Hamilton - Jacobi method for a particle in a magnetic field

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Vrbic
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Homework Statement


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)## .

Homework Equations

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.
 
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Vrbic said:

The Attempt at a Solution


##S_t(t)t=a_0t##
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?
##U_x^2+(W_y+2kx)^2=a (+b-b)##,
Doesn't the factor of (b-b) make the right side zero? Maybe just a typo.
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?
This looks correct to me.
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.
Yes
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.
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:
TSny said:
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.]
Ok, thank you very much for comments.