# D'Alembertian symbol $\Box$

The D'Alembertian is a generalization of the Laplacian operator to a space of arbitrary dimension and metric.

Where does the D'Alembertian symbol $\Box$ come from?

According to Wikipedia it has to do with 4-dimensional spaces, but the authors don't cite sources and I don't see why the relation.

• The laplacian in 3d is sometimes denoted by a triangle (sitting on its base: see en.m.wikipedia.org/wiki/Laplace_operator). In 4d, it becomes a square. – ZeroTheHero Apr 27 '18 at 0:28
• Why don't a diamond then? Maybe if you know where was this notation used for the first time I can find the context. – Ernesto Iglesias Apr 27 '18 at 0:33

For brevity we will introduce the following symbol, by setting, $$\Delta\mathrm U -\mathrm K_0\frac{d^2\mathrm U}{dt^2}=\Box\mathrm U$$
Here $\mathrm K_0=1\,/\,c^2$ (p. 336). It didn’t immediately catch on: it appears without comment in Poincaré’s (1906, p. 132) (cited by Wikipedia), but not in the book where Lorentz named it (1909, p. 17):
the operation $\Delta-\frac1{c^2}\frac{\partial^2}{\partial t^2}$ may be given the name of the Dalembertian (...) in commemoration of the fact that the mathematician d’Alembert was the first to solve a partial differential equation (...) which contains this operation, or rather the operation $\frac{\partial^2}{\partial x^2}-\frac1{c^2}\frac{\partial^2}{\partial t^2}$
In (1910, p. 175) Lewis (of photon fame) was still suggesting the notation $$\diamondsuit^2= \frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} + \frac{\partial^2}{\partial x_3^2} + \frac{\partial^2}{\partial x_4^2}\qquad(x_4=ict),$$ but Sommerfeld (1910, p. 661) and von Laue (1911, p. 71) chose again $\Box$ (pronounced “Delta”).