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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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Posts: 32
From: uk
Registered: 2/21/08
Re: Einstein's factor of 2 in starlight deflection
Posted: Aug 20, 2011 2:45 PM
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The simplest deflection model to follow is the orbital path. There is
one important difference in that the speed of our light or matter
particle has to be 1/(2^0.5) of escape speed but as this a theoretical
enquiry it shouldn't matter. Also, because orbital speed is slower by
this factor the free fall effect is just g and not g/2. This means
that a beam of, say, protons, travelling at orbital speed, will wrap
itself around the Schwarzschild perimeter in orbital procession along
the lines of the following numerical principle.
After travelling one Planck length the beam will 'fall' one Compton
wavelength, 1.105110504x10^-72m. As the time unit at this scale is
2r, or 5.9381812x10^3m and the Planck length at 8.1x10^-35m, divided
into this is 7.330336x10^37 units of 'fall' then 7.330336x10^37
multiplied by 1.10511044x10^-72m is 8.1x10^-35m. The total deflection
attributable to gravity over a 2r period of orbit is just one Planck
length. But one glance at a drawn geometrical model will show
otherwise. So what's going on? Take a look at what is going on during
basic free fall. Whatever the local rate of free fall it operates as
follows: call the rate of fall x then:
in the first Planck unit of time it will 'fall' a distance of x, that
is Compton length/2 in the following simple way:
Ist Planck time unit,1^2=1+0=1, 2nd, (2^2)=4=3+1. 3rd, (3^2)=9=5+4.
4th, (4^2)=16=7+9. 5th, (5^2)=25=9+16........and so it carries on, the
rate of increase is added to by a factor of 2 at each stage. It goes
1+3+5+7+9...and the reason it does so is because of the gravitational
accelerative process. But in the orbital procedure there is no
gravitational accelerative process during the course of deflection,
neither is there in non orbital procedures. What does happen,
however, is tangential sloping. At each interval of one Planck unit
of length travelled, the direction of our beam is redirected toward
the tangent on the point of contact it happens to be situated at any
given time. This repointing of direction replicates the effect of
freefall adding to 'distance fallen' and substituting a second
factor. Thus, our formula becomes {(7.330336x10^37)^2}
{1.10511044x10^-72}=5.93818x10^3m,2r or local c. Just like
gravitational acceleration.

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