This is something that's puzzled me for a while. Say you have a proton with quarks uud, with the u quarks having +2/3 charge and the d quark -1/3. Doesn't this make quite a neat fit for keeping the quarks confined within the proton, in terms of the electrostatic forces between the three quarks. At what point, and for what reason did it become necessary to invoke gluons to provide the extra 'colour charge' for ensuring quark confinement. Why is the ordinary electrostatic force not enough?

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    $\begingroup$ "Doesn't this make quite a neat fit for keeping the quarks confined within the proton" - While I can understand your reasoning for a neutron since it is neutral, I don't see it for a proton. On another note, color is required to fix the statistics if I'm not mistaken. $\endgroup$ Commented Jun 17, 2017 at 12:37
  • $\begingroup$ Even if the electrostatic interaction could hold individual protons together (which it can't), you'd still need the strong force to hold all the mutually repelling protons in a nucleus together. $\endgroup$
    – tparker
    Commented Jun 17, 2017 at 15:17
  • $\begingroup$ I was thinking more along the lines that the charges of the individual quarks and the directions along which those fractional charges act may be enough to explain the residual strong force. As for the fundamental strong force I really don't see why the fractionally charged quarks aren't enough to hold the hadrons together. Perhaps you do? $\endgroup$
    – Sam Cottle
    Commented Jun 17, 2017 at 15:32

2 Answers 2


The uncertainty principle tells us that the more tightly you confine an object the greater the uncertainty in its energy. For example you can use this in a hand waving sort of way to calculate the size of a hydrogen atom from its ground state energy or vice versa.

And when we do this we find the electrostatic force is simply not strong enough to confine the quarks inside a nucleus. If we had three quarks interacting just by the electrostatic force we'd get an object roughly the same size as an atom. The only way to get a bound state as small as a nucleus is to have a vastly stronger force binding the quarks together. That vastly stronger force is of course the strong force.


There is another issue related to the need for color charge and the associated gauge fields.

I'll quote from section 9.1 The colour degrees of freedom of "Gauge Theories in Particle Physics":

For a baryon made of three spin-1/2 quarks, the original non-relativistic quark model wavefunction took the form

$$\psi_{3q} = \psi_{space}\psi_{spin}\psi_{flavour}\qquad\qquad (9.1)$$

It was soon realised (e.g. Dalitz 1965) that the product of these space, spin and flavour wavefunctions for the ground state baryons was symmetric under interchange of any two quarks.


But we saw in $\S 4.5$ that quantum field theory requires fermions to obey the exclusion principle - i.e. the wavefunction $\psi_{3q}$ should be antisymmetric with respect to quark interchange. A simple way of implementing this requirement is to suppose that the quarks carry a further degree of freedom, called colour, with respect to which the 3q wavefunction can be antisymmetrised, as follows.


With the addition of this degree of freedom, we can certainly form a three-quark wavefunction which is antisymmetric in colour by using the antisymmetric symbol $\varepsilon_{\alpha\beta\gamma}$, namely

$$\psi_{3q,\mathrm{colour}} = \varepsilon_{\alpha\beta\gamma}\psi^\alpha\psi^\beta\psi^\gamma$$

and this must be then be multiplied into (9.1) to give the full 3q wavefunction.

  • $\begingroup$ Good point, explains the necessity of color SU(3) for the gauge symmetry. $\endgroup$
    – anna v
    Commented Jun 19, 2017 at 19:29

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