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The equation describing the force due to gravity is $$F = G \frac{m_1 m_2}{r^2}.$$ Similarly the force due to the electrostatic force is $$F = k \frac{q_1 q_2}{r^2}.$$

  1. Is there a similar equation that describes the force due to the strong nuclear force?

  2. What are the equivalent of masses/charges if there is?

  3. Is it still inverse square or something more complicated?

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From the study of the spectrum of quarkonium (bound system of quark and antiquark) and the comparison with positronium one finds as potential for the strong force $$V(r) = - \dfrac{4}{3} \dfrac{\alpha_s(r) \hbar c}{r} + kr$$ where the constant $k$ determines the field energy per unit length and is called string tension. For short distances this resembles the Coulomb law, while for large distances the $k\,r$ factor dominates (confinement). It is important to note that the coupling $\alpha_s$ also depends on the distance between the quarks.

This formula is valid and in agreement with theoretical predictions only for the quarkonium system and its typical energies and distances. For example charmonium: $r \approx 0.4 \ {\rm fm}$. So it is not as universal as eg. the gravity law in Newtonian gravity.

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  • $\begingroup$ . +1 This potential makes more physical sense for quarks since it includes both the QED-like $-1/r$ and the confining $+kr$. $\endgroup$
    – dbrane
    Apr 11 '11 at 20:48
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    $\begingroup$ Nice. Of course, the "breaking of the flux tube" has no classical or semi-classical analogue, making this formulation better for hand waving than calculation. $\endgroup$ Apr 11 '11 at 21:19
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    $\begingroup$ This is fine for the quark-quark interaction, but people reading this answer should be careful not to interpret it as a nucleon-nuclon interaction. $\endgroup$
    – user4552
    Jun 13 '13 at 16:56
  • $\begingroup$ @Johannes I edited your comma to a decimal dot - it was a bit confusing to English speakers - I'm assuming you mean "nought point four fm". Interestingly, in my own land, the Australian engineering drawing standard used to use a comma for the decimal marker too and I still do privately in handwritten calcs because a dot is too easy to lose track of - the silliest notation ever for something so significant as the decimal marker - I assume this is why Europe and eng. standards use the comma. However, I never use it in English language communication as it definitely confuses people. $\endgroup$ Nov 9 '13 at 5:18
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    $\begingroup$ Could you specify exactly what distance $r$ refers to? $\endgroup$ Feb 26 '18 at 1:43
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At the level of quantum hadron dynamics (i.e. the level of nuclear physics, not the level of particle physics where the real strong force lives) one can talk about a Yukawa potential of the form

$$ V(r) = - \frac{g^2}{4 \pi c^2} \frac{e^{-mr}}{r} $$

where $m$ is roughly the pion mass and $g$ is an effective coupling constant. To get the force related to this you would take the derivative in $r$.

This is a semi-classical approximation, but it is good enough that Walecka uses it breifly in his book.

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No, there is none such equation. Reason is that these equations are highly classical and invalid in both relativistic (there is an action at a distance, incompatible with finite speed of light) and quantum mechanical regime (distances strong force is important at are quite microscopic). Also, strong force is confining, meaning you can't ever observe individual color charged particles (color is a property associated with strong force), so there can't really be a macroscopic equation for them.

You obviously need at least quantum mechanics to account for strong force, because distances are so tiny (on the scale of nucleus or smaller). But it turns out you need relativity too. The complete theory which incorporates both QM and relativity is called quantum field theory and individual forces are described by QFT Lagrangians which essentially tell you which particles interact with which other particles (e.g. photons with electrically charged particles, gluons with color charged particles, etc.). This is the fundamental theory and the electric force equation you described can be derived from it in classical (both non-QM and nonrelativistic) limit. As for gravitation law, that too can be derived but from different theory, namely general relativity.

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  • $\begingroup$ Thanks, I guess that the goal is to formulate gravity in the same language of quantum field theory? That is the goal of stringtheory and other such unification theories? $\endgroup$
    – ergodicsum
    Apr 11 '11 at 18:21
  • $\begingroup$ @ergodicsum: yep, pretty much. (Either that, or formulate the standard model in the language of GR, or formulate both in some new theoretical framework yet to be discovered) $\endgroup$
    – David Z
    Apr 11 '11 at 18:28
  • $\begingroup$ @ergodicsum: that would be the intuitive proposition, right. But it turns out gravity doesn't play well with QFT in the way other forces do. So the language will probably be of some other theory (e.g. string theory) from which both QFT and GR can be derived in some limits. $\endgroup$
    – Marek
    Apr 11 '11 at 18:28
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    $\begingroup$ Well, it's still true that at short distances, much shorter than a Fermi, and in the non-relativistic limit, the strong force is still governed by the Coulomb's law. It's not a terribly useful limit for the strong force but it is misleading to suggest that the strong force is something "entirely different". $\endgroup$ Apr 11 '11 at 18:38
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    $\begingroup$ No, there is none such equation. This claim is much too strong. Quark-quark potentials like the one given in Johannes' answer are not the ultimate answer, but neither are they useless. They have considerable predictive value for a variety of complex phenomena. For example, you can use them in the Schrodinger equation to get quite accurate predictions of the spectrum of charmonium. (Charmonium is not very relativistic.) For example, the energy spectrum is predicted to go like $n^{2/3}$, which holds up quite well against experiment: arxiv.org/abs/hep-ph/0608103 $\endgroup$
    – user4552
    Jun 13 '13 at 17:02
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Let me add one obvious thing: There is an exact equation for the strong force. It is what Gross, Politzer and Wilczek got the Nobel prize for. It is called quantum chromodynamics (QCD). Google it or look it up in Wikipedia, and you can see the Lagrangian for QCD, and compare it to the Lagrangian for electrodynamics.

Of course, you could argue about the similarities and differences of a Lagrangian, and a force equation, such as your two examples.

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The strong force as seen in nuclear matter

The nuclear force, is now understood as a residual effect of the even more powerful strong force, or strong interaction, which is the attractive force that binds particles called quarks together, to form the nucleons themselves. This more powerful force is mediated by particles called gluons. Gluons hold quarks together with a force like that of electric charge, but of far greater power.

Marek is talking of the strong force that binds the quarks within the protons and neutrons. There are charges, called colored charges on the quarks, but protons and neutrons are color neutral. Nuclei are bound by the interplay between the residual strong force , the part that is not shielded by the color neutrality of the nucleons, and the electromagnetic force due to the charge of the protons. That also cannot be simply described. Various potentials are used to calculate nuclear interactions.

Simplicity and similarity of form for all forces comes not in the formalism of forces, but as Marek said the formalism of quantum field theory.

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Strong force holds up and down quarks together into a proton or neutron. It's really the nuclear force (or residual strong force) that keeps nucleons together in an atomic nucleus. The mass defect and therefore the nuclear binding energy is determined by the number of protons and neutrons in the nucleus. There are 5 terms that add up and contribute in the calculation of nuclear binding energy. It is called the Semi Empirical Formula of Nuclear Binding Energy.

See http://en.wikipedia.org/wiki/Nuclear_binding_energy#Semiempirical_formula_for_nuclear_binding_energy

For details check out - http://en.wikipedia.org/wiki/Semi-empirical_mass_formula

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    $\begingroup$ One of those terms is the Coulomb (i.e. electrostatic) term which has nothing to do with the strong interaction at all. $\endgroup$ Nov 12 '13 at 17:57
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I think that the earlier answers haven't adequately explained why there isn't a force law for the strong force. It's not because it's relativistic or quantum (electromagnetism is both of those). It's because it's nonlinear.

In the gravitational and electromagnetic formulas in the question, $q_1$ and $q_2$ (or $m_1$ and $m_2$) are the charges at exactly two points, and $r$ is the distance between those two points. The formulas can't be used if there are more than two charges. You can extend them to the general case by turning them into sums over pairs of points, but that only makes sense if the force is linear.

You can write down a two-particle force law for the strong force. It's $F=-V'(r)$ where $V(r)$ is the function in Johannes's answer. What you can't do is use that same force law when there are more particles – not because the law is wrong as such, but because you can't use linearity to generalize it.


Though there is no simple force law, you can build an intuition for the magnitude and direction of the strong force from these rules:

  1. Strong force field lines prefer to cluster in regions with a diameter of about 1 fm. They attract each other when farther apart and repel each other when closer (this is the nonlinearity).

  2. If the total strong charge in an enclosed region of space isn't "zero" (color singlet), then there have to be strong-force field lines crossing the boundary of the region. (This is the strong-force version of Gauss's law.)

  3. The energy density of the field is roughly constant where there are field lines and zero elsewhere.

  4. The force is minus the gradient of the potential.

If there are two charges in an otherwise empty universe, then applying "Gauss's law" to any region containing one particle and not the other tells you that there have to be field lines crossing the space between the particles. The lines can follow any path, but the lowest-energy configuration is that in which they're all in a straight "flux tube" with a diameter of about 1 fm. In that configuration, the only way to reduce the field energy is to bring the charges closer together, so there is a force in that direction, i.e., of each particle directly toward the other. The energy density of the tube per unit distance is constant, so the strength of the force is independent of the length of the tube. (It's about 10,000 newtons; that's the $k$ in Johannes's answer.)

If there are three or more charges, no proper subset of which is color neutral, then again there have to be field lines connecting all of the particles. The shortest, hence lowest energy, way to connect them is a Steiner tree. The force on each charge is toward the nearest vertex of the Steiner tree, and the magnitude of it is again a constant 10,000 N if the separations are large enough.

This is only an approximation, but it's not too bad. In figure 4 of this paper, which shows pentaquarks simulated in lattice QCD, you can clearly see the Steiner-tree structure.

Note that it's very common to see diagrams with incorrect flux tube geometries. For example, all of the configurations in this illustration are wrong except the single tube in the $u\overline c$ meson.


This approximation ignores pair production of quarks. The practical effect of pair production is that the flux tubes connecting charges can't get very long, because it's energetically cheaper to create quarks which allow them to "break".

Quark-like particles with high rest masses, charged under a strong-like force, could be connected by much longer flux tubes. Such particles are known as "quirks" and there are a few papers about them, such as this one which proposes searching for them at the LHC.


I said that the energy density of the field is zero where there are no field lines, but that isn't quite true: in principle it's nonzero out to infinity, but it dies off exponentially.

"Gauss's law" says that there aren't (or at least needn't be) flux tubes connecting different hadrons, but there is a residual strong force between hadrons, which dies off exponentially with distance. It's weak enough that it can (as far as I know) be treated as linear, and there is a "pairwise" force law for it like the gravitational and electrostatic laws. That's $F=-\nabla V$ where $V$ is the function in dmckee's answer.

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