# Different Masses of Pions

• ahaanomegas

#### ahaanomegas

The charged pion particles ($\pi^+$ and $\pi^-$) have different masses from the neutral pion particle ($\pi^0$). Why? It's not like the pions are nuclei that they have binding energy.

Actually it IS all about the binding energy. They are made of a pair of quarks bound together by the strong nuclear force. I am not sure what makes the neutral pion the lighter one though.

The pi0 has a greater mass than the pi+ or pi- due to the electromagnetic interaction.
The quark and anti-quark in the pion have opposite charges in the pi0, but the same sign of charge in the charged pions.

Pion mass is generated by electromagnetism and quark masses.

In QCD with massless quarks the pions would be Goldstone modes of spontaneously broken chiral SU(2) isospin symmetry and would therefore be exactly massless (plus small electromagnetic corrections).

By the way, I find hard to believe that the mass of the pion is going to be zero when d and u go to zero. They are already almost zero, and the mass of the pion is still one hundred MeVs, above the muon mass. Moreover, the argument needs f_pi, and once the pion mass goes under the muon mass, it becomes stable (if m_e=m_d=m_u=0).

My unproven opinion is that when the masses of the quarks go to zero, the mass of the pion go down until it becomes equal to the mass of the muon.

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.
Agreed!

The only open question is the order of magnitude of the electromagnetic correction. The el.-mag. interaction is usually ignored when considering isospin and chiral symmetry breaking.

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.

¿also with a massless electron?

It is only that I find very strange to have a f_pi that I can not measure.

Imagine other limit: m_e=m_u=0, m_d=(3 MeV + 6 MeV) * 105/140..

(assuming that 3 and 6 MeV are the real world quark masses, the 105 is mass of muon, the 140 mass of the real world pion).

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In the limit of massless quarks the pion is a Goldstone boson and must also be massless. This has been confirmed in lattice QCD.

Although I have been told this many times, I still don't get it. If the pion mass would be zero, does not this imply that the binding energy of the strong interaction is also zero? This is where the mass comes from after all. So would pions even be bound particles in such circumstances? QCD confuses me. 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. Or rather I suppose it implies that free quark masses are strictly infinite and it is the QCD binding energy which lowers the mass of the bound state to the observed number. So ok it could happen that the binding energy exactly takes the mass of the bound state to zero by some symmetry magic.
Perhaps someone can chime in with a more coherent story.

If the pion mass would be zero, does not this imply that the binding energy of the strong interaction is also zero?
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 like

$$3 m_\text{pion} \simeq 2 m_\text{proton} \simeq 6 m_\text{quark}$$

which is wrong! The mass is due to "field energy".

So would pions even be bound particles in such circumstances?
Yes.

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.
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).

Quarks + gluons form a highly non-perturbative field configuration. The energy of the bound states (e.g. proton) is stored in the field. Even this is somehow missleading as it's not a classical field, therefore it should be taken with care.