6 ms·

"Falling into the nucleus", as stated in that Stack Exchange question, is a classical (as in "wrong" :) ) way to look at it. The 1s orbital of the electron is

by euyyn 10y ago

"Falling into the nucleus", as stated in that Stack Exchange question, is a classical (as in "wrong" :) ) way to look at it. The 1s orbital of the electron is spherically symmetric, with non-zero probability of the electron being right in the center of the atom (so within the nucleus), and decreasing probability of it being farther away. So e.g. a good approximation of the He atom is of two neutrons, two protons, and two electrons, all in exactly the same position (with the electrons having "more spread"). At the end of the day, there are interactions that are seen in Nature and interactions that aren't. We codify the patterns as conservation of lepton number, baryon number, etc. If the numbers of what you have add up to zero, they can annihilate. These conservation laws also accommodate other weird things, like beta decay, in which a neutron decays ("splitting") into a proton, an electron, and an anti-neutrino. It's not that the neutron is "composed" of those three things. But its total energy is higher than the sum of the energy at rest of those three particles: Whenever that's the case and the conservation laws allow it, we see it happen naturally.

paulmd 10y ago
> At the end of the day, there are interactions that are seen in Nature and interactions that aren't. That's provably false. Every reaction occurs in nature, even the vanishingly improbable ones, just at an extremely low reaction coefficient. They're still there, just at a probability of 10^-14 or whatever.
- pranjalv123 10y ago
  What? There are interactions that can never occur because they're forbidden by the laws of physics - you'll never violate conservation of [energy, baryon number, momentum, etc.] even with probability 10^-14.
  - paulmd 10y ago
    Sure, technically correct is the best kind of correct I suppose. But if the reaction is the least bit reversible (and most are) then the reverse reaction is still proceeding even though the forward reaction is stronger. It's just doing so at an incredibly low rate, and the products are likely to be immediately converted by the forward reaction. You are looking at averages and claiming they hold for every single event in a stochastic simulation. Taking the example of entropy - it's not impossible that entropy decreases in a system, it's just less likely than it increasing. On a stochastic level, entropy decreases all the time, it's just that on average it increases more than it decreases. Reactions work the same way. You're not actually making chemicals react as a singular act, you're creating a forward reaction that occurs more rapidly than the reverse reaction. If we had a hypothetical chemical Maxwell's Demon - you could "bottle" up the tiny bits of those outputs from the reverse reaction before they underwent the forward reaction again. https://en.wikipedia.org/wiki/Maxwell's_demon https://en.wikipedia.org/wiki/Maxwell's_demon
    - euyyn 10y ago
      You're assuming one can expect the knowledge of chemical reactions to apply to particle interactions, and that is not the case. For example particle accelerators can shoot particles against some target one by one, and so there's no need to look at averages. Some interactions never occur, to the best of our measuring ability (i.e. highest energies and number of repetitions). Some others are right away forbidden by laws much stronger than entropy in thermodynamics, in the sense that they're not averages, but mathematical derivations off the symmetries of the universe.
- tokenizerrr 10y ago
  I'm very interested in seeing that proof.
  - deleted 10y ago
    [deleted]
  - paulmd 10y ago
    > The concentrations of reactants and products in an equilibrium mixture are determined by the analytical concentrations of the reagents (A and B or C and D) and the equilibrium constant, K. The magnitude of the equilibrium constant depends on the Gibbs free energy change for the reaction.[2] So, when the free energy change is large (more than about 30 kJ mol−1), then the equilibrium constant is large (log K > 3) and the concentrations of the reactants at equilibrium are very small. Such a reaction is sometimes considered to be an irreversible reaction, although in reality small amounts of the reactants are still expected to be present in the reacting system. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, for example, as does carbon dioxide (volatile) in the reaction https://en.wikipedia.org/wiki/Reversible_reaction https://en.wikipedia.org/wiki/Reversible_reaction Assuming you do not outright lose some reactants from the system, the reverse reaction is still occurring. However, because the reaction constant is so small, the resulting product is highly probable to essentially immediately undergo the forward reaction. However, it will still be present in some equilibrium - just an incredibly low one, in the ratio of the reaction rate coefficients. You'll have 10^14 times as much of the forward reaction, or whatever.
    - euyyn 10y ago
      Chemical reactions, ignoring those with nuclear interactions, are only bound by the law of conservation of energy and the decrease of entropy. A chemist looks at them in the macro scale, in which macroscopic properties like temperature and pressure are set, and thus each molecule's energy isn't a fixed value (instead being taken off a set, according to Maxwell's probability distribution). This, and the probabilistic nature of the law of entropy, leads to all allowed reactions happening somewhere at the micro scale, even if extremely rarely. If you look at the micro scale, though, and instead of having a soup of molecules you deal with single molecules, whether a reaction is possible or not isn't a probabilistic thing anymore. The energies your molecules have are actual numbers, and if they don't add up, you won't have a reaction.
neikos 10y ago
As someone who hasn't studied this (yet) does that mean that the reason protons and electrons don't annihilate is because their Baryon Number wouldn't sum up to zero? (With Protons having 1 and Electrons 0 if I understood correctly)
- frumiousirc 10y ago
  Neutron has a baryon number of 1 and so it is conserved in your example. Even in normal matter, some elements undergo Electron Capture (p + e- -> n + ν). But baryon number conservation is an asserted symmetry and there is no fundamental reason it holds. Finding proton decay would demonstrate that it does not but so far there is no evidence that protons (or bound neutrons) spontaneously decay.
- euyyn 10y ago
  That's what we currently observe, yes. The particles of matter in the Standard Model are quarks (that join to form protons and neutrons among other things) and leptons (electrons, muons, taus, and neutrinos). The number of {quarks - antiquarks}, as well as that of {leptons - antileptons}, is conserved. If you start with a proton and an electron, you have 3 quarks and 1 lepton, so whatever the interaction you have to end up with 3 quarks and 1 lepton (plus any number of quark-antiquark and lepton-antilepton pairs). E.g. for beta decay: neutron (3 quarks) -> proton (3 quarks) + electron (lepton) + antineutrino (antilepton)

by euyyn 10y ago

paulmd 10y ago

> At the end of the day, there are interactions that are seen in Nature and interactions that aren't. That's provably false. Every reaction occurs in nature, even the vanishingly improbable ones, just at an extremely low reaction coefficient. They're still there, just at a probability of 10^-14 or whatever.

pranjalv123 10y ago
What? There are interactions that can never occur because they're forbidden by the laws of physics - you'll never violate conservation of [energy, baryon number, momentum, etc.] even with probability 10^-14.
- paulmd 10y ago
  Sure, technically correct is the best kind of correct I suppose. But if the reaction is the least bit reversible (and most are) then the reverse reaction is still proceeding even though the forward reaction is stronger. It's just doing so at an incredibly low rate, and the products are likely to be immediately converted by the forward reaction. You are looking at averages and claiming they hold for every single event in a stochastic simulation. Taking the example of entropy - it's not impossible that entropy decreases in a system, it's just less likely than it increasing. On a stochastic level, entropy decreases all the time, it's just that on average it increases more than it decreases. Reactions work the same way. You're not actually making chemicals react as a singular act, you're creating a forward reaction that occurs more rapidly than the reverse reaction. If we had a hypothetical chemical Maxwell's Demon - you could "bottle" up the tiny bits of those outputs from the reverse reaction before they underwent the forward reaction again. https://en.wikipedia.org/wiki/Maxwell's_demon https://en.wikipedia.org/wiki/Maxwell's_demon
  - euyyn 10y ago
    You're assuming one can expect the knowledge of chemical reactions to apply to particle interactions, and that is not the case. For example particle accelerators can shoot particles against some target one by one, and so there's no need to look at averages. Some interactions never occur, to the best of our measuring ability (i.e. highest energies and number of repetitions). Some others are right away forbidden by laws much stronger than entropy in thermodynamics, in the sense that they're not averages, but mathematical derivations off the symmetries of the universe.
tokenizerrr 10y ago
I'm very interested in seeing that proof.
- deleted 10y ago
  [deleted]
- paulmd 10y ago
  > The concentrations of reactants and products in an equilibrium mixture are determined by the analytical concentrations of the reagents (A and B or C and D) and the equilibrium constant, K. The magnitude of the equilibrium constant depends on the Gibbs free energy change for the reaction.[2] So, when the free energy change is large (more than about 30 kJ mol−1), then the equilibrium constant is large (log K > 3) and the concentrations of the reactants at equilibrium are very small. Such a reaction is sometimes considered to be an irreversible reaction, although in reality small amounts of the reactants are still expected to be present in the reacting system. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, for example, as does carbon dioxide (volatile) in the reaction https://en.wikipedia.org/wiki/Reversible_reaction https://en.wikipedia.org/wiki/Reversible_reaction Assuming you do not outright lose some reactants from the system, the reverse reaction is still occurring. However, because the reaction constant is so small, the resulting product is highly probable to essentially immediately undergo the forward reaction. However, it will still be present in some equilibrium - just an incredibly low one, in the ratio of the reaction rate coefficients. You'll have 10^14 times as much of the forward reaction, or whatever.
  - euyyn 10y ago
    Chemical reactions, ignoring those with nuclear interactions, are only bound by the law of conservation of energy and the decrease of entropy. A chemist looks at them in the macro scale, in which macroscopic properties like temperature and pressure are set, and thus each molecule's energy isn't a fixed value (instead being taken off a set, according to Maxwell's probability distribution). This, and the probabilistic nature of the law of entropy, leads to all allowed reactions happening somewhere at the micro scale, even if extremely rarely. If you look at the micro scale, though, and instead of having a soup of molecules you deal with single molecules, whether a reaction is possible or not isn't a probabilistic thing anymore. The energies your molecules have are actual numbers, and if they don't add up, you won't have a reaction.

neikos 10y ago

As someone who hasn't studied this (yet) does that mean that the reason protons and electrons don't annihilate is because their Baryon Number wouldn't sum up to zero? (With Protons having 1 and Electrons 0 if I understood correctly)

frumiousirc 10y ago
Neutron has a baryon number of 1 and so it is conserved in your example. Even in normal matter, some elements undergo Electron Capture (p + e- -> n + ν). But baryon number conservation is an asserted symmetry and there is no fundamental reason it holds. Finding proton decay would demonstrate that it does not but so far there is no evidence that protons (or bound neutrons) spontaneously decay.
euyyn 10y ago
That's what we currently observe, yes. The particles of matter in the Standard Model are quarks (that join to form protons and neutrons among other things) and leptons (electrons, muons, taus, and neutrinos). The number of {quarks - antiquarks}, as well as that of {leptons - antileptons}, is conserved. If you start with a proton and an electron, you have 3 quarks and 1 lepton, so whatever the interaction you have to end up with 3 quarks and 1 lepton (plus any number of quark-antiquark and lepton-antilepton pairs). E.g. for beta decay: neutron (3 quarks) -> proton (3 quarks) + electron (lepton) + antineutrino (antilepton)