# I Lorentz transformation and its Noether current

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1. Jul 20, 2017

### Ken Gallock

Hi.
On page16 of David Tong's lecture note(http://www.damtp.cam.ac.uk/user/tong/qft.html), there is a topic about Noether current and Lorentz transformation.
I want to derive $\delta \mathcal{L}$, but during my calculation, I encountered this:
\begin{align}
\delta \mathcal{L}&=\dfrac{\partial \mathcal{L}}{\partial \phi}\delta \phi+\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\mu (\delta \phi)\\
&=\dfrac{\partial \mathcal{L}}{\partial \phi}(-\omega^\rho_{~\sigma}x^\sigma \partial_\rho \phi)+
\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\mu
(-\omega^\rho_{~\sigma}x^\sigma \partial_\rho \phi).
\end{align}
The second term,
$$\partial_\mu(-\omega^\rho_{~\sigma}x^\sigma \partial_\rho \phi)$$
is a troubling term for me. Since there is $x^\sigma$ and $\partial_\mu$, I thought I have to derivate $x^\sigma$ like $\partial_\mu x^\sigma$. But if I do so, it doesn't match with the result in the textbook.
Am I supposed not to derivate $x$? or am I missing something?

Thanks.

2. Jul 20, 2017

### Orodruin

Staff Emeritus
Exactly which equation are you trying to reproduce?

3. Jul 20, 2017

### Ken Gallock

I want eq(1.53):
$$\delta \mathcal{L}=-\omega^\mu_{~\nu}x^\nu\partial_\mu \mathcal{L}.$$

4. Jul 20, 2017

### Orodruin

Staff Emeritus
This is a direct consequence of the Lagrangian being a scalar field.

5. Jul 20, 2017

### Ken Gallock

Is it different when I'm dealing with spinor fields?

6. Jul 20, 2017

### samalkhaiat

You get the same result, if you differentiate and remember to use the fact that your field is a scalar and $\omega_{\mu\nu} = -\omega_{\nu\mu}$:
$$-\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi} \left(\omega^{\rho}{}_{\sigma} x^{\sigma} \ \partial_{\rho}\phi \right) + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi ) } \left( \omega^{\rho}{}_{\sigma}x^{\sigma} \ \partial_{\rho} \partial_{\mu} \phi + \partial^{\rho}\phi \ \omega_{\rho \mu}\right) .$$ Arrange the terms to get $$\delta \mathcal{L} = - \omega^{\rho}{}_{\sigma} x^{\sigma} \ \partial_{\rho}\mathcal{L} - \omega_{\rho \mu} \ \partial^{\rho} \phi \ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi )} .$$ Now, for a scalar field, you have $$\omega_{\rho \mu} \ \partial^{\rho} \phi \ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi )} \propto \omega_{\rho \mu} \ \partial^{\rho} \phi \ \partial^{\mu}\phi = 0 ,$$ because of the $\omega_{\rho \mu} = - \omega_{\mu\rho}$. So, you are left with $$\delta \mathcal{L} = - \omega^{\rho}{}_{\sigma} x^{\sigma} \ \partial_{\rho}\mathcal{L} = - \omega^{\rho}{}_{\sigma} \ \partial_{\rho} \left( x^{\sigma} \mathcal{L}\right) .$$

The Lagrangian will still be scalar, but the transformation of the (spinor) field will no longer be $\delta \psi = -\omega^{\mu}{}_{\nu}x^{\nu}\partial_{\mu}\psi .$ You have to account for the spin of the field by including an appropriate spin matrix $\Sigma$: $$\delta \psi_{a} = -\omega^{\mu}{}_{\nu}x^{\nu}\ \partial_{\mu}\psi_{a} + \omega_{\rho \sigma} (\Sigma^{\rho\sigma})_{a}{}^{c} \ \psi_{c} .$$

Last edited: Jul 20, 2017
7. Jul 21, 2017

### Ken Gallock

Thanks!
I got the same result!