For objects with a speed much lower than the speed of light, you can simply add and subtract velocities based on their vector orientation to one another. I'm aware that this does not apply to relativistic objects. A result of this is that the relativistic velocity-addition formula is needed to calculate the relative velocities, which is described as:

$u = \frac{u'+v}{1+\frac{u'v}{c^{2}}}$

Wherein, if I'm correct, $u$ is the object's velocity relative to reference frame $S$, $u'$ is an object's velocity relative to a reference frame $S'$ and $v$ is the velocity of reference frame $S'$ relative to reference frame S.

Where do I go wrong in my thinking, or what am I missing?: Say for example, if there are two spaceships: Spaceship A, which has a speed of $c/2$, and B has a speed of $c/3$. And we want to know what the speed of A measured in B. B is "stationary" and thus the "net" speed of spaceship A in the frame of B can be approximated by $c/6$, which is obviously wrong since were talking about relativistic velocities.

So I would think that I would need the formula described above in which $u' = c/6$ (the speed relative to reference frame $S'$, ($(\frac{3}{6} - \frac{2}{6}$)c) and $v = c/2$. Substituting this gives: $u = \frac{c/6 +c/2}{7/6} = \frac{4}{7}c$, which is bigger than the original speed of spaceship A. I have a strong feeling that I'm doing some conceptual things wrong.

  • 1
    $\begingroup$ Your formula should be $u = \frac{u'+v}{1+\frac{u'v}{c^{2}}}$ $\endgroup$
    – RC_23
    Dec 7, 2022 at 0:32
  • $\begingroup$ The relativistic velocity formula applies at all velocities. But the denominator being close to 1 is then basically below measurable thresholds. $\endgroup$
    – Triatticus
    Dec 7, 2022 at 1:02
  • $\begingroup$ @Triatticus Oke clear, thank you. So the numerator is basicly the classical way of velocity addition? $\endgroup$
    – measurepvm
    Dec 7, 2022 at 10:13
  • $\begingroup$ It is in a sense, as the answerer confirms. $\endgroup$
    – Triatticus
    Dec 7, 2022 at 14:45

1 Answer 1


The Newtonian relative speed is $c/2 - c/3$. The relativistic formula should have the same numerator as the Newtonian formula, and the denominator should use the same operator (addition or subtraction):

$$\frac{c/2 - c/3}{1 - \frac{(c/2)(c/3)}{c^2}} = c/5$$

$c/5$ is the speed of ship A in the rest frame of ship B.

To rederive the speed of A relative to the lab frame:

$$\frac{c/3 + c/5}{1 + \frac{(c/3)(c/5)}{c^2}} = c/2$$

Taking the Newtonian relative speed $c/6$ and plugging it into the special-relativistic formula will give you nonsensical results.

  • $\begingroup$ So when two object's are moving towards each other the denominator makes sure the speeds are relatively higher in the frame of on of them? And for two object's moving in the same direction visa versa? Is this why both de denominator and numerator change signs based on the vector direction of both objects? $\endgroup$
    – measurepvm
    Dec 7, 2022 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.