00 z 88 M. 9-9. Find the relation between the rate of change of angular position of a particle in a circular orbit With respect to proper time and the Schwarzschild radius of the orbit. Compare With (9.46) 123 124 CHAPTER 9. THE GEOMETRY OUTSIDE A SPHERICAL STAR Solution: For a circular orbit of radius R, the proper angular velocity is @_r (17‘ _ RE ' We therefore only have to determine 6 from the condition that a circular orbit lies at the minimum of the effective potential 5’er = 0 8T R This gives £2 2 M R2 R — 3M so that -l 2 §$=(%§)5(1*%4> This is faster than dqfi/ alt, given in (9.46). A clock on the circulating particle runs slow compared to a clock at infinity both because it is moving (time dilation) and because it is in a lower gravitational potential. The proper period is therefore less than the period in t. The proper speed must therefore be greater. Note that there can be no circular orbits with R < 3M. The problem can also be done by computing @-@fi alT — dt ClT and computing dgb/dt from (9.46) and dt/d’r from (9.21). 9-10. Find the linear velocity of a particle in a circular orbit of radius R in the Schwarzschild geometry that would be measured by a stationary observer stationed at one point on the orbit. What is its value at the ISCO? Solution: There are several ways of doing this problem, each involving projecting the four—velocity of the particle onto the orthonormal basis of the observer. It’s perhaps simplest to calculate the energy the observer measures and relate that to the velocity. From (7.53) and (5.44) m RP-0F—#* 1 pub W (l 124 PROBLEM 9.11 125 where p : mu is the particle’s four-momentum, uobs is the stationary observer’s four-velocity, and we have assumed a rest mass m for the particle. The components of uobs are in (9.16). The components of the four-velocity of the particle are in (9.47) and (9.48). The result is 3M % 2M -% H = 1—) <1——> 2 E gttmu uObS < R R ( ) Then computing V from (1) gives M 2% 2M % v=<—> <1——> - 3 R < ) At the ISCO, V 2 1/2. 9-11 A small perturbation of an unstable circular orbit will grow exponentially in time. Show that a displacement 6? from the unstable maximum of the Schwarzschild will grow initially as 67“ cc 6”“ where 7' is the proper time along the particle’s trajectory and 7* is a constant. Evaluate 7* Explain its behavior as the radius of the orbit approaches 6M. Solution: In the neighborhood of its maximum at radius rmax, the effective potential Vggfi‘) behaves 1 d2Veff V950.) : Veff (rmax) 'l' ‘2‘ < (172 ) (67‘)2 + . . . Tmax where 67" E r — rmax. Denote (czl2l/gff/dr2),max by —K2 since it is negative at the maximum of the potential. Eq. (9.29) becomes % (dglf — :- K2(6r)2 = 0. There are growing and decaying solutions to this upside down harmonic oscillation. The growing solution behaves as (57" oc exp (KT) 125 ...
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HW6Soln - 120 CHAPTER 9. THE GEOMETRY OUTSIDE A SPHERICAL...

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