The charged pion particles ([itex]\pi^+[/itex] and [itex]\pi^-[/itex]) have different masses from the neutral pion particle ([itex]\pi^0[/itex]). Why? It's not like the pions are nuclei that they have binding energy.
Agreed!In the limit of massless quarks the pion is a Goldstone boson and must also be massless. This has been confirmed in lattice QCD.
In the limit of massless quarks the pion is a Goldstone boson and must also be massless. This has been confirmed in lattice QCD. If the pion had a mass less than the muon, it would be perfectly happy decaying into an electron instead.
In the limit of massless quarks the pion is a Goldstone boson and must also be massless. This has been confirmed in lattice QCD.
No. The mass generated by QCD is dominated by non-perturbative effects, not by the mass of the individual particles. If quark mass would be relevant you would have something likeIf the pion mass would be zero, does not this imply that the binding energy of the strong interaction is also zero?
Yes.So would pions even be bound particles in such circumstances?
Binding energy is no reasonable concept in QCD b/c it applies only if the bound state is approximately 'the sum of its pieces', e.g. 'deuteron = proton + neutron + small el.-mag. field corrections', but in QCD the individual pieces do not exist in isolation (color confinement) and the bound state is by no means approximately the sum of three quarks (even if this was the original idea in the naive quark model).I suppose strictly the binding energy of these particles is infinite (since it takes infinite energy to rip the quarks apart) and that doesn't mean they should have infinite mass.