0
$\begingroup$

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 $A$. ($v_{BA}<c$) 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 $c/{\sqrt{2}}<v_{BA}<c$, $v_{AB}$ 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – harshit54 Jan 13 at 11:01
  • $\begingroup$ I am afraid not! $\endgroup$ – Mohammad Javanshiry Jan 13 at 11:03
  • $\begingroup$ 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. $\endgroup$ – harshit54 Jan 13 at 11:05
  • $\begingroup$ 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. $\endgroup$ – Mohammad Javanshiry Jan 13 at 11:13
  • $\begingroup$ Okay. Sorry for my misunderstanding. $\endgroup$ – harshit54 Jan 13 at 11:14
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ For God's sake, forget about that Lorentz transformation for a couple of days! I just want to know if an asymmetric relativity is empirically tenable. $\endgroup$ – Mohammad Javanshiry Jan 13 at 15:37
  • 3
    $\begingroup$ That is a rather excessive response. The Lorentz transform is empirically verified, so all empirically tenable theories must be empirically equivalent to the Lorentz transform $\endgroup$ – Dale Jan 13 at 15:53
  • $\begingroup$ Even if the Lorentz transformation is experimentally correct, this does not mean that other possible transformations are not or cannot be empirically verified. $\endgroup$ – Mohammad Javanshiry Jan 13 at 15:59
  • $\begingroup$ Besides, there are still some ambiguities with, e.g., the length contraction in SR. I think there is not any performed experiment that has directly verified/detected the length contraction. $\endgroup$ – Mohammad Javanshiry Jan 13 at 16:06
  • $\begingroup$ “this does not mean that other possible transformations are not or cannot be empirically verified.” It does mean that any such transforms must be empirically equivalent to the Lorentz transform. I.e. they might redefine v, but when you go to measure velocities using e.g. Doppler radar, then you will get u=-v. Also, remember this site is for mainstream physics $\endgroup$ – Dale Jan 13 at 16:12

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.