Let's define the global variation of a Lagrangian density as

$$\begin{align}\delta_{\epsilon}\mathcal{L} &= \frac{\partial\mathcal{L}}{\partial \psi}\delta_{\epsilon}\psi +\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\psi)}\partial_{\mu}(\delta_{\epsilon}\psi) \\&= \epsilon(x)\biggr(\frac{\partial\mathcal{L}}{\partial \psi}\delta\psi +\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\psi)}\partial_{\mu}(\delta\psi)\biggr) +\partial_{\mu}\epsilon(x)\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi \\&= \epsilon(x)\biggr(\frac{\partial\mathcal{L}}{\partial \psi} -\partial_{\mu}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\biggr)\delta\psi + \epsilon(x)\partial_{\mu}\biggr(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi\biggr) +\partial_{\mu}\epsilon(x)\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi \\&= \epsilon(x)\partial_{\mu}\biggr(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi\biggr) +\partial_{\mu}\epsilon(x)\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi \\&= \epsilon(x)\partial_{\mu}J^{\mu} + \partial_{\mu}\epsilon(x)\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\psi)}\delta\psi\end{align}\tag{1}$$

Where we assumed that Lagrangian is on-shell. What is the Noether current in local symmetry transformations?


1 Answer 1


If you have a variation $\delta\psi(x)=\varepsilon(x) h(x)$, then you can bake the spacetime-dependent parameter $\varepsilon$ into $h$ with no change. The Noether current is the same as in the usual case.

The case to which the second Noether theorem applies is when a field variation is parametrized by jets of an auxiliary vector bundle. In local coordinate language this means that$$ \delta\psi^i=Q^i_A[\psi]\lambda^A+Q^{i,\mu}_A[\phi]\lambda^A_{,\mu}+\dots+Q^{i,\mu_1...\mu_k}_A[\phi]\lambda^A_{,\mu_1...\mu_k}, $$ where the $\lambda^A=\lambda^A(x)$ are some arbitrary functions on spacetime and the coefficients $Q^{i,\mu_1...\mu_l}_A$ ($0\le l\le k$) are in general field-dependent.

The difference is with respect to the "ordinary" case is that when the symmetry variation is of this form, you can integrate by parts the variation of the Lagrangian with respect to the parameters $\lambda^A$ and get an expression of the form $$ 0=\delta\mathcal L=\mathcal N_A\lambda^A +d_\mu\left[\sum_k S^{\mu,\mu_1...\mu_k}_A\lambda^A_{,\mu_1...\mu_k}\right]. $$ If the expression in the brackets is denoted $\mathcal S^\mu$, then this expression means we get the relations $$ \mathcal N_A=0,\quad d_\mu\mathcal S^\mu=0 $$ as off-shell identities. Moreover, the exactness of a certain Spencer complex implies that (because $\mathcal S^\mu$ depends linearly on the jets of some vector bundle) there is some expression $\mathcal U^{\mu\nu}$ skew-symmetric in $\mu,\nu$ and also depending linearly on $\lambda$ and its derivatives such that $\mathcal S^\mu=d_\nu\mathcal U^{\mu\nu}$ and this is valid globally (i.e. it is not a Poincaré lemma type situation).

But this requires the symmetry variation $\delta\psi$ to have the specific form as given above with the parameters $\lambda$ being arbitrary. The symmetry variation being spacetime-dependent doesn't really mean anything, unless it depends on some spacetime function arbitrarily.

EDIT: Regarding the question

However, when you emphasize that there has to be spacetime-function dependence for things to be a non-trivial case, do you mean that the variation family $\hat\psi (x,s)$ might further depend on some spacetime function $f\in C(M)$?

in the comments.

Here's the thing. The variations we use for implementing Hamilton's principle is conceptually different from the variations we use in Noether's theorem.

In Hamilton's principle we perform variations over a single field configuration. In other words, we have the action $S[\psi]$ as a functional on some field space and we want to ask the question "Is the field configuration $\psi$ an extremal of the action?". To answer this, we consider a variation $\delta\psi^i(x)$ about the specific field configuration $\psi^i(x)$, which can be written the same way as it is done in the OP, by first considering a one-parameter deformation $\hat\psi (x,s)$ (with $\psi(x)=\hat\psi (x,0)$) of the specific field configuration and then differentiating with respect to $s$ at $s=0$.

When we state and prove Noether's theorem, we do not use variations about a single a priori fixed field configuration $\delta\psi^i$, we instead use a "variation field" $\delta\psi^i=\Xi^i[\psi]$ which depends on the field configuration itself and in principle can be applied to any field configuration. The difference is that the variation $\delta\psi^i=\frac{d \hat\psi (x,s)}{d s}|_{s=0}$ applied in Hamilton's principle is a "tangent vector to field space" while the variation $\delta\psi^i=\Xi^i[\psi]$ used in Noether's theorem is a "vector field" in field space.

Generally, the "variation field" depends on jets, eg. $$\delta\psi^i=\Xi^i(x,\psi(x),\partial\psi(x))$$(with possibly higher order dependencies as well).

As such, the infinitesimal variation used on Noether's theorem is not necessarily integrable into a finite flow in field space and might not be possible to be written as $\delta \psi=d\hat \psi / ds$.

If $\delta\psi^i=\Xi^i(x,\psi(x))$, then the $\Xi^i$ are the components of a vertical vector field on the field bundle and as such it can be integrated into a flow. But if the variation field depends derivatively on the field then the best you can do is to construct the evolution equation $$\frac{\partial \hat\psi^i}{\partial s}(s,x)=\Xi^i(x,\hat\psi(x,s),\partial\hat\psi (x,s)), $$ which may or may not have solutions (by the Cauchy-Kovalevskaya theorem, it has analytic solutions if the coefficients are analytic).

The long story short is that what is needed for the second Noether theorem to apply is that the "variation field" $$ \delta\psi^i=\Xi^i(x,\psi(x),\partial\psi(x),\dots,\partial^r\psi(x),\lambda(x),\partial\lambda(x),\dots,\partial^s\lambda(x)) $$ to depend on an additional parameter or sets of functional parameters $\lambda^A$ linearly and differentially and these functions be be arbitrary (i.e. the variation is a symmetry separately for each of these functions). But in this case the variation may or may not have an integrated version, and even if it has an integrated version it might depend on the parameters $\lambda$ in a very complicated, possibly nonlocal manner.

  • $\begingroup$ @Supergravity Edited the answer $\endgroup$ Oct 14, 2023 at 17:38
  • $\begingroup$ Thanks for your beautiful answer. $\endgroup$
    – user375448
    Oct 14, 2023 at 18:19

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