# Equation 2.16 in Srednicki

1. Feb 26, 2010

### LAHLH

Hi,

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

$$U(\Lambda)^{-1} M^{\mu\nu} U(\Lambda)=\Lambda^\mu{}_{\rho}\Lambda^\nu{}_{\sigma} M^{\rho\sigma}$$

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

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

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

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

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

Thanks for any help

2. Feb 26, 2010

### LAHLH

I have no idea how the metric has found its way into the generator commutation relation, v confused

3. Feb 26, 2010

### George Jones

Staff Emeritus
4. Feb 26, 2010

### LAHLH

Ah thanks so much, I did do a search for Srednicki but didn't see that thread, thanks again.