# Is this general symmetrisation of the canonical energy-momentum tensor correct?

There are multiple posts that exist already (such as here, here, here and here) about various specifics of the relationships between the canonical stress-energy-momentum tensor of a field theory,

$$T_{\mu\nu} = \eta_{\mu\nu} \mathcal{L} - \sum_{a} \frac{\partial \mathcal{L}}{\partial (\partial^\mu \varphi_a)} \partial_\nu \varphi_a,$$

using the $$(-,+,+,+)$$ signature, and the Hilbert stress-energy-momentum tensor of a field theory coupled to gravity,

$$T_{\mu\nu} = \frac{-2}{\sqrt{-g}} \frac{\partial(\mathcal{L} \sqrt{-g})}{\partial g^{\mu\nu}} = g_{\mu\nu} \mathcal{L} -2 \frac{\partial \mathcal{L}}{\partial g^{\mu\nu}},$$

but I feel rather lost reading the answers within. I know it is possible to modify the canonical SEM tensor by adding a divergenceless term to it, yielding a new SEM tensor which is still a conserved current for each $$\nu$$:

$$T_{\mu\nu} \rightarrow T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu}.$$

This post says that a particular choice of $$\chi_{\lambda\mu\nu}$$ makes the new SEM tensor symmetric:

$$\chi_{\lambda\mu\nu} = K_{\lambda\mu\nu} + K_{\mu\nu\lambda} + K_{\nu\mu\lambda},\quad K_{\lambda\mu\nu} = -\frac{i}{2} \sum_{a,b} \frac{\partial \mathcal{L}}{\partial (\partial^\lambda \varphi_a) } (J_{\mu\nu})_{ab} \varphi_b$$

where the $$(J_{\mu\nu})_{ab}$$ "are the representations of the Lorentz algebra under which the fields $$\varphi_a$$ transform".

1. Do the representations $$(J_{\mu\nu})_{ab}$$ take the forms written in this answer?
2. Is it correct that this choice always makes the new SEM tensor symmetric?
3. Does this choice yield the Hilbert SEM tensor after setting $$g_{\mu\nu}=\eta_{\mu\nu}$$?

EDIT: I just went through this process for the electromagnetic Lagrangian,

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$$

First, the canonical SEM tensor takes the form

$$T_{\mu\nu} = -\frac{1}{4}\eta_{\mu\nu}F_{\sigma\rho}F^{\sigma\rho} + F_{\mu\sigma}\partial_\nu A^\sigma .$$

Looking at the form of $$M_{\mu\nu}$$ in the answer I linked, I don't think it can contribute as there is an explicit $$x^\mu$$ dependence, so I just used

$$(J_{\mu\nu})^{\alpha\beta} = -i(\eta_\mu^{\,\,\alpha} \eta_\nu^{\,\,\beta} - \eta_\mu^{\,\,\beta} \eta_\nu^{\,\,\alpha}),$$

yielding

$$K_{\lambda\mu\nu} = \frac{1}{2}(F_{\lambda\mu}A_\nu - F_{\lambda\nu}A_\mu) \quad \Rightarrow \quad \chi_{\lambda\mu\nu} = F_{\lambda\mu}A_\nu.$$

We can now get the modification to the SEM tensor:

$$\partial^\lambda \chi_{\lambda\mu\nu} = F_{\sigma\mu}\partial^\sigma A_\nu,$$

where we make use of the equations of motion, $$\partial_\mu F^{\mu\nu} = 0$$, to cancel the first term. The SEM tensor becomes

$$T_{\mu\nu} = -\frac{1}{4}\eta_{\mu\nu}F_{\sigma\rho}F^{\sigma\rho} + F_{\mu\sigma}F_\nu^{\,\,\sigma},$$

which is the symmetric, standard form of the electromagnetic SEM tensor, equal to the Hilbert SEM tensor after setting $$g_{\mu\nu}=\eta_{\mu\nu}$$. Since $$M_{\mu\nu}$$ didn't play a role, I could possibly conclude that $$\chi_{\lambda\mu\nu}$$ is zero for a theory containing only scalar fields.

• Both the answers you linked to are mine and I can confirm that the answer to questions 1 and 2 is yes (WITHOUT $M_{\mu\nu}$ for 2.). The answer to the third is not always yes since the two tensors can differ up to improvement terms (there are many improvement terms which preserve symmetry). The canonical energy-momentum tensor is always symmetric for scalars. Jan 23 at 10:36
• @Prahar Thanks! Why do we not include $M_{\mu\nu}$? I can see that it doesn't work, but is there a deeper reason? As for the improvement terms you mention, are these added to the modified canonical tensor to get the Hilbert tensor, or the other way round, and do you know of an example? Jan 23 at 21:30
• $M_{\mu\nu}$ is a derivative operator that gives us the Lorentz transformation action on the coordinates. That part of the transformation is already accounted for in the original (asymmetric) $T_{\mu\nu}$. The extra term that is added keeps track of the spin indices of the field and their transformation. Jan 23 at 21:33
• The improvement terms are added to make the canonical e.m. tensor symmetric, NOT to get the GR stress tensor. As I mentioned earlier, the symmetric EM tensor (obtained by adding improvement terms) may not be the same as the GR tensor -- they two may differ up to other improvement terms. Jan 23 at 21:34
• Ah, I see that now. I'm still a bit confused about these "improvement terms" though. I think you are talking about extra terms other than $\partial^\lambda \chi_{\lambda\mu\nu}$ above? In the case of the EM tensor, they matched without any extra additions. Jan 24 at 0:07