Solve relativistic differential force equation for velocity

[psal]

Homework Statement

I proved that a relativistic 1D force is
F = \gamma³*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}}

Homework Equations

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

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

 

[LCKurtz]

Homework Statement

I proved that a relativistic 1D force is
F = \gamma^3m\frac{dVx}{dt} = m \frac {dVx}{dt}\frac 1{(1 - (v/c)^2)^{3/2}}

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.

 

[vela]

Homework Statement

I proved that a relativistic 1D force is
$$F = \gamma^3 m \frac{dV_x}{dt} = m \frac{dV_x}{dt} \frac{1}{[1 - (v/c)^2]^{3/2}}.$$ 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
$$v(t) = \frac{at}{\sqrt{1 + \left(\frac{at}{c}\right)^2}}$$

Homework Equations

\begin{align*}
a &= \frac{dV_x}{dt} \frac{1}{[1 - (v/c)^2]^{3/2}} \\
\beta &= \frac{v}{c} = \sin \theta \\
\cos \theta &= \sqrt{1 - \beta^2}
\end{align*}

The Attempt at a Solution

$$d\beta = \cos\theta\,d\theta$$
$$a(t) = \frac{c\cos\theta\,d\theta}{\cos^2\theta} = \frac{c\,d\theta}{\cos\theta}$$
I don't really know what to do from here to arrive at the answer.

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.