5
$\begingroup$

General solutions to the wave equation in $\mathbb R^3$,

$\partial_{tt}\phi(t, \mathbf r) = c^2\Delta \phi(t,\mathbf r)$

can be obtained by first splitting off the time component, e.g. with a Fourier transform or separation of variables, leading to the Helmholtz equation $(\Delta+k^2)\psi(\mathbf r)=0$. General solutions to the Helmholtz equation can then be found in various coordinate systems:

  • Cartesian coordinates lead to a superposition of exponential functions ("plane-wave expansion"),
  • spherical coordinates lead to to spherical harmonics (multipole expansion),
  • cylinder coordinates lead to Bessel functions, etc.

I'm wondering whether there is a coordinate-free general solution.

A coordinate-free "wave equation" is

$(d^\star d+dd^\star) \alpha = 0$

(see, for example, Frankel, The Geometry of Physics ed 1, Ch. 14). $\alpha$ is a one-form in Minkowski space, $d$ is the exterior derivative ("differential"), and $d^\star$ is the co-differential.

The existence of such a coordinate-free expansions would imply that the exponential, spherical harmonics, Bessel functions, etc., are related by simple coordinate transformations, which they are not. So the answer must be no. Or am I missing something?

$\endgroup$
6
$\begingroup$

In fancy "coordinate-free" language, the solutions to the wave equation $(d^* d + d d^*) \phi = 0$ are called harmonic $0$-forms. Hodge's theorem states that for certain compact manifolds, harmonic forms are in correspondence with cohomology classes. Thus for $0$-forms the statement is that the number of independent harmonic functions is the number of connected components of your manifold, since the harmonic forms are just constant on each component.

Unfortunately, this theorem doesn't apply to $\mathbb{R}^3$, as it isn't compact.

The existence of such a coordinate-free expansions would imply that the exponential, spherical harmonics, Bessel functions, etc., are related by simple coordinate transformations, which they are not.

Of course they're related, these just represent different bases for the same space of solutions. (Specifically, the basis that is useful in each corresponding coordinate system.) For example, taking the Fourier transform of a Bessel solution manifestly writes it in terms of exponentials. The standard plane wave expansion does precisely the reverse.

In the fancy language, you don't need to think about this... but there's also not much you can say.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I do have a strong hunch that the question doesn't "make sense". But I don't understand why. Hodge's theorem does not lead to any sort of expansion, so I don't know what to do with that theorem relative to my question. And a Fourier transform is not a coordinate transformation (on the manifold under consideration). $\endgroup$ – mcmayer Apr 8 at 4:25
  • $\begingroup$ @mcmayer On the first point, I just meant to say that Hodge's theorem isn't useful at all on $\mathbb{R}^3$, to illustrate that thinking in a "coordinate-free" way for this particular application doesn't help much. $\endgroup$ – knzhou Apr 8 at 4:26
  • $\begingroup$ @mcmayer On the second point, it's not merely a coordinate transformation. The set of harmonic functions is an infinite-dimensional vector space. The plane wave and Bessel pick out different bases for this space, so you must also apply a change of basis. $\endgroup$ – knzhou Apr 8 at 4:27
  • $\begingroup$ What is the theory called that formalizes this? It sounds like there's some connection to these jet bundle constructions. $\endgroup$ – mcmayer Apr 9 at 8:20

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.