In Srednicki's book he says that: The Noether current plays a special role if we can find a set of infinitesimal field transformations that leaves the lagrangian unchanged, or invariant. In this case, we have $\delta\mathcal{L}=0$, and we say that the lagrangian has a continuous symmetry. From eq(22.7), $$\partial_{\mu}j^{\mu}(x) = \delta\mathcal{L}(x)-\frac{\delta S}{\delta \varphi_{a}(x)}\delta \varphi_a(x)\tag{22.7}$$ then we have $\partial_\mu j^\mu=0$ $\textbf{whenever the field equations are satisfied,}$ and we say that the Noether current is conserved.

Here is my question: after choosing an appropriate set of infinitesimal transformations $\delta\varphi_a$, we have already rendered $\mathcal{L}(x)$ unchanged. As a corollary the second term in eq. (22.7),as a whole, equals to zero since $S=\int d^4 x\mathcal{L}(x)$, so $\partial_\mu j^\mu=0$. But Srednicki says the field equations have to be satisfied, which means $\frac{\delta S}{\delta \varphi_{a}(x)}=0$ holds for every index a. I don't know why I'm wrong.


1 Answer 1


It is not true that $$\delta\mathcal{L}(x)~=~0\text{ off-shell}\qquad\Rightarrow\qquad \frac{\delta S}{\delta \varphi_{a}(x)}\delta \varphi_a(x)~=~0\text{ off-shell}.$$ For a simple counterexample take $\mathcal{L}=\frac{1}{2}\partial_{\mu}\varphi_{a} \partial^{\mu}\varphi_{a}$ and $\delta\varphi_{a}(x)=\epsilon_a$.

  • $\begingroup$ Oh I see, thanks $\endgroup$ Commented Jul 9, 2019 at 9:30

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