How Do Dirac Gamma Matrices Satisfy Their Anticommutation Relations?

[McLaren Rulez]

Homework Statement

Given that $\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}*1$ where $1$ is the identity matrix and the $\gamma$ are the gamma matrices from the Dirac equation, prove that:

$\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}*1$

Homework Equations

$g^{\mu\nu}\gamma_{\nu}=\gamma^{\mu}$ and $g_{\mu\nu}\gamma^{\nu}=\gamma_{\mu}$

The Attempt at a Solution

I'm not sure what to start with. I tried expressing the terms of the relation to be proved as follows

$\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=g_{\mu\alpha}\gamma^{\alpha}g_{\nu \beta}\gamma^{\beta}+ g_{\nu\beta}\gamma^{\beta}g_{\mu\alpha}\gamma^{ \alpha }$

but that isn't going anywhere. So how do I approach this?

 

[dextercioby]

Hmm, just replace mu and nu with their possible values and see what you get. Don't forget that the metric tensor is diagonal (probably diag(+,-,-,-)).

 

[George Jones]

Given that $\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}*1$

Another way to do it is to multiply both sides of this equation by $g_{\alpha \mu} g _{\beta \nu}$.

 

[McLaren Rulez]

Thank you George Jones! That did the trick nicely. Using $g_{\mu\alpha}g^{\alpha\nu}=\delta^{\mu}_{\nu}$ the result follows easily.

dextercioby, thank you for replying. I think your method also works but I must assume the metric is diag(1, -1 , -1, -1) which is not always the case right?