Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Motivation: In odd dimensions, solutions to the wave equation: $u_{tt}(x,t)=\nabla u(x,t)$, $u_t(x,0)=0$, $u(x,0)=f(x)$, ($t\geq 0, x\in \mathbb{R}^n$) have the nice property that the value of $u(x,t)$ only depends on the values $f(y)$ with $|y-x|=t$. For even dimensions, the value $u(x,t)$ depend on all the values $f(y)$ with $|y-x|\leq t$. A consequence of this is, that when you switch a light bulb on and then off (in 3D), there will be a light wave traveling with the speed on light (D'oh) and behind the wave, there will be total darkness. But when you throw a rock into a pond (with a 2D surface), there will be lots of wave traveling outwards from where the rock hit the water and, in theory, the water will never a still again.

Question: Can anyone give an intuitive explanation of this difference between odd and even dimensions?

share|cite|improve this question
This is called Huygens' principle. See although I wouldn't call the explanation there "intuitive". – Michael Lugo Nov 3 '10 at 20:13
Related Phys.SE question: – Qmechanic Aug 3 '14 at 19:42
up vote 14 down vote accepted

I had this question sometime ago and was shown the explanation in Balazs 1954 paper "Wave propagation in even and odd dimensional spaces". The singularities of the integrands in the solution vary based on the dimension giving rise to this effect.

  1. When n is even, the integrand has a pole in the complex plane, so the contour has to be deformed, and the result refers to only one time value -- that of the pole.
  2. When n is odd, the integrand has a branch point, and the contour cannot encircle the singularity. Consequently the solution contains contributions from all times along the branch cut.
share|cite|improve this answer

This is a very good question. Kevin Brown once gave a very detailed answer to it which I will not bother to reproduce here.

A not-exactly precise way to think about it is this: a wave propagating in $n$ dimensions can be thought of as a wave propagating in $n+1$ dimensions, but one degree of freedom removed. More precisely: if $f$ solves the wave equation $(\partial_t^2 - \triangle) f(t,x) = 0$ on $\mathbb{R}^d$, then it also solves the equation $(\partial_t^2 - \partial_y^2 - \triangle) f(t,y,x) = 0$ on $\mathbb{R}^{d+1}$, if you assume that $f(t,y,x) = f(t,0,x) = f(t,x)$ for all $y$. (It is constant in the $y$ direction, so any derivative in that direction is 0.)

Now, if a wave were to propagate in, say, 5 dimensions with the property you mentioned. Then in 4 dimensions, the points "inside the light cone" is reachable in 5 dimensions as on the light cone. Or, in other words, assume $f(t,x)$ is a solution to wave equation in 4 dimensions. And let $f(t,y,x)$ be the trivial extension with one dimension added. So $f(t,y,x)$ solves the wave equation in 5 dimensions. Suppose at time $t = 0$, a light bulb turns on at the origin $x = 0$ in 4 dimension space. This corresponds to a bank of lightbulbs turning on in 5 dimensional space along the $y$ axis. The five dimensional principle says that a point at coordinates $(y,x)$ will be illuminated by a lightbulb at $(y_0,0)$ at time $t$ if $t^2 = x^2 + (y-y_0)^2$. But for every $x, y$ such that $|x| < |t|$, you can find two values of $y_0$ such that $t^2 = x^2 + (y-y_0)^2$. And so some light must lag behind. (In the study of partial differential equations, this is also known as the method of descent.)

But by this argument, there should still be some light lagging behind for 3 dimensions also! The claim, however, is this: due to the nature of the propagation, there is a destructive interference when you drop two dimensions. That you have destructive interference when you drop two dimensions, but not such problem when you drop only one dimension, is not at all obvious intuitively (at least to me). It however does fall out of the expression for the fundamental solution to the wave equation.

share|cite|improve this answer
@Ben: thanks for the edit. You are absolutely right of course, but unfortunately that perhaps means I've copied the wrong link when I wrote that answer; and 4 years down the line I am not sure whether there was another detailed answer by Baez that I had in mind. – Willie Wong Aug 19 '14 at 8:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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