Problem: linear dep. of time in hamilton

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Homework Help Overview

The discussion revolves around solving a Hamiltonian problem involving the function h = [p^2 / (2m)] + mAxt, where m and A are constants. The initial conditions are specified as t=0, x=0, and p=mv. Participants are exploring the application of Hamilton's principal function to approach the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the Hamilton-Jacobi equation and its relevance to the problem. There are attempts to express the solution in terms of separated variables, with some questioning the feasibility of this approach due to mixed terms involving both x and t. Others express confusion about how to proceed from the derived equations.

Discussion Status

The discussion is ongoing, with various participants offering hints and suggestions for approaching the problem. Some have shared their attempts and partial solutions, while others are still seeking clarity on the method to apply. There is no explicit consensus on the best approach yet.

Contextual Notes

Participants note that the problem may not require time separation, as indicated in the referenced textbook. There is also mention of initial conditions and the challenge posed by mixed terms in the equations derived.

Asle
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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


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

Daniel.
 
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 don't have to separate 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??)
 
Try to move the "t" dependence (time derivative+function) in the RHS and then try a solution
[tex]S(x,t)=T(t)X(x)[/tex]
Hopefully it works.

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

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

Daniel.
 
Last edited:
Thanks for the replies, but I am still not closer to obtaining the solution.

Anyone who has solved a similar problem ??
 
  • #10
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
Asle said:
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


Thanks in Advance
[tex]{dp\over dt}=-{\partial H\over\partial x}=-mAt[/tex]

and that's pretty easy to integrate.
 
  • #12
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. :smile:

Daniel.
 

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