Timeline for Would a sea level rise (on the other side of the Earth) if you would dip your finger into the ocean?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2013 at 21:07 | comment | added | Xurtio | Pascal's law assumes the fluid is incompressible. It also requires the fluid be confined. | |
| Aug 31, 2013 at 20:31 | comment | added | babou | I am trying to understand the relation between the spatial decay of compression and Pascal's law in steady state. Can you say a bit more on how this decay is evaluated ? | |
| Jul 25, 2013 at 10:12 | vote | accept | sabiland | ||
| Jul 21, 2013 at 22:42 | comment | added | Xurtio | I mentioned only spatial decay, not temporal decay. | |
| Jul 21, 2013 at 20:51 | comment | added | babou | I see... I was wondering because you stated that you "assumed the question was about the steady state solution". But Navier-Stokes is about fluid dynamics. So I did not see how there could be a decaying compression. | |
| Jul 21, 2013 at 18:39 | comment | added | Xurtio | A formal treatment would rely on Navier Stokes and volume viscocity, but just thinking of the bulk modulus for simplicity as you draw large concentric rings around the point where you put your finger in. The expression (dV/V) dV is change in volume, V is total volume. As you consider larger and larger concentric circles, dV stays the same, but V is getting larger, so the ratio dV/V is getting smaller, thus dP (change in pressure) is getting smaller as you go further away from the perturbation point. | |
| Jul 20, 2013 at 22:57 | comment | added | babou | Can you give some more details to explain how "the region of compression would decay as you moved further away ..." ? I am not sure I understand how this decay operates. Is that the compression created by the finger doing work against the water? | |
| Jul 18, 2013 at 10:52 | history | answered | Xurtio | CC BY-SA 3.0 |