- #36

#### Hans de Vries

Science Advisor

Gold Member

- 1,091

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**Classical Distance Ratio Formula**

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This one is the most “physical” yet. It’s also the simplest

and most accurate expression involving lepton mass ratios.

(fully within experimental range and exact to 0.0000073%)

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[tex]\frac{d_{crad}}{d_{spin}} \ \ = \ \ \frac{m_e}{m_\mu} \ (1 \ + \ \ a \ \frac{m_e}{m_\tau} \ ) , \ \ \ \ \ \ \ \ \ \ \ \ \ where \ \ a = \frac{1}{\frac

{1}{2}(1+\frac{1}{2})} = \frac{4}{3}[/tex]

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It relates the ratio of two “classical distances” with the

lepton mass ratio’s. Since distances scale (inversely)

proportional to mass we may expect the distance ratio

to be directly proportional to the lepton masses as well.

So far so good.

The distance ratio is dependent only on the fine structure

constant and consequently can be calculated with a very

high precision:

1/206.68905011 (69) _ dist. ratio from fine structure const.

1/206.689035 (19) ___ dist. ratio calculated with mass ratios

1/206.7682838 (54) __ electron/muon mass ratio

1/3477.48 (57)_______ electron/tau mass ratio

The electron/muon ratio already equals the distance

ratio to within 0.038% A simple correction term using

the electron/tau ratio then brings it to 0.0000073%

The distance ratio is the same for all spin ½ particles.

The classical distances are:

**dcrad:___ Classical (Electron) Radius**

=====================================

A lepton and anti-lepton spaced at this distance,

(which is inversely proportional to their mass),

have a (negative) potential energy which is equal

to their rest mass energy.

**dspin:___Classical (Electron) Spin ½ Orbital**

=====================================

A lepton and anti-lepton spaced at this distance,

(which is inversely proportional to their mass),

and orbiting at the frequency corresponding to their

rest mass, have an angular momentum of a spin ½

particle: [itex]\sqrt{(\frac{1}{2}(1+\frac{1}{2}))} \ \hbar[/itex]

It turns out that the velocity at which the leptons

would have to orbit here is equal for all spin ½ particles.

It’s a dimension less constant when written as v/c and

the solution of the following equation:

[tex]\frac{v^2/c^2}{\sqrt{1-v^2/c^2}} \ \ = \ \ \sqrt{\frac{1}{2}(1+\frac{1}{2})}[/tex]

Where the right hand term corresponds to the angular

momentum. When solved it gives:

[tex]\frac{v}{c} \ \ = \ \ v_{spin\frac{1}{2}} \ \ = \ \ 0.754141435281767 \ \ = \ \ \sqrt{\frac{1}{8}\sqrt{57} - \frac{3}{8}}[/tex]

The relation between the distance ratio and the fine

structure constant is:

[tex] \frac{d_{crad}}{d_{spin}} \ \ = \ \ \frac{1}{2}\alpha / v_{spin\frac{1}{2}} [/tex]

So that we can write exact to within 0.0000073%

[tex]\alpha \ \ = \ \ 2 \ v_{spin\frac{1}{2}} \ \ \frac{m_e}{m_\mu} \ (1 \ + \ \frac{1}{\frac{1}{2}(1+\frac{1}{2})} \ \frac{m_e}{m_\tau} \ ) \ \ [/tex]

The following CoData 2002 values were used so you

can test it yourself. (Don’t forget to apply a factor of

[itex]\frac{1}{\sqrt{1-v^2/c^2}} [/itex] to the mass when calculating the angular

momentum)

fine structure const 0.07297352568 ____ 0.00000000024

Planck constant ____ 6.626 0693e-34 ___ 0.0000011e-34 J s

speed of light _____ 299792458 ________ (exact) m/s

electron mass ______ 9.1093826e-31 ____ 0.0000016e-31 kg

speed of light _____ 299792458 ________ (exact) m/s

electric constant __ 8.854187817e-12 __ (exact) F/m

electric charge ____ 1.60217653e-19 ___ 0.00000014e-19 C

clas_electron_rad __ 2.817940285e-15 __ 0.000000028e-15 m

These "classical distances" are loaded with theoretical problems.

The classical electron radius has been used to try to cut of

the electric field close to the electron in the hope to avoid

the infinity problem of the [itex]1/r^2[/itex] field: Doesn’t work.

The spin angular momentum is much to high to fit into any

guess of the electron size. The angular orbit here is much

larger then electron radius! The electric force is more then

40 times higher then the centrifugal force to allow such an

orbit, et-cetera.

However, as a side remark, I've been looking at not so very

different scenarios were it seems that it may be possible to

bring the spin back into classical EM field theory without the

use of any new physics.

Regards, Hans

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