1
$\begingroup$

When I looked at the Noether's theorem(here, we discuss the field case), there are two ways to derive it. One way is to assume that the variation of Lagrange $\delta L$ is exactly equal to derivative of some function, say $\partial_\mu F^{\mu}$, then $$j^{\mu}=\frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}Y_a(\varphi)-F^{\mu}(\varphi) ,$$ where $Y_a$ is a small variation in $\varphi$. On the other hand, one can start by varying in coordinates and $\varphi$. Then one can derive $$j^{\mu}=\left[\frac{\partial \mathcal L}{\partial(\partial_\mu \varphi)}\partial_\nu\varphi-\delta_\nu^\mu\mathcal L\right]\delta x^\nu-\frac{\partial \mathcal L}{\partial (\partial_\mu\varphi)}\tilde{\delta}\varphi . $$ Here $\tilde{\delta}{\varphi}:=\varphi'(x')-\varphi(x)$ the difference in both $x$ and $\varphi$.

However, I have no idea how to connect these two charges, even though they should be the same. For the first kind of derivation, see here pp.17-18. For the second derivation, see here Field-theoretical derivation.

Thanks.

$\endgroup$
1

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.