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

How can i get from

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


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

  • $\begingroup$ Related: physics.stackexchange.com/q/28535/2451 $\endgroup$ – Qmechanic May 21 '13 at 21:01
  • $\begingroup$ 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. $\endgroup$ – Aftnix May 21 '13 at 21:04

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.

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

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