Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm stuck in following calcualtion from sredniki's QFT book.(Its actually in the solution manual)

How can i get from

$$\delta\omega_{\rho\sigma}(g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma}) $$

to

$$ \frac{1}{2}\delta\omega_{\rho\sigma}(g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma} -g^{\rho\mu}M^{\sigma\nu} + g^{\sigma\nu}M^{\mu\rho}) $$

share|improve this question
    
Related: physics.stackexchange.com/q/28535/2451 –  Qmechanic May 21 '13 at 21:01
    
That question and the answers given are a bit formal. Thanks for noticing it. I'm going through it and also try to understand it. –  Aftnix May 21 '13 at 21:04

1 Answer 1

up vote 1 down vote accepted

Since $\delta \omega_{\rho\sigma}$ is antisymmetric, the part of what's inside the parentheses symmetrized over $\rho$ and $\sigma$ will vanish because if you contract antisymmetric indices with symmetric ones, you get zero. So you can anti-symmetrize over that pair of indices inside the parentheses without changing anything, which is precisely what's inside the parentheses in the second line:

\begin{align} \delta \omega_{\rho \sigma} (g^{\sigma \mu} M^{\rho \nu} - g^{\rho \mu} M^{\sigma \nu}) &= \delta \omega_{\rho \sigma} \bigg[ \frac{1}{2} (g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma} -g^{\rho\mu}M^{\sigma\nu} + g^{\sigma\nu}M^{\mu\rho}) \\ & \qquad \qquad +\frac{1}{2} (g^{\sigma\mu}M^{\rho\nu} - g^{\rho\nu}M^{\mu\sigma} + g^{\rho\mu}M^{\sigma\nu} - g^{\sigma\nu}M^{\mu\rho}) \bigg] \end{align}

Here, I've just split the $gM$ factor into parts antisymmetric and symmetric in $\rho \sigma$. (You can check that the terms inside the brackets cancel directly to give the same thing as the left-hand side.) It's an easy exercise to check that for a general symmetric tensor $T^{\rho\sigma}$, you have $\delta \omega_{\rho \sigma}T^{\rho\sigma}=0$. Adding more indices to $T$ doesn't change that, so the second term inside the brackets just drops out, and you get your result.

share|improve this answer
    
Thanks for the answer. Yes $\delta\omega$ is anti-symmetric. –  Aftnix May 21 '13 at 21:26
    
Sure thing. I've added a little more explanation. Don't forget to accept it if this satisfies you. :) –  Mike May 21 '13 at 21:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.