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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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 haroldj.l.jones@gmail.com Posts: 67 Registered: 3/17/12
Re: Einstein's factor of 2 in starlight deflection
Posted: Jul 12, 2014 12:40 PM

Using an orbiting particle in place of the deflecting and escaping lightbeam from
a star's gravitational field becomes a handy template in the understanding of starlight deflection.

The simplest templates are best constructed at the Schwarzschild limit and scale because then all deflections scale down to neat radians and half radians. So the following deductions will be assumed to be at the Schwarzschild scale whatever the mass.

Looking at a satellite as it travels around its orbital path you will notice that it seems to do the following: first draw a circle, representing the orbital path, then at a point (a) on the circumference measure off a tangent measuring one radius forming the straight line (a) to (b) or ab. Then from point (a), in the same direction, measure an arc of one radius length at point (c) on the circumference. (a) to (b) would be the straight line the satellite would travel
without the pull of gravity. (a) to (c) is the shape of its journey under the influence of gravity. The gap between point (b) and point (c) measures a curve
of half a radius length. If you draw a tangent from point (c) and intersect the line ab at point (d) the the angle formed at bdc is equal to one radian. This rule does not just apply to an orbiting particle but equally applies to the freely escaping lightbeam; one radian, half a radius.

Given how little we understand of the true nature of space or gravity the best we can do is study current wisdom on the influence of interaction at the Planck scale. In this way I have chosen a Schwarzschild radius for a Sunlike star of
2,9689x10^3m and a Planck radius value of 4.05049x10^-35m.

Remember that the basic principle of the Planck length is that its mass has a Compton wavelength equal to its own Schwarzschild diameter, 2r.
At the Planck scale we can visualise on the above orbital template two equal circles rotating around one another. Maybe they swap roles, at one they are waves and another they are mass with gravitational pull. If one were to suddenly lose its gravitational pull the other would fly off at the same tangent principle as explained above. The gap between bc still being half a radius and
therefore a quarter of Compton wavelength.

Looking at that simplistic analogy consider this. The Sunlike star has a mass of just under 2x10^30kg, say, 1.99985x10^30kg. That gives it a Compton wavelength
of 1.10519x10^-72m. A quarter of this is 2.76298x10^-73m.

is 7.3299284x10^37. The square of 7.3299284x10^37 is 5.3278513x10^75. Multiply this by a quarter of the Compton wavelength, 2.76298x10^-73 and you come up with
1.48449x10^3m, half a radius again, which is the spatial gap bc at this scale.

Then multiply 7.3299284x10^37 by 2.76298x10^-73m and you get 2.025245x10^-35m
which is half a Planck radius. What you can gather from this is that the gravitational influence amounts to a series of pulses or nudges at the Planck scale. In this case 7.3299284x10^37 nudges. What you will also gather is that in reality the total amount that our satellite shifts, radius by radius, never
amounts to more than half a Planck radius. Everything else is down to Euclid.

Date Subject Author
8/20/11 h.jones
8/20/11 h.jones
8/24/11 h.jones
9/3/11 h.jones
9/10/11 DotProduct Quantum
10/9/11 h.jones
10/27/11 h.jones
1/7/12 h.jones
8/22/12 haroldj.l.jones@gmail.com
12/3/12 haroldj.l.jones@gmail.com
7/12/14 haroldj.l.jones@gmail.com
7/14/14 haroldj.l.jones@gmail.com