# Solve relativistic differential force equation for velocity

1. Feb 6, 2014

### psal

1. The problem statement, all variables and given/known data
I proved that a relativistic 1D force is
F = $\gamma$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) = $\frac{a*t}{\sqrt{1 + \frac{at}{c}2}}$

2. Relevant equations

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

3. The attempt at a solution
dβ = cosθdθ
a(t) = $\frac{c*cosθdθ}{cos2θ}$ = $\frac{cdθ}{cosθ}$
I don't really know what to do from here to arive at the answer

Last edited: Feb 6, 2014
2. Feb 6, 2014

### LCKurtz

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
3. Feb 6, 2014

### vela

Staff Emeritus
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.