<p>There are different factors that make water a media for waves. For transversal waves the viscosity and cohesion of molecules are essential. Even if the water is assumed not to evaporate. Try to imagine pushing a boat forward in a liquid with low viscosity. The propeller will just easily skid without generating any wave or traction.<br>
However for P waves a liquid with low or no viscosity would still work as long as it can contract and expand under stress. If you in a crude way look at a differential cube of that liquid omega= (k/m)^1/2. K is the water stiffness index and m is its mass.</p>

CC BY-SA 3.0
2016-06-20T21:58:52.837Z
2016-06-20T21:58:52.837Z
kamran
<p>Water molecules, as many people know, are polarized, and so water molecules tend to have an attractive force between them.  But how would the waves in water behave if this attraction no longer was present, but everything else was the same?</p>

CC BY-SA 3.0
2016-06-17T05:06:33.660Z
How big are the waves you are talking about? Are you strictly talking about capillary waves? Anything bigger than that will not be affected in any measurable way as other forces are larger and take precedence. (i.e. the restoring force is gravity.)
Isopycnal Oscillation
CC BY-SA 3.0
2016-06-17T17:51:43.847Z
I know next to nothing about waves in fluids, so I guess I'm asking about whatever type of wave allows the most striking difference. :P
Ben Sandeen
CC BY-SA 3.0
2016-06-18T22:22:27.680Z
@BenSandeen: without attractive forces water will evaporate and attain pressure-equilibrium with the surrounding gas. No water waves anymore.
AtmosphericPrisonEscape
CC BY-SA 3.0
2016-06-20T09:28:59.767Z
@AtmosphericPrisonEscape Interesting!  As for why that would happen; would it be just due to the fact that since the water molecules would have no affinity for each other, they are just as "happy" to be adjacent to each as they are to be dispersed throughout the air?
Ben Sandeen
CC BY-SA 3.0
2016-06-20T17:03:42.443Z
@BenSandeen: It's their mutually attractive forces that distinguish them on the thermodynamic level from the upper lying gas. If you take those away, the water molecules are just balls with velocity bouncing-off each other. Then they just bounce around like the upper lying gas. This gas still has internal density waves of course, but no surface and therefore no surface waves.
AtmosphericPrisonEscape
CC BY-SA 3.0
2016-06-20T18:44:49.773Z
@AtmosphericPrisonEscape: That makes sense.  But could my question be plausibly considered purely hypothetically, assuming that the water doesn't simply evaporate into the air but otherwise behaves like there's no attractive force?  I.e.: What happens when one considers only the mechanics of the waves?
Ben Sandeen
<ocean><water><waves><hypothetical><wave-modeling>
How would waves (in a fluid) behave without intermolecular attraction?
Ben Sandeen
# How would waves (in a fluid) behave without intermolecular attraction?

(Asked by: username_0 on 2016-06-17)

Water molecules, as many people know, are polarized, and so water molecules tend to have an attractive force between them. But how would the waves in water behave if this attraction no longer was present, but everything else was the same?

**Tags:** ocean, water, waves, hypothetical, wave-modeling

> **Comments:**
> **username_1**: How big are the waves you are talking about? Are you strictly talking about capillary waves? Anything bigger than that will not be affected in any measurable way as other forces are larger and take precedence. (i.e. the restoring force is gravity.)  
> **username_0**: I know next to nothing about waves in fluids, so I guess I'm asking about whatever type of wave allows the most striking difference. :P  
> **username_2**: @username_0: without attractive forces water will evaporate and attain pressure-equilibrium with the surrounding gas. No water waves anymore.  
> **username_0**: @username_2 Interesting!  As for why that would happen; would it be just due to the fact that since the water molecules would have no affinity for each other, they are just as "happy" to be adjacent to each as they are to be dispersed throughout the air?  
> **username_2**: @username_0: It's their mutually attractive forces that distinguish them on the thermodynamic level from the upper lying gas. If you take those away, the water molecules are just balls with velocity bouncing-off each other. Then they just bounce around like the upper lying gas. This gas still has internal density waves of course, but no surface and therefore no surface waves.  
> **username_0**: @username_2: That makes sense.  But could my question be plausibly considered purely hypothetically, assuming that the water doesn't simply evaporate into the air but otherwise behaves like there's no attractive force?  I.e.: What happens when one considers only the mechanics of the waves?  

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## 1 answers

### Answer by username_3 on 2016-06-20. Score: 1 | Accepted answer

There are different factors that make water a media for waves. For transversal waves the viscosity and cohesion of molecules are essential. Even if the water is assumed not to evaporate. Try to imagine pushing a boat forward in a liquid with low viscosity. The propeller will just easily skid without generating any wave or traction.  
However for P waves a liquid with low or no viscosity would still work as long as it can contract and expand under stress. If you in a crude way look at a differential cube of that liquid omega= (k/m)^1/2. K is the water stiffness index and m is its mass.


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