# Equation 2.14 in Srednicki

Science Advisor
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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

## Answers and Replies

StatusX
Homework Helper
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 in to 2.12 gives your answer.

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Science Advisor
Homework Helper
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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 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

malawi_glenn
Science Advisor
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how is

$${\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho$$ equal to $${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$ ?

The index structure inside the U should be $${}^a{}_b$$ and outside U nothing (full contracted), so what we can write $${\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho$$ 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 $${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$ ?

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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?

Fredrik
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how is

$${\Lambda_\nu}^\mu {\Lambda^\rho}_\sigma {\delta \omega^\nu}_\rho$$ equal to $${\Lambda^\nu}_\mu {\Lambda^\rho}_\sigma \delta \omega_{\nu\rho}$$ ?
It isn't. Let's write $\Lambda^{-1}\Lambda'\Lambda=\Lambda''$. You're trying to find the first order term of $U(\Lambda'')=U(1+\delta\omega'')=I+\frac i 2 \delta\omega''_{\mu\nu}M^{\mu\nu}$.

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

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George Jones
Staff Emeritus
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Everything is OK.

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

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