Difference between revisions of "Talk:Linear Theory of Ocean Surface Waves"
(3 intermediate revisions by 2 users not shown) | |||
Line 6: | Line 6: | ||
This is clearly wrong, considering that the ocean can contain standing waves (i.e. two waves with the same amplitude and wavelength moving in the opposite directions). There will be moments when the displacement for a standing wave is zero everywhere and the formula would evaluate to zero. However, the waves still contain energy because of the kinetic energy that now is because of the motion in the water. The formula does indeed calculate an energy, but it is only the potential energy of the waves and does not contain the kinetic energy. --[[User:Kri|Kri]] 22:11, 15 October 2010 (UTC) | This is clearly wrong, considering that the ocean can contain standing waves (i.e. two waves with the same amplitude and wavelength moving in the opposite directions). There will be moments when the displacement for a standing wave is zero everywhere and the formula would evaluate to zero. However, the waves still contain energy because of the kinetic energy that now is because of the motion in the water. The formula does indeed calculate an energy, but it is only the potential energy of the waves and does not contain the kinetic energy. --[[User:Kri|Kri]] 22:11, 15 October 2010 (UTC) | ||
+ | |||
+ | Your point is not correct because for linear waves there is a change from potential to kinetic energy and there is a theory that states that these are equal over one period. The averaging is with respect to time. However, the point is very subtle because we have assumed some kind of stationarity. I have tried to include this in the explanation. | ||
+ | |||
+ | written by Meylan 22:02, 17 October 2010 (UTC) | ||
+ | |||
+ | :That's exactly what I wrote but in other words, there is a change from potential to kinetic energy, but the formula does ''not'' include the kinetic energy, at least not if the angle brackets only denotes an average over a large surface (it doesn't really say over what the average is taken). | ||
+ | |||
+ | :You say that they are equal over one period. What is it that you mean are equal? The time integral of the two forms of energies over one period? And how long is one period when you have a superposition of multiple waves with different wavelengths? It is written in the article that the average kinetic and potential energies (I guess over a large spatial distribution and over a long period of time) have to be equal, but this is true even if the state is composed of multiple wavelengths. | ||
+ | |||
+ | :I can agree with the equation if the angle brackets denotes the average both over a large surface and over a long period of time, since the expression for a stationary wave has nodal points (where the displacement is zero) both in space and in time, but I think it should be more clear in the article what the brackets denote (I'm still not sure). --[[User:Kri|Kri]] 15:48, 11 December 2010 (UTC) | ||
+ | |||
+ | ::Ok, whit a second look I saw that the brackets denoted a time average, but I will do some additional clarifications in the article. --[[User:Kri|Kri]] 15:56, 11 December 2010 (UTC) | ||
+ | |||
+ | Your questions are interesting and I appreciate them. |
Latest revision as of 01:01, 14 December 2010
Equation for water energy incorrect
Hi, one thing that is stated in the article is that the energy of ocean waves in Joule per square meter is given by the equation
[math]\displaystyle{ E = p_w \ g \lt \zeta^2 \gt }[/math]
This is clearly wrong, considering that the ocean can contain standing waves (i.e. two waves with the same amplitude and wavelength moving in the opposite directions). There will be moments when the displacement for a standing wave is zero everywhere and the formula would evaluate to zero. However, the waves still contain energy because of the kinetic energy that now is because of the motion in the water. The formula does indeed calculate an energy, but it is only the potential energy of the waves and does not contain the kinetic energy. --Kri 22:11, 15 October 2010 (UTC)
Your point is not correct because for linear waves there is a change from potential to kinetic energy and there is a theory that states that these are equal over one period. The averaging is with respect to time. However, the point is very subtle because we have assumed some kind of stationarity. I have tried to include this in the explanation.
written by Meylan 22:02, 17 October 2010 (UTC)
- That's exactly what I wrote but in other words, there is a change from potential to kinetic energy, but the formula does not include the kinetic energy, at least not if the angle brackets only denotes an average over a large surface (it doesn't really say over what the average is taken).
- You say that they are equal over one period. What is it that you mean are equal? The time integral of the two forms of energies over one period? And how long is one period when you have a superposition of multiple waves with different wavelengths? It is written in the article that the average kinetic and potential energies (I guess over a large spatial distribution and over a long period of time) have to be equal, but this is true even if the state is composed of multiple wavelengths.
- I can agree with the equation if the angle brackets denotes the average both over a large surface and over a long period of time, since the expression for a stationary wave has nodal points (where the displacement is zero) both in space and in time, but I think it should be more clear in the article what the brackets denote (I'm still not sure). --Kri 15:48, 11 December 2010 (UTC)
- Ok, whit a second look I saw that the brackets denoted a time average, but I will do some additional clarifications in the article. --Kri 15:56, 11 December 2010 (UTC)
Your questions are interesting and I appreciate them.