Damping of Gravity Waves
- Fluid Mechanics 2.10
Surface Layer
Consider gravity waves on the free surface of a viscous incompressible fluid. Let $k$ be the wave number, $\lambda=2\pi/k$ the wavelength, and $\omega$ the frequency. We assume that the amplitude is small and that
\[ \frac{\nu}{\lambda^2}\ll\omega. \]
Equivalently, the viscous diffusion length during one period is much smaller than the wavelength:
\[ \delta=\sqrt{\frac{2\nu}{\omega}}\ll\lambda. \]
Thus the main motion is the potential motion of an ideal fluid. Viscosity only produces a thin rotational layer near the free surface, together with a slow decay of the wave amplitude.
There is an important difference from a solid boundary. At a solid wall, viscosity must change the tangential velocity from its ideal value to zero, which produces large velocity gradients. At a free surface, there is no no-slip condition. The viscous correction is only needed to make the tangential stress vanish. Therefore the leading dissipation may be computed directly from the ideal potential flow in the bulk.
Energy Dissipation
For an incompressible viscous fluid, the rate of viscous loss of mechanical energy is
\[ \dot E_{\mathrm{mech}} =-\frac{\mu}{2} \int \left( \pdv{v_i}{x_k}+\pdv{v_k}{x_i} \right)^2\,dV. \]
In the leading approximation the flow is potential:
\[ \b{v}=\grad\phi. \]
Hence
\[ \pdv{v_i}{x_k}=\pdvs{\phi}{x_i}{x_k}, \]
and the dissipation becomes
\[ \dot E_{\mathrm{mech}} =-2\mu \int \left( \pdvs{\phi}{x_i}{x_k} \right)^2\,dV. \]
For a deep-water wave propagating in the $x$ direction, with the undisturbed surface at $z=0$ and the fluid occupying $z<0$, the velocity potential has the form
\[ \phi=\phi_0e^{kz}\cos(kx-\omega t+\alpha). \]
It satisfies
\[ \laplacian\phi=0,\qquad z<0, \]
and decays as $z\to-\infty$. Since each second derivative gives a factor of order $k^2$, a direct calculation gives
\[ \overline{\dot E}_{\mathrm{mech}} =-8\mu k^4\int\overline{\phi^2}\,dV. \]
The mean kinetic and potential energies of a small gravity wave are equal. Therefore the mean total mechanical energy may be written as twice the mean kinetic energy:
\[ \overline{E}_{\mathrm{mech}} =\rho\int\overline{v^2}\,dV =\rho\int\overline{ \left(\pdv{\phi}{x_i}\right)^2 }\,dV. \]
For the above potential,
\[ \int\overline{ \left(\pdv{\phi}{x_i}\right)^2 }\,dV =2k^2\int\overline{\phi^2}\,dV. \]
Thus
\[ \overline{E}_{\mathrm{mech}} =2\rho k^2\int\overline{\phi^2}\,dV. \]
Damping Coefficient
Let the wave amplitude be proportional to $e^{-\gamma t}$. Then the energy, being quadratic in the amplitude, is proportional to $e^{-2\gamma t}$. Hence
\[ 2\gamma =-\frac{\overline{\dot E}_{\mathrm{mech}}} {\overline{E}_{\mathrm{mech}}}. \]
Using the expressions above,
\[ 2\gamma =\frac{8\mu k^4}{2\rho k^2} =4\nu k^2. \]
Therefore
\[ \gamma=2\nu k^2. \]
For deep-water gravity waves,
\[ \omega^2=gk. \]
Thus the damping coefficient may also be written as
\[ \gamma=2\nu\frac{\omega^4}{g^2}. \]
The result is small under the assumption $\nu k^2\ll\omega$. It is therefore a correction to the inviscid wave motion, not a change of the leading dispersion relation.
Arbitrary Viscosity
When the viscosity is not small, one must solve the linearized equations directly. For a two-dimensional disturbance proportional to $e^{ikx-i\omega t}$, the equations are
\[ \begin{aligned} \pdv{v_x}{t} &=-\frac{1}{\rho}\pdv{p}{x} +\nu\left(\pdvn{2}{v_x}{x}+\pdvn{2}{v_x}{z}\right), \nl \pdv{v_z}{t} &=-\frac{1}{\rho}\pdv{p}{z} +\nu\left(\pdvn{2}{v_z}{x}+\pdvn{2}{v_z}{z}\right)-g, \nl \pdv{v_x}{x}+\pdv{v_z}{z}&=0. \end{aligned} \]
At the free surface the tangential and normal stresses vanish, to first order in the wave amplitude:
\[ \sigma_{xz} =\mu\left(\pdv{v_x}{z}+\pdv{v_z}{x}\right)=0, \]
\[ \sigma_{zz} =-p+2\mu\pdv{v_z}{z}=0. \]
Solving the linear system with decay as $z\to-\infty$ gives the dispersion relation
\[ \left(2-\frac{i\omega}{\nu k^2}\right)^2 +\frac{g}{\nu^2k^3} =4\sqrt{1-\frac{i\omega}{\nu k^2}}. \]
In the weak-viscosity limit this gives
\[ \omega=\pm\sqrt{gk}-2i\nu k^2. \]
Thus the imaginary part reproduces the damping coefficient found from the energy method. In the opposite limit,
\[ \nu k^2\gg\sqrt{gk}, \]
the motion is no longer an oscillatory wave with slow damping. The disturbance decays aperiodically. The slowest decay has the order
\[ \omega\simeq -\,i\frac{g}{2\nu k}. \]
