The free, dispersionless wave equation on \((ct,\textbf x)\in\textbf R\times \textbf R^d\) is:
\[\frac{\partial^2\psi}{\partial (ct)^2}-\frac{\partial^2\psi}{\partial|\textbf x|^2}=0\]
or more simply as \(☐^2\psi=0\), where the d’Alembert operator is defined by \(☐^2:=\frac{\partial^2}{\partial (ct)^2}-\frac{\partial^2}{\partial|\textbf x|^2}=\partial^{\mu}\partial_{\nu}=\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}\) with the usual metric \(\eta=\text{diag}(1,-1,-1,-1)\) defining the hyperbolic geometry of Minkowski spacetime. In \(d=1\) spatial dimension, d’Alembert discovered an \(\text{SO}(2)\) linear transformation on \(1+1\)-dimensional spacetime \(\textbf R\times\textbf R\) which substantially simplified the form of the d’Alembert operator \(☐^2\). Specifically, consider the spacetime coordinates \((u,v)\) rotated \(45^{\circ}\) counterclockwise from \((ct, x)\) via:
\[\begin{pmatrix}u\\v\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix} ct\\ x\end{pmatrix}\]
Then the claim is that \(☐^2=2\frac{\partial^2}{\partial v\partial u}\) is a composition of directional derivatives along the orthogonal “light-like” worldlines in spacetime. This follows from two observations. First, notice that the \(1+1\)-dimensional d’Alembert operator \(☐^2\) is separable via a difference of squares into two transport operators \(☐^2=\left(\frac{\partial}{\partial(ct)}+\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial(ct)}-\frac{\partial}{\partial x}\right)\) (this trick only works in \(d=1\) spatial dimension!). Then, it suffices to show that \(\frac{\partial}{\partial v}\) corresponds to the first factor and likewise that \(\frac{\partial}{\partial u}\) corresponds to the second factor. The chain rule asserts that partial derivative operators transform under the transpose of the Jacobian:
\[\begin{pmatrix}\partial/\partial u\\ \partial/\partial v\end{pmatrix}=\left(\frac{\partial(ct,x)}{\partial(u,v)}\right)^T\begin{pmatrix} \partial/\partial(ct)\\ \partial/\partial x\end{pmatrix}\]
On the other hand, it is clear that the inverse Jacobian \(\frac{\partial(u,v)}{\partial(ct,x)}=\left(\frac{\partial(ct,x)}{\partial(u,v)}\right)^{-1}\) is given by (as it must, being the linearization of a map!):
\[\frac{\partial(u,v)}{\partial(ct,x)}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\]
So because \(\frac{\partial(u,v)}{\partial(ct,x)}\in\text{SO}(2)\triangleleft\text{O}(2)\), inverting it is precisely the same as transposing it. The involutory nature of transposition then yields the final result:
\[\begin{pmatrix}\partial/\partial u\\ \partial/\partial v\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix} \partial/\partial(ct)\\ \partial/\partial x\end{pmatrix}\]
consistent with the usual formulas for directional derivatives. From here, it is straightforward to directly integrate the free, dispersionless wave equation to see that the most general free, dispersionless wave \(\psi\in\ker(☐^2)\) in \(d=1\) spatial dimension is the superposition of a left travelling wave \(\psi_{\leftarrow}(x+ct)\) and a right travelling wave \(\psi_{\rightarrow}(x-ct)\), i.e. \(\psi(ct,x)=\psi_{\leftarrow}(x+ct)+\psi_{\rightarrow}(x-ct)\).