Origin of the Maxwell energy-momentum tensor?

[Khrapko]

Electrodynamics force is f_i=F_{ik}j^k=F_{ik}\partial_j F^{jk}. I claim that the only way to obtain the Maxwell energy-momentum tensor T_i^j=-F_{ik}F^{jk}+\delta_i^jF_{kl}F^{kl}/4 is to write the force as a divergence: f_i=-\partial_jT_i^j.

[Petr Mugver]

The energy-momentum tensor of any field must have zero divergence.

[Khrapko]

The energy-momentum tensor of any field must have zero divergence.
It is a widely-spread delusion.

[Tomsk]

It can also be derived from Noether's theorem, it is the conserved current of translations x \rightarrow x+a.

[Khrapko]

It can also be derived from Noether's theorem, it is the conserved current of translations x \rightarrow x+a.
It is a widely-spread delusion as well. The Noether's theorem gives various energy momentum tensors, but not the Maxwell tensor.
Sorry, tex works badly.
Canonical Lagrangian L_1L_1=-F_{ij}F^{ij}/4 gives canonical tensor
T_1{}T_1{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4
Dirac’s Lagrangian L_2 L_2=-F_{ij}F^{ij}/4-(\partial_iA^i)^2/2 gives T_2{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4-\partial_iA^j\partial_kA^k+\delta_i^j(\partial_kA^ k)^2/2,
Vector Lagrangian L_3=-\partial_iA^j\partial^iA_j/2 gives T_3{}_i^j=-\partial_iA_k\partial^jA^k+\delta_i^j\partial_kA_l \partial^kA^l,
Soper’s Lagrangian [1] L_4=-F_{ij}F^{ij}/4-A_ij^i gives T_4{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4+\delta_i^j A_kj^k. But Soper was mistaken: he obtain a false tensor T_f{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4+A_ij^j [2].
[1] D. E. Soper, Classical Field Theory (N.Y.: John Wiley, 1976).
[2] R.I. Khrapko, Professor Soper's mistake http://khrapkori.wmsite.ru/ftpgetfile.php?id=43&module=files

[Petr Mugver]

The fact that the energy momentum tensor you wrote can be obtained by Noether's theorem is NOT a "delusion": take the formalism of general relativity, take the usual electromagnetic field lagrangian, vary not only the A fields but also the metric, and finally impose the flat space-time metric... what will you get?

[Khrapko]

Emmy Noether did not intend to use curvilinear coordinates. And why we must prefer T_i^j=-F_{ik}F^{jk}+\delta_i^jF_{kl}F^{kl}/4 to T_1{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4, or others?
By the way, variation of coordinates does not give spin tensor!

[Petr Mugver]

It's just a conservation law. The symmetric tensor is preferred because it is...symmetric, I guess. Variation of coordinates (and fields) by Lorentz transformations DOES give you the spin tensor. I don't know what Noether intended or not, I havent't read her biography, but using her theorem you get a lot of conservation laws (all, as far as I know) of a lagrangian field theory, regardless if the lagrangian describes a drum, a water wave, a Higgs boson or a general relativistic matter system.

[Tomsk]

You're right that there's a problem with the normal derivation from Noether's theorem, the tensor

T_1{}_i^j=-\partial_iA_kF^{jk}+\delta_i^jF_{kl}F^{kl}/4

is not gauge invariant. But this can be solved with a couple of tricks. One is to perform a gauge transformation when you vary A, instead of

A_\mu (x) \rightarrow A_\mu (x+a) = A_\mu (x) + a^\nu \partial_\nu A_\mu

you can subtract a gauge term like this

A_\mu (x) \rightarrow A_\mu (x) + a^\nu \partial_\nu A_\mu - \partial_\mu (a^\nu A_\nu) = A_\mu (x) + a^\nu F_{\nu\mu}

because a is constant, and this gives you the standard gauge invariant stress energy tensor.

[Khrapko]

But this can be solved with a couple of tricks.
Sorry, I do not understand your tricks. They gives T_5{}_i^j=-\partial_iA_kF^{jk}-a^l\partial_iF_{lk} F^{jk}+\delta_i^jF_{kl}F^{kl}/4 (Sorry, tex works badly)
My thought is the Lagrange formalism with the Noether's theorem cannot give the electromagnetic energy-momentum tensor. And the deriving of this tensor, which is described in all textbooks, is a mistake. See [1]
[1] R.I.Khrapko, “Mechanical stresses produced by a light beam,” J. Modern Optics, 55, 1487-1500 (2008) http://khrapkori.wmsite.ru/ftpgetfile.php?module=files&id=9

[Tomsk]

Sorry but I don't understand how you got that expression. I may have messed up, I don't know. I got the standard Maxwell stress energy tensor from the general equation for a Noether current.

If you have some fields \phi_a (x) and a lagrangian \mathcal{L}(\phi_a (x),\partial_\mu \phi_a (x)), and if the action is invariant under some symmetry, you can transform the fields like this:
\phi_a (x) \rightarrow \phi_a (x) + \epsilon^\alpha \Phi_{a\alpha}(x)
where epsilon is an infinitesimal parameter and Phi encodes the transformation in terms of phi. The lagrangian transforms like
\mathcal{L} \rightarrow \mathcal{L} + \epsilon^\alpha \partial_\mu \Lambda^\mu_\alpha
So that the action is invariant. Then the Noether current is
j^\mu_\alpha = \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_a)}\Phi_{a\alpha} - \Lambda^\mu_\alpha
To get the Maxwell stress energy tensor, set \phi_a = A_\mu so that a is a spacetime index, and set \epsilon^\alpha = a^\nu where a is a constant infinitesimal vector so that alpha is also a spacetime index. Then to get the standard Maxwell stress energy tensor we require \Phi_{a\alpha} = F_{\mu\nu} which I got by Taylor expanding A as normal, A_\mu (x) \rightarrow A_\mu (x) + a^\nu \partial_\nu A_\mu then subtracting a gauge term, which doesn't affect the lagrangian, so it won't affect Lambda either. That gives a^\nu \Phi_{\mu\nu} = a^\nu \partial_\nu A_\mu - a^\nu \partial_\mu A_\nu = a^\nu F_{\mu\nu} I think that works but I'm not 100%, if you found a mistake let me know...

[Khrapko]

Usually, when deriving the Noether current (energy-momentum or angular momentum tensor), they use an infinitesimal coordinate transformation rather than transformation of field (the field satisfies Euler-Lagrange equations, i.e. field equations). Please present your calculation in details as an attachment.