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

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?