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Problem: linear dep. of time in hamilton

  1. Feb 5, 2005 #1
    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
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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

    Daniel.
     
  4. Feb 5, 2005 #3
    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??)
     
  5. Feb 5, 2005 #4

    dextercioby

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    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.
     
  6. Feb 5, 2005 #5
    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
     
  7. Feb 5, 2005 #6

    dextercioby

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    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.
     
  8. Feb 5, 2005 #7
    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
     
  9. Feb 5, 2005 #8

    dextercioby

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    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: Feb 5, 2005
  10. Feb 6, 2005 #9
    Thanks for the replies, but I am still not closer to obtaining the solution.

    Anyone who has solved a similar problem ??
     
  11. Feb 10, 2005 #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.
     
  12. Feb 10, 2005 #11

    krab

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

    and that's pretty easy to integrate.
     
  13. Feb 10, 2005 #12

    dextercioby

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    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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