3 ms·

> If I understand this correctly, the "right hand" means antiparticles? Do I have that right? No, it's about chirality: https://en.wikipedia.org/wiki/Chiralit

by T-A 5y ago

> If I understand this correctly, the "right hand" means antiparticles? Do I have that right? No, it's about chirality: https://en.wikipedia.org/wiki/Chirality_(physics) https://en.wikipedia.org/wiki/Chirality_(physics)

AnimalMuppet 5y ago
OK, then: Why would chirality determine whether a particle can interact with the weak force? (ELI5, probably, or at least ELI20, if you can.)
- SaberTail 5y ago
  This is more of an empirical thing. We observe effects such as parity violation that can be best explained by an interaction that only works on left-handed particles.
  - goatlover 5y ago
    So the theoretical thing would be to account for the observed effects.
- T-A 5y ago
  Ultimately it's an experimental fact - a very surprising one when discovered: https://en.wikipedia.org/wiki/Wu_experiment https://en.wikipedia.org/wiki/Wu_experiment In theoretical terms, the equation used to describe all massive fermions known at the time (like the electron) is built out of four-component quantities called Dirac spinors: https://en.wikipedia.org/wiki/Dirac_spinor https://en.wikipedia.org/wiki/Dirac_spinor A Dirac spinor can be viewed as composed of two simpler, two-component quantities called Weyl spinors: https://en.wikipedia.org/wiki/Weyl_equation#Weyl_spinors https://en.wikipedia.org/wiki/Weyl_equation#Weyl_spinors You can try to describe a fermion using only one Weyl spinor, but then it turns out that you can't build mass terms (unless you're willing to violate special relativity). The massless equation you can write down with a Weyl spinor has two plane wave solutions with opposite helicity, left and right. A Dirac spinor combines a left-handed and a right-handed Weyl spinor; the mass term of the Dirac equation "mixes" them, in the sense that if you start out with a Dirac spinor having its left-handed component set to 0, the mass term will cause it to grow at a rate proportional to mass. If you set the mass to exactly zero, you're left with two uncoupled Weyl equations, one for the left-handed component, one for the right-handed one. Knowing this, and faced with the experimental fact that weak interactions make a distinction between left and right, you write down separate weak interaction terms for the left- and right-handed components of your Dirac spinors. Eventually it turns out that the simplest choice, having only the left-handed components participate in those interactions, is the best fit to experiment: https://en.wikipedia.org/wiki/Weak_isospin#Relation_with_chirality https://en.wikipedia.org/wiki/Weak_isospin#Relation_with_chi... ¯\_(ツ)_/¯
  - AnimalMuppet 5y ago
    Thank you for a clear, detailed response. (It would be more clear if I understood the math behind spinors...) So, returning to a previous question: If I understand this correctly, it means that if we have a kind of nucleus that undergoes beta decay, and we built the exact same nucleus except out of antimatter, it would not undergo beta decay. Is that correct? Has anyone actually done that experiment? Or is it beyond our ability to construct things out of antimatter?
    - T-A 5y ago
      > if we have a kind of nucleus that undergoes beta decay, and we built the exact same nucleus except out of antimatter, it would not undergo beta decay. Is that correct? No. Until 1964 you would have been told that applying CP, i.e. the combination of C transformation (charge conjugation, i.e. change the signs of all charges) and P transformation (parity, i.e. swap left and right) would result in exactly the same decay rate. You would get anti-neutrinos (which are right-handed) instead of neutrinos, but that would be the only difference. Then it turned out that CP symmetry is also violated by weak interactions, though far from as neatly as just P: https://en.wikipedia.org/wiki/CP_violation https://en.wikipedia.org/wiki/CP_violation
      - AnimalMuppet 5y ago
        Then I don't understand something. (Or several somethings...) I thought you said that going to antimatter meant that you reversed chirality, and that right-handed chirality meant that a particle did not take part in the weak interaction. I interpreted that to mean that it didn't take part at all, not with any probability. But the CP violation article seems to be saying that CP violation is a matter of differences in probabilities. By the way, that article talks about anti-neutrinos, which from your initial post, I thought you were saying were impossible to detect if they existed? I'm confused... can you make any of this clearer? I'm mis-understanding something - can you tell what?
        T-A 5y ago

by T-A 5y ago

AnimalMuppet 5y ago

OK, then: Why would chirality determine whether a particle can interact with the weak force? (ELI5, probably, or at least ELI20, if you can.)

SaberTail 5y ago
This is more of an empirical thing. We observe effects such as parity violation that can be best explained by an interaction that only works on left-handed particles.
- goatlover 5y ago
  So the theoretical thing would be to account for the observed effects.
T-A 5y ago
Ultimately it's an experimental fact - a very surprising one when discovered: https://en.wikipedia.org/wiki/Wu_experiment https://en.wikipedia.org/wiki/Wu_experiment In theoretical terms, the equation used to describe all massive fermions known at the time (like the electron) is built out of four-component quantities called Dirac spinors: https://en.wikipedia.org/wiki/Dirac_spinor https://en.wikipedia.org/wiki/Dirac_spinor A Dirac spinor can be viewed as composed of two simpler, two-component quantities called Weyl spinors: https://en.wikipedia.org/wiki/Weyl_equation#Weyl_spinors https://en.wikipedia.org/wiki/Weyl_equation#Weyl_spinors You can try to describe a fermion using only one Weyl spinor, but then it turns out that you can't build mass terms (unless you're willing to violate special relativity). The massless equation you can write down with a Weyl spinor has two plane wave solutions with opposite helicity, left and right. A Dirac spinor combines a left-handed and a right-handed Weyl spinor; the mass term of the Dirac equation "mixes" them, in the sense that if you start out with a Dirac spinor having its left-handed component set to 0, the mass term will cause it to grow at a rate proportional to mass. If you set the mass to exactly zero, you're left with two uncoupled Weyl equations, one for the left-handed component, one for the right-handed one. Knowing this, and faced with the experimental fact that weak interactions make a distinction between left and right, you write down separate weak interaction terms for the left- and right-handed components of your Dirac spinors. Eventually it turns out that the simplest choice, having only the left-handed components participate in those interactions, is the best fit to experiment: https://en.wikipedia.org/wiki/Weak_isospin#Relation_with_chirality https://en.wikipedia.org/wiki/Weak_isospin#Relation_with_chi... ¯\_(ツ)_/¯
- AnimalMuppet 5y ago
  Thank you for a clear, detailed response. (It would be more clear if I understood the math behind spinors...) So, returning to a previous question: If I understand this correctly, it means that if we have a kind of nucleus that undergoes beta decay, and we built the exact same nucleus except out of antimatter, it would not undergo beta decay. Is that correct? Has anyone actually done that experiment? Or is it beyond our ability to construct things out of antimatter?
  - T-A 5y ago
    > if we have a kind of nucleus that undergoes beta decay, and we built the exact same nucleus except out of antimatter, it would not undergo beta decay. Is that correct? No. Until 1964 you would have been told that applying CP, i.e. the combination of C transformation (charge conjugation, i.e. change the signs of all charges) and P transformation (parity, i.e. swap left and right) would result in exactly the same decay rate. You would get anti-neutrinos (which are right-handed) instead of neutrinos, but that would be the only difference. Then it turned out that CP symmetry is also violated by weak interactions, though far from as neatly as just P: https://en.wikipedia.org/wiki/CP_violation https://en.wikipedia.org/wiki/CP_violation
    - AnimalMuppet 5y ago
      Then I don't understand something. (Or several somethings...) I thought you said that going to antimatter meant that you reversed chirality, and that right-handed chirality meant that a particle did not take part in the weak interaction. I interpreted that to mean that it didn't take part at all, not with any probability. But the CP violation article seems to be saying that CP violation is a matter of differences in probabilities. By the way, that article talks about anti-neutrinos, which from your initial post, I thought you were saying were impossible to detect if they existed? I'm confused... can you make any of this clearer? I'm mis-understanding something - can you tell what?
      - T-A 5y ago