Ripples in Ponds and the Wave Equation
Steven E. Landsburg
(Posted separately to sci.math.research)
When you drop a pebble in a pond and see a series of ripples, are
those ripples a) part of the solution to the 2-dimensional wave
equation on the surface of the water, or b) the result of vertical
oscillations of the water emanating from points below where the pebble
was dropped?
The remainder of this post is a more elaborate version of the above
question.
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, which
I've quoted at http://www.landsburg.com/appsmr.txt . The 3-dimensional
result generalizes to any odd dimension and the 2-dimensional result
to any even dimension, but here I'll be concerned only with the 2
and 3 dimensional cases.)
Many years ago, I had a conversation with an assistant professor
here at Rochester, in which he observed that sounds---that is,
disturbances in the three-dimensional air---propagate according
to the 3-d wave equation, which is why, when you hum for one
second, I hear the humming (after a slight delay) for exactly one
second. By contrast, a pebble tossed into a pond creates a
disturbance in the two-dimensional surface, which propagates
according to the 2-d wave equation, which is why the initial wave
crest is followed by a series of ripples. If sound waves were
followed by analogous ripples, the world would be a very noisy
place.
That assistant professor has left academics and I can find no trace
of him on the World Wide Web. However, in preparing this post, I've
found two web sites and one paper (by Sigurdur Helgason) that
make essentially the same assertion; I wouldn't be surprised if
the Helgason paper inspired my assistant professor and both web sites.
I've recently been engaged in an extended conversation with a
physicist who doubts that the ripple story is correct. His guess
---based, I am sure, on far better intuition than mine---is this:
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.)"
So my big question is:
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?
--
Steven E. Landsburg
http://www.landsburg.com/about2.html
Timo Nieminen
> When you drop a pebble in a pond and see a series of ripples, are
> those ripples a) part of the solution to the 2-dimensional wave
> equation on the surface of the water, or b) the result of vertical
> oscillations of the water emanating from points below where the pebble
> was dropped?
Why not both?
> 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.
Sure. Consider the case of a circularly symmetric monochromatic wave in
2D. We know a simple solution:
A = H0(kr) exp(-iwt), H0 is a Hankel function of the 1st kind.
In cylindrical coordinates, we have
(1/r) (d/dr) ( r dA/dr ) = - k^2 A
for the circularly symmetric Helmholtz equation, and with
H_n = (-1)^n H_[-n] and x^n H_[n-1](x) = (d/dx) ( x^n H_n(x) )
it's easy to show that the solution satisfies the equation. What does this
tell us about circular ripples from an impulse source?
It's easy to show that the plane (line?) wave case
exp( ikx - iwt ) is also a solution.
For the physically relevant case of capillary waves (i.e., small ripples,
or surface-tension waves), the speed must depend on the surface tension
and the density (the equivalent for 1D waves on a string would be the
tension and linear mass density, and the 3D equivalent, the elasticity and
density).
However, whereas the 1D and 3D cases give us
v = sqrt(T/density) = const,
the 2D case gives us
v = sqrt(kT/density).
So, even in the linear limit, we have dispersion, and lots of it.
For an impulse source, we have an infinitely broad white spectrum. We can
find the effect of each frequency component using H0(kr) exp(-iwt), but
the dependence of v on k means that v varies from 0 to infinite. So, we
expect to see ripples over some finite time for an impulse source.
Dropping a rock in won't be an impulse source, and we could perhaps model
this as a rectangular pulse excitation, which will lead to a spectrum with
some peak wavelength. Find the group velocity at the wavelength, and we
know how fast the peak of the bunch of ripples will move at. But the other
components will be there, and we can see them.
So, to answer your initial (and final, but I won't repeat them
here) question(s), I'd say that both a) and b) are true, where a) can be
the linear wave-equation, but can't be the dispersion-free wave equation.
If you prefer to restrict "wave equation" to constant v, then only b) is
true.
A common case is water waves. The details can be found in good books on
hydrodynamics, such as Lamb, Hydrodynamics. The above details are for
capillary waves (i.e., surface tension ripples), which are analogous to
the dispersion-free elastic waves we get in 1D and 3D. You can add gravity
waves (i.e., large water waves) and get more realistic behaviour over a
larger range of amplitudes (and depths). Note the very different
dispersion relation for gravity waves.
Your physical intuition in
> http://www.landsburg.com/appsmr.txt
explains why the cylindrically symmetric, z-invariant 3D case agrees with
the 2D case, but I think we need to keep in mind that the 2D case behaves
the way it does for quite different reasons.
> I've recently been engaged in an extended conversation with a
> physicist who doubts that the ripple story is correct. His guess
> ---based, I am sure, on far better intuition than mine---is this:
>
> 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.
We get ripples even in the linear limit. This is covered, e.g., in Lamb.
> 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.)"
I would be very suspicious of this reason. Firstly, you need to worry
about absorption - water is _much_ more viscous than air. Secondly, the
dispersion is severe.
More likely reasons are that there is little background noise when out
fishing on a calm day, and the water will reflect the sound waves - you're
not listening to a point source of sound, but to a point source + large
reflector.
--
Timo