# On a trick to derive the Noether current

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

I) Let there be given a local action functional

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

with the Lagrangian density

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

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

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

and

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

[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.

References:

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.

• 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, $k^\mu=k^\mu(x).$
– jj_p
Feb 19, 2014 at 20:21
• $k^{\mu}$ could in general depend on the fields $\phi(x)$ (and derivatives thereof to second order). Feb 19, 2014 at 20:27
• List of corrections to the answer (v3): 1. "eg. (7)" should be "eq. (7)". 2. The last eq. (6) should be eq. (10)....[Done.] Jul 4, 2014 at 7:18
• 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? Dec 3, 2016 at 12:14
• Eq. (8) is up to possible boundary terms. Dec 3, 2016 at 19:24