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Confusion with the Gordon identity

  1. Feb 17, 2016 #1
    For the Gordon identity

    $$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}u_{s}(\textbf{p}) = \bar{u}_{s'}(\textbf{p}')[(p'+p)^{\mu} -2iS^{\mu\nu} (p'-p)_{\nu}]u_{s}(\textbf{p}) $$

    If I plug in $\mu$=5, what exactly does the corresponding $(p'+p)^{5}$ represent?
    4 vectors can only have 4 components so is this just an exponential?

  2. jcsd
  3. Feb 17, 2016 #2
    That does not make sense. Why would you plug in ##\mu = 5##, and what would that be supposed to mean?
  4. Feb 17, 2016 #3


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    Gold Member

    It is a 4vector...
    [itex] ( p' + p ) ^\mu = p'^\mu + p^\mu [/itex]...
    the first notation is shorter... :smile:

    [itex]\mu[/itex] is an index taking values 0,1,2,3...

    Don't get confused with the [itex]\gamma^5[/itex]... it is not [itex]\gamma^\mu[/itex] with [itex]\mu=5[/itex], but it's a different object...
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