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

From Noether theory we can define the canonical energy–momentum tensor as $$$$T_{\mu\nu}\equiv\frac{\partial\mathcal{L}}{\partial(\partial^\mu\phi)}\partial_\nu\phi-\eta_{\mu\nu}\mathcal{L}.$$$$ $$T_{\mu\nu}$$ satisfies $$$$\partial^\mu T_{\mu\nu}=0.$$$$ For example, $$T_{\mu\nu}$$ of Dirac field $$\mathcal{L}=\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$ is $$$$(T_\text{D})_{\mu\nu}=i\bar{\psi}\gamma_\mu\partial_\nu\psi-\eta_{\mu\nu}\mathcal{L}.$$$$ But in some$$^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.

## 2 Answers

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 $$\partial_\mu T^{\mu\nu}=0$$.

• Hmm... Is $\partial_\mu(i\bar{\psi}\gamma^\mu\partial^\nu\psi)=0$? Feb 11 at 13:17
• 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. Feb 11 at 13:25
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.

• 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... Feb 11 at 14:43