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Conserved charge from Lorentz symmetry 
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#1
Apr1114, 11:57 AM

P: 53

I am trying to derive the conserved charge from the symmetry of the action under Lorentz transformations, but I am doing something wrong.
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 


#2
Apr1114, 12:58 PM

Sci Advisor
P: 910

The general form of Noether current is [tex]J^{ \mu } = \frac{ \partial \mathcal{L} }{ \partial ( \partial_{ \mu } \phi ) } \delta \phi + \delta x^{ \mu } \mathcal{L} ,[/tex] where [itex]\delta \phi ( x ) = \bar{ \phi } ( x )  \phi ( x )[/itex]. When you rewrite the current in terms of the energy momentum tensor, you get [tex]J^{ \mu } = \frac{ \partial \mathcal{L} }{ \partial ( \partial_{ \mu } \phi ) } \delta^{ * } \phi ( x )  T^{ \mu }{}_{ \nu } \delta x^{ \nu } ,[/tex] where now the variation in the field is given by [tex]\delta^{ * } \phi ( x ) = \bar{ \phi } ( \bar{ x } )  \phi ( x ) = \bar{ \phi } ( x + \delta x )  \phi ( x ) .[/tex] When you expand to 1st order in [itex]\delta x[/itex] you find that the two variations are related by [tex]\delta^{ * } \phi ( x ) = \delta \phi ( x ) + \delta x^{ \mu } \partial_{ \mu } \phi ( x ) .[/tex] For scalar field [itex]\delta^{ * } \phi ( x ) = 0[/itex], this implies [itex]J^{ \mu } = \delta x^{ \nu } T^{ \mu }{}_{ \nu }[/itex]. 


#3
Apr1114, 01:23 PM

P: 53

I understand what you say. How is it then consistent with
http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf Eqs. (1.51) and (1.52)? 


#4
Apr1114, 01:44 PM

Sci Advisor
P: 910

Conserved charge from Lorentz symmetry
[tex]\bar{ \phi } ( \bar{ x } ) = \phi ( x ) , \ \ \Rightarrow \ \ \delta^{*} \phi = 0 .[/tex] Use [itex]x = \Lambda^{  1 } \bar{ x }[/itex] and rename the coordinates [itex]x[/itex], (i.e. drop the bar from [itex]\bar{ x }[/itex]), you get [itex]\bar{ \phi } ( x ) = \phi ( \Lambda^{  1 } x )[/itex]. Now, if you expand this, as they did, you find that [itex]\delta \phi =  \delta x^{ \mu } \partial_{ \mu } \phi[/itex] 


#5
Apr1114, 01:52 PM

P: 53

but why in my formula you use [itex] \delta^* \phi \equiv \phi'(x')\phi(x) [/itex] and in the article it is [itex] \delta \phi = \phi'(x) \phi(x) [/itex]?



#6
Apr1114, 02:04 PM

Sci Advisor
P: 910




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