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
+
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 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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 Fall '09
 Work, General Relativity, Calculus of variations, line element, Schwarzschild

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