I believe the question title is sufficiently explanatory - who was the first mathematician to announce/discover that

$$\nabla^2\left(\frac{1}{r}\right) = - 4\pi\delta^{(3)}(\mathbf{r})$$

where $r = || \mathbf{r}||^{1/2}$ and $\delta^{(3)}( \cdot )$ is the 3-dimensional delta-function?

While I'm tempted to think it was Gauss (as most proofs use the divergence theorem, which I believe was discovered by Gauss), I haven't been able to find any resources which confirm this, so does anyone have an answer? (Additionally, a source would be quite nice)

  • 3
    $\begingroup$ maybe Paul Dirac $\endgroup$ – smiley06 Jun 6 '13 at 16:26
  • $\begingroup$ @smiley06 While that wouldn't surprise me at all (and makes some sense considering which of my supervisors initially showed me this), searching for this result while mentioning Dirac hasn't helped bring anything up unfortunately. $\endgroup$ – Andrew D Jun 6 '13 at 16:38

Such questions tend to be hard to answer, because people may begin using a concept, and understand it quite well, before the concept has a name, notation or a definition. The statement $\Delta(1/r)=\delta_0$ has two parts:

  1. $\Delta(1/r)=0$ outside of singularity $r=0$ (easy)
  2. $\Delta(1/r)$ is $-4\pi $ times the $\delta$ function at $0$ (whatever that means)

The first part was already known to Laplace. See the paper by G. Green, "An essay on the application of mathematical analysis to the theories of electricity and magnetism II", J. Reine Angew. Math., 44 (1852) pp. 356–374. (You can find it in online archives of Journal für die reine und angewandte Mathematik.) After defining $$V=\int \frac{\rho'\,dx'\,dy'\,dz'}{r'}$$ Green says on page 357:

Laplace has shown, in his Mc. Celeste, that the function $V$ has the property of satisfying the equation $$0=\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}$$

I leave it for someone else to find where exactly in Mécanique céleste Laplace did that.

Detailed computations for the first part are also found in a 1839 memoir by Gauss "Allgemeine Lehrsatze in Beziehung Auf Die Im Verkehrten Verhaltnisse Des Quadrats Der Entfernung Wirkenden Anziehungs- Und Abstossungs-Krafte", which you can read here. On page 8 Gauss says

Die bekannte Gleichung $$\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=0$$ gilt also für alle Punkte des Raumes, die ausserhalb der wir kenden Massen liegen.

Now let's turn to the singularity issue. It seems that Laplace did not understand it at all, because he did not exclude it from the domain in which he claimed Laplace's equation to hold. See page 54 of historical comments to the aforementioned memoir by Gauss:

Dieselbe ist von Laplace in der bei Art. 2 und 3 citirten Arbeit, sowie in der Mecanique Celeste aufgestellt, doch ohne die nöthige Einschrän kung auf Punkte ausserhalb der Masse.

On page 54 we also read that Poisson was the first to point out the oversight of Laplace. He understood what the inhomogeneous term should be, but could not prove it rigorously.

Poisson hat zuerst (Bulletin de la societe philomatique 1812 t. 3 p. 388) darauf aufmerksam gemacht, dass die Laplace'sche Gleichung nur für Punkte ausserhalb der Masse gültig sei, und er hat dann für Punkte der Masse die Laplace'sche Gleichung durch die nach ihm selbst genannte ersetzt; doch ist es ihm nicht gelungen, diese Gleichung auf strenge Art zu be weisen. Dies blieb vielmehr Gauss in der vorliegenden Darstellung vorbehalten.

Wikipedia article on Poisson concurs that the 1839 proof by Gauss was first.

The statement with $-4\pi$ (sans not yet known $\delta$) appears on page 12:

... der Werth von $\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}$ äqual wird dem Producte ans $-4\pi$ in die in $O$ stattfindende Dichtigkeit.

  • $\begingroup$ I should add that I learned of these sources from the EoM article Newton potential. $\endgroup$ – 40 votes Jul 14 '13 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.