# If the strong nuclear force is stronger than electrostatic repulsion, why don't nuclei collapse into a point?

Today in class we were discussing the strong nuclear force, and our teacher was explaining about how the strong nuclear force counteracts the repulsion force between protons in a nucleus.

When asked about the relative strength of the two forces in question, she said that "The strong nuclear force is the strongest force of nature, and is infinitely stronger than the repulsion force between the protons". Now, if that were true, how would the atom remain in equilibrium, because if I'm correct, Equilibrium is achieved when the net force on a body is zero. However, in this situation, that doesn't seem to be the case. Could someone elaborate on this apparent contradiction?

• I would strongly disagree with that statement (bad puns are great). Obviously the strong force is not infinitely times greater than the repulsion force. In order for that to be true, there would either have to be zero force repelling the protons or infinite strong force holding them together. I'm afraid neither is the case. Even a cursory search on wikipedia should tell you that the strong force is only 137 times greater than the EM force. Go back to your prof and tell her you disagree – Jim Jul 26 '16 at 13:11
• The origin of the strong force is the fact that the electromagnetic theory as it is known now demands the nucleus to fall apart. It isn't the case, so something had to be invented. – bright magus Jul 26 '16 at 13:12
• Are you sure that your teacher said 'strong force is infinitely stronger than repulsive force between protons? Is she said that, then its wrong. – AMS Jul 26 '16 at 13:16
• @Jim That is a pretty weak argument. The gravity of this question makes me wonder about how strongly you are trying to, em, help, given the bad pun in your answer. Together, unified, we can do better; we can lift up a standard of high quality comments, break the confinement of off topic comments, be grander than ourselves, stop stringing people along, and model better behavior. In theory, QED, but more importantly: OP, does simply pointing out that the ratio isn't actually infinite sufficient to answer your question, or are you wondering how the protons don't collide? – Yakk Jul 26 '16 at 15:29
• @Yakk I sentence you to walk the Planck. – Don Branson Jul 26 '16 at 15:41

First, the strong force acts on scales where our classical idea of forces as something that obeys Newton's laws breaks down anyway. The proper description of the strong force is as a quantum field theory. On the level of quarks, this is a theory of gluons, but on scales of the nucleus, only a "residual strong force", the nuclear force remains, which can be thought of as being effectively mediated by pions.

Now, a force mediated by pions is very different from one mediated by photons, for the simple reason that pions are massive. Massive forces do not, in their classical limit, follow a pure inverse square law, but yield the more general Yukawa potential, which goes as $\propto \frac{\mathrm{e}^{-mr}}{r^2}$ where $m$ is the mass of the mediating particle. That is, massive forces fall off far faster than electromagnetism.

So this makes it already difficult to tell what the "strength" of a force exactly is - it depends on the scale you are looking at, as Wikipedia's table for the strengths of the fundamental forces rightly acknowledges. However, in no sense is the strong force "infinitely stronger" than the electromagnetic force - it is simply much stronger than it, sufficient to keep nuclei together against electromagnetic repulsion.

Now, the person who said that it is "infinitely stronger" might have had something different in mind which is not actually related to the strength of the force but to its fundamentally quantum mechanical nature: Confinement, the phenomenon that particles charged under the fundamental (not the residual) strong force cannot freely exist in nature. When you try - electromagnetically or otherwise - to separate two quarks bound by the strong force, then you will never get two free quarks. The force between these two quarks stays constant with increasing distance, it does not obey an inverse square law at all, and in particular the energy to being on of the two quarks to infinity is not finite. At some point, when you have invested enough energy, there will be a spontaneous creation of a new quark-antiquark pair and you will end up with two bound quark systems, but no free quark. In this sense, one might say that the strong force is "infinitely stronger", but crucially this is not the aspect of the strong force that keeps nuclei together; the theory of pions shows no confinement.

• The equilibrium question: shouldn't mentioning exclusion pressure be part of an answer? – Yakk Jul 26 '16 at 15:15
• @Yakk Yes, if one wanted to do a full account of all the forces at work, exclusion would add to the electromagnetic repulsion. But the question appeared to me to be about a misconception that the strong force is "infinitely strong" - I presume OP has no problem with accepting that finitely strong forces (no matter whether there are two, three, four, or even more at work) can achieve equilibrium. – ACuriousMind Jul 26 '16 at 15:23
• Nice answer, I just would like to point out that it is not the force between two quarks that grows linearly with distance. It is the energy of the pair. – Diracology Jul 26 '16 at 15:59
• @Diracology: Ahh, right. The force is constant, I guess. – ACuriousMind Jul 26 '16 at 16:01
• This is still a bit short of the posed question, though. (i) The strong force doesn't need to be infinitely stronger than electromagnetic repulsion, it just needs to be stronger at all positions, and (ii) Yukawa potentials die off past a certain point, but the question concerns what happens in the $r\to0$ limit, and there the Yukawa potential is essentially equivalent to a Coulombic force. When two protons approach, at some point the strong attraction turns off and they must begin to repel past an equilibrium position - that's what the question is asking for. – Emilio Pisanty Jul 27 '16 at 0:56

To first approximation, nuclei do collapse into a point. The diameter of a nucleus is typically about $10^{-5}$ the diameter of an atom, which means the nucleus occupies something like $10^{-15}$ of the atomic volume. If your atom were the size of a house, the nucleus, to scale, would be the size of a grain of salt (and yet containing 99.95% of the atom's mass).

In nuclear physics it makes more sense to talk about energy than it does to talk about force. The two concepts are closely related. Two protons separated by a distance $r$ have an "electrical" interaction energy $$U_E = +\alpha\frac{\hbar c}{r}.$$ Here $\hbar c = 200\rm\,MeV\,fm$ relates distance and energy. The fine structure constant, $\alpha \approx 0.0073 \approx 1/137$, represents the strength of the electrical interaction. The positive sign, and the $r$ on the bottom, mean that bringing two protons closer together stores energy in the electric field between them. If you find yourself with two protons at rest some distance apart, they'll tend to evolve to reduce the energy stored in the electric field by moving yet further apart; this is the sense in which the electric interaction is repulsive.

The electric field isn't the only place where interacting protons can store energy. There is also the pion field, whose interaction energy is given by a Yukawa potential, $$U_\pi = -\alpha_\pi \frac{\hbar c}r \times e^{-m_\pi r}.$$ This is mostly the same as the electrical potential, but the differences are important:

• The sign is negative: protons liberate energy from the pion field by approaching each other, so the force is attractive.

• There's a different coupling constant, $\alpha_\pi$. If I'm converting units correctly, the pion coupling constant is something like $\alpha_\pi\sim100$: the energy associated with the "pion exchange" interaction may be as much as ten thousand times stronger than the energy associated with a "photon exchange" interaction.

• Critically, there's an exponential factor that depends on the mass of the pion, $m_\pi$. To be proper this should have some factors of $\hbar$ and $c$ to make the argument of the exponential dimensionless; alternatively we can be cleverer about units and not have any distracting cruft. This means there is a length scale $r_0 \propto 1/m_\pi$ beyond which the pion interaction totally dies away, but for very short distances the interaction looks like electricity with a different coupling constant.

These differences address a little of your confusion. The strong interaction is much stronger than electromagnetism, but not infinitely stronger. But your title question still remains: why doesn't the nucleus collapse exactly to a point? That's because we still haven't exhausted the forest of different ways that two protons can store energy. The next two that are important are the rho and omega fields, \begin{align} U_\rho &= +\alpha_\rho \frac{\hbar c}r \times e^{-m_\rho r}, \\ U_\omega &= +\alpha_\omega \frac{\hbar c}r \times e^{-m_\omega r}. \end{align}

Like the pion field, these meson interactions "turn off" when the nuclei separate from each other, but the distances are shorter than for the pion because the masses are larger. So the interaction between two protons has at least three different regions:

1. Distant protons are repelled by the electrical interaction.
2. Very close-together protons are repelled by vector meson interactions.
3. At intermediate distances, protons can be attracted to each other by pion interactions.

Here's a plot using the potentials we've discussed so far:

This simple model reproduces several real features of real nuclei:

• In uniform nuclear matter, nucleons are separated by about 1.2 fm.
• The energy required to remove one nucleon from a nucleus is typically about 10 MeV.
• In very heavy nuclei, the length scale for nuclear attraction is shorter than the diameter of the nucleus; such nuclei are unstable. You can't tell from this plot, but the total interaction energy becomes positive --- that is, electrostatic repulsion wins out over strong attraction --- at a separation of about 12.5 fm. Uranium-235, famously unstable, has diameter 14 fm. Uranium is unstable because protons at one end of the nucleus are repelled by, rather than attracted to, protons at the other end of the nucleus.
• I would like to point out that at the time I wrote my answer, the title question was "origin of strong force" and the question (to me) was rather more about what "the strong force is infinitely stronger" might actually mean. I'm not convinced the current revision preserved the intent of the question. (Nevertheless, your answer is useful, +1) – ACuriousMind Jul 28 '16 at 8:32