Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

What is the dominant cause for ocean waves at a beach? Are they the result of wind/pressure difference? If so, the waves do seem to exist in similar intensity even during relative quiet times of the day.

Is there a simple mathematical model that we can quickly explain the intensity/frequency of waves with? Does the strength of the waves (say the variance and mean of the amplitude of waves) relate to a simple physical quantity (temperature, off shore wind, pressure difference)?

share|cite|improve this question

2 Answers 2

up vote 3 down vote accepted

Yes, primarily wind. It's called the Kelvin-Helmholtz instability. Strong winds in an area will excite a range of wavelengths, the longer wavelengths will go faster according to the deep water dispersion relation ( speed proportional to square root of wavelength). So if you see a train of waves with decreasing wavelength over time, you could in principle infer a common point of origin (this is a common textbook or qualifying exam type problem).

share|cite|improve this answer
Thankyou for pointing me in the right direction. – New Horizon Jun 2 '11 at 18:07

The normal dominant cause of most waves you see (i.e., not a Tsunami) is frictional force between the wind and water over very large distances.

Even though the wind may not be blowing at your particular beach, it is ALWAYS blowing somewhere out in the ocean. The waves produced by this effect can (and do) travel thousands of miles away from the source.

share|cite|improve this answer
As stated they can come from thousands of kilometers away, and they travel time can be several days. There are complicated wave forcasts that are coupled to weather models. There are some complications not covered by the more simple theories, such as rogue waves, which theory says should be exceedingly rare, but aren't so rare after all. – Omega Centauri Jun 1 '11 at 1:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.