# Do strong and weak interactions have classical force fields as their limits?

Electromagnetic interaction has classical electromagnetism as its classical limit. Is it possible to similarly describe strong and weak interactions classically?

• Commented Oct 6, 2012 at 19:06
• Commented Mar 11, 2023 at 3:47

The particles that communicate the Weak interaction, i.e W Bosons and Z bosons are massive. So unlike Electromagnetism which is communicated by massless particles(Photons), the weak interaction has a very short range.

For Massive particles the Potential of interaction falls as

$V(x) = -K \frac{1}{r} e^{-m r}$

The range of this force is approximately equals 1/m.

The strong force is communicated by gluons which as massless, So one can wonder if they have a long ranged interaction like EM. However this effect called colour confinement comes into play. The full explanation of colour confinement is rather technical, and requires a lot study. But roughly speaking, It says that colour charges do not exist in Isolation. If you wish to Separate quark anti quark pairs in mesons, to a distance R one requires an energy proportional to R. So one requires an infinite amount of energy to separate mesons into their constituent quarks. Because all particles that we see, Baryons, mesons and so on are not coloured. Strong force is also a very short ranged force. (a few femto meters as quoted from WIKI). So there are no long ranged classical fields associated with strong and weak forces.

• At low energies, you can approximate the strong force as being transmitted by pions rather than gluons. Commented Oct 6, 2012 at 19:20
• Yes agreed, Pions are also massive hence have a range for interaction. Commented Oct 6, 2012 at 19:23
• @Prathyush The Coulomb potential turns out to be (roughly) Fourier transform of the scattering amplitude (for example, of electron-electron scattering) at the tree-level. Is there a similar way to calculate the weak interaction potential by computing electron-neutrino scattering amplitude in electroweak theory at the tree-level and taking the Fourier transform?
– SRS
Commented Oct 7, 2018 at 15:13

The electrostatic force between the electron and the proton (in classical terms) varies as $1/r^2$ so when the electron and proton are separated by a large distance the force goes to 0 therefore at large distance the electron and proton become free particles. Note that when the electron and proton are very close the force between them increases to infinity. Also note that for the electrostatic force there are two charges: $+$ and $-$.

The strong force is called a color force because there are 3 different kinds of color "charges" in Quantum Chromodynamics (QCD), unlike the case of electrostatics which only has 2 charges. (There are no "real" colors, or course, physicists just use the term color since there are 3 primary colors which matches up with the names needed for the 3 different "charges" of quarks.) One of the reasons why the strong color force between two or three quarks is different than the electrostatic force between an electron and proton is that the force carriers of the strong force (the massless gluons) are also colored therefore the gluons are also strongly attracted to each other. Whereas in the electrostatic case, the force carriers (virtual photons) are uncharged so two virtual photons do not attract each other.

The color force between two (or three) quarks is quite different than the electrostatic force between two charges. In a very simplified model you can think about the force between two quarks as varying like $r$ or $r^2$. First of all note that when $r\rightarrow 0$ the force goes to 0. This is the asymptotic freedom of the color force which was discovered in 1973 and for which Gross, Wilczek and Politzer were awarded the Nobel Prize in Physics in 2004. This means that at very high energies (and short distances) the quarks act like free particles and the color force is small.

However when $r\rightarrow \infty$ the force goes to $\infty$. This model of a force that increases with distance is another statement of the principle of color confinement in QCD. The fact that gluons will interact with other gluons with the same strong color force that attracts quarks is thought to be the reason for color confinement. So, if you start with the three different colored quarks bound into a colorless proton and if you try to pull one of the quarks out of the proton, it will take more and more force and thus more and more energy as you pull the quark out. Thus, as you try to separate the quark out of the proton, at some point when enough energy has been added to the system it becomes energetically favorable to create a new pair of quarks ($q\bar{q}$) in the region between the quark and the residual "proton". Now the the newly created $\bar{q}$ will be attracted to the quark that is being pulled out of the proton whereas the other newly created $q$ will be pulled back into the proton which will then constitute a normal proton again with 3 quarks. Meanwhile the $q$ that is being pulled out and the newly created $\bar{q}$ will become bound together as a meson - therefore the attempt to pull a quark out of a proton will result in a final state that has a meson and a proton. This is called color confinement - because you can never separate a single colored quark (or gluon) out of a proton or other hadron - all composite particles must be colorless - either a $q\bar{q}$ that is colorless (a meson) or three differently colored $q$'s that create a colorless proton or hadron. This strong color force is responsible for binding 3 quarks into hadrons (like protons or neutrons) or a $q\bar{q}$ into mesons.

Now when protons and neutrons are bound together in a nucleus, even though the proton and neutron as a whole is colorless, when they are close to each other a residual part of the color force will attract the proton and neutron together. This can be modeled as the exchange of $\pi$ mesons between the nucleons and since the pion has a mass, this will result in a short range force that will vary as:

$F(r)=\tfrac{\pm K}{r^2}e^{-mr} \ \ \$ where $m$ is the mass of the pion.

This residual color force is responsible for nuclear binding.

Now the weak interactions are mediated by the $W$ and $Z$ mesons which are much heavier than the pion by a factor of about 600 ($m_\pi \approx 130-135 MeV$ but $m_W \approx 91 GeV$ and $m_Z \approx 80 GeV$). Thus the weak force will also be of the form:

$F(r)=\tfrac{\pm K}{r^2}e^{-mr} \ \ \$ where $m$ is the mass of the $W$ or $Z$.

Now the coupling constant $K$ is about the same as the electromagnetic coupling constant, but since the range of the force is so small, it is a very weak force. In fact there are no known bound states that are held together by the weak force. The weak force mostly changes one type of particle into another type of particle. For example an electron can be converted into a neutrino ($\nu$) by a $W$ meson and one type of quark can change into another type of quark via a $W$ meson. This, for example, is how a free neutron decays into a proton plus an electron and a neutrino:

It is this ability to change the types of particles through the weak interactions that is most significant for "force", the fact that the range of the force is so small is one of the reasons why the weak interaction force is so weak and the actual "force" part of the weak force is largely insignificant since it does not result in any bound states.

• This is a nice answer in terms of giving background information, but it doesn't seem to directly address the point of the question. The question asks why there is no classical limit, not why the force is short range.
– user4552
Commented Nov 21, 2016 at 16:58

I think there are really two separate issues here. One is the range of the forces, and the other is the existence of a classical limit.

Basically, being able to write down a Lagrangian density isn't the same thing as being able to describe the classical theory that is the counterpart of a quantized system. In particular, it seems like this can't possibly work for unstable particles. For example, the Lagrangian density for muon decay has a constant in it, $G_F$, the Fermi coupling constant. The half-life of the muon goes like $h/G_F^2$. In the classical limit $h\rightarrow0$, the half-life goes to zero, so the classical theory of muons is a theory with no muons in it.

So you can't have a classical field theory of the weak force, simply because the W and Z are unstable.

The strong force is completely different. Gluons are massless and stable. Although they're self-interacting, so are gravitons, and there is a classical field theory of gravity. It's not completely obvious to me that we don't ever have a classical field corresponding to the strong force.

For example, take the case of two heavy nuclei scattering inelastically, below but close to the Coulomb barrier. The process is classical in the sense that the de Broglie wavelengths of the two nuclei are small compared to the sizes of the nuclei. Far below the Coulomb barrier, you get Rutherford scattering, which is completely classical -- you can describe it using Newton's laws. Closer to the Coulomb barrier, the nuclei can approach one another sufficiently closely for the strong force to act, but there is still an elastic scattering channel, which I think should be describable in purely classical terms.

• "So you can't have a classical field theory of the weak force, simply because the W and Z are unstable." I'm not sure I buy this argument. The W and Z particles, quantised excitations that they are, go away in the classical limit so there is no reason to talk about their lifetimes in the first place. I can write down the Yang-Mills field equations and, with suitable gauge fixing, solve them in a completely classical way treating fields as c-numbers. I'll have to think on your muon example since the "classical" limit is a Grassmann-number valued field theory which is hard to interpret. Commented Jun 13, 2013 at 16:19