In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity:
Prove that any spacetime transformation $\left(y^0,y^1,y^2,y^3\right)\leftrightarrow \left(x^0,x^1,x^2,x^3\right)$ that preserves intervals, that is, such that
$$\left(dy^0\right)^2-\left(dy^1\right)^2-\left(dy^2\right)^2-\left(dy^3\right)^2=\left(dx^0\right)^2-\left(dx^1\right)^2-\left(dx^2\right)^2-\left(dx^3\right)^2$$
is linear (assuming that the origins of both coordinates coincide). That is, show that $\frac{\partial y^i}{\partial x^j}=L_j^i$ is constant throughout spacetime (that is, show that $\frac{\partial L_j^i}{\partial x^k}=0$).
Thus far all I've been able to prove is that $g_{ij}L_p^iL_q^j=g_{pq}$ (where $g_{ij}$ is the metric tensor of special relativity) and that $\frac{\partial L_j^i}{\partial x^k}=\frac{\partial L_k^i}{\partial x^j}$. Any further ideas?
Answer
In hindsight, here is a short proof.
The metric $g_{\mu\nu}$ is the flat constant metric $\eta_{\mu\nu}$ in both coordinate systems. Therefore, the corresponding (uniquely defined) Levi-Civita Christoffel symbols
$$ \Gamma^{\lambda}_{\mu\nu}~=~0$$
are zero in both coordinate systems. It is well-known that the Christoffel symbol does not transform as a tensor under a local coordinate transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation,
$$\frac{\partial y^{\tau}}{\partial x^{\lambda}} \Gamma^{(x)\lambda}_{\mu\nu} ~=~\frac{\partial y^{\rho}}{\partial x^{\mu}}\, \frac{\partial y^{\sigma}}{\partial x^{\nu}}\, \Gamma^{(y)\tau}_{\rho\sigma}+ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}. $$
Hence all the second derivatives are zero,
$$ \frac{\partial^2 y^{\tau}}{\partial x^{\mu} \partial x^{\nu}}~=~0, $$
i.e. the transformation $x^{\mu} \to y^{\rho}=y^{\rho}(x)$ is affine.
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