# Conserved quantity for a particle in Magnetic Field

1. Apr 13, 2013

### Siberion

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:

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

Prove that $$Q={L} \cdot{B} + \frac{e}{2}((r \times B)(r \times B))$$ 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,

$$Px = mx' - ey/2$$
$$Py = my' + ex/2$$
$$Pz = mz'$$

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

$$H = \frac{m}{2} (x'^2 + y'^2 + z'^2)$$ 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:

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

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:

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

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. Apr 13, 2013

### TSny

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.

3. Apr 13, 2013

### Siberion

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.