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

86 Solutions to exercisesThe integrals can now be done, since they simply invert the differentiation by x.Evaluating the integrands at the end points of the integrals gives equation (2.20):dt return= 1 − 1dt2 (1 + sin θ)h +(t 0 ) + sin θh + [t 0 + (1 − sin θ)L]0+ 1 2 (1 − sin θ)h +(t 0 + 2L).2. If we Taylor-expand this equation in powers of L about L = 0, the leadingterm vanishes, and the first-order term is:dt return= L sin θ(1 − sin θ)ḣ + (t 0 ) + L(1 − sin θ)ḣ + (t 0 ),dt 0= L cos 2 θḣ + (t 0 ).This is just what was required. The factor of cos 2 θ comes, as we saw above, fromthe projection of the TT field on the x-coordinate direction.3. All the terms cancel and there is no effect on the return time.Exercise (b)This is part of the calculation in the previous example. All we need is the segmentwhere the light travels from the distant end to the centre:and so dt out /dt 0 is:t out = t 0 + 1 2 cos2 θ∫ L0h + [t 0 + L − (1 + sin θ)x]dxdt outdt 0= 1 + 1 2 (1 − sin θ)[h +(t 0 − sin θ L) − h + (t 0 + L)].Exercise (c)This question is frequently asked, but not by people who have done thecalculation. The answer is that the two effects occur in different gauges, notin the same one. So they cannot cancel. The apparent speed of light changes inthe TT gauge, but then the positions of the ends remain fixed, so that the effect isall in the coordinate speed. In a local Lorentz frame tied to one mass, the ends domove back and forth, but then the speed of light is invariant.Exercise (d)To first order we haveR µ αβν = Ɣµ αν,β − Ɣµ αβ,ν ,Ɣ µ αβ,ν = 1 2 ηµσ (h σβ,αν + h σα,βν − h αβ,σ ν ), (i) (7.13)Ɣ µ αν,β = 1 2 ηµσ (h σν,αβ + h ασ,νβ − h αν,σβ ). (ii) (7.14)