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Photons have no surface potential

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Steven Sesselmann

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Photons have no surface potential

PostThu Feb 26, 2015 2:51 pm

Hi All,

After successfully applying GP theory to solve the centripetal vs Electric force of the H Atom, ( viewtopic.php?f=5&t=33 ) I turned my attention to the photon.

Initially I was quite frustrated as my first calculations predicted nonsense, but after chewing on it for a few days it now comes together nicely.

What was not obvious to me at first, is that a photon rotates, i.e. it has angular momentum, this solved the problem.

First of all w start by defining a photon as a wave, with a peak and a trough distributed equipotentially around the baseline. Where the baseline is a geodesic radiating outwards from the observers ground potential (more on that later)

So let's first look at the simple case of a photon with 2 eV energy travelling horizontally at ground potential.

(a 3 eV photon is visible orange light and has a frequency of 483.6 THz.)

First it is necessary to show that the photon has angular momentum, so we have defined the wave is symmetrical with respect to the baseline, so a 2 eV wave will have a crest at 2 eV, a baseline at 1 eV and a Trough at 0 eV.

So with respect to the observer the difference in potential between the trough and the crest means according to GP that the two half waves travel at different speeds.

\[ \Delta v_{crest} = c(\frac{\Delta \phi}{\Phi}) = c(1/938000000) = +0.32 m/s\]

and

\[ \Delta v_{trough} = c(\frac{\Delta \phi}{\Phi}) = c(-1/938000000) = -0.32 m/s\]

So we see from the above equations that the trough and crest propagate at different speeds, and from this we can now work out what the coulomb force should be in order for the centripetal force and the coulomb force to balance.

(will insert calculation here later)

Now that we have shown that the photon rotates, and that it is symmetrical around the baseline, we can also prove that it's surface potential is zero. Just imagine a dumbell with equal and opposite charges rotating symmetrically on it's axis, it's net charge distribution on the surface will be zero.

Now according to GP theory any mass with a surface potential of zero must propagate at;

\[ \Delta v_{trough} = c(\frac{\Delta \phi}{\Phi}) = c( 0 - 930 MV/938 MV) = 0.991c \]

This might at first seem contradictory, but this slight deviation from the speed of light is due to the redshift experienced by observers at ground potential. In fact that's as fast as we will ever observe light here on Earth.

It does not matter at which potential a photon is initially generated, as long as the wave is symmetric around the baseline (with respect to the observer), it will have a zero surface potential, and will travel at the local speed of light along a geodesic line.

So there you have it, photons with any energy travel at the same speed, as they by definition have a zero surface potential, and posess intrinsic angular momentum, which is a function of it's energy.

Addendum:
Now if the observer changes potential relative to the photon emitting source, we see that the relationship between the observer and the photon changes, on a small scale this will follow euclidean geometry, but on a large scale it will follow the geometry of general relativity. Put simply, light will appear to propagate in a straight line, but in fact when we concider the observer to always be at rest, light clearly curves.


Steven

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Photon
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Steven Sesselmann
Only a person mad enough to think he can change the world, can actually do it...
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Steven Sesselmann

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Re: Photons have no surface potential

PostWed Dec 09, 2015 8:22 pm

Here is a link to a youtube video I made in response to a question on Facebook. I attempt to explain the difference between bosons and fermions.

Correction: In the beginning I incorrectly describe the perpendicular axis as the "potential energy" axis, it should simply be "potential axis"

https://youtu.be/sxWXMW-GbtE

Steven
Steven Sesselmann
Only a person mad enough to think he can change the world, can actually do it...

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