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4-velocity, the mass shell and potential energy

  1. Apr 6, 2008 #1
    4-velocity, the "mass shell" and potential energy

    For a free relativistic particle we have the condition


    putting [tex]p^0 = c^{-1}E[/tex] we may write this as

    [tex]p^\mu p_\mu = m^2c^2[/tex]

    And since [tex]p^\mu = m\frac{dx^\mu}{d\tau} = mu^\mu[/tex] we may express the same condition as

    [tex]u^\mu u_\mu = c^2[/tex]

    At least, this is true for a free particle. Is this condition affected by the presence of a potential?

    In the presence of a potential [tex]A^\mu[/tex] the momentum-energy relation becomes

    [tex](p^\mu-A^\mu)(p_\mu-A_\mu) = m^2c^2[/tex]

    doesn't it? But then in that case, we are talking about the canonical momentum and we no longer have [tex]p^\mu = mu^\mu[/tex]. But is [tex]p^\mu = mu^\mu+A^\mu[/tex]? So that we still have [tex]u^\mu u_\mu = c^2[/tex]? Can't find this in my texts and just wanted someone to confirm I understand it correctly.
  2. jcsd
  3. Apr 6, 2008 #2
    Always, [tex]u^\mu u_\mu = c^2[/tex]. Mechanical momentum is m times that. Canonical momentum is determined from what the Lagrangian of the system is.
  4. Apr 6, 2008 #3
    If you look in Jackson's text Classical Electrodynamics you will find something very close.

  5. Apr 8, 2008 #4
    Thanks, guys.
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