{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw1soln - Homework 1 solution Physics 776 Spring 2005...

This preview shows pages 1–2. Sign up to view the full content.

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 + 2 M r dT ) 2 - r 2 d Ω 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 let’s 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 Ω 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

hw1soln - Homework 1 solution Physics 776 Spring 2005...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online