More on frames of reference and coordinates in GR I have read other questions concerning this subject, and by now I believe that in order to solve a gravitational problem in GR, one has to basically abandon the notion of frames of reference. However, I cannot get my intuition to align with this, and I would like someone could explain me the logic and the intuition behind (part of) an answer to the following problem:

Two free falling masses are to be employed to measure the effect of a gravitational wave by monitoring the time delay in successive laser pulses going from one mass to the other. What is the maximal time delay?

Of course, the question provides a diagonal metric describing this wave pertubatively. Its exact form, though, is irrelevant here. The part of the solution that bothers me is the following:

We want to integrate the trajectory of the photon, and the first step is to impose the "infinitesimal" spacetime interval to be lightlike. Now, given that we ask that these masses lie on an $xy$ plane perperdicular to the propagation direction $z$ of the wave, we can impose that $dz^2 = 0 = z$.

This is what bothers me. Intuintively I can see why the photon should travel along a geodesic with constant $z$, and that this could be easily attributed the coordinate value $0$, but how do we formally justify this? Is the statement "$dz=0$" a rigorous one, or rather a shorthand notation meaning that the $z$ component along the geodesic is constant? Can we justify this fact a priori? How? If the geodesic deviation equation in this case tells us that the separation vector between the geodesics followed by each mass has zero $z$ component, can we employ this to justify a priori that the path followed by the light pulse has constant $z$?
 A: You can choose $\frac{dz}{d\tau}=0$ as an initial condition to the geodesic equation (where $\tau$ is the proper time of the particle). Based on the metric you've described, you should find that the geodesic equation for the $z$ coordinate will reduce to $\frac{d^2 z}{d\tau^2}=0$, so if the particle started off with zero velocity in the $z$ direction, that will remain true throughout the motion.
A: A metric for a plus-polarised, gravitational wave, travelling along the z-axis, is diagonal with elements of $g_{00}=-1$, $g_{11} = 1+h$, $g_{22}=1-h$ and $g_{33}=1$ in Cartesian coordinates, where $h$ is a small, time-dependent perturbation.
The Lagrangian can be written
$$L = \frac{1}{2} g_{\mu\nu} \frac{\textrm{d} x^{\mu}}{\textrm{d} \lambda}\frac{\textrm{d} x^{\nu}}{\textrm{d} \lambda},$$
$$L =\frac{1}{2}\left(-\dot{t}^2 +(1+h)\dot{x}^2 + (1-h)\dot{y}^2 + \dot{z}^2\right)\ ,$$
where the dot represents differentiation with respect to $\lambda$, any affine parameter along the path.
The Euler-Lagrange equation for $z$ then gives
$$\ddot{z} = 0\ .$$
Thus if you start a photon off in the xy plane with $\dot{z}=0$, it stays in the $xy$ plane and so $dz=0$.
