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Unformatted text preview: Homework 1 solution, Physics 776, Spring 2005 Painlev e-Gullstrand coordinates The line element for the unique spherically symmetric, vacuum solution to the Einstein equation can be written as ds 2 = dT 2- ( dr + s 2 M r dT ) 2- r 2 d 2 (1) Note that a surface of constant T is a flat Euclidean space. 1. Which value of r corresponds to the event horizon? Give a clear and pre- cise explanation of your answer, using the properties of the metric extracted directly from the above expression (i.e. without reference to some other co- ordinate system, for example). To eliminate clutter lets adopt units with 2 M = 1. It is convenient to first expand out the square: ds 2 = (1- r- 1 ) dT 2- 2 r- 1 / 2 dT dr- dr 2- r 2 d 2 . Let me give two different ways to locate/define the horizon in this context. ( a ) The surface r = 1 is null: the angular directions on it are spacelike but the T-translation direction is null. Therefore this surface describes an out- going spherical congruence of light rays whose cross-sectional area remains fixed. (This is called a marginally outer-trapped surface.) ( b ) No causal sig- nal inside r = 1 can exit this region. To see why, recall that a causal curve has ds 2 0. When r < 1 the dT 2 term is negative, so a causal curve must have dTdr < 0. If the curve is future oriented 1 , dT > 0, so dr < 0, i.e. it must go to smaller values of r , and in particular cannot escape to the region with r > 1. Thus r = 1 = 2 M is a causal horizon....
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This document was uploaded on 12/29/2011.
- Fall '09