Noether's theorem for scale invariance [duplicate]

Question

When we have the Lagrangian $$\mathcal{L} = \frac{1}{2} \partial _\mu \phi\partial^\mu \phi \tag{1} $$ We have a symmetry given by $$x^\mu\mapsto e^\alpha x^\mu, \qquad\phi\mapsto e^{-\alpha} \phi.\tag{2}$$ I'm struggling to find the Noether charge for this symmetry. The formula is $$j^\mu=\frac{\partial \mathcal{L}}{\partial\partial_\mu\phi}\delta\phi-k^\mu\tag{3}$$ where $$\delta \phi=-\phi \tag{4}$$ in this case, but I can't find $k^\mu$ such that $$\delta \mathcal {L}=\partial _\mu k^\mu .\tag{5}$$

Answer 1

If u want to compute the Noether currents, u can do as follows:

$$x'^u=x^u+\delta x^u \quad \delta x^u=e^aE_a^u$$ $$\phi'=\phi+\delta\phi \quad \delta\phi=e^aX_a$$ $$J_a^u=[\eta_p^uL-\frac{dL}{dd_u\phi}d_p\phi]E_a^p+\frac{dL}{dd_u\phi}X_a$$

So in your case results:

$$E^u=x^u \quad X=-\phi$$ $$J^u=\frac{1}{2}d_p\phi d^p\phi x^2-d_v\phi d^u\phi x^v-(d^u\phi)\phi$$ Using Euler-Lagrange equation $d_ud^u\phi=0$ it semplifies to

$$J^u=-d_v\phi d^u\phi x^v-(d^u\phi)\phi$$

that u can verify it is conserved