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Conserved quantity for a particle in Magnetic Field

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a particle of mass m and electric charge e moving in a uniform magnetic field given by B = Bẑ. Then the Lagrangian is given by:

    [tex]L = \frac{m}{2}(x'^2 + y'^2 + z'^2) + \frac{Be}{2}(xy' - yx')[/tex]

    Prove that [tex]Q={L} \cdot{B} + \frac{e}{2}((r \times B)(r \times B))[/tex] is a constant of motion, where L = r x p is the angular momentum.

    3. The attempt at a solution

    I proceeded to calculate the conjugated momenta,

    [tex]Px = mx' - ey/2[/tex]
    [tex]Py = my' + ex/2[/tex]
    [tex]Pz = mz'[/tex]

    Then I calculated the Hamiltonian. A lot of terms vanished and I ended up with just kinetic energy.

    [tex]H = \frac{m}{2} (x'^2 + y'^2 + z'^2)[/tex] which is a conserved quantity.

    I was expecting to also end up with a term for the potential. Is this caused by the fact that the magnetic field does no work on the particle so it doesn't affect the total energy?

    I expressed the hamiltonian in terms of the generalized momenta:

    [tex]H= \frac{m}{2}(Px^2 + Py^2 + Pz^2 + \frac{(Be^2)(x^2+y^2)}{4}+eB(yPx-xPy))[/tex]

    I tried re-arranging Q in terms of the vector position r = xî + yĵ + zk , while p would correspond to m(x' + y' + z')

    After doing dot and cross product operations, I ended up with the following expression for Q:

    [tex]Q = (xy' - yx')B + (e{B^2}/2) (x^2 + y^2)[/tex]

    which doesn't depend on z nor z', which gives me trouble when relating terms to the hamiltonian.

    Is there any flaw in my procedure? I've tried rearranging terms, playing with algebra, but I don't come up with anything satisfactory.

    Also, is the quantity Q a well known quantity? Is there something too obvious I'm missing here?

    Thanks for your help. It is really, really appreciated.
    Last edited: Apr 13, 2013
  2. jcsd
  3. Apr 13, 2013 #2


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    In a uniform magnetic field, the scalar potential ##\phi## can be chosen to be zero. The vector potential ##\textbf{A}## can be chosen as ##\textbf{A} = \textbf{B} \times \textbf{r}/2 ##, so you could express the Hamiltonian in terms of the vector potential if you wished.

    It all looks ok to me. Try taking the time derivative of ##Q## and see if it equals zero.
  4. Apr 13, 2013 #3
    Thanks a lot TSny, according to what I calculated, using E-L equations, its derivative indeed equals to zero. I got lost trying to rearrange the Hamiltonian, I should have checked the derivative of Q at first.
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