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Using (13.3), we nd R a bcd = c = c
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1 2 h bd 2 h de 2 h eb = ae c e + ae c b ae c d 2 x x x x x x
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2 2 2 1 a f h bc a f hc f af h f b + 2 xd x f xd xb xd xc
Now the index e in the rst expression is a dummy index. Let s relabel it as d in each term ae ae ae 2 h bd 2 h bd = a f c f x c x e x x 2 2hd f h de = a f c b x c x b x x 2 2h f b h eb = a f c d x c x d x x
The last term will cancel, allowing us to write
1 = a f 2
2hd f 2hc f 2 h bc 2 h bd + c b d f d b x c x f x x x x x x
(13.4)
Now the Ricci tensor is found using Rab = R c acb . Using R a bcd = c a bd d a bc , we have R a b = c c bd d c bc . If we de ne the d Alembertian operator W = 2 t 2 2 2 2 + 2+ 2 x2 y z = ab a b
Gravitational Waves
along with h = cd h cd , then the Ricci tensor can be written as 1 2hcb 2h 2hca Rab = + a c W h ab a b 2 x b x c x x x x The Ricci scalar is given by R= 2 h cd Wh xc xd (13.6) (13.5)
We can put these results together to write down the Einstein tensor for the linearized theory. To simplify notation, we will represent partial derivatives using commas; i.e., 2 h cd = h cd ,cd xc xd and 2h = h ,ab xa xb
Using this notation, the Einstein tensor is 1 G ab = h c a,bc + h c b,ac Whab h ,ab ab h cd ,cd + ab Wh 2 (13.7)
Traveling Wave Solutions
We now de ne the following function, known as the trace reverse in terms of h ab : 1 ab = h ab ab h 2 The trace of this function is ab = ac c b Now note that
c ab h = ac b h
(13.8)
h ab = ac h c b
Gravitational Waves
Putting these results together, we obtain 1 c ac c b = ac h c b ac b h 2 1 c c b = h c b b h 2 Now we take the trace by setting b c: 1 c c c = h c c c h 2 We are working in four dimensions. Therefore the trace of the Kronecker delta is c = 4. Setting = c c and h = h c c , we nd the trace to be c 1 1 c = h Tr c h = h (4) h = h 2h = h 2 2 (13.9)
This is why ab is known as the trace reverse of h ab . Now let s replace h ab by the trace reverse in Einstein s equation. Using (13.7) together with (13.8) and (13.9), we have 1 G ab = h c a,bc + h c b,ac Whab h ,ab ab h cd ,cd + ab Wh 2 1 1 c 1 c 1 = c a,bc + a h ,bc + c b,ac + b h ,ac W ab + ab h h ,ab 2 2 2 2 1 ab cd ,cd ab cd h ,cd + ab Wh 2 Let s rearrange the terms in this expression 1 1 c 1 c G ab = c a,bc + c b,ac W ab ab cd ,cd + a h ,bc + b h ,ac 2 2 2 W 1 1 ab h h ,ab ab cd h ,cd + ab Wh 2 2
Notice that
Gravitational Waves
1 c 1 a h ,bc = h ,ba 2 2 1 1 c b h ,ac = h ,ab 2 2 Now partial derivatives commute, so 1 h ,ba = 1 h ,ab and 2 2 1 c 1 c a h ,bc + b h ,ac = h ,ab 2 2 Inserting this into the Einstein tensor, we can cancel a term reducing it to 1 G ab = c a,bc + c b,ac W ab ab cd ,cd W 2 1 ab cd h ,cd + ab Wh 2 1 1 1 = c a,bc + c b,ac W ab ab cd ,cd + ab Wh ab cd h ,cd 2 2 2 Now ab W h = ab cd h ,cd . This cancels the last two terms and so using the trace reverse the Einstein tensor becomes 1 G ab = c a,bc + c b,ac W ab ab cd ,cd 2 (13.10) 1 ab h 2
To obtain the wave equations, we perform a gauge transformation. This is a coordinate transformation that leaves R a bcd , Rab , and R unchanged if we consider terms only to rst order in . The coordinate transformation that will do this is x a x a = x a + a where a is a function of position and a ,b coordinate transformation changes h ab as (13.11) 1. It can be shown that this
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