# Conserved charge from Lorentz symmetry

1. Apr 11, 2014

### spookyfish

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
$$J^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)} \delta\phi - T^{\mu \nu}\delta x_\nu$$
For an infinitesimal Lorentz transformations
$$\Lambda^\mu _\nu = \delta^\mu_\nu +\omega ^\mu_\nu$$
I get
$$\delta x_\nu = \omega_{\nu \sigma} x^\sigma, \quad \delta \phi = -\omega^\mu _\nu x^\nu \partial_\mu \phi$$
This gives
$$J^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)}(-\omega^\nu _\sigma x^\sigma \partial_\nu \phi) -T^{\mu \nu}\omega_{\nu \sigma} x^\sigma$$
where
$$T^{\mu \nu} = \frac{\partial \cal L}{\partial(\partial_\mu \phi)} \partial^\nu \phi -\cal L \eta^{\mu \nu}$$
so
$$J^\mu = -\omega_{\nu \sigma} x^\sigma \left(2\frac{\partial \cal L}{\partial(\partial_\mu \phi)}\partial^\nu \phi - \cal L \eta^{\mu \nu} \right)$$
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. Apr 11, 2014

### samalkhaiat

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

Last edited: Apr 11, 2014
3. Apr 11, 2014

### spookyfish

4. Apr 11, 2014

### samalkhaiat

Very consistent. The field in (1.51) is a scalar field.
$$\bar{ \phi } ( \bar{ x } ) = \phi ( x ) , \ \ \Rightarrow \ \ \delta^{*} \phi = 0 .$$
Use $x = \Lambda^{ - 1 } \bar{ x }$ and rename the coordinates $x$, (i.e. drop the bar from $\bar{ x }$), you get $\bar{ \phi } ( x ) = \phi ( \Lambda^{ - 1 } x )$. Now, if you expand this, as they did, you find that $\delta \phi = - \delta x^{ \mu } \partial_{ \mu } \phi$

5. Apr 11, 2014

### spookyfish

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

6. Apr 11, 2014

### samalkhaiat

Look at it again. In the article, he uses the current which does not involve T, this is why he used $\delta \phi$. If you want to start with the form of $J^{ \mu }$ which involves the tensor $T^{ \mu \nu }$, then you must use $\delta^{ * }\phi$, as I explained in my first post.