Consider a field $\phi$ which transforms as $\phi\rightarrow\phi+\delta\phi$ and say $X\left(\phi\right)=\delta\phi$ is a symmetry of your Lagrangian, which under that transformation changes by a total derivative $\delta L=\partial_{\mu}F^{\mu}$. Noether's theorem tells us there's a conserved current given by,

$$ j^{\mu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}X\left(\phi\right)-F^{\mu}\left(\phi\right). $$

Now, when considering spacetime translations we have

$$ x^{\nu}\rightarrow x^{\nu}-\varepsilon^{\nu} $$

$$ \phi\left(x\right)\rightarrow\phi\left(x\right)+\varepsilon^{\nu}\partial_{\nu}\phi\left(x\right) $$

$$ L\rightarrow L+\varepsilon^{\nu}\partial_{\nu}L\left(x\right) $$

$$ \left(j^{\mu}\right)_{\nu}=\frac{\partial L}{\partial\left(\partial_{\mu}\phi\right)}\partial_{\nu}\phi\left(x\right)-\delta_{\nu}^{\mu}L\equiv T_{\,\nu}^{\mu} $$

I'm following these well-known QFT notes by D. Tong, page 14. The thing I didn't understand in this derivation is how the last term, $\delta_{\nu}^{\mu}L$, comes about. I don't understand how $\partial_{\nu}L\left(x\right)$ becomes a $\delta_{\nu}^{\mu}L$. This is most certainly something very simple but it's driving me crazy. Can you help me clarify this?

  • $\begingroup$ For the corresponding derivation via Noether's theorem of energy conservation in classical mechanics (as opposed to field theory), see e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Jun 11, 2018 at 11:50
  • $\begingroup$ You've noticed that $\partial_\mu (F^\mu)_\nu\equiv \partial_\nu L$. From this, you should be able to solve for $(F^\mu)_\nu$. $\endgroup$ Jun 11, 2018 at 14:51

1 Answer 1


It appears because you have four directions for possible translations. So Noether's theorem tells you you will have four currents. If you do the derivation you will realize that before the last equation you wrote, everything will be contracted with $\epsilon^\nu$, but since this translation is arbitrary, you will be allowed to factor it away. Then as you can see from the previous equation: $$L\rightarrow L + \epsilon^\nu \partial_\nu L = L + \epsilon^\nu \delta_\nu^\mu \partial_\mu L$$ so you get that in Tong's notation $F^\mu_\nu(x) = \delta^\mu_\nu L(x)$.

  • $\begingroup$ Oh these subtleties... You're right, I had to go through the derivation to see why. Thank you. $\endgroup$
    – johani
    Jun 11, 2018 at 12:22

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