# FeaturedB Black holes squishing Earth

1. Oct 2, 2018

### DaveC426913

This may be condensed matter physics topic, but I'm looking for a layperson answer.

Scares of the CERN accelerator creating a black hole that swallows Earth are in the news once again.
https://www.newsweek.com/earth-shrunk-tiny-hyperdense-sphere-particle-accelerators-1145940
From 10 years ago:
https://www.newscientist.com/article/dn13555-particle-smasher-not-a-threat-to-the-earth/

I know it is not possible for an accelerator to produce such a black hole in practice. But surely it's impossible even in principle.

Surely, a particle accelerator creating a tiny black hole that could grow to swallow the Earth would violate the law of conservation of energy.

Whatever object was created would only have as much energy in it as was supplied. I mean, you can't have a free lunch here.

Contrarily, an atom bomb brings atoms together that already have energy in the form of bonds that hold the heavy elements together; all the bomb is doing is releasing that energy.

Where would a tiny black hole get the energy to destroy the Earth? Is it a wholly exothermic phenomenon? i.e. the energy is already there in the atoms, and a particle accelerator is simply releasing it, allowing atoms to fall together and coalesce at the singularity?

The implication of that is that all mass exists in a state of instability, on one side of an "energy hill" - the hill preventing it from collapsing into a BH - and all we have to do is just crush it enough to release that inherent energy? That seems wrong.

Again, contrarily, the universe can make black holes easily enough because it is effectively an open system; there is always enough energy, and occasionally a whole bunch of it can end up in one place.

Is my thinking sound?

2. Oct 2, 2018

3. Oct 2, 2018

### DaveC426913

Thanks. But this is really an academic question on my part about the physics. Not concerned about why this is in the news.

4. Oct 2, 2018

### Staff: Mentor

Then it should be in Quantum Physics, shouldn't it?

... and the quoted article sheds some light on the unsuccessful search for strangelets.

5. Oct 2, 2018

### PeroK

A tiny black hole would have no more gravity than the particles that created it. Black holes, in general, have the same gravity as the stars that created them. It's a myth that they suck in everything through some sort of super-gravity.

What is different about a black hole is that if you fall into one, you do not stop by hitting the surface of a star. In the case of the star, you collide with its surface. In the case of the black hole, you continue to fall and experience greater gravity - but only after you get closer than was possible with the star.

6. Oct 2, 2018

### Staff: Mentor

A microscopic black hole would rapidly evaporate into Hawking Radiation. Its lifetime would be very short, and proportional to the mass. A black hole with the mass of two protons, would have a lifetime of ??? (nanoseconds???) I wager that others here on PF can give us the exact number.

7. Oct 2, 2018

### DaveC426913

Yep. This I know. I have a history (including diagrams!) of explaining this to curious forum members.

Yes.

I'm just trying to figure out whether it's safe to say it can't happen because there simply isn't an Earth-swallowing-black-hole's amount of energy available.

Perhaps another way to phrase the question is: how massive must a black hole be before it can result in a runaway reaction?

I guess if you had a micro BH and just kept feeding it matter, it would always be able to grow. Which means my premise is faulty.

8. Oct 2, 2018

### DaveC426913

I think I've got it now.

The energy required to crush the Earth is already present in the form of gravitational potential energy. Every atom that is not at the CoG wants to fall to the CoG. So, if a cavity is formed by the BH eating what's around it (granting the BH lives long enough), more matter will fall toward it, which will then make it available to be consumed by the BH.

i.e. assuming the BH does not immediately evaporate*, it will result in a runaway reaction, consuming the Earth. So, my premise is faulty.

*But that is not a good assumption.

9. Oct 3, 2018

### sophiecentaur

I think this is the basic answer to the question. There are many stars and planets that orbit black holes and which are in no danger of being 'sucked in'.

10. Oct 3, 2018

### DaveC426913

While true, that is a very different scenario from a BH on/in Earth.

Material around BH in space has plenty of lateral motion, keeping matter from falling directly toward the BH.

The matter of Earth starts off stationary wrt the BH; the first thing it's going to do, when it can, is fall straight down toward the BH. And all the rest of the Earth's matter is poised just above that, waiting to fall straight down too (actually, explode straight down, since it's under a huge amount of compression).

Essentially, a BH in the Earth doesn't need to have any gravity at all. The Earth's matter will come straight to it.

OK, I've corrupted the scenario slightly. I'm now describing a BH at the centre of the Earth, as opposed to one in a CERN lab near the surface.

Still, the point remains - a BH on Earth is embedded in matter that is stationary wrt to it - not in an orbit.

Last edited: Oct 3, 2018
11. Oct 3, 2018

### Staff: Mentor

Moderator's note: the thread has been reopened.

12. Oct 3, 2018

### Staff: Mentor

Yes, and this does make a significant difference.

The reason is that there is a theorem called Buchdahl's Theorem, which says, in effect, that no matter can be in hydrostatic equilibrium if it is closer to a black hole's horizon than 9/8 of its Schwarzschild radius. Any matter that is present within that radius must fall into the hole. (Note that we are assuming the matter does not contain things like rocket engines that can provide thrust in the absence of hydrostatic equilibrium; for the case under discussion this should be a good assumption. )

So if a black hole is embedded in matter, it is guaranteed to gain mass; and as it gains mass, the radius within which matter must fall into it grows, so the process is self-reinforcing and will continue as long as there is matter close enough to the hole.

For a hole of sufficiently small mass, the rate of mass gain from the above process should be smaller than the rate of mass loss via Hawking radiation. However, even then there are some possible caveats. First, it's not clear exactly how Hawking radiation is supposed to work for a hole embedded in matter; all of the theoretical work on Hawking radiation assumes a hole surrounded by vacuum. But let's suppose that the hole being embedded in matter doesn't significantly affect the rate of Hawking radiation. That still leaves a second point: the radiation can't escape to infinity, because the hole is surrounded by matter. What will happen is that the matter surrounding the hole will heat up (to roughly the Hawking radiation temperature). It's not clear that all of the energy radiated as Hawking radiation will actually end up escaping, instead of being trapped in the surrounding matter and ultimately falling back into the hole.

13. Oct 3, 2018

### Staff: Mentor

The response that @DaveC426913 gave to this is correct: while it's true, it's a very different scenario from the one he was proposing. (See my previous post just now for some more on his scenario.)

14. Oct 3, 2018

### Staff: Mentor

Even with the lab at the surface, there will still be plenty of matter within 9/8 of the Schwarzschild radius, so the hole will gain mass (assuming Hawking radiation is small enough not to counterbalance this--see post #12). The difference between the lab at the surface and the BH at the center is that, at the surface, the hole will start to move: matter can't stop it from moving because it just swallows matter in its path. At least as a first approximation, the hole will assume an elliptical orbit about the Earth's center with an apogee equal to the Earth's radius and a perigee that I haven't calculated but which will be determined by its tangential velocity at the lab at the surface (and which will be pretty far inside the Earth since that tangential velocity is going to be a lot less than orbital velocity at the surface). Whereas the BH at the center will just sit there as it accretes mass.

15. Oct 4, 2018

### LURCH

Ive heard it said elsewhere that there are collisions taking place on the Earth every day with energies greater than those being generated at the LHC. I believe this was a reference to cosmic rays. So, even if it is possible for Earth-devouring black holes to be generated in a collider, it is far less probable than the chance that one of these tiny terrors will simply happen on its own some day. The fact that this has not yet occurred points to the likelihood that it cannot. But, even if our planet’s continued existence is merely a matter of probability, that probability is hardly effected at all by the very few collisions happening in accelerators around the world.

Therefore,The experiments at CERN either
a) do not constitute any risk at all, or
b) do not appreciably add to a risk that exists independently of that facility.

(Maybe that explains Dark Matter!)

16. Oct 4, 2018

### sophiecentaur

The question remains as to how long such a process might take. It would presumably be a bit 'runaway' so it could well be over before we could have time to notice it and to worry about it. The matter in the vicinity of the BH would be molten so it would fall down with no mechanical support.

17. Oct 4, 2018

### jbriggs444

If you removed the earth and replaced it with a black hole of equivalent mass, you'd be talking about [very] roughly ten minutes of free fall time to arrive at the singularity. So that's one simple lower bound on how long the process would take. I would be expecting billions (trillions? More?) of years to smoosh the earth into a black hole that starts with a sub-atomic size.

Just because something is fluid, that does not remove all mechanical support. The water in your bathtub does not drain instantly.

18. Oct 4, 2018

### sophiecentaur

Sufficient time for the Human Race to go completely loopy and try for a Star Ship escape then?

19. Oct 4, 2018

### Staff: Mentor

The relevant formula is:

$T(M) = 5120 \frac{\pi G^2 M^3}{\hbar c^4}$

So for two protons the evaporation time will be about 3 x 10-96 seconds.

For one gram of matter, about 8 x 10-26 seconds. I wonder how that compares to the reaction rate of a typical nuclear explosion?

20. Oct 4, 2018

### Staff: Mentor

Thank you @gneill. I would like to elaborate on that because I find this whole scenario unrealistic.

The key is to focus on the initial mass of the BH. Before it reaches appreciable mass, it must begin with the collision of two hadrons. (Events with more than two particles collapsing simultaneously are far less likely.)

So two protons collide, form a BH, and evaporate in 3x10-96 seconds.

The energy released in an explosion of 2*1.67*10−27 kg*c2 = 3*10-10 joules.

The average power of the energy release is 1077 GW. I did not compute the power density, or the flux of escaping radiation.

Would the BH suck in more mass before complete evaporation? Well, the time is very short. The density of mass in the collider beam is very small. And the radiation pressure on incoming particles would either deflect them, or slow them down enough to miss the 3x10-96 second window. Given these extreme numbers, I expect that classical calculations totally fail. We would need QED and perhaps GR to calculate the behavior of nearby particles in that time window with those conditions. My guess is that the probability of absorbing even one more particle is very small.

Does that sound realistic?