Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 50 / 4 / b

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

Let the symmetric metric perturbation $h^{\alpha \beta }$ be varied with compact support. Integrating the mixed derivative product by parts makes its integral equal to that of $A_{\sigma }{A}^{\sigma }$. Thus the supplied massless Fierz-Pauli action has, up to a boundary term, densityFor example, the equivalence of the mixed term follows from commuting flat-space partial derivatives after moving one derivative off $h^{\rho \sigma }$. Here $\delta h={\eta }_{\alpha \beta }\delta h^{\alpha \beta }$ and $\delta A^{\nu }={\partial }_{\mu }\delta h^{\mu \nu }$.

Integrating each variation by parts, the four terms contribute respectivelyThe symmetrization in the second term is required because the varied field is symmetric. Comparing this coefficient with the Einstein tensor perturbation above givesThus arbitrary compactly supported variations give $\delta G_{\alpha \beta }=0$, exactly the vacuum Linearized Einstein equations. The minus sign and overall normalization of the action do not change those equations; boundary conditions justify the discarded total derivatives.