Pressure inside an infinite ocean?

Equations of state

The "ocean" should be described by an equation of state that relates thermodynamic variables (e.g. density $\rho$, temperature $T$, pressure $p$, etc.) to one another. Different equations of state are valid in different regimes; some are valid at high temperatures and pressure, while others are valid at low ones. I think it's not particularly worth going over them here, but the usual (approximate) one for liquid at reasonable temperatures and pressures involves just density: $$\rho =\text{constant}$$ that is, liquid water is essentially incompressible. By looking at Bernoulli's equation (which works for incompressible fluids), we can see that in the case of a static fluid in a vanishing gravitational field, the ratio of pressure to density is constant, and therefore the pressure must be constant, too. Note that when I say "constant", I mean that a quantity does not change much if you change other, related, thermodynamic quantities. Change the pressure of water slightly and you won't significantly change its density.

If we were to increase the ambient pressure inside the fluid, the assumption of incompressibility would be invalid and we would have to pick a different equation of state.

Pressure and pressure gradients

I think there are two quantities that may have been mixed up here: fluid pressure and the pressure gradient inside the fluid. I'm not aware of any realistic equation of state - certainly not one for liquid water - that involves vanishing pressure for a fluid with non-zero density. Pressure typically arises from interparticle interactions, and as the density is decidedly non-zero, there's no reason for that pressure to go away.

Now, what will go away is the pressure gradient $\overrightarrow{\mathrm{\nabla }}p$. In a fluid in hydrostatic equilibrium experiencing a gravitational acceleration $\overrightarrow{g}$, the pressure gradient is $$\overrightarrow{\mathrm{\nabla }}p=\rho \overrightarrow{g}$$ and in our case, as $\overrightarrow{g}=0$, so is the pressure gradient.

A solid object

Now, say we have a cubical box of side length $L$ embedded in this fluid. All sides of the box will experience a force of magnitude $F=p{L}^2$, where $p$ is pressure; the force on each side is pressure times the area of a side, which follows from thinking about pressure as force per unit area. However, because there's no pressure gradient, these forces oppose each other equally, and so there will be no net force on the box.

The box may be deformed by the pressure exerted on it on all sides, but it will not experience a net force in any direction.

Cosmology

Several users (Yakk, Acccumulation, and Joe Bloggs) have noted that this universe will likely not remain uniform, but will suffer one of two gravitational instabilities:

• It may collapse in on itself in a Big Crunch if the density is larger than the critical density, which is actually quite low. If the critical density is the same as in our universe, then this universe is far too dense, and will almost certainly collapse.
• Small perturbations, like the quantum fluctuations seen in the early universe, will eventually cause structure formation in much the same way that our own universe first formed inhomogeneities. The characteristic scale of these fluctuations is $${\lambda }_J=c_s\sqrt{\frac{\pi }{G\rho }}$$ where $G$ is the gravitational constant and $c_s$ is the speed of sound. In this universe, $c_s\approx 1481\text{ m/s}$, and so $\lambda \approx {10}^7\text{ m}$, which seems rather large; the characteristic growth time will be $\tau \approx {10}^3\text{ s}$.

posted almost 6 years ago

HDE 226868‭

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