What holds the nucleus together? In a nucleus there are several protons, all of which are positive. Why don't they push themselves apart? It turns out that in nuclei there are, in addition to electrical forces, nonelectrical forces, called nuclear forces, which are greater than the electrical forces and which are able to hold the protons together in spite of the electrical repulsion. The nuclear forces, however, have a short range — their force falls off much more rapidly than $1/r^2$. And this has an important consequence. If a nucleus has too many protons in it, it gets too big, and it will not stay together. An example is uranium, with 92 protons. The nuclear forces act mainly between each proton (or neutron) and its nearest neighbour, while the electrical forces act over larger distances, giving a repulsion between each proton and all of the others in the nucleus. The more protons in a nucleus, the stronger is the electrical repulsion, until, as in the case of uranium, the balance is so delicate that the nucleus is almost ready to fly apart from the electrical repulsion. The energy which is liberated is the energy of the atomic bomb. This energy is usually called "nuclear" energy, but it is really "electrical" energy released when electrical forces have overcome the attractive nuclear forces.

Chapter 1, Volume II, The Feynman Lectures on Physics

If it's this difficult for nucleons to stay together, how did the first nucleus even form?

To make a nucleus, multiple protons and neutrons have to be brought together. It is reasonable to assume that the nucleons are at least several diameters of protons apart before any nucleus is formed. Since the nuclear forces fall off much faster than the electrical forces over distances, there will always be electrical repulsion preventing the component protons from coming together (unless they are already really close which is unlikely). What brings the protons together to form nuclei?


3 Answers 3


You need very high temperatures so that there is enough energy to overcome the electrical repulsion. This can happen in a few ways. One is a few minutes after the Big Bang, which produced an initial amount of mostly hydrogen and helium. After that the heavier nuclei were created in fusion processes in stars and in the supernova explosions.


Well, initially (at long distance), your different nuclei will push each other apart because of electric repulsion. This forms kind of an "Energy barrier", which should be overcome to achieve nuclear fusion. Basically, you need to provide the "push" necessary to get the nuclei close enough so that the nuclear forces can do their job.

It's like you have two people really not into getting each other at first. But, Under the recommandation of a common friend pushing them together, they might come to realize that this guy is really cool after all, once you get to know him.

The most natural way to provide this push is Simply thermal Energy: at very high temperatures, each particle will move around real fast in every directions; so fast that they might not be stopped by the electric repulsion, and overcome the barrier. And once they get to that point, they can realize that it's a pretty nice place to stay because that nuclear attraction is getting real strong.

This is basically what is happening Inside the sun (nuclear fusion of hydrogen into helium), and what powers it (because the process gives off a lot of heat). Little helpers like the quantum tunneling effect come into play as well, so that this thing achievable at lower temperatures than it would from pure classical predictions. But I think this is not necessary to elaborate on this to answer your conceptual question :)

EDIT: To answer your comment, Here is how you could do a (very) simplified estimation of the likelyhood of 2 particles bumping into each other. To make things easy, let's first assume that you are at a temperature so high that the electronic interactions can be disregarded (Energy barrier negligible compared to average kinetic Energy)

Then you will be in the situation of a random brownian motion, where the particles have random velocities, given by a probability distribution P(v) to have veolicty v (this distribution is of course determined by your temperature and statistical mechanics, cf Maxwell-Boltzmann distribution. It can be assumed isotropic). The probability of collision will depend on the Density of your gas of particles. Let's say you have N particles in a Vtot volume.

A spherical particle of radius r Moving at Velocity v in a time dt will cover an additional volume of $$dV = 2 \pi r^2vdt$$ (this is the cylindrical "trace" of the path of the particle) The probabilty to find another particle in there (probability that a particle was on its path = probability that they bumped into each other) is Simply given by: $$V_{particle}N/V_{tot}*dV=4/3\pi r^3*N/V_{tot}*dV$$ Integrating on all possible velocities, we get :

$$(\frac{8N\pi^2r^5}{3V_{tot}} \intop\nolimits_{-\infty }^{\infty } vP(v)dv )dt = \frac{8N\pi^2r^5}{3V_{tot}}v^*dt$$

where v* is the average Velocity. You may then Simply integrate over time.

Of course, taking into account the electric interactions would modify the result and make the Velocity distribution much more complex to estimate (many-body problem, basically). You would also have to take into account that a particle with an "infinite Velocity" (or just high enough), may not be trapped by the nuclear interaction, which would add one layer of complexity.

Well, in a nutshell: this is all a gross approximation in order to avoid the quantum many-body problem kinda fun. But I Believe this "infinite temperature" limit can be enlightening to probe how high temperatures can enable particles to get close and eventually stick.

  • $\begingroup$ What is the likelihood of two highly energetic nucleons heading towards each other with the correct speed and direction? Is this the explanation for the origins of most of the nuclei (not count the heavier nuclei) that are there today? $\endgroup$
    – Yashas
    Jan 3, 2019 at 5:51
  • $\begingroup$ that is a standard calculation that I do not know how to do! but yes, this is the origins of all the H and He isotopes that exist today. Recommend you read Weinberg, "The First Three Minutes", excellent description of this whole process. $\endgroup$ Jan 3, 2019 at 6:11
  • $\begingroup$ I updated my answer to include a way to estimate this probability you are asking about. Honestly, I lack knowledge on cosmolgy to assert with confidence that this the way how everything was initially made... But in a corner of my mind, I kinda sorta thought it was, you know... $\endgroup$ Jan 3, 2019 at 6:35

For neutrons, which are electrically neutral, there is no Coulomb repulsion keeping them away from the nucleus. As such, it's quite "easy" for neutrons to be absorbed into a nucleus*, which is why neutron radiation tends to make materials radioactive - the neutrons are absorbed into the nuclei of the material, which convert into a different, possibly-unstable isotope. So the argument presented only applies in cases where both nuclear fragments/nucleons are positively charged (meaning that neither of them are free neutrons, since there are currently no confirmed bound multi-neutron systems).

In that case, the basic answer, with some caveats, is that the kinetic energy of the colliding bodies brings them close enough together to make the attractive nuclear force** significant.

In collisions like this, the two objects start with some initial kinetic energy, far enough away from each other to be considered "at infinity" as far as the electric fields are concerned. At the point of closest approach, the two objects have no kinetic energy (in the center-of-mass frame), so pretty much all*** of the kinetic energy initially possessed by the objects is now converted into electric potential energy $U=\frac{kq_1q_2}{r}$. A higher potential energy corresponds to a smaller separation between the two (repelling) objects, and a higher initial kinetic energy corresponds to a higher final potential energy. Therefore, the higher the initial kinetic energy, the smaller the final separation of the two objects. Get them close enough together, and the attractive nuclear force takes over.

For example, we can do this calculation with two protons. Let's assume that they collide head-on$^\dagger$, and that they need to be separated by at most 1 fm for the attractive nuclear force to take over. We can then equate the sum of the kinetic energies to the Coulomb potential energies $2K=\frac{ke^2}{r}$. Plugging in the aforementioned values, we require $K=719$ keV. Dividing this by Boltzmann's constant $k_B$ gives an associated temperature of 8.3 billion Kelvin! This is obviously much too high a temperature for even the cores of stars, so this should be puzzling.

The solution to this puzzle is twofold: first, we have to realize that the associated temperature we have derived is assuming that the average kinetic energy$^{\dagger\dagger}$ of an ensemble of protons is above the threshold for fusion. This is not necessary for fusion to, for example, power the Sun; fusion releases enough energy that even a relatively low rate of fusion would be sufficient to counteract gravitational collapse. A gas of protons at a lower temperature will still have a sparse, very high-energy "tail" due to the way the Maxwell-Boltzmann distribution works, so as long as there is a sufficient "tail," fusion can be found in ensembles of much lower temperature.

However, when the comparisons are made between the Sun's temperature and the required fusion rate, we still come up short. This is when it was realized that protons are not classical particles. In particular, the distances involved with fusion are of the same scale as those involved with quantum tunneling! It turns out that the proton can occasionally "tunnel through" the highest part of the Coulomb barrier, significantly lowering the average energy required for fusion and finally giving figures that agree with observations.

So the real answer is that nucleosynthesis happens when there are 1) nucleons in a dense enough concentration to ensure frequent collisions, and 2) high enough average temperatures that the high-energy tail of the Maxwell-Boltzmann distribution can tunnel through the Coulomb barrier. Generally, you see these conditions in the cores of stars and in the early universe. It's no coincidence that these are the places where nucleosynthesis is observed to happen.

*The word "easy" is in quotes here because, in general, it's still relatively rare for this to happen at most energies. This is because, though there is nothing preventing an on-target neutron from entering the nucleus, the nucleus itself is still a very small target when compared to the rest of the atom. As such, you need a high density of incident neutrons, or a high density of target nuclei, for this to occur with any regularity at most energies (there is an exception to this - some isotopes have a neutron capture resonance at a particular energy which essentially makes the nucleus seem much bigger than it actually is, and thus makes neutron capture much more likely than it would otherwise be). These conditions generally only occur in the cores of stars, in nuclear reactors, and in the early universe, which is where most nucleosynthesis happens, so this makes sense.

**I refer to the force holding the nucleus together as the "attractive nuclear force" to differentiate it from the force that holds quarks together inside the nucleons, which is the strong nuclear force. It is true that the strong nuclear force is ultimately responsible for the interactions that generate the attractive nuclear force, but the attractive nuclear force is best seen as an emergent residue of the strong nuclear force, and behaves quite differently from its more fundamental counterpart.

***The electromagnetic field carries away some energy due to the accelerating charges, but at the energies we're talking about this is a very small correction and can be ignored for our purposes.

$^\dagger$If they don't collide head-on, the required kinetic energy will be higher, but generally not more than a factor of 10 higher unless the situation is quite extreme, so this is not a terrible assumption to make.

$^{\dagger\dagger}$Well, average to within a factor of 2 or 3, anyway.


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