
Re: Einstein's factor of 2 in starlight deflection
Posted:
Sep 10, 2011 8:11 AM


On Sep 3, 6:47 pm, "H.Jones" <h.jone...@googlemail.com> wrote: > On Aug 23, 12:21 pm, "H.Jones" <h.jone...@googlemail.com> wrote: > > > To obtain g at Planck time we merely divide our current SI G by the > > Planck frequency squared which is the same thing as multiplying G by > > Planck time squared. (6.6712139x10^11)/ > > (3.7007618x10^42)^2=4.87105x10^96. > > {(4.87105x10^96)x(2x10^30kg)}/(2.9690906x10^3m)^2=1.1051104x10^72m, > > g at one Planck time unit per time unit. > > Incidentally, if you take a look at this Planck G, 4.87105x10^96, > > divide by the Sun's Compton wavelength, 1.1051104x10^72, you wind up > > with the reciprocal of 2.2687315x10^23 which is the frequency of the > > Compton wavelength of the proton. > > The interesting thing about the Sun/proton mass content of the G/ > Planck frequency squared situation is what does it mean and where does > it lead to? Central to the basic principle is C/G. C because the > area under discussion concerns the kilogram/second and G because we > need to find it. All magnitudes of mass have a Schwarzshild diameter > and a Compton wavelength. They mirror one another because they are > equidistant, differentially, from the Planck length; one is bigger > than the Planck length and the other is smaller. In this situation > these magnitudes of mass have a twin where the situations are > reversed, the Schwarzschild diameter is equal to the other's Compton > wavelength and vice versa. The opposite to the kilogram, for instance, > is the planck mass squared. The opposite of the mass represented by G/ > C is h/4 or 1.65651887x10^34kg, and so on. If we divide 2x10^30kg by > 1.672623x10^27kg, the Sun and proton masses, we arrive at > 1.1957267x10^57. We can arrive at a fairly accurate assessment of the > radius by dividing this figure by (c^2)/h, the frequency of the > kilogram's Compton wavelength, which provides us with > (2.9690906x10^3m)^2. So. 2.9690906x10^3m can be considered a fairly > reliable approximation of the Schwarzschild radius of a mass, > 2x10^30kg. Now, the radius of our own kg/second system is, obviously, > C/2, therefore, the analogue of 1.1957267x10 ^57 is [(C/2)^2][c^2/ > h]=3.047659486x10^66 which is equal to 1.0097113x10^35kg, the current > approximation of the kilogram/second timescale mass, divided by > 3.313071538x10^32kg, which is, in this context, the analogue of the > proton mass. If we travel on a little further, somewhat higher than > our own kilogram/second time scale mass, until we reach an analogue to > the proton mass of h/4, 1.65651887x10^34kg, we get a timescale mass > of 2x1.009721668x10^37kg. This is represented by the ratio > 1.672623x10^27, the proton mass, over h/4, which equals > 1.00972167x10^7. Multiply this by the Sun's mass and we arrive at the > figure above. . . > We know that the proton's Compton wavelength is 1.32141x10^15m. We > know that (C/2)/2.9690906x10^3m=5.048556x10^4. > 1.32141x10^15x5.048556x10^4=6.6712139x10^11, or G. Or else, the > Compton wavelength of the proton's kilogram/second time scale's mass > analogue, 3.31307153x10^32kg, is also equal to G or 6.6712139x10^11.
Hi, just thought I'd help, as I have the answer to your paradox.
What you are really asking is;
Why this quantity 2.9690906x10^3m
Why does the SR of our sun help approximate G?
It's as simple as this.
A mass of 2*10^30kg is required for the volume of mass to have Hydrogen evenly spaced at the first Bohr orbit.
5.29*10^11.
Do the math;
Sun radius / (5.29*10^11) = protons per radius.
(4/3)*pi*protons per radius^3 = protons in sun.
Sun Mass / protons in sun = proton mass.
Hence your wavelength, 1.32141x10^15m
etc etc etc

