# A Simple question on Noether's first theorem

I am trying to understand Noether's first theorem in field theory and have read several references on the subject. They are all pretty clear except on one issue that all but one share:

These references (e.g. Peskin) all end up with a formula like:

$$\delta S = \int dx\ \partial_\mu \mathcal{J}^\mu \tag{1}$$

which they interpret as:

$$\partial_\mu \mathcal{J}^\mu = 0 \tag{2}$$

Here is my question:

The contribution in equation (1) is a surface term, where all the action happens at infinite distances - it says nothing about the bulk of space-time. And it can presumably be set to zero by adjusting field variations on the boundaries. Thus it should not lead to a conservation property in the bulk.

The one reference that does not use this inference is the excellent book by Weinberg, where he derives the theorem assuming that the epsilon parameter that governs the global symmetry actually depends on the coordinate $$x$$. He gets:

$$\delta S = \int dx\ \mathcal{J}^\mu \partial_\mu \epsilon \tag{3}$$

which can be integrated by parts to derive the Noether current in the bulk. Which is fine by me and indicates (perhaps) Weinberg's distaste for the "surface term" derivation. But is he still dealing with a global symmetry?

I fear that, given the prominence of Noether's theorem in field theory, the answer to my question will be embarrassingly simple but if anyone can help me out, I would be grateful.

P.S. There are many questions on Noether's theorem in physics questions sites. I have done the due diligence of looking at these sites and I do not see my query.

• By "Peskin," do you mean pages 17-18 in Peskin & Schroeder's Introduction to QFT? If so, then the thing they denote $\mathcal{J}^\mu$ isn't the conserved current. The current is the thing denoted $j^\mu$ in equation (2.12). The result $\partial_\mu j^\mu=0$ is a straighforward consequence of assuming that the fields satisfy the Euler-Lagrange equations and that the variation satisfies $\delta L = \partial^\mu\Lambda^\mu\delta\epsilon$ for some $\Lambda^\mu$, which Peskin & Schroeder denote $\mathcal{J}^\mu$ even though it is not the current. The current is shown in (2.12), denoted $j^\mu$. Oct 28, 2021 at 1:01
• Thanks for the suggestion. I was aware of the two currents. However, your boldface “not” made me question an assumption I was making and I am now OK with the derivation. Basically, I was thinking of everything in terms of the action but Noether’s theorem is derived on variations of the lagrangian density with necessary surface terms derived from the variations. Thanks! Oct 29, 2021 at 2:29

1. the global quasisymmetry holds for every spacetime integration region $$V$$,
This is explained in my Phys.SE answer here, which also explains the trick of using an infinitesimal $$x$$-dependent parameter $$\epsilon(x)$$ while discussing a global (=$$x$$-independent) quasisymmetry.