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Trajectory of a particle in a field (special relativity)

  1. Apr 24, 2010 #1
    Hi, I'm new on this forum and I would like to say hello to everybody!
    I have a problem with homework from my "Basics of theoretical phisics" class.

    1. The problem statement, all variables and given/known data
    I have to find a trajectory of a particle in field of force:

    [tex] F = - \frac{\alpha}{x^2} [/tex]

    2. Relevant equations

    I was said to use:

    [tex] F = \dot{p} = \frac{dp}{dt} [/tex] and [tex] p=mv\gamma [/tex]

    where [tex]\gamma[/tex] is: [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} [/tex] and m is rest mass.


    3. The attempt at a solution

    [tex]\dot{p} = \frac{dp}{dt}=m \frac{d (v\gamma)}{dt} = m (\dot{v}\gamma + v \dot{\gamma}) [/tex]

    [tex] \dot{\gamma} = -\frac{1}{2} (1-\frac{v^2}{c^2})^{-3/2} \frac{2v}{c} \dot{v} = \dot{v} \frac{v}{c^2} \gamma^3 [/tex]

    [tex] \dot{p} = m(\dot{v} \gamma + v \dot{v} \frac{v}{c^2} \gamma^3) = m \dot{v} \gamma (1+ \frac{v^2}{c^2} \gamma^2)= m \dot{v} \gamma^3 [/tex]

    [tex] v=\dot{x} , m \dot{v} \gamma^3 = m \ddot{x} \gamma^3 = -\frac{\alpha}{x^2} [/tex]

    [tex] \ddot{x} x^2 \gamma^3 = -\frac{\alpha}{m} [/tex]

    Now I have a problem with that equation. Any ideas how to solve it and get [tex] x(t) [/tex]?
     
  2. jcsd
  3. Apr 24, 2010 #2

    Cyosis

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    This is similar to a classical particle in a gravitational field which cannot be solved for x(t). Adding the gamma factor to it doesn't make things prettier. I don't see how you can solve this for x(t). Are you sure you have to use this method?
     
  4. Apr 24, 2010 #3

    gabbagabbahey

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    I think you'll want to use the work-energy theorem and conservation of energy instead...

    Edit: I suppose since you've come this far with the force method, you may as well continue....just multiply both sides of your equation by [tex]\frac{\dot{x}}{c^2x^2}[/tex] and integrate w.r.t. [itex]t[/itex]. You won't be able to get an solution in the form [itex]x(t)=\ldots[/itex], but you should be able to get an implicit solution
     
    Last edited: Apr 24, 2010
  5. Apr 25, 2010 #4
    Since the equation doesn't depend on [tex]t[/tex] explicitly, I used a substitution:
    [tex] \dot{x}=u(x)[/tex]
    [tex]\ddot{x}=\frac{du}{dt}=\frac{du}{dx}\frac{dx}{dt}=u'\dot{x}=u'u[/tex]

    After separating variables and integrating, I've got:

    [tex]\dot{x}=c\sqrt{1-\frac{c^4m^2x^2}{(\alpha+Dmx)^2}[/tex] where [tex]D[/tex] is a constant.
    I think I cannon do much more, tried to solve this in Mathematica, but got a nightmarish outcome.

    I don't really have an idea how to use it. Can you at least tell me how to start?
     
    Last edited: Apr 26, 2010
  6. Apr 25, 2010 #5

    gabbagabbahey

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    You've now got a separable 1st order ODE...surely you know how to solve that:wink:


    EDIT: Shouldn't you have [itex]m^2[/itex] in the numerator, and not [itex]m^4[/itex]?

    You'll end up with the same DE as above, but a little quicker. Energy is conserved here, so you have [itex]E=\gamma mc^2+U(x)[/itex] where the total energy [itex]E[/itex] is just some constant, and [itex]U(x)[/itex] is the potential energy of the particle due to the conservative force field

    [tex]U(x)=-\int \textbf{F}\cdot d\textbf{r}=\alpha\int \frac{dx}{x^2}=-\frac{\alpha}{x}[/tex]

    So,

    [tex]\gamma=\left(1-\frac{\dot{x}^2}{c^2}\right)^{-2}=\frac{E}{mc^2}-\frac{\alpha}{mc^2 x}[/tex]

    Solve for [itex]\dot{x}[/itex] and you should end up with the same DE you have above.
     
    Last edited: Apr 25, 2010
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