On Fri, Apr 26, 2013 at 12:11 AM, Harry Veeder <[email protected]> wrote:
A few experiments conducted before this showed
> ambiguous evidence: two protons emerged from the decay but one
> couldn’t tell that the protons had not been thrown out one at a time
> or both at the same time randomly from the whole Ne-18 or from a
> single lump.
>
A diproton is something to think about. It gives rise to or is indirectly
related to the following novice questions:
- In 3He, which is stable, electrostatic repulsion is felt between two
nucleons, and the strong interaction is felt equally between all nucleons.
In deuterium, which is stable, there is no electrostatic repulsion, and
the strong interaction is felt equally between both nucleons. In a
diproton, which is unstable, there is electrostatic repulsion, and
presumably the strong interaction is in affect to the same extent as
between the nucleons in a deuterium atom. Is the lack of stability of the
diproton due to a slight imbalance between electrostatic repulsion and the
residual strong force, or is it due to the combination of valence quarks
between the two nucleons not being quite right?
- Within a nucleon there are uncounted sea quarks and only three tiny
valence quarks, and these valence quarks are responsible for the residual
strong interaction with other nucleons. Why are there countless sea quarks
and exactly three valence quarks left over, above and beyond them? It
feels a little like having 3 in the numerator and infinity in the
denominator. What is the hidden structure that presumably keeps the
numerator and denominator in balance?
- A nucleus that decays, such as a diproton or tritium, is an unstable
nucleus transitioning to a more stable state. Perhaps there will be a beta
decay (tritium) or a fission (the diproton). Is there an analytical way to
work out the anticipated stability of a given set of nucleons in advance,
or is this something that can only be determined experimentally?
Eric
