
Re: SONNTAG! Symmetries of Nature 'n' Truth about Gravity(& Planck Units).
Posted:
Sep 8, 2012 12:55 PM


It is accepted that the Rydberg energy is electromagnetic and that of the deuterium atom is nuclear. If we consider a theoretical potential energy for the deuterium atom of 1.503278583x10^10 J(mc^2 where m=proton) then we have a ratio between the two of (1.503278583x10^10)/(2x2.179874166x10^18)= 3.448086x10^7. 3.448086x10^7 x 3.62994678(the quantum adjustor)=2.503273756x10^8. c/2.503273756x10^8=1.19760157, the Rydberg adjustor, if you like. 1.19760157(c/2)=1.795159592x10^8kg, the Rydberg mass. These values are nominally constant. Whatever G you choose they will be nominally the same as long as we use c & h, nominally equal to the SI system's. If you use, for example, 6.67261319x10^11 as your G then the correct extrapolation will tell you that such a model has a timescale mass of 1.00949965x10^35 mass units, a Planck mass of 1/3.66555236x10^7 and a Planck radius of 4.05084018x10^35 units of length. The G(c/2) product will be 1.00019955x10^2. Multiplying 1.00019955x10^2 by the Rydberg adjustor, 1.19760173, gives us 1.197840711x10^2, the Rydberg Gm product of this particular mass model. Unfortunately, it is not the SI's kilogramsecond timescale model because of what follows. If we divide 1.197840711x10^2, above, by this particular system's version of the Planck radius, 4.05084018x10^35, we arrive at 2.95701794x10^32, this system,s version of the Rydberg multiplier, but a quick look at the SI system's Rydberg multiplier, 2.95676257x10^32, shows this to be out by a ratio of 1.0000863. Further extrapolation will show that the true value of the mass unit of this system is greater by 1.0000863 than the kg, but, a unit of length will be shorter than our meter by this amount. The total mass content of the timescale mass, however, will be less than our timescale mass by the same amount. The reason why it is possible to work this out is as follows: If we divide (c/2) by the Rydberg Gm product, say 1.197633931x10^2, we arrive at 1.251603076x10^10. If we divide the Planck frequency, 3.700693x10^42, by the Rydberg multiplier, 2.95676257x10^32, we also get 1.25160307x10^10. Which, of course, is the same as for (c/2)/1.197633931x10^2. We can also work out that the Rydberg Gm product is greater than Gc/2 by 1.1976017, the Rydberg adjustor. If we take another look at the opening paragraphs, the ratio, 3.448086x10^7 x 3.62994678= 2.503273756x10^8; halve this and you get 1.251636878x10^8. If you look 56 lines above you will see that the Planck frequency divided by the Rydberg multiplier equals 1,25160307x10^10. It is a shadow of the similar looking 1.251636878x10^8. The difference is that one has been worked out using c/2 and the other uses G. But, because of the numerical interrelation with certain numbers we can find out the same information without G, using c/2, as we can with knowledge of G. What is also interesting is there is no numerical interruption in the gradient between the Bohr atom and the deuterium atom.

