Chladni figures are formed by sprinkling salt on a vertically vibrating membrane, f.e. [1.

Some of them are purported to be projections of hydrogen atom orbitals f.e. 1s, 2s, 2p, 3d, etc., e.g. 2.

One could bring it down to that the vibration patterns of Chladni figures are, as known, two-dimensional standing waves (this part is well established) and also to that -according to the Copenhagen (Born) interpretation- the hydrogen electron distribution also follows a (three-dimensional) standing wave pattern.

So far, so good, but what about the mathematical reasoning lying behind?

Could we claim that the two-dimensional Schrodinger equation for the hydrogen atom (i.e. the radial associated Laguerre polynomial, Ln) and the Chladni figures equation (i.e. the Bessel function of the first kind, Ja) are linear to each other? Any thoughts on the mathematical reasoning would be highly welcome.

Here is the complete list of relations between L and J I have come across: 3 and 4 and 5 and 6 (P being the associated Legendre polynomial which is turned to a constant in a 2D problem) but does anyone see a valid explanation in any of these relations?

  • 3
    $\begingroup$ The shapes are not the same. $\endgroup$
    – fqq
    Commented Aug 31, 2021 at 15:02

1 Answer 1


After you separate out the time dependence (solution: oscillations), you get equations of the form

$$\nabla^2 f(\vec r) = Ef(\vec r)$$

where $f$ is something: displacement, a wave function. When the radial part is separated out, the disk (or drum head), looks like (a form of Bessel's equation):

$$ rf''(r) + f'(r)-Krf(r) = 0 $$

Meanwhile, the hydrogen atom can be reduced to Kummer's equation:

$$ xf''(r)+(b-x)f'-af = 0 $$

Similar, but different. In each case, the solutions are standing waves (but not sine and cosine). The azimuthal parts are identical: waves going in a circle: $e^{\pm im\phi}$ with integer $m$. Moreover, the more nodes in $r$, the larger $m$ can be.

So they look similar, as they are standing waves, but on different shapes, with different spatial weighting.

A deeper study of the mathematical similarities would probably be found in hypergeometric functions, but that may be a math stack exchange question.

One final physics note: when one says "hydrogen orbitals", we generally think of Laguerre polynomials and spherical harmonics, thanks to the $SO(3)$ symmetry of the Coloumb hamiltonian. The problem also has a hidden $SO(4)$ symmetry that makes it separable in parabolic coordinates. The solutions are different orbitals that are equally valid....atoms don't care how we separate Schrödinger's equation. (This is why there is a double degeneracy: $SO(3)$ leads to degeneracy in $m$, while $SO(4)$ means the energy doesn't depend on $l$....only $n$ matters.).

  • $\begingroup$ "When the radial part is separated out, the disk (or drum head), looks like (a form of Bessel's equation):" Can you show or prove that? Or provide a reference to it? I'd have thought a drumhead would be different because the membrane is clamped at $r=R$, so $f(R)=0$ which is different for a plate. $\endgroup$
    – Gert
    Commented Aug 31, 2021 at 18:45
  • $\begingroup$ @Gert I looked up wikipedia on waves on a drum. If boundary conditions are a problem, then my answer applies to the clamped Chladni set up. Regarding Bessel's equation: I am operating under the assumption that any 2nd order equation solved by Bessel's functions is a Bessel's equation. $\endgroup$
    – JEB
    Commented Aug 31, 2021 at 19:13
  • $\begingroup$ "then my answer applies to the clamped Chladni set up" I don't think there is such a thing: the plates are excited at their boundary, so they can't be clamped there. But there is some similarity between Chladni plates and SE solutions, that's true. $\endgroup$
    – Gert
    Commented Aug 31, 2021 at 19:53
  • 1
    $\begingroup$ The transverse vibrations of a rigid plate (as used in Chladni figures) differ from those of a drumhead because the restoring force is due to the plate's rigidity, as opposed to the drumhead's tension. In particular, the analog of the Helmholtz equation for a thin rigid plate is $\nabla^4 f = \alpha f$ (i.e., the biharmonic operator), with $\alpha > 0$. The solutions still work out to be Bessel functions, but you have to include modified Bessel functions in your modes as well. $\endgroup$ Commented Aug 31, 2021 at 20:11
  • $\begingroup$ @MichaelSeifert Ah yes, you are right about the biharmonic operator. I remember it now. $\endgroup$
    – Gert
    Commented Aug 31, 2021 at 22:14

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