Origin of the Maxwell energy-momentum tensor?

1. Aug 11, 2010

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: $$f_i=-\partial_jT_i^j$$.

2. Aug 11, 2010

Petr Mugver

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

3. Aug 11, 2010

Khrapko

4. Aug 12, 2010

Tomsk

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

5. Aug 12, 2010

Khrapko

It is a widely-spread delusion as well. The Noether's theorem gives various energy momentum tensors, but not the Maxwell tensor.
Canonical Lagrangian L_1$$L_1=-F_{ij}F^{ij}/4$$ 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

Last edited: Aug 12, 2010
6. Aug 12, 2010

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?

7. Aug 12, 2010

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!

8. Aug 12, 2010

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.

9. Aug 12, 2010

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.

10. Aug 29, 2010

Khrapko

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

Last edited: Aug 29, 2010
11. Aug 30, 2010

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...

12. Aug 31, 2010

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.