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Unformatted text preview: PROBLEM 12.13 167 But since gm, 2 0 at 7' 2 2M and gm, 2 1, we conclude nr 2 0. A normal
vector is therefore,
71" 2 (1,0,0, 0) . This is null because
11 ~ n 2 gm,(n”)2 2 0 atr22M. 1213. a) An observer falls feet ﬁrst into a Schwarzschild black hole looking down
at her feet. Is there ever a moment when she cannot see her feet? For
instance, can she see her feet When her head is crossing the horizon? If
so What radius does she see them at? Does she ever see her feet hit the
singularity at r 2 0 assuming she remains intact until her head reaches
that radius? Analyze these questions With and Eddington—Finkelstein or Kruskal diagram. b) (Is it dark inside a black hole .9) An observer outside sees a star collapsing
to a black hole become dark. But would it be dark inside a black hole
assuming a collapsing star continues to radiate at a steady rate as measured by an observer on its surface? Solution: 167 168 CHAPTER 12. BLACK HOLES a) The ﬁgure above shows an EddingtonFinkelstein diagram with schematic
world lines of the observer’s head and feet. At a given t her feet are at a
smaller radius than her head because she is falling in feet ﬁrst. Radial rays
emitted originating at her feet are shown. (There segments of the light rays illustrated in Figure 12.2.) There is no instant when she is not receiving a light ray from her feet. She
sees them always. When her head crosses the horizon she sees her feet at
the same radius, because the horizon is generated by light rays. When her
head hits the singularity she still sees light from her feet that was emitted earlier but is falling into the singularity as well. (Some students interpret
this question to ask if she sees her feet when they hit the singularity. But there is no invariant meaning to “when” ) b) It will be light inside a black hole for the same reason. Just replace the
world line of feet in the above picture with the world line of the star’s surface. 12—14. [C] Once across the event horizon of a black hole, What is the longest
proper time the observer can spend before being destroyed in the singularity? Solution: The path of longest proper time should be a geodesic from r = 2M
to r = 0. Using (9.26) we have for the elapsed proper time T Z 12; d'r/(dr/dr) = — 2:4 d7" [62“ (“‘52) (1~ 2 /02M d7" [ﬁ—(Hé) (#4)] Written this way it is clear that nonzero values of 6 and 6 only decrease the
prOper time from E = e : 0. That geodesic therefore gives the longest time 1 One of the author’s students characterized this result as “The more you struggle,
the shorter your life.” 168 PROBLEM 12.15 169 1215. [C] A spaceship Whose mission is to study the environment around
black holes is hovering at a Schwarzschild coordinate radius R outside a
spherical black hole of mass M. To escape back to inﬁnity, crew must eject
part of the rest mass of the ship to propel the remaining fraction to escape
velocity. What is the largest fraction f of the rest mass that can escape to
inﬁnity? What happens to this fraction as R approaches 2M? Solution: The key to this problem is to recognize that in the ejection event,
energymomentum is conserved, but rest mass is not necessarily conserved.
Let m and u be the rest mass and the four—velocity of the rocket hovering at radiusR.
2M“%
0‘: 1~~— 0 0.
u 7707] Let mesc, um, and mej, uej be the corresponding quantities for the escaping
and ejected fragment. The minimum four—velocity for escape corresponds to an orbit with e = 1 and t 2 0 and is [cf. (9.35)] 2M—1 2M§
0‘: 1__ __
%m k R) ﬁ<R>’Qﬂ Conservation of threemomentum implies 2%
R U7. __ _ meSC (2M)%
eJ R
assuming that the fragment is ejected in the radial direction. Then imposing
u  u = —1, the fourvelocity of the ejected fragment has a time component M44 2M
“312(1‘?) {1"}? i 2 _ 7‘
0 —— mesc < > + mejuej OI‘ 170 CHAPTER 12. BLACK HOLES Conservation of energy t _ t t
mu — meSCueSC + mejuej then gives 1
2M % mesc mej 2M mesc 2 2
<1—~—>———< M > 1——— 1 
R m m R mej
The largest fragment that can escape is the largest value of meSC/m that can
satisfy this relation as mej / m ranges from zero to l Plotting the function for a few cases shows that mesc/m is maximized when mej = 0, i.e. all the rest
mass is turned into energy. Then _ (1 — 2M/R)%
m 1+(2M/R)% ' This vanishes when R 2 2M. 12—16. In Section 9.2 we derived the formula (9.20) for gravitational redshift
from a stationary observer. We started from the conservation law (8.32) arising
from the time translation symmetry of the Schwarzschild geometry. Use similar
techniques to derive for the redshift of light emmitted radially from a star in
free fall collapse as a function of the time 253 the radiation is received by a
distant observer. Compare your result with (12.11). Eq. (9.20) held for non
radial radiation as well. Do you expect that for radiation from the surface
of a collapsing star? (Note that in (9.20) R was the radius of the stationary
observer emitting the radiation, While in the example that led to (12.11), R
was the location of the observer receiving the radiation.) Solution: It’s simplest to use Schwarzschild coordinates to work out this
problem, although the computation in Eddington—Finkelstein coordinates is
similar. Suppose that p is the energy momentum four vector of a photon
emitted radially by the freely falling observer with frequency an. Thus [cf (5.83)] (12* = —p.u/h (1) 170 186 CHAPTER 13. ASTROPHYSICAL BLACK HOLES the Newtonian result gives for the predicted linear velocity Vpred: 0 GM 1/2 1” 1/2 km
ereﬁ’amim) 2542i?) ? Here the velocity of light has been introduced in a way to make computation
in geometrized units easy and the ﬁnal result has been evaluated with the data given in the problem.
A rough estimate of an angular radius of the orbit of the star 80—1 in Figure 13.4 is .15 ”. The resulting predicted linear velocity is Vpred z 1400 km/s . The observations show that SO—l moved an angular distance a z .15” over a
time T of four years. Assuming the orbit is transverse to the line of sight the observed angular velocity is
ad T
where a is measured in radians. Evaluating this gives Vbbs : VobS z 1600 km/s. This is unexpectedly good agreement for such a rough estimate. 135. What is the mass of a black hole formed at the beginning of the
universe that would explode by the Hawking process at the time the universe
becomes transparent to radiation — approximately 400,000 years after the big bang. Solution: From (13.19) a black hole which evaporates at a time t* has a mass
at an earlier time of 3 E
M 2 ——————— h * — .
(t) 15, 36071' (t t) The problem asks for the mass at t = 0 with t* = 3 X 105 yr. The above
expression is in units where M and t* are measured in units of cm, and h = 186 PROBLEM 13.6 187 62” z 10‘66 cm2. The conversion factors are 3 x 1010 cm/s and 3.15 x 107 s/yr.
Putting these together, we have M(0) =' 3 x 1016 cm : 4 x 1012 g where we have used [cf Appendix A] .742 X 10—28 cm/g. M(0) is about the
mass in a rock cube with side 100 m. 136. Estimate how long an electronpositron pair created in a vacuum
ﬂuctuation can last assuming that the ﬂuctuation can violate energy conservation
for a time At consistent with the energytime uncertainty principle AEAt > h Solution: The ﬂuctuation in energy is AE' = 2771802 The time is
h __ (1.05 X 10‘27erg  s) 2mecz " 2(9.11 >< 10—28g) (3 x 1010 cm/s)2 ~ 6 ><1O'22 s (!) At~ 137. Estimate the distance at which the energy received at Earth from
an exploding primordial black hole in the last one second of its life would
be comparable to that received from a nearby star in the same period. (For
deﬁniteness take the star to have the luminosity of the Sun and be 10 pc away.) Solution: Let M0 and T0 denote the present mass and temperature of a
black hole going to explode 1 sec after the present time to. As Example 13.2
showed, to have survived from the beginning of the universe to explode now
M0 ~ 1014 g. Correspondingly, from (13.18), To N 1012 K. The present
luminosity of the black hole L0 can be roughly estimated by L0 2 (0T3) (4mg) where a is the StephanBoltzmann constant and R0 ~ 2GM0/02 is the present
size of the black hole. Putting in the numbers gives R0 ~ 10—14 cm , L0 ~ 1017 erg/s. 187 ...
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This note was uploaded on 12/01/2011 for the course PHYS 5523 taught by Professor Kennefick during the Fall '11 term at Arkansas.
 Fall '11
 Kennefick
 Theory Of Relativity

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