3 added 4 characters in body edited Jul 4 '13 at 2:36 user7757 For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\delta \phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$$$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi^*-i \alpha (\partial_{\mu}\phi^*)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$$$$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}\alpha J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$ For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\delta \phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$ For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\delta \phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi^*-i \alpha (\partial_{\mu}\phi^*)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}\alpha J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$ 2 added 7 characters in body edited Jul 4 '13 at 2:29 user7757 For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\phi^*=-i\alpha(x)\phi$$$$\delta \phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*-m^2 \delta(\phi* \phi))=\partial_{\mu}J^{\mu}$$$$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*)-\phi^* \partial^{\mu}\phi$$$$J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$ For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*-m^2 \delta(\phi* \phi))=\partial_{\mu}J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*)-\phi^* \partial^{\mu}\phi$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$ For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\delta \phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*)-m^2 \delta(\phi* \phi)=\partial_{\mu}J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*-\phi^* \partial^{\mu}\phi)$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$ 1 answered Jul 4 '13 at 2:24 user7757 For a local gauge transformation $$\delta \phi = i \alpha(x)\phi$$, and $$\phi^*=-i\alpha(x)\phi$$. These equations imply $$\delta(\partial_{\mu}\phi)=i(\partial_{\mu} \alpha)\phi+i \alpha (\partial_{\mu}\phi)$$ and $$\delta(\partial_{\mu}\phi^*)=-i(\partial_{\mu} \alpha)\phi-i \alpha (\partial_{\mu}\phi)$$ Therefore, $$\delta \mathcal{L}=\delta(\partial_{\mu}\phi \partial^{\mu}\phi^*-m^2 \delta(\phi* \phi))=\partial_{\mu}J^{\mu}$$, where $$J^{\mu}=i(\phi \partial^{\mu}\phi^*)-\phi^* \partial^{\mu}\phi$$ Setting $$\delta S=\int \delta \mathcal{L}=0$$, one gets $$\partial_{\mu}J^{\mu}=0$$