(ed.). Gravitational waves (IOP, 2001)(422s).

72 Source calculationsIn this case the trace-free moment ˜M ij differs from M ij only by a constant,so we can use these values for M ij to calculate the field and luminosity.As an example of calculating the field, let us compute ¯h TTxx as seen by anobserver at a distance r from the system along the y-axis, i.e. lying in the planeof the orbit. We first need the TT part of the mass-quadrupole moment, fromequation (6.27):M TTxx = M xx − 1 2 (M xx + M zz ).However, since M zz = 0, this is just M xx /2. Then from equation (6.28) we find¯h TTxx =−2 µ r (R)2 cos[2(t − r)]. (7.3)Similarly, the result for the luminosity isL gw = 325 µ2 R 4 6 . (7.4)The various factors in these two equations are not independent, because theangular velocity is determined by the masses and separations of the stars. Whenobserving such a system, we cannot usually measure R directly, but we can infer from the observed gravitational-wave frequency, and we may often be able tomake a guess at the masses (we will see below that we can actually measure animportant quantity about the masses). So we eliminate R using the Newtonianorbit equationR 3 = m 1 + m 2 2 . (7.5)If in addition we use the gravitational-wave frequency gw = 2, we get¯h TTxx =−2 1/3 Å5/3 gw2/3cos[ gw (t − r)], (7.6)r4L gw =5 × 2 1/3 (Å gw) 103 , (7.7)where we have introduced the symbol for the chirp mass of the binary system:Å := µ 3/5 (m 1 + m 2 ) 2/5 .Notice that both the field and the luminosity depend only on Å, not on theindividual masses in any other combination.The power represented by L gw must be supplied by the orbital energy,E =−m 1 m 2 /2R. By eliminating R as before we find the equationE =− 12 5/3 Å5/3 2/3gw .This is remarkable because it too involves only the chirp mass Å. By setting therate of change of E equal to the (negative of the) luminosity, we find an equationfor the rate of change of the gravitational-wave frequency˙ gw =12 × 21/3Å 5/3 11/3gw . (7.8)5