Problem 7.4.1
problem id: per25
Obtain the equation of motion for photon in metrics ($ds^2=a^2(\eta)[(1+2\Phi)d\eta^2-(1-2\Phi)\delta_{ij}dx^idx^j]$) in linear approximation in $\Phi$.
Equation of geodesics describing the propagation of radiation in space with arbitrary curvature can be represented as: $$ \frac{dx^\alpha } {d\lambda } = p^\alpha,\; \frac{dp_\alpha } {d\lambda} = \frac{1} {2}\frac{\partial g_{\gamma \delta }} {\partial x^\alpha }p^\gamma p^\delta, $$ where $\lambda$ is an arbitrary affine parameter, which is taken along geodesic. Since photon has zero mass, the first integral of this equation is: $$ p^\alpha p_\alpha = g^{\mu \nu }p_\mu p_\nu = 0. $$ Using this relation, zero component can eliminated from the equations of geodesics: $$ p^0 = \frac{1} {a^2}\left( \sum\limits_{i = 1}^3 p_i^2 \right)^{1/2} = \frac{p}{a^2} $$ and $$ p_0 = \left( 1 + 2\Phi\right)p. $$ Hence, $$ \frac{dx^i} {d\eta } = \frac{p^i} {p^0} = \frac{ - \frac{1} {a^2}\left( 1 + 2\Phi \right){p_i}} {p^0} = r^i\left( 1 + 2\Phi\right), $$ where $r^i = - \frac{p_i}{p^0}$. Expressing $p^i$ and $p^0$ through $p$ and substituting metrics into the second equation while keeping only linear terms in $\Phi$ one obtains: $$\frac{dp_\alpha } {d\eta} = \frac{1} {2}\frac{\partial g_{\gamma \delta }} {\partial x^\alpha }\frac{p^\gamma p^\delta } {p^0} = 2p\frac{\partial \Phi } {\partial x^\alpha }. $$