Explicitly writing out Cartan s rst structure equation for d r , we nd that d r = = in VS .NET

Create QR Code ISO/IEC18004 in VS .NET Explicitly writing out Cartan s rst structure equation for d r , we nd that d r = =

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(6.13)
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The Einstein Field Equations
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Comparing this with (6.12), which has only basis one forms t and r in the expression, we guess that the only nonzero term is given by r t t and conclude that
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(6.14)
Using (6.10), we expand the left-hand side to get
r t
tt
tr
t
Comparing this with (6.14) shows that the only nonzero Ricci rotation coef cient in this expansion is
tr
b t
We now proceed to nd other nonzero Ricci rotation coef cients that are related to this one via symmetries. Recall that
ab
r t
ba ,
0,
(6.15)
This means that
t r t r
b r . t t t rt
Expanding +
t r rr
t r,
we have
t r
This tells us that
t rr
b . t
Moving to the nal basis one form, we have R R dt d + dr d t r (6.16)
d = d (R(t, r ) d ) = =
1 R b(t, r ) r 1 R t + e R t R r 1 R b(t, r ) 1 R t + e r R t R r
The Einstein Field Equations
Using Cartan s structure equation, we can write d =
t
r
(6.17)
Comparing this with (6.16), we conclude that
1 R R t
1 R b(t,r ) e R r
(6.18)
Using (6.10) to expand each term, we nd the following nonzero Ricci rotation coef cients:
t
1 R R t
r
1 R b(t,r ) e R r
(6.19)
As an aside, note that ca b = a cb . This means that any terms that match on the rst two indices must vanish. For example,
r
=
r
r
r
r
r
Now let s apply the symmetries listed in (6.15) to nd the other nonzero terms. For the rst term, we get
=
t
t
t
t
= t t
1 R R t
(6.20)
This leads us to the conclusion that
1 R R t
The Einstein Field Equations
r
Using (6.15), we see that
= =
r
.
So we have
1 R b(t,r ) e R r 1 R b(t,r ) = e R r
We now proceed to use Cartan s second structure equation to nd the components of the curvature tensor in the noncoordinate basis. Recall that the curvature two forms were de ned via
a b
1 a c Rb c d d 2
(6.21)
We will solve two terms and leave the remaining terms as an exercise. Setting a = r and b = t in (6.21) gives
r t
r t
r t
t t
t r
r t
(6.22)
We begin by calculating d d
using (6.14) and recalling that r = eb(t,r ) dr :
=d =
b r t
b b(t,r ) dr e t b t
2 b b(t,r ) e dt dr + t 2
eb(t,r ) dt dr +
b b b(t,r ) e dr dr t r
Since dr dr = 0, this simpli es to d
r t =
2 b b(t,r ) e dt dr + t 2 2b + t 2 b t
b t
eb(t,r ) dt dr (6.23)
t r
The Einstein Field Equations
The remaining terms in (6.22) all vanish:
r r t r
t t t
=0 =0 =
(since (since
t r
t r
= 0) = 0) (since = 0)
1 R b(t,r ) 1 R e =0 R r R t
Therefore (6.22) reduces to
r t
r t
2b + t 2
b t
t r
(6.24)
To nd the components of the curvature tensor, we apply (6.21), which in this case gives
r t
= = = =
1 r R c d 2 t cd 1 r 1 R t t r t r + R r t r t r t 2 2 1 r 1 R t t r t r R r t r t t r 2 2 1 R r t t r R r t r t t r 2
Now we use the at space metric of the local frame a b = diag( 1, 1, 1) to raise and lower indices, and then apply the symmetries of the curvature tensor to write R r t r t = r r Rr t r t = Rr t r t = Rr t t r = r r R r t t r = R r t t r Therefore, we have
r t
1 1 R r t t r R r t r t t r = R r t t r + R r t t r t r 2 2
= R r t t r t r
The Einstein Field Equations
Comparison with (6.24) leads us to conclude that 2 b + t 2
t .
R r t t r =
b t
Let s calculate the curvature two form
t t t
Using (6.21), we have
=d =d =d
t t
1 R + R t 1 R + R t
t r
t r
Now d 1 R R t R d t =d 1 R R d R t
=d = = and
2 R 2 R dt d + dr d t 2 t r e b(t,r ) 2 R r 1 2 R t + 2 R t R t r
t r
b r e b(t,r ) R t R r e b(t,r ) b R r R t r
The Einstein Field Equations
Putting these results together, we obtain
1 2 R t e b(t,r ) 2 R r e b(t,r ) b R r + R t 2 R t r R t r b R 1 2 R t e b(t,r ) 2 R r = + 2 R t R t r t r =
t
Again, to nd the components of the curvature tensor, we write out
1 t R cd c 2
to obtain 1 t 1 1 R r r + R t r r + R t t t 2 2 2 1 t t + R t 2 1 1 R t r R t r r + R t t R t t t = 2 2 = R t r r + R t t t
t
Therefore, we conclude that 1 2 R R t 2 e b(t,r ) R b R 2 R t r t r
t R t =
R t r =
All together, the nonzero components of the curvature tensor are 2 b + t 2 b t
R r t t r = R t t = R r r =
= R r t r t R t r = e b(t,r ) R b R 2 R t r t r = R t r
1 2 R = R t t , R t 2 e 2b(t,r ) R
b R 2b(t,r ) 2 R R b e 2 + t t r r r
(6.25)
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