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My Modern Physics textbook by Taylor states that when nucleons are less than 1 fm apart, there is a strong repulsive force between them. I am fairly certain that it is not the Pauli Exclusion Principle that the book is talking about. So if this is true, what is the mechanism for such a force?

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  • $\begingroup$ The "hard-core" repulsion sets the density of nuclear matter, so I believe that you are seeing the onset of Pauli exclusion (presumably between the partons making up the nucleons). $\endgroup$ – dmckee Sep 21 '13 at 14:56
  • $\begingroup$ @dmckee: that's exactly what I thought. Taylor is plotting the nuclear potential function stating, from my understanding, that it is a nuclear phenomena from nucleon-nucleon scattering. Furthermore, several pages later he explicitly talks about spin effects as if it was separate issue altogether. $\endgroup$ – Lisa Lee Sep 21 '13 at 15:39
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    $\begingroup$ The thing is that the nucleon--nucleon interaction is not the "real" strong force (which is mediated by the exchange of glouns and manifests down among the partons), but the "residual strong force" (what is left over after the binding of partons into baryons). And because it is an effective theory it does not require intellectual purity. A full description of the residual strong force with PEP in it's proper place is very complex. So your book is handing you a comfortable and useful lie to be replaced with a more complete theory later on. We do that a lot. $\endgroup$ – dmckee Sep 21 '13 at 15:54
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    $\begingroup$ @dmckee: The "hard-core" repulsion sets the density of nuclear matter Not true. The density of nuclear matter is set by a variety of factors. It's perfectly possible to reproduce the actual density of nuclear matter using an interaction without a hard core, such as a Skyrme interaction. See, e.g., arxiv.org/abs/1001.5377 and arxiv.org/abs/nucl-th/0607002 . $\endgroup$ – Ben Crowell Sep 21 '13 at 17:21
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    $\begingroup$ @dmckee: You can make all sorts of models of the nucleon-nucleon interaction. You've given an example of a model that gets the right result using a hard core. I've given an example that gets the right result using an interaction without a hard core. Since the result can be obtained with or without the hard core, it's wrong to attribute the result to a hard core. The reason that nuclear matter doesn't collapse to infinite density is fundamentally the zero-point motion required by the Heisenberg uncertainty principle. $\endgroup$ – Ben Crowell Sep 21 '13 at 17:50
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From Wikipedia:

The nuclear force is only felt among hadrons. At small separations between nucleons (less than ~ 0.7 fm between their centers, depending upon spin alignment) the force becomes repulsive, which keeps the nucleons at a certain average separation, even if they are of different types. This repulsion is to be understood in terms of the Pauli exclusion force for identical nucleons (such as two neutrons or two protons), and also a Pauli exclusion between quarks of the same type within nucleons, when the nucleons are different (a proton and a neutron, for example).

(Emphasis added.)

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  • $\begingroup$ at what distance does PEP start to take affect with nucleons? Also, please my comment above and please free to enlighten me. $\endgroup$ – Lisa Lee Sep 21 '13 at 15:41
  • $\begingroup$ @LisaLee: at what distance does PEP start to take affect with nucleons? It's not an effect that has a range. What's happening is that the nucleons are composite, so it's an approximation to think of them as pointlike fermions. See physics.stackexchange.com/questions/75403/… $\endgroup$ – Ben Crowell Sep 21 '13 at 17:53
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Nucleons get close to each other to form nuclei so this blanket statement of repulsion needs qualification and I am puzzled why the Pauli exclusion principle is involved.

Protons will repulse protons at one fermi due to the same charge repulsion. That is the reason that in order to bind into nuclei of more than one nucleon neutrons are necessary so that the repulsive electric force is overcome by the much stronger attractive force of the strong interactions and a stable solution exists , which are black lines in this wikipedia diagram of nuclides. All the other isotopes decay because there exist lower energy states than the one bound by the strong interactions between the nucleons (protons and neutrons). In general for high Z more than twice the number of neutrons are needed to overcome the same charge repulsion from the protons and get a stable nucleus.

The only place I have seen the Pauli exclusion principle invoked in connection to nucleons is in justifying why all the nucleons of a nucleus do not end up on one energy state, the lower one ( similar to why all the electrons in an atom do not end up in the lowest ground state), but fill up consecutive energy layers.

The Pauli exclusion principle is involved in the basic explanation of the shell model for nuclear energy states. The evidence for shell structure in the nucleus was surprising at the outset, because a dense collection of strongly interacting particles should be bumping into each other all the time, resulting in redirection and perhaps loss of energy for the particles. The Pauli principle effectively blocks the loss of energy because only one nuclear particle can occupy a given energy state (with spin 1/2, neutrons and protons are fermions.) In this dense collection of matter, all the low energy states will fill up. This means that the particles cannot take part in interactions which would lower their energy, because there are no lower energy states they can go to. Scattering from an external particle which raises the energy of a nucleon can happen, but scattering which lowers an energy level is blocked by the exclusion principle.

Now if he is talking of scatterings at high energy, proton proton scattering for example, that probe small distances, deep inelastic scattering showed the hard core existence of quarks. Again I have difficulty to think how the Pauli exclusion would contribute to this hard core since the energy states in scattering are continuous and fermions could easily be accommodated in different energy states but within the Heisenberg uncertainty principle indistinguishable.

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Protons and neutrons are different Protons contain one elementary charge and neutrons electric charges with no net charge. The consequence is that there exists an electric attraction between a proton and a neutron equilibrated by the magnetic repulsion between nucleons. The electromagnetic interactions between nucleons are, except for the Coulomb repulsion between protons completely ignored by the nuclear scientists.

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  • $\begingroup$ The nuclear fine and hyperfine splitting (i.e. magnetic effects between the nucleons) are not only treated in the literature, but treated in most introductory books on nuclear physics. $\endgroup$ – dmckee Mar 9 '14 at 14:49

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