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?

  • $\begingroup$ 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.) $\endgroup$ Jun 17, 2016 at 5:06
  • $\begingroup$ 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 $\endgroup$ Jun 17, 2016 at 17:51
  • $\begingroup$ @BenSandeen: without attractive forces water will evaporate and attain pressure-equilibrium with the surrounding gas. No water waves anymore. $\endgroup$ Jun 18, 2016 at 22:22
  • $\begingroup$ @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? $\endgroup$ Jun 20, 2016 at 9:28
  • $\begingroup$ @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. $\endgroup$ Jun 20, 2016 at 17:03

1 Answer 1


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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