# 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.

• Related Math.SE question: math.stackexchange.com/q/8794/11127 – Qmechanic Aug 3 '14 at 18:47
• Maybe from quantum mechanical considerations one can show this, I mean from the fact that the quantum concept of a photon considers Huygens principle as a necessity. There seems to be a proof for it here: mathpages.com/home/kmath242/kmath242.htm – Ellie Aug 3 '14 at 19:14
• @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. – Ellie 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 --- ex-moderator kitten 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

## 4 Answers

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.

• +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. – user4552 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. – user4552 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
• From a purely physical point of view, does Huygens' principle completely fail in even dimensions? or rather a modified version of it would still hold? If we say in 3D that the propagation (inside a sphere) speed is always $c$ and never below then for 2D (simplified pebble in water case) the modified principle would be: the wave propagates with $v \leq c$ and the direction of propagation isn't exactly defined by the normal to the surface anymore. – Ellie Aug 4 '14 at 13:27

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.

This is a more detailed version of @tparker's answer.

Suppose $\phi(x,t)$ is a spherically symmetric solution to the wave equation satisfying the initial conditions

$$\phi(x,0)=0$$ $${\partial \phi \over\partial t}(x,0)=f(x)$$

Then we have:

Theorem 1: If the number of dimensions is odd, $\phi(x,t)$ is completely determined by the values of $f$ on the past light cone'' of $(x,t)$.

Theorem 2: If the number of dimensions is even, $\phi(x,t)$ depends on the values of $f$ both on and inside the past light cone of $(x,t)$.

More precisely, let $M(x,t)$ be the mean value of $f$ on the sphere of radius $t$ around $x$. Then Theorems 1 and 2 follow from:

Theorem 1$'$: If the number of dimensions is odd, $\phi(x,t)=t M(x,t)$.

Theorem 2$'$: If the number of dimensions is even, then $$\phi(x,t)=\int_0^t s M(x,s) / \sqrt{t^2-s^2} ds$$

So in evenly many dimensions, the mean value of f on every sphere of every radius from $0$ to $t$ contributes to the solution, while in oddly many dimensions, only the sphere of radius $t$ contributes. In particular, in evenly many dimensions (but not in oddly many) an initial disturbance at the origin can have effects at $x$ long after the initial wave crest has passed.

Theorems $1'$ and $2'$ are not too difficult to prove, but it might be more illuminating to consider the underlying intuition. Namely:

Given initial data for (say) a 2-dimensional wave, we can create initial data for a 3-dimensional wave by using the same data and making it independent of the third coordinate, which I'll call $z$.

Now if we solve the 3-dimensional problem, we should get a solution independent of $z$; restricting to the plane, we've solved our 2-dimensional problem.

Under this operation, if our initial data are concentrated near the origin for the 2D-problem, they'll be concentrated all along the $z$-axis for the 3-D problem. So from every point along the $z$-axis, we get an expanding 3-dimensional sphere of non-zero wave.

Now consider a point $P$ in the plane. Every one of our vertical array of expanding spheres will eventually pass through point $P$. That's why there will be ongoing non-zero wave values at point $P$ (and explains exactly why it's everything inside the past light cone, not just on the light cone, that matters at a given event).

• Regarding the last comment on a $z$ independent solution, that is a nice way of thinking, but I think it's not clear why the same logic wouldn't apply going from 4d down to 3d. – KF Gauss May 7 '17 at 2:58
• I don't think your physical intuition reason is correct, for the reason given by @user157879. Your picture suggests that there is some dimension where the influence is only on the light cone, and that for all lower dimensions, you also get influence inside the light cone, which is not correct. In general, dimensionally reducing a theory in the way you describe does not always reproduce the same theory in a lower dimension - it often introduces new fields as well. – tparker Sep 29 '17 at 19:40
• @tparker: I think --- but am not certain --- that the intuition I gave is correct but incomplete. The missing part is, as you say, to explain why we can't extend the same intuition more than one dimension downward. I believe --- but again am not certain --- that you can explain this by arguing that interference somehow magically erases this effect when you go down from dimension $n$ to dimension $n-2$, though I have at the moment no good intuitive story for why you should expect this. – WillO Sep 29 '17 at 21:33
• I think I might have figured this out. The Green's function for the $(4+1)D$ wave equation is indeed sign-definite strictly inside the light cone, but it has a delta-function of the opposite sign exactly on the light cone. See eq. (36) of aapt.scitation.org/doi/abs/10.1119/1.17230. – tparker Jan 2 '18 at 6:32
• As improbable as it sounds, I suspect that when dimensionally reducing from $(4+1)d$ to $(3+1)d$ space, there's a magical conspiracy such that after the initial positive wavefront passes, its infinite-duration negative wake is continuously exactly canceled by successive positive wavefronts arriving from sources that are ever further-away in the fourth spatial dimension. – tparker Jan 2 '18 at 6:33

Huygen's principle is basically equivalent to the fact that the Green's function $G(s)$ for the wave equation only has support at $s = 0$, where $s$ in the invariant spacetime interval. In other words, signals can only propagate exactly on the light cone and not inside the light cone - they travel at the speed of light/sound without leaving a "wake" behind them. The fact that this property only holds in odd spatial dimensions is a fairly straightforward exercise in complex contour integration, demonstrated e.g. in https://link.springer.com/article/10.1007%2FBF02903572.

• Your answer succinctly says how Huygens principle breaks down, but saying this occurs in even dimensions because the math says so isn't very enlightening. – KF Gauss May 7 '17 at 3:02
• @user157879 True, but it answers the question. The OP didn't ask for intuition, they just asked for the explanation. – tparker May 7 '17 at 3:05
• If you say so... – KF Gauss May 7 '17 at 3:07