So the wave amplitude is always less than, or of the order of, the Newtonian potential φr. Why then can we detect h but not φr itself; why can we hope to find waves from a supernova in a distant galaxy without being able to detect its presence gravitationally before the explosion? The answer lies in the forces involved. The Newtonian tidal gravitational force on a detector of size l0 at a distance r is about φrl0/r2, while the wave force is hl0ω2 (see Eq. (9.45)). The wave force is thus a factor φ0(ωr)2 (φ0r/R)2 larger than the Newtonian force. For a relativistic system (φ0 0.1) of size 1 AU ( 1011 m), observed by a detector a distance 1023 m away, this factor is 1022. This estimate, incidentally, gives the largest distance r at which we may still approximate the gravitational field of a dynamical system as Newtonian (i.e. neglecting wave effects): r = R/φ0, where R is the size of the system.

The estimate in Eq. (9.100) is really an optimistic upper limit, because it assumed that all the mass motions contributed to Djk. In realistic situations this could be a serious overestimate because of the following fundamental fact: spherically symmetric motions do not radiate. The rigorous proof of this is discussed in Ch. 10, but in Exer. 40, § 9.7 we derive it from linearized theory, Eq. (9.101) below. It also seems to follow from Eq. (9.82a): if T00 is spherically symmetric, then Ilm is proportional to δlm and –I lm vanishes. But this argument has to be treated with care, since Eq. (9.82a) is part of an approximation designed to give only the dominant radiation. We would have to show that spherically symmetric motions would not contribute to terms of higher order in the approximation if they were present. This is in fact true, and it is interesting to ask what eliminates them. The answer is Eq. (9.77): conservation of energy eliminates ‘monopole’ radiation in linearized theory, just as conservation of charge eliminates monopole radiation in electromagnetism.

The danger of using Eq. (9.82a) too glibly is illustrated in Exer. 31e, § 9.7: four equal masses at the corners of a rotating square give no time-dependent Ilm and hence no radiation in this approximation. But they would emit radiation at a higher order of approximation.

Exact solution of the wave equation

Readers who have studied the wave equation, Eq. (9.64), will know that its outgoing-wave solution for arbitrary Tμν is given by the retarded integral

We let the origin be inside the source and we suppose that the field point xi is far away,

and that time derivatives of Tμν are small. Then, inside the integral, Eq. (9.101), the dominant contribution comes from replacing R−1 by r−1:

This is the generalization of Eq. (9.74). Now, by virtue of the conservation laws Eq. (9.77)

Tμν ,ν = 0,

we have

'

¯

i.e. the total energy and momentum are conserved. It follows that the 1/r part of h0μ is time independent to lowest order, so it will not contribute to any wave field. This generalizes Eq. (9.80). Again, see Exer. 32, § 9.7. Then, using Eq. (9.81), we get the generalization of Eq. (9.83):

As before, we can adopt the TT gauge to get

9.4 T h e e n e rg y c a r r i e d a wa y b y g ra v i t a t i o n a l wa v e s

Preview

We have seen that gravitational waves can put energy into things they pass through. This is how bar detectors work. It stands to reason, then, that they also carry energy away from their sources. This is a very important aspect of gravitational wave theory because, as we shall see, there are some circumstances in which the effects of this loss of energy on a source can be observed, even when the gravitational waves themselves cannot be detected. There are a number of different methods of deriving the formula for this energy loss (see Misner et al. 1973) and the problem has attracted a considerable amount of effort and has been attacked from many different points of view; see Damour (1987), Schutz (1980a), Futamase (1983), or Blanchet (2006). Our approach here will remain within the simple case of linear theory and will make the maximum use of what we already know about the waves.

In our discussion of the harmonic oscillator as a detector of waves in § 9.2, we implicitly assumed that the detector was a kind of ‘test body’, where the influence on the gravitational wave field is negligible. But this is, strictly speaking, inconsistent. If the detector acquires energy from the waves, then surely the waves must be weaker after passing through the detector. That is, ‘downstream’ of the detector they should have slightly lower amplitude than ‘upstream’. It is easy to see how this comes about once we realize that in § 9.2 we ignored the fact that the oscillator, once set in motion by the waves, will radiate waves itself. We solved this in § 9.3 and found, in Eq. (9.88), that waves of two frequencies will be emitted. Consider the emitted waves with frequency , the same as the incident wave.

¯

at any point a distance r away. (We call it δhxx to indicate that it is small compared to the incident wave.) It is a simple matter to get the total radiated field by adding up the contributions due to all the oscillators. In Fig. 9.7, consider a point P a distance z downstream from the plane of oscillators. Set up polar coordinates (ω˜ , φ) in the plane, centered at Q beneath P. A typical oscillator O at a distance ω˜ from Q contributes a field, Eq. (9.113), at P, with r = (ω˜ 2 + z2)1/2. Since the number of such oscillators between ω˜ and ω˜ + dω˜ is 2π σ ω˜ dω˜ , the total oscillator-produced field at P is

But we may change the integration variable to r,

w˜ dw˜ = r dr,

obtaining

If σ were constant, this would be trivial to integrate, but its value would be undefined at r = ∞. Physically, we should expect that the distant oscillators play no real role, so we adopt the device of assuming that σ is proportional to exp (−εr) and allowing ε to tend to zero after the integration. The result is

So the plane of oscillators sends out a net plane wave. To compare this to the incident wave, we must put Eq. (9.115) in the same TT gauge (recall that Eq. (9.83) is not in the TT gauge), with the result (Exer. 42, § 9.7)

If we now add this to the incident wave, Eq. (9.107), we get the net result, to first order in R,

This reduction must be responsible for the decrease in flux F downstream. Dividing Eq. (9.111) by Eq. (9.119) and using Eqs. (9.48) and (9.49) to eliminate R and φ gives the remarkably simple result

This is our key result. It says that a change δA in the amplitude A of a wave of frequencychanges its flux F (averaged over one period) by an amount depending only on , A, and δA. The oscillators helped us to derive this result from conservation of energy, but they have dropped out completely! Eq. (9.120) is a property of the wave itself. We can ‘integrate’ Eq. (9.120) to get the total flux of a wave of frequency and amplitude A:

(again, angle brackets denote an average over one period), and since there are only two

This form is invariant under background Lorentz transformations, but not under gauge changes. Since one polarization can be transformed into another by a background Lorentz transformation (a rotation), Eq. (9.122) applies to all polarizations and hence to arbitrary plane waves of frequency . In fact, since it gives an energy rate per unit area, it applies to any wavefront, either plane waves or the spherical expanding ones of § 9.3: we can always look at a small enough area that the curvature of the wavefront is not noticeable. The generalization to arbitrary waves (no single frequency) is in Exer. 43, § 9.7.

The reader who remembers the discussion of energy in § 7.3 may object that this whole derivation is suspect because of the difficulty of defining energy in GR. Indeed, we have not proved that energy is conserved, that the energy put into the oscillators must equal the decrease in flux; we have simply assumed this in order to derive the flux. Our proof