Gravity Waves

Consider a stationary body of water, of uniform depth $d$ , located on the surface of the Earth. This body is assumed to be sufficiently small compared to the Earth that its unperturbed surface is approximately planar. Let the Cartesian coordinate $z$ measure vertical height, with $z=0$ corresponding to the aforementioned surface. Suppose that a small amplitude wave propagates horizontally through the water, and let ${\bf v}({\bf r},t)$ be the associated velocity field.

Because water is essentially incompressible, its equations of motion are

$$\nabla\cdot{\bf v} = 0,$$ (11.1)

$$\rho\,\frac{\partial {\bf v}}{\partial t}+ \rho\,({\bf v}\cdot\nabla)\,{\bf v} = - \nabla p -\rho\,g\,{\bf e}_z + \mu\,\nabla^{\,2}{\bf v},$$ (11.2)

where $\rho$ is the (uniform) mass density, $\mu$ the (uniform) viscosity, and $g$ the (uniform) acceleration due to gravity. (See Section 1.14.) Let us write

$$p({\bf r},t) = p_0-\rho\,g\,z+p_1({\bf r},t),$$ (11.3)

where $p_0$ is atmospheric pressure, and $p_1$ the pressure perturbation due to the wave. Of course, in the absence of the wave, the water pressure a depth $h$ below the surface is $p_0+\rho\,g\,h$ . (See Chapter 2.) Substitution into Equation (11.2) yields

$$\rho\,\frac{\partial {\bf v}}{\partial t} \simeq - \nabla p_1 + \mu\,\nabla^{\,2}{\bf v},$$ (11.4)

where we have neglected terms that are second order in small quantities (i.e., terms of order $v^{\,2}$ ).

Let us also neglect viscosity, which is a good approximation provided that the wavelength is not ridiculously small. [For instance, for gravity waves in water, viscosity is negligible as long as $\lambda\gg (\nu^2/g)^{1/3}\sim 5\times 10^{-5}\,{\rm m}$ .] It follows that

$$\rho\,\frac{\partial {\bf v}}{\partial t} \simeq - \nabla p_1.$$ (11.5)

Taking the curl of this equation, we obtain

$$\rho\,\frac{\partial \mbox{\boldmath$\omega$}}{\partial t} \simeq 0,$$ (11.6)

where $\omega$ $=\nabla\times {\bf v}$ is the vorticity. We conclude that the velocity field associated with the wave is irrotational. Consequently, the previous equation is automatically satisfied by writing

$${\bf v}= -\nabla\phi,$$ (11.7)

where $\phi({\bf r},t)$ is the velocity potential. (See Section 4.15.) However, from Equation (11.1), the velocity field is also divergence free. It follows that the velocity potential satisfies Laplace's equation,

$$\nabla^{\,2}\phi = 0.$$ (11.8)

Finally, Equations (11.5) and (11.7) yield

$$p_1 = \rho\,\frac{\partial \phi}{\partial t}.$$ (11.9)

We now need to derive the physical constraints that must be satisfied at the water's upper and lower boundaries. It is assumed that the water is bounded from below by a solid surface located at $z=-d$ . Because the water must always remain in contact with this surface, the appropriate physical constraint at the lower boundary is $v_z\vert _{z=-d}=0$ (i.e., the normal velocity is zero at the lower boundary), or

$$\left.\frac{\partial \phi}{\partial z}\right\vert _{z=-d} = 0.$$ (11.10)

The water's upper boundary is a little more complicated, because it is a free surface. Let $\zeta$ represent the vertical displacement of this surface due to the wave. It follows that

$$\frac{\partial \zeta}{\partial t} = \left.v_z\right\vert _{z=0}=-\left.\frac{\partial \phi}{\partial z}\right\vert _{z=0}.$$ (11.11)

The appropriate physical constraint at the upper boundary is that the water pressure there must equal atmospheric pressure, because there cannot be a pressure discontinuity across a free surface (in the absence of surface tension--see Section 11.11). Accordingly, from Equation (11.3), we obtain

$$p_0 = p_0 -\rho\,g\,\zeta + \left.p_1\right\vert _{z=0},$$ (11.12)

or

$$\rho\,g\,\zeta=\left. p_1\right\vert _{z=0},$$ (11.13)

which implies that

$$\rho\,g\,\frac{\partial \zeta}{\partial t} = -\rho\,g\left.\frac{... ...l z}\right\vert _{z=0}=\left.\frac{\partial p_1}{\partial t}\right\vert _{z=0},$$ (11.14)

where use has been made of Equation (11.11). The previous expression can be combined with Equation (11.9) to give the boundary condition

$$\left.\frac{\partial \phi}{\partial z}\right\vert _{z=0} = -g^{\,-1}\left.\frac{\partial^{\,2}\phi}{\partial t^{\,2}}\right\vert _{z=0}.$$ (11.15)

Let us search for a wave-like solution of Equation (11.8) of the form

$$\phi({\bf r},t) = F(z)\,\cos(\omega\,t-k\,x).$$ (11.16)

This solution actually corresponds to a propagating plane wave of wave vector ${\bf k}=k\,{\bf e}_x$ , angular frequency $\omega$ , and amplitude $F(z)$ (Fitzpatrick 2013). Substitution into Equation (11.8) yields

$$\frac{d^{\,2} F}{dz^{\,2}} - k^{\,2}\,F= 0,$$ (11.17)

whose independent solutions are $\exp(+k\,z)$ and $\exp(-k\,z)$ . Hence, a general solution to Equation (11.8) takes the form

$$\phi(x,z,t) = A\,{\rm e}^{\,k\,z}\,\cos(\omega\,t-k\,x) + B\,{\rm e}^{-k\,z}\,\cos(\omega\,t-k\,x),$$ (11.18)

where $A$ and $B$ are arbitrary constants. The boundary condition (11.10) is satisfied provided that $B = A\,\exp(-2\,k\,d)$ , giving

$$\phi(x,z,t) = A\left[{\rm e}^{\,k\,z}\ + {\rm e}^{-k\,(z+2\,d)}\right]\cos(\omega\,t-k\,x),$$ (11.19)

The boundary condition (11.15) then yields

$$A\,k\left(1-{\rm e}^{-2\,k\,d}\right)\cos(\omega\,t-k\,x) = A\,\frac{\omega^{\,2}}{g}\left(1+{\rm e}^{-2\,k\,d}\right)\cos(\omega\,t-k\,x).$$ (11.20)

which reduces to the dispersion relation

$$\omega^{\,2} = g\,k\,\tanh(k\,d).$$ (11.21)

The type of wave described in this section is known as a gravity wave.