# Equation 2.14 in Srednicki

1. May 5, 2009

### 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

2. May 5, 2009

### 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):

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

Last edited: May 5, 2009
3. May 5, 2009

### nrqed

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

Makes perfect sense !!!

4. Jul 6, 2009

### malawi_glenn

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

Last edited: Jul 6, 2009
5. Jul 6, 2009

### malawi_glenn

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

6. Jul 6, 2009

### Fredrik

Staff Emeritus
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 $(\delta\omega'')_\mu{}_\sigma=\Lambda_\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega^\nu{}_\rho=\Lambda^\nu{}_\mu\Lambda^\rho{}_\sigma\delta\omega_\nu{}_\rho$.

Last edited: Jul 6, 2009
7. Jul 6, 2009

### George Jones

Staff Emeritus
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.

8. Jul 6, 2009

### malawi_glenn

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