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Many derivations of the Lorentz transformations assume they must be linear maps on $\mathbb R^4$, where we identify the components of $\mathbb R^4$ with orthogonal coordinate systems associated to each observer. I've heard the following argument that linearity follows from the principle of relativity: In any inertial reference frame, an object with no forces on it follows a straight line through spacetime. Thus, Lorentz transformations must preserve straight lines, and hence they are linear.

However, the above argument is flawed because objects can only move on timelike paths, so strictly speaking we should only assume that straight lines with $\|\vec v\|<c$ remain straight. I.e., if $T$ is a Lorentz transformation and

$$L_{\vec v,\vec s}:=\{(ct,\vec s+\vec vt)\,|\,t\in\mathbb R\},$$

we insist that for all $\vec v,\vec s\in\mathbb R^3$ with $\|\vec v\|<c$, we have

$$(1)\qquad T(L_{\vec v,\vec s})=L_{\vec v',\vec s'}$$

for some $\vec v',\vec s'\in\mathbb R^3$.

Is $(1)$ enough to imply linearity of $T$? If not, what about in conjunction with the second postulate? Again if not, can we extend our usage of the principle of relativity to cover all straight lines, not just timelike ones?

Note that unlike here, I am not assuming the spacetime interval is preserved between all pairs of points. It is argued here that linearity follows from (a particular version of) homogeneity and continuity. I'd prefer to lose the continuity assumption (despite its naturalness) and replace it with a more straightforward application of the principle of relativity.

Qmechanic
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WillG
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  • Said another way: if $ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$ is defined in one inertial frame $K$, and we perform a coordinate transform to another (inertial!) $K'$ by setting $t = t(t',x',y',z')$ etc..., plug it into $ds^2$, and then treat the $(t',x',y',z')$ as parametrized by the $ds'^2 = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2$ in $K'$, why does this imply $ds^2 = ds'^2$ must hold? Why isn't it more generally $ds^2 = (\frac{ds}{ds'})^2 ds'^2$ where $(\frac{ds}{ds'})^2 = a(v) \neq 1$, which allows non-linear transforms? See this. – bolbteppa Nov 29 '21 at 07:36
  • This perspective immediately lets you see how to generalize to GR by considering what fails when $K'$ is non-inertial on combining the conclusions with the principle of equivalence, and the arguments relate back directly to the postulates. – bolbteppa Nov 29 '21 at 07:44

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