Yukawa force vs Nuclear force

Question

I have read these questions:

Are Neutrons and anti-Neutrons attracted to each other over distance?

Where John Rennie says:

Neutrons and anti-neutrons repel each other with a Yukawa force mediated by pion exchange.

Is the long range neutron-antineutron interaction repulsive or attractive?

Where Luboš Motl says:

It follows that a neutron and an antineutron Yukawa-repel, too.

All along, I have thought that it is the residual strong force, that is the nuclear force, that is mediated by pions, that describes the interactions of the nucleons inside the nucleus. This means that according to wiki:

The nuclear force binds nucleons into atomic nuclei.

<p>The nuclear force is powerfully attractive between nucleons at distances of about 1 femtometre (fm, or 1.0 × 10−15 metres), but it rapidly decreases to insignificance at distances beyond about 2.5 fm. At distances less than 0.7 fm, the nuclear force becomes repulsive.</p>

So now I am a little bit confused, is the nuclear force the same as the Yukawa force, or is it different? Is it that the Yukawa force is the short distance version of the nuclear force?

Quesion:

1. Is the nuclear force the same as the Yukawa force or are they different?

Answer 1

Yukawa force is any force that is described by the potential of the form $V = k \frac{e^{-\lambda r}}{r}$. The nuclear force can be approximately described by such potential (with $\lambda \sim m_\pi)$, so it's an example of Yukawa force.

Answer 2

The nucleon-nucleon interaction is very complicated, and meson-exchange potentials are simple models (not derived from QCD) of the low energy regime. However, one-pion (or multi-pion) exchange is special, because the pion is the lightest state in QCD, and the one-pion exchange therefore rigorously describes the longest range part of the interaction. This can be formalized and systematically improved using chiral effective field theory.

The one-pion exchange interaction between two nucleons is $$ V_{NN}= \frac{m_\pi^2}{12\pi} \frac{g_A^2}{2f_\pi^2} (\sigma_1\cdot\sigma_2)(\tau_1\cdot\tau_2) \frac{e^{-m_\pi r}}{r} + (LS-coupling) $$ This interaction depends on the spin and isospin of the two nucleons. It is attractive in $I=0,S=1$ and $I=1,S=0$, repulsive in $I=S=0,1$. Two neutrons have to have $I=1$, so the interaction between two neutrons is attractive if the total spin is 0.

The nucleon-nucleon interaction can be related to the nucleon-anti-nucleon interaction using G-parity, $G=C\exp(-i\pi T_2)$, a combination of $C$-conjugation and isospin. The G-parity of the pion is negative, so the one-pion $N\bar{N}$ interaction is $$ V_{N\bar{N}}= - V_{NN} $$ but the two-pion amplitude has the same sign, etc. A neutron and an anti-neutron can have both $I=0$ and $I=1$ (as opposed to $nn$ which is always $I=1$). Staying with $I=0,S=1$, the interaction is now repulsive, but the $I=1,S=0$ part is attractive.

Note that if one construct phenomenological potentials using many meson exchanges, then the strong $N\bar{N}$ interaction is on average more attractive then the $NN$ interaction, see for example Buck et al.

Answer 3

Yukawa's theory is one model of the nuclear force. There are many such models. For a review, see Ruprecht Machleidt (2014), Scholarpedia, 9(1):30710..