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If we try to separate two quarks bound into a meson or a hadron, the energy in the gluon field eventually will be large enough to spawn a quark-antiquark pair.

How far can we stretch that gluon field before it "snaps"? What's the average distance? I can't seem to find it anywhere.

Calmarius
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2 Answers2

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From lattice calculations (see String Tension of Quark-Anti-Quark Pairs in Lattice QCD) it has been found that the string tension of the quarks, in the case of pions, is given by $$ \sqrt{\sigma}\sim460\ \mathrm{MeV} $$ which is equivalent to a length of $\sim 2.7\ \mathrm{fermi}$.

In the case of the charmonium ($\bar c c$), the tension (see Charmonium potential from full lattice QCD) is close to $$ \sqrt{\sigma}\sim394\ \mathrm{MeV} $$ that is, a lenght of $3.1\ \mathrm{fermi}$.

For a bottomonium state ($\bar b b$), the best estimate I've found (see Bottomonium states versus recent experimental observations in the QCD-inspired potential model) is $$ \sqrt\sigma\sim 410\ \mathrm{MeV} $$ i.e., $3\ \mathrm{fermi}$.

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From the HyperPhysics site on Quarks:

It is postulated that it may actually increase with distance at the rate of about 1 GeV per fermi. A free quark is not observed because by the time the separation is on an observable scale, the energy is far above the pair production energy for quark-antiquark pairs. For the U and D quarks the masses are 10s of MeV so pair production would occur for distances much less than a fermi. You would expect a lot of mesons (quark-antiquark pairs) in very high energy collision experiments and that is what is observed.

From what I can gather, the actual increase in energy per fermi is not known to a high accuracy because the range is so incredibly small. Considering that the strong nuclear force is only able to participate in interactions at distances of less than $10^{-15}$ m it is understandable that no exact numbers exist for the maximum separation two quarks can have.

Jaywalker
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