5 ms·

> Are we still wrongly talking about billiard balls? No. Particles are (approximately-)localized excitations in the corresponding field. Think waves, not wate

> Are we still wrongly talking about billiard balls? No. Particles are (approximately-)localized excitations in the corresponding field. Think waves, not water balloons.

hinkley 4y ago
When two waves interact it's more like an elastic collision, isn't it? From the article it wasn't clear to me if these extra high-energy particles they were seeing as fuzz in the data (which are heavier than a photon) are actually unexplained mass or a situation of conservation of energy meets special relativity (kinetic energy -> mass). If you put enough energy into separating quarks, I'm told you get extra quarks. So an energetic system where the masses don't add up doesn't seem like an epoch defining mystery to me. So what are we missing?
- hither_shores 4y ago
  > So an energetic system where the masses don't add up doesn't seem like an epoch defining mystery to me. Bound states aren't really made of their constituents in a classical sense. A proton is a particular configuration of the quark fields (really it's more complicated than this), but not a simple sum of quark particle states. And in particular, its mass doesn't have to be the sum of the masses of particle states.
  - hinkley 4y ago
    The mass+energy still have to sum up though, don't they?
    - hither_shores 4y ago
      The mass of what? There aren't literally three quarks inside a proton.
      - hinkley 4y ago
        Circling back to my original question, is the system exhibiting behavior that does not agree with special relativity (vs Newtonian physics)?
        hither_shores 4y ago
        With special-relativistic classical mechanics? Of course. With special-relativistic quantum mechanics? No.
bowsamic 4y ago
Actually the basic particle created by the creation operator for a mode is totally delocalised. Applying the field operator to get a localised particle is really an integral over many particles
- hither_shores 4y ago
  For a free field, yes. But the entities we actually interact with are ... interacting. Mode expansion doesn't work here. Electrons, photons, etc. aren't really "particles" in the sense of ordinary quantum mechanics at all.

> Are we still wrongly talking about billiard balls? No. Particles are (approximately-)localized excitations in the corresponding field. Think waves, not water balloons.

hinkley 4y ago

When two waves interact it's more like an elastic collision, isn't it? From the article it wasn't clear to me if these extra high-energy particles they were seeing as fuzz in the data (which are heavier than a photon) are actually unexplained mass or a situation of conservation of energy meets special relativity (kinetic energy -> mass). If you put enough energy into separating quarks, I'm told you get extra quarks. So an energetic system where the masses don't add up doesn't seem like an epoch defining mystery to me. So what are we missing?

hither_shores 4y ago
> So an energetic system where the masses don't add up doesn't seem like an epoch defining mystery to me. Bound states aren't really made of their constituents in a classical sense. A proton is a particular configuration of the quark fields (really it's more complicated than this), but not a simple sum of quark particle states. And in particular, its mass doesn't have to be the sum of the masses of particle states.
- hinkley 4y ago
  The mass+energy still have to sum up though, don't they?
  - hither_shores 4y ago
    The mass of what? There aren't literally three quarks inside a proton.
    - hinkley 4y ago
      Circling back to my original question, is the system exhibiting behavior that does not agree with special relativity (vs Newtonian physics)?
      - hither_shores 4y ago
        With special-relativistic classical mechanics? Of course. With special-relativistic quantum mechanics? No.

bowsamic 4y ago

Actually the basic particle created by the creation operator for a mode is totally delocalised. Applying the field operator to get a localised particle is really an integral over many particles

hither_shores 4y ago
For a free field, yes. But the entities we actually interact with are ... interacting. Mode expansion doesn't work here. Electrons, photons, etc. aren't really "particles" in the sense of ordinary quantum mechanics at all.