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An elementary definition of the stress-energy tensor $T_{\mu\nu}$ in terms of the Lagrangian for flat spacetime is $$T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}(\partial^\nu\phi)-\eta^{\mu\nu}\mathcal{L}.$$ For translational symmetries of spacetime both $T^{\mu\nu}$ is conserved: $$\partial_\mu T^{\mu\nu}=0.$$

But $T_{\mu\nu}$ is not always automatically symmetric in the indices $\mu,\nu$. Therefore, one considers a modified stress-energy tensor $\Theta_{\mu\nu}$ defined as $$\Theta^{\mu\nu}:=T^{\mu\nu}+\partial_\kappa A^{\kappa\mu\nu}$$ where $A^{\kappa\mu\nu}$ is an arbitrary tensor antisymmetric in the indices $\kappa,\mu$. This too satifies $$\partial_\mu\Theta^{\mu\nu}=0.$$

  • What is the need of defining a symmetric stress-energy tensor? What happens if I continue to work with $T^{\mu\nu}$ instead of the modified tensor $\Theta^{\mu\nu}$?

  • It is also unclear how one would choose the tensor $A^{\kappa\mu\nu}$ uniquely to symmetrize $T^{\mu\nu}$ and find a unique $\Theta^{\mu\nu}$.

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    $\begingroup$ First, we don't need the symmetrimized tensor be unique, we can choose arbitrary second order tensor whose divergence is zero to symmetrimize energy-stress tensor. Second, the reason we want to symmetrimize energy-stress tensor is due to the consideration of conservation law of angular momentum. Please check the last chapter of Classical Mechanics by Goldstein. $\endgroup$
    – GK1202
    Apr 10, 2020 at 5:53
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    $\begingroup$ @y255yan Answers should not belong to the comments. Consider expanding that into a proper answer. $\endgroup$
    – MannyC
    Apr 10, 2020 at 7:16
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    $\begingroup$ Related: physics.stackexchange.com/q/68564/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Apr 10, 2020 at 9:08
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    $\begingroup$ See Belinfante–Rosenfeld stress–energy tensor in Wikipedia. A motivation is to agree with the stress energy tensor used in general relativity. $\endgroup$
    – Valter Moretti
    Jun 26 at 14:27

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An asymmetrical energy-momentum tensor implies orbital angular momentum non-conservation. This is why a trick is used to symmetrize such EM tensors. My take on this is that instead the Lagrangian should be corrected.

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    $\begingroup$ I do not understand well: the (conserved) total angular momentum is obtained by the Noether theorem. It is always conserved even if the stress energy tensor is not symmetric. Could you expand your answer? $\endgroup$
    – Valter Moretti
    Jun 26 at 14:18
  • $\begingroup$ @ValterMoretti I added 'orbital'. You are right that the total angular momentum is conserved for a sufficently symmetric lagrangian. $\endgroup$
    – my2cts
    Aug 1 at 11:38

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