h.jones
Posts:
32
From:
uk
Registered:
2/21/08


Re: SONNTAG! Symmetries of Nature 'n' Truth about Gravity(& Planck Units).
Posted:
Feb 27, 2012 1:21 PM


THE SCHWARZSCHILD RADIUS:
Nassim Haramein's theory of the proton having a discrete mass of 8.9x10^11kg is interesting because it exhibits some aspects of the Schwarzschil/Compton opposites phenomena I have recently been mailing about. The main difference being that the opposite of the proton has a diameter equal to the proton's Compton wavelength of 1.32141x10^15m making its mass 4.45039x10^11kg not 8.9x10^11kg. Therefore, its radius is half the wavelength and its Gm product 29.6906036. The theory seems to suggest that the mass has turned itself inside out in some way. There are some suggestions that can go some way to explain why.
In gravitational free fall onto a Schwarzschild surface measured in Planck time units, the rate of fall is half the Compton length(of the gravitating body)per Planck time unit. So in the case of the Sun the rate of fall would be half the Compton length of the Sun per Planck time unit by the falling object. But this rule only applies down to and including a Planck sphere. After that the rule no longer applies. The Schwarzschild radius of a proton is 2.483176x10^54m. Half its Compton wavelength is 6.60705x10^16m, some 2.66072482x10^38 times bigger. If an object was subjected to this amount of acceleration it would have to travel faster than light. Clearly, a new set of rules is needed. What about relativistic transfer? what about the primordial black holes that were abundant after the Big Bang? These were estimated to be around the size of the proton opposite. The surface g at the Planck limit is 1.109439913x10^51m, which must be equivalent to c. If the primordial black hole's surface g exceeded this it would go into relativistic meltdown or gravitational relativistic spacetime contraction. If radius contracted so would surface area, so would length, so would time. A four dimension meltdown. For a 2.66x10^38 mass contraction we would need a 4.038776x10^9 contraction rate to effect that.
The Rydberg energy is 2.179874166x10^18J. We're not sure what the proton nucleus poential is. What the nucleus does do, however, is exert a force on the wave structure of another particle. The mass time equivalence of a wave can be whittled down to h/c^2 or 7.3725x10^51kg. 2.95676257x10^32x7.3725x10^51=2.17987x10^18J. 2.95676257x10^32 is c^3 multiplied by the Rydberg constant. So something in the nucleus is generating all that power, 2.95676x10^32.
In recent posts mention has been made of the constant 3.62994678 fom the Gn model. If we take half the proton wavelength, 6.060705x10^16m and divide by 3.62884678 we get 1.82015064x10^16m. If we divide this into the Rydberg energy of 2.179874166x10^18J we get the figure 1.1973731534x10^2. This is the discrete Gm equivalence of the nucleus. So our formula could be:
(1.1973731534x10^2)(7.37250325x10^51)/4.05049049x10^35m= 2.179874166x10^18J. The radius or distance, here, is measured by the Planck radius. The situation.here, is nucleus/wave structure which could be construed as zero separating distance. As nothing can be considered smaller than the Planck radius it must be Planck.

