## If it quacks like a parabola…September 21st, 2011 Patrick Stein

I am working on a game idea that involves (special) relativistic mechanics instead of classical mechanics. Working out the details that I needed was easy enough if I assumed that:

• ships had a maximum speed at which they could move relative to my base frame
• ships could instantly go from stopped to maximum speed or vice-versa

I didn’t like those assumptions at all. So, I started playing with the equations for relativity. In classical mechanics, the rate-of-change of velocity equals the force you’re applying divided by your mass: $\frac{dv}{dt} = \frac{F}{m}$.

In special relativity, your mass increases with velocity. So, that equation becomes: $\frac{d\left(\frac{v}{\sqrt{1-v^2}}\right)}{dt} = \frac{F}{m_0}$ (assuming units where the speed of light is 1 unit of distance per 1 unit of time and $m_0$ is your rest-mass).

For the purposes of this post, I’m going to assume the simplest initial conditions: you start motionless and at the origin. For ease of notation, let $a = \frac{F}{m_0}$. Solving the above differential equation to get a formula for velocity and solving the resulting differential equation to get the distance $x$ you’ve travelled in my base frame by time $t$, the answer comes out to: $x(t) = \frac{1}{a}\left(-1 + \sqrt{1+a^2t^2}\right)$.

I have solved this problem at least thirty times in the past two months. Sometimes I used the simple initial conditions as above. Sometimes I did it in all of its gory details (including the messy case where the applied force is not aligned with the current velocity).

I got the same answer (well, plus the extra mess when I did the full-on problem) every way that I tried it.

So, why did I do it over and over again?

If this were classical mechanics, the end equation would have been $x(t) = \frac{1}{2}at^2$. And, I know that for low velocities, the classical mechanics answer should be almost identical to the special relativity answer. And, there was no way that I thought $\frac{1}{a}\left(-1 + \sqrt{1+a^2t^2}\right) \approx \frac{1}{2}at^2$.

I knew what the graph $x = \sqrt{a^2t^2}$ looked like when $t \ge 0$. It is a straight line. It doesn’t look much like the parabola $x = \frac{1}{2}a t^2$ at all.

My assumption was that since $x = \sqrt{a^2t^2}$ was a straight line for $t \ge 0$, then $x = \sqrt{1 + a^2t^2}$ would be a straight line shifted up one unit and bent (concave-down) a little bit like the graph of $x = \sqrt{at}$ is bent.

Boy was I wrong. Here is a plot of the two together (created with fooplot). The red line is the classical mechanics version. The black line is the relativistic version. Here, the force is such that the body is accelerating at a rate of the speed of light per second so they’ve already gotten up to around 28,000 miles per second before you can see any separation in the graphs here.

Definitely, I can see the resemblance. Now, to fix my intuition about square-roots.