(special relativity)Trajectory under constant ordinary force

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle is subject to a constant force F on +x direction. At t = 0, it is located at origin with velocity vo in +y direction.

    2. Relevant equations

    Determine the trajectory of the particle. x(t),y(t),z(t)

    3. The attempt at a solution

    [tex]\vec{p}= \int \vec{F} dt[/tex]
    [tex]\vec{p} = \vec{F}t + constant[/tex]

    At t=0
    [tex]\vec{p} = \gamma mv_{o}[/tex]
    [tex]\vec{p} = Ft\hat{x} + \gamma mv_{o}\hat{y}[/tex]

    what should I do next? should I integral over [tex]p_{x}[/tex] and [tex]p_{y}[/tex] separately? Is that so, what are the exact steps?

    I tried using
    [tex] p_{x} = \frac{mu_{x}}{\sqrt{1-\frac{u^{2}_{x}}{c^{2}}}} = Ft[/tex]
    [tex]x(t) = \frac{mc^{2}}{F} (\sqrt{1+\frac{Ft}{mc}^{2}} -1) [/tex]

    what about y(t)? it seems it is a linear with t. I don't know where I get wrong.
    Last edited: May 11, 2010
  2. jcsd
  3. May 11, 2010 #2


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    [itex]\gamma[/itex] involves [itex]u^2[/itex], not just the x-component of the velocity:wink:
  4. May 11, 2010 #3
    Do you mean I have to use
    [tex]\frac{mu_{x}}{\sqrt{1-\frac{u^{2}_{x}+u^{2}_{y}}{c^{2}}}} = Ft[/tex]
    [tex]\frac{mu_{y}}{\sqrt{1-\frac{u^{2}_{x}+u^{2}_{y}}{c^{2}}}} = \gamma mu_{o}[/tex]

    and solve these coupling equations?
    Last edited: May 11, 2010
  5. May 11, 2010 #4


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  6. May 11, 2010 #5
    Do I have another way, because there will be messy if I solve the coupling and then take the integral.
    Would Lorentz's transform help me to simplify the process?
  7. May 11, 2010 #6
    Using L.T. s.t. [tex]\bar{S}[/tex] is moving in y-direction with [tex]u_{o}[/tex]

    [tex]\bar{F_{x}}=\gamma F[/tex]


    Is this correct?
    Last edited: May 11, 2010
  8. May 11, 2010 #7
    I think my L.T. transform got some problems since the answer is too simple..
    Can anyone give a hand?
  9. May 12, 2010 #8
    I just sketch the method I used.
    First use
    [tex]\frac{d\gamma mu_{x}}{dt}=F_{x}=F[/tex]
    [tex]\frac{d\gamma mu_{y}}{dt}=0[/tex]
    [tex]\frac{d\gamma mu_{z}}{dt}=0[/tex]

    Then shows z is always zero
    Hence the [tex]u^{2}[/tex] in the [tex]\gamma[/tex] becomes [tex]u_{x}^{2}+u_{y}^{2}[/tex]

    Afterwards, integrated out [tex]u_{x}[/tex] and [tex]u_{y}[/tex]
    decouple the equation s.t. one involves only [tex]u_{x}[/tex] and other [tex]u_{y}[/tex]
    integrate it again, you will get the answer
    Last edited: May 12, 2010
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