# Srednicki chapter 22: continuous symmetries and conserved current

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.

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 $$\delta\varphi_{a}(x)=\epsilon_a$$.

• Oh I see, thanks – hehehehehehe Jul 9 at 9:30