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Raise perturbation indices with the background Minkowski metric. To first order, the inverse metric is $g^{ik}={\eta }^{ik}-ϵh^{ik}+O({ϵ}^2)$ and the Levi-Civita connection isQuadratic connection products in the Riemann curvature tensor are of order ${ϵ}^2$. Substitution into the paper's curvature convention givesThe signs follow the convention on the cover sheet; reversing the curvature definition would reverse this expression.

The gauge freedom is the freedom to identify points of the perturbed spacetime with points of the flat background in slightly different coordinates. Under the passive infinitesimal change $x^{\prime i}=x^i+ϵ{\xi }^i(x)$, the metric transformation law gives the linearized coordinate gauge transformationThus a nonzero perturbation can partly represent a coordinate change rather than physical curvature. Substituting this variation into the displayed curvature formula givesEach third derivative has a partner differing only by the order of its commuting partial derivatives. This explicitly proves gauge invariance of the linearized Riemann tensor. The restriction to first order matters: a coordinate change acts on nonzero background curvature as well, but the background here is flat.

Define the trace-reversed metric perturbation and d'Alembert operator byIn four dimensions $\bar{h}=-h$ and $h_{ik}={\bar{h}}_{ik}-\frac{1}{2}{\eta }_{ik}\bar{h}$. Contract the curvature formula using $R_{ik}=R^m{}_{imk}$:up to $O({ϵ}^2)$. Consequently the Einstein tensor isChoose Lorenz gauge in linearized gravity, ${\partial }^k{\bar{h}}_{ik}=0$. It can be imposed locally by solving $\square {\xi }_i={\partial }^k{\bar{h}}_{ik}$, since the gauge transformation changes this divergence by $-\square {\xi }_i$. The remaining coordinate freedom satisfies $\square {\xi }_i=0$. In this gauge,This is the requested Linearized Einstein equations in the stated sign convention. Their divergence requires ${\partial }^k{T}_{ik}=0$ to this order.

For the photon beam, introduce $u=t-x$ and the constant null covector ${\ell }_i=(1,-1,0,0)$, so ${\ell }^i=(1,1,0,0)$. The stress-energy tensor isIt is trace-free and conserved: ${\ell }^i{\partial }_i\rho (u)=0$. A beam-adapted particular solution, with no independently added homogeneous gravitational field, isThe equality follows because its trace is zero, and the Lorenz condition holds because ${\ell }^k{\partial }_{k}H=0$. The field equation becomes a transverse Poisson equation:For the transversely uniform source in the question, one particular solution is $H=4\pi G\rho (u)(y^2+z^2)$. Thus the sourced components can be chosen asBoundary conditions and homogeneous solutions determine additional freedom; the source does not uniquely specify every component in every gauge. In particular, taking $H$ to depend only on $t-x$ would give $\square H=0$ and cannot solve the equation for nonzero $\rho$. The uniform idealization is not an asymptotically flat finite beam, and its quadratic particular solution is a weak-field description only where $|ϵH|\ll 1$.

Finally consider another photon with tangent proportional to ${\ell }^i$. Its tangent remains null because $h_{ik}{\ell }^i{\ell }^k=0$. Moreover $h_{ik}{\ell }^k=0$, and since $\ell$ is constant, the connection contraction isThe geodesic equation is therefore satisfied by a straight affinely parametrized ray along the beam. Hence co-propagating photons experience no deflection from the beam's gravitational field. This is parallel photon beams in linearized gravity; it does not assert absence of forces on counter-propagating photons or effects from unrelated homogeneous gravitational perturbations.