5 ms·

It would sink into the center of the earth and then oscillate around the center until friction stops it dead center. Materials in the earth's crust and mantle a

It would sink into the center of the earth and then oscillate around the center until friction stops it dead center. Materials in the earth's crust and mantle are not strong enough to stop that mass from sinking ever deeper. But that assumes that this spoon of matter were stable in that state, which it isn't. That kind of density can only be held up with some force like gravity keeping up the pressure. The gravity of that spoonful is insufficient, so the internal pressure will drive apart the neutrons. The effects would probably look like a very very large nuke or large asteroid impact.

ben_w 3y ago
Very large asteroid, too big for a nuke: https://www.wolframalpha.com/input?i=%280.782343+MeV+%2F+neutron+mass%29+*+5e12+kg+in+megatons+of+TNT+equivalent+ https://www.wolframalpha.com/input?i=%280.782343+MeV+%2F+neu... ~89 billion megatons of TNT equivalent
- raattgift 3y ago
  What motivates your first factor? 0.782343 MeV is the free neutron beta decay; where in the solar system are the free neutrons minutes after they are magically teleported to terrestrial ground zero as a something like a (degenerate, possibly ultra-relativistic) Fermi gas? I think most attempts to arrive at an answer will end up somewhere between half and virtually all of them being "not very close" (~ light-minutes) away, and that's assuming one corrects for the differences in escape velocities. (The equatorial escape velocity of a spinning neutron star is in tenths of the speed of light, thus the sobriquet "relativistic star"). Without this correction, it is likely the bulk of the expanding drop of Fermi gas just exits the atmosphere in milliseconds (timed by terrestrial stopwatches), with time dilation extending the mean lifetime of the free neutrons in the drop comparably to the extended lifetime of atmospheric muons from cosmic rays. The bulk of the beta decays happen at a distance from terrestrial ground zero best measured in astronomical units. If we play Star Trek transporter games such that the neutrons arrive at ground zero at rest in local East-North-Up coordinates, you'd want to know the internal kinetic energy (KE) density of the (pure-)neutron star, which will be in the range of 20-40 for x in 10^{x} J m^-3. The 10^25ish or even 10^30ish joules of KE will be released from our several cm^3 spoonful practically all at once and practically omnidirectionally from ground zero (so again, most free neutron decays happen at ~ AU distances from ground zero because they'll zip right through the atmosphere). The expansion of the suddenly unpressurized gas of neutrons will make a mess, particularly the fraction that slams into and through the ground. Part of the mess is neutron scattering physics, and I have no expertise there, but I would guess there wouldn't be any free neutrons near ground zero (and probably not within the solid Earth) in ~minutes. Additionally, one might compare the R-process <https://en.wikipedia.org/wiki/R-process https://en.wikipedia.org/wiki/R-process> for kilonovas in which a binary neutron star collision ejects high-neutron-density matter which decompresses pretty spectacularly, forming lots of heavy elements. To summarize, I think the free neutron decay timescale (mean lifetime ~ 15 minutes, multiply by ln 2 if you prefer half-life) is simply too long after the neutron star material is teleported to Earth: any free neutrons that haven't been absorbed into heavy nuclei likely will be millions of kilometres away from ground zero when they decay.
  - ben_w 3y ago
    > I think most attempts to arrive at an answer will end up somewhere between half and virtually all of them being "not very close" (~ light-minutes) away Mean free path of free neutrons moving past normal matter is only in the order of centimetres, exactly how many centimetres depends on the neutron energy and the specific nuclei it's interacting with, but still order of centimetres. Given the relative masses, I can assume the air above will be exploded out of the way; but the half going down will have all of the earth as a moderator… and also serve as a neutron-absorbing backstop that will probably increase the actual yield. I'm also ignoring any binding energy between the neutrons. I'm basically treating them as disconnected from the first moment, which may be a terrible idea, but AFAIK nobody actually knows how long a macroscopic combination of this scale would remain stable for.
    - raattgift 3y ago
      I don't know enough about neutron physics to comment usefully on your mean free path logic, but I do know that solar eruptive activity can launch relativistic neutrons at Earth which can be detected even at sea level using scintillators, and that mountaintop detection has been around since the early 1980s. Shibata 1994, Propagation of Solar Neutrons <https://sci-hub.se/https://doi.org/10.1029/93JA03175 https://sci-hub.se/https://doi.org/10.1029/93JA03175>, §4.2.1 (Fig 3) higher energy neutrons get further into the atmosphere, so I don't think the atmosphere is much of a barrier for the comparable (MeV-GeV) teleported neutron-star neutrons. We seem to agree that free neutrons don't stay free neutrons when they slam into the solid earth. I too wanted to think about neutrons as a non-self-interacting gas, but that just doesn't work: Meyer 1994, https://ned.ipac.caltech.edu/level5/Sept01/Meyer/Meyer3.html https://ned.ipac.caltech.edu/level5/Sept01/Meyer/Meyer3.html (Paragraph beginning with, "Only the strong gravity of the neutron star keeps such matter from exploding apart." Cold in this context is partly explained in the preceding paragraph; in inner regions the matter is a degenerate gas meaning the particle kinetic energy becomes dependent on the density or equivalently pressure becomes independent of temperature; even at enormous pressures, degenerate gases don't hold much thermal energy -- that was practically all radiated away when the NS was young. Our teleporting (of inner region matter) therefore engages a very low-entropy r-process. Outer regions are just too complicated and varied for a HN comment. The crust is thin -- a few to a few hundred metres or so compared to an NS radius of ~ 10km. It's also much less dense, so is a small fraction of the NS mass, and thus maybe not a target for our teleportation. Here's a 180-page open access review: https://link.springer.com/article/10.12942/lrr-2008-10 https://link.springer.com/article/10.12942/lrr-2008-10 Pesky electrons and protons complicating things.
Eddy_Viscosity2 3y ago
I've heard that black hole matter is stable at any size, so how much difference in density is there between neutron star matter and black hole matter where it crosses the threshold of stability from its own gravity?
- deleted 3y ago
  [deleted]
- johndunne 3y ago
  Black hole's aren't matter, they're pure gravitational binding energy. A neutron star becomes a black hole when the neutrons pushing against each other can't push back at the gravitational forces (neutron degeneracy pressure) and the neutrons do something we're not sure of... but whatever happens, they're crushed down into something smaller than a neutron star; into a singuality and we see the result.. a black hole. Eternal darkness for the poor neutrons; this bit gives me chills.
  - AnimalMuppet 3y ago
    > Black hole's aren't matter, they're pure gravitational binding energy. Could you expand on this a bit? What exactly do you mean, and what is your basis for saying that it's true?
    - ben_w 3y ago
      In GR, black holes have only three distinguishable properties: mass, charge, and angular momentum. If you have one made from matter, one from antimatter, and one from sufficiently concentrated light, all three are indistinguishable. As I'm not a physicist, I wouldn't risk phrasing this as "pure gravitational binding energy" just in case this has a specific and different meaning. I read the interior of an event horizon immediately causes problems with quantum mechanics' no-cloning rule, so I suspect the actual problem here is "QM and GR are fighting again" and we can't get any answer until we've resolved that.
    - johndunne 3y ago
euroderf 3y ago
35th-century science fair project gone very, very wrong.

ben_w 3y ago

Very large asteroid, too big for a nuke: https://www.wolframalpha.com/input?i=%280.782343+MeV+%2F+neutron+mass%29+*+5e12+kg+in+megatons+of+TNT+equivalent+ https://www.wolframalpha.com/input?i=%280.782343+MeV+%2F+neu... ~89 billion megatons of TNT equivalent

raattgift 3y ago
What motivates your first factor? 0.782343 MeV is the free neutron beta decay; where in the solar system are the free neutrons minutes after they are magically teleported to terrestrial ground zero as a something like a (degenerate, possibly ultra-relativistic) Fermi gas? I think most attempts to arrive at an answer will end up somewhere between half and virtually all of them being "not very close" (~ light-minutes) away, and that's assuming one corrects for the differences in escape velocities. (The equatorial escape velocity of a spinning neutron star is in tenths of the speed of light, thus the sobriquet "relativistic star"). Without this correction, it is likely the bulk of the expanding drop of Fermi gas just exits the atmosphere in milliseconds (timed by terrestrial stopwatches), with time dilation extending the mean lifetime of the free neutrons in the drop comparably to the extended lifetime of atmospheric muons from cosmic rays. The bulk of the beta decays happen at a distance from terrestrial ground zero best measured in astronomical units. If we play Star Trek transporter games such that the neutrons arrive at ground zero at rest in local East-North-Up coordinates, you'd want to know the internal kinetic energy (KE) density of the (pure-)neutron star, which will be in the range of 20-40 for x in 10^{x} J m^-3. The 10^25ish or even 10^30ish joules of KE will be released from our several cm^3 spoonful practically all at once and practically omnidirectionally from ground zero (so again, most free neutron decays happen at ~ AU distances from ground zero because they'll zip right through the atmosphere). The expansion of the suddenly unpressurized gas of neutrons will make a mess, particularly the fraction that slams into and through the ground. Part of the mess is neutron scattering physics, and I have no expertise there, but I would guess there wouldn't be any free neutrons near ground zero (and probably not within the solid Earth) in ~minutes. Additionally, one might compare the R-process <https://en.wikipedia.org/wiki/R-process https://en.wikipedia.org/wiki/R-process> for kilonovas in which a binary neutron star collision ejects high-neutron-density matter which decompresses pretty spectacularly, forming lots of heavy elements. To summarize, I think the free neutron decay timescale (mean lifetime ~ 15 minutes, multiply by ln 2 if you prefer half-life) is simply too long after the neutron star material is teleported to Earth: any free neutrons that haven't been absorbed into heavy nuclei likely will be millions of kilometres away from ground zero when they decay.
- ben_w 3y ago
  > I think most attempts to arrive at an answer will end up somewhere between half and virtually all of them being "not very close" (~ light-minutes) away Mean free path of free neutrons moving past normal matter is only in the order of centimetres, exactly how many centimetres depends on the neutron energy and the specific nuclei it's interacting with, but still order of centimetres. Given the relative masses, I can assume the air above will be exploded out of the way; but the half going down will have all of the earth as a moderator… and also serve as a neutron-absorbing backstop that will probably increase the actual yield. I'm also ignoring any binding energy between the neutrons. I'm basically treating them as disconnected from the first moment, which may be a terrible idea, but AFAIK nobody actually knows how long a macroscopic combination of this scale would remain stable for.
  - raattgift 3y ago
    I don't know enough about neutron physics to comment usefully on your mean free path logic, but I do know that solar eruptive activity can launch relativistic neutrons at Earth which can be detected even at sea level using scintillators, and that mountaintop detection has been around since the early 1980s. Shibata 1994, Propagation of Solar Neutrons <https://sci-hub.se/https://doi.org/10.1029/93JA03175 https://sci-hub.se/https://doi.org/10.1029/93JA03175>, §4.2.1 (Fig 3) higher energy neutrons get further into the atmosphere, so I don't think the atmosphere is much of a barrier for the comparable (MeV-GeV) teleported neutron-star neutrons. We seem to agree that free neutrons don't stay free neutrons when they slam into the solid earth. I too wanted to think about neutrons as a non-self-interacting gas, but that just doesn't work: Meyer 1994, https://ned.ipac.caltech.edu/level5/Sept01/Meyer/Meyer3.html https://ned.ipac.caltech.edu/level5/Sept01/Meyer/Meyer3.html (Paragraph beginning with, "Only the strong gravity of the neutron star keeps such matter from exploding apart." Cold in this context is partly explained in the preceding paragraph; in inner regions the matter is a degenerate gas meaning the particle kinetic energy becomes dependent on the density or equivalently pressure becomes independent of temperature; even at enormous pressures, degenerate gases don't hold much thermal energy -- that was practically all radiated away when the NS was young. Our teleporting (of inner region matter) therefore engages a very low-entropy r-process. Outer regions are just too complicated and varied for a HN comment. The crust is thin -- a few to a few hundred metres or so compared to an NS radius of ~ 10km. It's also much less dense, so is a small fraction of the NS mass, and thus maybe not a target for our teleportation. Here's a 180-page open access review: https://link.springer.com/article/10.12942/lrr-2008-10 https://link.springer.com/article/10.12942/lrr-2008-10 Pesky electrons and protons complicating things.

Eddy_Viscosity2 3y ago

I've heard that black hole matter is stable at any size, so how much difference in density is there between neutron star matter and black hole matter where it crosses the threshold of stability from its own gravity?

deleted 3y ago
[deleted]
johndunne 3y ago
Black hole's aren't matter, they're pure gravitational binding energy. A neutron star becomes a black hole when the neutrons pushing against each other can't push back at the gravitational forces (neutron degeneracy pressure) and the neutrons do something we're not sure of... but whatever happens, they're crushed down into something smaller than a neutron star; into a singuality and we see the result.. a black hole. Eternal darkness for the poor neutrons; this bit gives me chills.
- AnimalMuppet 3y ago
  > Black hole's aren't matter, they're pure gravitational binding energy. Could you expand on this a bit? What exactly do you mean, and what is your basis for saying that it's true?
  - ben_w 3y ago
    In GR, black holes have only three distinguishable properties: mass, charge, and angular momentum. If you have one made from matter, one from antimatter, and one from sufficiently concentrated light, all three are indistinguishable. As I'm not a physicist, I wouldn't risk phrasing this as "pure gravitational binding energy" just in case this has a specific and different meaning. I read the interior of an event horizon immediately causes problems with quantum mechanics' no-cloning rule, so I suspect the actual problem here is "QM and GR are fighting again" and we can't get any answer until we've resolved that.
  - johndunne 3y ago

euroderf 3y ago

35th-century science fair project gone very, very wrong.