@sonhouse saidA black hole that small would have to be compressed way beyond neutron matter or perhaps even quark matter, unless it just dips into another universe.
https://phys.org/news/2020-07-scientists-planet-primordial-black-hole.html
That would be an interesting find indeed.
A BH that small, wouldn't it be emitting detectable Hawking radiation, on its way out as a black hole?
Anyone know how small they get before it totally evaporates?
A neutron star that size would definitely "evaporate", but so violently that it could turn Earth into blazing hot dust.
11 Jul 20
@bunnyknight saidAll blackholes have a singularity within them. The mathematics indicates infinite density: mass/volume, with mass>0 and volume=0. Is that really the physics of the matter as well? I doubt it. To solve the riddle I think requires a theory that unifies general relativity theory with quantum mechanics.
A black hole that small would have to be compressed way beyond neutron matter or perhaps even quark matter, unless it just dips into another universe.
A neutron star that size would definitely "evaporate", but so violently that it could turn Earth into blazing hot dust.
@bunnyknight saidWhile tremendous pressure (provided by gravity) is required to force electrons to merge with protons to create neutrons, I think a fist-sized chunk of neutronium somehow broken off a neutron star would last a very long time. What would make it evaporate precipitously?
A black hole that small would have to be compressed way beyond neutron matter or perhaps even quark matter, unless it just dips into another universe.
A neutron star that size would definitely "evaporate", but so violently that it could turn Earth into blazing hot dust.
It takes a huge amount of pressure to turn graphite into diamonds, but once the carbon atoms are arranged in a diamond crystalline lattice, it is very stable even when the pressure is removed.
11 Jul 20
@soothfast saidI think "infinite" is a dangerous word to use; in the real world there may not be anything infinite at all, except in our imagination. In case of a black hole, we really have no idea of their actual diameter, assuming it even resembles any matter at all, since it's well hidden beneath the event horizon.
All blackholes have a singularity within them. The mathematics indicates infinite density: mass/volume, with mass>0 and volume=0. Is that really the physics of the matter as well? I doubt it. To solve the riddle I think requires a theory that unifies general relativity theory with quantum mechanics.
@soothfast saidFrom what I've read, neutronium behaves more like a compressed spring than a compressed diamond. The instant that gravity ceases to hold it in place, those neutrons explode into a cloud of normal matter with glorious violence. Even the subatomic particles instantly change from one to another at some quantum level.
While tremendous pressure (provided by gravity) is required to force electrons to merge with protons to create neutrons, I think a fist-sized chunk of neutronium somehow broken off a neutron star would last a very long time. What would make it evaporate precipitously?
It takes a huge amount of pressure to turn graphite into diamonds, but once the carbon atoms are arranged in a diamond crystalline lattice, it is very stable even when the pressure is removed.
@bunnyknight saidWhat you read may be right. I used to know that isolated neutrons have an average lifespan of 15 minutes and then undergo beta decay. A fist-sized chunk of neutronium would have a decent gravitational field, but not one strong enough to keep this decay from occurring irreversibly. The gravity would hold the assemblage of neutrons in a spherical shape for a very brief spell, I suspect, but the decay process would introduce a rapidly increasing number of protons and electrons. It would become a small ball of extremely hot ionized hydrogen in short order, probably in an explosive manner.
From what I've read, neutronium behaves more like a compressed spring than a compressed diamond. The instant that gravity ceases to hold it in place, those neutrons explode into a cloud of normal matter with glorious violence. Even the subatomic particles instantly change from one to another at some quantum level.
https://en.wikipedia.org/wiki/Neutronium
https://en.wikipedia.org/wiki/Beta_decay
Perhaps DeepThought could provide deeper thoughts on the matter.
11 Jul 20
@bunnyknight saidI don't think any physical model has validity under conditions that require division by zero. It means a new and better model must needs be devised.
I think "infinite" is a dangerous word to use; in the real world there may not be anything infinite at all, except in our imagination. In case of a black hole, we really have no idea of their actual diameter, assuming it even resembles any matter at all, since it's well hidden beneath the event horizon.
12 Jul 20
@soothfast saidIt's 3.50 in the morning, so I'll comment more in the tomorrow. Briefly, a ball of neutronium that isn't gravitationally bound would emit electrons like no-one's business and undergo massive spontaneous nuclear fission. As an empirical argument, in the absence of an overriding gravitational field, neutronium is unstable because we see nucleii with roughly even numbers of protons and neutrons. If neutronium were stable it would be the ground state. We know states with too many neutrons aren't stable because we see tritium, which is unstable with a half-life of about 12 years, helium-3 which is the stable state, but we don't see Lithium-3 (3 protons) or neutronium-3 (3 neutrons).
What you read may be right. I used to know that isolated neutrons have an average lifespan of 15 minutes and then undergo beta decay. A fist-sized chunk of neutronium would have a decent gravitational field, but not one strong enough to keep this decay from occurring irreversibly. The gravity would hold the assemblage of neutrons in a spherical shape for a very brief spe ...[text shortened]... en.wikipedia.org/wiki/Beta_decay
Perhaps DeepThought could provide deeper thoughts on the matter.
With diamond it requires a lot of temperature and pressure to form, but once that energy has been provided the result is stable, the stability of the state isn't necessarily indicative of the difficulty in forming it. Besides, have you never seen a diamond burn?
12 Jul 20
@DeepThought
If it did turn out to be a grapefruit sized black hole, how can you calculate the mass of one that size? I guess you could manipulate the Schwarzchild radius. 2GM/C^2
so that would be M=2G/c^2*SR.
Does that sound right? I was thinking of how large the original star would have been squashed to grapefruit size, so you would have to deal with estimates of original density to work that one out.
12 Jul 20
@sonhouse saidrₛ = 2GM/c²
@DeepThought
If it did turn out to be a grapefruit sized black hole, how can you calculate the mass of one that size? I guess you could manipulate the Schwarzchild radius. 2GM/C^2
so that would be M=2G/c^2*SR.
Does that sound right? I was thinking of how large the original star would have been squashed to grapefruit size, so you would have to deal with estimates of original density to work that one out.
M = rₛc²/2G
The Earth's Schwartzshild radius is of the order of 8mm. The radius of a grapefruit is about 4 to 5 times that so the corresponding mass would be ~10³¹ kg. Such a black hole would be primordial and not the result of a stellar collapse. Black holes start at about 2.17 solar masses and there simply hasn't been enough time for solar mass black holes to undergo Hawking evaporation since the start of the universe so such an object would have to be a relic of the Planck era.
13 Jul 20
@sonhouse saidThe Wikipedia page gives us the temperature of a black hole as seen by an asymptotic observer:
https://phys.org/news/2020-07-scientists-planet-primordial-black-hole.html
That would be an interesting find indeed.
A BH that small, wouldn't it be emitting detectable Hawking radiation, on its way out as a black hole?
Anyone know how small they get before it totally evaporates?
T = ℏc³ / 8πGkM = ℏc / 4πkrₛ
I put in some numbers for a black hole of 1 solar mass and Schwartzschild radius 2950m the temperature as measured by an asymptotic observer is 61 nano-kelvin. For an object with a radius of 4 cm (about the size of a grapefruit) the mass would be about 4.5 times that of Earth and the temperature comes out at 4.5 millikelvin.
To be at the same temperature as the cosmic microwave background radiation, 2.7 kelvin, the black hole would have to have a mass about 3/4 of a percent of the Earth's and a Schwartzschild radius of 67 microns.
The expected lifetime of a black hole with 1 solar mass is of the order of 10^67 years, the grapefruit sized object would only last a mere 10^52 years. A black hole with a temperature equal to the cosmic microwave background would have a life time of the order of 10^44 years. So none of these objects could evaporate in the life time of the universe so far. Further to that since we'd expect them to be absorbing the microwave background they'll be accreting, not evaporating - with the odd effect that as they absorb energy they cool and therefore have a negative heat capacity at constant pressure.
[1] https://en.wikipedia.org/wiki/Hawking_radiation
13 Jul 20
@DeepThought
Oh, a bit more complex than I thought.
So black holes of any size will tend to lower the background temperature of the universe?
One thing about absorbing radiation, it seems to me there would be a cutoff in the frequency it could absorb, so if you condsider a 4cm BH as if it were a receiving antenna, a dipole with that length would be sensitive to about 9 gigahertz or so. So a BH would adsorb 9 or 10 Ghz and up in frequency with no upper limit.
That would seem to me to imply say a 300 Mhz signal, 1 meter wavelength, would not be absorbed by that BH, or at least only a very small portion would be absorbed since the wavelength would be some 25 times bigger so it just wouldn't fit inside that collector area.
13 Jul 20
@soothfast saidThe binding energy of a nucleus is estimated by the Liquid Drop model [1]. There's a volume term, a surface area term, a coulomb term which is zero for neutronium, an asymmetry term, which is just a constant for neutronium, and a pairing term, which is of the order of 1 MeV and we can just ignore for a large enough object. This gives us:
What you read may be right. I used to know that isolated neutrons have an average lifespan of 15 minutes and then undergo beta decay. A fist-sized chunk of neutronium would have a decent gravitational field, but not one strong enough to keep this decay from occurring irreversibly. The gravity would hold the assemblage of neutrons in a spherical shape for a very brief spe ...[text shortened]... en.wikipedia.org/wiki/Beta_decay
Perhaps DeepThought could provide deeper thoughts on the matter.
B(A) = a_v A - a_s A^(2/3) - a_a
a_v ~ 15.8 MeV
a_s ~ 18.3 MeV
a_a ~ 23.2 MeV
Solving the cubic equation gives B(A) = 0 for A ~ 2.035 which is in accordance with observation. A quasi-bound state of two neutrons has been observed (to my surprise). Note that the pairing term will be important here.
The binding energy per nucleon is then:
B(A)/A = a_v - a_s A^(-1/3) - a_a/A
This is positive for bound states and negative for unbound states, the total energy of the nucleus is:
E = mₙc² N + mₚ c² Z - B(A, Z)
where mₙ and mₚ are the masses of the neutron and proton respectively. So the stability of neutronium depends on whether gravity can override the weak force and prevent beta decay. For small nucleii it can't, as we know. Nuclear fission is driven by the coulomb term, so there should be a threshold value where the gravitational binding energy overrides the potential energy release from beta decay. I'll have a think about how to estimate that.
[1] https://en.wikipedia.org/wiki/Semi-empirical_mass_formula
