5 ms·

Is there an ELICU (college undergrad) as to why 50 to 130 is forbidden but below and above is allowed?

saagarjha 7y ago
How’s the one the article mentions? > A pair-instability supernova happens when the core grows so hot that light begins to spontaneously convert into electron-positron pairs. The light’s radiation pressure had kept the star’s core intact; when the light transforms into matter, the resulting pressure drop causes the core to rapidly shrink and become even hotter, further accelerating pair production and causing a runaway effect. Eventually the core gets so hot that oxygen ignites. This fully reverses the core’s implosion, so that it explodes instead. For cores with a mass between about 65 and 130 times that of our sun (according to current estimates), the star is completely obliterated. Cores between about 50 and 65 solar masses pulsate, shedding mass in a series of explosions until they drop below the range where pair instability occurs. Thus there should be no black holes with masses in the 50-to-130-solar-mass range.
- changoplatanero 7y ago
  What if two blackholes with 50 solar masses collide into each other?
  - lostmsu 7y ago
    AFAIK, nothing prevents that. You'd want blackholes with 100 solar masses though, as the collision will apparently shed significant part of the mass in gravitational waves.
    - TCR19 7y ago
      So does that mean during a collision between blackholes mass actually does/can escape a blackhole? I thought mass couldn't escape, at least not in any observable way that we can detect. [I don't know a ton about astrophysics, so apologies if this question doesn't make sense].
      - saagarjha 7y ago
        Mass (or rather, energy) escaping a black hole through gravitational waves has been detected experimentally. Similarly, Hawking radiation allows for black holes to emit energy as well (though this is due to quantum effects near the event horizon).
      - ncmncm 7y ago
        No, it cannot. The only energy that can escape in the collision comes from the kinetic energy of the two bodies, and from their colliding accretion discs, and energy bound up in magnetic fields outside the holes. It is interesting that the kinetic energy of a black hole is technically outside the hole. Energy radiated as a consequence of pair production on the event horizon would fluctuate as the surface changed shape and area, but not much.
        saagarjha 7y ago
        > It is interesting that the kinetic energy of a black hole is technically outside the hole. How so?
        raattgift 7y ago
        It's not. Even in Newtonian mechanics kinetic energy is an observer-dependent quantity; in General Relativity we care about the stress-energy-momentum tensor, and you can slice that up into kinetic and potential energies as you wish by applying coordinates. The vast majority of the stress-energy in the immediate neighbourhood of an accreting black hole formed by stellar collapse is found inside the horizon, even if there is a substantial accretion structure. General Relativity doesn't make any predictions about the forms stress-energy-momentum can take; we get that from matter theories like classical (but relativistic) Maxwell's equations, quantum electrodynamics, or the full Standard Model. (One also runs into "toy" or "test" forms of matter -- various idealized space-filling fluids or dusts, mainly, that approximate matter in the large: huge numbers of stars or huge numbers of galaxies, for example). So General Relativity also has nothing much to say about kinetic energy vs potential energy: just that they each must contribute to the Einstein Field Equations and that typically means being encoded in the stress-energy tensor. The perfectly elastic bouncing of microscopic particles of a gas each bouncing in one dimension between opposite sides of the inside of a gas container produces a beautiful relationship between the kinetic energy of a gas and its pressure. https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressure_and_kinetic_energy https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressu... (PV = 2/3 K) Assuming isotropic pressure at time t, we encode an identical contribution into the pressure components of the stress-energy tensor (the green diagonals here https://en.wikipedia.org/wiki/Stress–energy_tensor#/media/File:StressEnergyTensor_contravariant.svg https://en.wikipedia.org/wiki/Stress–energy_tensor#/media/Fi... ) for every point within the gas cylinder. The inner regions of massive stars have a lot of pressure; collapsars like neutron stars have even more pressure deep within them. In a runaway collapse that leads to the formation of a black hole, pressure typically dominates the stress-energy tensor, driving the formation of the event horizon. By contrast, a Schwarzschild black hole is a vacuum solution of the Einstein Field Equation, meaning that the stress-energy tensor is everywhere zero. Thus there is no pressure. The source of the "central" mass inside the horizon of a Schwarzschild is best thought of as gravity self-gravitating, or if you like, "it's just defined as curvature alone until you throw a test object in". If we perturb the Schwarzschild black hole by throwing neutral test objects through the horizon, the stress-energy tensor inside the horizon must be somewhere nonzero, but once through the horizon (ignoring quantum effects) the nonzero stress-energy stays in there, and everywhere outside the stress-energy tensor returns to zero. An astrophysical black hole by stellar collapse locks up stress-energy in the same way: once it's inside the event horizon, it stays there (ignoring quantum effects, principally Hawking radiation). Stress-energy outside the horizon might cross the horizon in various ways, or it might form some long lived arrangement sufficiently far from the horizon. You wouldn't say that a white dwarf partner in a white-dwarf/black-hole binary forms "the kinetic [or other] energy of a black hole", would you? If not, then neither does any matter near -- but outside -- the horizon. The no-hair theorem(s) mean(s) that in general you cannot distinguish between an uncharged, zero-angular-momentum black hole formed by stellar collapse (or black hole mergers) and a Schwarzschild (vacuum) black hole (say, formed primordially from nothing but gravitational radiation), even in binaries. A binary of uncharged, no-angular-momentum black holes may be two Schwarzschild BHs or two astrophysical BHs or one of each. The momentum-energy that leaves the binary cannot come from the stress-energy tensor of a Schwarzschild black hole, because it's zero everywhere in the horizon. If no-hair is true, it can't come from the stress-energy tensor in the interior of a black hole formed by stellar collapse, either. A very large primordial BH will have had a bunch of things fall into it (if nothing else, lots of CMB photons), but can still have essentially no stress-energy inside. Primordial black holes are not especially crazy: that's one possible way to explain supermassive black holes ( https://en.wikipedia.org/wiki/Supermassive_black_hole#Formation https://en.wikipedia.org/wiki/Supermassive_black_hole#Format... and note it's possible that primordial black holes can start with and retain essentially zero angular momentum). A pair of primordial SMBHs, each near the centre of mass of merging galaxy clusters, may have eaten a bunch of stellar masses worth of dust and gas in their lifetimes, but not nearly enough to account for the gravitational radiation that will be emitted late in their inspiral, let alone during the merger and ringdown. Instead, it is the angular momentum of the binary system (as a whole, since in this sketch neither BH rotates) that must power the gravitational radiation during the inspiral.
  - sandworm101 7y ago
    The chances of two random black holes ever colliding with each other (not a binary-pair situation) is astronomically rare. On cosmic scales, black holes are microscopic. They are far smaller than stars of similar masses, but have we ever seen two random stars just collide? Two black holes colliding would be like two bullets colliding if fired randomly from guns on different continents. The likelihood is so low that any detection within our sphere of observation would be very suspect.
    - phkahler 7y ago
      We have not seen stars collide, but black hole mergers seem to be a regular thing now that we have a detector. I would posit the difference is that you have to be looking at a the stars at the time of collision, but for black holes we have an omnidirectional detector with high SNR.
      - sandworm101 7y ago
        We have seen binary pairs collide, objects that were orbiting each other and who in all probability started their lives as a pair of objects and share a common formation story. We have not seen two non-associated stars born in different places randomly cross paths, which is the sort of merger necessary to form some sizes of black holes.
        Retric 7y ago
        Binary’s don’t always start from the same nebula. Over millions of years stars move around quite a bit and can form or break apart binary’s.
        sandworm101 7y ago
        To form a binary, for one star to capture another, requires some careful planning. (A binary pair grabbing a passing star and exchanging momentum.) They don't just get close and start orbiting each other. If a star was captured by another, it wouldn't be a tight pair that may one day merge. They would be very far apart, so far that merger would take far longer than the age of the universe.
- SkyBelow 7y ago
  It kinda hand waves over a number of parts and is more simplified than what I was looking for. To give a very very simplified summary of my reading and the questions I have remaining, it seems to be there are two equations for what is going on here. One is the force of gravitational and light radiation collapse that results in the initial increase in heat and the other is for the force of oxygen explosion. At under 50 solar mass, the oxygen explosion is 0 and so it forms. At 50 to 65 solar mass, the oxygen explosion is enough to throw off mass but not enough to obliterate the star, which eventually moves it to the 50 solar mass range. At 65 to 130 solar mass, the oxygen explosion is violent enough to destroy the star without a black hole. At 130+ solar mass, the oxygen explosion is too weak to overcome the gravitational collapse force and a black hole forms (but masses this large seem to be rare). What I'm wondering is why are those the specific numbers. Is it really just a case of 'take the equations, plug in the numbers, and this is what you get', or is there some explanation that is easier to conceptualize. For example, the 50 cut off seems to be that is the threshold needed to even have oxygen ignite. But for 50 to 65, why isn't the explosion enough to destroy the star? At this point, the force from gravity holding the star together would be less than at 65+ solar mass, so why isn't the core obliterated? Is it because there is a different sort of oxygen explosion that only happens under the force of 65+ solar mass that is much stronger than the one that happens at 50 solar mass? And as for the 130 threshold, shouldn't the more solar mass mean the more oxygen to explode, so shouldn't the force of the explosion continue to be higher than the force of the gravitational collapse? The article clearly claims this isn't the case, but doesn't explain why. P.S. Now that I'm reading over it again, I think there might be three forces and I misunderstood the direction of the light radiation force that invalidates all of the above. It appears the light radiation is pushing outward, same direction as the oxygen explosion. If I consider all three forces it might be the model I'm looking for.
  - saagarjha 7y ago
    > Now that I'm reading over it again, I think there might be three forces and I misunderstood the direction of the light radiation force that invalidates all of the above. It appears the light radiation is pushing outward, same direction as the oxygen explosion. If I consider all three forces it might be the model I'm looking for. Yes: radiation pressure is directed outwards (this is how main-sequence stars maintain hydrostatic equilibrium against gravity, which tends inwards).
jschwartzi 7y ago
Because the black hole can't form from a star of 50 to 130 solar masses but it can, for example, start at 40 solar masses, collapse into a black hole, and then accumulate another 20 solar masses over its lifetime to become a black hole of 60 solar masses. Edit: the mechanics of this are left as an exercise to the reader.

How’s the one the article mentions? > A pair-instability supernova happens when the core grows so hot that light begins to spontaneously convert into electron-positron pairs. The light’s radiation pressure had kept the star’s core intact; when the light transforms into matter, the resulting pressure drop causes the core to rapidly shrink and become even hotter, further accelerating pair production and causing a runaway effect. Eventually the core gets so hot that oxygen ignites. This fully reverses the core’s implosion, so that it explodes instead. For cores with a mass between about 65 and 130 times that of our sun (according to current estimates), the star is completely obliterated. Cores between about 50 and 65 solar masses pulsate, shedding mass in a series of explosions until they drop below the range where pair instability occurs. Thus there should be no black holes with masses in the 50-to-130-solar-mass range.

changoplatanero 7y ago
What if two blackholes with 50 solar masses collide into each other?
- lostmsu 7y ago
  AFAIK, nothing prevents that. You'd want blackholes with 100 solar masses though, as the collision will apparently shed significant part of the mass in gravitational waves.
  - TCR19 7y ago
    So does that mean during a collision between blackholes mass actually does/can escape a blackhole? I thought mass couldn't escape, at least not in any observable way that we can detect. [I don't know a ton about astrophysics, so apologies if this question doesn't make sense].
    - saagarjha 7y ago
      Mass (or rather, energy) escaping a black hole through gravitational waves has been detected experimentally. Similarly, Hawking radiation allows for black holes to emit energy as well (though this is due to quantum effects near the event horizon).
    - ncmncm 7y ago
      No, it cannot. The only energy that can escape in the collision comes from the kinetic energy of the two bodies, and from their colliding accretion discs, and energy bound up in magnetic fields outside the holes. It is interesting that the kinetic energy of a black hole is technically outside the hole. Energy radiated as a consequence of pair production on the event horizon would fluctuate as the surface changed shape and area, but not much.
      - saagarjha 7y ago
        > It is interesting that the kinetic energy of a black hole is technically outside the hole. How so?
        raattgift 7y ago
        It's not. Even in Newtonian mechanics kinetic energy is an observer-dependent quantity; in General Relativity we care about the stress-energy-momentum tensor, and you can slice that up into kinetic and potential energies as you wish by applying coordinates. The vast majority of the stress-energy in the immediate neighbourhood of an accreting black hole formed by stellar collapse is found inside the horizon, even if there is a substantial accretion structure. General Relativity doesn't make any predictions about the forms stress-energy-momentum can take; we get that from matter theories like classical (but relativistic) Maxwell's equations, quantum electrodynamics, or the full Standard Model. (One also runs into "toy" or "test" forms of matter -- various idealized space-filling fluids or dusts, mainly, that approximate matter in the large: huge numbers of stars or huge numbers of galaxies, for example). So General Relativity also has nothing much to say about kinetic energy vs potential energy: just that they each must contribute to the Einstein Field Equations and that typically means being encoded in the stress-energy tensor. The perfectly elastic bouncing of microscopic particles of a gas each bouncing in one dimension between opposite sides of the inside of a gas container produces a beautiful relationship between the kinetic energy of a gas and its pressure. https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressure_and_kinetic_energy https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#Pressu... (PV = 2/3 K) Assuming isotropic pressure at time t, we encode an identical contribution into the pressure components of the stress-energy tensor (the green diagonals here https://en.wikipedia.org/wiki/Stress–energy_tensor#/media/File:StressEnergyTensor_contravariant.svg https://en.wikipedia.org/wiki/Stress–energy_tensor#/media/Fi... ) for every point within the gas cylinder. The inner regions of massive stars have a lot of pressure; collapsars like neutron stars have even more pressure deep within them. In a runaway collapse that leads to the formation of a black hole, pressure typically dominates the stress-energy tensor, driving the formation of the event horizon. By contrast, a Schwarzschild black hole is a vacuum solution of the Einstein Field Equation, meaning that the stress-energy tensor is everywhere zero. Thus there is no pressure. The source of the "central" mass inside the horizon of a Schwarzschild is best thought of as gravity self-gravitating, or if you like, "it's just defined as curvature alone until you throw a test object in". If we perturb the Schwarzschild black hole by throwing neutral test objects through the horizon, the stress-energy tensor inside the horizon must be somewhere nonzero, but once through the horizon (ignoring quantum effects) the nonzero stress-energy stays in there, and everywhere outside the stress-energy tensor returns to zero. An astrophysical black hole by stellar collapse locks up stress-energy in the same way: once it's inside the event horizon, it stays there (ignoring quantum effects, principally Hawking radiation). Stress-energy outside the horizon might cross the horizon in various ways, or it might form some long lived arrangement sufficiently far from the horizon. You wouldn't say that a white dwarf partner in a white-dwarf/black-hole binary forms "the kinetic [or other] energy of a black hole", would you? If not, then neither does any matter near -- but outside -- the horizon. The no-hair theorem(s) mean(s) that in general you cannot distinguish between an uncharged, zero-angular-momentum black hole formed by stellar collapse (or black hole mergers) and a Schwarzschild (vacuum) black hole (say, formed primordially from nothing but gravitational radiation), even in binaries. A binary of uncharged, no-angular-momentum black holes may be two Schwarzschild BHs or two astrophysical BHs or one of each. The momentum-energy that leaves the binary cannot come from the stress-energy tensor of a Schwarzschild black hole, because it's zero everywhere in the horizon. If no-hair is true, it can't come from the stress-energy tensor in the interior of a black hole formed by stellar collapse, either. A very large primordial BH will have had a bunch of things fall into it (if nothing else, lots of CMB photons), but can still have essentially no stress-energy inside. Primordial black holes are not especially crazy: that's one possible way to explain supermassive black holes ( https://en.wikipedia.org/wiki/Supermassive_black_hole#Formation https://en.wikipedia.org/wiki/Supermassive_black_hole#Format... and note it's possible that primordial black holes can start with and retain essentially zero angular momentum). A pair of primordial SMBHs, each near the centre of mass of merging galaxy clusters, may have eaten a bunch of stellar masses worth of dust and gas in their lifetimes, but not nearly enough to account for the gravitational radiation that will be emitted late in their inspiral, let alone during the merger and ringdown. Instead, it is the angular momentum of the binary system (as a whole, since in this sketch neither BH rotates) that must power the gravitational radiation during the inspiral.
- sandworm101 7y ago
  The chances of two random black holes ever colliding with each other (not a binary-pair situation) is astronomically rare. On cosmic scales, black holes are microscopic. They are far smaller than stars of similar masses, but have we ever seen two random stars just collide? Two black holes colliding would be like two bullets colliding if fired randomly from guns on different continents. The likelihood is so low that any detection within our sphere of observation would be very suspect.
  - phkahler 7y ago
    We have not seen stars collide, but black hole mergers seem to be a regular thing now that we have a detector. I would posit the difference is that you have to be looking at a the stars at the time of collision, but for black holes we have an omnidirectional detector with high SNR.
    - sandworm101 7y ago
      We have seen binary pairs collide, objects that were orbiting each other and who in all probability started their lives as a pair of objects and share a common formation story. We have not seen two non-associated stars born in different places randomly cross paths, which is the sort of merger necessary to form some sizes of black holes.
      - Retric 7y ago
        Binary’s don’t always start from the same nebula. Over millions of years stars move around quite a bit and can form or break apart binary’s.
        sandworm101 7y ago
        To form a binary, for one star to capture another, requires some careful planning. (A binary pair grabbing a passing star and exchanging momentum.) They don't just get close and start orbiting each other. If a star was captured by another, it wouldn't be a tight pair that may one day merge. They would be very far apart, so far that merger would take far longer than the age of the universe.
SkyBelow 7y ago
It kinda hand waves over a number of parts and is more simplified than what I was looking for. To give a very very simplified summary of my reading and the questions I have remaining, it seems to be there are two equations for what is going on here. One is the force of gravitational and light radiation collapse that results in the initial increase in heat and the other is for the force of oxygen explosion. At under 50 solar mass, the oxygen explosion is 0 and so it forms. At 50 to 65 solar mass, the oxygen explosion is enough to throw off mass but not enough to obliterate the star, which eventually moves it to the 50 solar mass range. At 65 to 130 solar mass, the oxygen explosion is violent enough to destroy the star without a black hole. At 130+ solar mass, the oxygen explosion is too weak to overcome the gravitational collapse force and a black hole forms (but masses this large seem to be rare). What I'm wondering is why are those the specific numbers. Is it really just a case of 'take the equations, plug in the numbers, and this is what you get', or is there some explanation that is easier to conceptualize. For example, the 50 cut off seems to be that is the threshold needed to even have oxygen ignite. But for 50 to 65, why isn't the explosion enough to destroy the star? At this point, the force from gravity holding the star together would be less than at 65+ solar mass, so why isn't the core obliterated? Is it because there is a different sort of oxygen explosion that only happens under the force of 65+ solar mass that is much stronger than the one that happens at 50 solar mass? And as for the 130 threshold, shouldn't the more solar mass mean the more oxygen to explode, so shouldn't the force of the explosion continue to be higher than the force of the gravitational collapse? The article clearly claims this isn't the case, but doesn't explain why. P.S. Now that I'm reading over it again, I think there might be three forces and I misunderstood the direction of the light radiation force that invalidates all of the above. It appears the light radiation is pushing outward, same direction as the oxygen explosion. If I consider all three forces it might be the model I'm looking for.
- saagarjha 7y ago
  > Now that I'm reading over it again, I think there might be three forces and I misunderstood the direction of the light radiation force that invalidates all of the above. It appears the light radiation is pushing outward, same direction as the oxygen explosion. If I consider all three forces it might be the model I'm looking for. Yes: radiation pressure is directed outwards (this is how main-sequence stars maintain hydrostatic equilibrium against gravity, which tends inwards).

jschwartzi 7y ago

Because the black hole can't form from a star of 50 to 130 solar masses but it can, for example, start at 40 solar masses, collapse into a black hole, and then accumulate another 20 solar masses over its lifetime to become a black hole of 60 solar masses. Edit: the mechanics of this are left as an exercise to the reader.