Can the $\eta_{\mu\nu}\mathcal{L}$ term in canonical energy–momentum tensor be omitted?

Question

From Noether theory we can define the canonical energy–momentum tensor as \begin{equation} T_{\mu\nu}\equiv\frac{\partial\mathcal{L}}{\partial(\partial^\mu\phi)}\partial_\nu\phi-\eta_{\mu\nu}\mathcal{L}. \end{equation} $T_{\mu\nu}$ satisfies \begin{equation} \partial^\mu T_{\mu\nu}=0. \end{equation} For example, $T_{\mu\nu}$ of Dirac field $\mathcal{L}=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$ is \begin{equation} (T_\text{D})_{\mu\nu}=i\bar{\psi}\gamma_\mu\partial_\nu\psi-\eta_{\mu\nu}\mathcal{L}. \end{equation} But in some$^1$ books and papers, I see the authors omit the $\eta_{\mu\nu}\mathcal{L}$ term in $T_{\mu\nu}$. The reason is that Noether theory holds only "on shell". This means we've used EOM. Thus $\mathcal{L}=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi=\bar{\psi}\times\text{EOM}=0$ so we omit the $\eta_{\mu\nu}\mathcal{L}$ term.

My question is

1. Can the $\eta_{\mu\nu}\mathcal{L}$ term be omitted? Obviously if we omit the $\eta_{\mu\nu}\mathcal{L}$ term then $T_{\mu\nu}$(made up of the rest term) does not satisfy $\partial^\mu T_{\mu\nu}=0$.

2. If the $\eta_{\mu\nu}\mathcal{L}$ term here (for Dirac field) can be omitted. Can this term be omitted for other Lagrangian, for example, the Maxwell Lagrangian $-\frac14F^2$?

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$^1$ For example https://arxiv.org/abs/1905.08113. There is no $\eta_{\mu\nu}\mathcal{L}$ term in (57). Another example is on Wikipedia, at the end of the "Belinfante–Rosenfeld and the Hilbert energy–momentum tensor" part.

Answer 1

1. The continuity equation $d_{\mu} T^{\mu\nu}\approx 0$ only holds on-shell, so often one is only interested in the canonical SEM tensor on-shell. Then one can disregard terms that vanish on-shell.

2. For Lagrangian densities that vanish on-shell, see e.g. this and this Phys.SE posts.

Answer 2

Remember that Noether only says that $\partial_\mu T^{\mu\nu}=0$ when the equations of motion are satisfied. Consequently, if the $L$ term vanishes when we satisfy the equations of motion (as it does for the Dirac equation) then we can omit it, and what is left still satisfies $\partial_\mu T^{\mu\nu}=0$.