The Linearized Metric in Visual Studio .NET

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The Linearized Metric
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We begin by considering a metric that differs from the at Minkowski metric by a small perturbation. If we take to be some small constant parameter (| | 1), then we can write the metric tensor as gab = ab + h ab (13.1)
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where we neglect all terms of order 2 or higher since is small. Our rst step in the analysis will be to write down the form of the various quantities such as the Christoffel symbols, the Riemann tensor, and the Ricci tensor when we write the metric in this form. Since we will drop all terms that are order 2 or higher, these quantities will assume fairly simple forms. Ultimately, we will show that this will allow us to write the Einstein eld equations in the form of a wave equation. We might as well take things in order and so we begin with the Christoffel symbols. We will work in a coordinate basis and so we compute the following:
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1 ad g 2
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gdb gbc gcd + d b x x xc
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To see how this works out, let s consider one term gbc bc h bc = d ( bc + h bc ) = + d x x xd xd
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Gravitational Waves
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Now the Minkowski metric is given by ab = diag (1, 1, 1, 1) and so bc = 0. We can pull the constant outside of the derivative and so xd h bc gbc = d d x x To obtain the form of the Christoffel symbols, we need to know the form of g ab . We begin by observing that we can raise indices with the Minkowski metric; i.e., h ab = ac bd h cd
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c We also recall that the metric tensor satis es gab g bc = a (note this is also true for the Minkowski metric). The linearized form of the metric with raised indices will be similar but we assume it can be written as g ab = ab + ah ab , where a is a constant to be determined. Now ignoring terms of order 2 , we nd c a = ( ab + h ab ) bc + a h bc
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= ab bc + bc h ab + a ab h bc + a 2 h ab h bc = ab bc + bc h ab + a ab h bc
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c = a + bc h ab + a ab h bc
For this to be true, the relation inside the parentheses must vanish. Therefore, we have a ab h bc = bc h ab Let s work on the left-hand side: a ab h bc = a ab be c f h e f
e = a a c f h e f
= a c f h a f Now, notice that the index f is repeated and is therefore a dummy index. We rename it b and stick the result back into a ab h bc = bc h ab to obtain a bc h ab = bc h ab
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Note that we used the fact that the metric is symmetric to write cb = bc . We conclude that a = 1 and we can write g ab = ab h ab We now return to the Christoffel symbols. Using (13.2) together with h bc and ignoring terms of second order in , we nd xd
a bc
(13.2)
gbc xd
1 = g ad 2 = g ad 2 = =
gdb gbc gcd + d b x x xc h db h bc h cd + d b x x xc h db h bc h cd + d b x x xc h db h bc h cd + xd xb xc
ad h ad 2 1 ad 2 h ad 2
Dropping the second-order terms, we conclude that in the linearized theory we have
1 = ad 2
h db h bc h cd + d b x x xc
(13.3)
We can use this expression to write down the Riemann tensor and the Ricci tensor. The Riemann tensor is given by R a bcd = c
a bd
d
Looking at the last two terms, e bd a ec and e bc a ed , in comparison with (13.3) shows that these terms will result in terms involving 2 , and so we obtain the simpli ed expression R a bcd = c
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