I have a question regarding the physical significance of the canonical energy momentum tensor $T_\nu ^\mu$ in the context of classical field theory. It is defined as

$T_\nu ^\mu = \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \Phi^I)} \partial_\nu \Phi^I - \delta_\nu ^\mu \mathcal{L} $, where $\Phi^I$ is the set of all relevant scalar, vector, or tensor fields in the Lagrangian, and $I$ is the corresponding indices. For example, if we consider the Lagrangian for the free gauge field in electrodynamics $\mathcal{L}_A = - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}$, where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, we have that $\Phi^I = A_\mu$. Accordingly, the (covariant) canonical energy momentum tensor is

$T_{\mu \nu} = \frac{1}{4}\eta_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta} - F_{\mu \alpha} \partial _ \nu A^ \alpha$.

It satisfies that $\partial _\mu T_\nu ^\mu = 0$ provided that the system, which it describes, is independent of translations in spacetime. This is a consequence of Noethers theorem.

My problem: On the one hand, it is used to define the canonical 4-momentum by

$P_\nu = \int T_{\nu}^0 d^3 x $,

i.e. it is used to define some physical observables of the system. Also, the canonical Hamiltonian $\mathcal{H}$ density is defined by $\mathcal{H}=T_{0}^0$

On the other hand, $T_\nu ^\mu$ is not in general gauge invariant - but physical observables must be gauge invariant. If the gauge field is time independent, i.e. $\partial_0 A^\alpha = 0$ for each $\alpha$, then $T_0 ^0$ (and thus also the energy) can be gauge independent, as is the case for $T_0 ^0$ for free electrodynamics defined above.

However, in general for time-dependent gauge field $T_\nu ^\mu$ is not gauge-invariant, which must imply that the canonical Hamiltonian density is not in general gauge independent and thus not observable.

My question: Is it possible to show that the 4-momentum is always gauge invariant and thus observable despite the fact that $T_\nu ^\mu$ is not?

If this is not possible, what even is the significance (and relevance) of $T_\nu ^\mu$ and the canonical 4-momentum?


First §: perhaps mention that you are talking about electromagnetism and show or mention which lagrangian you assume. Belinfante, I assume. Otherwise, well put.

Second §: the Noether tensor derived from the Belinfante tensor is not gauge invariant and asymmetric. Could you explain what you mean by "in general"? I don't know any Noether energy-momentum density that is gauge invariant. There is none.

Now your question. In order for the four momentum to be gauge invariant, the underlying density must be. The Noether expression is not. The Belinfante expression is needed. However, the latter is a physically not completely justifiable, ad hoc modification of the former. If you by all means want a gauge invariant theory you have to live with this. If not, take a look at my paper at https://arxiv.org/abs/physics/0106078.

| cite | improve this answer | |
  • $\begingroup$ I have tried to address your comments. I want to add that I am not talking about the Belifante tensor (which is in fact symmetric and gauge invariant). I am talking about the Noether energy momentum tensor. I have provided an example of a gauge invariant Noether energy momentum tensor. Good point! The gauge invariant Belifante tensor is defined from the Noether tensor such that the energies match for trivial topology - and this only makes sense if the energy defined by the Noether tensor is in fact observable. That is what is bugging me here, because the Belifante tensor is widely used. $\endgroup$ – Simon G. Aug 9 '19 at 17:54
  • $\begingroup$ Correction: "I have provided an example of a gauge invariant Noether energy momentum density". $\endgroup$ – Simon G. Aug 9 '19 at 18:14
  • $\begingroup$ The tensor you give is not gauge invariant and is asymmetric. $\endgroup$ – my2cts Aug 9 '19 at 18:29
  • $\begingroup$ You asked for a "Nother energy momentum density", which I took to mean the Hamiltonian density. In static free electrodynamics, it is gauge invariant, which allows for the energy to be as well. If you meant the Noether tensor, you might be right that it is never gauge invariant except for fields constant in time and space. Can you answer why we even consider canonical momentum as defined by the Noether tensor if it is not an observable quantity? $\endgroup$ – Simon G. Aug 9 '19 at 18:39
  • $\begingroup$ The Noether theorem is sound. Once you pick a hamiltonian it gives you the conserved currents straight away. That is why Noether currents are of essential importance. In the case of electromagnetism physicists are not happy with the results and modify them. They consider this permissible because the physical consequences are very small. $\endgroup$ – my2cts Aug 9 '19 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.