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Dynamics of a particle in a potential U(x) = f|x|

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data
    The one-dimensional hamiltonian [itex]H=c|p| + f|x|[/itex] describes a relativistic particle in an attractive potential [itex]U(x) = f|x|[/itex].
    (a) Obtain the dynamical equations and integrate them explicitly for the initial conditions [itex]x=x_{0}>0, p=0[/itex]. Show from first principles that the motion is periodic with period 4E/fc, where [itex]E=f/x_{0}[/itex] is the initial energy.


    2. Relevant equations
    The dynamical equations can be found by the usual method:

    [itex]\dot{x}=[H,x]_{PB}=\frac{\partial H}{\partial p}[/itex]
    [itex]\dot{p}=-[H,p]_{PB}=-\frac{\partial H}{\partial x}[/itex]



    3. The attempt at a solution
    I can get the dynamical equations from the formalism shown in (2), but I am not sure how I should interpret the initial conditions and then integrate. I assume I am just integrating the following WRT t:

    [itex]\dot{x}= c\frac{p}{|p|}[/itex]
    [itex]\dot{p}= -f\frac{x}{|x|}[/itex]

    The anti-derivative of the first is just [itex]p sgn(p) + c[/itex] and the second is similarly [itex]x sgn(x) + c[/itex].

    The phase space portrait would be that of a rhombus with ever increasing "radii", but I am not entirely sure how that helps. Any suggestions would be helpful.
     
  2. jcsd
  3. Apr 16, 2012 #2
    It would be incorrect to straightforwardly integrate each equation, because x and p are both functions of time -- so the RHS of each equation is an implicit function of time. The typical approach for Hamiltonian problems, to get the equation of motion, is to take the time derivative of the first equation, which would give you [itex]\dot{p}[/itex] on the RHS, and then substitute what you know from the second equation into the first to get a differential equation for x explicitly in terms of time.
     
  4. Apr 18, 2012 #3
    I am not sure why I made that mistake. But if I were to do as you suggest I end up with something similar to this:

    [itex]x''= c(-f (\frac{x}{|x|}-1)/|p|)[/itex]

    so there is still a dependence of x'' on |p|.
     
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