172
11 Linearized General Relativity and Gravitational Waves
See (11.99) in Appendix 3, for the electromagnetic analog of this.
It is interesting to observe that the solution (11.51) obtained to study gravitational
waves also contains the solution for the static field (11.21) of a mass distribution,
which we discussed in connection with the classical limit. To see this we set μ =
ν = 0 and see immediately that ¯
h 00 is consistent with (11.20) and (11.21) since the
integral of the density is the mass M. In abbreviated notation we thus have,
¯
h 00 = −
4G
c 2 r
T 00 d
3 x
= −
4G M
c 2 r
.
(11.52)
It is always useful to have such a “sanity check.”
Equation (11.51) is a complete (but approximate) solution to the problem we
posed, the metric field from a small and distant source with a known energy
momentum distribution. However it has two aspects that require further attention.
First, it is not in the most convenient form for typical sources, and secondly it is
clearly not in the traceless transverse gauge form that is convenient for studying
motion in a detector system. Dealing with these requires some straight-forward but
somewhat lengthy algebra.
To begin we will show that if either of the subscripts in (11.51) is zero the integral
is constant in time and thus not relevant for wave analysis. Our tool for showing
this is the zero divergence of the energy-momentum tensor, that is the conservation
of energy-momentum. We set μ = 0 and differentiate the integral in (11.51) with
respect to the retarded time to obtain
∂
∂t
T
0ν d
3 x
=
T
0ν ,0 d
3 x
.
(11.53)
Consider first ν = 0 so the zero divergence of the energy-momentum tensor implies
T
0β ,β = T
00 ,0 + T
0i ,i = 0, T
00 ,0 = −T
0i ,i ,
(11.54)
and, with the help of Gauss’s Theorem, we may evaluate (11.53) for as a surface
integral,
∂
∂t
T
00 d
3 x
= −
T
0k ,k d
3 x
= −
S
T
0k dS k .
(11.55)
But the source is of limited extent by assumption, so we can choose the surface S
to be outside the source region, and the surface integral is thus zero and does not
correspond to gravitational waves. The same manipulations show that for ν = j the
integral (11.53) is also zero. Only the space components of the energy-momentum
tensor produce waves. This is consistent with our previous result (11.52).
Having disposed of the μ = 0 parts of the solution (11.51) we are left with, in
obvious abbreviated notation,
11 Linearized General Relativity and Gravitational Waves
See (11.99) in Appendix 3, for the electromagnetic analog of this.
It is interesting to observe that the solution (11.51) obtained to study gravitational
waves also contains the solution for the static field (11.21) of a mass distribution,
which we discussed in connection with the classical limit. To see this we set μ =
ν = 0 and see immediately that ¯
h 00 is consistent with (11.20) and (11.21) since the
integral of the density is the mass M. In abbreviated notation we thus have,
¯
h 00 = −
4G
c 2 r
T 00 d
3 x
= −
4G M
c 2 r
.
(11.52)
It is always useful to have such a “sanity check.”
Equation (11.51) is a complete (but approximate) solution to the problem we
posed, the metric field from a small and distant source with a known energy
momentum distribution. However it has two aspects that require further attention.
First, it is not in the most convenient form for typical sources, and secondly it is
clearly not in the traceless transverse gauge form that is convenient for studying
motion in a detector system. Dealing with these requires some straight-forward but
somewhat lengthy algebra.
To begin we will show that if either of the subscripts in (11.51) is zero the integral
is constant in time and thus not relevant for wave analysis. Our tool for showing
this is the zero divergence of the energy-momentum tensor, that is the conservation
of energy-momentum. We set μ = 0 and differentiate the integral in (11.51) with
respect to the retarded time to obtain
∂
∂t
T
0ν d
3 x
=
T
0ν ,0 d
3 x
.
(11.53)
Consider first ν = 0 so the zero divergence of the energy-momentum tensor implies
T
0β ,β = T
00 ,0 + T
0i ,i = 0, T
00 ,0 = −T
0i ,i ,
(11.54)
and, with the help of Gauss’s Theorem, we may evaluate (11.53) for as a surface
integral,
∂
∂t
T
00 d
3 x
= −
T
0k ,k d
3 x
= −
S
T
0k dS k .
(11.55)
But the source is of limited extent by assumption, so we can choose the surface S
to be outside the source region, and the surface integral is thus zero and does not
correspond to gravitational waves. The same manipulations show that for ν = j the
integral (11.53) is also zero. Only the space components of the energy-momentum
tensor produce waves. This is consistent with our previous result (11.52).
Having disposed of the μ = 0 parts of the solution (11.51) we are left with, in
obvious abbreviated notation,