B Could a Particle Accelerator Create a Black Hole That Destroys Earth?

[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?

 

[fresh_42]

I'm looking for a layperson answer.

A layman's opinion:
Rees needs publicity to sell his book and others take the chance to fill their magazines.
https://www.sciencealert.com/cosmol...30-ft-by-particle-accelerator-experiments-lhc

A bit sad in my opinion, but he's not the first and won't be the last who sells his soul for profit.

 

[DaveC426913]

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

 

[fresh_42]

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

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

 

[PeroK]

Where would a tiny black hole get the energy to destroy the Earth?

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.

 

[anorlunda]

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.

 

[DaveC426913]

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.

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

A microscopic black hole would rapidly evaporate into Hawking Radiation.

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.

 

[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.

 

[sophiecentaur]

A tiny black hole would have no more gravity than the particles that created it.

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'.

 

[DaveC426913]

There are many stars and planets that orbit black holes and which are in no danger of being 'sucked in'.

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.

 

[PeterDonis]

Moderator's note: the thread has been reopened.

 

[PeterDonis]

a BH on Earth is embedded in matter that is stationary wrt to it

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.

 

[PeterDonis]

There are many stars and planets that orbit black holes and which are in no danger of being 'sucked in'.

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.)

 

[PeterDonis]

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

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.

 

[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!)

 

[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.

 

[jbriggs444]

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.

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.

 

[sophiecentaur]

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.

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

 

[gneill]

A black hole with the mass of two protons, would have a lifetime of ?

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?

 

[anorlunda]

The relevant formula is:

T(M)=5120πG2M3ℏc4T(M)=5120πG2M3ℏc4T(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.

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?

 

[jbriggs444]

Sufficient time for the Human Race to go completely loopy

Don't have to wait for that.

 

[gneill]

Does that sound realistic?

Looks reasonable to me.

 

[DaveC426913]

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.

But the centre of theEarth is under pressure. A bathtub at the centre of the Earth would "drain" explosively - in microseconds -cwith such a pressure differential.

Why would it take billions/trillions of years? (Presumably, the vast majority of that time would be spent at subatomic size, with the last macro-scale gobbling happening in just moments.)
With matter crushing down on it at millions of atmospheres, why would it take so long to grow?

 

[anorlunda]

Looks reasonable to me.

LOL I meant does it sound like a realistic scenario for squishing the Earth?

 

[jbriggs444]

But the centre of theEarth is under pressure. A bathtub at the centre of the Earth would "drain" explosively - in microseconds -cwith such a pressure differential.

An ordinary bathtub with a one inch radius spigot would drain quickly under such a pressure differential, certainly. But an Earth-sized bathtub draining through a sub-atomic orifice? Let's see that calculation.

We already know that we have a 10 minute lower bound.

Edit: @PeterDonis has more precisely quoted this particular lower bound at approximately 20 minutes.

 

[PeterDonis]

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

The relevant energy is not the rest energy of two protons, but the energy in the lab frame of the collision. For the LHC, that's now about 13 TeV IIRC. That's about 13,000 proton masses, which increases the time by a factor of about $10^{12}$. That doesn't change the qualitative conclusion, but I think it's worth noting the correct numbers.

 

[gneill]

LOL I meant does it sound like a realistic scenario for squishing the Earth?

It sounds like a reasonable argument against creating self-sustaining microscopic black holes.

 

[PeterDonis]

My guess is that the probability of absorbing even one more particle is very small.

I would agree; for a hole of the size that could hypothetically be produced by the LHC, the evaporation time is so short that it should dominate anything else in the dynamics.

 

[PeterDonis]

the centre of theEarth is under pressure. A bathtub at the centre of the Earth would "drain" explosively - in microseconds -cwith such a pressure differential.

No, it wouldn't, because the matter starts out at rest and it will take time for it to cover the distance to the center. Roughly speaking, if we assume that the matter has zero viscosity for this purpose (since it's all going down the hole at the center so matter just crossing the horizon won't "push back" against matter behind it), the time for the matter at the Earth's surface to reach the center and get swallowed by the hole should be about 20 minutes--one fourth of the free-fall orbit time.

 

[fresh_42]

I would agree; for a hole of the size that could hypothetically be produced by the LHC, the evaporation time is so short that it should dominate anything else in the dynamics.

But aren't the products of a scattering process still very fast, say near $c$ such that there would be enough time to hit the wall?

 

[PeterDonis]

aren't the products of a scattering process still very fast, say near $c$ such that there would be enough time to hit the wall?

If we assume a collision of equal energy particles moving in opposite directions (which AFAIK is the normal setup in an experiment like the LHC), a black hole that was produced could have zero momentum. That doesn't happen with normal collision products because the energy of the products is so much larger than their rest energy that they have to be moving very fast. But a collision that produced a black hole could have all of the collision energy converted to rest energy of the hole.

Also, even if the hole were moving at close to $c$, its evaporation time is so short that it wouldn't be able to move very far--only about $10^{-75}$ meters, based on the numbers in post #26.

 

[gneill]

I would agree; for a hole of the size that could hypothetically be produced by the LHC, the evaporation time is so short that it should dominate anything else in the dynamics.

Not to mention that the Schwarzschild radius of such a black hole would be on the order of 10-121 nm, far smaller than the size of a typical atom. Presumably the BH could pass through a given atom without eating anything much as a comet can pass through our solar system without hitting a planet.

 

[PeterDonis]

Not to mention that the Schwarzschild radius of such a black hole would be on the order of 10-121 nm

I don't think it's quite that small. If $M$ is 10,000 proton masses, or about $10^{-23}$ kg, then $2GM / c^2$ gives $2 \times 6.67 \times 10^{-11} \times 10^{-23} / 9 \times 10^{16}$, or about $10^{-50}$ m, which is $10^{-41}$ nm. Still very, very small compared to the sizes of atoms or even nuclei, of course.

Another wrinkle to consider is that a black hole of mass that small--well under the Planck mass--might not even be possible, depending on how quantum gravity turns out.

 

[gneill]

I don't think it's quite that small. If $M$ is 10,000 proton masses, or about $10^{-23}$ kg, then $2GM / c^2$ gives $2 \times 6.67 \times 10^{-11} \times 10^{-23} / 9 \times 10^{16}$, or about $10^{-50}$ m, which is $10^{-41}$ nm. Still very, very small compared to the sizes of atoms or even nuclei, of course.

Aurgh. Thanks for catching that. I'd just done the calculation for the apparent volume enclosed by the Schwarzschild radius and picked up that number thinking it was the radius calculation I'd done next to it. Very sloppy on my part.

Another wrinkle to consider is that a black hole of mass that small--well under the Planck mass--might not even be possible, depending on how quantum gravity turns out.

That's an interesting thought; Nature protecting itself from random micro black holes gobbling up all the matter.

 

[fresh_42]

That's an interesting thought; Nature protecting itself from random micro black holes gobbling up all the matter

Or to add another fantasy: It happened so often before until the right setup has been left.

 

[DaveC426913]

No, it wouldn't, because the matter starts out at rest and it will take time for it to cover the distance to the center. Roughly speaking, if we assume that the matter has zero viscosity for this purpose (since it's all going down the hole at the center so matter just crossing the horizon won't "push back" against matter behind it), the time for the matter at the Earth's surface to reach the center and get swallowed by the hole should be about 20 minutes--one fourth of the free-fall orbit time.

Mayhap, but that only applies once the BH is macro-scale. For most of the BH' life, the distance to atoms is on the atomic scale. i.e. the fall time would be in picoseconds.

 

[jbriggs444]

Mayhap, but that only applies once the BH is macro-scale. For most of the BH' life, the distance to atoms is on the atomic scale. i.e. the fall time would be in picoseconds.

When the black hole is micro-scale, it consumes even less.

 

[PeterDonis]

For most of the BH' life, the distance to atoms is on the atomic scale. i.e. the fall time would be in picoseconds.

I don't think this is true for a hole that is actually going to grow (i.e., it won't evaporate before it accretes enough mass to push the evaporation time far out into the future), at least not within any kind of ordinary matter like that of the Earth. I can't resist analyzing this in more detail, so here goes.

First, let's write down a general formula for the fall time to a hole of mass $M$ from a distance $R$. We can use the Newtonian equations for a point source of gravity for this because, as we'll see, the Schwarzschild radius of the holes we'll be discussing is so much smaller than the distances we'll be calculating that we can ignore relativistic corrections, since they will only be significant for a tiny fraction of the fall time. The Newtonian free fall time to a point mass $M$ from a starting radius $R$ is

$$
t = \frac{\pi}{2} \frac{R^{\frac{3}{2}}}{\sqrt{2GM}} = 1.36 \times 10^5 \sqrt{\frac{R^3}{M}}
$$

where I have plugged in all of the numerical factors, in SI units (so $R$ is in meters and $M$ is in kilograms, and $t$ comes out in seconds).

Next, the formula for the Hawking evaporation time of a black hole of mass $M$ (as has already been posted, I'm posting it again for convenience) is

$$
T = 5120 \frac{\pi G^2 M^3}{\hbar c^4} = 8.4 \times 10^{-17} M^3
$$

where again I have plugged in all of the numerical factors in SI units (so again $M$ is in kilograms, and $T$ comes out in seconds).

Evaluating the inequality $t < T$ gives us a first cut at a relationship between $M$ and $R$, i.e., within what radius $R$ surrounding matter would have to be at to be able to fall into a hole of mass $M$ before it evaporates. This relationship is

$$
R < 7.25 \times 10^{-15} M^{\frac{7}{3}}
$$

So, for example, a 1 kilogram hole would have to have surrounding matter within $7.25 \times 10^{-15}$ meters to have any of it fall in before the hole evaporates. Since that distance is roughly the size of an atomic nucleus, such a hole would evaporate well before any adjacent atoms had time to fall in; only if the hole happened to be created inside an atomic nucleus (or inside an object with density comparable to that of an atomic nucleus, such as a neutron star) would anything have a chance to fall in prior to evaporation.

To get $R$ to roughly the size of an atom, i.e., roughly $10^{-10}$ meters, requires a hole of about $59.4$ kg. So a hole with mass less than that would not have a chance of pulling in adjacent atoms before evaporating.

For a hole to have a significant chance of accreting enough matter to keep itself from evaporating, it probably needs to have a mass within the radius $R$ that is comparable to its own mass. If we assume Euclidean spatial geometry (which is a good approximation for a planet like the Earth or an ordinary star like the Sun, but might not be for something like a neutron star), then the mass inside radius $R$ is $\frac{4}{3} \pi R^3 \rho$; so if we want that mass to be the same as the mass $M$ of the hole, we plug the formula for $R$ in terms of $M$ above (using equality instead of less than) into the Euclidean formula, which, after some algebra, gives

$$
M = 9.25 \times 10^6 \frac{1}{\rho^{\frac{1}{6}}}
$$

So for a density of $1$ (the density of air, roughly), $M$ would be $9.25 \times 10^6$ kg, or 9250 metric tons. The $R$ corresponding to this is about 130 meters. For a density equal to the average density of Earth, about $5500$, $M$ reduces to about $2.2 \times 10^6$ kg, or 2200 metric tons. The $R$ corresponding to this is about $4.6$ meters. And the fall time $t$ for this $R$ and $M$ is about $5.9$ seconds. (Note that this hole has a Schwarzschild radius of about $3.3 \times 10^{-21}$ meters, still much smaller than an atomic nucleus.)

So if we imagine a black hole of mass 2200 metric tons that somehow got created inside a ball of Earthlike-density matter with the same mass, and radius 4.6 meters, it would take almost 6 seconds for the hole to swallow all of the mass; and for most of that time, the distances that matter was falling to reach the hole would not be microscopic. Even more so, if we imagine such a hole created at the center of the Earth, it would take 6 seconds for it to swallow all of the matter in a sphere 4.6 meters in radius around it; but it would take about 20 minutes (per my previous post) to swallow the entire Earth, and for almost all of that time, the distances would not be microscopic. (Note that we can't just plug the Earth's radius and 2200 metric tons for $M$ into the formula for $t$ above; we would need to figure out a formula for $t$ if $M$ is changing with time as the hole accretes matter. That's more than I have the time or patience for right now.)

 

[DaveC426913]

I don't think this is true for a hole that is actually going to grow (i.e., it won't evaporate before it accretes enough mass to push the evaporation time far out into the future)

Yes, I concede that evaporation puts the nail in the coffin of a subatomic runaway black hole.

I started this thread because I was thinking that evaporation doesn't seem to provide sufficient warm & fuzzies that it can't happen. I'll just have to take Hawking's word for it that it really happens at the rate predicted. (Has this been experimentally verified?)

I was looking for an answer more concrete than "it could grow - IF it weren't stopped by this other mysterious phenomenon". I was looking for a "it can't happen at all" kind of answer to mollify friends who read these stories and ask.

So for a density of $1$ (the density of air, roughly), $M$ would be $9.25 \times 10^6$ kg, or 9250 metric tons. The $R$ corresponding to this is about 130 meters. For a density equal to the average density of Earth, about $5500$, $M$ reduces to about $2.2 \times 10^6$ kg, or 2200 metric tons.

Is that a reasonable estimate of the pressure at the centre of the Earth?

But I guess, when we're talking about BH's, a few orders of magnitude in density is insignificant.

 

[DaveC426913]

When the black hole is micro-scale, it consumes even less.

Which might just balance out against the fact that the fall-time is virtually zero.
But I'm not certain which way it would go.

Even if it's only consuming a few atoms at a time, if that occurs every pico-second, how fast might it grow?

 

[PeterDonis]

I'll just have to take Hawking's word for it that it really happens at the rate predicted. (Has this been experimentally verified?)

No, nor is it expected to be any time soon. The evaporation rate from any hole we can observe (stellar mass or larger) is many, many orders of magnitude too small to measure. And we aren't likely to be able to manufacture a hole small enough that we could measure its evaporation rate any time soon.

Is that a reasonable estimate of the pressure at the centre of the Earth?

I assume you mean the density at the center--no, it's not, I was just using the Earth's average density for a rough estimate. Since the threshold mass goes as the inverse sixth root of the density, the difference between the Earth's average density and the density at the center of the Earth doesn't make much difference in this calculation.

Even if it's only consuming a few atoms at a time, if that occurs every pico-second, how fast might it grow?

Consider the hole for which $M$ is just large enough that an adjacent atom can fall in before it evaporates; this is the hole with mass $59.4$ kg as calculated in post #38. The number of atoms that can be adjacent, if they are packed as a solid, is of order unity; let's suppose it's 10, as an upper limit. Let's suppose each atom has mass of about $10^{-25}$ kg (this is an atomic weight somewhere around that of iron). Then the hole sucks in $10^{-24} / 59.4$ or a fraction of about $10^{-26}$ its own mass; this will not increase its evaporation time by enough to matter.

This is the reason why I said in post #38 that, in order to grow, the hole has to have mass large enough that a mass on the order of its own mass is within the radius $R$ calculated using the formula in that post; that will ensure that enough mass falls into keep the hole from evaporating. That mass is, as I calculated, around 2200 metric tons, or about 37,000 times the mass of the 59.4 kg hole discussed just above. Since the fall time goes like the inverse square root of $M$, this means the fall time for adjacent atoms into this hole will be about 200 times shorter than the fall time for adjacent atoms into the 59.4 kg hole. So something like an atom per 1/100 of a picosecond is how quickly a hole needs to suck in adjacent atoms in order to grow fast enough to keep itself from evaporating.

 

[DaveC426913]

I assume you mean the density at the center

No, I meant pressure. The atoms are not merely falling, they have a huge weight on top of them. They're being crammed into the void left by atoms that are consumed by the BH.

 

[LURCH]

Or to add another fantasy: It happened so often before until the right setup has been left.

That’s what I’m saying; we’ve found a new theory for dark matter. The universe is mostly filled with teensey-weensey black holes that used to be planets. (Lol)

Come to think of it, we may have explained the Fermi Paradox at the same time. That’s it, I’m rushing over to the sci-fi Forum to inspire some aspiring author!

 

[PeterDonis]

I meant pressure. The atoms are not merely falling, they have a huge weight on top of them.

Ah, ok. I didn't take any pressure effects into account since I was only looking at very rough orders of magnitude. I also didn't take into account that as the hole evaporates, its Hawking radiation exerts pressure outward.

 

[PeterDonis]

The atoms are not merely falling, they have a huge weight on top of them. They're being crammed into the void left by atoms that are consumed by the BH.

I'm not sure this is right. The force between the atoms in the normal case is there because the situation is static: none of the atoms can free-fall. If a void is created by the BH at the center, and the atoms start falling into it, the force between the atoms should vanish, at least to a first approximation (assuming a perfectly symmetrical fall process). The atoms on top could only "push" the atoms below if for some reason the atoms below were falling slower than the atoms on top; but that shouldn't be the case, at least to a first approximation.

 

[anorlunda]

If we go back to the original premise of this thread, that a BH could form in the LHC at CERN. In that scenario, the BH must be built one particle at a time. But as I showed in #20, the 2 proton BH event is explosive, not implosive. @PeterDonis analysis in #38, considered only free fall, and not the repulsive forces of radiation pressure. Even if the BH is embedded in ordinary matter, the net forces are initially repulsive rather than attractive. So I see no credible evolution to build a massive BH, adding one particle at a time.

Must we trust that Hawking was correct? If micro black holes did not evaporate, then we would find it difficult to explain why we don't find micro primordial black holes everywhere, or to explain the genesis of all the elementary particles at the origin of the universe instead of genesis of primordial black holes. If there was a non-zero probability of two particles forming a BH in the genesis, and if that was irreversible, then why not all particles becoming BH? I view the absence of primordial BH as an argument that the two-particle BH event (if it exists at all) must be time reversible.

 

[fresh_42]

I was looking for a "it can't happen at all" kind of answer to mollify friends who read these stories and ask.

I think you can say:
"Our best experts in the field have calculated various scenarios and none led to a result which was even close to a disaster."

 

[jbriggs444]

I'm not sure this is right. The force between the atoms in the normal case is there because the situation is static: none of the atoms can free-fall. If a void is created by the BH at the center, and the atoms start falling into it, the force between the atoms should vanish, at least to a first approximation (assuming a perfectly symmetrical fall process). The atoms on top could only "push" the atoms below if for some reason the atoms below were falling slower than the atoms on top; but that shouldn't be the case, at least to a first approximation.

Why would one not look at this in the Newtonian approximation as an example of a choked flow? You have a volumetric flow rate which can be calculated in terms of velocity multiplied by cross-sectional area.

Velocity scales as the square root of gravitational potential. [Newtonian] potential scales with inverse radius. So we are talking $v \propto \frac{1}{\sqrt{r}}$. Cross section scales as inverse radius squared. So we are talking $a \propto r^2$

In the absence of a pressure gradient, the volumetric flow rate at radius r is then given by $VFR(r) \propto r^{1.5}$. But since the equilibrium mass flow rate must be equal everywhere, we either need a pressure gradient to slow down the flow up top to match the choked flow rate at the bottom or a density gradient to accomplish the same thing. A density gradient implies a pressure gradient anyway.

Or am I missing something? [Obviously I'm missing that it's not a perfect equilibrium and that the Newtonian approximation breaks down near the bottom]

Edit: Boneheaded math mistake rectified.

 

[PeterDonis]

Why would one not look at this in the Newtonian approximation as an example of a choked flow?

Because the black hole at the center acts as an infinite sink, from a Newtonian standpoint. Matter goes in but never comes out; it gets destroyed in the singularity inside the hole. So there is no "choke point" in the flow. A more technical way of putting it is that the presence of the black hole means that the continuity equation is violated (flow lines all go into the hole, none come out).

 

[PeterDonis]

Even if the BH is embedded in ordinary matter, the net forces are initially repulsive rather than attractive.

They are for a hole with mass on the order of the energies available in the LHC, yes; you calculated that in post #20.

For the much larger holes I have been considering, let's see. The Hawking radiation power is

$$
P = \frac{\hbar c^6}{15360 \pi G^2 M^2} = 3.562 \times 10^{32} \frac{1}{M^2}
$$

We can convert this to a flux (in Watts per square meter) at a one-atom distance (which I am taking to be $10^{-10}$ meters) by dividing by the area of a 2-sphere with that radius, $4 \pi \times 10^{-20}$:

$$
F = 2.835 \times 10^{51} \frac{1}{M^2}
$$

We can then convert this to an effective energy density, and hence (since we're dealing with light) an effective pressure, by dividing by $c$ to obtain

$$
p = 9.455 \times 10^{42} \frac{1}{M^2}
$$

For $M$ equal to 2200 metric tons (or $2.2 \times 10^{6}$ kg), this works out to $p = 1.95 \times 10^{30}$ Pascals. This is indeed still huge, certainly large enough to keep matter from falling into the hole while it evaporates, despite my previous calculations (which assumed free falling matter).

From the above formula, if we wanted a pressure no greater than, say, atmospheric pressure at the Earth's surface, about $10^5$ Pascals, we would need a hole with mass $9.66 \times 10^{18}$ kg. This is about the mass of a moderately sized asteroid. Note that this hole has a Schwarzschild radius of $1.43 \times 10^{-8}$ meters, so at this point we would have to change the analysis anyway since the hole is larger than the one-atom distance I assumed above. But at any rate it seems clear that for the hole's Hawking radiation pressure not to easily prevent adjacent matter from falling in, the hole has to be of at least "small astronomical body" mass.