Proving Feynman Slash Identity: 2a\cdot b

[nicksauce]

Homework Statement

I am trying to prove that \displaystyle{\not} a \displaystyle{\not} b + \displaystyle{\not} b \displaystyle{\not} a = 2a\cdot b using the relation \{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}

Homework Equations

The Attempt at a Solution

If I work backwards,
<br /> 2a\cdot b = 2a_{\mu} g^{\mu \nu} b_{\nu}<br /> = a_{\mu}(\gamma^{\mu}\gamma^{\nu})b_{\nu} + a_{\mu}(\gamma^{\nu}\gamma^{\mu})b_{\nu}

The first term is \displaystyle{\not} a \displaystyle{\not} b but the second term doesn't seem to look like \displaystyle{\not} b \displaystyle{\not} a. Am I missing something here?

 

[Ben Niehoff]

You're missing the fact that a_\mu and b_\nu are ordinary numbers, and so commute with everything.