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I'm trying to understand which assumption are necessary to prove the invariance of the spacetime interval $$\Delta s^2=c^2\Delta t^2-\Delta \mathbf{x}^2$$ in special relativity. The postulates of special relativity are:

  • Principle of relativity

  • speed of light is constant in every frame of reference

From the second postulate is evident that $$ds'^2=0 \iff ds^2=0.$$ Then from the answer Invariance of spacetime interval directly from postulate we may see how this, and the fact that the two infinitesimal are of the same order, leads to $$ds'^2=ads^2.$$ In some other answers (e.g.Proving invariance of $ds^2$ from the invariance of the speed of light) it is pointed out that $ds'^2=ads^2$ arises from the fact that my coordinate transformation is linear.

My question is: what is the assumption I use to prove $ds'^2=ads^2$? Do I need linearity and if so where do I use it? And what about the principle of relativity?

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  • $\begingroup$ You need linearity of coordinates transformations because a free particle moving in spacetime should follow a straight line in any inertial frame (relativity theory should be compatible with the principle of inertia). You also need homogeneity of spacetime, and isotropy at each point of spacetime (which implies homogeneity). $\endgroup$ – Cham Oct 20 '18 at 13:40
  • $\begingroup$ Thank you for your answer. I know I need linearity so that "a non-accelerated motion is mapped into another non accelerated motion". But I don't see how linearity plays a role into the specific issue of the invariance of the spacetime interval between two inertial frames. $\endgroup$ – Luthien Oct 20 '18 at 13:45
  • $\begingroup$ If the metric takes a specific form (Minkowski) in a given inertial frame, then it should obviously have the same form in all other inertial frames since these frames are all equivalent. You can't distinguish the frames from their metric. $\endgroup$ – Cham Oct 20 '18 at 13:51
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    $\begingroup$ For my personnal notes on relativity, I wrote 4 pages on this problem a long time ago. If you can read French, I could send them to you by email. It's essentially the same demonstration as the one found in Landau/Lifchitz's old book. You could also check Ohanian's book : Gravitation and spacetime. I think there's a nice demonstration in there. $\endgroup$ – Cham Oct 20 '18 at 14:15
  • $\begingroup$ Related : Proof that spacetime interval is invariant $\endgroup$ – Frobenius Oct 20 '18 at 16:06
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I think there are two ways to argue this:

  1. via the principle of relativity: the metric should be the same in all inertial frames, thus $ds^2 = ds'^2$ immediately. (However it's not immediate that the metric should be Minkowski)

  2. via the constancy of the speed of light: $ds^2=0$ if and only if $ds'^2=0$, then using linearity as in Valter Moretti's linked answer we must have $ds'^2 = a ds^2$. One can then argue that $a=1$.

Note that linearity of Lorentz transformations is required in the second path. One can show that homogeneity implies that Lorentz transformations are affine. In this case this is sufficient for linearity because we only care about differences $\Delta x$, which are insensitive to an added constant.

As to the question: where do I use linearity? the answer is in Valter Moretti's proof to the theorem linked above.

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  • $\begingroup$ So, correct me if I'm wrong: the fact that that the two infinitesimal are of the same order plus the constancy of the speed of light make me say that $ds′2=ads2$. Using homogeneity of space (THUS linearity) I can set $a=1$. So in this proof my assumptions are: constancy of the speed of light and homogeneity of space (i.e. linearity), I don't need to use the principle of relativity. Am I right? $\endgroup$ – Luthien Oct 21 '18 at 12:23
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    $\begingroup$ Costancy of the speed of light plus linearity (or infinitesimals of the same order) gives $ds'^2=a ds^2$. The reason why you can set $a=1$ on the other hand has nothing to do with linearity, but rather with homogeneity and isotropy. This is explained by Landau in the first few pages of The Classical Theory of Fields. $\endgroup$ – John Donne Oct 21 '18 at 12:45

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