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

Solutions to exercises 85For this problem we need the xx-component because the photon is propagatingalong this direction, and we will always stay at z = 0, so we haveh TTxx = cos 2 θh + (t − x sin θ).We see that for this geometry the wave amplitude is reduced by a factor of cos 2 θ.Generalizing the argument in the text, the relation between time and positionfor the photon on its trip outwards along the x-direction is t = t 0 + x, wheret 0 is the starting time. The analogous relation after the photon is reflected ist = t 0 + L + (L − x), since in this case x decreases in time from L to 0. If weput these into the equation for the linearized corrections to the return time, we get{ ∫ Lt return = t 0 + 2L + 1 2 cos2 θ h + [t 0 + (1 − sin θ)x]dx0∫ L}× h + [t 0 + 2L − (1 + sin θ)x]dx .0This expression must be differentiated with respect to t 0 to find the variationof the return time as a function of the start time. The key point is how tohandle differentiation within the integrals. Consider, for example, the functionh + [t 0 + (1 − sin θ)x]. It is a function of a single argument,ξ := t 0 + (1 − sin θ)xso derivatives with respect to t 0 can be converted to derivatives with respect to xas followsdh += dh + dξ= dh +dt 0 dξ dt 0 dξ ;dh +dx = dh + dξdξ dx = (1 − sin θ)dh +dξ .It follows thatdh += dh +/(1 − sin θ).dt 0 dxOn the return trip the factor will be −(1 + sin θ). So when we differentiate wecan convert the derivatives with respect to t 0 inside the integrals into derivativeswith respect to x. Taking account of the factor cos 2 θ = (1 − sin θ)(1 + sin θ) infront of the integrals, the result isdt returndt 0= 1 + 1 2 (1 + sin θ) ∫ L+ 1 2 (1 − sin θ) ∫ L00dh +dx [t 0 + (1 − sin θ)x]dxdh +dx [t 0 + 2L − (1 + sin θ)x]dx.