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Here is what I have understood about the strong force: This is a close range force which acts on two charges. The way it works is that a charge releases a meson which attracts the two charges together and this force is created by a potential given by $$V_{\text{Yukawa}}(r) = -g^2\frac{e^{-\alpha m r}}{r}.$$ (This is from the Wikipedia page). Please correct me if my understanding is wrong.

Now my first question is I know that $g^2$ and $\alpha$ are scaling constants but what values should I put for them to accurately describe the strong force?

My second question is what is $r$? On the Wikipedia page it says, "$r$ is the radial distance to the particle" but what is "the particle?" Are they refering to the meson? If so, I thought the meson is moving so won't the radial distance be changing?

Qmechanic
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Dan
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1 Answers1

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Well, nuclear binding is a complex, multifaceted, business, and the Yukawa potential is a framework conceptual cornerstone for it. For the virtual (~metaphorical; unreal; computational; quantum mechanical) meson which intermediates the "force", normally a pion, with mass m = 138 MeV, α =1, as follows from the pion propagator, outlined in WP.

The mass in the exponent typifies the intermediating meson, not the two nucleons bound by the potential, at a distance r from each other.

But as you see from the linked SP article, there are actually several mesons involved, involving several masses and coupling strengths g, resulting in a complex picture, with lots of parameters, masses, signs, etc, all to be determined by a plethora of experiments. They are normally of order 1, but can vary substantially with the nucleus in question.

The takeaway is that the picture to be emulated, in the back of the reader's mind, is the EM Coulomb potential, with a massless photon, m =0, so infinite range, and coupling strength e, the elementary electric charge. This morphs to the Yukawa with its range shortened, and its coupling strength enhanced enormously, but I'd doubt specific values of such parameters, available in reviews, would help you much.

Edit in response to comment questions

As always in HEP and nuclear physics, the natural units employed in the nondimensionalization set $\hbar \to 1$, $c\to 1$, so that masses are measured in MeV, while distances, such as r, are measured in 1/MeV. You may readily deduce then that $\frac{1}{\hbox{MeV}}= 197.3 \hbox{fm} \sim 2\cdot 10^{-13}$m . In your conventions, α =1 and g ~ 10 are dimensionless.

Cosmas Zachos
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