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Lagrangian for a particle in a uniform magnetic field

  1. Jul 1, 2012 #1
    Any advice on deriving the lagrangian for a particle in a magnetic field?

    [itex]L=\frac{1}{2}mv^{2} + \frac{q}{c}\mathbf{v}\mathbf{.} \mathbf{A}[/itex]

    I've been searching through Griffith's, Jackson, and google to no avail.
    Can we start from the lorentz force and work backwards?

    Thanks for the help.
     
  2. jcsd
  3. Jul 5, 2012 #2

    gabbagabbahey

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    The closest I've seen to a strict derivation of the E-M Lagrangian is to use a little vector calculus to rewrite the Lorentz Force Law (in Gaussian units) as

    [tex]\mathbf{F} = -\nabla\left[qV - \frac{q}{c}\mathbf{v}\cdot\mathbf{A}\right] -q\frac{d\mathbf{A}}{dt}[/tex]

    And from there, postulate a generalized potential of [itex]U(q,\dot{q},t) = qV - \frac{q}{c}\mathbf{v}\cdot\mathbf{A}[/itex] and then (and this is very important!) check that the Euler-Lagrange equations result in the correct force law, regardless of your gauge choice.
     
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