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Derivation of Velocity from Acceleration in Special Relativity

  1. Jul 18, 2014 #1
    I got stuck in deriving the velocity of the particle from the acceleration equation. Here are the details of the problem.

    The acceleration of a particle with a relativistic momentum is

    [itex]\vec{a} = \frac{\vec{F}}{\gamma m} - \frac{\vec{v}}{\gamma m c^2}\left(\vec{F} \cdot \vec{v}\right)[/itex].

    By integrating the above equation, the velocity of the particle can be calculated if the force applied and the mass of the particle are given. The greek letter [itex]\gamma[/itex] in this case is [itex]\left(1 - \frac{v^2}{c^2}\right)^{-1/2}[/itex]

    Considering that a constant force [itex]\it{F}[/itex] is applied to a mass [itex]\it{m}[/itex] initially at rest, the velocity of the particle is

    [itex]\int \vec{a} dt = \int \left[\frac{\vec{F}}{\gamma m} - \frac{\vec{v}}{\gamma m c^2}\left(\vec{F} \cdot \vec{v}\right)\right] dt \\

    \vec{v} = \int \left[\frac{\vec{F}}{\gamma m} - \frac{\vec{v^2}}{\gamma m c^2}\left(\vec{F} \right)\right] dt[/itex].

    Since [itex]F[/itex] and [itex]m[/itex] are constants in time, they are removed from the integral. Gamma is taken out from the two terms. Then,

    [itex]\vec{v} = \frac{\vec{F}}{m} \int \frac{1}{\gamma} \left(1 - \frac{v^2}{c^2}\right) dt[/itex].

    Using the definition of gamma, the integrand is

    [itex]\vec{v} = \frac{\vec{F}}{m} \int \left(1 - \frac{v^2}{c^2}\right)^{3/2} dt[/itex].

    This is where I get stuck. I don't know how to integrate it since the integrand is not a direct function of [itex]t[/itex] but of [itex]v[/itex] which is a function of [itex]t[/itex]. I tried having [itex]v = at[/itex], but I don't know if this is correct.
  2. jcsd
  3. Jul 19, 2014 #2


    Staff: Mentor

    You cannot substitute v=at because that only works if a is constant, which it is not here. Personally, I would start with your last equation, differentiate both sides with respect to t, and solve the resulting differential equation. Are you familiar with differential equations, or do you have access to a math package like Mathematica?
  4. Jul 19, 2014 #3
    You have the differential equation
    [tex]\frac{{dv}}{{dt}} = \frac{F}{m} \cdot \left( {1 - \frac{{v^2 }}{{c^2 }}} \right)^{\frac{3}{2}}[/tex]
    that can be solved by separation of variables.
  5. Jul 19, 2014 #4


    User Avatar
    Science Advisor

    You have to know F(x,v,a,t). If F is constant, or a function of only t, then [itex] m\gamma^3 dv=Fdt[/itex]
    can be integrated.
  6. Jul 19, 2014 #5
    Yes. I am familiar with the mathematics. The resulting differential equation will be the one shown by DrStupid. By separating the variables,

    [itex]\frac{dv}{dt} = \frac{F}{m} \cdot \left(\frac{c^2 - v^2}{c^2}\right)^{3/2} = \frac{F}{m c^2} \cdot \left(c^2 - v^2\right)^{3/2} \\
    \left(c^2 - v^2\right)^{-3/2} dv = \frac{F}{m c^2} dt \\
    \int \left(c^2 - v^2\right)^{-3/2} dv= \int \frac{F}{m c^2} dt \\
    \frac{v}{c^2 \sqrt{c^2 - v^2}} = \frac{F t}{m c^2}.[/itex]

    Could you please check the mathematics? After this, I just need to solve for [itex]v[/itex], right?

    Would you be suggesting that [itex]dt = \frac{m \gamma^3}{F} dv[/itex]? If I were to substitute this to my last equation, wouldn't be the constants [itex]F[/itex] and [itex]m[/itex] cancel, so does [itex]\gamma[/itex]? I would be left out with the integral of [itex]dv[/itex] which will result into [itex]v[/itex].
  7. Jul 20, 2014 #6
    [itex]\frac{dv}{dt} = \frac{F}{m} \cdot \left(\frac{c^2 - v^2}{c^2}\right)^{3/2} = \frac{F}{m c^3} \cdot \left(c^2 - v^2\right)^{3/2}[/itex]

    should be more accurate
  8. Jul 20, 2014 #7
    Oh, yes, I forgot about the 3rd-power-exponent. Thanks a lot!
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