6 ms·

That the total mass of neutrinos is commensurate with / exceeds all the hadronic mass is rarely mentioned. With so many interactions emitting neutrinos, does ev

by ncmncm 5y ago

That the total mass of neutrinos is commensurate with / exceeds all the hadronic mass is rarely mentioned. With so many interactions emitting neutrinos, does everything end up as neutrinos, finally (neglecting expansion)? Or does something consume neutrinos and yield net hadronic stuff?

PaulHoule 5y ago
Particles like protons and electrons can never turn entirely to neutrinos because electric charge is conserved. Protons and bound neutrons seem to be the only stable configuration of quarks & gluons, ultimately there is quark charge that is conserved that prevents quarks from going away unless protons really do decay... which would have to happen at a very low rate if it does.
- dataflow 5y ago
  As a layman I never understood how the stability/decay of a neutron could be worse when it is unbound than bound. Does it even make sense to call a neutron a "particle" if its internal stability becomes worse when there is nothing around it? Is there any layman explanation of this that might vaguely resemble reality? The only way I can imagine this making sense is to just imagine neutrons and protons as dissolving together (liquid drop model?) rather than as being individual particles. And even then, I thought that model is inaccurate for explaining other phenomena.
  - looklookatme 5y ago
    This is a massive oversimplification and anthropomorphizes elementary particles, but hopefully satisfies your curiosity. The reason for the worsening stability is because a free neutron has more potential energy than one bound to a proton. In the same way that a muon will "preferably" decay into an electron by interacting with a W-, if you were able to have a down quark floating in space by itself (you can't, for unrelated reasons,) that down quark would "want" to decay into a much lighter up quark by exchanging a W-. In both of these cases, the elementary particle loses mass through the interaction. Inside the free neutron, that's what happens, too. A down quark interacts with a W- to transform itself into an up quark; the neutron is now a proton, with the W- carrying away some of the mass and electric charge. However, neutrons and protons are more complex than just quarks. They're also big bundles of gluons, exchanging colour charge amongst each other. These gluons have mass and you need energy to create them. They actually account for most of the mass of a neutron/proton. A bound neutron and proton are exchanging these gluons amongst each other, too. Suppose you had a bound neutron and proton. If the neutron were to decay into a proton, the electric charge of the particles (now both +1) would cause them to repel; you can imagine either of two things might happen: the protons will now move apart _or_ they'll stick together. In the former, you need kinetic energy and in the latter, you're going to need more gluons to hold the whole thing together. Both of those cases requires more energy (==mass) than the neutron would "lose" by converting to a proton. So, if you can imagine the potential energy landscape, the bound neutron is at a local minimum (lower than a free neutron)
    - dataflow 5y ago
      Thanks a lot for the explanation! I see what you're saying. It raises some follow-up questions for me, if you don't mind: Why can't the W-'s kinetic energy be smaller than whatever the neutron would lose by converting to a proton? There's no quantization preventing that here, right? My guess here is that it would not have (for the lack of a better term) sufficient "escape velocity", meaning that it would fall back into the neutron again, but wouldn't that imply (a) this is more like a dynamic equilibrium than a static one, where neutrons turn into protons and back into neutrons repeatedly, and (b) quantum uncertainty (Heisenberg, tunneling, etc.) should still mean that bound neutrons should still decay once in a while, and perhaps (c) a neutron next to a proton might randomly "swap" places once in a while if a W- from one gets pulled into the other one? (And if any of these is the case, then where do they draw the line to declare that a bound neutron is "stable"?)
      - looklookatme 5y ago
        I think I see what you're asking, but I'm not sure I can give a satisfactory (and correct) answer, here but I think you're on the right track: > Why can't the W-'s kinetic energy be smaller... That W- is a virtual particle - it can have any energy (according to some probability distribution) and you're correct that it doesn't _have_ to observe conservation of energy as long as that violation only happens for a very short period of time, which is precisely Heisenberg - there are other numbers it must conserve, such as electric charge. > (b) quantum uncertainty (Heisenberg, tunneling, etc.) should still mean that bound neutrons should still decay once in a while We do see this; this is what beta decay is. > (c) a neutron next to a proton might randomly "swap" places once in a while In general, this is correct. With limitations on what quantum numbers need to be conserved, these events are all constantly happening all the time. When writing out the equations for the system, you don't really have a term describing a proton and a term describing a neutron; you have a complicated mess of _all possible_ interactions and each elementary particle is described as a sort of "mix" of all the things it could be. So, in that sense, a bound neutron and proton is it's own thing - there's plenty of examples of these quasiparticles throughout physics and the definition of "stable" is quite application specific.
  - ncmncm 5y ago
    The neutron isn't really a particle at all, but a seething mass of quarks and gluons crawling over one another, some trying desperately to get free and the others even more vigorously clawing them back, with fur (color-charge) and blood (fractional electrostatic charge) flying.
philipov 5y ago
I believe no matter what else, everything eventually ends up as black holes.
- codedokode 5y ago
  And given that in the beginning there was a Big Bang, we can assume that this is what happens when the black hole becomes large enough.
  - PaulHoule 5y ago
    This paper has a mind-blowing idea, that topological defects like magnetic monopoles, cosmic strings, domain walls actually breed https://arxiv.org/abs/hep-th/9402115 https://arxiv.org/abs/hep-th/9402115 and far far away more of them are being created and inflating, once you far enough way you are in something that looks like our universe, somewhere out there inflation is stopping and the big bang is happening there. It's easy in my mind looking at the classical black hole picture and wondering what happens around the singularity to wonder if inflation gets re-started inside a black hole and maybe in that old picture where you can go into a rotating black hole and come out in the "asymptotically flat spacetime" it is a universe made by that mechanism. Of course I think the classical black hole picture might be "not even wrong" and you might run into more than one firewall (such as the inflation zone) on the way there.
- ChrisLomont 5y ago
  Black holes evaporate back into particles via Hawking radiation.
  - ncmncm 5y ago
    But if there is nothing but black holes, the particles end up falling back in. Take that, Hawking!
    - planck01 5y ago
      Fun idea! But space keeps on expanding. So at some point most mass will be in black holes, but every black hole will be so far from every other black hole that even light will travel forever without ever being able to reach another.

by ncmncm 5y ago

PaulHoule 5y ago

Particles like protons and electrons can never turn entirely to neutrinos because electric charge is conserved. Protons and bound neutrons seem to be the only stable configuration of quarks & gluons, ultimately there is quark charge that is conserved that prevents quarks from going away unless protons really do decay... which would have to happen at a very low rate if it does.

dataflow 5y ago
As a layman I never understood how the stability/decay of a neutron could be worse when it is unbound than bound. Does it even make sense to call a neutron a "particle" if its internal stability becomes worse when there is nothing around it? Is there any layman explanation of this that might vaguely resemble reality? The only way I can imagine this making sense is to just imagine neutrons and protons as dissolving together (liquid drop model?) rather than as being individual particles. And even then, I thought that model is inaccurate for explaining other phenomena.
- looklookatme 5y ago
  This is a massive oversimplification and anthropomorphizes elementary particles, but hopefully satisfies your curiosity. The reason for the worsening stability is because a free neutron has more potential energy than one bound to a proton. In the same way that a muon will "preferably" decay into an electron by interacting with a W-, if you were able to have a down quark floating in space by itself (you can't, for unrelated reasons,) that down quark would "want" to decay into a much lighter up quark by exchanging a W-. In both of these cases, the elementary particle loses mass through the interaction. Inside the free neutron, that's what happens, too. A down quark interacts with a W- to transform itself into an up quark; the neutron is now a proton, with the W- carrying away some of the mass and electric charge. However, neutrons and protons are more complex than just quarks. They're also big bundles of gluons, exchanging colour charge amongst each other. These gluons have mass and you need energy to create them. They actually account for most of the mass of a neutron/proton. A bound neutron and proton are exchanging these gluons amongst each other, too. Suppose you had a bound neutron and proton. If the neutron were to decay into a proton, the electric charge of the particles (now both +1) would cause them to repel; you can imagine either of two things might happen: the protons will now move apart _or_ they'll stick together. In the former, you need kinetic energy and in the latter, you're going to need more gluons to hold the whole thing together. Both of those cases requires more energy (==mass) than the neutron would "lose" by converting to a proton. So, if you can imagine the potential energy landscape, the bound neutron is at a local minimum (lower than a free neutron)
  - dataflow 5y ago
    Thanks a lot for the explanation! I see what you're saying. It raises some follow-up questions for me, if you don't mind: Why can't the W-'s kinetic energy be smaller than whatever the neutron would lose by converting to a proton? There's no quantization preventing that here, right? My guess here is that it would not have (for the lack of a better term) sufficient "escape velocity", meaning that it would fall back into the neutron again, but wouldn't that imply (a) this is more like a dynamic equilibrium than a static one, where neutrons turn into protons and back into neutrons repeatedly, and (b) quantum uncertainty (Heisenberg, tunneling, etc.) should still mean that bound neutrons should still decay once in a while, and perhaps (c) a neutron next to a proton might randomly "swap" places once in a while if a W- from one gets pulled into the other one? (And if any of these is the case, then where do they draw the line to declare that a bound neutron is "stable"?)
    - looklookatme 5y ago
      I think I see what you're asking, but I'm not sure I can give a satisfactory (and correct) answer, here but I think you're on the right track: > Why can't the W-'s kinetic energy be smaller... That W- is a virtual particle - it can have any energy (according to some probability distribution) and you're correct that it doesn't _have_ to observe conservation of energy as long as that violation only happens for a very short period of time, which is precisely Heisenberg - there are other numbers it must conserve, such as electric charge. > (b) quantum uncertainty (Heisenberg, tunneling, etc.) should still mean that bound neutrons should still decay once in a while We do see this; this is what beta decay is. > (c) a neutron next to a proton might randomly "swap" places once in a while In general, this is correct. With limitations on what quantum numbers need to be conserved, these events are all constantly happening all the time. When writing out the equations for the system, you don't really have a term describing a proton and a term describing a neutron; you have a complicated mess of _all possible_ interactions and each elementary particle is described as a sort of "mix" of all the things it could be. So, in that sense, a bound neutron and proton is it's own thing - there's plenty of examples of these quasiparticles throughout physics and the definition of "stable" is quite application specific.
- ncmncm 5y ago
  The neutron isn't really a particle at all, but a seething mass of quarks and gluons crawling over one another, some trying desperately to get free and the others even more vigorously clawing them back, with fur (color-charge) and blood (fractional electrostatic charge) flying.

philipov 5y ago

I believe no matter what else, everything eventually ends up as black holes.

codedokode 5y ago
And given that in the beginning there was a Big Bang, we can assume that this is what happens when the black hole becomes large enough.
- PaulHoule 5y ago
  This paper has a mind-blowing idea, that topological defects like magnetic monopoles, cosmic strings, domain walls actually breed https://arxiv.org/abs/hep-th/9402115 https://arxiv.org/abs/hep-th/9402115 and far far away more of them are being created and inflating, once you far enough way you are in something that looks like our universe, somewhere out there inflation is stopping and the big bang is happening there. It's easy in my mind looking at the classical black hole picture and wondering what happens around the singularity to wonder if inflation gets re-started inside a black hole and maybe in that old picture where you can go into a rotating black hole and come out in the "asymptotically flat spacetime" it is a universe made by that mechanism. Of course I think the classical black hole picture might be "not even wrong" and you might run into more than one firewall (such as the inflation zone) on the way there.
ChrisLomont 5y ago
Black holes evaporate back into particles via Hawking radiation.
- ncmncm 5y ago
  But if there is nothing but black holes, the particles end up falling back in. Take that, Hawking!
  - planck01 5y ago
    Fun idea! But space keeps on expanding. So at some point most mass will be in black holes, but every black hole will be so far from every other black hole that even light will travel forever without ever being able to reach another.