In 1998, Japanese scientists discovered that neutronos (ultimatons) are not weightless; the weight of a neutrono is a tiny fraction of that of an electron.
This note represents a feeble attempt to understand the structure of an electron. We are vaguely aware that some investigations were done about quarks; two ups and one down form a proton and two downs and one down a neutron, etc. and there are other particles. But they are subatomic particles. It appears that physicists are still busy breaking up these particles, but not electrons. So this page includes idle speculations about the structure of an electron. The fundamental elements are called ultimatons here. We may require a gigantic microscope(?) or supercollider yet to be invented.
How are ultimatons positioned in a typical electron? We have used the following assumptions to build an electron:
No. Because of the huddling proclivity, even if all ultimatons are temporarily positioned on the surface only, they will jump to the core and fill it. This implies that these ultimatons have more than one layer inside an electron.
Specifically, suppose that 100 ultimatons are sprinkled on the surface of a hypothetical hollow globe. Then the distance between two adjacent ultimatons is much smaller than two ultimatons positioned at diametrically opposite side of the globe. Huddling proclivity insures that in this case they would jump to the core.
Huddling proclivity implies that there are at least two layers of ultimatons within an electron. There might be more than two layers.
Huddling proclivity also implies that an electron is a compact globe. How compact it should be is somewhat ambiguous.
2. How many ultimatons are located in the upper hemisphere?
Consider the upper hemisphere of an electron. Here, the desigination of upper and lower hemisphere of an electron is arbitrary. By the symmetry assumption, there are exactly 50 ultimatons in the upper and lower hemispheres. One can argue that some ultimatons stay on the border lines between the upper and lower hemispheres. However, in this case, all you have to do is rotate the electron a few degrees around the nucleus so that these borderline ultimatons belong to either the upper hemisphere or lower hemisphere. This finding, in itself, does not appear to be so important. However, it implies that no ultimaton occupies the dead center of an electron.
If an ultimaton occupied the center of an electron, excluding this center, there would be only 49 ultimatons in the upper hemisphere, and an equal number of ultimatons in the lower hemisphere, resulting in a total of 99 ultimatons. If the electron has 50 on one hemisphere and 49 in the other hemisphere, excluding the center, then it is not symmetric. Thus, symmetry assumption precludes the possibility that there is a central ultimaton within an electron.
3. Can there be two ultimatons in the first layer (or center)?
No. Surrounding them with the remaining 98 ultimatons would not make the group a true globe. The line connecting the inner two gives the electron an axis, and other surrounding ultimatons (not revolving) could make the set symmetric around this imaginary axis, but not in other directions.
4. Can there be three ultimatons in the first layer?
No. The three ultimatons in equal distance will define an equilateral triangle. The remaining 97 ultimatons might define a symmetric group, and the plane defined by the triangle might provide the axis of symmetry, but the electron thus formed may not necessarily be symmetric in every direction.
Moreover, since 97 is an odd number, one of them must also be on the same plane as the equilateral triangle to make the group symmetric around this plane. Such a flat inner structure is not likely to make the globe a round sphere on its surface. It will show on the outside.
To offset this flat inner strcutre, it would require two other flat planes, orthogonal, or perpendicular to the first plane. Hence, it would require at least 16 ultimatons in the inner layer.
5. Can there be four ultimatons in the first layer?
Perhaps. That is a possibility, because four ultimatons equally spaced (not on a plane) define a tetrahedron, as shown in Picture 1. The tip of the red stick indicates the imaginary center of the tetrahedron. In this case, one can derive four spokes emanating from the center to the four vertices. (Five is also a possibility, and I have not been able to rule out this case. In this case, connecting the ultimatons with straight lines will yield a hexahedron.)
Given that the innermost layer is a tetrahedron, how many ultimatons can be attached in the second layer? It is not clear. Consider the following representation of a spoke:
(0) -- 1 -- 3 -- 5 -- 7 -- 9,
where the parenthsis ( ) refers to the center, the first element denotes the number of the ultimatons in the first layer, the second denotes that of the second layer, and so on. This spoke indicates that it has none in the center, one in the first layer, and three in the second layer, etc.
In this case, each spoke contains 25 ultimatons, and since there are four spokes from the center, there would be exactly 100 ultimatons. I have not been able to construct a globe with this structure because of limitations of tools (The sticks do not bend). One other problem is that there are five layers, perhaps too many.
In a hexahedron, as shown in Picture 2, all lines connecting the ultimatons have the same distance.
From the empty center, one can draw five lines to the five innermost ultimatons. Each of these lines from the empty center can be further extended to add more ultimatons. This hexahedrons with five ultimatons can be represented by the following:
(0) -- 1 -- 3 -- 6 -- 10.
Picture 3 shows this spoke with only three layers, the last layer being 10 (not shown). The tip of the red stick indicates the empty nucleus of the electron.
This hexahedron is a stable symmetric collection of ultimatons. In this case, on each spoke, there are a total of 20 ultimatons. Since there are five spokes, there would be exactly 100 ultimatons. One conceptual problem with this model is that there are still four layers, and my sons have not been able to construct this model because of limited tools (again, the sticks are not flexible). Perhaps a computer program could do this, and Douglas is working on this. Anyway this is a possibility.
Since August 11, 1998.