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Charged particle in EM field

  1. Feb 17, 2012 #1
    If you go to the relativistic Hamiltonian, what allows us to go from
    [tex]\dot{\vec{P}} = - \frac{\partial \mathcal{H}}{\partial \vec{x}} = e (\vec{\nabla} \vec{A}) \cdot \dot{\vec{x}} - e \vec{\nabla} \phi [/tex]
    to
    [tex]\frac{d}{d t}\left(\frac{m \dot{\vec{x}}} {\sqrt {1 - \frac{\dot{\vec{x}}^2}{c^2}}}\right) = e \vec{E} + e \dot{\vec{x}} \times \vec{B}[/tex]

    Is there some identity with
    [tex]\frac{\partial A^i}{\partial x^i}[/tex]
    that turns it into a curl that I'm missing?
     
  2. jcsd
  3. Feb 18, 2012 #2
    You should try using both Hamilton equations to find the relation between P and the velocity. Then get rid of P.

    For a somewhat different Hamiltonian - see the attached gif form Cordinalesi, "Classical mechanics for physics graduate students".

    Cordinalesi p. 251.gif
     
    Last edited: Feb 18, 2012
  4. Feb 18, 2012 #3
    Both Hamilton's equations are used to get to the first line. I'm looking at the RHS and wondering how to go from the RHS of eqn 1 to the RHS of eqn 2. I'm having trouble coming up with an identity that will give me a curl and divergence of the vector potential from its form in eqn 1.
     
  5. Feb 18, 2012 #4
    The LHS is not the same, so the RHS are also different - not a surprise.
     
  6. Feb 18, 2012 #5
    I'm missing something. Why are the two LHS not the same? P is the relativistic momentum, so dP/dt is the LHS, no?
     
  7. Feb 18, 2012 #6
    There are two kinds of momenta: kinematic (or "kinetic") momentum and canonical momentum. When there is an electromagnetic potential present, they are not equal. You can find the difference from your Hamiltonian. P there is the canonical momentum, not the kinetic momentum.
     
  8. Feb 19, 2012 #7
    Ah, I see. So yeah, like you said earlier, I can get a relationship between x'' and p', then go through a whole bunch of tedious algebra to get the result. :(
     
  9. Feb 19, 2012 #8
    Sometimes it is like that. We have to go through the whole bunch of tedious algebra to get, at the end, simple result. In such cases we are entled to think: "There must be a simple method!". And often, though not always, there is indeed one. In this case, I personally, do not know, right now, a simple method. Maybe someone knows it? Probably using Lagrangian formalism, rather than Hamiltonian one, is a simpler method. But you have asked about the Hamiltonian ....
     
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