Assume someone can accelerate at $9.8 \frac{m}{s^2}$, or 1g, from their perspective, indefinitely (in a single direction). I want to compute how long it would take them to reach a given velocity from the reference frame of an outside observer.
A friend posed a question about how long it would take a theoretical rocket to travel light years without killing the occupants and I, having not touched special relativity in years, tried to work that out and failed.
Wikipedia tells me that the proper acceleration, $a^0$, is the acceleration in the inertial reference frames that instantaneously have the same velocity as the body, and that seems like the thing I'm constraining to 1g.
Given that's the case, Wikipedia gives $a^0=\frac{a}{(1-\frac{u^2}{c^2})^{\frac32}}$ which I rearranged as $a=a^0(1-\frac{u^2}{c^2})^{\frac32}$. I plugged in $a=\frac{du}{dt}$ and tried to solve: $$\begin{align*} \frac{du}{dt}&=a^0(1-\frac{u^2}{c^2})^{\frac32}\\ \frac{du}{dt}&=g(1-\frac{u^2}{c^2})^{\frac32}\\ du\,(1-\frac{u^2}{c^2})^{\frac23}&=g\,dt\\ \int(1-\frac{u^2}{c^2})^{\frac23}\,du&=\int g\,dt\\ \int(1-\frac{u^2}{c^2})^{\frac23}\,du&=gt \end{align*}$$ The last line would depend on initial condition, but if we assume we start at rest, then I think the $+C$ just drops out.
Thus it seems like if I just do the integral on the left, I'd have an expression in which I can plug in $u$ and get $t$ out.
So I chucked the whole thing into Mathematica. It told me the integral was some hypergeometric thing. But then I tried plugging in $c$ for $u$, expecting it to blow up, since I don't think the rocket would be able to ever reach $c$, but to my surprise it spit out $t=0.7$ years, which doesn't seem right.
Clearly I'm forgetting something very important (or just math-ing very wrong). Does anyone know what's up here? Thanks.
