4
$\begingroup$

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.

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.

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?
$\endgroup$
  • $\begingroup$ I don't understand the question. It seems to have several unrelated ideas. One part is about the interaction between neutrons and antineutrons. One is about the range of the strong force between hadrons. And a third part is about the distinction between a Yukawa force and the strong force between hadrons. $\endgroup$ – Ben Crowell May 26 at 0:11
  • $\begingroup$ Related: physics.stackexchange.com/questions/288357/… $\endgroup$ – Lewis Miller May 26 at 1:59
1
$\begingroup$

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..

$\endgroup$
  • $\begingroup$ Is it the short distance version of the nuclear force? $\endgroup$ – Árpád Szendrei May 26 at 19:15
  • 1
    $\begingroup$ No. It is a model of the nuclear force for any distance. Notice that for large distances the model says that the force essentially vanishes due to the exponential cutoff. $\endgroup$ – user26872 May 26 at 19:49
  • $\begingroup$ are there other models of the nuclear force? You are saying that it is one model. $\endgroup$ – Árpád Szendrei May 26 at 19:56
  • 1
    $\begingroup$ @ÁrpádSzendrei Yes. The Yukawa interactions are generally two-particle in nature, i.e. between one nucleon and another. There are also (for example) mean-field models, where one nucleon essentially interacts with a potential formed due to the average of all of the other nucleons at once. The typical mean-field potential is the Woods-Saxon potential (see e.g. en.wikipedia.org/wiki/Woods%E2%80%93Saxon_potential). $\endgroup$ – probably_someone May 26 at 23:25
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is it the short distance version of the nuclear force? $\endgroup$ – Árpád Szendrei May 26 at 19:15
  • $\begingroup$ @ÁrpádSzendrei Not quite. It is a shorter-distance description than simply accounting for Coulomb repulsion. It is a longer-distance description than a model which also takes into account the exchange of heavier mediators than pions, such as $\rho$ and $\omega$ mesons. $\endgroup$ – probably_someone May 26 at 23:27
  • $\begingroup$ @probably_someone can you please elaborate this in an answer? I am a little confused now. Seems like you know the answer. Nuclear force is using heavier mesons too, but the Yukawa not? $\endgroup$ – Árpád Szendrei May 26 at 23:45
  • $\begingroup$ @ÁrpádSzendrei You're asking a really broad question, one which is explained over many chapters of a nuclear physics textbook. In general, "the nuclear force" is the phenomenon in nature that we can perform experiments to measure and characterize, and which is described by many different classes of models. It doesn't correspond to a particular model, and currently no single model is uniformly better than all other models at describing the nuclear force under all conditions. Nuclear physics isn't like particle physics - there isn't really a unifying framework like the Standard Model. $\endgroup$ – probably_someone May 26 at 23:49
  • 1
    $\begingroup$ @ÁrpádSzendrei The issue is not that the Standard Model "has no jurisdiction" in nuclear physics. The issue is that, due to QCD being non-renormalizable, we currently cannot feasibly perform analytical calculations of nuclear forces using the fundamental interactions in the Standard Model (numerical calculations using lattice QCD are slowly becoming more doable, but these are mainly dependent on the state of the available computing power). It's not that it doesn't describe these interactions, but we don't currently know how to feasibly demonstrate that it does with current technology. $\endgroup$ – probably_someone May 27 at 0:03
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Nitpick: It’s actually $({{\tau }_{1}}\cdot {{\tau }_{2}})({{s}_{1}}\cdot \nabla )({{s}_{2}}\cdot \nabla )(Yukawa)$, which gives a contact term plus a tensor term. $\endgroup$ – Bert Barrois May 28 at 13:38
  • $\begingroup$ @BertBarrois Yes, thanks. 1) I suppressed the delta-function, because from the modern (EFT) point of view it is not observable (it renormalizes other short range operators). 2) I only wrote the central term, the non-central (tensor force) part is what I meant by "LS-coupling". $\endgroup$ – Thomas May 28 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.