# Homework Help: Problem: linear dep. of time in hamilton

1. Feb 5, 2005

### Asle

Hello

I am having a litte problem solving h=[ p^2 / (2m) ] + mAxt
where m, A are constants. initial conditions: t=0, x=0, p= mv

Supposedly sol this with Hamiltons principal function.
A hint for start would be nice

2. Feb 5, 2005

### dextercioby

What have you done so far??Can u obtain the Hamilton-Jacobi equation??

Daniel.

3. Feb 5, 2005

### Asle

I am really at lost in this problem.

(1/2m)(dS/dx)^2 - mAxt + dS/dt = 0, where all d are partial derivatives
This should be the correct equation, but how to solve it?
Supposedly this is a case where you dont have to seperate t
(Goldstein, 445 and prob 8 479, third edition)

But I still fail to se how to proceed from this point..
(or maybe I should use another apporth??)

4. Feb 5, 2005

### dextercioby

Try to move the "t" dependence (time derivative+function) in the RHS and then try a solution
$$S(x,t)=T(t)X(x)$$
Hopefully it works.

Daniel.

5. Feb 5, 2005

### Asle

Thanks for the tip, but I am just having a problem with that..
its -mAxt, so its a function of both x and t.

I have the solution:
S = x(alpha) + 0.5 mAxt^2 + (1/40)mA^2 t^5 - (1/6)A(alpha)t^3 - (1/2m)(alpha)^2t

6. Feb 5, 2005

### dextercioby

Yes,i've searched for the problem (#8,pag.479) in Goldstein and it's true that in tex,page 445 he says that this problem can be solved without time-separation.Unfortunately,i don't see how...It simply evades me...Just for curiosity,how did u find the solution u posted?

Daniel.

7. Feb 5, 2005

### Asle

I am currently taking a course in classical mechanics and my professor have handed out a lot of problems, some with solution.

But how do you solve the S(x,t)=X(x)T(t) in this case.
Remeber its (dS/dx)^2 and thus yields: (dX/dx)^2T^2
And the term containing both x and t..?
Should it go with T or X or both?

I now the procedyre from QM(Sch. Eq.)
But in that case its simple to seperate

8. Feb 5, 2005

### dextercioby

I'm afraid you can't draw a parallel with the Schrödinger's equation,simply because that is linear in space derivatives
$$(\frac{\partial S}{\partial x})^{2}\neq \frac{\partial^{2}S}{\partial x^{2}}$$

Yes,it's precisely that mixed term which prevents a separation writing the solution as a product of functions...

Daniel.

Last edited: Feb 5, 2005
9. Feb 6, 2005

### Asle

Thanks for the replies, but I am still not closer to obtaining the solution.

Anyone who has solved a similar problem ??

10. Feb 10, 2005

### Asle

Solution

Found the solution.
If anyone passes by:

Write S = W(x) + W(t)x + f(t)
and calculate (dS/dx) and then square it.

Set up and solve for terms just contaoing x and then the ones containing t.

And with initial conditions you obtain the solution.

11. Feb 10, 2005

### krab

$${dp\over dt}=-{\partial H\over\partial x}=-mAt$$

and that's pretty easy to integrate.

12. Feb 10, 2005

### dextercioby

Yes,Krab,but the idea was for her/him to integrate the H-J equation,which indeed is equivalent to integrating Hamilton's eqns,but it's still something different.
I'm glad he/she finally found a way.

Daniel.