 ## Book 6 - Derivation of General Formula for Constants of the Cosmology and Quantum Mechanics - part II

23 September 2017 AD, the feast of St Pio and St Linus

Andrew Yanthar-Wasilik

Finding General Formula for Equations from the previous book, “Book 5 Integer Formula for Dimensionless Coupling Constants of Fundamental Forces”.

This time general term of constant similar to the fine structure constant, alpha, αE, let’s name it just alpha, α, will be derived.

It is quite possible, that there are also other constants, we do not know yet, so the general term alpha, α, seems appropriate.

In Part I of the Book 6 the Exponent Main, ExpM was derived:

ExpM = (A/B)C

✠Now, continuing further derivation from Book 5, we get part D (Equation D is the exponent needed to calculate the value of the function at (x=D), represented by partial sequence:

D16 = 16 + ExpM16

D17 = 17 + ExpM17

D1 + 1 + ExpM1

General formula for exponent Dxis:

Dx = [ x + ExpMx ]

or just:

✠Dx = [ x + ExpM ]     (Eqn. D)

Having exponent D we may get the value of transcendent function at point x:

✠FT( x = D ) = ( C0) * ( π/ e )D    (Eqn. FT)

✠Next step is getting the Partial Exponent ExpP, we have three terms of the the sequence:

ExpP16 = ( 16 + (24 / 24) ) / ExpM16

ExpP17 = ( 17 + (27 / 24) ) / ExpM17

ExpP1 = ( 1 + (-21 / 24) ) / ExpM1

...,1,...,16, 17, ... are just x

...,(-21 / 24),...,(24 / 24),(27 / 24),... are equal to exponent

y = ((3)/(24))*x + (-(24)/(24)) = (3*x − 24)/(24) = (x − 8)/(8)

of (Eqn. A)

Ax = ( C0)((x − 8)/(8))     (Eqn. A)

Adding these two terms

x + y = x + (x − 8)/(8) = (8x + x − 8)/(8) = (9x − 8)/(8)

This sum has to be divided by Exponent Main, ExpM, to get value of Exponent Partial, ExpP:

✠ExpPx = ((9x − 8))/(8ExpM)      (Eqn. EP)

✠Last sequence will be the denominator of

( αE )( − 1 ⁄ 2) = ( FT(x) / ( 8 + 2 * (24/24))) ExpP     (Eqn. αE)

which is

For constant C 16: ( 8 + 2 * (24/24))

For constant C 17: ( 9 + 2 * (24/27))

For constant C 1: ( -7 + 2 * (-24/21))

First part of this sum is gives partial sequence:

...,-7,...,8,9,...

and this is equal to ( x - 8 )

Second part of the sum gives partial sequence:

...,2 * (-24/21),...,2 * (24/24),2 * (24/27),...

We have already calculated this. It is equal to

EpartB = 2 * ((8)/(x − 8)) = ((16)/(x − 8))

Adding both terms gives:

( x - 8 ) + ((16)/(x − 8)) = (x2 − 16x + 80)/(x − 8)

Taking reciprocal of this term we can avoid quotient and use product instead in (Eqn αE)

✠Finally, the General Formula for for ( α )( − (1)/(2)) is:

✠( αx )( − (1)/(2)) = {[(C0)(πe)(x + ExpM)]*[(x − 8)/(x2 − 16x + 80)]}[(9x − 8)/(8ExpM)]      (Eqn. α)

To get (α)1simply square the previous equation

To get ( α) simply take reciprocal of the previous equation (for those not much in the math).

Getting the shorter formula for ExpM and αis probably possible but very difficult. I am not sure if I can do that, maybe some professional mathematicians can derive it.

Now, this General Equation for any x allows calculating any value of alpha.

Next book, Book 7 will show coupling constant of gravity force (two candidates) and, switching to Quantum Mechanics Neutrino Mixing Angles and Weinberg Angle for weak interactions.

Quarks will follow later.

Andrew Yanthar-Wasilik

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