Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 3

Write $g_{\mu \nu }={\eta }_{\mu \nu }+ϵh_{\mu \nu }$ and retain first-order terms. The quadratic Christoffel products in the supplied Ricci formula drop out. The wave-coordinate condition becomes the Lorenz gauge in linearized gravityThe linearized Ricci tensor and scalar then satisfySubstitution into $G_{\mu \nu }=8\pi T_{\mu \nu }$ givesIf the paper denotes the trace-reversed variable itself by $h_{\mu \nu }$ in its displayed equation, this is exactly that convention.

Under $h_{\mu \nu }^{\prime }=h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }$, trace reversal givesTaking a divergence yieldsApplying $\square$ to the transformed field likewise produces only derivatives of $\square \xi$. Therefore $\square {\xi }^{\mu }=0$ preserves both the gauge condition and the sourced wave equation. These are the residual gauge transformations.

Let $s=k_{\rho }{x}^{\rho }$. In vacuum,A nonzero localized profile cannot have $H^{\prime \prime }=0$, because an affine function does not decay at both ends. HenceDecay removes the integration constant, so the second condition is equivalently $k^{\mu }{H}_{\mu \nu }=0$. Thus a nontrivial plane gravitational wave in linearized gravity has a null wavevector and transverse amplitude.

A spatial rotation and rescaling put the future null vector in the form $k^{\mu }=(1,0,0,1)$, so the profile depends on $z-t$. Transversality gives four linear relations among the ten symmetric components. The four residual gauge functions satisfying $\square {\xi }^{\mu }=0$ remove the time and longitudinal components; the remaining trace can be removed by the residual transformation indicated in the question. The resulting transverse-traceless gauge isThe two arbitrary functions are the plus and cross gravitational-wave polarizations. They are the two physical degrees of freedom left after the four gauge conditions and four residual coordinate freedoms are removed.