On The Fine Structure Constant

We have the following numerology,

$$ a_1 = \alpha = 0.007297352569311, $$

and,

$$ 1/a_1 = 137.03599908348923, $$

and,

$$ a_2 = 1 - 137*\alpha = 0.0002626980043929361, $$

and,

$$ 1/a_2 = 3806.6524422630514, $$

and,

$$ a_3 = 1 - 3806*a_2 = 0.00017139528048515373, $$

and,

$$ 1/a_3 = 5834.466370190515, $$

and,

$$ a_4 = 1 - 5834*a_3 = 0.00017139528048515373, $$

and,

$$ 1/a_4 = 12510.375853474035, $$

and,

$$ a_5 = 12510*a_4 = 1, $$

Hence,

$$ a_4 = 1/12510, $$

and,

$$ a_3 = (1-1/12510)/5834 $$

and,

$$ a_2 = (1-(1-1/12510)/5834)/3806 $$

and,

$$ \alpha^* = a_1 = (1 - (1-(1-1/12510)/5834)/3806)/137. $$

This means that this expansion produces the value,

$$ \alpha^* = 0.007297352569310999, $$

which is as close as it can get given the exactness of the measured value of the fine struture constant.

So what's the meaning of this. Well \(\alpha\) has the property that it creates a uniform loop distribution verry efficiently. Consider,

$$ \cos(\alpha \phi). $$

When the loop 1 time (\(\phi \to 2\pi\) the extra position is \(2\pi \alpha\) and we will at each loop fill up the circle uniformly. Now at the 137 turn we have

$$ \cos(2\pi \alpha 137) = \cos(2\pi (1-a_2)). $$

As we see we will continue to effectively make an even more uniform distribution with the fraction \(1/137/3806\) and so it continuous and this explains that \(\alpha\) has to do with reaching a uniform distribution in a effective way.

Let's analyse this a bit more, write a random number

$$ b = 1/(137 + R) $$

with \(R \in Re[0,1)\). Then

$$ 137b = 137/(137 + R) = 1 - R/(137+R) \approx 1 - R/137 $$

So according to the progression in R = 1/25 or such, not entirely unlikely but the probability for this is less than 5% and I assumed as an Hypothesis before checking this expansion having a hunch that we should be on the extreme end here without actually checking it first. 5854 seam to be just due to chance and that's fine as this figure varies and can't be decided as checked by doing a sensitivity analysis. So it is statistically significant that \(\alpha\) may be tuned for getting a uniform circle or some other reason.

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