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Topic: Einstein's factor of 2 in starlight deflection
Replies: 12   Last Post: Jul 14, 2014 3:17 PM

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DotProduct Quantum

Posts: 5
Registered: 9/9/11
Re: Einstein's factor of 2 in starlight deflection
Posted: Sep 10, 2011 8:11 AM
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On Sep 3, 6:47 pm, "H.Jones" <> wrote:
> On Aug 23, 12:21 pm, "H.Jones" <> 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
> equi-distant, 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.


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

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