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Equation 2.14 in Srednicki

  1. May 5, 2009 #1

    nrqed

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    He defines

    [tex] U(1 + \delta \omega) \approx 1 + \frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu} [/tex]

    Then he considers

    [tex] U(\Lambda^{-1} \Lambda' \Lambda) [/tex]

    with [itex] \Lambda' = 1 + \delta \omega' [/itex]
    He then says that

    [tex] U(\Lambda^{-1} \Lambda' \Lambda) \approx \delta \omega_{\mu \nu} \Lambda^{\mu}_{\, \, \rho} \Lambda^{\nu}_{\, \, \sigma} M^{\rho \sigma} [/tex]


    I don't see why this is true. (by the way, I assume the [itex] \omega [/itex] is actually meant to be [itex] \omega'[/itex] ). I don't see how the [itex] \Lambda^{-1} \Lambda [/itex] turned into the expression on the right.

    thanks
     
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  3. May 5, 2009 #2

    StatusX

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    We have:

    [tex] \Lambda^{-1} \Lambda' \Lambda = {(\Lambda^{-1})^\mu}_\nu {\Lambda'^\nu}_\rho {\Lambda^\rho}_\sigma [/tex]

    Now, from 2.5 we know [itex]{(\Lambda^{-1})^\mu}_\nu = {\Lambda_\nu}^{\mu} [/itex], so this becomes (expanding [itex]\Lambda'[/itex]):

    [tex] = {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma \left( {\delta^\nu}_\rho + {\delta \omega^\nu}_\rho \right) [/tex]

    So that the linear term is (with both indices down):

    [tex] {\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho} [/tex]

    Plugging in to 2.12 gives your answer.
     
    Last edited: May 5, 2009
  4. May 5, 2009 #3

    nrqed

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    Thank you very much StatusX for all your help, it is very much appreciated.

    Makes perfect sense !!!

    Thanks again for your help
     
  5. Jul 6, 2009 #4

    malawi_glenn

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    how is

    [tex]
    {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho
    [/tex] equal to [tex]
    {\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}
    [/tex] ?

    The index structure inside the U should be [tex]{}^a{}_b[/tex] and outside U nothing (full contracted), so what we can write [tex]
    {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho
    [/tex] equal to as is:
    [tex]\Lambda^{\nu \mu}{\Lambda^\rho}_\sigma {\delta \omega _\nu}_\rho [/tex] and then what? how can one conclude that the linear term with "both indices down" is [tex]
    {\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}
    [/tex] ?
     
    Last edited: Jul 6, 2009
  6. Jul 6, 2009 #5

    malawi_glenn

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    I mean, is it an "argument by analogy" or is there a more profound way to find these things out?
     
  7. Jul 6, 2009 #6

    Fredrik

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    It isn't. Let's write [itex]\Lambda^{-1}\Lambda'\Lambda=\Lambda''[/itex]. You're trying to find the first order term of [itex]U(\Lambda'')=U(1+\delta\omega'')=I+\frac i 2 \delta\omega''_{\mu\nu}M^{\mu\nu}[/itex].

    You already know that [itex](\delta\omega'')^\mu{}_\sigma=\Lambda_\nu{}^\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho[/itex]. That implies that [itex](\delta\omega'')_\mu{}_\sigma=\Lambda_\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho=\Lambda^\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega_\nu{}_\rho[/itex].
     
    Last edited: Jul 6, 2009
  8. Jul 6, 2009 #7

    George Jones

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    Everything is OK.

    [tex]U \left( \Lambda^{-1} \Lambda ' \Lambda \right) = U \left( 1 + \delta \Omega \right) = 1 + \frac{i}{2 \hbar} \delta \Omega_{\mu \nu} M^{\mu \nu},[/tex]

    where

    [tex]\delta \Omega = \Lambda^{-1} \delta \omega \Lambda.[/tex]

    Consequently,

    [tex]{\delta \Omega^\mu}_\sigma = {\Lambda_\nu}^\mu \delta {\omega^\nu}_\rho {\Lambda^\rho}_\sigma[/tex]

    and

    [tex]\delta \Omega_{\mu \sigma} = {\Lambda^\nu}_\mu \delta \omega_{\nu \rho} {\Lambda^\rho}_\sigma .[/tex]

    I hope I haven't screwed up the indicies too much.
     
  9. Jul 6, 2009 #8

    malawi_glenn

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    Thanx Fredrik and George, I had those two things in mind, you confirmed them :-)
     
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