Lack of symmetry of the canonical stress-energy tensor

Why in the general case of classical field theory canonical stress-energy tensor doesn't have symmetry of the permutation of the indices?

For explanation, let's have a "derivation" of an expression for tensor by using symmetry of translations.

For translations we have $$x'^{\nu} = x^{\nu} + \omega^{\nu } \Rightarrow X^{\nu}_{\alpha } = \delta^{\nu}_{\alpha}, \quad Y_{k, \alpha} = 0,$$ so Noether current has following expression, $$J^{\mu}_{\alpha} = -\left(\frac{\partial L}{\partial (\partial_{\mu}\Psi_{k})}\partial_{\alpha} \Psi_{k} - \delta^{\mu}_{\alpha}L\right),$$ and it isn't symmetrized.

Does it have some fundamental meaning?

• – Qmechanic Jun 30 '13 at 0:30

Symmetry of the canonical energy-momentum tensor can be related to the spin of the object(s) that contribute to it (in other words, the representation of the Lorentz group under the fields transform). Note that the canonical EM tensor is obtained by using the Noether's procedure for translational symmetry $$T_{\mu\nu} = \sum\limits_r \frac{\delta {\cal L}}{\delta \left( \partial^\mu \phi_r \right)} \partial_\nu \phi_r - g_{\mu\nu} {\cal L}$$ This expression is clearly not symmetric. However, we can comment about its (non-)symmetry by looking at the conserved quantity corresponding to Lorentz transformation. By the Noether's procedure, we can show that this is $$M_{\mu\nu\alpha} = T_{\mu\alpha} x_\nu - T_{\mu\nu} x_\alpha - \frac{\delta {\cal L}}{ \delta \left( \partial^\mu \phi_r \right) } \left( J_{\nu\alpha} \right)^{rs} \phi_s$$ Here, $\left( J_{\nu\alpha} \right)^{rs}$ is the representation of the Lorentz algebra under which the set of fields $\phi_r$ transforms. Conservation of this quantity implies $$0 = \partial^\mu M_{\mu\nu\alpha} = T_{\nu\alpha} - T_{\alpha\nu} - \partial^\mu \left( \frac{\delta {\cal L}}{ \delta \left( \partial^\mu \phi_r \right) } \left( J_{\nu\alpha} \right)^{rs} \phi_s \right)$$ This implies $$T_{\nu\alpha} - T_{\alpha\nu} = \partial^\mu \left( \frac{\delta {\cal L}}{ \delta \left( \partial^\mu \phi_r \right) } \left( J_{\nu\alpha} \right)^{rs} \phi_s \right)$$ Non-symmetry of the stress-energy tensor is an indication that the fields that are contributing to it transform non-trivially under the Lorentz group. In particular, the canonical EM tensor is symmetric only if the theory contains only scalars.
The way I like to think about the process of "symmetrizing the EM tensor" is the following. The canonical EM tensor does not contain any "spin information" and one needs the angular momentum tensor for that information. However, a symmetrized EM tensor is essentially defined to "absorb" in the spin information of the field content so that the angular momentum tensor is no longer needed (it contains more information?). The reason I think of this like this is that in terms of the symmetric EM tensor we can define another conserved quantity $${\tilde M}_{\mu\nu\alpha} = {\tilde T}_{\mu\alpha} x_\nu - {\tilde T}_{\mu\nu} x_\alpha$$ Since ${\tilde T}_{\mu\nu}$ is symmetric the tensor above is trivially conserved and does not contain any new information. However, this modified angular momentum tensor still generates all the conserved quantities as $M_{\mu\nu\alpha}$. It then seems to me that ${\tilde T}_{\mu\nu}$ already contains information about the conserved angular momentum.
This also reconciles with the fact that the symmetrized EM tensor is often the same as one obtains by varying the metric, usually defined as $$T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu} }$$ Since the metric (gravitation) couples to all particles in a universal fashion, the above definition of the EM tensor should involve spin information as well, and therefore, must be equal to (or atleast closely related to) ${\tilde T}_{\mu\nu}$ described above.