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I Lorentz transformation and its Noether current

  1. Jul 20, 2017 #1
    Hi.
    I'd like to ask about the calculation of Noether current.
    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. jcsd
  3. Jul 20, 2017 #2

    Orodruin

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    Exactly which equation are you trying to reproduce?
     
  4. Jul 20, 2017 #3
    I want eq(1.53):
    $$
    \delta \mathcal{L}=-\omega^\mu_{~\nu}x^\nu\partial_\mu \mathcal{L}.
    $$
     
  5. Jul 20, 2017 #4

    Orodruin

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    This is a direct consequence of the Lagrangian being a scalar field.
     
  6. Jul 20, 2017 #5
    Is it different when I'm dealing with spinor fields?
     
  7. Jul 20, 2017 #6

    samalkhaiat

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

    The Lagrangian will still be scalar, but the transformation of the (spinor) field will no longer be [itex]\delta \psi = -\omega^{\mu}{}_{\nu}x^{\nu}\partial_{\mu}\psi .[/itex] You have to account for the spin of the field by including an appropriate spin matrix [itex]\Sigma[/itex]: [tex] \delta \psi_{a} = -\omega^{\mu}{}_{\nu}x^{\nu}\ \partial_{\mu}\psi_{a} + \omega_{\rho \sigma} (\Sigma^{\rho\sigma})_{a}{}^{c} \ \psi_{c} .[/tex]
     
    Last edited: Jul 20, 2017
  8. Jul 21, 2017 #7
    Thanks!
    I got the same result!
     
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