Null Tetrads and the Petrov in VS .NET

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Null Tetrads and the Petrov
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Next we use g ab to raise indices. We already know what l a is. The components of n a are given by n a = g ab n b n u = g ub n b = g uu n u + g uv n v = n v = 1 n v = g vb n b = g vu n u + g vv n v = n u H n v = 1 1 H H = H 2 2
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It can be easily veri ed that the remaining components are zero. The nal result is that 1 n a = 1, H, 0, 0 2 A similar exercise shows that 1 m a = (0, 0, 1, i) 2 1 and m a = (0, 0, 1, i) 2 (9.28) (9.27)
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We are going to need the Christoffel symbols for this metric so that we can calculate covariant derivatives. These are fairly easy to calculate; we will compute two of them explicitly. Now
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v xu
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1 vd g ( x gdu + u gd x d gxu ) 2
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There is only one nonconstant term in the metric with lowered indices, guu = H . Therefore, we can drop the last two terms and this reduces to
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1 H 1 vu g ( x guu ) = 2 2 x 1 H 1 vu g y guu = 2 2 y
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Substitution of y for x yields a similar result
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v yu
All together, the nonzero Christoffel symbols for the metric are
v xu
1 H , 2 x 1 H , 2 u
yu x
1 H 2 y = 1 H , 2 x
y uu
(9.29) = 1 H 2 y
Null Tetrads and the Petrov
With this information in hand we can calculate the spin coef cients and the Weyl scalars. The spin coef cients can be calculated from = b n a m a l b , = b n a m a m b = b l a m a n b , = = = b l a m a m b , = b l a m a l b , = = b l a m a m b = b n a m a n b , = b n a m a m b ,
1 b la n a l b b m a m a l b , 2 1 b la n a m b b m a m a m b , 2
1 b la n a n b b m a m a n b 2 1 b la n a m b b m a m a m b 2
Now b l a = 0 and so we can immediately conclude that = = = = 0. As a result, we see there is no shear, twisting, or divergence for the null congruence in this case. In addition, one can see that = = = = 0. The problem is starting to look tractable; we are going to have to worry about only very few terms. Let s consider , which will turn out to be nonzero. We will start by writing out the summation implied by the formula. However, note that we aren t going to have to write down every term. We are saved by the fact that m has only x and y components while n has only u and v components. Proceeding, we have = b n a m a n b = u n x m x n u u n y m y n u v n x m x n v v n y m y n v The last two terms drop out. Let s look at the very last term to see why. Remembering that n y = 0, we have v n y = n y v
d yv n d
yv n u
yv n v v
Having a look at the Christoffel symbols (0.29), we see that u yv = So we can ignore these terms. Now the rst two terms work out to be u n x = u n y = n x u n y u
d xu n d
= 0.
= =
xu n v
= =
1 H 2 x 1 H 2 y
yu n d
yu n v
Null Tetrads and the Petrov
Putting these results back into the expression for , we nd that = u n x m x n u u n y m y n u = 1 H 2 x 1 2 1 H 2 y 1 i 2 (9.30)
1 = 2 2
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