After five days of silence, I can reveal a bit more about my work on how the Sun works. To some extent this a neat jump from my post about photosynthesis, as we need to know how the photons are created that are captured by plants. I still have more questions than answers, but the basic ideas are beginning to materialize.
Firstly, we need to consider the immense heat and pressure at the core of the sun, where most fusion takes place. The temperature of the core of the sun is 15.5 million Kelvin/°Celsius (unit irrelevant at this scale), the pressure 265 billion bar and the density of the mixture of hydrogen (more specifically protons), helium and trace other elements is 150 g/ml. Initially these numbers don't seem to mean anything, but these can be used to evaluate the nature of the physical conditions in solar fusion.
First of all, from the kinetic theory of gases we can calculate that 15.5 million Kelvin means that the hydrogen plasma moves at the speed of 620 000 m/s, or 0.21 % the speed of light. This also means that the hydrogen plasma has a mass 0.21 % higher than at 0 K. However, this extra mass isn’t just some abstract energy, but concrete increase in the length of string, that allows the supramolecular shell to rotate.
Next, the density of 150 g/ml might not mean anything without context. However, this is 150 times denser than water and almost 8 times denser than gold and almost 7 times denser than osmium, the densest substance on earth. So how can anything be denser than gold or osmium? Well, all of us have heard that most of visible matter is empty space. However, if one does not understand the concept of the orbital being a rotating sphere of infinitesimally small particles moving at the speed of light, it might be hard to understand what separates these near empty shells from each other.
Even in the solar core, most of the space is still empty. However, now we deal with a crusts that aren’t the size of a supramolecular orbital, that can be as large as several micrometers, but the orbital of hydrogen, which is immensely smaller. We know the distance of two hydrogens in H2 is 74 pm (pm = one millionth of a millionth of a meter). If we assume that the hydrogen orbital are two spheres, the 74 pm is split into 2 x 37 pm. The volume of these spheres is
Knowing that the mass of a hydrogen atom is
we can estimate that the dense packing of these orbitals would be roughly 32 g/ml. This is still five times smaller than the density of the solar core, but not hugely off. Knowing that over half of the mass of the solar core is made of helium, with an atomic mass four times larger than hydrogen, the estimate is not as far off anymore. This indicates also that there is a possibility that the supramolecular orbitals of proton are no longer separated from each other, but already overlapping. This would point that the idea of overlapped orbitals in my last post isn’t completely crazy.
However, in my last post I entertained the idea that fusion takes place when two orbital collide. However, based on this approach it seems more likely that the two supramolecular orbitals of proton are already overlapping when a collision takes place. This means that there is still a collision that stops the supramolecular orbital of proton from rotating, but the unspooling of the stopped orbital impacts two orbitals that had been rotating in synchrony.
To illustrate the process of overlapping orbitals, I went again for some blender magic. First, I made a mirror image of the hydrogen orbital and overlayed it with the original orbital. These reflect the unfused proton and neutron orbitals. Then I rotated them so that four loops would intersect both a the center and at the edges of the orbitals.
Then I colored pairs of loops so that it is possible to follow the path of the Planck spheres on the orbitals. I also added arrows at critical points to help with this. Yello/yellow-green color depicts the neutron orbital and blue/turquoise depict the proton orbital.
To help a bit further, I took the illustration in blender and drew the same scheme in PowerPoint. As this x-z projection hides half of the orbitals, I drew both layers: the upper layer above and the lower layer below.
If one observes closely, the orbitals no longer pass the intersections at a right angle. This complicates the calculation a bit. At least for me, the above scheme shows that when the orbitals fuse, the rather complicated shape of the orbitals does not remain, but rather the complicated orbital rearranges and relaxes into two circular orbitals at right angles, just as noted in the last post.
However, only when I drew the orbitals in Blender, I noticed that the neighboring intersection in the orbitals are twisted. I don’t know the exact reason for this, but it is related to the tilting of the proton and neutron orbitals. This might be relevant to the generation/emission of the positron in the reaction.
While this approach is far from clear or sure, it seems that the length of the circular arcs in the scheme above is just the right amount longer than the length of the full circle orbital after fusion to account for the binding energy in the fusion of protons into deuterium.
It might seem that I am getting close to something that could be written up as a manuscript, much in the same vein as the Counterevidence paper. However, the only feasible way for this would be if I was able to conclusively show that the geometrical approach truly explained the binding energy of deuterium. Intuitively I feel that I am on the right track. However, to get from the Blender/PowerPoint scheme to forming new equations and using them to calculate orbital lengths isn’t that easy.
The last time I did anything like this was in the spring. That time the only way to solve the equations were to find trigonometric shortcuts with which avoided mathematics I knew I didn’t know how to use. There’s no guarantee such shortcuts exist for this scheme.
Of course, If someone else would like to try a crack at this, be my guest. The basic equations can be found in the supplementary information file in the Counterevidence paper. If you manage to turn these rough ideas into measurable predictions, you’ll be in good company. In 1939, Hans Bethe did the work using ‘standard’ quantum mechanics and received the 1967 Nobel prize in physics for it. If even part of the results obtained with wave functions can be replicated with the trigonometric approach and deterministic strings traveling at the speed of light, there’s a good chance that there might be a Nobel prize waiting at the other end.
Of course, this is with the major caveat that the basic idea must be right. If my whole theoretical construct is wrong, this will be relegated to the long list of ‘almost theories of everything’.
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