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This question is somewhat historic.

Let $\Theta_{\mu\nu}$ denote the canonical stress-energy tensor of some matter field $\psi$ in special relativity.

It is often stated that the reason why we want this tensor to be symmetric, so that 4-angular momentum is conserved. One can arrive at this (erroneous) result even without Noether's theorem, as then the infinitesimal 4-momentum of a spatial cube $d^3x$ centered on $x$ is $$ dP^\mu(x)=\Theta^{0\mu}(x)d^3x, $$ and hence the infinitesimal 4-angular momentum of the same spatial cube with respect to the origin is $$ d J^{\mu\nu}=x^\mu dP^\nu-x^\nu dP^\mu=(x^\mu\Theta^{0\nu}-x^\nu\Theta^{0\mu})d^3x, $$ implying that the density for 4-angular momentum is $$ \mathcal J^{\mu\nu\rho}=x^\nu\Theta^{\mu\rho}-x^\rho\Theta^{\mu\nu}. $$

However we obviously want this to be conserved, so one calculates the divergence to obtain $$ \partial_\mu\mathcal J^{\mu\nu\rho}=\Theta^{\nu\rho}-\Theta^{\rho\nu},$$ hence the conclusion that if 4-angular momentum is to be conserved then the stress-energy tensor must be symmetric.

One can also obtain the same result from Noether's theorem, but only if the matter field is treated as a scalar field.

However if it is also considered that the matter field $\psi$ takes values in a representation space of $\mathfrak{so}(3,1)$, the Lie algebra of the Lorentz group, and in this representation, the generators are $\Sigma_{\mu\nu}$ (skew in the indices), then from Noether's theorem, the proper current obtained is $$ \mathcal M^{\mu\nu\rho}=\mathcal J^{\mu\nu\rho}+\mathcal S^{\mu\nu\rho}, $$ with $$ \mathcal S^{\mu\nu\rho}=\mathcal P^\mu\Sigma^{\nu\rho}\psi, $$ where $$ \mathcal P^\mu=\frac{\partial\mathcal L}{\partial\partial_\mu\psi} $$ are the Lagrangian momenta.

Checking the conservation law for this current gives $$ \partial_\mu\mathcal M^{\mu\nu\rho}=2\Theta^{[\nu\rho]}+\partial_\mu\mathcal S^{\mu\nu\rho}, $$ so we can see that the antisymmetric part of the canonical stress-energy tensor is generated by the divergence of the spin current density $\mathcal S$, and angular momentum is only conserved if we consider both "orbital" and "spin" angular momentum together.


Hence, the conclusion that the stress-energy tensor must be symmetric due to angular momentum conservation is wrong and is based on ignoring the spin of the fields.

Of course, the Einstein-Hilbert stress-energy tensor is symmetric by design, and one can arrive at the Einstein Field Equation $$ G_{\mu\nu}=8\pi G T_{\mu\nu} $$ without a Lagrangian formalism, and this equation does imply that the stress-energy tensor must be symmetric, however...

  • It seems to me - and this is just intuition - that even before GR was properly formulated the stress-energy tensor was considered to be having to be symmetric.
  • If we consider unsymmetric stress-energy tensors, what stops one from having $T_{\mu\nu}=\Theta_{(\mu\nu)}$ on the RHS of the EFE?

So my question is, what was the reason that physicists originally wanted symmetric stress-energy tensors?

Was it only GR?

Was it that they never realized that the internal Lorentz transformations also need to be considered for conservation?

Is there any other "proper" reason besides GR for wanting symmetric tensors? (Here I am obviously ignoring the fact that the canonical tensor suffers from other maladies too, like being gauge-dependent for EM/YM)

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