Lorentz transformations are often defined to be linear. But suppose instead we only consider transformations that preserve the spacetime interval. Is it possible to prove that those transformations are affine (mapping midpoints to midpoints) mathematically?

That is to say, given the map $d:R^4 \times R^4 \to R$ defined by $$d((t_{1},x_1,y_1,z_1),(t_2,x_2,y_2,z_2)) = (t_1-t_2)^2 - (x_1-x_2)^2 - (y_1-y_2)^2 - (z_1-z_2)^2.$$

And given a diffeomorphism $D$ of $R^4$ that preserves $d$. Can we show that $D$ maps midpoints to midpoints? An analogous theorem for euclidean spaces (Mazur-Ulam) suggests the affirmative. But the proof does not seem to carry over.

Ideally, I would like a proof that is comprehensible to the average undergraduate.


We shall write $\eta$ for the the Minkowski metric in $\mathbb{R}^{1+k}$ with $k\in\mathbb{N}$. That is, for points $x$ and $y$ in the spacetime, we have $$\eta_{\mu,\nu}\,x^\mu\,y^\nu=x^0y^0-x^1y^1-x^2y^2-\ldots-x^ky^k\,,$$ where we use the Einstein convention (in other words, $\eta$ is the matrix $\text{diag}(1,-1,-1,\ldots,-1)$ in which there are $k$ occurrences of $-1$). Now, the distance between $x$ and $y$ is $$d(x,y)=\eta_{\mu,\nu}\,(x-y)^{\mu}\,(x-y)^{\nu}\,.$$ Suppose that $T$ is an isometry of the spacetime with respect to $d$. Write $\partial_\mu$ for the partial derivative with respect to the $\mu$-th coordinate. We want to show that $T$ is an affine transformation. That is, $\partial_\mu T$ is constant for every $\mu$.

The isometry condition $d\big(T(x),T(y)\big)=d(x,y)$ leads to $$\eta_{\rho,\sigma}\,\big(T(x)-T(y)\big)^\rho\,\big(T(x)-T(y)\big)^\sigma=\eta_{\rho,\sigma}\,(x-y)^\rho\,(x-y)^\sigma\,.$$ Taking the partial derivative with respect to $x^\mu$ and using the symmetry of $\eta$, we have $$2\,\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(T(x)-T(y)\big)^\sigma=2\,\eta_{\rho,\sigma}\,\delta^\rho_\mu\,(x-y)^\sigma=2\,\eta_{\mu,\sigma}\,(x-y)^\sigma\,,$$ where $\delta$ is the Kronecke delta. Hence, $$\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(T(x)-T(y)\big)^\sigma=\eta_{\mu,\sigma}\,(x-y)^\sigma\,.$$

Now, taking the partial derivative with respect to $y^\nu$ yields $$-\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(\partial_\nu T(y)\big)^\sigma=-\eta_{\mu,\sigma}\delta^\sigma_\nu=-\eta_{\mu,\nu}\,,$$ or $$\eta_{\mu,\nu}=\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(\partial_\nu T(y)\big)^\sigma$$ for all spacetime locations $x,y$.

Finally, $$ \begin{align} \eta_{\rho,\sigma}\big(\partial_\mu T(x)\big)^\rho\left(T(x)-T(y)-\partial_\nu T(y)\cdot (x-y)^\nu\right)^\sigma=&\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(T(x)-T(y)\big)^\sigma \\ &-\left(\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(\partial_\nu T(y)\big)^\sigma\right)(x-y)^\nu \\ =&\eta_{\rho,\sigma}\,\big(\partial_\mu T(x)\big)^\rho\,\big(T(x)-T(y)\big)^\sigma-\eta_{\mu,\nu}\,(x-y)^\nu \\ =&0\,. \end{align}$$ Because $T$ is an isometry, its Jacobian matrix $\left[\left(\partial_\mu T(x)\right)^\nu\right]_{\mu,\nu\in\{0,1,2,\ldots,k\}}$ is invertible. Hence, the vectors $v_\mu(x)\in\mathbb{R}^{1+k}$ defined by $\left(v_\mu(x)\right)^\nu:=\left(\partial_\mu T(x)\right)^\nu$ span the whole $\mathbb{R}^{1+k}$ (technically, this should be the tangent space $\mathcal{T}_x\mathbb{R}^{1+k}$, but due to flatness of the space, we may identify $\mathcal{T}_x\mathbb{R}^{1+k}$ as $\mathbb{R}^{1+k}$). Note that the equation $$\eta_{\rho,\sigma}\big(v_\mu(x)\big)^\rho\left(T(x)-T(y)-\partial_\nu T(y)\cdot (x-y)^\nu\right)^\sigma=0$$ implies that $$\eta_{\rho,\sigma}u^\rho\left(T(x)-T(y)-\partial_\nu T(y)\cdot (x-y)^\nu\right)^\sigma=0$$ for any vector $u \in\mathbb{R}^{1+k}$. Since $\eta$ is nondegenerate, $$\left(T(x)-T(y)-\partial_\nu T(y)\cdot (x-y)^\nu\right)^\sigma=0\,.$$ Ergo, $$T(x)=T(y)+\partial_\nu T(y)\cdot(x-y)^\nu$$ for all spacetime locations $x,y$. Taking derivative with respect to $x^\mu$, we obtain $$\partial_\mu T(x)=\partial_\nu T(y)\,\delta^\nu_\mu=\partial_\mu T(y)\,.$$ Hence, $\partial_\mu T$ is constant and the result follows.

There is an even simpler argument not assuming differentiability of $T$. Let $S(x):=T(x)-T(0)$ for all $x$. Then, $S(0)=0$ and $S$ is still an isometry. For every $x,y$, $\eta_{\mu,\nu}\,\big(S(x)\big)^\mu\,\big(S(x)\big)^\nu=\eta_{\mu,\nu}\,x^\mu\,x^\nu$, $\eta_{\mu,\nu}\,\big(S(y)\big)^\mu\,\big(S(y)\big)^\nu=\eta_{\mu,\nu}\,y^\mu\,y^\nu$, and $\eta_{\mu,\nu}\,\big(S(x)-S(y)\big)^\mu\,\big(S(x)-S(y)\big)^\nu=\eta_{\mu,\nu}\,(x-y)^\mu\,(x-y)^\nu$. With some algebraic manipulation, we get $$\eta_{\mu,\nu}\,\big(S(x)\big)^\mu\,\big(S(y)\big)^\nu =\eta_{\mu,\nu}\,x^\mu\,y^\nu\,.$$ By choosing $n$ distinct values of $y$ such that $S(y)$'s are linearly independent, we can solve for $S(x)$ as a linear function of $x$ (where the nondegeneracy of $\eta$ is once used again). Thus, $S$ is a linear transformation, so $T$ is an affine transformation.

This shorter argument can be used for non-flat metric tensors as well. It can be applied to determine every isometry of $\mathbb{R}^{r+k}$ with respect to a metric of signature $(r,k,0)$, where $r,k\in\mathbb{N}$, if the images if some $r+k+1$ points are known. (Not every collection of $r+k+1$ works.)

P.S. This argument works for any flat metric tensor on any $\mathbb{R}^{r+k}$, where $r,k\in\mathbb{N}$, with signature $(r,k,0)$, not just the Minkowski metric. That is, any isometry of $\mathbb{R}^{r+k}$ with respect to a nondegenerate flat metric is always an affine transformation.


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