I am having trouble understanding a calculation done by V. Mukhanov in his book "Physical Foundations of Cosmology". In the beginning of the chapter 9, the following arguments are stated:
- The state of a single photon with a given polarization at conformal time $\eta$ is completely characterized by its position in the space $x^{i}(\eta)$ and its 3-momentum $p_{i}(\eta)$, where $i=1,2,3$ is the spacial index.
- The frequency of a photon with 4-momentum $p_{\alpha}$ measured by an observer having the 4-velocity $u^{\alpha}$ is equal to the scalar product of these vectors: $\omega=p_{\alpha}u^{\alpha}$
- Let us consider two observers $O$ and $\tilde{O}$, who are at rest with respect to two different frames related by the coordinate transformation $\tilde{x}^{\alpha}=x^{\alpha}+\xi^{\alpha}$. In the rest frame of each observer, the zeroth component of the 4-velocity can be expressed through the metric by using the relation $g_{\alpha\beta}u^{\alpha}u^{\beta}=g_{00}(u^{0})^{2}=1$.
I understand that, through these arguments, one can obtain the following expression for the frequency of a same photon \begin{equation*} \omega=p_{\alpha}u^{\alpha}=\frac{p_{0}}{\sqrt{g_{00}}}\quad\text{and}\quad\tilde{\omega}=\frac{\tilde{p}_{0}}{\sqrt{\tilde{g}_{00}}} \end{equation*} But, what I can not figure out is how to get to the following expression: \begin{equation*} \tilde{\omega}=\omega\left(1+\frac{\partial\xi^{i}}{\partial\eta}l^{i} \right) \end{equation*} where \begin{equation*} l^{i}\equiv -\frac{p_{i}}{p}\quad\text{and}\quad p^{2}=\sum (p_{i})^{2} \end{equation*} How do you get to that expression?
