# 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

It is a widely-spread delusion.

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.
Sorry, tex works badly.
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.

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