We all know that as waves approach the shallow shores, the waves begin to form a characteristic shape. The upper portion of these breaking waves appears to curl forward and downwards over the bottom segment of the wave, before breaking into "white wash". The image below illustrates what this characteristic shape looks like:

enter image description here

So why do waves form this characteristic breaking shape as they approach the shallow shores?

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    $\begingroup$ At a very basic level, the bottom of the wave gets slowed by bed friction, causing the steepness of the wave's slope to increase, and there's a limit to how steep a wave can be before it falls over. But I'm sure somebody will give a much fuller answer. $\endgroup$ Apr 22, 2014 at 11:20
  • $\begingroup$ @Geodude Though bed friction is significant in shallow water, it is not quite what causes the waves to shoal. See my answer below. $\endgroup$ Apr 23, 2014 at 1:16
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    $\begingroup$ @SimonW The effect of friction is mostly confined to the boundary layer, the viscous effects will diffuse into the top layer altering its vorticity, but that contribution is relatively small. Check IRO-bot. $\endgroup$ Apr 23, 2014 at 3:23
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    $\begingroup$ By the way, the reason a wave breaks is because the maximum horizontal fluid velocity due to the wave, U, that is the speed that an orbiting particle possesses as the wave advances, is greater than the wavespeed, c. So if the Froude number = U/c > 1 the orbital velocity of a fluid particle exceeds the wave speed and the wave topples over. $\endgroup$ Apr 23, 2014 at 3:51
  • $\begingroup$ I stand corrected - thank you :-) In particular, I had never made the connection to Froude number, although it seems obvious now that you mention it. $\endgroup$ Apr 23, 2014 at 5:19

2 Answers 2


The physical process you describe is known as wave shoaling.

At the basic level, waves propagating into shallow water become shorter and higher, and consequently, steeper. In shallow water, the water particles near the crest move forward faster than those below them. Similarly, the particles near the trough move backward faster than those above them. This causes strong shearing of the near-surface body of water, eventually forming a plunging breaker, or a surf wave. For small-slope (linear) and inviscid (no friction) waves, the above is a consequence of the bottom boundary condition for the water velocity to be zero at the sea floor.

There are two fundamental and related properties of water waves that contribute to shoaling. One is the wave frequency $\omega$ remaining constant as the depth $d$ decreases. Think of this as the conservation of wave crests at any fixed point. However, the wavenumber $k$ (wavelength $\lambda$) must increase (decrease) with decreasing depth, as per the dispersion relationship of water waves (neglecting the effects of viscosity):

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

where $g$ is gravitational acceleration. In shallow water, the dispersion relationship reduces to:

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

and phase speed $C_p$ and group velocity $C_g$ are both proportional to the square root of the water depth:

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

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

Individual wave crests propagate with phase speed $C_p$. Wave groups and wave energy propagate with group velocity $C_g$. Thus, waves entering shallow water become shorter and slower. The second important property leading to shoaling is the conservation of wave energy flux, which is proportional to the group velocity and wave energy:

$$\dfrac{\partial(C_{g}E)}{\partial x}=0$$

Because the group velocity decreases, the wave energy (read: height) must increase (locally). This causes the waves to grow in height as they enter shallow water. As pointed out by @IsopycnalOscillation in a comment above, the separation of water particles from the wave crests happen because the individual orbital velocities at the top of the crest exceed by far the phase speed of the wave.

Although bottom friction is significant and non-negligible in shallow water, it does not cause shoaling. Mathematically, wave shoaling can occur for completely inviscid and small-slope (linear) waves that propagate into water of decreasing depth.

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    $\begingroup$ +1 for mention of dispersion. $\endgroup$
    – Neo
    Apr 23, 2014 at 1:16

One thing that I would add is a discussion of a physical parameter that is a simple measure of whether a wave is going to topple or not. The Froude number is defined as the ratio of the maximum absolute fluid velocity, $U$, and the wavespeed, $c$:

$F = U/c$

Because of the mechanism explained by IRO-bot,

In shallow water, the water particles near the crest move forward faster than those below them. Similarly, the particles near the trough move backward faster than those above them.

the highest velocities are encountered at the top of the wave and therefore a good rule of thumb is, if the surface fluid velocity aproaches the wave speed, i.e $F \to 1$, then the wave will topple over.

The question is what values of $F$ give rise to that perfect tube feeling that surfers yearn for and so perfectly exemplified by your picture.

Of course, since $F > 1$ implies that the wave amplitude is large, linear theory is no longer helpful. We may have to wait until we can get some definite answers from DNS 3D simulations of oceanic scale wave breaking phenomena.


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