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Let $O$ and $\overline{O}$ be inertial frames, $A$ and $B$ be events, the interval between $A$ and $B$ be $(\Delta t, \Delta x, \Delta y, \Delta z)$ in $O$ and $(\overline{\Delta t}, \overline{\Delta x}, \overline{\Delta y}, \overline{\Delta z)}$ in $\overline{O}$.

Is the linearity of the correspondence between $(\Delta t, \Delta x, \Delta y, \Delta z)$ and $(\overline{\Delta t}, \overline{\Delta x}, \overline{\Delta y}, \overline{\Delta z)}$ a postulate in the special relativity theory?

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marked as duplicate by John Rennie special-relativity Jan 9 '18 at 6:55

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  • $\begingroup$ Well I don't think so. The linear transformations namely Lorentz transformations are special because under them the laws of electromagnetism are invariant. But, one can say that as we deal with only inertial frames in special relativity, switching between them is done using linear transformations. However the postulates of special relativity, I think, makes no such assumptions. $\endgroup$ – Naman Agarwal Jan 9 '18 at 6:08
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  • $\begingroup$ @NamanAgarwal Couldn't it (or something that implies it) be an implicit assumption in special relativity? I was wondering about this question myself mainly because of GR. For example, isn't the statement that inertial space-time coordinates are flat everywhere an implicit assumption in SR? $\endgroup$ – SpiralRain Jan 9 '18 at 6:29
  • $\begingroup$ LT (Lorentz transformation) shows a linearity in the coordinates as a consequence of the principles of SR (special relativity). Another consequence is that the metric of IRF's (inertial reference frame) is flat. For a demonstration, refer to my answer to the question: "Distance formula in Euclidean space vs. Spacetime Interval - why is one Pythagorean and one not?". $\endgroup$ – Michele Grosso Jan 9 '18 at 17:02
  • $\begingroup$ Here the link: physics.stackexchange.com/questions/371338/… $\endgroup$ – Michele Grosso Jan 10 '18 at 15:01