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Conserved charge from Lorentz symmetry

by spookyfish
Tags: charge, conserved, lorentz, symmetry
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spookyfish
#1
Apr11-14, 11:57 AM
P: 43
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
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samalkhaiat
#2
Apr11-14, 12:58 PM
Sci Advisor
P: 883
Quote Quote by spookyfish View Post
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
Exactly what is it that you are doing here? You want to substitute stuff in a formula to get the same formula back?
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].
spookyfish
#3
Apr11-14, 01:23 PM
P: 43
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)?

samalkhaiat
#4
Apr11-14, 01:44 PM
Sci Advisor
P: 883
Conserved charge from Lorentz symmetry

Quote Quote by spookyfish View Post
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)?
Very consistent. The field in (1.51) is a scalar field.
[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]
spookyfish
#5
Apr11-14, 01:52 PM
P: 43
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]?
samalkhaiat
#6
Apr11-14, 02:04 PM
Sci Advisor
P: 883
Quote Quote by spookyfish View Post
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]?
Look at it again. In the article, he uses the current which does not involve T, this is why he used [itex]\delta \phi[/itex]. If you want to start with the form of [itex]J^{ \mu }[/itex] which involves the tensor [itex]T^{ \mu \nu }[/itex], then you must use [itex]\delta^{ * }\phi[/itex], as I explained in my first post.


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