1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solve relativistic differential force equation for velocity

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data
    I proved that a relativistic 1D force is
    F = [itex]\gamma[/itex]3*m*dVx/dt = m * dVx/dt * 1/ (1 - (v/c)2)3/2

    Then, "This is a separable differential equation that can be solved using a trig
    substitution. Use this (or some other technique that works) to show that the velocity is given by
    v(t) = [itex]\frac{a*t}{\sqrt{1 + \frac{at}{c}2}}[/itex]

    2. Relevant equations

    a = [itex]\frac{dVx}{dt}[/itex] * [itex]\frac{1}{(1-\frac{v}{c}3/2}[/itex]
    β = [itex]\frac{v}{c}[/itex] = sinΘ
    cosΘ = [itex]\sqrt{1 - β2}[/itex]

    3. The attempt at a solution
    dβ = cosθdθ
    a(t) = [itex]\frac{c*cosθdθ}{cos2θ}[/itex] = [itex]\frac{cdθ}{cosθ}[/itex]
    I don't really know what to do from here to arive at the answer
    Last edited: Feb 6, 2014
  2. jcsd
  3. Feb 6, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think I fixed the Tex in your first equation. Don't use the sup and /sup tags in a tex expression. If you want an exponent of 3/2 just use ^{3/2} in the Tex. You also don't need the * for multiplication. You can use \cdot if you really want a multiplication sign. I will leave it to you to fix the rest if you desire. Also you can preview your posts before posting to see if the Tex is working.
    Last edited: Feb 6, 2014
  4. Feb 6, 2014 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I take it you're using ##V_x## and ##v## to represent the same thing. Don't do that. Pick one. It also looks like you're supposed to assume the acceleration ##a## is constant.

    After the substitution, you have
    $$a\,dt = \frac{c \cos\theta \, d\theta}{(1-\sin^2\theta)^{3/2}}.$$ If you simplify that, you don't get what you got. Your last line, in particular, doesn't make sense. You shouldn't have a ##d\theta## all alone in the equation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted