In my field theory class we recently derived Noether's theorem: We consider a infinitessimal transformation $\phi \to \phi + \epsilon \,\delta\phi$ of our field which preserves action i. e. $\delta S = 0$. This last condition is supposedly equivalent to $\delta \mathcal{L}$ being a divergence, i. e. $\delta \mathcal{L} = \epsilon\,\partial_\mu I^\mu$. Then you can expand \begin{align} \delta \mathcal{L} &= \frac{\partial \mathcal{L}}{\partial \phi} \epsilon \, \delta \phi + \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \epsilon \partial_\mu(\delta\phi) \\ &= \left( - \partial_\mu \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} + \frac{\partial \mathcal{L}}{\partial \phi}\right) \epsilon \, \delta \phi + \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi\right) \epsilon \end{align}

The first term vanishes by the eqns. of motion and so we get $$ \partial_\mu \left( \frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi - I^\mu\right) = 0 $$ i.e. $\frac{\partial\mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi - I^\mu$ is a conserved current.

Two things confuse me here:

We didn't use that the transformation is supposed to be a symmetry of the system (that is $\delta S = 0$). In a point-particle setting (that is a field that only depends on time) we can have $L = T - V$ with $\partial V / \partial q \neq 0$, but we can look at a translation $q \to q + \epsilon$ which gives us $$ L \to L + \epsilon \frac{\partial}{\partial t} \left( - t \frac{\partial V}{\partial q} \right) $$ so $L$ is only changed up to a "divergence" and the resulting conserved quantity is $p + t \frac{\partial V}{\partial q}$, which is indeed conserved, even though our system was not space-homogeneous.

The other source of confusion (which I guess is related to the first) is this argument that $\delta S = 0$ is equivalent to $\delta \mathcal{L} = \epsilon \partial_\mu I^\mu$. In $\mathbb{R}^n$, every scalar function can be written as a divergence, so this doesn't seem to add up. Is $I^\mu$ maybe supposed to be a function of the fields only?


2 Answers 2


There are at least 2 issues with OP's discussion (v2):

  1. One should properly distinguish between total and explicit spacetime derivatives, cf. e.g. my Phys.SE answer here.

    In particular, an infinitesimal quasisymmetry of the Lagrangian (density), means by definition that the infinitesimal variation is a total (space)time divergence.

  2. Note that not all terms in a Lagrangian (density) [or variations thereof] are total (space)time divergences.


My bad, I see now what happened, and as @Qmechanic noted above, Lagrangians are invariant under the addition of total derivatives.


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