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= r sin d
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Applying (9.10), we have the following relations: l= t + r dt + dr = 2 2
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Null Tetrads and the Petrov
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n= m= t r dt dr = 2 2 r d + ir sin d + i = 2 2
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r d ir sin d i = m= 2 2 In many cases, you will see the vectors written in terms of components with respect to the coordinate basis, i.e., v a = v 0 , v 1 , v 2 , v 3 or in this case v a = v t , v r , v , v or v a = v t , vr , v , v . Using this notation, we have 1 1 la = (1, 1, 0, 0) , n a = (1, 1, 0, 0) 2 2 1 1 m a = (0, 0, r, ir sin ) , m a = (0, 0, r, ir sin ) 2 2 With this in mind, let s verify (9.8). Starting with gtt in (9.11), we have gtt = lt n t + lt n t m t m t m t m t = = Moving on, we nd grr = lr n r + lr n r m r m r m r m r = 1 2 1 2 + 1 2 1 2 1 2 1 2 + 1 2 1 2
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1 1 + =1 2 2
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1 1 = = 1 2 2 r g = l n + l n m m m m = = r 2 r r
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r2 r2 = r 2 2 2
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ir sin 2 ir sin 2
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g = l n + l n m m m m = 2
= r 2 sin2 We can raise and lower the indices of the null tetrad using the metric gab . To see this, we use (9.9) together with the inner product relations de ned in (9.2), (9.3), (9.4), and (9.5). Let s consider one case. We have g ab = l a n b + l b n a m a m b m b m a What we would like is that l a = g ablb . So let s just put it there on both sides: g ablb = l a n b lb + l b n a lb m a m b lb m b m a lb Now we rearrange terms a bit and use the fact that A B = Ab Bb to write g ablb = l a n b lb + n a l b lb m a m b lb m a m b lb = l a (l n) + n a (l l) m a (m l) m a (m l) The inner product relations tell us the only term that survives is l n = 1. Therefore, we obtain the desired result l a = g ablb . EXAMPLE 9-4 Using g ab , nd l a , n a , and m a , and verify the inner product relations given in (9.2), (9.4), and (9.5). SOLUTION 9-4 Now the metric with raised indices is given by 1 0 g ab = 0 0 0 1 0 0 0 0 r12 0 0 0 0
1 r 2 sin2 1 2
We can proceed in the usual way. For l a = g ablb and la = have l t = g tt 1 2 1 = (+1) 2
(1, 1, 0, 0), we
1 = 2
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l r = grr 1 2 1 = ( 1) 2 1 = 2
1 l a = (1, 1, 0, 0) 2 Therefore, we have l l = l a la = For n a =
1 2
1 2
1 2
1 + 2
1 2
1 1 =0 2 2
(1, 1, 0, 0), we obtain n t = g tt 1 2 1 = (+1) 2 1 = 2 1 1 = ( 1) = 2 2
1 n r = grr 2 1 n a = (1, 1, 0, 0) 2 And so we have n n = n a n a = Now we compute l n = l a na = 1 2 1 2
1 2
1 2
1 2
1 2 =
= 0.
1 + 2
1 2
1 1 + =1 2 2
So far, so good. Now let s check the complex basis vectors. For m a we nd m = g m = 1 r2 r2 r 1 sin2 1 = 2r ir sin 2 = i 2r sin
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