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.

This Wikipedia page explains that for each of the four main gamma matrices $\gamma^{\mu}$, you can find the covariant matrices $\gamma_{\mu}$ with the equation $\gamma_{\mu} = \eta_{\mu\nu}\gamma^{\mu}$. But that formula doesn't make any sense for $\gamma^5$ because $\eta_{\mu\nu}$ does not have that many indices. So what is $\gamma_5$?

share|improve this question
add comment

2 Answers

The 'five' in $\gamma_5$ is not a Lorentz index, so it doesn't make sense to lower or raise it. It can be defined in different ways, one convention is: $$\gamma_5 = \frac{i}{24}\epsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} = \frac{i}{24}\epsilon^{\mu\nu\rho\sigma}\gamma_{\mu}\gamma_{\nu}\gamma_{\rho}\gamma_{\sigma}$$, where epsilon is the totally antisymmetric tensor.

share|improve this answer
2  
Well, $\gamma_5$ may be a Lorentz index. All the matrices including $\gamma_5$ can be used to produce the generators of the Lorentz group in 5 i.e. 4+1 spacetime dimensions because the give gamma matrices still anticommute with each other, and square to $\pm 1$. Of course that we may define $\gamma_5$ and $\gamma^5$ to be either the same thing or the same thing with the opposite sign, to suit any conventions. –  Luboš Motl May 31 '13 at 17:44
add comment

$\gamma^5$ would be a 'measurer of parity', which is sensible to changes in orientation due to coordinate transformations. $\gamma^5 := i\gamma^0\gamma^1\gamma^2\gamma^3 $

if you try defining $ \gamma_5 := i\gamma_0\gamma_1\gamma_2\gamma_3 $ you probably end up with $\gamma^5=-\gamma_5$ because $\det \eta=-1$

(I haven't done the calculations, so I might be missing an extra sign or the like)

share|improve this answer
add comment

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.