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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)?

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    $\begingroup$ Proper time is not a property of waves or particles, it is a property of paths in spacetime. That paths that represent motions with the speed of light (which is indeed predicted by the theory of electromagnetism) have zero proper time is a direct consequence of the way the Minkowski metric works. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 6, 2014 at 17:21
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/122735/2451 $\endgroup$
    – Qmechanic ♦
    Commented Jul 6, 2014 at 17:21
  • $\begingroup$ @ACuriousMind: Thank you for this information. I agree that Maxwell equations are taking into account Minkowski spacetime as they are Lorentz-invariant. They describe time and distance of waves, but do they also take into account the fact that the spacetime interval of electromagnetic waves is zero? - Also, a photon is not aging and the age seems to be a property of particles. $\endgroup$
    – Moonraker
    Commented Jul 6, 2014 at 17:36
  • $\begingroup$ Though there is no proper time, I belive that it's possible to define an affine parameter to an photon. $\endgroup$
    – Hydro Guy
    Commented Jul 6, 2014 at 17:51

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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 $\tau = - \int ds^2 = 0$.

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  • $\begingroup$ Thank you. As I understand you do not derive Tau directly from Maxwell equations, but Maxwell equations are referring to the spacetime interval equation? $\endgroup$
    – Moonraker
    Commented Jul 6, 2014 at 17:55
  • $\begingroup$ The proper time $\tau$ is the path length as measured in the Minkowski metric (for a discussion why this is indeed proper time, see here). Maxwell's equations do not refer to anything, it simply turns out that they are Lorentz invariant. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 6, 2014 at 17:59

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