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Commutation relation

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Given a Poincaré transformation, Lorentz+translation, I have to find the Poincaré generators in the scalar field representation and then prove that the commutation relations.

    I've done the first part but I can't prove the commutation relations.

    2. Relevant equations

    [tex]P_{\mu}=i\partial_{\mu}[/tex]

    [tex]M_{\mu\nu}=i\left(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\right)[/tex]


    3. The attempt at a solution

    For example for the mixed commutator after doing some straight-forward algebra

    [tex]\left[M_{\mu\nu},P_{\rho}\right]=i^{2}\left[\left(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\right),\partial_{\rho}\right]=\left[\partial_{\rho},\left(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\right)\right]=\partial_{\rho}x_{\mu}\partial_{\nu}-\partial_{\rho}x_{\nu}\partial_{\mu}[/tex]

    Now if we recall the definition of the generator of the translations [tex]
    P_{\mu}=i\partial_{\mu}\implies\partial_{\mu}=\frac{P_{\mu}}{i}=-iP_{\mu}
    [/tex]
    [tex]\left[M_{\mu\nu},P_{\rho}\right]=\partial_{\rho}x_{\mu}\partial_{\nu}-\partial_{\rho}x_{\nu}\partial_{\mu}=\partial_{\rho}x_{\mu}\left(-iP_{\nu}\right)-\partial_{\rho}x_{\nu}\left(-iP_{\mu}\right)=i\left(\partial_{\rho}x_{\nu}P_{\mu}-\partial_{\rho}x_{\mu}P_{\nu}\right)[/tex]

    I know the results of the commutators from the Poincaré algebra so [tex]
    \partial_{\rho}x_{\mu}=g_{\rho\mu}[/tex] but I don't understand it. I thought that
    [tex]
    \partial_{\rho}x_{\mu}=\delta_{\rho\mu}[/tex]

    Any help in order to prove the penultimate relation ? Because I don't know how to go from
    [tex]
    \partial_{\rho}x_{\mu}=\delta_{\rho\mu}[/tex] to
    [tex]
    \partial_{\rho}x_{\mu}=g_{\rho\mu}[/tex]

    Thanks
     
  2. jcsd
  3. Mar 15, 2015 #2

    Orodruin

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    This is not true. What is true is ##\partial_\rho x^\mu = \delta_\rho^\mu##, which is not the same thing.
     
  4. Mar 15, 2015 #3
    @Orodruin Can you develop it a little more pls ? Because I have problems with this little quibbling of index notation and such
     
  5. Mar 15, 2015 #4

    Orodruin

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    The coordinate with covariant index does not fulfil the relation you quoted. It only holds for a contravariant index and when you lower it using the metric you get the relation you were looking for. Placing it on a more appropriate level is impossible if you do not specify exactly which part you are having problems with and what your current level of understanding is.
     
  6. Mar 15, 2015 #5
    So basically [tex]\partial_{\rho}x_{\nu}=\partial_{\rho}g_{v\alpha}x^{\alpha}=g_{v\alpha}\partial_{\rho}x^{\alpha}=g_{v\alpha}\delta_{\rho}^{\alpha}=g_{\nu\rho}=g_{\rho\nu}[/tex]

    Any mistake ?
     
  7. Mar 15, 2015 #6

    Orodruin

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    Just beware that things will not work out like this once you start looking at GR and the metric becomes coordinate dependent ... The assumption here is that you are working with the Minkowski metric.
     
  8. Mar 15, 2015 #7
    Yes, indeed I'm using the Minkowski metric. The part where I have problems, as I just realized right now, is the difference between [tex]\partial_{\alpha}x_{\nu}[/tex]
    [tex]\partial_{\rho}x^{\alpha}[/tex]
    [tex]\partial^{\rho}x_{\alpha}[/tex]
    [tex]\partial^{\rho}x^{\alpha}[/tex]

    If you know any book for dummies like me where this topic is covered I would appreciate that. Thank you for your time
     
  9. Mar 15, 2015 #8

    Orodruin

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    The only thing to realise is essentially that ##x^\mu## is different from ##x_\mu## and that
    $$
    \partial_\mu x^\nu \equiv \frac{\partial x^\nu}{\partial x^\mu}$$
    and there really is not much more to it. You can work it all out from there.
     
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