# I Why can we not produce a "giant" nucleus?

#### PeterDonis

Mentor
What is the degeneracy pressure?

#### hilbert2

Gold Member
No, it would collapse. A static object made of ordinary matter can't remain static at a radius below 9/8 of the Schwarzschild radius for its mass.
That was what I meant, it would stay confined in the form of a black hole, with the electric field produced by the protons probably extending outside that BH.

#### mfb

Mentor
Neutron stars have no protons so there is no Coulomb repulsion to cancel. However, Coulomb repulsion is not the only effect involved. There is also degeneracy pressure, which is what balances gravity in a neutron star.
Neutron stars have protons - just not as many as neutrons. The interior has many neutrons and a few protons and electrons. The outer parts have nuclei and even regular atoms.

#### PeterDonis

Mentor
Neutron stars have protons
But few enough that their Coulomb repulsion is negligible compared to the neutron degeneracy pressure. Also, the protons are balanced by electrons, which cancels at least part of the Coulomb repulsion anyway.

The outer parts have nuclei and even regular atoms.
But again, the effects of these are negligible compared to neutron degeneracy pressure.

#### TeethWhitener

Gold Member
It is the van der Waals force between neutral matter. This is still more long range (namely $O(r^{-6})$) than the exponentially decaying strong force.
The $O(r^{-6})$ part of the van der Waals potential is the attractive portion of it (induced dipole-induced dipole). The repulsive portion is far more complicated, but an exponential works pretty well as an approximation.

#### A. Neumaier

The $O(r^{-6})$ part of the van der Waals potential is the attractive portion of it (induced dipole-induced dipole). The repulsive portion is far more complicated, but an exponential works pretty well as an approximation.
The repulsive force between neutral atoms is the exchange force due to the Pauli principle. It grows for $r\to 0$ only like $r^{-1}$. An exponential is a traditional but incorrect approximation.

But I just realized that all this onöy applies to atoms and not to nuclei. It is still relevant for the formation of neutron stars since gravitation must overcome the repulsive force between atoms....

#### Xforce

But why in the Earth we only see a finite number of neutrons in atoms?What is the condition to exist a "giant" atom?
Actually, when gravity is holding them together instead of strong nuclear force (I’m talking about neutron stars) that’s the condition of having a giant atom... because gravity decays much slower than nuclear forces over distances. However, try not to make it too big because it will collapse and becomes a black hole.

#### fxdung

What is an exchange force as Prf Neumaier said?

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