I'm completely new to this. I was trying to derive the conserved current of the following Lagrangian density: $$ \mathcal L(\phi, \partial_\mu \phi) = \frac{1}{2}\partial_\mu \phi \partial^\mu\phi + \mathcal L_{\text{int}}(\phi) $$ If we: $$ x^\mu \to x^\mu + \delta x^\mu $$ $$ \phi(x^\mu + \delta x^\mu) \approx \phi + \delta x^\lambda \partial_\lambda\phi $$ $$ \mathcal L_{\text{int}}(\phi + \delta x^\mu \partial_\mu\phi) \approx \mathcal L_{\text{int}} + \delta x^\lambda\frac{\partial \mathcal L_{\text{int}}}{\partial(\partial_\lambda \phi)} = \mathcal L_{\text{int}} $$ Hence we get (to first order): $$ \mathcal L(\phi + \delta \phi) \approx \frac{1}{2}\partial_\mu \phi \partial^\mu\phi + \mathcal L_{\text{int}}(\phi) + \partial_\mu \phi \partial^\mu(\delta x^\lambda \partial_\lambda\phi) $$ $$ \mathcal L(\phi + \delta \phi) = \mathcal L(\phi) + \partial_\mu \phi \partial^\mu(\delta x^\lambda \partial_\lambda\phi) $$
Finding the action: $$ S[\mathcal L(\phi + \delta \phi)] = S[\mathcal L(\phi)] + \int d^4 x\; \left [ \partial_\mu \phi \partial^\mu(\delta x^\lambda \partial_\lambda\phi) \right ] $$ Setting the variation of S to zero: $$ \delta S[\mathcal L] = \int d^4 x\; \left [ \partial_\mu \phi \partial^\mu(\delta x^\lambda \partial_\lambda\phi) \right ] = 0 $$ But I do not know if my reasoning is good? Should the conserved current be a vector? And, could you made this current independent of $\delta x^\lambda$?
