From: columbus@strident.think.com (Michael Weiss)
Newsgroups: sci.math,sci.physics
Subject: Re: More on Huygens' principle
Date: 30 Dec 92 11:58:47
Organization: Thinking Machines Corporation, Cambridge MA, USA
Distribution: world
	<25DEC199222030714@zeus.tamu.edu>
NNTP-Posting-Host: strident.think.com
In-reply-to: dwr2560@zeus.tamu.edu's message of 25 Dec 1992 22:03 CST

I wrote:

   If one assumes that Huygens' principle holds in 3 spatial dimensions, one
   can deduce that it should fail in 2.  Hadamard called this argument the
   "method of descent".  Briefly: consider a pulse disturbance at t=0 along
   the entire z-axis.  It creates spreading waves that obviously will have no
   dependence on z, and hence satisfy the 2-dimensional wave equation.  Now we
   apply Huygens' principle in 3 dimensions--- that is, we add up spherical
   waves spreading from each (0,0,z) (i.e., integrate over z).  An observer
   positioned at (x,y,0) will of course "hear" the pulse at t = sqrt(x^2+y^2),
   due to the spherical wave spreading out from (0,0,0) but will also hear
   something at t = sqrt(x^2 + y^2 + z^2), thanks to the spherical wave from
   (0,0,z). QED (I was hoping there was a similarly intuitive argument for why
   Huygens' should work for 3 dimensions, but I guess not.)

Dave Ring replies:

   I feel like something is missing. Wouldn't a similar line of reasoning
   show that HP in 5 dimensions implies no HP in 3?

Good question!  PDE's aren't really my thing, and I don't have an answer
that satisfies me.  I do have a few thoughts (see below), but
perhaps someone who really understands this stuff will clear up the mystery.

First: the general solution to the wave equation in one dimension with in
initial conditions u(x,0) = v(x), u_t(x,0) = w(x) is

                                    /x+t
    u(x,t) = v(x+t)/2 + v(x-t)/2 +  |      w(s) ds
                                    /x-t

This makes sense intuitively, once you realize that an "explosion" at x=t=0
(i.e., v(x) identically zero and w(x) a delta function) will propagate like
so:
                       ------------
                   <== |          | ==>
          -------------           -----------------

as Herbert Kranzer's post pointed out (just decompose the delta pulse into
two "step function" waves, travelling in opposite directions).  The v's
give travelling waves in opposite directions that propagate "without
echoes", and the integral represents the effect of the delta pulses.

Here is a loose statement of Huygens' principle in three dimensions:

    An observer at time t will feel the effect of what was happening at
    time 0 integrated over a sphere of radius t, centered at the observer.

More precisely:

    The solution to the 3-dimensional wave equation with initial conditions
    u(x,y,z,t) = v(x,y,z), u_t(x,y,z,t) = w(x,y,z) is


			/  w(x,y,z)               d   /  v(x,y,z)
   u(x,y,z,t) = (1/4pi) |  -------- dS + (1/4pi) ---  |  -------- dS
			/     t                   dt  /      t
			S                             S


where S is the sphere of radius t, centered at (x,y,z) (and d stands for a
partial derivative).  The w term makes sense intuitively if we think of w
as being composed of delta pulses, and assume that the effect dies off as
1/distance.  This in turn makes sense if we assume that the energy per unit
volume of the wave is proportional to u^2--- which is certainly true for
the usual examples.  The v term can be derived from the w term, making use
of the fact that if u solves the wave equation, then so does u_t.

Hadamard's "method of descent" shows how Huygens' principle fails if we
descend from 3 dimensions to 2 (by assuming that w and hence u has no
dependence on z).  Dave Ring asked about descending from 5 dimensions to 3.
To warm up for it, I thought about descending from 3 to 1.  So assume w has
no dependence on x or y.

Consider a delta pulse at t=0 along the entire z=0 plane.  To make things a
little more intuitive, let's replace this with a pulse along the slab 0 < z
< epsilon, with epsilon infinitesimal.  (In other words, w is constant
inside the slab, and zero outside the slab.)  Place an observer at distance
r from the xy-plane.  So the observer feels the effects of what was
happening at t=0 on a sphere of radius t around it.

Now, for t > r, this sphere intersects the slab in a zone of height
epsilon--- for t=r it's a spherical cap.  But the area of such a zone is
simply 2pi*t*epsilon--- a nice geometrical fact discovered, I believe, by
Archimedes, and featured in one of Martin Gardner's columns.  And so the
area we're integrating over is proportional to t, and so the integral is
constant.  So we get the one-dimensional case back.

For descending from 5 to 3 dimensions, you need the analog to the above
formula for u.  John Baez gave one formula in his post, and I looked up
some others.  I think I could carry through the calculations and verify
that HP persists as you descend from 5 to 3 dimensions, but I have no
intuitive feel for what's going on behind the symbol pushing.  So I
quit here (hopefully passing the ball to someone else).

Xref: sparky sci.math:17621 comp.text.tex:14577