# What does it mean for waves to “feel” the bottom?

While typically waves are said to "feel" the bottom when the depth of the water is less than half the wavelength, what does it mean for the waves to "feel"?

Furthermore, why does this happen at the depth of half the wavelength?

• I've never heard this terminology before, but here's an example of usage. Interesting question! – naught101 Oct 9 '14 at 0:48
• – milancurcic Oct 9 '14 at 3:17

In water wave physics, when we say that the wave "feels" the bottom, we mean that the water depth affects the properties of the wave.

The dispersion relationship for water waves is:

$$\omega^2 = gk \tanh{(kd)}$$

where $\omega$ is the wave frequency, $k$ is the wavenumber, $d$ is the mean water depth, and $g$ is gravitational acceleration. We distinguish "shallow" and "deep" water waves by the value of $kd$, which includes both the wavenumber and the water depth:

• Shallow water waves when $kd < 0.3$;
• Intermediate water waves when $0.3 < kd < 3$;
• Deep water waves when $kd > 3$.

Thus, a long swell wave may act as a shallow water wave in depths of 10 m, but also a very short wave may act as a deep water wave in depths of 1 m.

How are these limits for $kd$ obtained? The tangent hyperbolic function has some convenient properties for the limiting values of its argument:

For deep water waves, $kd$ is very large, so $\tanh{(kd)} \rightarrow 1$. The dispersion relationship then reduces to:

$$\omega^2 = gk$$

Phase and group speeds are then:

$$C_p = \dfrac{\omega}{k} = \sqrt{\dfrac{g}{k}}$$

$$C_g = \dfrac{\partial \omega}{\partial k} = \dfrac{1}{2}\sqrt{\dfrac{g}{k}}$$

Notice that $C_p$ and $C_g$ are not a function of water depth, thus it is said that deep water waves don't "feel" the bottom.

On the other hand, for shallow water waves, $kd$ is small (approximately 0.3 or less), and $\tanh{(kd)} \rightarrow kd$. The dispersion relationship is now:

$$\omega^2 = gk^2d$$

and phase and group speeds are functions of water depth:

$$C_p = \dfrac{\omega}{k} = \sqrt{gd}$$

$$C_g = \dfrac{\partial \omega}{\partial k} = \sqrt{gd}$$

So, why is it said that the waves "feel" the bottom at the water depth of half wavelength?

$$kd = k \dfrac{\lambda}{2} = k \dfrac{2\pi}{2k} = \pi$$

and this is approximately the value below which the regime transitions from deep water to intermediate water, i.e. $\tanh{(kd)} \approx 1$ does not hold anymore.

Feel the bottom refers to the fact that the wave-induced velocity field extends all the way from the top of the water column to the bottom of the water column. When the wave "feels the bottom" it means that there is some interaction with the bottom boundary. A very thin boundary layer develops at the bottom where vorticity is generated due to the velocity field interacting with the bed roughness. The vorticity can diffuse up into the interior of the fluid and is responsible for turbulent activity inside the boundary layer. Due to the persistent nature of waves, their action is thought to play an important role in the redistribution of particulate matter in the nearshore, such as nutrients, larvae, sediment and even contaminants.

The value of 1/2 depth is somewhat arbitrary but provides a good estimate based on theory (see IRO-bot's answer). This picture exemplifies it pretty well.

• +1 Yes, this figure is a very neat summary. – milancurcic Oct 9 '14 at 3:25
• It's really nice to see two answers from different perspectives - one with mathematical theory, one with physical / intuitive explanation. However, I'd dispute the assertion that the figure seems to make that all shallow water waves break. – Semidiurnal Simon Oct 9 '14 at 10:09
• The L/2 makes a kind of intuitive sense, but why L/20? Is that just a rough estimate? – naught101 Oct 10 '14 at 2:58
• @naught101 sorry that comment didn't come out right: That is the point when $kd \approx tanh(kd)$. That's because we know that for shallow water $kd << 1$ and that the $\tanh$ approximation says that $\tanh(x) = x$ as $x \to 0$ i.imgur.com/jPc1V0T.png – Isopycnal Oscillation Oct 10 '14 at 7:26