The principle of relativity and why Inertial frames attribute the same velocity to one another

Question

In introductory texts introducing relativity, it is always assumed that frames measure the same velocity for each other. For example if frame S' moves at velocity v with respect to respect, then S moves at velocity -v in the frame of S'. This is intuitive, but in a theory where we can't assume lengths and times as staying the same between frames, how can we assume velocities being the same.

Most textbooks explain this away with the principle of relativity (first postulate), that laws of physics stay the same between frames. I feel I don't really understand this principle or how it answers my question so I suppose my question is about the specifics of the principle of relativity at large.

note: I understand other questions very similar to this have been asked before, however none of the answers I found really explained this specific question.

Answer 1

The following is not a proof from first principles, but merely a result assuming that the special relativity is consistently represented on a spacetime diagram with its Minkowskian spacetime geometry.

The short answer is "similar triangles".

• Consider Euclidean geometry.
  The measure of the signed-angles satisfy $$\measuredangle TOQ = -\measuredangle UOZ,$$ which can be realized using Euclidean-arc-length of the unit-circular-arc intercepted by the two rays $OT$ and $OU$, or the area of the unit-circular sector intercepted by the two rays $OT$ and $OU$.
  
  Using trigonometry, with slope $m=\frac{OPP}{ADJ}=\tan\phi$, we have $$\frac{TQ}{OT}=\tan\phi_{TOQ} = -\tan\phi_{UOZ}=-\frac{UZ}{OU}.$$ ($TQ$ is tangent to the circle of radius $OT$. So, $TQ$ is perpendicular to $OT$.
  $UZ$ is tangent to the circle of radius $OU$. So, $UZ$ is perpendicular to $OU$.)
  

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• Now, consider Minkowskian spacetime-geometry.
  The measure of the signed-rapidities (where I would like an arc of a hyperbola in the angle-symbol) satisfy $$\angle TOQ = -\angle SOZ,$$ which can be realized using Minkowski-arc-length of the unit-Minkowoski-circular-arc (an arc of the unit-hyperbola) intercepted by the two rays $OT$ and $OS$, or the area of the unit-Minkowski-circular sector intercepted (a sector of the unit-hyperbola) by the two rays $OT$ and $OS$.
  
  Using hyperbolic trigonometry, with velocity $v=\frac{OPP}{ADJ}=\tanh\theta$, we have $$\frac{TQ}{OT}=\tanh\theta_{TOQ} = -\tanh\theta_{SOZ}=-\frac{SZ}{OS}.$$ ($TQ$ is tangent to the Minkowski-circle of radius $OT$. So, $TQ$ is Minkowski-perpendicular to $OT$.
  $SZ$ is tangent to the Minkowski-circle of radius $OS$. So, $SZ$ is Minkowski-perpendicular to $OS$.)
  
  You can verify the equality by counting the diamonds along the indicated segments. (The "rotated graph paper" helps draw the diamonds for "nice" velocities [whose doppler-factor is rational, like $v=(3/5)c$ whose $k=2$].)