11.5: Water

Water is pretty complicated stuff. It wets things. It has viscosity. It forms whirlpools and eddies and has nonlinear turbulent motions that we cannot hope to understand using the techniques that we have at our disposal. In this section, we consider a somewhat idealized fluid, that we will call “dry water” (after Feynman) that has none of this complicated structure. It has three features that we will keep in common with the real thing. It has mass density. It has surface tension, and it is nearly incompressible. Let’s see how it waves.

Imagine an infinite universe full of an incompressible, frictionless liquid. This will allow us to see the consequences of the incompressibility in a simple, qualitative way. Consider the analog of a plane sound wave in such a system. That is, for example, a plane wave traveling in the \(x\) direction (with \(k_{y} = k_{z} = 0\)) with longitudinal displacements in the \(x\) direction. If the liquid is truly incompressible, the \(k_{x}\) must be zero for this wave, because any longitudinal displacement must be accompanied by compressions and rarefactions of the medium. Thus, for such a plane wave, \(\vec{k} = 0\). There are no nontrivial plane waves in the infinite system! In general, we do not expect that all the components of the \(k\) vector must vanish, because even in an incompressible liquid, displacement in one direction is allowed if it is accompanied by appropriate motion in other directions. But what we have seen is that we cannot have a mode that has a real \(\vec{k}\) vector. That would be a plane wave, which we have seen is not compatible with incompressibility. Instead, we expect that the constraint \(k_{x} = 0\) will be replaced by a constraint on the rotation invariant length of the \(k\) vector, that \(\vec{k} \cdot \vec{k}=0\). If some of the components of the \(\vec{k}\) vector are imaginary, this can be satisfied for nonzero \(\vec{k}\).

Note that the condition \(\vec{k} \cdot \vec{k}=0\) is not exactly a dispersion relation, because it makes no reference to frequency. But it is the whole story for an infinite system of incompressible fluid. In fact, it is clear that there are no harmonic waves in the infinite system, because there is nothing to produce a restoring force. Even if there is a gravitational field, the pressure in the liquid just adjusts itself to cancel the effect of gravity. We can get a nontrivial dispersion relation only when there is a surface. The dispersion relation then depends on the physics of the surface. This would seem to violate our general principle that the dispersion relation is a property of the infinite system. What is happening is this. The relation, \(\vec{k} \cdot \vec{k}=0\) is really the only dispersion relation that makes any sense for the three-dimensional infinite system. When we introduce a surface, we have broken the translation invariance in the direction normal to the surface. This allows us to get a nontrivial dispersion relation for the two-dimensional system parallel to the surface.

Mathematics of Water Waves

Now let us try to make these considerations quantitative. As usual, we will label our fluid in terms of the equilibrium positions of its parts. Then call the displacement from equilibrium of the fluid that is at the point \(\vec{r}\) at equilibrium \[\epsilon \vec{\psi}(\vec{r}, t)\]

for some small \(\epsilon\). This means that the actual position of the water is⁶ \[\vec{R}(\vec{r}, t)=\vec{r}+\epsilon \vec{\psi}(\vec{r}, t) .\]

We can regard (11.109) as a kind of change of coordinates. It maps us from the equilibrium coordinates (a rather arbitrary label because the water is free to flow) to the physical coordinates that tell us where the water actually is. If the water is incompressible, which is a pretty good approximation, then a small volume element should have the same volume in equilibrium and in the physical coordinates. \[d R_{x} d R_{y} d R_{z}=d x d y d z .\]

This will be the case if the determinant of the Jacobian matrix equals 1: \[\operatorname{det}\left(\begin{array}{ccc}
\frac{\partial R_{x}}{\partial x} & \frac{\partial R_{x}}{\partial y} & \frac{\partial R_{x}}{\partial z} \\
\frac{\partial R_{y}}{\partial x} & \frac{\partial R_{y}}{\partial y} & \frac{\partial R_{y}}{\partial z} \\
\frac{\partial R_{z}}{\partial x} & \frac{\partial R_{z}}{\partial y} & \frac{\partial R_{z}}{\partial z}
\end{array}\right)=1 .\]

Because \(\epsilon\) is small, we can expand (11.111) to lowest order in \(\epsilon\), \[\begin{gathered}
=\operatorname{det}\left(\begin{array}{ccc}
1+\epsilon \frac{\partial \psi_{x}}{\partial x} & \epsilon \frac{\partial \psi_{x}}{\partial y} & \epsilon \frac{\partial \psi_{x}}{\partial z} \\
\epsilon \frac{\partial \psi_{y}}{\partial x} & 1+\epsilon \frac{\partial \psi_{y}}{\partial y} & \epsilon \frac{\partial \psi_{y}}{\partial z} \\
\epsilon \frac{\partial \psi_{z}}{\partial x} & \epsilon \frac{\partial \psi_{z}}{\partial y} & 1+\epsilon \frac{\partial \psi_{z}}{\partial z}
\end{array}\right) \\
=1+\epsilon \vec{\nabla} \cdot \vec{\psi}+\mathcal{O}\left(\epsilon^{2}\right) .
\end{gathered}\]

Thus \[\vec{\nabla} \cdot \vec{\psi}=0 .\]

(11.113) is very reasonable. It is the statement that the flux of displacement into or out of any region vanishes.⁷ This is what we expected from our qualitative discussion.

To see what this means for waves, let us also assume that there are no eddies. The mathematical statement of this is \[\vec{\nabla} \times \vec{\psi}=0 .\]

If we do not assume (11.114), angular momentum conservation becomes important and life becomes very complicated. You will have to wait for courses on fluid dynamics to learn more about it. With the simplifying assumption, (11.114), the displacement can be written as the gradient of a scalar function, \(\chi\), \[\epsilon \vec{\psi}=\epsilon \nabla \chi .\]

This simplifies our life enormously, because we can now deal with the scalar quantity, \(\chi\). Space translation invariance tells us that we can find modes of the form \[\chi=e^{i \vec{k} \cdot \vec{r}-i \omega t} ,\]

which gives a displacement of the form \[\epsilon \vec{\psi}=i \epsilon \vec{k} e^{i \vec{k} \cdot \vec{r}-i \omega t} .\]

The condition, (11.113) then becomes \[\vec{k} \cdot \vec{k}=0 ,\]

as anticipated in our qualitative discussion at the beginning of the section.

Depth

11-3
Let us now consider waves in an “ocean” of depth \(L\), ignoring frictional forces, eddies and nonlinearities. We will further restrict our attention to a two-dimensional situation. Let \(y\) be the vertical direction, and consider water waves in the \(x\) direction. That is, we will take \(k_{x}\) real, because we are interested in wave propagation in the \(x\) direction, and \(k_{y}\) pure imaginary with the same magnitude, so that (11.118) is satisfied. Then we assume that nothing depends on the other coordinate, \(z\). Having simplified things this far, we may as well assume that our ocean is a rectangular box. Then the modes of interest of the infinite system look like \[\chi_{\infty}(x, y, t)=e^{\pm i k x \pm k y-i \omega t} .\]

If the ocean has a bottom at \(y = 0\), then the vertical displacement must vanish at \(y = 0\). Then (11.115) implies that we must combine modes of the infinite system to get a \(\chi\) whose \(y\) derivative vanishes at \(y = 0\), to get \[\chi(x, y, t) \propto e^{\pm i k x-i \omega t} \cosh k y .\]

where \(\cosh\) is the “hyperbolic cosine.” defined by \[\cosh x \equiv \frac{e^{x}+e^{-x}}{2} .\]

Then from (11.115), we get \[\begin{gathered}
\psi_{x}(x, y, t)=\frac{\partial}{\partial x} \chi(x, y, t)=\pm i e^{\pm i k x-i \omega t} \cosh k y . \\
\psi_{y}(x, y, l)=\frac{\partial}{\partial y} \chi(x, y, l)=e^{\pm i k x-i \omega t} \sinh k y .
\end{gathered}

Before going further, note that we could extend these considerations by adding a \(z\) coordinate. Then (11.120) would become \[\chi(x, y, t) \propto e^{\left(\pm i k_{x} x \pm i k_{z} z\right)-i \omega t} \cosh k y\]

where \[k=\sqrt{k_{x}^{2}+k_{z}^{2}} .\]

These are the two-dimensional wave modes of the infinite ocean of depth \(L\). The \(y\) dependence is completely fixed by the boundary condition at the bottom and the condition \(\vec{k} \cdot \vec{k}=0\). The only interesting dependence, from the point of view of space translation invariance, is the dependence on \(x\) and \(z\).

Now, let us return to the rectangular ocean, and the \(z\)-independent modes, (11.122). If our ocean has sides at \(x = 0\) and \(x = X\), we must choose linear combinations of the modes, (11.122), such that the \(x\) displacement vanishes at the sides. We can do this for \(x = 0\) by forming the combinations \[\begin{aligned}
&\psi_{x}(x, y, t)=-\sin k x \cosh k y \cos \omega t , \\
&\psi_{y}(x, y, t)=\cos k x \sinh k y \cos \omega t .
\end{aligned}\]

Then if \[k=\frac{n \pi}{X} ,\]

the boundary condition at \(x = X\) is satisfied as well.

Figure \( 11.22\): The motion of an incompressible fluid in a wave.

Now we know the mathematics of the displacement of the dry water. Before we go on to discuss the dispersion relation, let us pause to consider what this actually looks like. Imagine that we put a regular rectangular grid of points in the water in equilibrium. Then in Figure \( 11.22\), we show what the grid looks like in the mode, (11.125) with \(n = 1\).

Each of the little rectangles in (11.22) was a square in equilibrium position (when \(\psi=0\)). Note the way incompressibility works. When the water is squeezed in one direction, it is stretched in the other. You can see this in motion in program 11-3.

Figure \( 11.23\): The surface of a water wave, with horizontal displacement suppressed.

Having stared at this, we can now forget about it for a while, and concentrate just on the surface. That is what matters for the dispersion relation. For ease of presentation in the diagrams below, we will exaggerate the displacement in the vertical \(y\) direction and forget about the displacement of the surface in the x direction (which won’t matter anyway). Then the wave looks like the picture in Figure \( 11.23\). We will use energy arguments to get the dispersion relation. There are three contributions to the total energy of the standing wave, (11.125) — gravitational potential energy, energy stored in surface tension, and kinetic energy. Let us consider them in turn.

Figure \( 11.24\): Water is removed from the rectangle in \(X\) − \(x\) and raised to the rectangle at \(x\).

Gravitational Potential

In the diagram in Figure \( 11.24\), you can see that the overall effect of the displacements in the mode (11.125) is to take a chunk of the water from \(X\) − \(x\), raise it by \(\epsilon \psi_{y}(x, L, t)\) (the vertical displacement of the surface), and move it over to \(x\). The volume of this chunk is \(W d x \in \psi_{y}(x, L, t)\) where \(dx\) is the length of chunk and \(W\) is the width in the \(z\) direction (into the paper). Thus the total gravitational potential is \[\begin{aligned}
V_{\text {grav }}=\rho g & \int d V \Delta h=\rho g W \int_{0}^{\frac{\pi}{2 k}} d x\left|\epsilon \psi_{y}(x, L, t)\right|^{2}+\mathcal{O}\left(\epsilon^{3}\right) \\
=\rho g W \int_{0}^{\frac{\pi}{2 k}} d x \epsilon^{2} \cos ^{2} k x \sinh ^{2} k L \cos ^{2} \omega t+\cdots \\
\quad=\frac{\pi}{4 k} \rho g W \epsilon^{2} \sinh ^{2} k L \cos ^{2} \omega t+\cdots .
\end{aligned}\]

Surface Tension

The energy stored in surface tension is \(W\) times the difference between the length of the surface and the equilibrium length (\(X\)). This requires that we be a little careful about the position of the surface, going back to (11.109). The position of the surface is \[R_{x}(x, t)=x+\epsilon \psi_{x}(x, L, t), \quad R_{y}(x, t)=\epsilon \psi_{y}(x, L, t) .\]

The length is then \[\int_{0}^{X} d x \sqrt{\frac{\partial R_{x}}{\partial x}^{2}+\frac{\partial R_{y}^{2}}{\partial x}} .\]

But \[\frac{\partial R_{x}}{\partial x}=1+\epsilon \frac{\partial}{\partial x} \psi_{x}, \quad \frac{\partial R_{y}}{\partial x}=\epsilon \frac{\partial}{\partial x} \psi_{y} .\]

Thus \[\begin{aligned}
& V_{\text {surface }}=T \times\left(\text { Area }-\text { Area }_{0}\right) \\
=& T W \int_{0}^{\frac{\pi}{k}} d x\left(\sqrt{\left(1+\epsilon \partial \psi_{x} / \partial x\right)^{2}+\left(\epsilon \partial \psi_{y} / \partial x\right)^{2}}-1\right) \\
=& T W \int_{0}^{\frac{\pi}{k}} d x\left(\epsilon \partial \psi_{x} / \partial x+\frac{1}{2}\left(\epsilon \partial \psi_{y} / \partial x\right)^{2}+\mathcal{O}\left(\epsilon^{3}\right)\right) .
\end{aligned}\]

The order \(\epsilon\) term in (11.131) cancels when integrated of \(x\), so \[\begin{gathered}
=T W \epsilon^{2} \int_{0}^{\frac{\pi}{k}} d x \frac{1}{2} k^{2} \sin ^{2} k x \sinh ^{2} k L \cos ^{2} \omega t+\cdots \\
\quad=\frac{\pi}{4 k} T W \epsilon^{2} k^{2} \sinh ^{2} k L \cos ^{2} \omega t+\cdots .
\end{gathered}\]

Kinetic Energy

The kinetic energy is obtained by integrating \(\frac{1}{2} m v^{2}\) over the whole volume of the liquid: \[\begin{gathered}
K E=\frac{1}{2} \rho \int d V \vec{v}^{2} \\
=\frac{1}{2} \rho W \int_{0}^{\frac{\pi}{k}} d x \int_{0}^{L} d y\left(\left(\epsilon \partial \psi_{x} / \partial t\right)^{2}+\left(\epsilon \partial \psi_{y} / \partial t\right)^{2}\right)
\end{gathered}\]

\[\begin{aligned}
&=\frac{1}{2} \rho W \epsilon^{2} \int_{0}^{\frac{\pi}{k}} d x \int_{0}^{L} d y \omega^{2} \sin ^{2} \omega t \\
&\cdot\left(\cos ^{2} k x \sinh ^{2} k y+\sin ^{2} k x \cosh ^{2} k y\right)
\end{aligned}\]

\[\begin{gathered}
=\frac{\pi}{4 k} \rho W \epsilon^{2} \int_{0}^{L} d y \omega^{2} \sin ^{2} \omega t\left(\sinh ^{2} k y+\cosh ^{2} k y\right) \\
=\frac{\pi}{4 k} \rho W \epsilon^{2} \int_{0}^{L} d y \omega^{2} \sin ^{2} \omega t \cosh 2 k y \\
=\frac{\pi}{8 k^{2}} \rho W \epsilon^{2} \omega^{2} \sinh 2 k L \sin ^{2} \omega t .
\end{gathered}\]

Dispersion Relation

The total of (11.127)-(11.135) is \[\begin{gathered}
V_{\text {grav }}+V_{\text {surface }}+K E=\frac{\pi}{4 k} \rho g W \epsilon^{2} \sinh ^{2} k L \cos ^{2} \omega t \\
+\frac{\pi}{4 k} T W \epsilon^{2} k^{2} \sinh ^{2} k L \cos ^{2} \omega t+\frac{\pi}{8 k^{2}} \rho W \omega^{2} \epsilon^{2} \sinh 2 k L \sin ^{2} \omega t+\cdots .
\end{gathered}\]

This must be constant in time, which implies \[\begin{aligned}
\omega^{2} &=\frac{2 \sinh ^{2} k L\left(g k+\frac{T}{\rho} k^{3}\right)}{\sinh 2 k L} \\
&=\left(g k+\frac{T}{\rho} k^{3}\right) \tanh k L
\end{aligned}\]

where \(\tanh\) is the “hyperbolic tangent,” defined by \[\tanh x \equiv \frac{\sinh x}{\cosh x}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} .\]

Note that in the twin limit of long wavelength and shallow water, the water waves become nondispersive — for \(k L \ll 1\), and \(\rho g k \gg T k^{3}\) — \(\tanh k L\) → \(kL\) \[\omega^{2} \approx g L k^{2} .\]

Gravity versus Surface Tension

The dispersion relation, (11.139), involves a competition between gravity and surface tension. For long wavelengths gravity dominates and the \(gk\) term is most important. For short wavelengths, surface tension dominates and the \(\frac{T k^{3}}{\rho}\) term is more important. The cross-over occurs for wave numbers of order \[k \approx k_{0}=\sqrt{\frac{\rho g}{T}} .\]

The cross-over wavelength is actually a familiar distance. There is a much more familiar process that involves a similar competition between gravity and surface tension. Consider a water drop on a low friction surface, such as a teflon frying pan. A very tiny drop is nearly spherical. But as the size of the drop increases, it begins to flatten out. Then when the drop increases above a critical size, the height of the drop does not increase. It spreads out with a fixed height, \(h\), as shown in cross-section in Figure \( 11.25\).

Figure \( 11.25\): The cross-section of a water droplet on a frictionless surface.

As with the dispersion relation, we can understand what is going on by considering the energy. The total energy of the drop is a sum of the gravitational potential energy and the energy due to surface tension. \[V_{\text {grav }} \approx \frac{1}{2} \rho g h v ,\]

where \(v\) is the volume of the drop and \[V_{\text {surface }} \approx \frac{T v}{h} .\]

The volume is fixed, so the equilibrium value of \(h\) minimizes the sum \[V_{\text {grav }}+V_{\text {surface }} \approx \frac{1}{2} \rho g h v+\frac{T v}{h} .\]

The minimum occurs for \[T=\frac{1}{2} \rho g h^{2} .\]

The measured surface tension of water is \(T \approx 72 \space \mathrm{dynes} / \mathrm{cm}\). This gives the familiar height of a water drop, \(h \approx 0.4 \mathrm{~cm}\). This height is related to \(k_{0}\) by \[k_{0}=\sqrt{\frac{\rho g}{T}} \approx \frac{\sqrt{2}}{h} .\]

_________________________
⁶Here we can take \(\psi\) to be dimensionless and let the parameter, \(\epsilon\), be a small displacement.
⁷Note, however, that for large \(\epsilon\), incompressibility is the nonlinear constraint, (11.111).