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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!! 95. 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 97. 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 fourvelocity. 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 —puobS 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+gA£)% 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. 99. 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). 910. 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
RP0F—#* 1
pub W (l 124 PROBLEM 9.11 125 where p : mu is the particle’s fourmomentum, uobs is the stationary observer’s
fourvelocity, and we have assumed a rest mass m for the particle. The
components of uobs are in (9.16). The components of the fourvelocity 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. 911 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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 Fall '11
 Kennefick
 Theory Of Relativity

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