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Feynman slash identity 
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#1
Nov2309, 03:49 PM

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P: 1,275

1. The problem statement, all variables and given/known data
I am trying to prove that [tex]\displaystyle{\not} a \displaystyle{\not} b + \displaystyle{\not} b \displaystyle{\not} a = 2a\cdot b[/tex] using the relation [tex]\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}[/tex] 2. Relevant equations 3. The attempt at a solution If I work backwards, [tex] 2a\cdot b = 2a_{\mu} g^{\mu \nu} b_{\nu} = a_{\mu}(\gamma^{\mu}\gamma^{\nu})b_{\nu} + a_{\mu}(\gamma^{\nu}\gamma^{\mu})b_{\nu}[/tex] The first term is [tex]\displaystyle{\not} a \displaystyle{\not} b[/tex] but the second term doesn't seem to look like [tex]\displaystyle{\not} b \displaystyle{\not} a[/tex]. Am I missing something here? 


#2
Nov2309, 04:31 PM

Sci Advisor
P: 1,588

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



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