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<p>Grahame,<br>
<br>
</p>
<p>regarding the radius of the electron, I think that it is well
defined.</p>
<p><br>
</p>
<p>Schrödinger evaluated "the size" of the electron in his famous
paper of 1930 by QM means, and his result was "about 4 * 10^-13
m". From my model there follows a more precise value which is R =
3.86 * 10^-13 m. <br>
What about the spin? If the mass of a particle is m =
h(bar)/(R*c) (this follows from my model) then you can reorder
it: m*R*c = h(bar). This is the classical definition of the
angular momentum. The result is constant for a lepton and for a
quark independent of the individual particle, but it has a factor
of 1/2 missing. The cause is that the mass follows here from a
special mechanism which is not reflected by the classical
understanding of a mass which is distributed within the particle.</p>
<p><br>
</p>
<p>If interested please look at my site "Origin of Mass". (
<a class="moz-txt-link-abbreviated" href="http://www.ag-physics.org">www.ag-physics.org</a> )<br>
</p>
<p><br>
</p>
<p>Best regards<br>
Albrecht</p>
<p><br>
</p>
<br>
<div class="moz-cite-prefix">Am 08.07.2016 um 11:45 schrieb John
Williamson:<br>
</div>
<blockquote
cite="mid:7DC02B7BFEAA614DA666120C8A0260C9147C1C26@CMS08-01.campus.gla.ac.uk"
type="cite">
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<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Yep it is,indeed
not so simple.<br>
<br>
Grahame, you say ...<br>
<br>
</span><span
style="font-family:Arial;mso-fareast-font-family:"Times
New Roman";
color:navy" lang="EN-US">Angular momentum is given by linear
tangential momentum multiplied by radius - so angular
momentum of the electron is mcR. Since mc is constant, R
must also be constant if angular momentum is invariant
(which I believe we agree it is)<br>
<br>
</span><span style="mso-fareast-font-family:"Times New
Roman";
mso-bidi-font-family:"Times New Roman""
lang="EN-US">Hmm, I kind of do and do not agree. This kind
of thing is (perhaps) part of the story, but anyway only a
part. Such a thing is, indeed A component of angular
momentum, but it is wholly inadequate to describe quantum
spin. It is the only component for simple models where a
something goes round and round in circles in ordinary space,
even so it immediately begs the question "what is R?" and
then the further question "what is m?" let alone the deeper
questions - why that R and why that m? and what is it
orbiting about and what is orbiting?<br>
<br>
Going to "what is R?. The R needs to be, in my view, at
least “complex”. I recently read your 1973 article Alex.
Very beautiful. 1973! Had we been aware of it at the time
I’m sure Martin and I would have referenced it as a possible
confinement scenario. There you recognize, correctly, the
huge angular momentum density and use that as an input. I
think the subsequent double covering problems<span
style="mso-spacerun:yes">
</span>and the sign change similar to those encountered by
other folk in trying to model stuff using the half-integer
Legendre polynomials, are best treated by going more complex
than complex, and using a proper non-commutative algebra.
Tricky, I know, but nature, (especially 3D rotations) IS
non-commutative.</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Coming back to
angular momentum and the underlying nature of spin. This IS
hard. No simple way round it. Properly, the momentum is
itself contains a division of space by time (the velocity).
It is properly a bi-vector. Further, the orbital angular
momentum (what Grahame is talking about), contains a
multiplication of this by a perpendicular vector (R ). That
is, properly, a tri-vector (the dual of a vector).<span
style="mso-spacerun:yes">
</span>Remember, torque and energy have the same SI units
(force times distance), but are quite different (energy is a
scalar, torque is a bi-vector). Same kind of thing needed
here in your thinking and visualisation (but worse). You can
think of the bi-vector ness (of the trivector) either in
your momentum or in your “radius” –either way hypercomplex.
Also – to go further you need to go to differential forms –
not just see this as just some mass m orbiting on some
(massless) stick of fixed length R! Sorry Grahame, but this
is what your model of angular momentum looks like to me.
Orbiting around what? What is orbiting and what is it
orbiting around?</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">This all sounds
pretty horrible, but it is not as bad as you think. The
Maxwell equations already contain much of this complexity,
and describe light well. One of the Maxwell equations IS the
(partial at least) tri-vector equation for the
electromagnetic fluid. Analysing this properly, with the
right extensions, does give an intrinsic angular momentum
density which can be integrated. I’m not very good at this
kind of thing, but that is just the kind of thing I’m trying
to do.</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">My new photon
wavefunction does, at least do this. As the energy varies
the curvature varies inversely to maintain the angular
momentum at hbar to arbitrary energies. Sticking this into
our electron model then gives a half-integral spin at
arbitrary energies (since it is a double-loop and
transforms, further, as a looping photon).</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Anyway gotta go ..
still dealing with fallout from the exams …</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
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</span></p>
<p class="MsoNormal"
style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span
style="mso-fareast-font-family:"Times New
Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Regards, John.</span><span
lang="EN-US"></span></p>
<br>
<font class="" face="Arial" color="#000080" size="2"><br>
</font>
<div style="font-family: Times New Roman; color: #000000;
font-size: 16px">
<hr tabindex="-1">
<div style="direction: ltr;" id="divRpF123027"><font
face="Tahoma" color="#000000" size="2"><b>From:</b>
General
[<a class="moz-txt-link-abbreviated" href="mailto:general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org">general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org</a>]
on behalf of Richard Gauthier [<a class="moz-txt-link-abbreviated" href="mailto:richgauthier@gmail.com">richgauthier@gmail.com</a>]<br>
<b>Sent:</b> Friday, July 08, 2016 6:13 AM<br>
<b>To:</b> Nature of Light and Particles - General
Discussion<br>
<b>Subject:</b> Re: [General] double photon cycle,
subjective v objective realities<br>
</font><br>
</div>
<div>
<div class="">Hello Grahame,</div>
<div class=""><br class="">
</div>
<div class=""> Unfortunately the situation is not so
simple. Neither of our published electron models includes
a specific photon model with its own spin, where this
photon model moves along the helical trajectory described
in our models. If that photon moving along the helical
trajectory has a spin that is is independent of the energy
of the photon (which is the nature of photons) then as the
photon's trajectory in the your double-loop constant
helical radius electron model gets more and more straight
with increasing electron speed, then the spin of this
circulating photon adds more and more to the spin 1/2 of
your electron model produced by its circling transverse
component of momentum mc at constant radius R. The result
is that a circulating spin 1 photon along your constant
radius R helical trajectory would give your electron model
a total spin of one and a half units of spin hbar at
highly relativistic velocities. A circulating spin 1/2
photon traveling along your constant radius R trajectory
would give your electron model a total spin of 1/2 + 1/2
= 1 unit of hbar of spin at highly relativistic
velocities. It is only if the radius R of the photon’s
helical trajectory decreases with increasing velocity to
become insignificant (compared to R in a resting electron)
at relativistic velocities that the spin of the electron
model at relativistic velocities will equal only the spin
of the photon composing the electron model. Ideally the
helically circulating photon model of the electron will
have longitudinal spin component 1/2 hbar at all
velocities of the electron model from very slow velocities
to very highly relativistic velocities. </div>
<div class=""><br class="">
</div>
<div class=""> I have an unpublished internally
superluminal (v=c sqrt(2) ) helically circulating spin-1/2
photon model whose spin remains 1/2 at all energies, which
may be suitable for modeling the electron. I described
this photon model in this email list in the past. The
radius of my published spin-1/2 charged-photon electron
model’s photon trajectory decreases as 1/gamma^2 with
increasing electron velocity, so this does not produce the
complication described above when the helical radius of
the photon’s trajectory is a constant R at all electron
velocities.</div>
<div class=""><br class="">
</div>
<div class=""> Richard</div>
<br class="">
<div>
<blockquote type="cite" class="">
<div class="">On Jul 7, 2016, at 1:00 AM, Dr Grahame
Blackwell <<a moz-do-not-send="true"
href="mailto:grahame@starweave.com" class=""
target="_blank">grahame@starweave.com</a>> wrote:</div>
<br class="Apple-interchange-newline">
<div class="">
<div class="" style="font-family:Helvetica;
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<font class="" face="Arial" color="#000080" size="2">Thanks
Richard,</font></div>
<div class="" style="font-family:Helvetica;
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</div>
<div class="" style="font-family:Helvetica;
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<font class="" face="Arial" color="#000080" size="2">That's
precisely what I've been trying to say, without in
any way resting on any generally-accepted results
that might be regarded as consequences of SR (and
so open to question).</font></div>
<div class="" style="font-family:Helvetica;
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</div>
<div class="" style="font-family:Helvetica;
font-size:12px; font-style:normal;
font-variant:normal; font-weight:normal;
letter-spacing:normal; line-height:normal;
orphans:auto; text-align:start; text-indent:0px;
text-transform:none; white-space:normal;
widows:auto; word-spacing:0px;
background-color:rgb(255,255,255)">
<font class="" face="Arial" color="#000080" size="2">If
we agree that the transverse momentum component of
the electron is a direct consequence of the
rotational component of its formative photon (as I
hope we do!) then that rotational component is
acting at radius R of the electron at that time
from its centre. Angular momentum is given by
linear tangential momentum multiplied by radius -
so angular momentum of the electron is mcR. Since
mc is constant, R must also be constant if angular
momentum is invariant (which I believe we agree it
is).</font></div>
<div class="" style="font-family:Helvetica;
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letter-spacing:normal; line-height:normal;
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</div>
<div class="" style="font-family:Helvetica;
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font-variant:normal; font-weight:normal;
letter-spacing:normal; line-height:normal;
orphans:auto; text-align:start; text-indent:0px;
text-transform:none; white-space:normal;
widows:auto; word-spacing:0px;
background-color:rgb(255,255,255)">
<font class="" face="Arial" color="#000080" size="2">Just
one further point: Richard, you refer to m as the
electron's invariant mass. If we regard mass as
that quality of an object that resists
acceleration (and so is proportional to the
instantaneous force required to induce an
instantaneous acceleration), then my research
indicates that the mass is<span
class="Apple-converted-space"> </span><em
class="">not</em><span
class="Apple-converted-space"> </span>invariant
- though it will appear so from measurements taken
within the electron's moving frame. My analysis
shows that objective mass varies with speed and
the relationship E = mc^2 is applicable only for
an objectively static object/particle. The m
referred to above, as I see it, is the objective
rest-mass of the electron (i.e. its mass when
objectively static), which corresponds to the
energy required to maintain the
formative structure of the electron (as opposed to
that required to maintain its linear motion).
This is of course constant.</font></div>
<div class="" style="font-family:Helvetica;
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</div>
<div class="" style="font-family:Helvetica;
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text-transform:none; white-space:normal;
widows:auto; word-spacing:0px;
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<font class="" face="Arial" color="#000080" size="2">Best
regards,</font></div>
<div class="" style="font-family:Helvetica;
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background-color:rgb(255,255,255)">
<font class="" face="Arial" color="#000080" size="2">Grahame</font></div>
<blockquote class="" type="cite"
style="font-family:Helvetica; font-size:12px;
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<div class="" style="font-style:normal;
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font-family:arial">
----- Original Message -----<span
class="Apple-converted-space"> </span></div>
<div class="" style="font-style:normal;
font-variant:normal; font-weight:normal;
font-size:10pt; line-height:normal;
font-family:arial;
background-color:rgb(228,228,228)">
<b class="">From:</b><span
class="Apple-converted-space"> </span><a
moz-do-not-send="true"
title="richgauthier@gmail.com"
href="mailto:richgauthier@gmail.com" class=""
target="_blank">Richard Gauthier</a></div>
<div class="" style="font-style:normal;
font-variant:normal; font-weight:normal;
font-size:10pt; line-height:normal;
font-family:arial">
<b class="">To:</b><span
class="Apple-converted-space"> </span><a
moz-do-not-send="true"
title="general@lists.natureoflightandparticles.org"
href="mailto:general@lists.natureoflightandparticles.org" class=""
target="_blank">Nature of Light and Particles -
General Discussion</a></div>
<div class="" style="font-style:normal;
font-variant:normal; font-weight:normal;
font-size:10pt; line-height:normal;
font-family:arial">
<b class="">Sent:</b><span
class="Apple-converted-space"> </span>Thursday,
July 07, 2016 6:42 AM</div>
<div class="" style="font-style:normal;
font-variant:normal; font-weight:normal;
font-size:10pt; line-height:normal;
font-family:arial">
<b class="">Subject:</b><span
class="Apple-converted-space"> </span>Re:
[General] double photon cycle, subjective v
objective realities</div>
<div class=""><br class="">
</div>
<div class="">Chip and Grahame,</div>
<div class=""> Lets be specific to the electron to
avoid unnecessary vagueness. The moving electron
(composed of a circulating photon) has a constant
transverse internal momentum component mc and a
longitudinal external momentum component p=gamma
mv. These two momenta add vectorially (by the
Pythagorean theorem) to give P^2 = p^2 + (mc)^2
where P=E/c is the momentum P=gamma mc of the
helically circulating photon of energy E = gamma
mc^2 that is the total energy of the linearly
moving electron, modeled by the helically moving
photon. This relationship is equivalent to the
relativistic energy-momentum equation for a moving
electron: E^2 = (pc)^2 + m^2 c^4 which,
substituting E=Pc, gives (Pc)^2 = (pc)^2 +
(mc^2) c^2 .. Dividing by c^2 gives P^2 = p^2 +
(mc)^2 as given above. So as the electron speeds
up, the transverse momentum component mc of the
electron’s total (internal plus external) momentum
P remains constant even for a highly relativistic
electron. The electron’s constant transverse
internal momentum component mc corresponds to (and
leads to a derivation of) the electron’s invariant
mass m.</div>
<div class=""> Richard</div>
<br class="">
<div class="">
<blockquote type="cite" class="">
<div class="">On Jul 6, 2016, at 10:18 AM, Dr
Grahame Blackwell <<a
moz-do-not-send="true"
href="mailto:grahame@starweave.com" class=""
target="_blank">grahame@starweave.com</a>>
wrote:</div>
<br class="Apple-interchange-newline">
<div class="">
<div class="" style="text-transform:none;
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<font class="" face="Arial" color="#000080"
size="2">Yes Chip,</font></div>
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</div>
<div class="" style="text-transform:none;
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<font class="" face="Arial" color="#000080"
size="2">Certainly the momentum of the
confined wave increases - but that
increased momentum should not ALL be
reckoned as ANGULAR momentum of the
electron.</font></div>
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</div>
<div class="" style="text-transform:none;
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<font class="" face="Arial" color="#000080"
size="2">We know that a component of the
momentum of that photon is linear - it's
the linear momentum of the electron in
motion. There is another component of
that photon that's orthogonal to that,
i.e. in the direction of the cyclic motion
of the photon. As the linear velocity of
the electron increases, the linear
component of the photon momentum increases
- however the orthogonal, cyclic,
component of that photon momentum does NOT
increase, since the 'pitch angle' of the
helical motion of that photon increases
with linear electron velocity, and so also
with photon frequency, so as to precisely
cancel out the effect of that increased
frequency in the resolved-component cyclic
direction.</font></div>
<div class="" style="text-transform:none;
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</div>
<div class="" style="text-transform:none;
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<font class="" face="Arial" color="#000080"
size="2">The angular momentum of the
electron, dictated by the angular momentum
contribution of the photon, does NOT
depend on the FULL momentum of the photon
- it ONLY depends on that component of the
photon that acts cyclically, i.e. the
component that's orthogonal to the linear
motion of the photon. That component
remains constant (as long as the radius of
the photon cycle remains constant).</font></div>
<div class="" style="text-transform:none;
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</div>
<div class="" style="text-transform:none;
background-color:rgb(255,255,255);
text-indent:0px; font-style:normal;
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<font class="" face="Arial" color="#000080"
size="2">For example, if an electron is
travelling with linear speed 0.6c then its
formative photon is travelling in a
helical path which, if we were to flatten
it out (as in relativistic energy-momentum
relation) we'd find that formative photon
having a linear motion component of 0.6c
and cyclic speed component of 0.8c. This
means that the ANGULAR momentum imparted
by the photon will only be 0.8 of that
which it would give if it were travelling
fully cyclically at speed c (as for a
static particle). Since the frequency of
the photon will be increased by a gamma
factor of 1/0.8 for such motion, the
decreased (0.8) contribution of momentum
for increased (1/0.8) frequency will be
exactly what it was for the static
particle.</font></div>
<div class="" style="text-transform:none;
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letter-spacing:normal; word-spacing:0px">
</div>
<div class="" style="text-transform:none;
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<font class="" face="Arial" color="#000080"
size="2">I hope that helps make things
clearer.</font></div>
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<font class="" face="Arial" color="#000080"
size="2">Best regards,</font></div>
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<font class="" face="Arial" color="#000080"
size="2">Grahame</font></div>
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