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Relativistic velocity derivation under constant force

  1. Jan 3, 2008 #1
    I've been attempting to derive the relativistic velocity equation (for an object accelerating with a constant force- [tex]\frac{Fct}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]) for near a month without any solution.

    I've derived an equation form the F=ma relation which includes a Sin function, so I know it's wrong. However, I haven't been able to determine why. Could someone point me in the right direction?

    My derivation:

    Using [tex]a=\frac{F}{m}[/tex] and [tex]a\equiv\frac{dv}{dt}[/tex]


    [tex]m dv= F dt[/tex] where [tex]m\equiv\frac{m_{o}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

    [tex]\frac{m_{o} dv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}= F dt[/tex]

    Integrate both sides

    [tex]m_{o}c Sin^{-1}(\frac{v}{c})=Ft + constant[/tex]

    Which gives an answer that doesn't make sense and obviously doesn't represent v(t)

    [tex]\frac{v}{c}=Sin(\frac{Ft + const}{m_{o}c})[/tex]

    Any idea where the error is, or how to get the correct expression? Any help is appreciated.
  2. jcsd
  3. Jan 3, 2008 #2
    You have to write

    [tex]\vec{F}=\frac{d\,\vec{p}}{d\,t},\quad \vec{p}=m\,\gamma\,\vec{v}, \quad \gamma=\frac{1}{\sqrt{1-v^2/c^2}} [/tex]

    Thus first integrate

    [tex]\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{p}}=\vec{F}\,t+\vec{p}_o[/tex]

    plug in [tex] \vec{p}=m\,\gamma\,\vec{v}[/tex] and solve for the velocity.
  4. Jan 4, 2008 #3


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    I assume you are talking coordinate values here, not "proper values", meaning the constant force will produce decreasing coordinate acceleration with time.

    Question: will such a constant (coordinate) force produce a constant proper acceleration, i.e., as measured by an accelerometer on board the spacecraft?
    Last edited: Jan 4, 2008
  5. Jan 4, 2008 #4
    I don't really understand the terminology here! It is not you by my English! :smile:

    The constant force [itex] \vec{F}[/itex] I mean is like the spacecraft is like a charge accelerating in a constant eletrical field. Is that what you mean by constant coordinate force?
  6. Jan 4, 2008 #5
    Wow... Thanks much, RC. I'm still not sure where my original math went wrong, but I'm okay with pondering that until I get the answer. My urgent task was to know how to derive the equation, which you have generously provided.

    Jorrie- Yes: this applies to accelerating charges in a constant electric field (very good guess RC- you're on fire tonight), so no proper measurements required.

    Regards to all,
  7. Jan 4, 2008 #6
    Newton's 2nd law must be written in the form

    [tex]\vec{F}=\frac{d\,\vec{p}}{d\,t}\Rightarrow \vec{F}=\gamma\,m\,\vec{a}+\gamma^3\,m\,\frac{\vec{v}\cdot\vec{a}}{c^2}\,\vec{v}[/tex]

    for the spatial compontets of the four-force.

    But this is a horrible point to start! :smile:
  8. Jan 4, 2008 #7
    This link gives the equations of a rocket with constant acceleration (as felt by the occupants of the rocket)

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    It does not show the derivations but it gives you something to cross check your answers against.

    This link gives a derivation of velocity from force. I’m not sure if the derivation is correct but the final answer agrees with the equations given by Baez.

    http://www.usna.edu/Users/physics/mungan/Scholarship/ConstantAcceleration.pdf [Broken]

    The author makes the observation “The fact that the passengers feel one gee means that the force F measured in frame R is mg. However, as proved by French Eq. (7-20), an observer in frame E measures the same (longitudinal) force, despite the fact that she measures neither the same (relativistic) mass nor acceleration of the rocket. “ which simplifies things a bit.

    Substituting F/m for a in the Baez equations(where m is the proper mass) and dividing by c we get:

    [tex]\frac{v}{c}= \frac{1}{\sqrt{1+(m c)^2/(Ft)^2}}[/tex]


    [tex]\frac{v}{c}= tanh\left(\frac{F T}{m c}\right)[/tex]

    The last equation is in terms of the proper time and becomes:

    [tex]\frac{v}{c}= tanh\left( sinh^{-1}\left(\frac{F t}{m c}\right)\right) [/tex] in terms of coordinate time.

    I hope that helps.
    Last edited by a moderator: May 3, 2017
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