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Imagine a universe full of water. A comfortable 22 degrees Celsius warm, density of about 998 grams per liter, and it fills up everything for as far as can be observed. There's nothing else; neither empty pockets nor denser planets. Just water.

This would be an extremely massive universe, but if I understand the physics correctly there would be no spontaneous black hole formation. The density and temperature would be totally uniform, and so would the gravity: every H2O molecule would be pulled in every direction at once, with the resultant force being zero. No movement in the water, no build-up of mass, so no black holes.

What if some people from a different dimension pay a visit though... They venture through the Phlebotinum portal, find themselves in their spacecraft-turned-submersible in this weird universe, and be adding their denser-than-water selves to the mix. Suddenly water is pulled slightly in their direction, increasing local pressure (and local density, and thus local mass) even further... cascade into a black hole begins!

Or would it? Water doesn't like being compressed, so there is a high amount of force required to make water locally denser. That means there's a hurdle to overcome before the gravitationally compressed water becomes sufficiently dense (compared to standard water) to have sufficient gravity of its own to continue compression, and eventually collapse into a black hole.

I think it may just hold up against the addition of a single spacecraft, which would have minute gravity of its own. But I cannot tell for sure.

Can you quantify what local density variation would still be allowable in a water-filled universe, without collapsing the lot into a black hole? Could, for example, this universe have an Earth-sized rocky planet in it? Or would the addition of one grain of sand be enough to begin a cascade?

Assume all of our known laws of physics apply, except this universe is not expanding or contracting. I am primarily interested in the short-term effects of a new mass added to the water universe, not whether this universe would suffer a big crunch eons into the future.

How this strange universe came to be is out of the scope of the question :-) (short version: simulated universe)

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – L.Dutch
    Nov 8 '21 at 11:53
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Perturbations in a medium resulting in the formation of black holes have been studied extensively in the context of primordial black holes, though to result from density perturbations in the early universe. There are plenty of analytical and numerical studies of the required amplitude of such perturbations $\delta_c=\delta\rho/\rho$ (e.g. Harada et al. 2016). Unfortunately, these largely focus on perfect fluids (with equations of state of the form $p=w\rho c^2$, with $p$ pressure, $\rho$ density and $w$ dimensionless) and the radiation-dominated era of the universe. Water isn't a perfect fluid and this universe isn't radiation-dominated (!), so unfortunately we can't invoke those.

It's been argued (Carr 1975) that perturbations leading to the collapse of a region and the formation of a primordial black hole would need to be on the order of the Jeans length, a quantity more commonly used when studying the collapse of molecular clouds into stars. The Jeans length is $$\lambda_J=\left(\frac{15k_BT}{4\pi Gm\rho}\right)^{1/2}$$ with $T$ and $\rho$ the temperature and density of the medium and $m$ the mass of its constituent particles - in this case, water molecules. Plugging your conditions into this yields a length of $\sim$1600 km - large by the standards of spacecraft but small by comparison to the typical lengths of molecular clouds and the early-universe perturbations that form primordial black holes.

(As a side note: There are two ways to think about the Jeans length based on two different derivations, which agree to within a factor of a few. One equates thermal and gravitational potential energy and says that beyond $\lambda_J$, gravity wins over thermal pressure. The other calculates the collapse time and then derives the distance over which a wave could propagate across the region of interest and back within that time to stabilize the mass. I prefer the latter, you can think about the criterion with either interpretation.)

From this argument - which I believe is applicable to your scenario - this spacecraft would not cause a black hole to form; its length is much less than $\lambda_J$. I would expect some black holes to form anyway due to natural random (possibly Gaussian-distributed) fluctuations in density, but this particular perturbation doesn't seem large enough to be problematic.

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  • $\begingroup$ So, a molecular cloud of some constant density & temperature has a critical radius where, above it, gravity always wins out--the 'Jean's length'? How does this relate to the length of spacecraft entering this watery universe? $\endgroup$
    – BMF
    Nov 13 '21 at 21:35
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Your space ship is limited to an inch long, unless you increase the cosmological constant a tad.

The critical density of the universe is 9.9E-30 g/mL. The density of your liquid water universe is (before any subsequent cosmology occurs) 1 g/mL. That means that your rho/rhoc is about 1E+29. Put into this StackExchange problem and that means k=1E+58 (H/c)^2, where I'm going to go out on a limb here and assume that question intends H to be the Hubble constant, 1/(4.55E17 s). H/c is about 7E-27 m, so this tots up to 50,000 / m^2 . Now not having had any coursework in this physics (sorry, should have mentioned that before) I'm not entirely how to interpret inverse length squared as a curvature, but taking a wild guess, ... I should listen to Logan, whose keyword Gaussian curvature is most helpful. The radius of a sphere (actually a hypersphere here) should simply be the inverse square root of the 5/cm^2 above, or 0.45 cm. The circumference of a circular cross-section is 0.45 cm * 2 * pi = 2.8 cm = just over 1 inch. The space ship had better be smaller than that, or it is going to have trouble parking. (Have you ever tried to convince an insurance adjuster you rear-ended yourself?)

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    $\begingroup$ The radius of a sphere is the square root of the reciprocal of its Gaussian curvature. So, no need to involve pi here. $\endgroup$ Nov 8 '21 at 23:25
  • $\begingroup$ It took me three times reading the answer to sorta understand it, but then I got the joke as well and I really appreciate it :D This is a very interesting property of or problem with my water universe. So universal topology is definitely something I need to look into, perhaps by tweaking the cosmological constant as you suggest. But the focus problem was black hole collapse. $\endgroup$
    – KeizerHarm
    Nov 8 '21 at 23:27
  • $\begingroup$ There is a certain degree to blur the boundary between a closed universe and a black hole, though I think they are not generally taken to be the same thing. See math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… . Again, I haven't actually taken a course in this. Still, it seemed close enough to write up here. $\endgroup$ Nov 9 '21 at 2:25
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We can work out the pressure on the space craft from the water.

(I'm assuming a static universe that is not expending)

Now for a small change in radius, $dr$, the resulting change in pressure, $dp$, (assuming its inside an incompressible material of density $\rho$) is

$$dp=\rho g(r)dr$$

Now $g(r)$ is the strength of the gravitational field at the radius $r$.

Now from the shell theorem the gravitation field, for an enclosed mass, $M_{enc}$ is

$$ g(r)=\frac{GM_{enc}}{r^2}=\frac{G(M_{space ship}+M_{water})}{r^2}$$

Now the approximate (accurate in the case of large $r$) mass of water is give by

$$M_{water}=\rho \frac{4}{3}\pi r^3$$

So the pressure change is

$$ dp=\rho \frac{G(M_{space ship}+\rho \frac{4}{3}\pi r^3)}{r^2}dr$$

which simplifies to $$ dp=\rho \left(\frac{GM_{space ship}}{r^2}+G\rho \frac{4}{3}\pi r\right)dr$$ from this we can see the problem as $r$ gets larger the increase in pressure grows. So for an infinitely large sphere of water the pressure would be infinite.

So I don't think your ship\universe would survive.

The collapse wouldn't happen before the ship arrived, as there would be no variances in the gravitational field.

hopefully that helps

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  • $\begingroup$ Exercise for you: compute the variation of $g$ inside an isolated spherical planet of a material of uniform density $\rho$. Then think that, for an infinite universe, any point is a good approximation of its center. $\endgroup$ Nov 7 '21 at 0:19
  • $\begingroup$ @AdrianColomitchi, in that case $g=G\rho \pi \frac{4}{3} r$ $\endgroup$
    – Nyra
    Nov 7 '21 at 0:29
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    $\begingroup$ " with a space ship in the universe it is no longer uniform" But the water that fills the universe still is uniform. Gravitational interaction is always additive, which meant the effect of two causes is the sum of the effects of each cause. You don't get to dismiss the $g$ as zero in the absence of the ship, but consider it (as following a linear law) when you introduce a grain-of-sand-sized-ship in the medium. The additivity is what guarantees the shell theorem, don't throw it away or your integrals will fail. $\endgroup$ Nov 7 '21 at 3:28
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    $\begingroup$ (3) At the end of the day the big issue is that Poisson's equation isn't guaranteed unique solutions when the mass distribution/gravitational potential doesn't decay to 0 quickly enough. So finding the gravitational field of a universe with uniform mass density within Newtonian gravity is an ill posed problem and saying that the answer is zero gravitational field is wrong which can be easily seen-- $\nabla \cdot \mathbf{0} = 0 \neq \rho_{water}$ $\endgroup$ Nov 7 '21 at 21:40
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    $\begingroup$ @AdrianColomitchi because the water has temperature, there are constant density fluctuations. I think to answer the question, one needs to find a tipping point where a mass of higher-density water becomes gravitationally self-sufficient against the overwhelming equalizing forces (gravitational + surrounding temperature differentials). $\endgroup$
    – BMF
    Nov 7 '21 at 23:18

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