HW8Soln - PROBLEM 12.13 167 But since gm, 2 0 at 7' 2 2M...

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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. 12-13. 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 Eddington-Finkelstein 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 non-zero 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 12-15. [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, energy-momentum 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 three-momentum implies 2% R U7. __ _ meSC (2M)% eJ R assuming that the fragment is ejected in the radial direction. Then imposing u - u = —1, the four-velocity 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 red-shift 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. 13-5. 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 10-16 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. 13-6. Estimate how long an electron-positron pair created in a vacuum fluctuation can last assuming that the fluctuation can violate energy conservation for a time At consistent with the energy-time 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~ 13-7. 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 Stephan-Boltzmann 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.

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HW8Soln - PROBLEM 12.13 167 But since gm, 2 0 at 7' 2 2M...

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