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".
- Do the representations $(J_{\mu\nu})_{ab}$ take the forms written in this answer?
- Is it correct that this choice always makes the new SEM tensor symmetric?
- 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.
