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  • $\begingroup$ 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? $\endgroup$
    – babou
    Commented Jul 20, 2013 at 22:57
  • $\begingroup$ 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. $\endgroup$
    – Xurtio
    Commented Jul 21, 2013 at 18:39
  • $\begingroup$ 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. $\endgroup$
    – babou
    Commented Jul 21, 2013 at 20:51
  • $\begingroup$ I mentioned only spatial decay, not temporal decay. $\endgroup$
    – Xurtio
    Commented Jul 21, 2013 at 22:42
  • $\begingroup$ 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 ? $\endgroup$
    – babou
    Commented Aug 31, 2013 at 20:31