
Re: gravity begins with the proton
Posted:
Jun 7, 2014 1:17 PM


The strong/gravitational force ratio of 2.660725x10^38, (Mpl/Mpr)^2, can be found by a simple route. {c(qa)29.6906036}/8=4.038778x10^9. (4.08778x10^9)^4 = 2.660725x10^38. G(Mpl)^2=hc/4=4.9661186x10^26. 29.6906036 is the GM product of a primordial black hole where the Schwarzschild diameter is equal to the Compton wavelength of the proton. (1/hc/4)^0.333r=1.0882306x10^9/4=c(qa), qa = 3.62994678. But, 2.660725x10^38 does not only determine the function of the microscopic world of nucleons and electrons but also the structures of stars. 2.660725x10^38 Planck masses comes to 7.259355x10^30kg. This the threshold mass that determines whether a collapsing star becomes a black hole or stays a neutron star.
It is not hard to see why this is so.
(2.660725x10^38)^1.5 proton masses equals 4.34x10^57 times 1.672623x10^27kg which is also 7.259355x10^30kg. The Schwarzschild radius of this particular star will be 1.077724141x10^4m. Divide this by half the Compton wavelength of the proton, 1.32141x10^15, and you get 1.6311729x10^19. The square of this is 2.660725x10^38. The cube is 4.34x10^57. This means that in this particular Schwarzschild Sphere will fit 4.34x10^57 Compton proton waves, assuming they are spherical, the same number as proton masses residing in the same star.
Not only that, 2.660725x10^38 proton masses become 4.45039x10^11kg, the mass of the primordial black hole. 7.259355x10^30kg/qa is a little under 2x10^30kg. This is around the mass of our sun and is the average mass for stars in general.
We see here the two ends of the Cosmic scale corresponding to a set of pure numbers determined by the Planck and proton masses and that cute little constant determined by the G(Mpl)^2 formula, 3.62994678, qa.
All of this can be found from two constants, c & h.
Work it out; we start off with space and energy and then fro this cosmic soup we get primordial black holes. The surface gravity approaches c but spacetime continuity cannot be maintained. Try to do a diagram where g approaches c; do a section from say 0.5c, in stages, down to 0.99c, using the Lorentz contraction formula. Break it up into five stages, 0.5,0.666r,0.75,0.9 to 0.99 and you'll see what I mean. No smooth pathway. When the radius contracts by 4.03877x10^9 its area conracts by the square of this and time slows down by a further dimension making a four dimensional collapse down to the proton.
The surface is 'charged' because it is still operating as 4.45039x10^kg within its own nucleus. That is why the gravitational force operating beyond the nucleus is 2.660725x10^38 times weaker. The wave structure transcends such problems and allows strong and electric forces such greater energies.

