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The Einstein Field Equations
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EXAMPLE 6-3 Using the results of Example 6-1, nd the components of the Einstein tensor in the coordinate basis. SOLUTION 6-3 We begin by calculating the components of the Ricci tensor. For now, we continue working in the orthonormal basis. Therefore, we use
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Ra b = R c a c b
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(6.26)
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The rst nonzero component of the Ricci tensor is
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Rt t = R c t ct = R t t t t + R r t r t + R t t
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Noting that R t t = R t t = R t t = Rt t = t t R t t = R t t and using the results of the previous example, we nd that
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Rt t = Next we calculate
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2 b t 2
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b t
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1 2 R R t 2
(6.27)
Rt r = R c t cr = R t t t r + R r t r r + R t r
The only nonzero term in this sum is R t r = R t r and so
Rt r = Next, we nd that
b R 2 R e b(t,r ) + R t r t r
(6.28)
r Rr r = R c r cr = R t r t r + Rr r r + R r r
= R t r t r + R r r = 2 b + t 2 b t
(6.29) + e 2b(t,r ) R b R 2b(t,r ) 2 R R b 2 + e t t r r r
The Einstein Field Equations
Finally, using the same method it can be shown that R = 1 2 R e 2b(t,r ) + R t 2 R b R 2b(t,r ) 2 R R b 2 + e t t r r r
(6.30)
The next step is to nd the Ricci scalar using R = a b Ra b together with (6.27), (6.29), and (6.30):
R = Rt t + Rr r + R =2 2 b b +2 2 t t
2 2 R e 2b(t,r ) b R 2b(t,r ) 2 R R b +2 2 + e R t 2 R t t r r r (6.31)
In the local frame, we can nd the components of the Einstein tensor using 1 G a b = Ra b a b R 2 For example, using (6.29) together with (6.31), we nd
G r r = Rr r = 2 b + t 2 = 1 R 2 b t
(6.32)
e 2b(t,r ) R
b R 2b(t,r ) 2 R R b 2 + e t t r r r 2 2 R e 2b(t,r ) +2 R t 2 R b R 2b(t,r ) 2 R R b e 2 + t t r r r
b 1 2 b 2 2 +2 2 t t 1 2 R R t 2
Using (6.30) and (6.31), we obtain
G = R = 1 R 2 b R 2b(t,r ) 2 R R b e 2 + t t r r r
1 2 R e 2b(t,r ) + R t 2 R
The Einstein Field Equations
b 1 2 b 2 2 +2 2 t t 2 b t 2 b t
2 2 R e 2b(t,r ) +2 R t 2 R
b R 2b(t,r ) 2 R R b e 2 + t t r r r
A similar exercise shows that the other nonzero components are G t t = G t r = 1 b R e 2b(t,r ) 2 R e 2b(t,r ) R b + R t t R r 2 R r r e b(t,r ) 2 R e b(t,r ) b R R t r R t r
To write the components of the Einstein tensor in the coordinate basis, we need to write down the transformation matrix a b . Using the metric ds 2 = dt 2 + e2b(t,r ) dr 2 + R(t, r ) d 2 , this is easy enough: 1 0 0 a 0 eb(t,r ) 0 (6.33) b = 0 0 R(t, r ) The transformation is given by G ab =
c a d b G cd
(6.34)
Note that the Einstein summation convention is being used on the right side of (6.34). However, since (6.33) is diagonal, each expression will use only one term from the sum. Considering each term in turn, using (6.33) and (6.34) we nd G tt =
t t t t Gtt
= ( 1)( 1)G t t b R 2b(t,r ) 2 R R b 2 + e t t r r r (6.35)
= G t t = G tr =
t t
e 2b(t,r ) R
r r G tr
= ( 1)(eb(t,r ) ) = 1 R
e b(t,r ) b R e b(t,r ) 2 R R t r R t r
(6.36)
2 R b R t r t r
G rr =
The Einstein Field Equations
r r r r Grr
e2b(t,r ) 2 R = R t 2 and nally G = =R
2 G
(6.37)
2b 2 t
b t b t
(6.38)
= R
2b + t 2
EXAMPLE 6-4 Using the results of Example 6-2, use the Einstein equations with nonzero cosmological constant < 0 to nd the functional form of eb(t,r ) and R(t, r ). SOLUTION 6-4 As stated earlier, the energy-momentum tensor for dust is given by Tab = u a u b . Since we are working in 2+1 dimensions, we will call u a the three velocity. It will be easiest to work in the co-moving frame. In this case, the three velocity takes on the simple form u a = u t , u r , u = (1, 0, 0). We can also use the local at space metric a b 1 = 0 0 0 0 1 0 0 1
(6.39)
Einstein s equations can then be written as G ab + a b = Ta b (6.40)
where is a constant. Since we are taking < 0, we will make this more explicit by writing = 2 for some 2 > 0. 1 2 Earlier we found that G r r = R tR . Using r r = 1 and Tr r = 0 with (6.40), we 2 nd 2 R + 2 R = 0 t 2 (6.41)