The Noether current corresponding to the transformation $\phi \to e^{i\alpha} \phi$ for the Klein-Gordon Lagrangian density
$$\mathcal{L}~=~|\partial_{\mu}\phi|^2 -m^2 |\phi|^2$$
by finding $\delta S$, and setting it to zero. The general formula for a global transformation is
$$j^{\mu}=\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\Delta \phi-\mathcal{J}^{\mu},$$
where $ \partial_{\mu} \mathcal{J}^{\mu}$ is the change in the Lagrangian density due to the transformation. (See Peskin section 2.2).
How do I find the Noether's current corresponding to a local transformation $\phi \to e^{i\alpha(x)}\phi$?
No comments:
Post a Comment