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In the following question the OP asked what is meant in chapter 27-4 of Feynman's Lectures on Physics Vol.II by the ambiguity in the location of electromagnetic field energy: Why is there ambiguity of the field energy?

@AccidentalFourierTransform answers the question with an example of how u and S may be redefined and how the location of the energy is ambiguous. It is also mentioned that u and S are contained in $T_{\mu \nu}$ which leads me to believe that their form somehow appears in or is derived from the matter Lagrangian.

My question is: When u and S are redefined, how does that change the E&M Lagrangian? Said differently, what changes in the E&M Lagrangian leave the equations of motion the same (in the absence of gravity) while redefining u and S?

My initial guess was that Feynman was talking about some kind of gauge transformation. However, as Feynman mentioned, the redefinitions would have measurable effects in a full theory of gravity. Therefore, I believe these redefinitions are not gauge transformations.

Thanks in advance for your help!

speckofdust
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    These ambiguities can occur when you add boundary terms to the action, though I'm sure there are other ways. (This was basically the first question I ever asked on this site, here.) – knzhou Jan 20 '18 at 15:40

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Redefining $u$ and $S$ does not even have to change the Lagrangian. One way of seeing this is by considering $T_{\mu \nu}$. If you don't demand it to be symmetric (which is an additional condition that you may need for full covariant consistency) you are free to redefine it by adding a term like $\partial_\sigma f^{\sigma \mu \nu}$ such that $f^{\sigma \mu \nu} = - f^{\mu \sigma \nu} $. In the case of electromagnetism, you can do this explicitly, by taking, for instance, $f^{\sigma \mu \nu} = F^{\sigma \mu} A^{\nu}$. You can check the details in section 12.10 of Jackson's book, Classical Electrodynamics. There you see that a term as this makes the difference between the symmetric stress energy tensor (which reproduces $u$ and $S$ as usual) and the canonical stress energy tensor, which reproduces these quantities up to a four divergence. The equations of motion will also stay the same.

secavara
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  • It is not immediately clear to me that adding or subtracting such terms is simply a redefinition which leaves the Lagrangian unchanged. I can see that subtracting the term from the canonical stress energy tensor is a procedure to get to the symmetric stress energy tensor. However, I would think that the Lagrangian and canonical momentum that give rise to one stress energy tensor versus the other, will differ. Perhaps as @knzhou mentions above, the two Lagrangians will differ by a boundary term? – speckofdust Jan 22 '18 at 18:12
  • Both the canonical and the symmetric stress energy tensors have very specific definitions in terms of the (same) Lagrangian. In the book, it seems like you have to compute first the canonical, then add a term to get the other. But later in GR you'll see that the symmetric one has a very specific definition in terms of the action. In fact, the symmetric one has an abundant number of advantages (see this). If you accept these definitions, then ultimately there is no need for surface terms. – secavara Jan 22 '18 at 18:43
  • I see your point that each stress energy can be defined from the same Lagrangian. What I am still confused about, is Feynman's claims that different choices for u and S will affect the behavior of a system when gravity is included. He mentions you could in principle do an experiment to determine which is the correct choice. I would expect the differences to somehow show up in the equations of motion, ie originate from differences in the Lagrangian. So I think I am looking to understand what is this class of matter Lagrangians that are all valid until we are able to perform this experiment? – speckofdust Jan 22 '18 at 19:22
  • Frankly I'd not pretend to say that I can fully read Feynman's intentions in his statement but my believe is that his concern is more directly related to absolute amounts of energy. As he mentions, these would affect the EOMs of the system of gravity+electromagnetic field. If you only keep the dynamics of classical electromagnetism, it is very unlikely you'll find measurable effects, and you will mostly care about relative amounts of energy. I see where you are going but I'm afraid you'll have to go beyond. Taking this into account, I think you might be able to ask a new question ... – secavara Jan 22 '18 at 19:47
  • ... , if I you don't find it out there already, regarding measurable effects (relativistic or even quantum) pointing to deviations in the way we compute/measure the absolute energy (if it can be defined anyway) of, well, anything, or in particular, the energy of the electromagnetic field. I think this is a fairly valid one in which people from different backgrounds could contribute with distinct points of view. – secavara Jan 22 '18 at 19:52