17  Surface waves 2

In this lecture we want to study the wave solution that we derived in the previous lecture. We will first look at the paths that the particles in the fluid follow. Then we will investigate under what condition the linear approximation that we made to simplify the surface conditions is valid. Finally we look at the dispersion in the surface waves.

You find content related to this lecture in the textbooks:

17.1 Pathlines

From the velocity potential \(\phi\) in Eq. 16.22 we can read off the components of the velocity field as \[\begin{split} u_x&=\partial_x\phi=A\omega e^{k\,y}\cos(kx-\omega t),\\ u_y&=\partial_y\phi=A\omega e^{k\,y}\sin(kx-\omega t). \end{split} \tag{17.1}\] The factor of \(e^{k\,y}\) shows us that the velocity decreases exponentially with depth.

We want to understand how the fluid particles move, i.e., we want to determine the pathlines. These satisfy \[ \frac{\underline{x}(t)}{dt}=\underline{u}(\underline{x}(t),t). \tag{17.2}\] Unfortunately these nonlinear equations are too complicated to solve analytically. So we linearise. We know from experience that, while the waves on the surface move over long distances, the particles in the fluid are not swept away with the wave but only move up and down and forth and back a little bit. So we write the position of the particle as \[ \underline{x}(t)=\bar{\underline{x}}+\hat{\underline{x}}(t), \tag{17.3}\] where \(\bar{\underline{x}}\) is the mean position and the \(\hat{\underline{x}}\) are assumed to be small. Substituting this into Eq. 17.2 and looking at the components gives \[\begin{split} \frac{d\hat{x}}{dt}&\approx A\omega e^{k\,\bar{y}}\cos(k\bar{x}-\omega t),\\ \frac{d\hat{y}}{dt}&\approx A\omega e^{k\,\bar{y}}\sin(k\bar{x}-\omega t). \end{split} \tag{17.4}\] This is easy to integrate to \[\begin{split} \hat{x}&\approx -A e^{k\,\bar{y}}\sin(k\bar{x}-\omega t),\\ \hat{y}&\approx A e^{k\,\bar{y}}\cos(k\bar{x}-\omega t). \end{split} \tag{17.5}\] This describes motion counterclockwise around a circle of radius \(A e^{k\,\bar{y}}\) around the mean position. When that radius is small, the pathlines are close to perfect circles. Closer to the surface where the radius gets bigger the circle will get deformed a bit.

17.2 Validity of linear approximation

Now that we have the solutions to the linearised surface conditions we can check under what circumstances the linearisation was a good approximation. For example in Eq. 16.9 we dropped the term \(u_x\partial_x\eta\). This is a good approximation only if the dropped term is much smaller in magnitude than the terms that remain. So we need \[ |u_x\partial_x\eta|\ll|u_y|. \tag{17.6}\] Substituting our solutions from Eq. 17.1 and Eq. 16.21 gives \[ |A\omega e^{k\,\eta}\cos(kx-\omega t)A\,k\sin(kx-\omega t)| \ll|A\omega e^{k\,\eta}\sin(kx-\omega t)|. \tag{17.7}\] We can cancel a sin from both sides. The factors of cosine can be dropped because it only takes values between \(-1\) and \(1\). So, after cancellations, the condition becomes \[ |A\,k|\ll 1. \tag{17.8}\] Written in terms of the wavelength \(\lambda\) this becomes \[ A\ll\frac{\lambda}{2\pi}. \tag{17.9}\] So the amplitude of the wave must be small compared to its wavelength. This turns out to be exactly the same condition that we also needed in our linearisation while deriving the equation for waves on a string in lecture 1.

We made other approximations while we linearised the surface conditions and we need to check whether they too are valid under the same circumstances. For example we replaced \(u_y(x,\eta,t)\) by \(u_y(x,0,t)\). This is a good approximation if \[ |u_y(x,\eta,t)-u_y(x,0,t)|\ll|u_y(x,0,t)|. \tag{17.10}\] Substituting our solution gives \[ |A\omega \left(e^{k\,\eta}-1\right)\sin(kx-\omega t)| \ll|A\omega \sin(kx-\omega t)|, \tag{17.11}\] which simplifies to \[ \left|e^{k\,\eta}-1\right|\ll1. \tag{17.12}\] This holds if \(|k\, \eta|\ll 1\) which in turn holds if \(|k\,A|\ll 1\).

I leave it up to you to go through the approximations that we made in Section 16.1.2 to verify that they too are good when \(|k\,A|\ll 1\).

17.3 Circular waves

You will have observed that when you drop a pebble into a pond, this creates waves that travel outwards from the disturbance with circular wave crests. We can easily find the solutions describing such circular waves by switching to polar coordinates \(r\), \(\theta\) for the \(x\)-\(z\) plane such that \(x=r\cos\theta\) and \(z=r\sin\theta\).

The circular waves we are looking for look the same in all directions, so we will be looking for solutions for the velocity potential \(\phi\) and the surface function \(\eta\) that are independent of the \(\theta\) coordinate: \[ \phi = \phi(r,y,t),~~~\eta = \eta(r,t). \]

The Laplace equation for the velocity potential \(\phi\) then becomes \[ 0 = \nabla^2\phi=\frac{1}{r}\partial_r^2(r\phi)+\partial_y^2\phi. \tag{17.13}\]

Introducing again the distance \(d\) from the surface, \(d(r,y,t)=y-\eta(r,t)\), the kinematic surface condition in these coordinates is \[\begin{split} \frac{Dd}{Dt}&=\partial_td+\underline{u}\cdot\underline{\nabla} d=\partial_td+u_r\partial_r d+u_y\partial_y d\\ &=-\partial_t\eta-u_r\partial_r \eta+u_y=0. \end{split} \tag{17.14}\] After linearisation this reduces to \[ \partial_t\eta(r,t)=u_y(r,0,t). \tag{17.15}\] This is the same as in the case of Carthesian coordinates, just with \(r\) taking the place of \(x\). Similarly the linearised dynamic surface condition becomes \[ \partial_t\phi(r,0,t)=-g\eta(r,t). \tag{17.16}\] Thus the only difference between the problem of circular waves compared to the problem of plane waves we studied earlier is that the \(\partial_x^2\phi\) term in the Laplace equation has been replaced by \(1/r \partial_r^2(r\phi)\). This motivates us to make the modified Ansatz \[ \phi(r,y,t) = f(y)\frac{1}{r}\sin(kr-\omega t). \] Substituting this into the Laplace equation -Eq. 17.13 again gives the equation \(f''-k^2f\) and imposing that the solution must stay finite as \(y\to-\infty\) singles out the solution \(f(y)=D\,e^{ky}\) for some constant \(D\). Thus \[ \phi(r,y,t)=D e^{ky}\frac{1}{r}\sin(kx-\omega t) \]

Substituting this velocity potential into the dynamical surface condition -Eq. 17.16 gives \[ \eta(r,t)=\frac{\omega\,D}{g}\frac{1}{r}\cos(kr-\omega t) \] and then the kinematic surface condition -Eq. 17.15 gives the dispersion relation \[ \omega = \sqrt{k\,g}. \] This is identical to the dispersion relation for the plane wave solution.