I Is Relativistic Action for a beam of light = zero?

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

Thanks in advance.
 
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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?
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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