# Does something like Pascal's Law apply in a Swimming Pool? If not, how does pressure spread?

I've been trying to wrap my head what happens when a force is exerted on the surface of a swimming pool / pond, or underneath it. Examples include a small boy cannon-balling into the deep end, or a fish flipping its tail under water. I assume the water is contained on all sides except the top.

Pascal's Law (as I understand it) suggests that if something like this were done to a closed system, the pressure of events like these would be equalized throughout the whole system instantaneously (or effectively instantaneously, for small enough systems), applying everywhere in the fluid immediately afterward. This may be wrong, but my understanding is that for a fish in a pipe full of water, closed on one end, and plugged with a piston-n-spring on the other, when it pushes with its tail it moves the full column of water between it and the piston to move; it effectively pushes on the piston at a distance, nearly instantly, moving forward as the piston moves back.

Now, what happens in the swimming pool? I assume the fish is not pushing a column of water to push off against a far wall behind it. What does the pressure gradient look like shortly after these events when they occur in a pool, or the water flow? Does the fish actually push water everywhere in the pool, or just in a tight locality around itself, or some combination? Does the cannonballer raise the pool water height a little bit everywhere all at once, the splash and wave a separate effect? I assume these are not the case, and that some sort of pressure gradient spreads out from the events, accelerating water and leading to flow; but I don't understand how fast this happens, or in what directions, or with what fall off. And at least somewhere in my head, it seems like pressure could move throughout the pool very quickly - limited by the speed of sound in water right?

Despite knowing little physics and calculus I'm hoping to program an approximating 2D simulation. Navier Stokes is destroying me; pictures of computed vector fields for similar examples - especially in time sequence - would be a holy grail for me at this point (all I can find is stuff related to Stokes Waves on Wikipedia, which assumes the system is already in a steady state). Otherwise something like hints towards specific computation at the level of a high school education would be appreciated. All answers welcome though, as I suspect even my basics may be wrong.

(For simplicity, assume my water is perfectly ideal, inviscid, irrotational, continuous fluid of even density and temperature, laminar flows, yadda yadda...)

Thanks.

• For the fish-in-a-tube, the piston would not move, because the water+fish is not very compressible. The water is heavy (just as heavy as the fish), and what the fish is doing is displacing itself forward by displacing the surrounding water backward. For the boy-jumping, you have a surface wave that propagates outward. – Mike Dunlavey Jul 22 '13 at 12:14
• I do not understand about fish-in-a-tube then. To displace fluid, fluid must move. Fluid movement means fluid velocity. Fluid velocity is induced by fluid pressure, specifically pressure gradient. But Pascal's Barrel says any pressure the fish creates is everywhere instantly; so there can be no pressure gradient, no flow, and thus no displacement. Where did I go wrong? – jdowdell Jul 22 '13 at 22:57
• I think there can be pressure gradient. The fish moves its tail, creating a pressure gradient moving water backward and the fish forward. Pascal's barrel does not say pressure is everywhere instantaneously the same. It is a fairly static system where you can approximate the pressure as being roughly the same throughout. That won't stop a fish from swimming around in the barrel. – Mike Dunlavey Jul 23 '13 at 2:09
• (If anyone was interested, as for the programming a simulation part, I found the magic phrases were "Lax-Wendroff shallow water", and/or "saint venant equations." If you go off after Navier Stokes or Euler's or 'modeling water' and your calculus isn't up to snuff you'll just drown in math and high level academic papers that are overkill for simple water models.) – jdowdell Jul 23 '13 at 5:45

If we assume continuity, infinite speed of sound, no viscosity, and laminar flow, then the key is in Bernoulli's equation. In general the pressure variation is very complicated, but to get some ideas how it works we can consider the pressure variation a very simple system that can be calculated analytically.
I came up with this simple system:

Let's say somewhere in the water far from the surfaces, suddenly a cavitation bubble of radius $R$ with pressure $P_0-\Delta P$ is produced. Where $P_0$ is the atmospheric pressure, let's assume that the scale of the system we are playing with is small enough that the pressure of undisturbed water anywhere is equal to the atmospheric pressure. Because of the pressure drop, water will start moving radially to fill the bubble as shown in the picture. If the flow is continue and laminar then we have the following flux relation

$v 4\pi r^2= constant$

$v=\frac{C}{r^2}$..........................................(1)

Now Bernoulli's equation gives

$P_0=P(r)+\frac{1}{2}\rho v^2=P(r)+\frac{1}{2}\rho \frac{C^2}{r^4}$

Because $v=0$ at a point far from the bubble. Boundary condition at $r=R$ gives

$P_0=(P_0-\Delta P)+\frac{1}{2}\rho \frac{C^2}{R^4}$

Eliminating $C$ and $\rho$ we get

$P(r)=P_0+\Delta P \frac{R^4}{r^4}$

As expected $r\rightarrow \infty$, $P(r)\rightarrow P_0$. Thus the pressure change fades away as we go farther from the source of disturbance.

In a more general case where the above assumptions still hold, the velocity profile of the flow equivalent to our eq.(1) is quite complicated. Eq. (1) can be replaced with a general flux equation which holds along a streamline

$vA=constant$..................................(2)

Where $A$ is the cross sectional area of a streamline portion. A typical flow's streamlines caused by a moving object look like this

We can view eq.(2) as an equation that holds in the moving object's frame. As we can see in the picture above, initially the streamlines are uniformly separated. Let's denote the cross sectional area of each slice of stream line portion as $A_0$, from here we know that if the cross sectional area of a stream line portion equals to $A_0$ then its pressure is unchanged or $P_0$. We can see that the streamlines near the object are denser, which means that their cross sectional area are smaller than $A_0$. Thus from eq.(2) we realize that the water is moving faster there, and Bernoulli's equation says that the pressure there is lower than $P_0$. As we move perpendicularly to the flow, away from the moving object the streamlines' cross sectional area get more and more similar to that of undisturbed ones and so does the pressure there. Therefore it can explain how the pressure disturbance decreases over distance from the source.