The answer to the OP's question is that the density of nuclear matter is limited by the Pauli exclusion principle and the Heisenberg uncertainty principle.
For a fixed N and Z, if you make a nucleus more compact, you make it more bound in terms of the strong nuclear force, which is attractive. However, the Heisenberg uncertainty principle results in an increase in kinetic energy. The total energy is minimized at a certain density.
For varying N and Z, you do not get an increase in density by adding more nucleons. The reason for this is the Pauli exclusion principle.
Azael said:
In the textbook(Introductory Nuclear Physics by Krane) I had in a nuclear physics class it says that the nucleon-nucleon interaction becomes repulsive at very small distances.
When I read that I assumed the strong force itself becomes repulsive at short distances.
But does it really mean the columb repulsion dominate over the strong force at extremely small distances?
jdog said:
Yes. The Strong force goes like exp(-ar)/r so it falls off very quickly when r is large, and the columb force (with it's 1/r^2 dependence) dominates when r goes to zero, So the strong force will only hold protons in place if they are close enough, and if they get too close, the columb repulsion takes over. i.e. it is in fact a stable equilibrium.
This is incorrect. The nuclear force is often modeled by an attractive force that drops off exponentially, *plus* a repulsive core at very small distances. This is what Krane is talking about. It has nothing to do with the coulomb repulsion. The coulomb repulsion does not blow up at r=0, because the charge distribution of each nucleon is not pointlike.
bjschaeffer said:
The main force that keeps protons from "merging" into each other is the so-called "tensor force" which is in fact the magnetic repulsion between the magnetic moments of the nucleons. The electrostatic repulsion between protons is not efficient enough.
This is incorrect. Electromagnetic interactions are not really relevant to this entire discussion, nor is it particularly relevant to make a distinction between the charged protons and the uncharged neutrons. The Coulomb energy in the liquid-drop model
http://en.wikipedia.org/wiki/Liquid_drop_model varies like Z^2, so it's negligible for light nuclei. Heavy nuclei don't have a different bulk density than light nuclei, so you can see that the electromagnetic interactions aren't strong enough to have a significant effect on this discussion.
marlon said:
But actually, you do not need to go THAT deep. I suggest you do some research. Try answering this question : what conservation laws need to be respected when a neutron decays into a proton and vice versa. Then you will understand why this decay does not happen "at random" in the nucleus.
marlon
This has nothing to do with the OP's question. The OP's question is a question about nuclear structure, not a question about beta decay.
physicophile said:
As I understand, the reason I can't put my hand through a wall is due to the electromagnetic repulusion of electron.
This is not really correct. It's more closely related to the Pauli exclusion principle. The bulk electrical force between your hand and the wall is zero, since both objects have zero charge density on a macroscopic scale. If you look at the microscopic scale, the electromagnetic interaction between your hand and the wall may actually be attractive, i.e., your hand may tend to stick to the wall.
physicophile said:
I guess my question can be rephrased why isn't baryonic matter allways an bose-enstien condestate.
Nuclear matter is cold and composed of fermions; you can't get a Bose-Einstein condensate with fermions. Condensed matter at normal temperatures can be composed of either fermionic or bosonic atoms, but you don't get a Bose-Einstein condensate because of the temperature.