Gravity Waves

Consider a stationary body of water, of uniform depth $d$, located on the surface of the Earth. Let us find the dispersion relation of a plane wave propagating across the water's surface. Suppose that the Cartesian coordinate $x$ measures horizontal distance, while the coordinate $z$ measures vertical height, with $z=0$ corresponding to the unperturbed surface of the water. Let there be no variation in the $y$-direction. In other words, let the wavefronts of the wave all be parallel to the $y$-axis. Finally, let $v_x(x,z,t)$ and $v_z(x,z,t)$ be the perturbed horizontal and vertical velocity fields of the water due to the wave. It is assumed that there is no motion in the $y$-direction.

Water is essentially incompressible (i.e., its bulk modulus is very large) (Batchelor 2000). Thus, any (low speed) wave disturbance in water is constrained to preserve the volume of a co-moving volume element. Equivalently, the inflow rate of water into a stationary volume element must match the outflow rate. Consider a stationary cubic volume element lying between $x$ and $x+dx$, $y$ and $y+dy$, and $z$ and $z+dz$. The element has two faces, of area $dy\,dz$, perpendicular to the $x$-axis, located at $x$ and $x+dx$. Water flows into the element through the former face at the rate $v_x(x,z,t)\,dy\,dz$ (i.e., the product of the area of the face and the normal velocity), and out of the element though the latter face at the rate $v_x(x+dx,z,t)\,dy\,dz$. The element also has two faces perpendicular to the $y$-axis, but there is no flow through these faces, because $v_y=0$. Finally, the element has two faces, of area $dx\,dy$, perpendicular to the $z$-axis, located at $z$ and $z+dz$. Water flows into the element through the former face at the rate $v_z(x,z,t)\,dx\,dy$, and out of the element though the latter face at the rate $v_z(x,z+dz,t)\,dx\,dy$. Thus, the net rate at which water flows into the element is $v_x(x,z,t)\,dy\,dz + v_z(x,z,t)\,dx\,dy$. Likewise, the net rate at which water flows out of the element is $v_x(x+dx,z,t)\,dy\,dz + v_z(x,z+dz,t)\,dx\,dy$. If the water is to remain incompressible then the inflow and outflow rates must match. In other words,

$$v_x(x,z,t)\,dy\,dz + v_z(x,z,t)\,dx\,dy = v_x(x+dx,z,t)\,dy\,dz + v_z(x,z+dz,t)\,dx\,dy,$$ (9.258)

or

$$\left[\frac{v_x(x+dx,z,t)-v_x(x,z,t)}{dx} + \frac{v_z(x,z+dz,t)-v_z(x,z,t)}{dz} \right]dx\,dy\,dz=0.$$ (9.259)

Hence, the incompressibility constraint reduces to

$$\frac{\partial v_x}{\partial x} + \frac{\partial v_z}{\partial z} = 0,$$ (9.260)

which is consistent with Equation (7.194) in the incompressible limit $K\rightarrow \infty$.^9.1 Incidentally, it is reasonable to neglect the finite compressibility of water in this investigation because sound waves (which require finite compressibility) propagate much faster than surface waves, and, therefore, do not couple to them. (See the two footnotes in this section.)

Consider the equation of motion of a small volume element of water lying between $x$ and $x+dx$, $y$ and $y+dy$, and $z$ and $z+dz$. The mass of this element is $\rho\,dx\,dy\,dz$, where $\rho=1.0\times 10^3\,{\rm m}^{-3}$ is the uniform mass density of water. Suppose that $p(x,z,t)$ is the pressure in the water, which is assumed to be isotropic (Batchelor 2000). The net horizontal force on the element is $p(x,z,t)\,dy\,dz - p(x+dx,z,t)\,dy\,dz$ (because force is pressure times area, and the external pressure forces acting on the element are directed inward normal to its surface). Hence, the element's horizontal equation of motion is

$$\rho\,dx\,dy\,dz\,\frac{\partial v_x(x,z,t)}{\partial t} = -\left[\frac{p(x+dx,z,t)-p(x,z,t)}{dx}\right]dx\,dy\,dz,$$ (9.261)

which reduces to

$$\rho\,\frac{\partial v_x}{\partial t} = - \frac{\partial p}{\partial x}.$$ (9.262)

[See Equation (7.192).] The vertical equation of motion is similar, except that the element is subject to a downward acceleration, $g\simeq 9.8\,{\rm m\,s}^{-2}$, due to gravity. Hence, we obtain

$$\rho\,\frac{\partial v_z}{\partial t} = -\frac{\partial p}{\partial z} - \rho\,g.$$ (9.263)

[See Equation (7.193).]

We can write

$$p = p_0-\rho\,g\,z+p_1,$$ (9.264)

where $p_0$ is atmospheric pressure (i.e., the air pressure just above the surface of the water), and $p_1$ is the pressure perturbation due to the wave. In the absence of the wave, the water pressure at a depth $h$ below the surface is $p_0+\rho\,g\,h$ (Batchelor 2000). Substitution into Equations (9.262) and (9.263) yields

$$\rho\,\frac{\partial v_x}{\partial t} = - \frac{\partial p_1}{\partial x},$$ (9.265)

$$\rho\,\frac{\partial v_z}{\partial t} = - \frac{\partial p_1}{\partial z}.$$ (9.266)

It follows that

$$\rho\,\frac{\partial^{\,2} v_x}{\partial z\,\partial t} -\rho\,\f... ...partial z\,\partial x} + \frac{\partial^{\,2} p_1}{\partial x\,\partial z} = 0,$$ (9.267)

which implies that

$$\rho\,\frac{\partial}{\partial t}\!\left(\frac{\partial v_x}{\partial z} -\frac{\partial v_z}{\partial x}\right) = 0,$$ (9.268)

or

$$\frac{\partial v_x}{\partial z} -\frac{\partial v_z}{\partial x} = 0.$$ (9.269)

(Actually, this quantity could be non-zero and constant in time, but this is not consistent with an oscillating wave-like solution.)

Equation (9.269) is automatically satisfied by writing the fluid velocity in terms of a velocity potential; that is,

$$v_x = \frac{\partial\phi}{\partial x},$$ (9.270)

$$v_z =\frac{\partial \phi}{\partial z}.$$ (9.271)

(See Section 7.12.) Equation (9.260) then gives ^9.2

$$\frac{\partial^{\,2}\phi}{\partial x^{\,2}} + \frac{\partial^{\,2}\phi}{\partial z^{\,2}} = 0.$$ (9.272)

Finally, Equations (9.265) and (9.266) yield

$$p_1 = -\rho\,\frac{\partial \phi}{\partial t}.$$ (9.273)

As we have just demonstrated, surface waves in water are governed by Equation (9.272), which is known as Laplace's equation. We now need to derive the physical constraints that must be satisfied by the solution to this equation at the water's upper and lower boundaries. The water is bounded from below by a solid surface located at $z=-d$. Assuming that the water always remains in contact with this surface, the appropriate physical constraint at the lower boundary is $v_z(x,-d,t)=0$ (i.e., there is no vertical motion of the water at the lower boundary), or

$$\left.\frac{\partial \phi}{\partial z}\right\vert _{z=-d} = 0.$$ (9.274)

The physical constraint at the water's upper boundary is a little more complicated, because this boundary is a free surface. Let $\zeta(x,t)$ represent the vertical displacement of the water's surface. It follows that

$$\frac{\partial \zeta}{\partial t} = \left.v_z\right\vert _{z=0}=\left.\frac{\partial \phi}{\partial z}\right\vert _{z=0}.$$ (9.275)

The physical constraint at the surface is that the water pressure is equal to atmospheric pressure, because there cannot be a pressure discontinuity across a free surface (in the absence of surface tension). Thus, it follows from Equation (9.264) that

$$p_0 = p_0 -\rho\,g\,\zeta(x,t) + p_1(x,0,t).$$ (9.276)

Finally, differentiating with respect to $t$, and making use of Equations (9.273) and (9.275), we obtain

$$\left.\frac{\partial \phi}{\partial z}\right\vert _{z=0} = -g^{-1}\left.\frac{\partial^{\,2}\phi}{\partial t^{\,2}}\right\vert _{z=0}.$$ (9.277)

Hence, the problem boils down to solving Laplace's equation, (9.272), subject to the physical constraints (9.274) and (9.277).

Let us search for a propagating wave-like solution of Equation (9.272) of the form

$$\phi(x,z,t) = F(z)\,\cos(\omega\,t-k\,x).$$ (9.278)

Substitution into Equation (9.272) yields

$$\frac{d^{\,2} F}{dz^{\,2}} - k^{\,2}\,F= 0,$$ (9.279)

whose independent solutions are $\exp(+k\,z)$ and $\exp(-k\,z)$. Hence, the most general wave-like solution to Laplace's equation is

$$\phi(x,z,t) = A\,{\rm e}^{\,k\,z}\,\cos(\omega\,t-k\,x) + B\,{\rm e}^{-k\,z}\,\cos(\omega\,t-k\,x),$$ (9.280)

where $A$ and $B$ are arbitrary constants. The boundary condition (9.274) is satisfied provided that $B = A\,\exp(-2\,k\,d)$, giving

$$\phi(x,z,t) = A\left[{\rm e}^{\,k\,z}\ + {\rm e}^{-k\,(z+2\,d)}\right]\cos(\omega\,t-k\,x),$$ (9.281)

The boundary condition (9.277) yields

$$A\,k\left(1-{\rm e}^{-2\,k\,d}\right)\cos(\omega\,t-k\,x) = A\,\frac{\omega^{\,2}}{g}\left(1+{\rm e}^{-2\,k\,d}\right)\cos(\omega\,t-k\,x),$$ (9.282)

which reduces to the dispersion relation

$$\omega^{\,2} = g\,k\,\tanh(k\,d).$$ (9.283)

The type of wave just described is generally known as a gravity wave. We conclude that a gravity wave of arbitrary wavenumber $k$, propagating horizontally through water of depth $d$, has a phase velocity

$$v_p = \frac{\omega}{k}=(g\,d)^{1/2}\left[\frac{\tanh(k\,d)}{k\,d}\right]^{1/2}.$$ (9.284)

Moreover, the ratio of the group to the phase velocity is

$$\frac{v_g}{v_p} =\frac{k}{\omega}\,\frac{d\omega}{dk}=\frac{1}{2}\left[1+\frac{2\,k\,d}{\sinh(2\,k\,d)}\right].$$ (9.285)

Finally, the velocity fields associated with a gravity wave of surface amplitude $a$ are

$$v_x(x,z,t) =a\,\omega\,\frac{\cosh[k\,(z+d)]}{\sinh(k\,d)}\,\sin(\omega\,t-k\,x),$$ (9.286)

$$v_z(x,z,t) =a\,\omega\,\frac{\sinh[k\,(z+d)]}{\sinh(k\,d)}\,\cos(\omega\,t-k\,x).$$ (9.287)

In shallow water (i.e., $k\,d\ll 1$), Equation (9.283) reduces to the linear dispersion relation

$$\omega = k\,\sqrt{g\,d}.$$ (9.288)

Here, use has been made of the small argument expansion $\tanh x\simeq x$ for $\vert x\vert\ll 1$. (See Appendix B.) It follows that gravity waves in shallow water are non-dispersive in nature, and propagate at the phase velocity $\sqrt{g\,d}$. On the other hand, in deep water (i.e., $k\,d\gg 1$), Equation (9.283) reduces to the nonlinear dispersion relation

$$\omega = \sqrt{k\,g}.$$ (9.289)

Here, use has been made of the large argument expansion $\tanh x\simeq 1$ for $x\gg 1$. We conclude that gravity waves in deep water are dispersive in nature. The phase velocity of the waves is $v_p=\omega/k = \sqrt{g/k}$, whereas the group velocity is $v_g=d\omega/dk = (1/2)\,\sqrt{g/k}= v_p/2$. In other words, the group velocity is half the phase velocity, and is largest for long wavelength (i.e., small $k$) waves.

The mean kinetic energy per unit surface area associated with a gravity wave is defined

$$K = \left\langle\int_{-d}^\zeta \frac{1}{2}\,\rho\,v^{\,2}\,dz\right\rangle,$$ (9.290)

where

$$\zeta(x,t) = a\,\sin(\omega\,t-k\,x)$$ (9.291)

is the vertical displacement at the surface, and

$$\langle \cdots\rangle = \int_0^{2\pi} (\cdots)\,\frac{d(k\,x)}{2\pi}$$ (9.292)

represents an average over a wavelength. Given that $\langle \cos^2(\omega\,t-k\,x)\rangle =\langle \sin^2(\omega\,t-k\,x)\rangle=1/2$, it follows from Equations (9.286) and (9.287) that, to second order in $a$,

$$K = \frac{1}{4}\,\rho\,a^{\,2}\,\omega^{\,2}\int_{-d}^0 \frac{\co... ...,d)}\,dz=\frac{1}{4}\,\rho\,g\,a^{\,2}\,\frac{\omega^{\,2}}{g\,k\,\tanh(k\,d)}.$$ (9.293)

Making use of the general dispersion relation (9.283), we obtain

$$K = \frac{1}{4}\,\rho\,g\,a^{\,2}.$$ (9.294)

The mean potential energy perturbation per unit surface area associated with a gravity wave is defined

$$U = \left\langle\int_{-d}^\zeta \rho\,g\,z\,dz\right\rangle + \frac{1}{2}\,\rho\,g\,d^{\,2},$$ (9.295)

which yields

$$U = \left\langle \frac{1}{2}\,\rho\,g\,(\zeta^{\,2}-d^{\,2})\righ... ...frac{1}{2}\,\rho\,g\,d^{\,2} = \frac{1}{2}\,\rho\,g\,\langle\zeta^{\,2}\rangle,$$ (9.296)

or

$$U = \frac{1}{4}\,\rho\,g\,a^{\,2}.$$ (9.297)

In other words, the mean potential energy per unit surface area of a gravity wave is equal to its mean kinetic energy per unit surface area.

Finally, the mean total energy per unit surface area associated with a gravity wave is

$$E = K+ U = \frac{1}{2}\,\rho\,g\,a^{\,2}.$$ (9.298)

This energy depends on the wave amplitude at the surface, but is independent of the wavelength, or the water depth.