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

Variational principles and the energy in gravitational waves 53own stress-energy tensor. To do this we have to expand the action out to secondorder in the metric perturbation,∫I[g µµ + h µν ] = R(g µν + h µν , g µν,α + h µν,α ,...) √ −g[g µν + h µν ]d 4 x∫= R(g µν ,...) √ ∫ √ δ(R −g)−g d 4 x +h µν d 4 x+ 1 2δg µν∫ ( ∂ 2 (R √ −g)∂g µν ∂g αβh µν h αβ + 2 ∂2 (R √ −g)∂g µν ∂g αβ,γh µν h αβ,γ+ ∂2 (R √ −g)h µν,τ h αβ,γ + 2 ∂2 (R √ )−g)h µν h αβ,γ τ d 4 x∂g µν,τ ∂g αβ,γ ∂g µν ∂g αβ,γ τ+ O(3).The first term is the action for the background metric g µν . The second termvanishes (see equation (5.2)), since we assume that the background metric is asolution of the Einstein vacuum equation itself, at least to lowest order.If we compare the above equation with equation (5.4), we can see that thethird term, complicated as it seems, is an effective ‘matter’ Lagrangian for thegravitational field. Indeed, if one varies it with respect to h µν holding g µνfixed (as we would do for a physical matter field on the background), then thecomplicated coefficients are fixed and one can straightforwardly show that onegets exactly the linear perturbation of the Einstein tensor itself. Its vanishing isthe equation for the gravitational-wave perturbation h µν . In this way we haveshown that, for a small amplitude perturbation, the gravitational wave can betreated as a ‘matter’field with its own Lagrangian and field equations.Given this Lagrangian, we should be able to calculate the effective stressenergytensor of the wave field by taking the variations of the effective Lagrangianwith respect to g µν , holding the ‘matter’field h µν fixed:withL (GW)√ −g = 132πT (GW)αβ √ −g = 2 ∂ L(GW) [g µν , h µν ] √ −g∂g αβ(5.11)( ∂ 2 (R √ −g)∂g µν ∂g αβh µν h αβ + 2 ∂2 (R √ −g)∂g µν ∂g αβ,γh µν h αβ,γ+ ∂2 (R √ −g)∂g µν,τ ∂g αβ,γh µν,τ h αβ,γ + 2 ∂2 (R √ −g)∂g µν ∂g αβ,γ τh µν h αβ,γ τ).(5.12)This quantity is quadratic in the wave amplitude h µν . It could be simplifiedfurther by integrations by parts, such as by taking a derivative off h αβ,γ τ . Thiswould change the coefficients of the other terms. We will not need to worry aboutfinding the ‘best’ form for the expression (4.12), as we now show.