7
$\begingroup$

It has been asked again and the answer was that there is a kind of resonance between the wave and the medium.

Does anybody know something more about this resonance?

$\endgroup$
3
  • 3
    $\begingroup$ I have a suspicion that dispersion in gravity waves in water is related to the fact that when modeling a wave as an oscillator, the restoring force (gravity) is not proportional to the displacement, as it would be if it were provided by a spring. Does anybody know if this is true? $\endgroup$ May 13 '21 at 15:57
  • 3
    $\begingroup$ Note from Niels: one of the world experts in gravity wave propagation in water just told me that the restoring force in water waves DOES go in proportion to amplitude (more or less) because that force looks like a pressure which scales with depth. He has a book coming out soon which I will attempt to read. $\endgroup$ May 13 '21 at 16:15
  • 2
    $\begingroup$ It is interesting to note that some surface water waves are potentially non-dispersive, and are called solitons. See here and here for instance $\endgroup$
    – Superbee
    May 16 '21 at 17:22
11
+250
$\begingroup$

Fluids have two basic modes of excitation, propagating (sound) modes with a dispersion relation $\omega^2\simeq c_s^2k^2$, and diffusive modes $w\simeq iDk^2$, where the speed of sound $c_s$ and the diffusivity $D$ depend on microscopic properties of the fluid (the equation of state and the transport coefficients). The propagating mode is longitudinal (direction of fluid velocity parallel to the direction of the mode), and the diffusive mode is transverse. The form of these dispersion relations is set by Galilean and rotational invariance (essentially, $\vec\nabla$ is a vector and the equations of motion must always have at least two spatial derivatives). The speed of sound is determined by the compressibility of the fluid. Note that the relation $\omega^2\simeq c^2_s k^2$ is approximate, valid in the long wave-length limit $k\to 0$. At higher order in $k$ there will be corrections that lead to dispersion and damping.

Fluids with a free surface subject to gravity also have surface modes. These modes have a preferred direction, so there is nothing that forbids equations of motion that are first order in space and second order in time, so that $\omega^2 \sim k$. How this comes about is explained in any text book on fluid mechanics. What one does is to write down an equation for the position of the free surface, which ensures that fluid particles don't leave the fluid, and an equation that describes the acceration of fluid particles by the balance of pressure and gravitational potential, the Euler or Bernoulli equation. The restoring force is gravity, and we find $$ \omega = \sqrt{gk} $$ so that the speed of sound $v=(\partial w)/(\partial k)\sim 1/\sqrt{k}$ depends on wave number $k$. This is what we call dispersive. In particular, long wave length modes (tsunamis) propagate fast. Note that this relation is also not exact. There are corrections due to the finite depth of the water (which limits the speed of tsunamis), due to damping, surface tension, and large amplitude.

Postcript: (Regarding resonances) I don't think surface waves are "resonant" in the usual sense of the word. A resonance is an excitation for which an internal frequency matches the frequency of an external perturbation. A pendulum, for example, has an internal frequency $\sqrt{g/l}$. For surface waves, there is no intrinsic length scale that sets the frequency, and the frequency is set by $k$, $\omega \sim \sqrt{gk}$.

There is one interesting limit in which it looks more like a resonance phenomenon. Consider long wavelength waves in shallow water. Now there is a relevant scale set by the depth of the water, $h$, and the frequency approaches $\omega\sim \sqrt{g/h}$.

$\endgroup$
2
  • $\begingroup$ This is nice answer (+1), but nothing about resonance mentioned by Larsa. $\endgroup$ May 16 '21 at 13:15
  • $\begingroup$ @AlexTrounev Thanks for bringing this up. I don't think that surface gravity waves are resonant in the usual sense of the word $\endgroup$
    – Thomas
    May 16 '21 at 15:33
9
$\begingroup$

I'll show how the dispersion relation for surface water waves can be derived.

The words about "resonance" in the body of the question are true in the broad sense that any wave can be regarded as a kind of resonance, loosely speaking. The question in the title is more specific, so I'll focus on that. A wave is called dispersive if its speed depends on its wavelength $\lambda$. In terms of the angular frequency $\omega=2\pi f$ and wavenumber $k=2\pi/\lambda$, this condition means that $\omega$ is not proportional to $k$. The goal is to understand why surface water waves have this property.

Inputs

Let $\mathbf{x}=(x,y,z)$ be a point in space, either on or below the surface of the water, and let $\mathbf{v}(\mathbf{x})$ be the velocity of the water at that point. Take $z$ to be the vertical direction, and let $h(x,y,t)$ be the height of the water's surface relative to what it would be with no waves. The wave is governed by two forces:

  • Gravity, which tries to keep the surface low.

  • Surface tension, which tries to keep the surface flat.

According to refs 1 and 2, the force of gravity is the dominant one for long wavelengths $\lambda\gg 2\text{ cm}$ (called gravity waves), and the surface tension dominates for short wavelengths $\lambda\ll 2\text{ cm}$ (called capillary waves). Both limits exhibit dispersion. I'll keep the wavelength $\lambda$ arbitrary, but I'll use these approximations:

  • The water is incompressible: $\nabla\cdot\mathbf{v}=0$. The density $\rho$ is constant and uniform.

  • The water is irrotational: $\nabla\times\mathbf{v}=0$.

  • The bottom of the body of water is flat (no underwater mountains).

  • The changes in height $h$ (the amplitude of the wave) are small enough to justify using the linear approximation. This approximation allows us to ignore the difference between the time-derivative of $\mathbf{v}$ at a fixed point in space and the "material" time-derivative of $\mathbf{v}$ at a point that moves along with the flow.

Analysis

The irrotational condition implies that we can write $\mathbf{v}$ as the gradient of a scalar function $\phi$: $$ \newcommand{\bfv}{\mathbf{v}} \newcommand{\bfx}{\mathbf{x}} \bfv=\nabla\phi. $$ Then the incompressibility condition implies $$ \nabla^2\phi=0. \tag{I} $$ At the surface, the $z$-component of $\bfv$ must match the time-derivative of the height-function $h$: $$ \nabla_z\phi=\frac{dh}{dt} \hspace{1cm} \text{at }z=h. \tag{H} $$ At the bottom of the body of water, the $z$-component of $\bfv$ must be zero: $$ \nabla_z\phi=0 \hspace{1cm} \text{at }z=-D, \tag{D} $$ where $D$ stands for "depth." At any point under the surface, the water's acceleration is determined by the forces of gravity and pressure: $$ \frac{d\bfv}{dt}+g\nabla z+\frac{\nabla P}{\rho}=0, $$ where $g$ is the acceleration of gravity, $P$ is the pressure, and $\rho$ is the density (which we're assuming is constant and uniform — the incompressibility condition). Note that $\nabla z$ is just a convenient way of writing the vertical unit vector. Write this equation in terms of $\phi$ to get $$ \nabla\left(\frac{d\phi}{dt}+g z+\frac{P}{\rho}\right)=0, $$ which implies $$ \frac{d\phi}{dt}+g z+\frac{P-P_0}{\rho}=0 $$ where $P_0$ is an integration constant that we can interpret as the pressure at the surface. The jump in pressure at the surface is determined by the surface tension: $$ P-P_0 = -T \nabla^2 h $$ where $\nabla^2 h$ measures the curvature of the surface and $T$ is a constant relating the tension to the degree of curvature. (Surface tension tries to make the surface flat, so more curvature means more tension.) Use that in the preceding equation to get $$ \frac{d\phi}{dt}+g h-\frac{T}{\rho}\nabla^2 h=0 \hspace{1cm} \text{at }z=h. $$ Recall that $h$ is independent of $z$, so this can also be written $$ \frac{d\phi}{dt}+g h-\frac{T}{\rho}(\nabla_x^2+\nabla_y^2) h=0 \hspace{1cm} \text{at }z=h. \tag{1} $$ To express this entirely in terms of $\phi$, take the time-derivative of (1) and use (H) to get $$ \frac{d^2\phi}{dt^2}+g \nabla_z \phi -\frac{T}{\rho}(\nabla_x^2+\nabla_y^2) \nabla_z \phi =0 \hspace{1cm} \text{at }z=h. \tag{2} $$ Now consider a wave moving in the $x$-direction: $$ \phi(\bfx)=f(z)\sin(\omega t-kx), \tag{A} $$ where $f$ is a function to be determined. Use this ansatz in (2) to get $$ -\omega^2 f+g f'+ \frac{T}{\rho}k^2 f' = 0 \hspace{1cm} \text{at }z=h, \tag{3} $$ where $f'(z)=df/dz$. Use the ansatz (A) in the incompressibility condition (I) to get $$ f''-k^2f =0, $$ and combine this with the bottom boundary condition (D) to infer $$ f(z)\propto \cosh(kz+kD). $$ Use this in (3) to get the final result $$ \omega^2 \approx \left(gk+\frac{T}{\rho}k^3\right)\tanh(kD), \tag{R} $$ where the small-amplitude approximation was used again to simplify the argument of the tanh function. This is the dispersion relation for surface water waves, under the approximations listed above.

Equation (R) shows that $\omega$ is generally not proportional to $k$, so surface water waves are generally dispersive: their speed depends on their wavelength. The speed (phase velocity) is $\omega/k$, so equation (R) shows that in the deep-water approximation ($\tanh(kD)\approx 1$), longer wavelengths travel faster when gravity is the dominant restoring force (gravity waves), and shorter wavelengths travel faster when surface tension is the dominant restoring force (capillary waves).


References:

  1. Kartashova (2009), "Nonlinear resonances of water waves," Discrete and Continuous Dynamical Systems Series B 12:607-621 (https://www3.risc.jku.at/publications/download/risc_3822/resonances_6.pdf)

  2. Mei (2004), "Chapter four: waves in water," notes for an MIT course in wave propagation (http://web.mit.edu/1.138j/www/material/chap-4.pdf)

$\endgroup$
2
  • 1
    $\begingroup$ This is very nice answer (+1), but also nothing about resonance mentioned by Larsa. :) $\endgroup$ May 16 '21 at 15:49
  • $\begingroup$ That's still OK in my book- a very nice answer, and a great derivation! -NN $\endgroup$ May 16 '21 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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