26 July 2024

Sts. Joachim and Anne 1^st century B.C.; St. Anne

 

In this article, I will calculate the one correct value of the anomalous magnetic moment of Muon g-2. We have six different values of Muon g-2; one theoretical, three from experiments posted by the Wikipedia article, and two newer ones, one from Fermilab and one world average. They are all different. Of course, only one can be correct. How do we do that? Using transcendentals gives a more precise result. I call these numbers ‘transcendentals’, because they contain somehow in them the laws of physics and other fields of science (it seems to be so), which I try to prove here in this section of posts. If all seven results (3 x g-2 and 4 x g-factors) will give the right results (and they do), how do you explain it? Instead of spending billions of $$$ on finding tiny numbers, one can obtain the same results almost for free (even I can afford it) and a lot faster. Again, a set of rules is required on how to use transcendentals in their pure form, and, multiplication/division tables and summation/subtraction tables. I will be working hard on it, although it seems to be impossible to understand the rules right now. Maybe in the far future, there will be some smart guys figuring that out if I do not succeed.

Anyway, here are the six different results and the links to publications.

https://en.wikipedia.org/wiki/Anomalous_magnetic_dipole_moment

https://en.wikipedia.org/wiki/Muon_g-2

https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.131.161802

https://muon-g-2.fnal.gov/

https://indico.cern.ch/event/838862/contributions/3609622/attachments/2054740/3445104/FPCP-2020-muon_g-2_Experiment.pdf

 

1. MUON g-2 THEORY = 0.00116591804(51)

2. E821 EXPERIMENT at Brookhaven National Laboratory = 0.0011659209(6)

3. EXPERIMENT ‘MUON G-2’ at Fermilab, year 2021 = 0.00116592040(54)

4. AN AVERAGE WITH EXISTING MEASUREMENTS = 0.00116592061(41)

5. FERMI NATIONAL LABORATORY = 0.00116592055(24)

6. COMBINED Brookhaven and Fermi Labs/World Average = 0.00116592059(22)

 

27 July 2024

St. Pantaleon 305 AD; Sts. Nathalia, Aurelius, Liliosa, Felix and George 852 AD;

St. Celestine I 432 AD

 

As I said, only one is correct (hopefully). This can be determined by using transcendentals. Only one value will be closest to or equal to the result. I wrote ‘hopefully’, because the multiplication/division and summation/subtraction tables are not complete yet. The first configuration, P-F-S, and the fourth configuration, F-S-P, are complete. The second configuration, P-S-F; third, F-P-S; fifth, S-P-F; and the sixth configuration, S-F-P are partly done (division/multiplication tables, i.e., constants going down). I did not compute the summation/subtraction for these four configurations. Neither are six triplets done. However, it should not be a problem, because calculations using given now, constants seem to be close to the experimental values, if not 100% accurate. Once I code FORTRAN programs, I will post all tables (four tables per set, six sets of pairs, and six sets of triplets, plus three singles, two sets of each). It is a bit too much to compute all of them by hand.

Here are the links to some of the tables (from previous articles):

https://luxdeluce.com/15-new-version-book-1.html

https://luxdeluce.com/17-table-of-transcendental-constants-going-down-from-constants-number-8-equal-to.html

https://luxdeluce.com/360-148-division-multiplication-cross-product-table-of-transcendental-constants-in-quantum-physics-and-cosmology.html

https://luxdeluce.com/359-147-subtraction-summation-dot-product-table-of-transcendental-constants-in-quantum-physics-and-cosmology.html

 

Quantum scientists claim an almost unprecedented accuracy of the experiments from the year 2023, 0.19 ppm, which is indeed astonishing (This is the result number six, ‘World Average’). I did a couple of calculations of these values. Experiment number six (World Average), experiment number four (Fermilab, an average, with math a bit complex), and others, but I did not like the mathematics behind it (it was too complicated), so, I will present here just the two above results.

Anyway, here are the results before the computations are presented.

The experiment number six (2023), World Average:

 

COMBINED Brookhaven and Fermi Labs/World Average = 0.00116592059(22)

 

And my calculations (exact, I claim), experiment number six:

 

g-2 muon = 0.00116592060375 (12 digits accuracy HP Prime Calculator)

 

Less accurate result, experiment number four, an average:

 

AN AVERAGE WITH EXISTING MEASURENENTS = 0.00116592061(41)

 

And, from my math (I did not like the math behind it, but the results were correct after all, experiment number four:

 

g-2 muon = 0.00116592060658 (12 digits accuracy HP Prime Calculator)

 

So, once I used the new constant from set number five, F3Vx, 5 Pair, the math became simple, and the result is exact as you can see above, in experiment number six:

 

g-2 muon = 0.00116592060375 (12 digits accuracy HP Prime Calculator)

versus experiment number six (world average):

COMBINED Brookhaven and Fermi Labs/World Average, g-2 muon = 0.00116592059(22)

 

How do we calculate that? A bit of inventive thinking is required. I will continue, in a similar way to g-2 TAU, which was the easiest to compute (constant from point A). That is how I chose the first constant. The second constant close to the value computed from point B is the constant from point C. The third and final one, from point E, is again as close as possible to the result in D. Only these three numbers and one person, one calculator or computer, and no funding are required to obtain the exact value of g-2 MUON (and other particles or states). Once the rules of computing are figured out (probably by using advanced computers), we will have access to new fields of precise knowledge, without arduous calculations and long, long scientific formulas. It may become a reality in the future.

Computing the anomalous magnetic moment of MUON, g-2 MUON, experiment number six.

 

Step A ◊

Choose constant F7T+①P, i.e., F7T from the summation/subtraction table, first set, pairs.

F7T+①P = 1.77841101582E-5, shift the decimal point (multiply by 10000) and add 1.0,

It becomes

A = (F7T+①P x 10000) + 1 = 1.177841101582

 

Step B ◊

Compute (A - g-2 MUON (from Fermilab results)

B = 1.177841101582 – (0.00116592059 x 1000) = 0.01192051158

 

 

Step C ◊

Choose a close-to-result B constant from the tables. It will be S5T+①P, i.e., S5T from the summation/subtraction table first set, pairs.

S5T+①P = 1936.66444048, shift the decimal point by 10000, then add one to the result.

It becomes

C = (S5T+①P / 10000) + 1 = 1.193666444048

 

Step D ◊

Divide C by (B x 100). It becomes:

D = C / (B x 100) = = 1.193666444048 / (0.01192051158 x 100) = 1.00135504759

 

Step E ◊

Find from the tables a close constant to the result D.

F3Vx⑤P, i.e., F3Vx from multiplication/division table, fifth set, pairs.

F3Vx⑤P = 1.35620229978E-20

Multiply by 10¹⁷ and add 1.0 as usual. It becomes:

E = ((F3Vx⑤P = 1.35620229978E-20) x 10E17) + 1.0 = 1.00135620229978

 

Step F ◊

Divide C by E

F = C / E = = 1.193666444048 / 1.00135620229978 = 1.19204978339

 

Step G ◊

We can calculate the exact value of g-2 MUON anomalous magnetic moment, using just three transcendentals, instead of a half a mile long equation.

g-2 MUON = (A – F / 100) / 1000 = (1.177841101582 – (1.19204978339 / 100)) / 1000 = 0.00116592060375

versus experiment number six:

g-2 MUON experiment = 0.00116592059(22)

 

The calculated result speaks for itself.

 

How to check if the result is correct?

Starting with step B, substitute the final result for g-2 MUON, and proceed through all the steps until step G. The final result equals the initial input. 

g-2 MUON =  0.00116592060375

 

Titian - The Descent of the Holy Ghost:

 

Veni Sancte Spiritus, Gregorian chant   >>>   https://www.youtube.com/watch?v=Z6hqAfsHURo