Intuitive explanation of the difference between waves in odd and even dimensions Motivation: In odd dimensions, solutions to the wave equation: $$u_{tt}(x,t)=\nabla u(x,t), \qquad u_t(x,0)=0, \qquad u(x,0)=f(x)$$ where $t \geq 0$ and $x \in \mathbb{R}^n$, 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 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 waves 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?
 A: 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.  


*

*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.

*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.

A: 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. 
