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Solving functions for S in a q-q* Hamilton-Jacobi diffeq

  1. Jun 3, 2016 #1
    1. The problem statement, all variables and given/known data
    68e84adeb694fd0f3b56217e4dc7352a.png

    2. Relevant equations


    3. The attempt at a solution
    So far I have a solution for a) as
    y%7D%20e%20B%20x%20-%20%5Cfrac%7B1%7D%7B2%7D%20m%20%28eBx%29%5E%7B2%7D%20-%20%28eBx%29%5E%7B2%7D.gif
    For b) I formulate the equation as
    20%28eBx%29%5E%7B2%7D%20+%20%28%5Cfrac%7B%5Cpartial%20S%7D%7B%5Cpartial%20t%7D%29%20%3D%200.gif
    and so far for c) I have
    gif.gif
    My main idea at the moment is that as the Lagrangian was not time dependent, the Hamiltonian will not be. Following on maybe I can say that when separating variables T is constant so I can separate it so
    gif.gif but this is as far as I could get, as I couldn't think of how to separate them.

    Any ideas, or are my problems due to a mistake in working?
     
  2. jcsd
  3. Jun 4, 2016 #2

    TSny

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    The last three terms do not have the correct dimesions. I believe you have dropped some factors of ##2## and ##m##. So, check your derivation.

    Alternately, you might see if you can first take the Lagrangian as given in terms of a general vector potential ##\bf{A}## and show that the Hamiltonian can be written nicely as ##H = \frac{1}{2m} (\mathbf{p}-e \mathbf{A}) \cdot (\mathbf{p}-e \mathbf{A})##. Then substitute the given expression for ##\mathbf{A}##.
    You probably didn't mean to have an equal sign after the ##T(t, \overline{x}, \overline{y})##

    You cannot conclude that ##T## is a constant. But you should be able to argue that ##\frac{\partial T}{\partial t}## is a constant.
    Also, you should hope that ##X## and ##Y## are not identically equal to zero. If they were, there would be no hope of finding the correct equations for ##x(t)## and ##y(t)##.
     
    Last edited: Jun 4, 2016
  4. Jun 6, 2016 #3
    My Hamilton was a mess from a mistake early on. After redoing it I get 7D%20-%203%20%28eBx%29%5E%7B2%7D%29%7D%7B2m%7D%20+%20%5Cfrac%7Bp_%7By%7D%28eBx%29%7D%7Bm%7D.gif .

    I can then formulate the hamiltonian jacobi again and get 03%28eBx%29%5E%7B2%7D%20%29%20+%20%5Cfrac%7B%5Cpartial%20S%7D%7B%5Cpartial%20t%7D%20%3D%200.gif

    Then (and I'm mostly going from an example in my lecture notes and the question hints) I can separate the variables out as gif.gif

    That can be substituted as
    gif.gif

    In my notes the example only contains two separate variables e.g. X and T. He then makes every non time dependent term equal the time dependent terms and says they are both equal to a constant. He then solves X and T wrt this constant.

    I'm not sure how to take this question further and solve for these functions.
     
  5. Jun 6, 2016 #4

    TSny

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    The numerical factor of -3 is not correct for the ##(eBx)^2## term. Also, I believe the last term has the wrong sign.
    Yes
    Here you want to use the usual type of reasoning when separating variables.
    Suppose you have an equation of the form ##f(x) + g(y)## = 0 that must hold for all values of ##x## and ##y## within some domain. Here, ##f(x)## is a function only of ##x## and ##g(y)## is a function only of ##y##. Then you can argue that ##f(x)## must be a constant: ##f(x) = C## for all ##x##. Moreover, ##g(y)## must also be a constant which is the negative of ##C##; i.e., ##g(y) = -C##. Try to construct such an argument.

    For your H-J equation, you have something of the form ##f(x,y) + \frac{dT(t)}{dt} = 0## that must hold for a range of values of ##x, y##, and ##t##. Use the type of argument mentioned above to deduce something about ##\frac{dT}{dt}##.
     
  6. Jun 6, 2016 #5
    Okay, so I must have assumed something wrong early on or have made a pretty bad mistake in my work (not unheard of)
    C%5C%20H%20%3D%20p_%7Bx%7D%20%5Cdot%7Bx%7D%20+%20p_%7By%7D%20%5Cdot%7By%7D%20-%20L%20%5C%5C.gif
     
  7. Jun 6, 2016 #6

    TSny

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    So far so good. Just a matter of simplifying.
     
  8. Jun 6, 2016 #7
    The Hamiltonian is the Legendre transformation of the Lagrangian:
    ## H = \sum_{i} \dot q_i \frac {\partial L} {\partial \dot q_i} - L = m(\dot x^2 + \dot y^2) + eBx\dot y - \frac {m} {2}(\dot x^2 + \dot y^2)- eBx\dot y##
    ## = \frac {1} {2m} (p_x^2 + p_y^2)##
    In this problem you need to find constants of motion. If you solve for the Lagrangian eqns. of motion, you will find:
    ## m\ddot x -eB\dot y = 0 →\frac {d} {dt}(m\dot x -eBy) = 0## and
    ## m\ddot y +eB\dot x = 0→\frac {d} {dt}(m\dot y +eBx) = 0##
    Thus we have the constants of motion:
    ##\tilde p_x =(p_x -eBy) ##,
    ##\tilde p_y =(p_y +eBx)##.
    We write the HJ equation with ## T(t,\tilde x,\tilde y) = \tilde xt## and substitute ## p_y = \tilde p_y - eBx## to get:
    ## (\frac {dX} {dx})^2 + ( \tilde p_y - eBx)^2 = 2m \tilde x##,
    ## X= \int dx\sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}##
    ## S= \int dx\sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2} - \tilde xt##
    ##p_x = \frac {\partial S} {\partial x} = \sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2} ##
    ##\tilde p_x = -\frac {\partial S} {\partial \tilde x} = -m \int dx\frac {1} { \sqrt { 2m \tilde x- ( \tilde p_y - eBx)^2}} +t##
    We now make the substitution in the integral:
    ## x' =\tilde p_y - eBx ##,
    ## dx' = -eBdx ##
    ##\tilde p_x = -\frac {\partial S} {\partial \tilde x} = \frac {m} {eB} \int \frac {dx'} { \sqrt { 2m \tilde x- x'^2}} +t = \frac {m} {eB}sin^{-1}(\frac{\tilde p_y - eBx} { 2m \tilde x}) + t##
    You can do the algebra to solve for x in terms of ##\tilde p_x ,\tilde p_y## and ##\tilde x##.
    To get y just use the constant of motion equation ## y = \frac {m\dot x - \tilde p_x} {eB}##
     
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