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Equation 2.16 in Srednicki

  1. Feb 26, 2010 #1
    Hi,

    I'm having a little troubling reaching this equation. I'm starting with 2.14 which is:

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

    Now letting [tex] \Lambda=1+\delta\omega [/tex] and using [tex] U(1+\delta\omega)=I+\frac{i}{2\hbar} \delta\omega_{\mu\nu}M^{\mu\nu} [/tex], I get:

    [tex] U(1+\delta\omega )^{-1} M^{\mu\nu} U(1+\delta\omega)=\Lambda^\mu{}_{\rho}\Lambda^\nu{}_{\sigma} M^{\rho\sigma} [/tex]

    [tex] => (I-\frac{i}{2\hbar} \delta\omega_{\alpha\beta}M^{\alpha\beta}) M^{\mu\nu}( I+\frac{i}{2\hbar} \delta\omega_{\xi\chi}M^{\xi\chi})=\Lambda^\mu{}_{\rho}\Lambda^\nu{}_{\sigma} M^{\rho\sigma} [/tex]

    [tex] => M^{\mu\nu}+\frac{i}{2\hbar} \delta\omega_{\xi\chi}M^{\mu\nu}M^{\xi\chi}-\frac{i}{2\hbar} \delta\omega_{\alpha\beta}M^{\alpha\beta}M^{\mu\nu}=\Lambda^\mu{}_{\rho}\Lambda^\nu{}_{\sigma} M^{\rho\sigma} [/tex]

    I'm not really sure where to go from here, I guess I can't simply say [tex] \Lambda^\mu{}_{\rho}\Lambda^\nu{}_{\sigma} M^{\rho\sigma}=M^{\mu\nu} [/tex] and cancel this from each side?
    Not sure how else I could get some that had [tex] \delta\omega [/tex] in every term otherwise, so I can equate the antisymmetric parts of there coefficients as Srednicki suggests.

    Thanks for any help
     
  2. jcsd
  3. Feb 26, 2010 #2
    I have no idea how the metric has found its way into the generator commutation relation, v confused
     
  4. Feb 26, 2010 #3

    George Jones

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  5. Feb 26, 2010 #4
    Ah thanks so much, I did do a search for Srednicki but didn't see that thread, thanks again.
     
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