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

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.