I like some 'science' in my 'science fiction', so I started crunching out the kinematic equations for some of the scenarios my characters are getting involved in, and ran smack dab into an issue. (Please excuse my formatting, I can't figure out sub/superscript notations)

$v_f = \sqrt{2ad}$, with $v_i$ assumed to be 'close enough' to zero. That's easy enough to crunch the math on.

Except... With sufficiently high values for $a$ or $d$, you start crowding, or even violating, the light speed limit.

And I can't find the equations to help handle relativistic distortion! I know they exist because I remember working with relativity when I took my physics class, years and years ago.

  • $\begingroup$ Why do you need this equation is my first question. Without the context it is hard to suggest the right equation to use --also your Newtonian equation is wrong (it is missing a square root over 2ad). $\endgroup$
    – Apoorv
    Jul 21, 2012 at 9:24
  • $\begingroup$ Sorry I forgot to add more context to the question. It's for a science-fiction story, in which I discovered my numbers had the potential to bump up against C -- which I know is a no-no. $\endgroup$
    – RonLugge
    Jul 22, 2012 at 1:41

2 Answers 2


I'm guessing you really just want the appropriate equations rather than an in depth treatment of relativistic acceleration. In that case read Phil Gibbs' article on the relativistic rocket. In particular, the velocity measured by an observer watching the rocket is:

$$ v = \frac{at} {\sqrt{1 + (at/c)^2}} $$

and the distance the rocket has travelled is:

$$ d = \frac{c^2}{a} \left(\sqrt{(1 + (at/c)^2} - 1\right) $$

Phil Gibbs doesn't give velocity as a function of distance, but it shouldn't be hard to use the second equation to substitute for $t$ in the first equation. That would give you velocity as a function of distance. Incidentally, as Monster Truck points out, the equation for distance is $v^2 = 2ad$ not $v = 2ad$, though I'm guessing that's just a typo.

Note that the variable $t$ in these equations is the time measured by the stationary observer watching the rocket. Because of the time dilation that happens at speeds near the speed of light, the time measured by the stationary observer and the the time measured by the crew of the rocket will not be the same. Because of this the equations give the velocity and distance of the accelerating rocket as measured by the stationary observer. The experience of the crew inside the rocket would be different.

In the unlikely event you feel the urge to find out how these equations are derived, borrow a copy of Gravitation by Misner, Thorne and Wheeler. The equations for the relativistic rocket are derived in chapter 6. However be warned that it's not easy going.

  • $\begingroup$ The article is actually by Philip Gibbs, with updates by Bill Woods and Don Koks, according to the credits at the top of the article. The web page (along with the rest of the Physics FAQ) is hosted and (according to the main page) maintained by Don Koks at the University of Adelaide, and a mirror is also hosted by John Baez at UC Riverside (hence your link to the page on John Baez's site). $\endgroup$
    – Matt
    Aug 18, 2014 at 20:47
  • $\begingroup$ The link to John Baez's article has been broken. Is this correct? $\endgroup$ Mar 29, 2017 at 17:26

I gues you want to simulate things in a game. In principle, you should make your dynamics Poincare-invariant. But the difficulty is that there is no nice $N$-particle dynamics with all the required properties (Jordan-Currie-Sudarshan theorem).

The 2-body case can be reduced to an external field problem by working in a common rest frame. But once you have more than two dynamic entities, you need to resort to approximations, not only for the numerics but also for the ideal dynamics.

Unfortunately, I don't know of any work on sensible relativistic approximations at speeds close to $c$, even in the 3-body case.

If you are always far from the speed of light, you can use the post-Newton approximation. http://www.math.ca/cjm/v1/cjm1949v01.0209-0241.pdf

  • $\begingroup$ Actually, it's a story -- not a game. But thanks for the response. $\endgroup$
    – RonLugge
    Jul 22, 2012 at 1:38

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