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Unformatted text preview: Homework 1 solution, Physics 776, Spring 2005 Painlev eGullstrand 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 Ttranslation direction is null. Therefore this surface describes an out going spherical congruence of light rays whose crosssectional area remains fixed. (This is called a marginally outertrapped 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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