If two different objects (for example two rockets) move in opposite direction at close to the speed of light (for example 0.8c and 0.9c), how do I calculate the difference in speed between the two (which in classical physics would be 0.8c + 0.9c)?

  • $\begingroup$ Related: physics.stackexchange.com/q/23625/2451 $\endgroup$
    – Qmechanic
    Jun 30, 2012 at 13:12
  • $\begingroup$ I Am probably wrong but the difference is how much faster one is moving compared to the other? Based on the mathmatical definition here mathsisfun.com/definitions/difference.html you subtract higher number from lower number and the answer is the difference $\endgroup$
    – Argus
    Jun 30, 2012 at 18:44
  • $\begingroup$ Am I confusing The grammar here such as asking what is the rate of egress and not the difference in speed? $\endgroup$
    – Argus
    Jun 30, 2012 at 18:47
  • $\begingroup$ @Argus Ah, yes, I meant the total speed of the rockets closing in, not the difference of speed. Thanks for pointing that out. $\endgroup$
    – Quispiam
    Jul 1, 2012 at 18:25
  • $\begingroup$ @alfredCentauri: yeah alfred seems to be on a roll that last few days kudos for your dedication to our site :) $\endgroup$
    – Argus
    Jul 1, 2012 at 18:31

2 Answers 2


First, let's be clear on the physical setup here. Suppose that, in the reference frame of the Earth, there is a rocket moving in one direction at 0.8 c and another rocket moving in the opposite direction at 0.9 c. In this reference frame, the distance between the two rockets is increasing at a rate of 1.7 c. That's OK because, in this frame, no thing is observed to be traveling faster than light.

However, to determine the speed of one rocket, as observed in the reference frame of the other rocket, we must use the relativistic velocity addition formula since we are combining speeds from two different reference frames:

$s = {v+u \over 1+(vu/c^2)}$

Where, in this case, $v$ is the velocity of one rocket in Earth's reference frame and $u$ is the velocity of Earth in the other rocket's reference frame.

  • $\begingroup$ Alfred Centauri. What a great name. $\endgroup$ Dec 12, 2018 at 15:08

A very useful concept in thinking about this issue is "rapidity". There is a slightly opaque (IHMO) article on this subject in Wikipedia, but in one dimension, it seems intuitively rather close to the old Star Trek "warp" notion used to describe speed.

The great thing is that it is additive between frames, and at low speeds v (as compared to the speed of light c) it reduces to (v/c), usually denoted beta.

The rapidity u of an object traveling at velocity v with respect to an observer is:

u = arctanh(v/c) = arctanh(beta).


v = c*tanh(u).

Here tanh is the hyperbolic tangent, and arctanh is its inverse function. As for tha analogous trig functions, tanh = sinh/cosh.

The rapidity u of a rocket (or other object) in a given frame is just the speed of that rocket, as would be calculated non-relativistically by an inertial guidance system on board the rocket itself, after it accelerated from rest to its given velocity v.

It is the speed you would naively calculate if you were on the ship, using just an accelerometer and the ship's clock, and knowing nothing of Relativity.

If your rapidity is 1, you are are traveling at ~0.761*c; this is very nearly the speed you would reach if you accelerated in a straight line at 1g for 1 year.

Rapidity 2 is nearly what you would reach if you accelerated at 1g for 2 years, when v would be ~ 0.964*c; etc. Call it "warp 2"!



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