Equation 2.14 in Srednicki

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  • #1
nrqed
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Main Question or Discussion Point

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
 

Answers and Replies

  • #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.
 
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  • #3
nrqed
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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.
Thank you very much StatusX for all your help, it is very much appreciated.

Makes perfect sense !!!

Thanks again for your help
 
  • #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] ?
 
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  • #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?
 
  • #6
Fredrik
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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] ?
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].
 
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  • #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.
 
  • #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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