A non-perturbative derivation of the exact value of the SU(2) coupling value g from the standard Electroweak Lagrangian itself.

[Hans de Vries]

http://chip-architect.com/physics/Electroweak_coupling_g.pdf


The value derived for g0 is simply 2/pi.


We assume that the observable coupling
constants e, g and g' contain both a charge
interaction term as well as a spin term:

e = ( 1 + spin_term) e0
g = ( 1 + spin_term) g0

charge_term = 1
spin_term = alpha( 1 + alpha/(2pi) )

Where 1+alpha/(2pi) is the magnetic moment.
This gives us the "charge-interaction-only"
coupling constants:

e0 = spin-less electromagnetic coupling e
g0 = spin-less electroweak coupling g

Which turn out to be both simple
mathematical values. With alpha=e^2/4pi)
we get:



1/ 139. 045 636 8 (1) = alpha0
1/ 139. 045 636 6 = exp(-pi^2/2)

0. 636 52 (47) = g0
0. 636 62 = 2/pi


Regards, Hans

[FrediFizzx]

Post by Hans de Vries
http://chip-architect.com/physics/Electroweak_coupling_g.pdf
The value derived for g0 is simply 2/pi.
We assume that the observable coupling
constants e, g and g' contain both a charge
e = ( 1 + spin_term) e0
g = ( 1 + spin_term) g0
charge_term = 1
spin_term = alpha( 1 + alpha/(2pi) )
Where 1+alpha/(2pi) is the magnetic moment.
This gives us the "charge-interaction-only"
e0 = spin-less electromagnetic coupling e
g0 = spin-less electroweak coupling g
Which turn out to be both simple
mathematical values. With alpha=e^2/4pi)
1/ 139. 045 636 8 (1) = alpha0
1/ 139. 045 636 6 = exp(-pi^2/2)
0. 636 52 (47) = g0
0. 636 62 = 2/pi

Hmm... Interesting. Seems like everyone has be going the other way for
alpha0. ;-) So you are saying the pure electric charge, e0, is actually
weaker once the spin term associated with vacuum polarization is
removed. I am trying to get a more mental picture of that. But why the
factor 2/pi^2? What could be generating this physically? Any
conjectures?

Best,

Fred Diether

[Hans de Vries]

Post by FrediFizzx

Post by Hans de Vries
Which turn out to be both simple
mathematical values. With alpha=e^2/4pi)
1/ 139. 045 636 8 (1) = alpha0
1/ 139. 045 636 6 = exp(-pi^2/2)
0. 636 52 (47) = g0
0. 636 62 = 2/pi

Dear Fred,

Post by FrediFizzx
So you are saying the pure electric charge, e0, is actually
weaker once the spin term associated with vacuum polarization is
removed. I am trying to get a more mental picture of that.

Indeed, note that both e and e0 here are the low
energy limits of the running coupling 'constant'.

The assumption is that spin also has a capability
to polarize the vacuum, however weaker than
the electric field by a factor of:

alpha( 1 + alpha/(2pi) ) = 0.0073058278...

The term between brackets suggest a coupling
with the magnetic moment, while the scale of
the coupling is the alpha at the left. This then
leads to a value of exp(-pi^2/2):

1/ 139. 045 636 8 (1) = alpha0
1/ 139. 045 636 6 = exp(-pi^2/2)

Exact to ~10 digits....

- - - - - - - - - - - - - - - - - - - - - -

Post by FrediFizzx
factor 2/pi^2? What could be generating this physically? Any
conjectures?

The whole quest is now to find out where this
could come from. What I'm trying to do, is to
get these values from a non-perturbative
treatment of QFT.

I think I may have succeeded here for the other
Electroweak coupling constant g0

Post by FrediFizzx

Post by Hans de Vries
0. 636 52 (47) = g0
0. 636 62 = 2/pi

Here we used the covariant derivative
(d/dx - ieA) with an actual sinusoidal
field A. This then leads to a frequency
modulated waveform of the electron in
the form of:

exp(-iEt + ieA sin (w t) /w )

which becomes:

exp(-iEt + ig/2 W sin (w t) /w )

in the case of the EW coupling constant g.


This corresponds to the frequency spectrum
of good old FM radio which is:

J0(m)f0 + J1(m)f1 + J2(m)f2 + J3(m)f3 ....

Where the Bessel function coefficients
now become:

J0(m) = amplitude of 0 photon absorption
J1(m) = amplitude of 1 photon absorption
J2(m) = amplitude of 2 photon absorption
J3(m) = amplitude of 3 photon absorption
.......
And 'm' denotes the "modulation factor"

These coefficients are unitary in both
amplitude as well as probability (!):

J0(m) + J1(m) + J2(m) + J3(m) .... = 1

|J0(m)|^2 + |J1(m)|^2 + J2(m)|^2 ..... = 1


Now, it comes: (!)
==============

The average number of photons absorbed is:

1*|J1(m)|^2 + 2*J2(m)|^2 + 3*|J3(m)|^2..... = 1/pi

At the zero energy level with a coupling
constant of 1. So there's an inherent factor
of 1/pi in the non-perturbative approach:

0.31831 = 1/pi
0.30282 = e
0.32058 = g/2

Note that the Lagrangian uses the values e
and g/2 where the factor 1/2 is historical.

Now, the ratio between g/2 and 1/pi is about
(1 + alpha) which is (within the limited
precision for g/2) equal to the ratio e/e0.

This relation is then what the paper is based on:

http://chip-architect.com/physics/Electroweak_coupling_g.pdf


Regards, Hans

[FrediFizzx]

Post by Hans de Vries

Post by FrediFizzx

Post by Hans de Vries
Which turn out to be both simple
mathematical values. With alpha=e^2/4pi)
1/ 139. 045 636 8 (1) = alpha0
1/ 139. 045 636 6 = exp(-pi^2/2)
0. 636 52 (47) = g0
0. 636 62 = 2/pi

Dear Fred,

Post by FrediFizzx
So you are saying the pure electric charge, e0, is actually
weaker once the spin term associated with vacuum polarization is
removed. I am trying to get a more mental picture of that.

Indeed, note that both e and e0 here are the low
energy limits of the running coupling 'constant'.
The assumption is that spin also has a capability
to polarize the vacuum, however weaker than
alpha( 1 + alpha/(2pi) ) = 0.0073058278...

OK, that is more clear now. It is polarization of the "vacuum" from
spin component.

Post by Hans de Vries
The term between brackets suggest a coupling
with the magnetic moment, while the scale of
the coupling is the alpha at the left. This then
1/ 139. 045 636 8 (1) = alpha0
1/ 139. 045 636 6 = exp(-pi^2/2)
Exact to ~10 digits....
- - - - - - - - - - - - - - - - - - - - - -

Post by FrediFizzx
factor 2/pi^2? What could be generating this physically? Any
conjectures?

The whole quest is now to find out where this
could come from. What I'm trying to do, is to
get these values from a non-perturbative
treatment of QFT.
I think I may have succeeded here for the other
Electroweak coupling constant g0

Post by FrediFizzx

Post by Hans de Vries
0. 636 52 (47) = g0
0. 636 62 = 2/pi

Here we used the covariant derivative
(d/dx - ieA) with an actual sinusoidal
field A. This then leads to a frequency
modulated waveform of the electron in
exp(-iEt + ieA sin (w t) /w )
exp(-iEt + ig/2 W sin (w t) /w )
in the case of the EW coupling constant g.
This corresponds to the frequency spectrum
J0(m)f0 + J1(m)f1 + J2(m)f2 + J3(m)f3 ....
Where the Bessel function coefficients
J0(m) = amplitude of 0 photon absorption
J1(m) = amplitude of 1 photon absorption
J2(m) = amplitude of 2 photon absorption
J3(m) = amplitude of 3 photon absorption
.......
And 'm' denotes the "modulation factor"
These coefficients are unitary in both
J0(m) + J1(m) + J2(m) + J3(m) .... = 1
|J0(m)|^2 + |J1(m)|^2 + J2(m)|^2 ..... = 1
Now, it comes: (!)
==============
1*|J1(m)|^2 + 2*J2(m)|^2 + 3*|J3(m)|^2..... = 1/pi

Yes, fantanstic if true. This is also more clear now.

Post by Hans de Vries
At the zero energy level with a coupling
constant of 1. So there's an inherent factor
0.31831 = 1/pi
0.30282 = e
0.32058 = g/2
Note that the Lagrangian uses the values e
and g/2 where the factor 1/2 is historical.
Now, the ratio between g/2 and 1/pi is about
(1 + alpha) which is (within the limited
precision for g/2) equal to the ratio e/e0.
http://chip-architect.com/physics/Electroweak_coupling_g.pdf

I am getting the impression here that you are on to an anomaly in
electronic charge (and electroweak) in addition to the magnetic moment
anomaly. But in the case of electronic charge, how might this be
experimentally determined?

Best,

Fred Diether