I have read this question:

Is there an equation for the strong nuclear force?

Weak force: attractive or repulsive?

And yes, I have read the comment, where is says that the weak and the strong forces do not have classical field theories, Still, in the answers, there is an equation for the strong force.


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 $$F = k \frac{q_1 q_2}{r^2}.$$

Here goes the strong force:

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.

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

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

But nothing mentions anything similar to the weak force, though, the weak force can be repulsive or attractive, see here:

Weak force: attractive or repulsive?

So there must be an interaction due to the weak force, and it has to have some kind of similar description.


  1. Is there a similar equation that describes the force due to the weak interaction?

1 Answer 1


This very issue turned up in a question about energy differences between enantiomers, but there, only neutral currents were involved. Regardless of whether W or Z bosons are exchanged, the $\tfrac{1}{r}$ potential gets replaced by a Yukawa $\tfrac{1}{r}\exp (-mr)$ , where m denotes the mass of the heavy boson. The W couples to a $V-A$ current, and the Z to other linear combinations, due to the Weinberg mixing angle. If the fermions are non-relativistic, a VV term such as $(\bar{\psi }{{\gamma }_{\mu }}\psi )(\bar{\psi }{{\gamma }^{\mu }}\psi )$ yields an unremarkable constant coefficient, but a parity-violating VA term introduces a factor of $(\mathbf{s}\cdot \mathbf{p})$ at the axial current vertex, and an AA term $(\mathbf{s}\cdot \mathbf{p})(\mathbf{{s}'}\cdot \mathbf{{p}'})$. The idea of a non-relativistic neutrino is laughable, so if you are interested in W exchange, you will need a better approximation for the axial current vertex.


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