The discussion of ripples in Chapter 4 of The Big Questions led to a series of discussions with a prominent physicist about the source of those ripples. I had assumed that the ripples on a pond form a solution to the two-dimensional wave equation; my correspondent suggested that they are largely the result of vertical oscillations of the water emanating from points below where the pebble was dropped. Further conversations with additional physicists have yielded no clear consensus on the issue. What follows is a (fairly technical) attempt to delineate the issues. First: In three dimensions, any spherically symmetric solution to the wave equation is of the form psi(r,t) = f(r-t)/r + g(r+t)/r. In two dimensions, this is false. In particular, consider initial conditions of the form psi(r,0) = 0 d psi(r,0)/dt = q(r) Then in three dimensions, the value of psi at a point (r,t) is determined completely by the values of q at points a such that (a,0) is on the past "light" cone of (r,t). By contrast, in two dimensions, the value of psi at a point (r,t) dependes on the values of q at points a such that (a,0) is either *on or inside* the past "light" cone of (r,t). These facts follow from the Asgeirsson Mean Value Theorem; I have included a precise statement as an appendix to this document, and the intuition behind the proof as a second appendix. There is, then, no question that a disturbance causes ripples in two dimensions and not in three dimensions---just as I've said in the book. But the remaining question is: Are these the ripples that you actually see on a real-world pond, or are they swamped by ripples from vertical oscillations? My prominent physicist friend believes the latter. His guess is the following: 1) Although the 2-d wave solution to the wave equation allows for "ripples" (by which I mean effects of anything outside the past light cone), these ripples are likely to be small and insignificant at distances far from the source on the scale of a wavelength. 2) The ripples we actually see on ponds are therefore not part of the 2-d solution, but instead the result of nonlinear vertical oscillations of the water at the center of the disturbance initiated by the dropped pebble. In partial support of this view, one of my correspondent's colleagues makes the following observation: "The reason you can hear the conversations of very distant fisherman so clearly when you're out fishing on a calm day, is that the sounds are carried by surface vibrations of the water (the intensity of which only diminishes as the inverse distance in 2 dimensions rather than the inverse square of the distance in 3 dimensions), which disagrees with your description of how sounds would get smeared out in 2D. (The conversations are not smeared out.)" This leaves me with four unanswered (or at least not fully answered) questions: Question 1: Why does a dropped pebble cause ripples? Is it (primarily) because these ripples are part of the solution to the 2-d wave equation, or (primarily) because of vertical oscillations in the water, or something else? I can imagine two ways to tackle this question. Method A is to actually solve the 2-d wave equation (at least numerically) for an appropriate disturbance and see what it looks like. Method B is to drop pebbles into ponds of various depths; it seems to me that if vertical oscillations were the culprit, then the ripples would look very different in a shallow pond than in a deep pond. Question 2: Has anyone carried out either of these methods? What are the results? Question 3: Would either of these methods in fact be definitive or is there something I'm overlooking? Question 4: Is there some other way to decide this question? I have posed these questions in online forums here and here, where you'll find some quite thoughtful replies. APPENDIX 1: PRECISE STATEMENT We seek a spherically symmetric solution to the wave equation for given initial conditions. Those conditions are: phi(x,0)=0 phi'(x,0)=f(x) (where ' means derivative with respect to t). The following are consequences of the Asgeirsson Mean Value Theorem: Theorem: In 3 dimensions, phi(x,t) is completely determined by the values of f on the past light cone of (x,t). Theorem: In 2 dimensions, 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 Theorem: In 3 dimensions, phi(x,t) is given by t M(x,t). Theorem: In 2 dimensions, phi(x,t) is given by an integral, where s ranges from 0 to t and the integrand is: s M(x,s) / Sqrt[t^2-s^2] In other words, in 2 dimensions, the mean value of f on every sphere of every radius from 0 to t contributes to the solution, while in 3 dimensions, only the sphere of radius t contributes. In particular, an initial disturbance at the origin can have effects at x long after the initial wave crest has passed. APPENDIX 2: THE INTUITION Given initial data for 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 (if I've got this right) 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.