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  Trace of six gamma matrices

+ 3 like - 0 dislike

I need to calculate this expression:
Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5}) $$ 
I know that I can express this as:
+g^{\nu\rho}\epsilon^{\mu\sigma\alpha\beta}-g^{\nu\sigma}\epsilon^{\mu\rho\alpha\beta}+g^{\nu\alpha}\epsilon^{\mu\rho\sigma\beta}\\&-&g^{\nu\beta}\epsilon^{\mu\rho\sigma\alpha}+g^{\rho\sigma}\epsilon^{\mu\nu\alpha\beta}-g^{\rho\alpha}\epsilon^{\mu\nu\sigma\beta}+g^{\rho\beta}\epsilon^{\mu\nu\sigma\alpha}\\&+&g^{\sigma\alpha}\epsilon^{\mu\nu\rho\beta}-g^{\sigma\beta}\epsilon^{\mu\nu\rho\alpha}+g^{\alpha\beta}\epsilon^{\mu\nu\rho\sigma}) \end{eqnarray}

So, some of this terms are the same and some vanish. My question is how to show that.
I know that:

\begin{eqnarray}Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5})=&-&4i(g^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}-g^{\mu\rho}\epsilon^{\nu\sigma\alpha\beta}+g^{\rho\nu}\epsilon^{\mu\sigma\alpha\beta}-g^{\alpha\beta}\epsilon^{\sigma\mu\nu\rho}\\&+&g^{\sigma\beta}\epsilon^{\alpha\mu\nu\rho}-g^{\sigma\alpha}\epsilon^{\beta\mu\nu\rho}) \end{eqnarray}
So only six terms survive, but how?

asked Nov 4, 2015 in Theoretical Physics by VIS123 (15 points) [ revision history ]
edited Nov 4, 2015 by VIS123

I don't have time to look into the details, but have you tried to use the identity $g^{\mu[\sigma}\epsilon^{\nu\rho\alpha\beta]}=0$? Since we are in 4 dimension, the antisymmetrisation of 5 indices is always 0.

Thanks @Jia Yiyang. I did that yesterday and got my result right.

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