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May 2, 2014 at 17:30 comment added Valter Moretti If you fix $\mu$ and $\nu$ $g^{\mu\nu}$ is a number, so it commutes with every matrix.
May 2, 2014 at 15:36 history tweeted twitter.com/#!/StackPhysics/status/462254362409304064
May 2, 2014 at 15:33 comment added camzor00 I don't see it, even with the hint..
May 2, 2014 at 14:59 comment added Qmechanic Hint to the question (v2): The commutation relation has an (implicit) identity matrix on the rhs: $\{\gamma^\mu,\gamma^\nu\} ~=~ 2g^{\mu\nu}~{\bf1}_{4\times 4}$.
May 2, 2014 at 14:50 history edited Qmechanic CC BY-SA 3.0
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May 2, 2014 at 14:45 history asked camzor00 CC BY-SA 3.0