1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Oh I see, thanks $\endgroup$ – hehehehehehe Jul 9 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.