Suppose, in whatever dimension and theory, the action $S$ is invariant for a global symmetry with a continuous parameter $\epsilon$.

The trick to get the Noether current consists in making the variation local: the standard argument, which doesn't convince me and for which I'd like a more formal explanation, is that, since the global symmetry is in force, the only term appearing in the variation will be proportional to derivatives of $\epsilon,$ and thus the involved current $J^\mu$ will be conserved on-shell:

$$ \delta S = \int \mathrm{d}^n x \ J^\mu \partial_\mu \epsilon .\tag{*}$$

This is stated, e.g., in Superstring Theory: Volume 1 by Green Schwarz Witten on page 69 and The Quantum Theory of Fields, Volume 1 by Weinberg on page 307.

In other words, why a term $$ \int \mathrm{d}^n x \ K(x) \ \epsilon(x)$$ is forbidden?

Taking from the answer below, I believe two nice references are

  1. theorem 4.1
  2. example 2.2.5

1 Answer 1


I) Let there be given a local action functional

$$ S[\phi]~=~\int_V \mathrm{d}^nx ~{\cal L}, \tag{1}$$

with the Lagrangian density

$$ {\cal L}(\phi(x),\partial\phi(x),x). \tag{2}$$

[We leave it to the reader to extend to higher-derivative theories. See also e.g. Ref. 1.]

II) We want to study an infinitesimal variation$^1$

$$ \delta x^{\mu}~=~\epsilon X^{\mu} \qquad\text{and}\qquad \delta\phi^{\alpha}~=~\epsilon Y^{\alpha}\tag{3}$$

of spacetime coordinates $x^{\mu}$ and fields $\phi^{\alpha}$, with arbitrary $x$-dependent infinitesimal $\epsilon(x)$, and with some given fixed generating functions

$$ X^{\mu}(x)\qquad\text{and}\qquad Y^{\alpha}(\phi(x),\partial\phi(x),x).\tag{4}$$

Then the corresponding infinitesimal variation of the action $S$ takes the form$^2$

$$ \delta S ~\sim~ \int_V \mathrm{d}^n x \left(\epsilon ~ k + j^{\mu} ~ d_{\mu} \epsilon \right) \tag{5}$$

for some structure functions

$$ k(\phi(x),\partial\phi(x),\partial^2\phi(x),x)\tag{6}$$


$$ j^\mu(\phi(x),\partial\phi(x),x).\tag{7}$$

[One may show that some terms in the $k$ structure function (6) are proportional to eoms, which are typically of second order, and therefore the $k$ structure function (6) may depend on second-order spacetime derivatives.]

III) Next we assume that the action $S$ has a quasisymmetry$^3$ for $x$-independent infinitesimal $\epsilon$. Then eq. (5) reduces to

$$ 0~\sim~\epsilon\int_V \mathrm{d}^n x~ k. \tag{8}$$

IV) Now let us return to OP's question. Due to the fact that eq. (8) holds for all off-shell field configurations, we may show that eq. (8) is only possible if

$$ k ~=~ d_{\mu}k^{\mu} \tag{9}$$

is a total divergence. (Here the words on-shell and off-shell refer to whether the eoms are satisfied or not.) In more detail, there are two possibilities:

  1. If we know that eq. (8) holds for every integration region $V$, we can deduce eq. (9) by localization.

  2. If we only know that eq. (8) holds for a single fixed integration region $V$, then the reason for eq. (9) is that the Euler-Lagrange derivatives of the functional $K[\phi]:=\int_V \mathrm{d}^n x~ k$ must be identically zero. Therefore $k$ itself must be a total divergence, due to an algebraic Poincare lemma of the so-called bi-variational complex, see e.g. Ref. 2. [Note that there could in principle be topological obstructions in field configuration space which ruin this proof of eq. (9).] See also this related Phys.SE answer by me.

V) One may show that the $j^\mu$ structure functions (7) are precisely the bare Noether currents. Next define the full Noether currents

$$ J^{\mu}~:=~j^{\mu}-k^{\mu}.\tag{10}$$

On-shell, after an integration by parts, eq. (5) becomes

$$ \begin{align} 0~\sim~~~~~&\text{(boundary terms)}~\approx~ \delta S \cr ~\stackrel{(5)+(9)+(10)}{\sim}& \int_V \mathrm{d}^n x ~ J^{\mu}~ d_{\mu}\epsilon \cr ~\sim~~~~~& -\int_V \mathrm{d}^n x ~ \epsilon~ d_{\mu} J^{\mu} \end{align}\tag{11}$$

for arbitrary $x$-dependent infinitesimal $\epsilon(x)$. Equation (11) is precisely OP's sought-for eq. (*).

VI) Equation (11) implies (via the fundamental lemma of calculus of variations) the conservation law

$$ d_{\mu}J^{\mu}~\approx~0, \tag{12}$$

in agreement with Noether's theorem.


  1. P.K. Townsend, Noether theorems and higher derivatives, arXiv:1605.07128.

  2. G. Barnich, F. Brandt and M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.


$^1$ Since the $x$-dependence of $\epsilon(x)$ is supposed to be just an artificial trick imposed by us, we may assume that there do not appear any derivatives of $\epsilon(x)$ in the transformation law (3), as such terms would vanish anyway when $\epsilon$ is $x$-independent.

$^2$ Notation: The $\sim$ symbol means equality modulo boundary terms. The $\approx$ symbol means equality modulo eqs. of motion.

$^3$ A quasisymmetry of a local action $S=\int_V d^dx ~{\cal L}$ means that the infinitesimal change $\delta S\sim 0$ is a boundary term under the quasisymmetry transformation.

  • $\begingroup$ One last comment, just to see if I got this straight: in your notation, we have $\frac{\delta K}{\delta \phi}=0.$ This implies, provided this generalized Poincare' lemma holds, that $\frac{\delta k}{\delta \phi}=0,$ which is (always) equivalent to $k=\partial_\mu k^\mu,$ and (of this I'd like a confirmation) this $k^\mu$ is field independent, $k^\mu=k^\mu(x).$ $\endgroup$
    – jj_p
    Feb 19, 2014 at 20:21
  • $\begingroup$ $k^{\mu}$ could in general depend on the fields $\phi(x)$ (and derivatives thereof to second order). $\endgroup$
    – Qmechanic
    Feb 19, 2014 at 20:27
  • $\begingroup$ List of corrections to the answer (v3): 1. "eg. (7)" should be "eq. (7)". 2. The last eq. (6) should be eq. (10)....[Done.] $\endgroup$
    – Qmechanic
    Jul 4, 2014 at 7:18
  • $\begingroup$ Dear Qmechanic, could you please extend a little on your point (IV.1)? 1. What do you mean by "localization"? 2. Does that mean that we can conclude that $k=0$ and so, by eq. (2.6) of Townsend's paper above, restricting only to first order derivatives, the fact that only $j^\mu d _\mu \varepsilon$ appears in the general variation? $\endgroup$
    – pppqqq
    Dec 3, 2016 at 12:14
  • 1
    $\begingroup$ Eq. (8) is up to possible boundary terms. $\endgroup$
    – Qmechanic
    Dec 3, 2016 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.