Here, two objects A and B are moving relative to each other. I use the Einstein velocity addition formula $v = {v_1 + v_2\over 1 + {v_1 v_2\over c^2}}$ to calculate the relative speed between A and B, where $v_1$ (for A) and $v_2$ (for B) are as measured by each observer in their own inertial frames and are as relative to the observer (say there is an observer C, then $v_1$ and $v_2$ are measured by C relative to C). The question is will the observers all calculate the same value $v$ in their own frames? In particular, will A and B agree on the same value $v$? Let's say one calculates $v = 0.1c$, will others also get $0.1c$?

Another thought is if A and B are stationary to each other as measured by C, will another observer D also see A and B stationary to each other?


Is this question related to the principle of relativity, i.e.(copied from wikipedia), if a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K.

So in the above exmaple, if C sees A and B are stationary to each other, but D doesn't agree, wouldn't this be a violation of the principle of relativity?

  • $\begingroup$ I have updated the question to clear any confusion in wording $\endgroup$ Apr 11, 2015 at 11:12
  • $\begingroup$ Please note: I don't mean if A's relative velocity to B and B's relative velocity to A are opposite vectors. In case you're interested, surprisingly, they are not. See en.wikipedia.org/wiki/Thomas_precession. But this is completely different level of discussion than special relativity. $\endgroup$ Apr 12, 2015 at 0:09
  • $\begingroup$ I understand that you are not asking whether the velocity that A sees B move with is equal and opposite to what B sees A move with. However, those are indeed opposite. Thomas precession is about boosts in different directions. In the frame of A, thing B moves with a particular velocity. In the frame of B, thing A moves with a particular velocity and they are indeed opposite to each other. $\endgroup$
    – Timaeus
    Apr 12, 2015 at 2:16

2 Answers 2


The velocity addition formula you cite does not quantify what you think it quantifies. I'm going to say see and when I do, I mean "computes relative to its own frame".

The velocity addition formula describes the following setup. Frame 0 sees a particle (particle 1) moving in a direction at speed $v_1$. Frame 1 sees that particle 1 at rest and frame 1 sees another particle (particle 2) moving in the same direction at speed $v_2$. One can then ask. What is the speed at which frame 0 sees particle 2 move at? And the answer is:


It does not compute the speed that particle1/frame1 one sees particle2/frame2 move. That is actually what the $v_2$ represents. And the super important thing to understand is that if you associate a frame with both the particles that are moving, then there are three frames involved. This is essential to answer your updated question.

How does this bear on the principle of relativity? What if frame0 sees the two particles moving at the same speed? Then $v_1=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$ and we find that $v_2$=0, which is as expected if the two particles have no relative velocity.

And if instead of frame0 you used a different frame, $v_2$ would still be zero, since $v_2$ is literally defined as the speed of particle2 in the frame of particle1 so every frame whatsoever agrees on the numerical value of $v_2$, in this case, zero. So if you didn't use frame0 and you used a different frame you just get $$\frac{w+v_2}{1+\frac{wv_2}{c^2}}=\frac{w+0}{1+\frac{w0}{c^2}}=w.$$ In other words the other frame sees particle1 and particle2 move at the same speed (in this case $w$), which is consistent with $v_2=0$, that they have no relative motion.

If you want a formula for the relative velocity with everything measured in a single frame, just solve the equation for $v_2$, like

$$v_2=\frac{v-v_1}{1-\frac{vv_1}{c^2}},$$ which indeed looks very very similar.

And then yes, all the numbers above (where $v$ and $v_1$ are measured in the same frame) can be used to compute $v_2$ in such a way as to give the same answer ($v_2$) for any frame.

  • $\begingroup$ Thank you very much for answering. I think my confusion arises from the difference between "observed" relative velocity and "calculated" relative velocity. In your last equation, $v$ and $v1$ are the "observed" velocity in that frame, $v2$ is the "calculated" relative velocity in the frame, which has the same value as the "observed" relative velocity in frame1. $\endgroup$ Apr 12, 2015 at 11:44
  • $\begingroup$ I invented the term "calculated" relative velocity to differentiate $v2$ from the "observed" value in frame 0, which would be measured by taking a rule and a stop watch and do $v2' = \Delta x/\Delta t$, and of course $v2 \neq v2'$. $\endgroup$ Apr 12, 2015 at 11:53
  • $\begingroup$ @stackoverflower If you measure $v$ and $v_1$ in frame 0, then you can calculate $v_2$ using my last formula which is indeed what frame1 measures for particle 2. However, in your original question you talked as if my first equation computed that. They are similar looking equations, but different equations. $\endgroup$
    – Timaeus
    Apr 12, 2015 at 19:41

Let's say you have 3 systems. $B$ moving relative to $C$ with velocity $u$ and $A$ moving relative to $C$ with velocity $v$, all along one axis.

$A$ will "measure" for the velocity of $B$: $$ u' = \frac{u-v}{1-\frac{uv}{c^2}} $$

While $B$ will "measure" for the velocity of $A$: $$ v' = \frac{v-u}{1-\frac{uv}{c^2}} = -u' $$

It holds true that the velocity of $B$ relative to $A$ is the same in terms of magnitude but opposite in terms of direction than the velocity of $A$ relative to $B$. If this were not true, you couldn't define a proper transformation for both systems, since the transformation $A\rightarrow_v B \rightarrow_{\tilde{v}} A'$ would not yield $A'=A$. The systems $A$ and $B$ should obey the same unique transformation rules. There is nothing special about them.

  • $\begingroup$ There is a misunderstanding of my question. In your example, the quantity I want to measure is "the speed of C relative to A", and whether the value is the same when measured by B in his own system, and by A in his own system. $\endgroup$ Apr 11, 2015 at 11:09
  • $\begingroup$ @stackoverflower: permutated the letters. I think it's clearer now. You can check for yourself if $v=u=0$ that a observer D will see $\tilde{v}=\tilde{u}$. $\endgroup$
    – image357
    Apr 11, 2015 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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