Is Relativistic Action for a beam of light = zero?

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SUMMARY

The discussion centers on the relativistic definition of Action and its application to a beam of light. It is established that the Lagrangian Density for an electromagnetic (EM) plane wave, derived from Maxwell's Equations, is identically zero when considering a source-free scenario. The conversation highlights that while the numeric value of Action may seem irrelevant, the equations of motion remain unchanged even if an arbitrary constant is added to the Lagrangian. The relativity of Maxwell's Equations and their Lorentz invariance are also emphasized as critical aspects of this topic.

PREREQUISITES
  • Understanding of Maxwell's Equations
  • Familiarity with Lagrangian Mechanics
  • Knowledge of Electromagnetic Field Theory
  • Concept of Lorentz Invariance
NEXT STEPS
  • Study the derivation of the Lagrangian Density from Maxwell's Equations
  • Explore the implications of Lorentz invariance in classical physics
  • Investigate the role of arbitrary constants in Lagrangian formulations
  • Examine the relationship between Action and equations of motion in relativistic systems
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Physicists, students of theoretical physics, and anyone interested in the foundations of electromagnetism and relativistic mechanics will benefit from this discussion.

LarryS
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TL;DR
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?"
 
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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?
 
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