In a relativistic context we do analyze the propagation characteristics of waves using the four-wave vector $k^\mu = (\frac{\omega}{c}, \vec{k})$ instead of considering $\vec{k}$ and $\omega$ separately because it's invariant under Lorentz transformations. Hence, it allows us to understand better wave-particle interactions when i.e. electrons move at relativistic speed in various materials.

It would be more accurate to focus on the frequency and propagation vector of the photon relative to the electron’s frame when you mentioned the photon's four-momentum in the lab $S$.

To determine how the non-fixed axis of rotation affects the kinetic energy of the cube, you need to consider the contributions from both rotational motions. The rotational kinetic energy for an object rotating with angular velocity $\omega$ and moment of inertia $I$ is given by $K = \frac{1}{2}I\omega^2$. Unfortunately, I don't have enough characters left here to write the whole explanation, but I hope my comments and my answer were useful :)

To say that it would apply to "whatever shape you might need" would be too generic, but yeah, the general approach works with various shapes (i.e.: spheres, cylinders, rectangular prisms..). However, it's obvious that the formula will change if you change the kind of shape you're working with, so keep that in mind.

You're welcome, and yes, you're correct.

You correctly recognized that $\delta V_n = 0$, essentially saying that the basis vectors $V_n$ don't change under the transformation, which is reasonable if you're dealing with a coordinate transformation. However, you made an error in applying the product rule in the first term, here's the correct version: $$ \delta g_{\mu n} = \lambda V^m\partial_m(B_\mu V_n) + \partial_{\mu}(\lambda V^m\delta_{mn})+\lambda\partial_nV^mB_{\mu}\delta_{m\mu} $$

I think that it depends on the specific theory under consideration. In General Relativity, where torsion is assumed to be zero by construction, this term you mentioned vanishes, even though in more general theories of gravity (such as Einstein-Cartan theory or teleparallel gravity, where torsion is non-zero) this term does not necessarily vanish and contributes to the dynamics of the theory

To demonstrate that light deviates from a straight path in an accelerated frame, you can use the equivalence principle, which states that the effects of gravity are locally indistinguishable from the effects of acceleration. This principle allows us to draw parallels between the behavior of light in an accelerated frame and in a gravitational field. If you want a more detailed explanation we could start a chatroom

No, future infinity doesn't have a past end because it represents everything that can happen infinitely far into the future. It's like the endlessly distant horizon on a highway stretching forever, and the past has already happened. It's a separate concept with its own "infinitely far back" point, which we call past infinity. To explain it more easily, imagine a number line stretching infinitely in both directions: future infinity would be at positive infinity ($+\infty$) on that line, while past infinity would be at negative infinity ($-\infty$). They are separate concepts, not connected.