I am assuming it is possible to calculate the mass of water if the volume is known. My assumption is that doing this for stagnant water, in a container perhaps, is easy.

However I want calculate, approximately, for a wave of water at sea.

What kind of information would I need if I wanted to calculate the mass or volume of a wave of water?

For example, if I know the space from one wave crest to the next and the period could I use that information?

So again how would I calculate the mass of a wave of water?

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    $\begingroup$ What do you mean by weight of a wave. I would say that waves have no mass. $\endgroup$ – jinawee Apr 14 '14 at 9:44
  • $\begingroup$ You need to define the volume which makes up your wave. Once you've done that all you need to do is calculate the integral. Since waves are more or less a surface phenomenon, at the very least you need to define how much of the water between the surface and the ocean floor is "contained" in the wave. $\endgroup$ – Carl Witthoft Apr 14 '14 at 13:03
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    $\begingroup$ A wave transports momentum and energy without transferring mass. Here's a tutorial on ocean waves: surflibrary.org/wavephysics.pdf $\endgroup$ – user6972 Apr 14 '14 at 17:37
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    $\begingroup$ Wouldn't the masses(or volumes) in the crests and troughs cancel each other, as you reason it? But it still does not make sense to find the mass or volume of a wave, beacause we usually study waves for other purposes. $\endgroup$ – Shubham Apr 14 '14 at 20:23
  • $\begingroup$ @jinawee how is it possible that a wave of water "have no mass"? $\endgroup$ – CharlieK Apr 14 '14 at 21:15

There has to be a few assumptions. Let's assume we are talking about a linear plane wave in relatively deep water. Because the the case where the bottom comes into play the upward hydrostatic force distorts the wave. Picking deep water or insuring the relative depth of d to L (d is average water depth and L is the wavelength of the wave) is $d/L > 0.5$. Compare the deep water case to the shallow water case and the particle motion in each condition (SWL is standing or still water level) is shown by the circles or ovals:

deep vs shallow

When a wave propagates from deep water offshore in to shallower water near-shore the wave length decreases, but at a slower rate than that at which the depth decreases. This shift to higher frequency also deforms the wave. The bottom is applying force on the wave which also lifts it up above the SWL until it becomes unstable and breaks.

Since in your comments you mention you're looking for momentum (not really mass), we have to talk about what that means in a wave. In fluid flow problems, some analyses are best carried out by energy considerations (e.g., head loss along a length of pipe) and some by momentum considerations (e.g., force exerted by a water jet hitting a wall). Similarly, for waves it is better to consider the flux of momentum for some problem analyses. For wave analyses, the flux of momentum is commonly referred to as the wave "radiation stress" which may be defined as "the excess flow of momentum due to the presence of waves" (Longuet-Higgins and Stewart, 1964).

The analysis of the momentum is based on the particle motion shown above and the computation is well illustrated on the wiki page for momentum flux including the boundary conditions. (Note that in this derivation gravity is defined as negative downwards). The end result for deep water waves is a momentum flux of

$\frac{dM}{dt}=-\frac{1}{4}\rho gA^2$

Where $\rho$ is the water density, g is gravity and A is the height of the wave (middle to crest).

As an aside about breaking waves they are very different conditions.

In the case of a breaking wave the common profile is shown below (not a sawtooth wave).


With some of the common terms defined as:

  • Vertical asymmetry: $\frac{a_c}{H}$
  • Slope asymmetry: 0.5(slope a + slope b)
  • Horizontal asymmetry (1): distance 1/distance 2
  • Horizontal asymmetry (2) distance 3/distance 4

The vertical asymmetry continuously increase as the wave shoals, reaching a maximum of between 0.62 and 0.74 at breaking. The slope and horizontal asymmetries also continuously increased as the wave shoaled; but, as opposed to vertical asymmetries steeper bottom slopes caused greater slope and horizontal asymmetries. And as the wave height increases the crest particle velocity will eventually equal the wave's phase velocity (or celerity $C=\frac{L}{T}$) and the wave will break.

Breaking waves fit into 4 basic category conditions/shapes:

waves breaking


Since it's the momentum of the wave that you are after, here's a good way to estimate its mass:

On the open ocean: Waves are approximately sinusoidal in shape. Take a trough - the lowest point for a wave - as the base. Estimate the height of the wave and perform an integral to estimate the cross-sectional area of a wave. Then multiply this by the approximate length of either the wave or the section of the wave you are interested in (try for the latter because ocean waves can be very long). Now that you have its volume, multiply by the density, ~$1000kg/m^3$. Voila! You have an estimated mass.

Close to shore: This is a bit trickier, waves close to shore are more of a sawtooth shape and they can end up, well, wave-shaped (the semi-arc that we all see surfers on). You can ask around on Math.SE probably for how to estimate the area of that shape, but still you only have to multiply by the length of the wave you want and then by the density. Probably easy to treat the wave as a wall of water if you want a tsunami of something of that sort.

In pretty much all cases, the bottom of the wave can be assumed to be the lowest trough. What you'll find in the end is that a lot of mass is in a wave.

  • $\begingroup$ Nit-pick: Waves don't have momentum because there is no mass transport (until they go non-linear and collapse). They have momentum flux. Here's a good article about it atmos-dynamics.damtp.cam.ac.uk/people/mem/papers/RECOIL/… $\endgroup$ – user6972 Apr 15 '14 at 21:19
  • $\begingroup$ @user6972 Thanks for an article reference, but I'm well aware of the idea of no mass transport. However, you tell the victims of a tsunami that waves transport no mass or have no momentum. See what they say $\endgroup$ – Jim Apr 16 '14 at 5:20
  • $\begingroup$ One more nit-pick, travelling waves transport energy. A wave that transports energy has a momentum that is proportional to the kinetic energy divided by the velocity of the wave $\endgroup$ – Jim Apr 16 '14 at 5:27
  • $\begingroup$ You're referring to "added mass" or "virtual mass" a renormalization technique used to estimate force. Force is usually broken into incident force, refraction force and radiation force. ocw.mit.edu/courses/mechanical-engineering/… $\endgroup$ – user6972 Apr 16 '14 at 20:18
  • $\begingroup$ @user6972 Exactly! That is why I think this is a justifiable answer; it is meant for estimating a force by providing a way of expressing the relevant mass of water. No one will argue that a transverse wave is not actually propagating mass, but I believe the OP is looking for a way to get the mass in order to express the magnitude of the potential force a wave could apply to something. Shock and awe type thing. For that purpose, virtual mass is effective $\endgroup$ – Jim Apr 16 '14 at 21:46

Under the assumptions of linear wave theory, there is no net transport of mass, but there is a net transport or energy and momentum. Because you are asking about sea waves, I believe these are appropriate assumptions for your question. For example, it would be the sort of analysis to perform to estimate the horizontal forces a wave train (e.g. tsunami) would impart on a body (a boat) floating in the water.

Some basic fluid mechanics and calculus is required to solve even simple cases, but you can find a fairly complete walk-through (for plane-waves at least) here: Wave Momentum Flux .


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