homework and exercises - Generalizing a relativistic kinematics formula for spatial-acceleration dependence

I'm starting from this expression

$$\alpha dt = \gamma^3 dv$$

where $\alpha$ is proper acceleration of a point particle, $dv$ and $dt$ are coordinate differentials of velocity and time, and $\gamma$ is the relativistic factor of the particle being subject to the acceleration

If $\alpha$ is constant, one arrives at the usual expression:

$$t_f - t_0 = \frac{1}{\alpha} (\frac{v_f}{\sqrt{1-\frac{v_f^2}{c^2}}} - \frac{v_0}{\sqrt{1-\frac{v_0^2}{c^2}}})$$

Now, I have some dependence of acceleration to position; $\alpha(x) = f(x)$ and I'm not sure how to integrate it in order to obtain a similar expression relating time, velocity and position. For instance, I tried the following for the left-side differential:

$$\alpha dt = \alpha(x) \frac{dt}{dx} {dx} = \int{ \frac{ \alpha(x) }{v(x)} dx }$$

and leaving the right-hand side untouched. When I do this I get a weird integral with velocity in both sides, and I'm not sure how to continue.

Update

so, after a while of staring at this, i noticed an error i was doing, and actually the 2nd derivative of acceleration should look like

$$\frac{d^2 x}{d \tau^2} = \gamma^4 \frac{d^2 x}{dt^2}$$

So far so good, but the approach on the comments doesn't seems to work. I try with a simple force potential: $\alpha(x) = -k x^3$, but is not clear how to work from it

$$-k x^3 = \gamma^4 \frac{d^2 x}{dt^2}$$

$$-k x^3 (1 - 2 \frac{1}{c^2} (\frac{dx}{dt})^2 + \frac{1}{c^4} (\frac{dx}{dt})^4 ) = \frac{d^2 x}{dt^2}$$

It looks like a (nonlinear) differential equation. I just want to be sure i'm on the right track. This is what it takes to solve this kind of problem? Or should some numerical integration/summation be enough? What throws me off in particular, is that if i would replace $-k x^3$ with just $\alpha_0$ (constant acceleration case), the $\gamma^4$ would still look pretty ugly and i wouldn't know how to solve the problem even in that case using the above expression, when we already know that a closed formula exists

Answer

Let's start with $$\frac{\text{d}v}{\left(1-v^2/c^2\right)^{3/2}} = \alpha\text{d}t = \frac{\alpha(x)}{v}\text{d}x.$$ As Clem suggested, multiply both sides by $v$, so that $$\int_{v_0}^{v}\frac{v'\text{d}v'}{\left(1-v'^2/c^2\right)^{3/2}} = \int_{x_0}^x\alpha(x')\text{d}x' = A(x),$$ which yields $$\frac{1}{\sqrt{1-v^2/c^2}} - \frac{1}{\sqrt{1-v_0^2/c^2}} = \frac{A(x)}{c^2},$$ or (assuming $v_0\geqslant 0$) $$v = c\frac{\sqrt{\left(A(x)/c^2+\gamma_0\right)^2-1}}{A(x)/c^2+\gamma_0} = \frac{\text{d}x}{\text{d}t},$$ with $$\gamma_0 = \frac{1}{\sqrt{1-v_0^2/c^2}}.$$ Thus $$\int_{x_0}^x\frac{A(x')/c^2+\gamma_0}{\sqrt{\left(A(x')/c^2+\gamma_0\right)^2-1}}\text{d}x' = c(t-t_0).$$ For a given $\alpha(x)$, you can solve this to get $x(t)$, and thus $A(t)=A(x(t))$, which will give you $v(t)$. If the acceleration is constant, then $A(x)=\alpha(x-x_0)$, and the expressions should reduce to the familiar formulae.