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The Canonical energy momentum tensor is given by $$T_{\mu\nu} = \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_s)} \partial_\nu \phi_s - g_{\mu\nu} {\cal L} $$ A priori, there is no reason to believe that the EM tensor above is symmetric. To symmetrize it we do the following trick.

To any EM tensor we can add the following term without changing its divergence and the conserved charges. $${\tilde T}_{\mu\nu} = T_{\mu\nu} + \partial^\beta \chi_{\beta\mu\nu} $$ where $\chi_{\beta\mu\nu} = - \chi_{\mu\beta\nu}$. The antisymmetry of $\chi$ in its $\mu\nu$ indices keeps the divergence of this new EM tensor zero. It also keeps the corresponding conserved charges the same.

Now even though $T_{\mu\nu}$ is not a symmetric tensor, it is possible to choose $\chi_{\beta\mu\nu}$ in such a way so as to make ${\tilde T}_{\mu\nu}$ symmetric. It can be shown that choosing

$$\chi_{\lambda\mu\nu} = - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right]$$ makes the new EM tensor symmetric. Here $(I_{\mu\nu})_{rs}$ is the representation of the Lorentz Algebra under which the fields $\phi_s$ transform.

Here's my question - Is it possible to obtain the symmetric EM tensor directly from variational principles by adding a total derivative term to the Lagrangian. In other words, by shifting ${\cal L} \to {\cal L} + \partial_\mu X^\mu$, and choosing $X^\mu$ appropriately, can be exactly get the shift in the EM tensor required, in order to make the EM tensor symmetric?

What I've done so far - It is possible to show that under a shift in the Lagrangian by a total derivative, one shifts the EM tensor by $T_{\mu\nu} \to T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu}$ where

$$X_{\lambda\mu\nu} = \frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} $$

What I wish to do next - I now have a differential equation that I wish to solve.

$$\frac{1}{2} \frac{\delta X_\lambda}{\delta (\partial^\mu \phi_r)} \partial_\nu \phi_r - \frac{1}{2} \frac{ \delta X_\mu }{\delta (\partial_\lambda \phi_r)} \partial_\nu \phi_r + X_\mu g_{\lambda\nu} - X_\lambda g_{\mu\nu} = - \frac{i}{2}\left[ \frac{\delta {\cal L}}{\delta (\partial^\mu \phi_r) } (I_{\nu\lambda})_{rs} \phi_s + \frac{\delta {\cal L}}{\delta (\partial^\lambda \phi_r) } (I_{\mu\nu})_{rs} \phi_s + \frac{\delta {\cal L} }{\delta (\partial^\nu \phi_r) } (I_{\mu\lambda})_{rs} \phi_s \right] $$

Any ideas on how to solve this?

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Related: physics.stackexchange.com/q/27048/2451 –  Qmechanic Sep 19 '12 at 14:57
    
Symmetrizability is equivalent with Lorentz invariance. Thus you need to assume that in your arguments. –  Arnold Neumaier Sep 19 '12 at 17:12
    
Can you explain that a bit more? I don't undestand what you're trying to say. Thanks! –  Prahar Sep 23 '12 at 17:00
    
Without the assumption of Lorentz invariance of the action, there is no symmetric e/m tensor, and the standard recipe fails. Lorentz invariance gives you additional properties that you must exploit in your derivation; otherwise you will not be able to arrive at the conclusion (since it might not hold). - If you reply to a comment, you should mention the name, like in @Prahar, so that the original commenter is informed. I noticed your comment only by chance (and hence very late). –  Arnold Neumaier Nov 1 '12 at 17:58

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