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(when going through this calculation, be careful with the overall minus sign). A similar exercise shows that the spin coef cients = = 0. With all the vanishing spin coef cients, the Newman-Penrose identity we need to calculate, 4 , takes on a very simple form. The expression we need is given by (9.23): = ( + ) (3 ) + 3 + + In this case, we have
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Recalling (9.13), this equation becomes 4 = m a a . Note, however, that is a scalar we need to calculate only partial derivatives. The end result is that
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= m a a = m x x + m y y 1 = 2 i + 2 = 1 4 1 x 2 2 1 y 2 2 + H H +i x y H H +i x y i 4 y 2 H 2 H +i 2 x y y
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1 2 H i 2 H 1 2 H + 4 x2 2 x y 4 y2
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and so we conclude that the Weyl scalar is given by
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(9.31)
In the exercises, if you are so inclined you can show that the only nonzero component of the Ricci tensor is given by
2 H 2 H + x2 y2
(9.32)
The fact that 4 is the only nonzero Weyl scalar tells us that the Petrov type is Type N, meaning that the principal null direction is repeated four times. Therefore, this metric describes transverse gravity waves.
Quiz
1. Using the Minkowski metric ab = diag(1, 1, 1, 1), A = (3, 0, 3, 0) is (a) timelike (b) not enough information has been given (c) null (d) spacelike Using the de nition of the complex vectors in the null tetrad, i.e., ma = j a + ika , 2 ma = j a ika 2
which one of the following statements is true (a) These are null vectors and that m m = 1. (b) These are timelike vectors and that m m = 1. (c) These are null vectors and that m m = 1. (d) These are null vectors and that m m = 0. 3. Consider Example 9-5. Which of the following relations is true (a) = b n a m a l b = 0 (b) = b n a m a l b = 1 b (c) = b n a m a l = 0 b (d) = b n a m a l = 1