3 ms·

ChatGPT just told me that ... the Schwarzschild radius for a quark-gluon plasma with a mass of 10^12 kg is approximately 1.485x10^-15 meters, which is extrem

ChatGPT just told me that ... the Schwarzschild radius for a quark-gluon plasma with a mass of 10^12 kg is approximately 1.485x10^-15 meters, which is extremely small. Extremely small is around the diameter of a proton. Which would mean that any quark star would a black hole. So this estimate must be wrong, right?

danparsonson 2y ago
Leaving aside the calculation (I don't know if it's correct), and the sense or not of relying on a fancy word-association machine for such calculations (I wouldn't), your Schwarzchild radius is even smaller than that and you're not a black hole - an object is only a black hole if it is entirely inside its Schwarzchild radius.
- delichon 2y ago
  If the Schwarzchild radius is reached within a larger mass, wouldn't that radius become inescapable and eventually consume the remainder while growing in proportion? Since I did a sanity check on the result which seems to have failed, why do you think that it's necessary to warn me not to rely on it? The hostility on this site for using LLMs as a tool for exploration is confusing to me. I also learn a lot by asking questions of highly unreliable humans. LLMs don't seem to be much worse than that.
  - ant6n 2y ago
    [flagged]
    - michaelcampbell 2y ago
      Please, no.
  - AnimalMuppet 2y ago
    > If the Schwarzchild radius is reached within a larger mass, wouldn't that radius become inescapable and eventually consume the remainder while growing in proportion? If I understand your question correctly, no. If I have a mass M that corresponds to a Schwarzchild radius r, but the mass occupies a larger radius R, no, it will not create a black hole, since the mass m within r is less than the critical mass M. The portion of M that is within r can still escape that radius.
  - javaunsafe2019 2y ago
    This is maybe cause ppl don’t want to rely on some machine hallucinating wrong answer when trying to answer fundamental questions but rather use real wisdom instead.
  - bell-cot 2y ago
    It's fairly well-accepted that tiny black holes are quite unstable: https://en.wikipedia.org/wiki/Hawking_radiation https://en.wikipedia.org/wiki/Hawking_radiation On a larger scale, kinda yes - a small-ish black hole would generally "feed" from any near-enough object. Though that feeding will generally be extremely energetic and messy, and fling most of the victim away at high velocity.
  - commodoreboxer 2y ago
    I don't have a problem with your post, but the hostility I see is mostly out of a real concern that people posting LLM comments could degrade discussion. It's probably also association, because a lot of people have started posting comments that just say "ChatGPT says..." and nearly the entirety of the body is just LLM output. Honestly, a lot of people are just sick of seeing anything about LLMs and don't like seeing it at all in threads that aren't already about it.
    - queuebert 2y ago
      Exactly. It's not appropriate for this type of question. LLMs can summarize the global internet sentiment about something pretty well, but they aren't the right architecture for answering precise scientific questions, nor will they ever be. It's like asking an old, experienced physicist to guess answers to things instead of solving equations.
      - ben_w 2y ago
        They're already better than I'd have expected possible, so I wouldn't say "never" — but also yes I agree with you, currently I'd compare their output to an off-the-cuff answer from an experienced professional (or an undergraduate trying hard).
      - turzmo 2y ago
        As a physicist I can say that unless ChatGPT is asked a definitional question, or a very common question (whose solution is likely described many places on the internet anyway), it is very likely to be wrong. Personally more than 90% of the questions I've asked it have been flat-out wrong so I stopped using it entirely.
  - layer8 2y ago
    You only have a black hole and an event horizon if the mass whose Schwarzschild radius you are calculating is completely contained within said radius.
  - noslenwerdna 2y ago
    As others have said, that's not what a Schwarzschild radius means. You would only get a black hole if all the matter is inside the radius. The Schwarzschild radius comes from the free space solution to Einstein's equation. It only holds in free space. So if you want to use it here, all the matter would need to be within Schwarzschild radius, and then you could calculate what happens outside that radius (again, in the matter free space outside). If you want to know how gravity behaves inside a star you have to use a different tool.
  - danparsonson 2y ago
    Lots of other replies so hopefully it makes sense but just to be clear: 'the' Schwarzchild radius isn't one specific number, it's based on the mass of an object. Every massive object has a different SR (also called its Event Horizon), and if you compress the object entirely to within that radius, it will become a black hole. It's just a measure of how small the object needs to be before light cannot escape its surface.
itishappy 2y ago
Neutron stars weigh in at about 1.4 solar masses, or 10^30kg, which gives a Schwarzschild radius of about 3mi. A neutron star can be viewed as a mass-just-shy-of-a-black-hole, so I'd expect it to be relatively similar in scale.
- pdonis 2y ago
  > Neutron stars weigh in at about 1.4 solar masses Observed neutron stars actually have a range of masses, from the 1.4 solar mass point (which is the maximum mass for a white dwarf, so one would expect to see neutron stars of about that mass that were just over the white dwarf limit) up to, IIRC, almost 3 solar masses for the largest one that has been observed. > A neutron star can be viewed as a mass-just-shy-of-a-black-hole Not in general, no. A neutron star just under the maximum mass limit for neutron stars, which is believed to be about 3 solar masses, could be sort of viewed this way, but even then it's not really correct, since there is nothing that forbids a black hole with a mass smaller than that from existing. It's just extremely unlikely that such a black hole could be formed by the collapse of a star, since such a collapse would be expected to stop at the neutron star stage (or at the white dwarf stage if the star was less than 1.4 solar masses).
  - itishappy 2y ago
    Excellent points! Galaxies have more mass than either, but behave very differently. I would correct my statement by saying it's the concentration of mass (density) that defines neutron stars and black holes.
    - pdonis 2y ago
      > it's the concentration of mass (density) that defines neutron stars and black holes. Not really, no. What defines a neutron star is that it is in hydrostatic equilibrium supported by neutron degeneracy pressure. If we include the quark-gluon plasma phase, we can just amend that to being supported by quark degeneracy pressure (and noting that "neutron" is just a special case of "quark" for this purpose). Whereas a normal star is in hydrostatic equilibrium supported by thermal pressure (with fusion reactions providing the heat source). And a galaxy is not in hydrostatic equilibrium at all, it's composed of stars in free-fall orbits. The average densities of the objects in all three of these cases are consequences of the above. A black hole is defined by having an event horizon and being vacuum. There is no well-defined concept of "density" for a black hole, nor is there "concentration of mass" in the sense of the hole being made of matter; it's not, it's vacuum. The "mass" of a black hole is a global feature of its spacetime geometry.
mr_mitm 2y ago
The result is correct. But I don't understand your conclusion. Where does that number 10^12 kg come from?

danparsonson 2y ago

Leaving aside the calculation (I don't know if it's correct), and the sense or not of relying on a fancy word-association machine for such calculations (I wouldn't), your Schwarzchild radius is even smaller than that and you're not a black hole - an object is only a black hole if it is entirely inside its Schwarzchild radius.

delichon 2y ago
If the Schwarzchild radius is reached within a larger mass, wouldn't that radius become inescapable and eventually consume the remainder while growing in proportion? Since I did a sanity check on the result which seems to have failed, why do you think that it's necessary to warn me not to rely on it? The hostility on this site for using LLMs as a tool for exploration is confusing to me. I also learn a lot by asking questions of highly unreliable humans. LLMs don't seem to be much worse than that.
- ant6n 2y ago
  [flagged]
  - michaelcampbell 2y ago
    Please, no.
- AnimalMuppet 2y ago
  > If the Schwarzchild radius is reached within a larger mass, wouldn't that radius become inescapable and eventually consume the remainder while growing in proportion? If I understand your question correctly, no. If I have a mass M that corresponds to a Schwarzchild radius r, but the mass occupies a larger radius R, no, it will not create a black hole, since the mass m within r is less than the critical mass M. The portion of M that is within r can still escape that radius.
- javaunsafe2019 2y ago
  This is maybe cause ppl don’t want to rely on some machine hallucinating wrong answer when trying to answer fundamental questions but rather use real wisdom instead.
- bell-cot 2y ago
  It's fairly well-accepted that tiny black holes are quite unstable: https://en.wikipedia.org/wiki/Hawking_radiation https://en.wikipedia.org/wiki/Hawking_radiation On a larger scale, kinda yes - a small-ish black hole would generally "feed" from any near-enough object. Though that feeding will generally be extremely energetic and messy, and fling most of the victim away at high velocity.
- commodoreboxer 2y ago
  I don't have a problem with your post, but the hostility I see is mostly out of a real concern that people posting LLM comments could degrade discussion. It's probably also association, because a lot of people have started posting comments that just say "ChatGPT says..." and nearly the entirety of the body is just LLM output. Honestly, a lot of people are just sick of seeing anything about LLMs and don't like seeing it at all in threads that aren't already about it.
  - queuebert 2y ago
    Exactly. It's not appropriate for this type of question. LLMs can summarize the global internet sentiment about something pretty well, but they aren't the right architecture for answering precise scientific questions, nor will they ever be. It's like asking an old, experienced physicist to guess answers to things instead of solving equations.
    - ben_w 2y ago
      They're already better than I'd have expected possible, so I wouldn't say "never" — but also yes I agree with you, currently I'd compare their output to an off-the-cuff answer from an experienced professional (or an undergraduate trying hard).
    - turzmo 2y ago
      As a physicist I can say that unless ChatGPT is asked a definitional question, or a very common question (whose solution is likely described many places on the internet anyway), it is very likely to be wrong. Personally more than 90% of the questions I've asked it have been flat-out wrong so I stopped using it entirely.
- layer8 2y ago
  You only have a black hole and an event horizon if the mass whose Schwarzschild radius you are calculating is completely contained within said radius.
- noslenwerdna 2y ago
  As others have said, that's not what a Schwarzschild radius means. You would only get a black hole if all the matter is inside the radius. The Schwarzschild radius comes from the free space solution to Einstein's equation. It only holds in free space. So if you want to use it here, all the matter would need to be within Schwarzschild radius, and then you could calculate what happens outside that radius (again, in the matter free space outside). If you want to know how gravity behaves inside a star you have to use a different tool.
- danparsonson 2y ago
  Lots of other replies so hopefully it makes sense but just to be clear: 'the' Schwarzchild radius isn't one specific number, it's based on the mass of an object. Every massive object has a different SR (also called its Event Horizon), and if you compress the object entirely to within that radius, it will become a black hole. It's just a measure of how small the object needs to be before light cannot escape its surface.

itishappy 2y ago

Neutron stars weigh in at about 1.4 solar masses, or 10^30kg, which gives a Schwarzschild radius of about 3mi. A neutron star can be viewed as a mass-just-shy-of-a-black-hole, so I'd expect it to be relatively similar in scale.

pdonis 2y ago
> Neutron stars weigh in at about 1.4 solar masses Observed neutron stars actually have a range of masses, from the 1.4 solar mass point (which is the maximum mass for a white dwarf, so one would expect to see neutron stars of about that mass that were just over the white dwarf limit) up to, IIRC, almost 3 solar masses for the largest one that has been observed. > A neutron star can be viewed as a mass-just-shy-of-a-black-hole Not in general, no. A neutron star just under the maximum mass limit for neutron stars, which is believed to be about 3 solar masses, could be sort of viewed this way, but even then it's not really correct, since there is nothing that forbids a black hole with a mass smaller than that from existing. It's just extremely unlikely that such a black hole could be formed by the collapse of a star, since such a collapse would be expected to stop at the neutron star stage (or at the white dwarf stage if the star was less than 1.4 solar masses).
- itishappy 2y ago
  Excellent points! Galaxies have more mass than either, but behave very differently. I would correct my statement by saying it's the concentration of mass (density) that defines neutron stars and black holes.
  - pdonis 2y ago
    > it's the concentration of mass (density) that defines neutron stars and black holes. Not really, no. What defines a neutron star is that it is in hydrostatic equilibrium supported by neutron degeneracy pressure. If we include the quark-gluon plasma phase, we can just amend that to being supported by quark degeneracy pressure (and noting that "neutron" is just a special case of "quark" for this purpose). Whereas a normal star is in hydrostatic equilibrium supported by thermal pressure (with fusion reactions providing the heat source). And a galaxy is not in hydrostatic equilibrium at all, it's composed of stars in free-fall orbits. The average densities of the objects in all three of these cases are consequences of the above. A black hole is defined by having an event horizon and being vacuum. There is no well-defined concept of "density" for a black hole, nor is there "concentration of mass" in the sense of the hole being made of matter; it's not, it's vacuum. The "mass" of a black hole is a global feature of its spacetime geometry.

mr_mitm 2y ago

The result is correct. But I don't understand your conclusion. Where does that number 10^12 kg come from?