# How to fill the gap in this derivation of proper acceleration without four vectors?

I've scoured this site, the web, and some books for derivations of the formula for proper acceleration as a function of coordinate acceleration ($$\alpha = \gamma^3 a$$) without four vectors or hyperbolic trig functions. I've found a number of them but they all skip so many steps that they're impossible for me to fully follow. The most complete attempt I've seen is this one (see the "3-vector approach" section), and I can follow all of it except for part of one step—where it implies that

$$\frac{d}{dt}\frac{u_x - v}{1-\frac{vu_x}{c^2}} = \frac{1}{\gamma^2\left(1-\frac{vu_x}{c^2}\right)^2}\frac{d^2x}{dt^2}$$

I can't see how to get from the left side to the right side, so can you either break it down for me or provide another derivation altogether? I realize that proper acceleration can also be derived using four vectors or rapidity but neither of those are yet intuitive enough for me to give me a complete understanding of proper acceleration and how to use it or to fully convince me that the formula $$\alpha = \gamma^3 a$$ is correct. I'm fine with assuming that all acceleration is in the $$x$$ direction and parallel to the velocity.

The next best thing would be an explanation of why the formula should make sense intuitively. The next would be a four vector approach that's simple and fully spelled out. For example, this one was a good start but the norm of the four acceleration it ends with doesn't seem to give the expected result (not that it's wrong; I'm probably just missing something).

The only variable in $$(u_x - v)/(1-\frac{vu_x}{c^2})$$ which depends on time is the particle's velocity in the unprimed frame, $$u_x \equiv dx/dt$$. So applying the quotient rule for derivatives, $$\frac{d}{dt}\frac{g(t)}{f(t)} = \frac{f(t)\frac{d}{dt}g(t) - g(t) \frac{d}{dt}f(t)}{f(t)^2}$$, $$\begin{equation} \frac{d}{dt}\frac{u_x - v}{1-\frac{vu_x}{c^2}} = \frac{1}{\left(1-\frac{vu_x}{c^2}\right)^2}\left[ \left(1-\frac{vu_x}{c^2}\right) \left( \frac{d^2 x}{dt^2} \right) - \left(u_x - v \right) \left( - \frac{v}{c^2} \frac{d^2 x}{dt^2} \right) \right] = \frac{1}{\left(1-\frac{vu_x}{c^2}\right)^2}\left[ \left(1-\frac{vu_x}{c^2}\right) - \left(u_x - v \right) \left( - \frac{v}{c^2} \right) \right]\frac{d^2 x}{dt^2} \\ = \frac{1}{\left(1-\frac{vu_x}{c^2}\right)^2}\left[ 1-\frac{vu_x}{c^2} + \frac{vu_x}{c^2} - \frac{v^2}{c^2} \right]\frac{d^2 x}{dt^2} \\ = \frac{1}{\gamma^2\left(1-\frac{vu_x}{c^2}\right)^2}\frac{d^2x}{dt^2}. \end{equation}$$
• Excellent, thank you! I did try the quotient rule but didn’t find the right way of managing the terms to get the $\frac{1}{\gamma^2}$. Apr 14 at 22:09