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


Re: Einstein's factor of 2 in starlight deflection
Posted:
Aug 20, 2011 2:45 PM


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.

