Mass of composite particles in a non abelian gauge theory.

rkrsnan

Is it possible to produce massive composite particles from a non abelian gauge theory of massless fermions? I know that if the quarks were massless, the pions will be massless too (goldstone bosons). But what about baryons? Will they be also massless? If so, can we make a general statement that composite particles from any non abelian gauge theory of massless fermions will always be massless? Is there some mathematical proof for that?

Vanadium 50

Baryons would not be massless.

rkrsnan

Thanks for the reply. So I guess, if we have only massless up and down quarks and chromodynamics, then at low energy we will have massive protons and neutrons interacting via the exchange of massless pios. A theory with massless particles is scale invariant. But if we generate massive composite particles, then the low energy theory will not be scale invariant. Is there any contradiction in that?

JustinLevy

I'm also confused at how bound states of fermions would arise.

Consider the abelian case first. Look at the ground state energy of hydrogen or positronium. It is proportional to the reduced mass of the charged particles. So if the mass goes to zero, it looks like there would be no bound state.

This also makes sense dimensionally. Given the dimensionless coupling strength alpha, the only length scale in the problem is the mass of the particles.

Where does this logic fail? Is it at least correct for the Abelion case, and the non-abelion case has a loop-hole?

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As an aside, if _all_ we had was fundamentally massless electrons, positrons, and photons ... at low energies, would the "effective" theory still give the electrons a mass? As the electron moves, the vacuum polarizes in its vicinity which must take energy. So would renormalization to low energies give the electron an effective mass anyway? Can a low energy theory actually have massless fermions?

genneth

Non-Abelian Yang-Mills theories are "asymptotically free", which means that at large energies, the coupling constant is finite. However, this also means that as you run the renormalisation towards the infrared, the coupling constant will logarithmically diverge. Around when the coupling becomes unity, perturbation theory will break down, and that will produce a new energy scale.

Frank Wilczek has some nice papers about this, such as: http://arxiv.org/abs/hep-ph/0201222

Vanadium 50

The scale is Lambda_QCD.

In QCD I can even make a bound state of entirely massless particles: glueballs.

the_house

Just to expand on what genneth Vanadium said, in case you're not familiar. It actually isn't true that a theory with only massless particles is necessarily scale invariant -- the obvious example being QCD. The QCD Lagrangian is scale invariant when the masses are set to zero, and therefore (as your intuition tells you) a classical field theory defined by this Lagrangian is scale invariant. However, the quantum theory does not actually respect this symmetry. An energy scale is dynamically generated, and we call such a scale Lambda_QCD.

You are correct that baryons couldn't have a non-zero mass in a scale invariant theory, but it turns out QCD with massless quarks isn't actually scale invariant, and there actually is an intrinsic length scale in the theory, even though it doesn't appear in the Lagrangian.