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Cosmology
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Using this as a guide, we write the spatial component of the metric in the present case in the following way: d 2 = e2 f (r ) dr 2 + r 2 d 2 + r 2 sin2 d 2 We could go through the long and tedious process of deriving the Christoffel symbols and the components of the Ricci tensor to nd e2 f (r ) . However, there is an easier method available. The similarity in form to the Schwarzschild metric means that we can quickly derive the function e2 f (r ) by using our previous results. Recalling our results for the Schwarzschild metric d2 + dr 2 d2 + dr 2
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e 2 (r )
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e 2 (r )
1 d 2 1 d 2 1 e 2 e e + + r dr r dr r2 1 d 2 1 d 2 1 e 2 + + e e r dr r dr r2
we can solve the problem by focusing on one term. It is the easiest to choose Rr r and work in the coordinate basis. We can write Rr r in the coordinate basis as d2 + dr 2
Rrr =
d dr
d dr
d dr
2 d r dr
(12.2)
We can obtain the equations in the present case by setting 0 and f . So for Rrr , we have Rrr = 2 df r dr (12.3)
Using (12.1), Ri jkl = K gik g jl gil g jk , we can obtain the Ricci tensor in terms of K and the metric. We have labeled the indices using (i, j, k, l) because we
Cosmology
are considering only the spatial part of the metric. The Ricci tensor is de ned as R jl = R k jkl We obtain the contraction via R k jkl = g ki Ri jkl = g ki K gik g jl gil g jk = K g ki gik g jl g ki gil g jk
k = K k g jl lk g jk k = K k g jl g jl
We are working in three dimensions, and the Einstein summation convenk tion is being used, and so k = 1 + 1 + 1 = 3. This means that the Ricci tensor for three dimensions in the constant curvature case is given by R jl = K 3g jl g jl = 2K g jl . We can use this with our results we obtained by comparison to the Schwarzschild solution to quickly nd the function e2 f. Now, looking at d 2 = e2 f (r ) dr 2 + r 2 d 2 + r 2 sin2 d 2 , we can write the metric as e2 f gij = 0 0 0 r2 0 0 0 2 2 r sin
Using R jl = 2K g jl together with (12.3) and grr = e2 f , we obtain the following differential equation: 2 df = 2K e2 f r dr Cross multiplying we obtain e 2 f d f = K r dr
Integration yields
Cosmology
r2 1 e 2 f = K + C 2 2 where C is a constant of integration. Solving we nd e2 f = 1 C Kr2 (12.4)
To nd the constant C, we can use the other components of the Ricci tensor. In the coordinate basis, R for the Schwarzschild metric is R = 1 e 2 + r e 2 ( ). Therefore in the present case we have, using 0 and f, R = 1 e 2 f + r e 2 f df dr
2 df r dr
Let s rewrite this, using our previous results. We found that therefore, r e 2 f df = r e 2 f K r e2 f = K r 2 dr
= 2K e2 f ;
Using this along with (12.4), we can rewrite R as R = 1 e 2 f + r e 2 f df = 1 C + K r 2 + K r 2 = 1 C + 2K r 2 dr
Now we use R jl = 2K g jl together with g = r 2 . We have R = 2K g = 2K r 2 . Therefore, we have 1 C + 2K r 2 = 2K r 2 or 1 C = 0 C = 1. The nal result is then e2 f = 1 1 Kr2
and we can write the spatial part of the metric as d 2 = dr 2 + r 2 d 2 + r 2 sin2 d 2 1 kr 2 (12.5)
The curvature constant K is normalized and denoted by k (we can absorb any constants into the scale factor). There are three cases to consider. If k = +1,
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