Corrections to above. > Another way of looking at the Hydrogen atom, other than the Rydberg > Multiplier method, is as follows: Because the proton is that much > smaller than the Planck mass the theoretical surface g is likely to be > higher than c(in equivalence) necessitating a rise in the event horizon > bringing its diameter nearer to the size of its Compton wavelength. > This results in a Lorentz transformation of the proton's Gm product to > 29.6906036. This multiplied by the proton mass, 1.672623x10^-27kg, is > 4.966118653x10^-26, which is equal to hc/4 and also equal to the Gm^2 > product of G times the Planck mass squared. The electron speed is > 2.187691412x10^6 ms so the electron Compton wavelength will be > 3.324918617x10^-10m; divide this into hc/4 and you get > 1.49360607x10^-16. Divide this by the Rydberg energy and you get > 137.035989/2, the reciprocal of the fine structure constant/2. We can > bypass this with hv, using the electron speed, above, we get > 1.4495808x10^-27. Divide this into hc and you get 137.03598, the reciprocal of the coupling constant for the electro-magnetic interaction. > > THE DEUTERIUM ATOM > > We can apply the same reasoning to the deuterium atom as follows: > One nucleon is ascribed an energy of 1.78204x10-13 J, a speed > equivalent to 1.03219017x10^7ms and a wavelength equivalence of > 3.83793982x10^-14m. (hc)/3.83793982x10^-14m equals > 5.175817116x10^-12. Divide this by the Deuterium nucleon energy of > 1.78204x10^-13 J and you get 29.04431, the deuterium nucleon analogue to 137.035989. > Using the hv principle above, hx1.03219017x10^7ms=6.839369997x10-27. > Divide this by 1.44958084x10^-27, above, and you get 4.718170781. > Multiply this by 29.04432, above, and you get 137.035989. This is > because c/v=29.04427537 and 29.6906036 divided by > 29.04427537=1.022253206; multiply this by 4 then > multiply by the proton mass and divide by the wavelength > 3.83793982x10-14m and you get 1.7820438x10^-13 J, the nucleon energy. The question is, what is the nature of the shovel that does the shovelling, the force? The answer is that we don't know. The transition from Bohr atom to the deuterium seems to be one of getting over the hurdle of repulsion between nucleons. Once that is achieved, given the closer proximity, the forces don't seem to be that fantastically different in magnitude. The likelihood is, or was, that in the beginning there was an explosive atmosphere where just about everything was operating at close to the speed of light, including gravity. Relativistic effects began to slow things down to the stable conditions we experience in the Solar System today but at the heart of the atom is an indication of the volatile nature of the Big Bang. Back then there was one force and it was very loud. The pacifying effects of relativity and the Lorentz transformation have created the orderliness of today's universe. Without them the Big Bang would still be on the rampage.