Is Relativistic Action for a beam of light = zero?

In summary, the relativistic action for a beam of light is considered to be zero due to the fact that light travels along null geodesics in spacetime, where the proper time experienced by a photon is zero. This implies that the standard formulation of action, which typically involves integrating the Lagrangian over proper time, results in a value of zero for massless particles like photons. Thus, while the action may not provide useful physical insights in the context of light, it aligns with the principles of relativity and the nature of light's motion.
  • #1
LarryS
Gold Member
354
33
TL;DR Summary
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.
 
Physics news on Phys.org
  • #2
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?
 
  • #3
Let's take a step back. Sans shouting. What makes you think the numeric value of the action has any meaning whatsoever?
 
  • #4
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?
 
  • #5
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?"
 
  • Like
Likes LarryS
  • #6
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?
 
  • Like
Likes LarryS

FAQ: Is Relativistic Action for a beam of light = zero?

What is the relativistic action in the context of physics?

The relativistic action is a quantity used in the principle of least action, which is a fundamental concept in physics. It is an integral over time of the Lagrangian, which encapsulates the dynamics of a system. For a particle, the action is typically the integral of the difference between kinetic and potential energy over time.

What does it mean for the action of a beam of light to be zero?

For the action of a beam of light to be zero means that the integral of the Lagrangian over the path of the light is zero. This usually implies that the light follows a path that makes the action stationary (a minimum or a saddle point), which in the case of light in a vacuum, corresponds to traveling along null geodesics in spacetime.

How is the action for a beam of light calculated in relativity?

In relativity, the action for light is calculated using the principle of least action applied to the electromagnetic field. For a beam of light, which follows a null path (where the spacetime interval is zero), the action can be expressed in terms of the electromagnetic field tensor. However, for a single photon or light beam, the action is typically zero because the path is null.

Why is the action for a beam of light considered to be zero in certain cases?

The action for a beam of light is considered to be zero in certain cases because light travels along null geodesics in spacetime. In these paths, the spacetime interval is zero, leading to the integral of the Lagrangian over the path also being zero. This is a result of the fact that light, in a vacuum, follows the path that extremizes the action, which in this case is zero.

What are the implications of a zero action for a beam of light in physics?

The implications of a zero action for a beam of light are significant in theoretical physics. It means that light travels along paths that are invariant under Lorentz transformations, which is a cornerstone of the theory of relativity. This property is crucial for understanding the behavior of light in various physical contexts, including gravitational lensing and the propagation of light in different media.

Similar threads

A Another action for the relativistic particle
Replies
12
Views
3K
B Constant Speed of Light: Low Speed Explained
Replies
18
Views
2K
A Action for a relativistic free particle
Replies
9
Views
3K
I Doubts about the relativistic description of electrical interactions
Replies
2
Views
860
B Sagnac effect for matter beams
2
Replies
40
Views
3K
I Action in Relativity: Non-Relativistic vs Relativistic Particles
Replies
8
Views
2K
I Ricci tensor from this action
Replies
4
Views
678
B Light Beams Attraction: e=mc^2 & Asymmetry
Replies
5
Views
1K
B Relativistic Velocity Addition: Is c Ever Exceeded?
Replies
16
Views
2K
I Can gravitational lensing detect Very small deviations in light propagation?
Replies
8
Views
845

Hot Threads

Recent Insights

Back
Top