Understanding the Equation 2.14 and its Application in Srednicki's Theory

[nrqed]

He defines

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

Then he considers

U(\Lambda^{-1} \Lambda' \Lambda)

with \Lambda' = 1 + \delta \omega'
He then says that

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


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

thanks

 

[StatusX]

We have:

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

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

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

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

{\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}

Plugging into 2.12 gives your answer.

 

[nrqed]

We have:

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

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

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

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

{\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}

Plugging into 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

 

[malawi_glenn]

how is

<br /> {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho <br /> equal to <br /> {\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho} <br /> ?

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

 

[malawi_glenn]

I mean, is it an "argument by analogy" or is there a more profound way to find these things out?

 

[Fredrik]

how is

<br /> {\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho <br /> equal to <br /> {\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho} <br /> ?

It isn't. Let's write \Lambda^{-1}\Lambda&#039;\Lambda=\Lambda&#039;&#039;. You're trying to find the first order term of U(\Lambda&#039;&#039;)=U(1+\delta\omega&#039;&#039;)=I+\frac i 2 \delta\omega&#039;&#039;_{\mu\nu}M^{\mu\nu}.

You already know that (\delta\omega&#039;&#039;)^\mu{}_\sigma=\Lambda_\nu{}^\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho. That implies that (\delta\omega&#039;&#039;)_\mu{}_\sigma=\Lambda_\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho=\Lambda^\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega_\nu{}_\rho.

 

[George Jones]

Everything is OK.

U \left( \Lambda^{-1} \Lambda &#039; \Lambda \right) = U \left( 1 + \delta \Omega \right) = 1 + \frac{i}{2 \hbar} \delta \Omega_{\mu \nu} M^{\mu \nu},

where

\delta \Omega = \Lambda^{-1} \delta \omega \Lambda.

Consequently,

{\delta \Omega^\mu}_\sigma = {\Lambda_\nu}^\mu \delta {\omega^\nu}_\rho {\Lambda^\rho}_\sigma

and

\delta \Omega_{\mu \sigma} = {\Lambda^\nu}_\mu \delta \omega_{\nu \rho} {\Lambda^\rho}_\sigma .

I hope I haven't screwed up the indicies too much.

 

[malawi_glenn]

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