# Hamilton - Jacobi method for a particle in a magnetic field

1. Sep 1, 2017

### Vrbic

1. The problem statement, all variables and given/known data
Hamiltonian of charged particle in magnetic field in 2D is $H(x,y,p_x,p_y)=\frac{(p_x-ky)^2+(p_y+kx)^2}{2m}$ where $k$ and $m$ are constant parameters. For separation of this system use $S=U(x)+W(y)+kxy+S_t(t)$. Solve Hamilton - Jacobi equation to get $x(t), y(t)$ .

2. Relevant equations

3. The attempt at a solution
When I substitute $p_i=\frac{dS}{dx_i}$ in the assignment and assumptation $S_t(t)t=a_0t$ (hamiltonian don't depend on time) I get
$U_x^2+(W_y+2kx)^2=a (+b-b)$,
where $a=2ma_0$, $b$ is separation constant and subscript means derivative. I separate it and I get
$W_y=b$
$U_x=\sqrt{a-(2kx+b)^2}$...my first question is, if it is good procedure and is it right?
Then I integrate but second terms is quite...complex, so I denote solution of this integral as $I(x;a;b)+d$, $c$ and $d$ are constants.
$W=by+c$
$U=I+d$
I suppose that constants $c$ and $d$ are unimportant because of they are just additive. So
$S=I+by+kxy-\frac{a}{2m}t$
and finally $x(t), y(t)$ I get from
$\frac{dS}{da}=e$ and $\frac{dS}{db}=f$ where $e,f$ are constants.
Please comment my procedure and also if exist some "better" prcedure when I could get exact $x(t),y(t)$ because from this integral $I$ it is impossible.
Thank you for an advice.

2. Sep 1, 2017

### TSny

Did you mean to have the factor of $t$ on the left side? Do you need a negative sign in front of the right side?
Doesn't the factor of (b-b) make the right side zero? Maybe just a typo.
This looks correct to me.
Yes
This looks good. It should not be hard to evaluate $\partial I / \partial a$ and $\partial I / \partial b$. [Edit: Differentiate "under the integral sign" and then do the integral.]

Last edited: Sep 1, 2017
3. Sep 5, 2017

### Vrbic

Ok, thank you very much for comments.