While reading a lecture note, I am quite stuck on the very first part of proving Noether's theorem. As a setup, they defined the transformation $\delta \phi_a(x) = X_a(\phi)$ is a symmetry if the Lagrangian changes by a total derivative, $\delta L = \partial_\mu F^\mu$ for some set of functions $F^\mu(\phi)$.

Here, I don't quite see the connection between the transformation being a symmetry and Lagrangian changing by a total derivative. Could someone please explain how those two conditions are connected in a more intuitive sense?


2 Answers 2


A transformation is a symmetry if it leaves the action $S$ invariant (not the Lagrangian). Since $$ S = \int \! d^4 x \, \mathcal{L}, $$ if $\mathcal{L}$ changes by a total derivative, you can use the $d=4$ equivalent of Gauss' theorem to produce a surface term which we say goes to 0 at the boundary of space-time. this is not obvious though and there do exist theories where this surface term can cause you issues, but in general we assume that once you integrate over the whole of $\mathbb{R}^{3,1}$, it disappears.

  • $\begingroup$ Some additional explanations : the equations of motion are given by $\delta S = 0$. Therefore, a transformation that changes the lagrangian only by a total derivative will preserve the equations of motion : it is a symmetry of the system. $\endgroup$ Apr 29, 2021 at 10:13

Such a transformation is a symmetry of the action but not of the Lagrangian. It leaves the equations of motion invariant but does not correspond to a Noether conserved current.

  • 1
    $\begingroup$ It does have an associated Noether current, namely $$j^{\mu} = \delta\phi_a \frac{\partial \mathcal L}{\partial (\partial_\mu \phi_a)} - F^\mu$$ $\endgroup$ Apr 29, 2021 at 10:15
  • $\begingroup$ @solublefish How is this a conserved Noether current? $\endgroup$
    – my2cts
    May 7, 2021 at 6:20
  • $\begingroup$ The variation of the Lagrangian is $$\delta \mathcal L = \delta\phi_a \frac{\partial \mathcal L}{\partial \phi_a} + \partial_\mu\delta\phi_a \frac{\partial \mathcal L}{\partial ( \partial_\mu \phi_a)}$$ When the Euler-Lagrange are satisfied, you have : $$\delta \mathcal L = \delta\phi_a \partial_\mu\frac{\partial \mathcal L}{\partial ( \partial_\mu \phi_a)} + \partial_\mu\delta\phi_a \frac{\partial \mathcal L}{\partial ( \partial_\mu \phi_a)} = \partial_\mu \left (\delta\phi_a \frac{\partial \mathcal L}{\partial ( \partial_\mu \phi_a)}\right)$$ $\endgroup$ May 7, 2021 at 8:31
  • $\begingroup$ Therefore, when $\delta \mathcal L = \partial_\mu F^\mu$, you see that $$j^\mu = \delta \phi_a \frac{\partial \mathcal L}{\partial(\partial_\mu \phi_a)} - F^\mu$$ is a conserved current. $\endgroup$ May 7, 2021 at 8:33
  • $\begingroup$ @SolubleFish This is not a comment but a different answer. $\endgroup$
    – my2cts
    May 7, 2021 at 9:26

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