# Why is Huygens' principle only valid in an odd number of spatial dimensions?

Apparently Huygens' principle is only valid in an odd number of spatial dimensions:

Why is this?

[EDIT] This is somewhat perplexing, since AFAIK it's pretty common to teach freshmen about double- and single-slit diffraction using a two-dimensional analysis and invoking Huygens' principle. Does this work only because there's an ignored third axis of translational symmetry?

I wonder if it's possible to gain insight by making a grid and doing sort of a finite-element analysis.

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Related Math.SE question: math.stackexchange.com/q/8794/11127 – Qmechanic Aug 3 '14 at 18:47
@Phonon: I don't think it has anything to do with quantum mechanics. Huygens' principle is classical. – Ben Crowell Aug 3 '14 at 19:27
@BenCrowell I'll try to see if one can crudely work this out, but from what I've found so far, the underlying reason seems to be purely mathematical. One source claims that by solving the wave equation for an even space dimension, one finds infinitely many velocities for the wave propagation. Whereas for odd-dimensions, the equations can always be reduced down to a polar one, which then results in a simple spherical wave propagating with unit velocity. Of course a rigorous mathematical proof must be backing such claims, but I'm eager to see whether there's any physical intuition behind it. – Phonon Aug 3 '14 at 19:39
Er ... and yet we do demos that depend on the validity of Huygen's principle using surface waves on water. Are we just getting away with a fast one because the amplitude dies away to soon too see the violation? Or don't surface waves count as two-dimensional for the purposes of this discussion? (I guess this is related to Ben's "Edit".) – dmckee Aug 4 '14 at 1:11
See also e.g. mathpages.com/home/kmath242/kmath242.htm - the difference between odd and even dimensions may be seen in the Taylor expansions. Effectively, one needs things like $(d/2)!$ in the denominator of the Taylor coefficients, by a recursive relationship, and those behave differently for even and odd $d$. – Luboš Motl Aug 4 '14 at 12:47

You should look at the form of the advanced fundamental solution of D'Alembert equation, built up in geodesically convex open sets including the source localized at the event $y$ and the test point localized at the enent $x$ receiving the wave generating by the source. The construction, at least for analytic manifolds with analytic metrics, is obtained by summing a nice series originally discovered by Hadamard (and handled by Riesz actually; there is indeed a wonderful paper in French by Riesz about this fantastic construction nowadays relating the heat kernel theory with QFT in curve spacetime). Hadamard-Riesz' results have been extended to the smooth case by sevaral modern authors (see Guenther's and Friedlander's textbooks). The series, if the dimension gives rise to a fundamental solution containing a term which is completely supported on the light cone emanating from $y$. Therefore, referring to this term only, the solutions of D'Alembert equation emitted by $y$ propagates along null geodesics to reach $x$ from $y$. This basically is Huygens' principle.

If the dimension is even and the manifold is not flat or the dimension is odd, further terms appear added to the one localized on the light cone. The underlying "mathematical phenomenon" is more or less the same, in flat spacetime, when adding a mass to D'Alembert operator thus passing to the Klein-Gordon equation which does not obey Huygens' principle.

The relevant point is that this further term is now supported inside the future light cone emanating from $y$. In this case there is a contribution to wave solutions emitted by $y$ propagating along timelike geodesics from $y$ to $x$, and Huygens' principle fails.

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+1, thanks, but although this may be very helpful to others, it's much too technical for me, and I'm only interested in the case of flat spacetime. – Ben Crowell Aug 3 '14 at 19:57
I am sorry but this is a technical issue! – Valter Moretti Aug 3 '14 at 20:05
Well, some technical issues can be understood in less technical terms, especially if one is willing to settle for less generality. – Ben Crowell Aug 3 '14 at 20:16
Yes you are right, but basically I am a mathematician, so we probably have different points of view. – Valter Moretti Aug 3 '14 at 20:18
"and the manifold is not flat" - I thought Huygens failed even in flat $2n+1$ spacetime. Is this not the case? – Chris White Aug 4 '14 at 6:39

I think this originated with Hadamard and his Method of Descent. See Lectures on Cauchys Problem in Linear Partial differential Equations--starting on page 7. His results were that waves in two dimensions did not propagate sharply, but had a wake (a tail, ..). Eg. a circular wave propagating in two dimensional space vs. a spherical wave propagating in three dimensional space where it would propagate cleanly without a wake.

Hadamard essentially took a slice through a cylindrical wave in three dimensions to get a circular wave in two dimensions (descending one dimension). People have taken propagating without a wake to be one criterion in satisfying Huygens Principle.

So this is the origin of 'why', if you accept Hadamard's results.

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