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Prove that a given Lagrangian is not T-V (EM field)

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a particle of mass m and electric charge e subject to a uniform electromagnetic field (E(x,t),B(x,t)). We must remember that the force they exert is given by:

    [tex]F = cE(x,t) + ex' \times B(x,t)[/tex]

    A principle of action that represents such particle subject to the Lorentz force is given by:

    [tex]I = ∫(\frac{m}{2}x'*x' + eA(x,t)x' + e{\phi}(x,t)dt[/tex]

    Prove that this action doesn't have a Lagrangian in the form T-V.

    3. The attempt at a solution

    Basically I would like to know what is the most appropriate way to demonstrate this in a formal manner.

    By inspection I can see that the first term corresponds to the kinetic energy of the particle, the second term corresponds to the vector potential whose curl leads to the magnetic field. The third term corresponds to the scalar potential from which we can obtain the electric field.

    I've already calculated the equations of motion for such lagrangian, and have proven that they indeed lead to the Lorentz force, but I'm kind of confused about how to approach this question.For me, the velocity dependent potential is the term which gives the lagrangian this so called "different form", but how could I formally prove it?

    Thanks in advance for your help! It is greatly appreciated.
     
  2. jcsd
  3. Apr 17, 2013 #2
    Hmm, I think I solved the problem.

    We can take the Lagrange equations in the form that follows from the D'Alembert principle, and supose that the generalized force is derivable not from the potential V but from a more general function U, which is usually called generalized potential. Then, after some arrangements, L = T-U
     
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