How Does a Magnetic Field Influence Hamiltonian Mechanics in Planar Motion?

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Homework Statement


A charged particle is constrained to move in a plane under the influence of a central force potential V=1/2kr^2 and a constand magnetic field B perpendicular to the plane, so that A=\frac{1}{2}\vec B \times \vec{\dot{r}}. We want to solve the problem with the Hamilton-Jacobi method, but my problem focuses elsewhere.


Homework Equations


The lagrangian is

L=\frac{1}{2}m\dot{r}^2+q\vec A \cdot \vec{\dot{r}} -\frac{1}{2}kr^2

I'm interested in the second term. If \vec B=B \hat z, then the term is written

\frac{qB}{2}(-y\dot{x}+x\dot{y})

or, using polar coordinates,

\frac{qB}{2}r^2\dot{\theta}

Now, the conjugate momentum is

p_{\theta}=\frac{\partial L}{\partial\theta}=...+\frac{qB}{2}r^2

If one tries to find the Hamiltonian, such as

H=p_r\dot{r}+p_{\theta}\dot{\theta}-L

finds that this term disappears, which is at least strange.

So, am I missing something fundamental here? Thanx in advance!
 
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Hi PhysiSmo,

Your initial calculation is correct in the sense that the term containing A does appear to drop out in the Hamiltonian. How then can the Hamiltonian know about the magnetic field? The key is to remember that the Hamiltonian is properly a function of position and momentum (not velocity), and this means you need to write all velocities in terms of momenta when using the Hamiltonian. You should find that A suddenly reappears in your Hamiltonian in a somewhat different way.

As an aside, you can convince yourself that this is a general fact about terms in the Lagrangian linear in the velocity: they seem to drop out of the Hamiltonian, but appear instead when solving for the velocities in terms of the canonical momenta.

Also, if you have the hand waving intuitive notion that the Hamiltonian is the energy, you might expect the A term to disappear since it has no interpretation in terms of energy.

Hope this helps!
 
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It really helps, indeed! Thank you Physics Monkey!
 
and the H-J eqn?
 
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