Dirac Gamma matrices question

  • #1

Homework Statement



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

[itex]\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}*1[/itex]


Homework Equations



[itex]g^{\mu\nu}\gamma_{\nu}=\gamma^{\mu}[/itex] and [itex]g_{\mu\nu}\gamma^{\nu}=\gamma_{\mu}[/itex]

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

[itex]\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 }[/itex]

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

Answers and Replies

  • #2
dextercioby
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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(+,-,-,-)).
 
  • #3
George Jones
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Given that [itex]\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}*1[/itex]
Another way to do it is to multiply both sides of this equation by [itex]g_{\alpha \mu} g _{\beta \nu}[/itex].
 
  • #4
Thank you George Jones! That did the trick nicely. Using [itex]g_{\mu\alpha}g^{\alpha\nu}=\delta^{\mu}_{\nu}[/itex] 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?
 

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