I am trying to derive the conserved charge from the symmetry of the action under Lorentz transformations, but I am doing something wrong.(adsbygoogle = window.adsbygoogle || []).push({});

Noether's theorem states that the current is

[tex]

J^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)} \delta\phi - T^{\mu \nu}\delta x_\nu

[/tex]

For an infinitesimal Lorentz transformations

[tex]

\Lambda^\mu _\nu = \delta^\mu_\nu +\omega ^\mu_\nu

[/tex]

I get

[tex]\delta x_\nu = \omega_{\nu \sigma} x^\sigma, \quad \delta \phi = -\omega^\mu _\nu x^\nu \partial_\mu \phi [/tex]

This gives

[tex]

J^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)}(-\omega^\nu _\sigma x^\sigma \partial_\nu \phi) -T^{\mu \nu}\omega_{\nu \sigma} x^\sigma

[/tex]

where

[tex]

T^{\mu \nu} = \frac{\partial \cal L}{\partial(\partial_\mu \phi)} \partial^\nu \phi -\cal L \eta^{\mu \nu}

[/tex]

so

[tex]

J^\mu = -\omega_{\nu \sigma} x^\sigma \left(2\frac{\partial \cal L}{\partial(\partial_\mu \phi)}\partial^\nu \phi - \cal L \eta^{\mu \nu} \right)

[/tex]

this is "almost" the correct form but I have this factor of 2 which is wrong. I will be happy to know where is my mistake

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# Conserved charge from Lorentz symmetry

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