Black holes squishing Earth

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  • #26
PeterDonis
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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.
 
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  • #27
gneill
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LOL :wink: 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.
 
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  • #28
PeterDonis
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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.
 
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  • #29
PeterDonis
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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.
 
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  • #30
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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?
 
  • #31
PeterDonis
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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.
 
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  • #32
gneill
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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.
 
  • #33
PeterDonis
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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.
 
  • #34
gneill
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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.
 
  • #35
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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.
 
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  • #36
DaveC426913
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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.
 
  • #37
jbriggs444
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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.
 
  • #38
PeterDonis
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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. :wink:

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.)
 
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  • #39
DaveC426913
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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.
 
  • #40
DaveC426913
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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?
 
  • #41
PeterDonis
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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 in to 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.
 
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  • #42
DaveC426913
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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.
 
  • #43
LURCH
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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!
 
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  • #44
PeterDonis
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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.
 
  • #45
PeterDonis
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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.
 
  • #46
anorlunda
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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.
 
  • #47
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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."
 
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  • #48
jbriggs444
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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.
 
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  • #49
PeterDonis
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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).
 
  • #50
PeterDonis
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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.
 
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