Is Relativistic Action for a beam of light = zero?

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Discussion Overview

The discussion centers around the concept of the relativistic definition of action, particularly in relation to a beam of light. Participants explore the implications of this definition and its relationship to the Lagrangian density derived from Maxwell's equations, questioning the significance of the numeric value of action in this context.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions whether the action for a beam of light is always zero under the relativistic definition of action.
  • Another participant challenges the clarity of the term "RELATIVISTIC definition of Action" and suggests looking into textbooks for further understanding.
  • A different participant raises a philosophical point about the numeric value of action and its meaning.
  • One participant states that the Lagrangian density for the electromagnetic (EM) field can be derived from Maxwell's equations and notes that for an EM plane wave, the Lagrangian density could be zero.
  • A follow-up question is posed regarding the impact of adding a constant to the Lagrangian on the equations of motion, implying that it may not matter.
  • Another participant clarifies that Maxwell's equations can be viewed as the Euler-Lagrange equations for the EM field Lagrangian and emphasizes the Lorentz invariance of these equations.

Areas of Agreement / Disagreement

Participants express differing views on the significance of the action's numeric value and whether it holds any meaning, indicating that multiple competing perspectives remain without consensus.

Contextual Notes

There are unresolved questions regarding the assumptions behind the relativistic definition of action and the implications of the Lagrangian density being zero for an EM plane wave.

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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