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Topic: SONNTAG! Symmetries of Nature 'n' Truth about Gravity(& Planck Units).
Replies: 9   Last Post: Oct 12, 2012 4:51 PM

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Registered: 3/17/12
Re: SONNTAG! Symmetries of Nature 'n' Truth about Gravity(& Planck Units).
Posted: Sep 8, 2012 12:55 PM
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It is accepted that the Rydberg energy is electro-magnetic 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 time-scale 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 kilogram-second time-scale 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 time-scale 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 5-6 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.

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