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The Yukawa Lagrangian

$$\mathcal{L}=-\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu}+\dfrac{1}{2}m^2A_\mu A^\mu-A_\mu J^\mu$$

can be used to derive the equations of motion

$$\Box A_\mu=J_\mu-m^2A_\mu$$

for the four-potential $A_\mu$. It turns out that if we have $J_\mu$ the current of one charge $e$ at rest at the origin, that is, $J_0(\mathbf{x})=e\delta^{(3)}(\mathbf{x})$ and $J_i=0$, then we can solve the equation by Fourier transform and get

$$A_0(r)=\dfrac{e}{4\pi r} e^{-mr}.$$

It is that then that this potential could explain how protons are bound in the nucleus.

How can we see that? Because honestly I don't see it. I mean, is it by computing the associated force?

In that case I computed it and we would have (assuming the force works as in electrostatics $F_Y = qA_0(r)$.

$$F_Y=\dfrac{e^2}{4\pi r^2}e^{-mr}+\dfrac{me^2}{4\pi r}e^{-mr}.$$

Now, why would this force explain how the protons are bound? I believe this has to do with some specific value for $m$ that allows this force to overcome the repulsion due to the coulomb force

$$F_C=\dfrac{e^2}{4\pi r^2},$$

but I don't know how to make this precise.

Gold
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  • See my answer to this question: https://physics.stackexchange.com/questions/288357/why-is-nuclear-force-spin-dependent/288499#288499 – Lewis Miller Apr 13 '17 at 13:11

2 Answers2

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Consider that two protons can never be bound gravitationally: the electro-gravitational force is \begin{align} \vec F &= \frac{Gm^2}{r^2}\left(-\hat r\right) + \frac{\alpha\hbar c}{r^2}\hat r \\&= \left(-{Gm^2}+ \alpha{\hbar c} \right)\frac{\hat r}{r^2} \\&\approx \left(-\frac23\times 10^{-38} + \frac{1}{137} \right) \frac{\hbar c \hat r}{r^2} \end{align} Because the functional form of the electric and gravitational forces is the same, the electric force is (much) stronger than the gravitational force at all distances, and the net electro-gravitational interaction between protons is always repulsive.

The Yukawa force, on the other hand, is proportional to $\exp \frac{-r}{r_0}$, where the length scale $r_0$ gets shorter as the mediating particle gets heavier. That means you can make the Yukawa coupling as strong as you like, but as long as your two protons are several $r_0$ apart there isn't any strength left to it. The Yukawa interaction is "local" in a way that the electric force is not.

rob
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  • Thanks for the answer. I understand your point, the Yukawa force has that exponential term that allows one to fix the strength of the force with the parameter $m$ right?Anyway, by the formula I derived isn't this Yukawa force always repulsive also? And in this Lagrangian, it is meant to have the electrostatic interaction built into it, so that the resulting force is the net force? – Gold Apr 13 '17 at 03:59
  • There are both attractive and repulsive Yukawa terms. Between nucleons, the pion-mediated interaction is attractive, while the heavier ω and ρ mesons mediate repulsive interactions. With lots of caveats (see e.g. the other answer by annav), this gives you nucleon-nucleon attraction at distances like $r\approx \hbar c/m_\pi c^2$, but "hard-core" repulsion at closer distances. – rob Apr 13 '17 at 12:27
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This answer is in addition to Rob's.

How can we see that? Because honestly I don't see it. I mean, is it by computing the associated force?

Nuclei belong to the quantum mechanical domain and have to be studied with quantum mechanical tools. Forces in the quantum mechanical domain are replaced by potentials in solving Schrodinger or Dirac or Klein Gordon equations.

Adding the attractive Coulomb force to the problem "electron + proton" fits the spectrum data of the Hydrogen atom. The solution of these equations represent bound states in the case of the Schrodinger equation solved for the Coulomb potential and fit to the experimental spectrum of the hydrogen atom , thus validating the quantum mechanical method. The Coulomb potential at the quantum mechanical level explains the observed data, and predict new successfully.

Adding an attractive potential to overcome the repulsive Coulomb potential in the equations for "proton" + "rest of the nucleus" problem, gives bound states leading to the shell model of the nucleus. .

The many body problem of interactions at the quantum mechanical level introduced the mathematics of Quantum Field Theory, where the forces are emergent from underlying interactions. The Yukawa potential is a step more sophisticated than the harmonic oscillator used for the shell model, :

Hideki Yukawa showed in the 1930s that such a potential arises from the exchange of a massive scalar field such as the field of a massive boson. Since the field mediator is massive the corresponding force has a certain range, which is inversely proportional to the mass of the mediator particle m {\displaystyle m} m.1 Because the approximate range of the nuclear force was known, Yukawa's equation could be used to predict the approximate rest mass of the particle mediating the force field, even before it was discovered. In the case of the nuclear force, this mass was predicted to be about 200 times the mass of the electron, and *this was later considered to be a prediction of the existence of the pion, before it was detected in 1947.*

Please note that all symmetric potentials expanded in a Taylor series have the harmonic oscillator as a first term, and thus the Yukawa potential includes the successes of the shell model in fitting the nuclear series.

anna v
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