# Why do two moving inertial frames of reference atribute the same velocity to each other?

In both relativity and Newtonian mechanics, it is a self-evident axiom according to which two inertial observers who move relative to each other measure the same velocity of, say, $$v$$. I want to know why this is so. One may say that our daily experience, easily verifies this postulate, however, to me, it is not convincing.

Assume that two objects $$A$$ & $$B$$ are at rest with respect to each other. Then, $$B$$ decides to approach $$A$$ at a constant velocity after undergoing an arbitrary acceleration. If $$A$$ measures the velocity of $$B$$ as $$v_{BA}$$, I can easily claim that $$B$$ measures the velocity of $$v_{AB}$$ for $$A$$ so that for non-relativistic velocities we always have: $$v_{BA} \approx v_{AB}=v$$
However, which experiments have verified the above equation thus far for when the velocities are relativistic?

For better compatibility with experiment, assume that $$B$$, who has already undergone an acceleration, can never exceed the speed of light as measured by $$A$$. ($$v_{BA}) Now, assume that $$v_{AB}$$ complies with the challenging equation below: $$v_{AB}=\gamma_{v_{BA}}v_{BA}$$ where $$\gamma_{v_{BA}}$$ is the traditional Lorentz factor. It is clear that if $$c/{\sqrt{2}}, $$v_{AB}$$ exceeds the speed of light so that for $$v_{BA}\approx c$$, $$v_{AB}$$ tends to infinity. If we want to disprove the odd equation above, we have to investigate the viewpoint of observer $$B$$ for when his velocity is a considerable portion of the speed of light as measured by $$A$$. According to our current technology, it seems impossible to set a human being in motion at a significant portion of the speed of light, and want him/her to tell us about their measurements as to the relative velocity of the surrounding objects especially the other observer ($$A$$)!

Although this deduction upsets the symmetry of special relativity, I am curious to know if there are some experiments against this idea.

• I am not familiar with special relativity. However isn't it true that two inertial observers moving relative to each other measure the same acceleration and not velocity. – harshit54 Jan 13 at 11:01
• I am afraid not! – Mohammad Javanshiry Jan 13 at 11:03
• Let's say that I move with $1m/s$ and you move at $2m/s$ in the same direction. If we observe a stationary object, you will measure $-2m/s$ and I will measure $-1m/s$ for the same object. – harshit54 Jan 13 at 11:05
• I think you misunderstood the question. There is no third object in my example. Assume that I move at $1m/s$ towards you. (Indeed, you attribute this velocity to me.) What velocity do I measure for you who approach me? This is the question. – Mohammad Javanshiry Jan 13 at 11:13
• Okay. Sorry for my misunderstanding. – harshit54 Jan 13 at 11:14

In addition to being intuitive and also required by the principle of relativity, this is actually fairly easy to derive directly from the Lorentz transform. For simplicity of notation I will suppress the y and z direction and do a Lorentz transform in matrix form using units where c=1. So if frame B is moving at velocity v with respect to frame A then the transform from A to B is $$\Lambda(v)= \left(\begin{array}{rr} {\gamma} & -{\gamma} v \\ -{\gamma} v & {\gamma} \end{array}\right)$$ where $$\gamma=(1-v^2)^{-1/2}$$. Now, the transform from B to A is $$\Lambda(u)$$ such that $$\Lambda(u)\Lambda(v)= \left(\begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array}\right)$$ The solution to this is $$\Lambda(u)=\Lambda(v)^{-1}=\Lambda(-v)$$. So the velocity of A with respect to B is equal and opposite of the velocity of B with respect to A.