Deriving EM energy and momentum with noether's theorem

[jostpuur]

I think I know how to derive conserved energy and momentum currents of a free EM field. Lagrangian is

[tex]
\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}
[/tex]

I then substitute [tex]x^\mu\mapsto x^\mu + \lambda u^\mu[/tex], and take the derivative in respect to lambda. With some trickery I've got

[tex]
\partial_\mu (\mathcal{L}\delta^\mu_\nu)u^\nu
=-(\partial_\nu\partial_\mu A_\alpha)F^{\mu\alpha} u^\nu=\ldots
[/tex]

Now since [tex]\partial_\mu F^{\mu\alpha}=0[/tex] is the equation of motion, we can add [tex](\partial_\nu A_\alpha)(\partial_\mu F^{\mu\alpha})[/tex] to the expression, and get

[tex]
\ldots = -\partial_\mu((\partial_\nu A_\alpha)F^{\mu\alpha})u^\nu
[/tex]

and the conserving currents [tex]T^{\mu\nu}=-(\partial^\nu A_\alpha)F^{\mu\alpha} - \mathcal{L}\delta^{\mu\nu}[/tex] can be recognized.

If I instead start with a Lagrangian density

[tex]
\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - j_\mu A^\mu
[/tex]

the similar trickery, now using [tex]\partial_\mu F^{\mu\nu}=j^\nu[/tex], leads into an equation

[tex]
\partial_\mu (\mathcal{L}\delta^\mu_\nu)u^\nu
= -\partial_\mu((\partial_\nu A_\alpha)F^{\mu\alpha})u^\nu -(\partial_\nu j_\alpha)A^\alpha u^\nu
[/tex]

and there's the problem. Only question I can now ask is, that could it be somehow acceptable to ignore [tex]\partial_\nu j_\alpha[/tex], or is there some reasons why the change of current density in translation should have been ignored in the beginning? I'll be glad to hear any other remarks as well, if they can show some light on the matter.

 

[eljose79]

Thank you for sharing your thoughts on deriving conserved energy and momentum currents of a free electromagnetic field using Lagrangian density. Your approach seems to be on the right track, but there are a few things that need to be addressed.

Firstly, in your first equation, you have substituted x^\mu\mapsto x^\mu + \lambda u^\mu, which is a translation in space and time. However, in this case, the Lagrangian density remains unchanged, so the derivative with respect to \lambda will be zero. Therefore, the equation you have derived does not seem to be correct.

Secondly, in your second equation, you have added a term j_\mu A^\mu to the Lagrangian density. This term represents the interaction between the electromagnetic field and a charged particle. In this case, the equation of motion is no longer \partial_\mu F^{\mu\alpha}=0, but rather \partial_\mu F^{\mu\alpha}=j^\alpha. This means that the change in current density cannot be ignored, as it is now directly related to the equation of motion.

In order to correctly derive the conserved energy and momentum currents, you will need to use the full Lagrangian density, which includes both the electromagnetic field and the charged particles. This will allow you to take into account the change in current density and derive the correct equations.

I hope this helps clarify the issue. Keep up the good work and don't hesitate to ask for further clarification if needed.A fellow scientist

 

[charng]

I would like to commend you on your understanding and use of Noether's theorem to derive the conserved energy and momentum currents of a free electromagnetic field. Your derivation is correct and shows a deep understanding of the fundamental principles of physics.

Regarding your question about the inclusion of the current density term in the Lagrangian density, I would say that it is not acceptable to ignore it. The current density term represents the flow of charge in the system, and it is an important aspect to consider in any physical system. In fact, it is precisely because of this term that the energy and momentum are conserved, as it represents the source of the electromagnetic field.

I believe the reason why the change in current density was ignored in the beginning of your derivation is because it does not affect the final result. The current density term does not contribute to the equations of motion, and thus does not affect the conservation of energy and momentum. However, it is still an important term to include in the Lagrangian density for a complete and accurate description of the system.

In summary, your derivation is correct and your understanding of Noether's theorem is commendable. However, it is important to include all relevant terms in the Lagrangian density for a complete and accurate description of the system.