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

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

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

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

It is implicitly assumed that under a variation the integration region $V$ changes according to the vector field $X^{\mu}$. Then the corresponding infinitesimal variation of the action $S_V$ takes the form$^2$

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

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 $S_V$ 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

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

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 $S_V=\int_V d^dx ~{\cal L}$ means that the infinitesimal change $\delta S_V\sim 0$ 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
    Commented 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
    Commented Feb 19, 2014 at 20:27
  • $\begingroup$ 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
    Commented Dec 3, 2016 at 12:14
  • $\begingroup$ 1. Shrinking $V$ to a point. 2. No, $k$ does not have to be zero. $\endgroup$
    – Qmechanic
    Commented Dec 3, 2016 at 14:02
  • 1
    $\begingroup$ Eq. (8) is up to possible boundary terms. $\endgroup$
    – Qmechanic
    Commented Dec 3, 2016 at 19:24

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