By jj_p

2014-02-19 10:18:36 8 Comments

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


@Qmechanic 2014-02-19 11:59:17

I) Let there be given a local action functional

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

with the Lagrangian density

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

[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$^1$

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

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

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

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

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

for some structure functions

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


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

[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

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

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

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

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

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

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

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

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

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

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.

@jj_p 2014-02-19 17:23:45

Thanks for the answer. The step I would like to be made more precise is that, if the Euler-Lagrange derivative of a functional is identically zero, then its lagrangian must be a total deivative. It is stated without proof also in (example 2.2.5) for the case of maps from $\mathbb{S}^1$

@Qmechanic 2014-02-19 18:05:53

I updated the answer.

@jj_p 2014-02-19 20:21:07

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, $k^\mu=k^\mu(x).$

@Qmechanic 2014-02-19 20:27:40

$k^{\mu}$ could in general depend on the fields $\phi(x)$ (and derivatives thereof to second order).

@pppqqq 2016-12-03 12:14:35

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?

@Qmechanic 2016-12-03 14:02:02

1. Shrinking $V$ to a point. 2. No, $k$ does not have to be zero.

@pppqqq 2016-12-03 19:18:01

2. Is that for the "up to possible boundary terms"? As it stands, eq. (8) for every measurable $V$ surely implies that $k=0$ almost everywhere, everywhere assuming $k$ continuous.

@Qmechanic 2016-12-03 19:24:23

Eq. (8) is up to possible boundary terms.

@LLang 2017-01-07 08:28:44

Why is it possible to assume that the variation of the action has the form of Eq. (5)?

@Qmechanic 2017-01-07 14:14:24

I plan an update.

@Arturo don Juan 2019-02-03 20:53:29

I don't quite understand how one can deduce (9) from (8), even after considering the fact that we can shrink the arbitrary integration region $V$ to a point. It seems like a point that I would normally (naively?) simply take to be true.

@Qmechanic 2019-05-01 07:47:24

Correction to the answer (v8) below eq. (10): integration by part should be integration by parts.

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