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 $$\delta \mathcal {L}=\partial _\mu k^\mu .\tag{5}$$
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$\begingroup$ Hint: The scale transformation has both vertical $\phi$ components & horizontal $x^{\mu}$ components. The horizontal contribution is missing in your Noether current formula. $\endgroup$– Qmechanic ♦Jan 6, 2019 at 15:14
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1$\begingroup$ Related: physics.stackexchange.com/q/270245/2451 $\endgroup$– Qmechanic ♦Jan 6, 2019 at 15:23
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$\begingroup$ What Qmechanic (probably) means by horizontal & vertical components is that your Eq. (3) is wrong: the transformation of $\phi(x)$ under (2) is $\phi(x) \to e^{-\alpha}\phi(e^\alpha x) \approx \phi(x) + \alpha \left[ -\phi(x) + x^\mu \partial_\mu \phi(x) \right] + \mathcal{O}(\alpha^2)$, i.e. $\delta\phi = - \phi + x^\mu \partial_\mu \phi$. You should be able to derive the Noether current from here. $\endgroup$– M.JoJan 7, 2019 at 13:13
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1 Answer
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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
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1$\begingroup$ Wow to such badly wording (u instead of you) and badly formated equations! $d$ instead of partial derivatives, $x^2$ and ugly tensorial indices ... are worth a -1 vote. $\endgroup$– ChamApr 30, 2019 at 2:09
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