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- Thread starter rkrsnan
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Vanadium 50

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Baryons would not be massless.

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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?

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

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Vanadium 50

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

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A theory with massless particles is scale invariant.

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.

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