What is the role of the classical equations of motion in the derivation of the Noether current?

Question

I am trying to understand a very fundamental statement from the Book: Condensed Matter Field Theory from A.Altland and B.Simons:

Suppose we have a transformation:

$$x^\mu \to (x^{\prime})^{\mu} = x^\mu + f^\mu_a \omega^a(x)$$ and $$\phi^i(x)\to (\phi^{\prime})^i =\phi^i(x) + F^i_a \omega^a(x)$$

then we can compute the action difference

$$\Delta S = \int_V d^m x^\prime \mathcal{L}(\phi^\prime(x^\prime),\partial_{x^\prime} \phi^\prime(x^\prime))-\int_V d^m x \mathcal{L}(\phi (x),\partial_x \phi (x))$$

where we can express everything in terms of $x$ by using the transformation formulas and the Jacobi determinant. So far so good. Now comes the first statement:

(1) "So far, we did not use the fact that the transformation was actually meant to be a symmetry transformation. By definition we are dealing with a symmetry if for a constant parameter $\omega^a$ (e.g. a uniform rotation or global translation etc.) the action difference vanishes."

Yes I get that.

(2)"In other words the leading contribution to the action difference must be linear in the derivatives $\partial_{x^\mu} \omega^a$"

According to this answer to the Phys.SE question On a trick to derive the Noether current we just artificially added a $x$ dependence in the variation parameter. Then suppose we would have a symmetry then

$$\Delta S \overset{!}{=} 0 = \int_V [...]_1 \omega^a + j^\mu_a \partial_\mu \omega ^a \overset{\omega^a \text{is constant}}{=} \omega^a \int_V [...]_1=0 \to [...]_1=\partial_\mu k^\mu_a$$

This expression for $[...]_1$ we can replace in the formula for $[...]_1$ and integrate by parts once to get $\Delta S = \int_V J^\mu_a \partial_\mu \omega^a $ where we assume that the variation on the boundary $\partial V$ vanishes and $J^\mu_a=j^\mu_a-k^\mu_a$. After expanding the action difference in the derivative of $\omega$ we identify the Noether current.

Now comes the tricky part:

(3) "For a general field configuration, there is not much to say about the Noether current. However, if the field $\phi$ obeys the classical equations of motion and the theory is symmetric, the Noether current in locally conserved, $\partial_\mu J^\mu_a=0$. This follows from the fact, for a solution $\phi$ of the Euler Lagrange equation the linear variation in any parameter must vanish."

Is it correct that they just mean that by integrating by parts we arrive at $\Delta S = -\int_V d^m x \partial_\mu J^\mu_a \omega_a$. Then we use that $\phi$ is classically conserved which means that any linear variation vanishes?

I.e. $\partial_\mu J\mu_a =0$ which is the continuity equation.

So the only difference between the symmetry condition and the condition that $\phi$ obeys the equation of motion is that

• Symmetry transformation $\to \Delta S \sim 0$ modulo boundary terms

• $\phi$ obeys equation of motion $\to \Delta S = 0$ since all linear variations vanish

Is that correct?

Answer 1

1. OP's reasoning is basically correct, except that it can be further relaxed to allow boundary terms in more places, cf. the notion of quasisymmetry. E.g. when one varies the action on-shell, there could in principle still be boundary terms because the infinitesimal variations in the context of Noether's theorem don't have to obey boundary conditions.

2. Concerning OP's title question, the main point is that the quasisymmetry transformation is an off-shell quasisymmetry of the action, while the conservation law (i.e. the continuity equation) only holds on-shell. In other words, the equations of motion only play a role in the latter.