Do the Maxwell equations yield the proper time of electromagnetic waves? I apologize in advance for possible errors in my premises as I have no precise knowledge of Maxwell equations. Proposals for the correction or even abandon of my question are welcome.
As Maxwell equations are a full description of electromagnetic waves I suppose that they also describe the time a wave takes from the place of its emission to the place of absorption. My question: As Maxwell equations are relativistically invariant (or at least compatible with special relativity), do they also yield the proper time (which is always zero for electromagnetic waves in vacuum)?
 A: The Maxwell equations don't need to "take into account" that the proper time of light-like paths is zero.
The definition of the Minkowski metric as
$$\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2 - \mathrm{d}y^2 - \mathrm{d}z^2$$
Lorentz invariance means that all physical laws are invariant under the isometries of this metric, which are the Lorentz group.
The Maxwell equations are manifestly Lorentz invariant if you simply write them as $\partial_\mu F^{\mu\nu} = 0$. They also give as a result that the speed of electromagnetic waves is $c$. Since they are Lorentz invariant, this is true in all Lorentz-related frames. Now, if you plug any path that represents motion with the speed $c$ into the metric, it is zero. Why? Parametrize the path as $t\mapsto (t,ct,0,0)$ (you can certainly Lorentz transform so that that is possible). Plug it in. Now,
$$\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}(ct)^2 = c^2\mathrm{d}t^2 - c^2\mathrm{d}t^2 = 0$$ 
And so the proper time along that path is $\tau = - \int ds^2 = 0$.
