Is Relativistic Action for a beam of light = zero?

Under the RELATIVISTIC definition of Action, is the Action for a beam of light always zero?

Thanks in advance.

LarryS said:

Under the RELATIVISTIC definition of Action, is the Action for a beam of light always zero?

What do you think the "RELATIVISTIC definition of Action" (not sure why you felt it necessary to shout) is? Have you tried looking in any textbooks or other references?

Let's take a step back. Sans shouting. What makes you think the numeric value of the action has any meaning whatsoever?

The Lagrangian Density for the EM field can be derived from Maxwell's Equations (via the EM Field Tensor). Also, from Maxwell's Equations, the electric and magnetic energy densities of an EM plane wave are equal. So, if you focus only on the EM plane wave, the Lagrangian Density for that would be identically zero. Or, am I not seeing something?

How would the equations of motion change if you added a constant to the Lagrangian?

When you answer "not at all", the next question is "then how can the numeric value matter?"

LarryS said:

The Lagrangian Density for the EM field can be derived from Maxwell's Equations (via the EM Field Tensor).

Actually, it's the other way around: Maxwell's Equations are the Euler-Lagrange equations for the EM field Lagrangian.

LarryS said:

from Maxwell's Equations, the electric and magnetic energy densities of an EM plane wave are equal

More precisely, a source-free EM plane wave.

LarryS said:

if you focus only on the EM plane wave, the Lagrangian Density for that would be identically zero.

Unless, as @Vanadium 50 says, you add an arbitrary constant, which has no effect on the equations of motion.

However, if we leave that aside, what, exactly, remains unanswered from your original question? Do you realize that Maxwell's Equations, and the Lagrangian they are derived from, are relativistic? That is, they are Lorentz invariant?