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Dirac equation and gamma factor

  1. Apr 12, 2012 #1
    I am reading about Dirac's equation for relativistic electron in Feynman's book "Quantum Electrodynamics". Factor [itex]\gamma =(1-v^2)^{-1/2}[/itex] (units c=1) is almost always presented in non quantum calculations of Special relativity. But in his book I also find it on page 44 in lecture "Relativistic invariance", when he shows Lorentz transformations of his matrix [itex]\gamma_{x,y,z,t}[/itex]
    1. But, I still ever do not understand, what is quantum analog of equation [itex]p=\gamma mv[/itex].
    2. Is this analog [itex]\alpha p[/itex], or only [itex]p[/itex]. [itex]\alpha[/itex] is matrix.
    3. I also wish examples which uses the above [itex]\gamma =(1-v^2)^{-1/2}[/itex] in calculations?
  2. jcsd
  3. Apr 13, 2012 #2


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    exponent137, There are at least three answers, depending on how complicated you want to get. If you only want the gamma factor, it's easy, the equation to use is the relativistic expression for energy, E = γmc2, implying γ = E/mc2.

    If you want to know what happens to p = γmv in the framework of the Dirac equation, it's a little more complicated. The difficulty is in how to define v, and requires a discussion of "zitterbewegung" and the "Newton-Wigner position operator" (which is usually denoted by X to distinguish it from the normal position operator x.)

    The velocity is dX/dt, and I quote the result from a big book on relativistic quantum field theory: "Within the manifold of positive energy solutions the time derivative of X is the operator pc2/E, which can be identified with the velocity of the particle."

    Sure enough, using the special relativity formulas, pc2/E = (γmvc2)/(γmc2) = v.

    If you really want the full expression, here it is! dX/dt = (pc2/E2) (βmc2 + α·p c), where α, β are the Dirac matrices.
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