Electro-Proton Mass Ratio
&
Fine Structure Constant

 T
he pure number values of both these familiar terms

=1836.152701(37)
=137.0359895(61) Codata 1986

cannot be 'mathematicals' (2.71828... 3.14159... etc.) because they are too large. Neither can they be simply related to astronomical pure numbers like 1039 because they are too small. Yet one or the other (or both) must be the case if the universe is a unique and necessary entity.

These numbers can be derived to some accuracy from equations related to the ground state of the hydrogen atom with relativistic corrections. This requires a new view of probabilistic expectation.

For present purposes we set the value of c, h-bar, proton mass, and Compton wavelength of the proton at 1 unit.

The force between an electron and proton in the ground state of hydrogen can then be given as a 'quasi-pure number' (a number with dimensions but derived from non-arbitrary pure numbers).

F = (2) / r = 3 2

Where r = h / () ; is the fine structure constant as a velocity ; and is the electron-proton mass ratio as the mass of the electron.

And

e2 / r2 = mv2 / r (from the equality of electrostatic and centrifugal force)

So the velocity of the electron in the ground state orbit is 1/137 of light velocity (c=1). And r = 1 / in proton Compton wavelengths.

On this scale the force is ~1.15 x E-13. (And on the unit scale the number is the same due to algebraic cancellations.) This number gives us reason to suspect that the force between one electron and one proton is related inversely to the square root of 1026 (the radius of the universe in unit lengths).

In view of the relationships given in #26 this is an obvious guess.

The question to be answered then is, "What are the most probable values of and given F and the form of the equation?".

A definition of energy is here necessary.
Energy is a distortion of the positional field or isotropic grid (potential), i.e. a departure from the default state; or the capacity to cause such a distortion by being in relative motion (kinetic).

Any localized distortion of a field possesses a position relative to all other points and distortions and hence must possess a positional field as well. Energy in the form of charge must therefore possess mass.

The problem concerns the amount of mass to be ascribed to 'unit' charge.
Clearly, the charge must have varying field density at varying distances from the charge center. If the charge center is defined as having finite density what is that density and at what distance will it possess one half or two times the central denstiy?

Thus, the shape of the charge gradient with respect to the defined gradient of the positional field and initial isotropic grid is indefinite and the mass to be assigned to it is also indefinite.

The proposed solution is as follows:

There are two factors to be considered in the equation 1.15 x E-13 = 3 2 : the parameters (,) and the exponents (3,2).

If we impartially assign the square root of F to each parameter the calculated values of and are

= F1/6 = 1/143
= F1/4 = 1/1716 (about a 5 % error)

But the exponents of and are unequal (3,2) so that an impartial assignment is unwarranted. should take more than F1/2 and less.

So that,

3 = Q(F1/2) and 2 = (F1/2) / Q

where Q is some unknown factor which changes the weight of each parameter in a probabalistic calculation and preserves all possible number pairs which generate 1.15 E-13 in the equation ( F = 3 2 ) .

Solving for Q:

/ = 3/2 is the exponential ratio justified as mathematically significant by the following limit equation.

|lim (ix > 0) { [ (x+ix) / x ]>w (-1) } / { [ x/(x + ix)]w (-1) } | = (>'w)/w

where i means increment (with due regard to Html text limitations).

Now, since / = / = / = 1 (in terms of exponential ratios),( i.e. by symmetric probability), any w / w reduces to / leaving excess 's as a determining factor in any Q value greater than 1.

 Thus, Q / (1/Q) = Q^2 = 1 (for all a^w / B^w ) and for a^3 / B^2 ............Q / (1/Q) .. . . = 1 = ...................aa / BB (Q) x [ Q / (1/Q) ].... = 3/2 = . . . (a) x [ aa / BB ] ............Q^3......... .....= 3/2 = .......... . ..aaa / BB therefore ; ................... Q = (3/2)^1/3

Click here to see a justification for the procedure given in the above table.

In general, for all real w greater than or equal to 1, in the form (>w)( w) = F,

Q = (>w / w)1 / (>w-w+2)

Q maximum is 1.3210997...= (3.591.../1)^(1/4.591...)
where 1.321...^3.591... = 3.591... / 1.321... = 2.718281828... = e

Go back to 19a

Applied to the original problem:

= [(3/2)(1/3) times F1/2]1/3 = 1 / 137.0317...
= [(2/3)(1/3) times F1/2]1/2 = 1 / 1836.2365...

giving accuracies of 1 part in 32,000 and 21,000.

The remaining error is taken up primarily by bound state relativistic effects.

.511 MeV / (.511 MeV - 13.6 eV) = 1.0000266
In eV, the electron loses the Rydberg energy (13.6eV) in falling to the ground state. Since the e/P mass ratio is between unbound particles, it is logical to assume that this amount of energy should be subtracted from the electron.

1 / [1- (1/137.036)2]1/2 = 1.0000266
This relativistic increase in mass due to the velocity of the electron in orbit is exactly equal to the Rydberg energy. Hence, the electron loses one Rydberg in its fall and one more from converting some of its rest mass into relativistic mass increase. Therefore, the calculated e/P mass ratio must be modified to

Eq.#1..........revised = (1-2)

where the relativistic mass increase is carried out twice to obtain (1-2).
Note: It makes no appreciable difference if one multiplies twice by 1.0000266 = 1.00005320071... or if one adds twice .0000266 = .00005320000... .

The previous equation can be used to calculate a revised force (F).

Frevised = rev4 / (2/32/3) ;..............from = (2/31/3 times F1/2)1/2

Then, revised = (3/21/3 times Frev1/2)1/3 = 1/137.0328... and

137.036 / 137.03238 = 1.0000263 which appears to be the same modifier used for r, i.e.

Eq.#2........arevised = / (1 - 2)1/2

, in terms of , can now be given by algebraic manipulation.

 r = [ 3/21/3 {r 4 / (2/32/3)}1/2 ]1/3

Substituting from the right hand side of equation #1 for r
and #2 for r in the above equation yields the following,

= [3/2 times 2/31/3] / [ (1-2)7/4 ]

Note: /r = r2 / 2 and that 1/2 mv2 = 1/2 mrv(r)2
Accuracy is now improved by 99% to 1 part in 2,000,000, i.e. if we put the Codata '86 value of in the equation we get =1836.1517...

Codata 1986 gives an accuracy estimate on the order of 1 part in 50,000,000 so that another smaller correction is almost certainly necessary. This may be due to the electron g-factor.