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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 $-\frac14F^2$?


$^1$ 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.

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2 Answers 2

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  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.

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  • $\begingroup$ Thank you. Actucally Klein-Gordon Lagragian can be written as $-\frac12\phi\square\phi$, spin 1 field Lagragian can be written as $\frac12A_\mu(\eta^{\mu\nu}\partial^2-\partial^\mu\partial^\nu)A_\nu$. It seems that for spin 0,1/2,1 the $\eta^{\mu\nu}\mathcal{L}$ term in energy–momentum tensor can all be omitted... $\endgroup$
    – Kernifan
    Feb 11, 2023 at 14:43
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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 $\partial_\mu T^{\mu\nu}=0$.

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  • $\begingroup$ Hmm... Is $\partial_\mu(i\bar{\psi}\gamma^\mu\partial^\nu\psi)=0$? $\endgroup$
    – Kernifan
    Feb 11, 2023 at 13:17
  • $\begingroup$ The Dirac action integral is $\int \bar \psi( \gamma^\mu \partial_\mu+ m)\psi d^dx$, the E of M is $( \gamma^\mu \partial_\mu+ m)\psi =0$, and the derivative of something that is identically zero zero is zero. $\endgroup$
    – mike stone
    Feb 11, 2023 at 13:25

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