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7 Gravitational Waves and the Binary Pulsars |
7.3.2 Energy and Momentum of a Plane Gravitational Wave
Let us consider the perturbed metric (7.3.2) and let us introduce a null coordinate system adapted to Minkowski space:
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− = u = x0 − x1; xi = x2, x3 |
(7.3.13) |
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+ = v = x0 + x1 |
(7.3.14) |
Correspondingly the Minkowskian metric takes the form:
ds2 = 2 dx+ dx− − dxi dxi = ημν dxμ dxν |
(7.3.15) |
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where: |
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ημν |
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(7.3.16) |
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A plane wave corresponds to the case where the metric deformation is a function only of one the light-cone coordinates u, v: for outgoing waves only of u, for incoming waves only of v. Since we are interested in outgoing waves we choose:
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hμν (x) = hμν (u) |
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(7.3.17) |
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Relying on the relations: |
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∂+ = |
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∂ |
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∂− = |
∂ |
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∂v |
∂u |
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∂+ = |
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∂ |
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= ∂−; |
∂− = η−+ |
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= ∂+ |
(7.3.18) |
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∂x− |
∂x+ |
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∂i = |
ηij |
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= −∂i |
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∂xj |
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from the Hilbert de Donder gauge condition: |
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∂μγμν = 0 |
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(7.3.19) |
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by using γμν = γμν (u) we obtain: |
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γ+ν = const = 0 |
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(7.3.20) |
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The last of the above equations is fixed by our physically chosen boundary conditions. At infinity, namely at very remote future times and in very distant space locations where the wave has not yet arrived, the metric is just Minkowski. Hence there is no constant part of γμν .
Next we recall the discussion of Chap. 5 about residual gauge transformations. The Lorentz covariant Hilbert de Donder gauge is not complete, since there exist
290 |
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7 Gravitational Waves and the Binary Pulsars |
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ω−i = − |
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(∂+hil − ∂i h+l ) dxl |
(7.3.29) |
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ωij = − |
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(∂i hj l − ∂j hil ) dxl |
(7.3.30) |
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2 |
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In the plane wave case denoting the derivative ∂/∂u by a dot we find:
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ω+− = ω−i = ωij = 0 |
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(7.3.31) |
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= − |
2 ˙ij |
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ω+i |
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h |
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dxj |
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(7.3.32) |
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or more explicitly: |
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+ |
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+ ˙ |
3 |
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ω |
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a dx |
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b dx |
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(7.3.33) |
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ω |
+ |
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− a˙ dx |
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b dx |
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Let us now consider the stress-energy tensor 3-form (7.3.10). Since the only nonvanishing component is ω+i , the second addend ωaf ωf b vanishes identically. Hence
td = |
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κ ωab ωcf Ef εabcd |
(7.3.34) |
Now (ab) must necessarily be (+i). Hence c must necessarily be 2 or 3 so that f = +, also necessarily. Therefore d cannot be anything else but −. Hence only t− is non-vanishing.
The final result is: |
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− = − κ |
+ |
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− = κ |
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+ |
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t |
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2 |
ω |
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E |
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(a)2 |
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dx |
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dx |
3 |
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du |
(7.3.35) |
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This result can be interpreted by recalling the encoding of the symmetric stressenergy in the stress-energy 3-form. This encoding was discussed in Chap. 5 and takes the following form:
td = Tdp ηpq εqrsuEr Es Eu |
(7.3.36) |
Applying the above relation to the stress energy 3-form of a plane wave as given by (7.3.35), we obtain:
t− = 2 · 3t−−η−+ε+23−E2 E3 E− = 6t−− dx2 dx3 du |
(7.3.37) |
Hence by comparison we conclude:
t−− = − |
01 = |
3κ |
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+ |
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t |
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(all other entries vanish) |
(7.3.38) |
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