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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).

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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}

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  • $\begingroup$ Excellent, thank you! I did try the quotient rule but didn’t find the right way of managing the terms to get the $\frac{1}{\gamma^2}$. $\endgroup$ Apr 14 at 22:09

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