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= =
r r
= =
=
r
And so, from (10.16) we conclude that
r t
r tt
t r
d (r ) t e dr
d (r ) = e dr
(10.19)
The Schwarzschild Solution
Using (10.17), we nd
r
r
r
= e
e (r ) r
(10.20)
and nally
r
(r )
r e (r ) r
r
r
= e
(10.21)
(r )
Solving for the Curvature Tensor
We now compute the components of the curvature tensor, using Cartan s second structure equation; i.e., ab = d ab + ac cb = 1 R abcd c d . We will ex 2 plicitly compute one of the curvature two forms. Consider r t = d r t + r c c . For the summation, we have t
r c
r r
c t
t
r t
t t t
r r
r t
r
t
r
t
since
t t
= =d =d =
r t
= 0. This leaves d (r ) t e dr
r t
d (r ) (r ) e dt dr d dr
d2 (r ) (r ) e dr dt + dr 2 d dr
e (r ) (r ) dr dt
d (r ) (r ) e dr dt dr
Using (10.7) to invert the differentials and write them in terms of the basis one forms, we arrive at the following expression:
r t
d2 + dr 2
d dr
d dr
d dr
e 2 (r ) r t
(10.22)
The Schwarzschild Solution
Now let s write out the curvature two form in terms of the components of the Riemann tensor, using the expression ab = 1 R abcd c d . Remember 2 that the Einstein summation convention is being used, so there is a summation over c and d. However, noticing that in (10.22) there is only one term, r t , we have to consider only two terms in the sum, those involving r t and t r . So we have
r t
1 r 1 1 R t cd c d = R r t r t r t + R r t t r t r 2 2 2
Now we use the fact that t r = r t to write this as
r t
1 r 1 1 R t r t r t R r t t r r t = R r t r t R r t t r r t 2 2 2
Using the same type of procedure we ve seen in earlier chapters, we can simplify this even further using the symmetries of the Riemann tensor: R r t t r = r r R r t t r = R r t t r = R r t r t = r r R r t r t = Rtr r t and so we have
r t
1 1 R r t r t R r t t r r t = R r t r t + R r t r t r t = R r t r t r t 2 2
Comparison with (10.22) shows that R r t r t = d2 + dr 2 d dr
d dr
d dr
e 2 (r )
All together, the nonzero components of the Riemann tensor are given by R r t r t R t t R r r R
d 2 d dr dr 1 d 2 t = R t = e r dr 1 d 2 = R r r = e r dr 1 e 2 = r2 d2 = + dr 2
d dr
e 2 (r )
(10.23)
The Schwarzschild Solution
All other nonzero components can be found using the symmetries of the Riemann tensor.
The Vacuum Equations
Now we can compute the components of the Ricci tensor to obtain the vacuum c equations. This is done relatively easily by computing Ra b = Ra cb . The rst term is easy to calculate Rt t = R t t t t + R r t r t + R t t + R t t
d2 = + dr 2
d dr
d dr
d dr
2 d 2 (r ) e r dr
(10.24)
By showing that R t r t r = R r t r t , you can see that d2 + Rr r = dr 2 Moving on we have R = R t t + R r r + R + R
d dr
d dr
d dr
2 d 2 (r ) e r dr
(10.25)
1 d 2 1 d 2 1 e 2 + + e e r dr r dr r2
(10.26)
R = R t t + R r r + R + R
1 d 2 1 d 2 1 e 2 e e = + + r dr r dr r2
(10.27)
We obtain the vacuum equations by setting each component of the Ricci tensor equal to zero [recall (10.1)]. There are only two vacuum equations we need in order to nd the functional form of (r ) and (r ). We use (10.24) and (10.25)
and set them equal to zero, which gives d2 + dr 2 d2 + dr 2 d dr d dr
The Schwarzschild Solution
d dr d dr
d dr d dr
2 d =0 r dr 2 d =0 r dr
(10.28)
(10.29)
We subtract (10.29) from (10.28) and get d d + =0 dr dr This implies that the sum of these functions is constant + = const = k We can use a trick so that the constant will vanish. We change the time coordinate to t tek . This means that dt dtek , dt 2 dt 2 e2k e2 dt 2 e2( +k) dt 2 . In other words, we have transformed + k and so + =0 = Now we use this to replace with in (10.29) and get d2 d 2 2 dr dr
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