Physics 139: Problem Set 9 solutions

Transcription

1 Physics 139: Problem Set 9 solutions ay 1, 14 Hartle 1.4 Consider the spacetime specified by the line element ds dt + ) dr + r dθ + sin θdφ ) Except for r, the coordinate t is always timelike and the coordinate r is spacelike. a) a) Find a transformation to new coordinates v, r, θ, φ) analogous to 1.1) that sets g rr and shows that the geometry is not singular at r. b) Sketch a t, r) diagram analogous to Figure 1. showing the world lines of ingoing and outgoing light rays and the light cones. c) Is this the geometry of a black hole? Put t v F r). The metric becomes ds + dv + F 1 ) F ) r dvdr ] dr + r dθ + sin θdφ ) where F df/dr. The choice F ) makes g rr and gives the metric ds which is non-singular at r. dv + dvdr + r dθ + sin θdφ ), 1

2 b) Integrating F, the explicit form for the coordinate transformation is t v r ) log r/ 1 + r, in which we ve chosen the constant of integration so that v at t r. Solving for ds, we see that light rays move on curves of v const and dv dr 1 /r Equivalently, in terms of t v r used to plot Eddington-Finkelstein diagrams, and d t dr 1 d t dr /r 5 t

3 c) This is a black hole. As seen in the figure, no light rays from inside r escape to outside r. Put differently, r is a stationary null surface generated by the light rays at r. Hartle 1.6 An observer decides to explore the geometry outside a Schwarzschild black hole of mass 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 3 and crossing it again? Orbits in the Schwarzschild geometry are characterized by values of e and l. For a given l 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 c.f. 9.34)] r max ) l ) ] 1/ l As l/ becomes large, this radius becomes smaller approaching r 3 as ) r max ] l In order not to fall into the black hole, e 1)/ should be just below the maximum of the potential at this radius, which is V eff r max ) 1 ) l + 54 The observer should therefore start with a very large l, with a large energy very nearly at the speed of light. He will then be able to explore down to the radius r 3. Further, since the energy is close to the maximum of the potential, the observer will spend a very long time in the vicinity of r 3, orbiting the black hole many times before escaping to infinity. Put differently, the observer should come as close as possible to the circular light ray orbit at r 3. Hartle 1.8 Can an observer who falls into a spherical black hole receive information about events which take place outside? Is there any region of spacetime outside the black hole which an interior observer cannot eventually see? Analyze these questions using a diagram like the one in Figure 1.6. An observer who falls into the black hole can, in principle, receive information in the shaded region below, between crossing the horizon at r and destruction at r. This includes information about events outside r but not all of the region outside r. 3

4 U 1 Hartle 1.14 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? The path of longest proper time should be a geodesic from r to r. Using 9.6) we have for the elapsed proper time τ dr dr/dr/dτ) dr e e 1 + l r 1 + l r ) 1 r ) 1 r )] 1/ )] 1/ Written this way, it is clear that non-zero values of l and e only decrease the elapsed proper time relative to the case with l e. That geodesic therefore gives the longest time Hartle 1.15 τ dr r 1) 1/ 1 dξ ξ1/ 1 ξ π A spaceship whose mission is to study the environment around black holes is hovering at a Schwarzschild coordinate radius outside a spherical black hole of mass. 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 approaches? 4

5 The key to this problem is to recognize that in the ejection event, energy-momentum is conserved, but the rest mass is not necessarily conserved. Let m and u be the rest mass and the four-velocity of the rocket hovering at radius. { u α 1 ) 1/,,, }. Let m esc, u esc, and, u ej be the corresponding quantities for the escaping and ejected fragments. The minimum four-velocity for escape corresponds to an orbit with e 1 and l and is c.f. 9.35)] u α esc Conservation of three-momentum implies { 1 ) 1, 1/,, } m esc ) 1/ + u r ej or u r ej ) 1/ assuming that the fragment is ejected in the radial direction. four-velocity of the ejected fragment has a time component Conservation of energy then gives 1 u t ej 1 ) 1/ ) 1 { 1 1 mu t m esc u t esc + u t ej m ) + mej ) { 1 m 1 Then imposing u u 1, the ]} 1/ ]} 1/ The largest fragment that can escape is the largest value of m esc /m that can satisfy this relation as /m ranges from to 1. By inspection, this fraction is maximized when, i.e., all the rest mass of the ejected fraction is converted to energy. Then ) 1 /) m max 1 + / which vanishes when. 5