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<p>Sorry, a little error below:</p>
<p>The period of the motion in the electron will not be reduced but
will be <u>extended </u>following <font size="-1">T' = T *
sqrt(1/(1--v<sup>2</sup>/c<sup>2</sup>))</font></p>
<p><font size="-1">Albrecht<br>
</font></p>
<br>
<div class="moz-cite-prefix">Am 09.01.2017 um 21:31 schrieb Albrecht
Giese:<br>
</div>
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<p><font size="-1">Hi Chip, hi All,<br>
</font></p>
<p><font size="-1">the problem of the limitation of the internal
speed in the electron is not complicated. It is the cause if
relativistic dilation.</font></p>
<p><font size="-1">If an electron is a particle which is built by
something which permanently orbits at c, then in case of the
motion of the electron, this internal speed will continue to
be c with respect to the external frame. If now the electron
moves into an axial direction with respect to the orbit at
speed v then the circular motion will turn into a helical
motion. If the motion on the helix is still c then the period
T of this motion will be reduced to some T' as given by
Pythagoras: T' = T * sqrt(1/(1--v<sup>2</sup>/c<sup>2</sup>)),
which by the way is the Lorentz factor of SRT.</font></p>
<p><font size="-1">If the electron moves into an arbitrary
direction with respect to the orbit, then the calculation of
more complicated but has the same result. I can give it if
there is a demand.<br>
</font></p>
<p><font size="-1">To the radius of the electron itself (and I
must apologize that I did not fully follow the preceding
discussion:</font></p>
<p><font size="-1">If the elementary charge e<sub>0</sub> in the
electron orbits at c then the magnetic moment of the electron
is classically µ = i*pi*R<sup>2</sup> where we insert for the
current i = e<sub>0</sub> * c/(2pi*R) . Then we get µ = c * e<sub>0</sub>*
R/2 . Now we can use the known value of the magnetic moment µ
to determine the radius R. The result of this is R = 3.86 * 10<sup>-13</sup>
m. <br>
</font></p>
<p><font size="-1">This result is in conflict with main stream as
the official physics says that the electron is point-like
(R<10<sup>-18 </sup>m). But it is in agreement with Erwin
Schrödinger. In his famous paper in which Schrödinger
evaluated the Dirac function, his result for the "size of the
electron" was "roughly about" R = 4 * 10<sup>-13</sup> m.
Schrödinger came to this result by pure QM considerations. And
then he makes a funny statement. He says in his paper: "We
know that the electron is point-like. So, there must be an
error in my calculation. But I cannot find this error". - I
think that not Schrödinger was in error but main stream is in
error. And this early result of Schrödinger confirms the
classical calculation which I have shown above.</font></p>
<p><font size="-1">Does this help the discussion?</font></p>
<p><font size="-1">Albrecht</font><br>
<br>
</p>
<div class="moz-cite-prefix">Am 09.01.2017 um 19:10 schrieb Chip
Akins:<br>
</div>
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<div class="WordSection1">
<p class="MsoNormal"><span style="color:black">Hi All<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">For those of
yoµu who hold the hard line that nothing can move faster
than c (a common interpretation of SR) the following is a
bit of speculation.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">If the energy
within the electron is all circulating at c, and the
electron is an extended particle, then the field lines
might look something like the following illustration
<o:p></o:p></span></p>
<p class="MsoNormal"><!--[if gte vml 1]><v:shapetype id="_x0000_t75" coordsize="21600,21600" o:spt="75" o:preferrelative="t" path="m@4@5l@4@11@9@11@9@5xe" filled="f" stroked="f">
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</v:shapetype><v:shape id="Picture_x0020_1" o:spid="_x0000_s1026" type="#_x0000_t75" style='position:absolute;margin-left:-16.55pt;margin-top:-111pt;width:131.45pt;height:124.5pt;z-index:251659264;visibility:visible;mso-wrap-style:square;mso-width-percent:0;mso-height-percent:0;mso-wrap-distance-left:9pt;mso-wrap-distance-top:0;mso-wrap-distance-right:9pt;mso-wrap-distance-bottom:0;mso-position-horizontal:absolute;mso-position-horizontal-relative:text;mso-position-vertical:absolute;mso-position-vertical-relative:text;mso-width-percent:0;mso-height-percent:0;mso-width-relative:margin;mso-height-relative:margin'>
<v:imagedata src="mailbox:///C:/Users/AL/AppData/Roaming/Thunderbird/Profiles/lthhzma2.default/Mail/pop3.strato-7.de/Inbox?number=270185650&header=quotebody&part=1.1.2&filename=image001.png" o:title="" />
<w:wrap type="tight"/>
</v:shape><![endif]--><!--[if !vml]--><img
src="cid:part1.AE27A6CC.2B065A63@a-giese.de"
v:shapes="Picture_x0020_1" height="166" hspace="12"
align="left" width="175"><!--[endif]--><span
style="color:black"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">At any rate,
the field lines would spiral outward from the center,
moving at c at all points.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">This structure
would not exhibit a specific frequency, or a finite set of
frequencies, but would contain any frequency one might
choose. So unless we can conceive of some mechanism which
would only make certain frequencies visible, or some
boundary conditions which would constrain the energy to a
specific radius. Then this approach is not useful in
discovering the electrons mysteries.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">In fact, if a
photon, or an EM wave if you prefer, can have a spin of
hbar, and has a momentum of <i>p=E/c</i>, then the radius
of action of this wave is <i>r = hbar/momentum</i>. Such
a wave then must have a transverse displacement velocity
of at least 3.489 times <i>c</i> in order for the wave to
exist in this form. Also, the internal wavefront must be
moving at the <i>sqrt(2) c. </i>So I think it must be
that some things simply move faster than <i>c </i>as
John Stewart Bell suggested. A more Lorentzian form of
relativity.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">Chip<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<div>
<div style="border:none;border-top:solid #E1E1E1
1.0pt;padding:3.0pt 0in 0in 0in">
<p class="MsoNormal"><b><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif">
General
[<a moz-do-not-send="true"
class="moz-txt-link-freetext"
href="mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org">mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org</a>]
<b>On Behalf Of </b>Dr Grahame Blackwell<br>
<b>Sent:</b> Sunday, January 08, 2017 4:10 PM<br>
<b>To:</b> Nature of Light and Particles - General
Discussion <a moz-do-not-send="true"
class="moz-txt-link-rfc2396E"
href="mailto:general@lists.natureoflightandparticles.org"><general@lists.natureoflightandparticles.org></a><br>
<b>Subject:</b> Re: [General] On particle radius<o:p></o:p></span></p>
</div>
</div>
<p class="MsoNormal"><o:p> </o:p></p>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif;color:navy">Hi
Chip,</span><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"> <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif;color:navy">Many
thanks indeed for your succinct and well-presented case
('succinct' is clearly a useful word in this discussion
- as well as a good strategy!).</span><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif;color:navy">I
need to go through this carefully and thoroughly and see
how it relates to my own understanding of the
situation. As we're all agreed, we all have things to
learn from each other and (here I DO agree with Vivian's
metaphor) each have some aspect of the elephant (in the
room?) to contribute. I'm really looking forward to
considering what you've said below and hopefully
assimilating it into a fuller understanding on my own
part of the issues that need to be taken into
consideration.</span><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"> <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif;color:navy">I'll
come back to you when I've processed it thoroughly (may
take a few days) and have some thoughts to offer.</span><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"> <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif;color:navy">Thanks
again,</span><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif;color:navy">Grahame</span><o:p></o:p></p>
</div>
<blockquote style="border:none;border-left:solid navy
1.5pt;padding:0in 0in 0in
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<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Arial",sans-serif">-----
Original Message ----- <o:p></o:p></span></p>
</div>
<div>
<p class="MsoNormal" style="background:#E4E4E4"><b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif">From:</span></b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif"> <a
moz-do-not-send="true"
href="mailto:chipakins@gmail.com"
title="chipakins@gmail.com">Chip Akins</a> <o:p></o:p></span></p>
</div>
<div>
<p class="MsoNormal"><b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif">To:</span></b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif"> <a
moz-do-not-send="true"
href="mailto:general@lists.natureoflightandparticles.org"
title="general@lists.natureoflightandparticles.org">'Nature
of Light and Particles - General Discussion'</a> <o:p></o:p></span></p>
</div>
<div>
<p class="MsoNormal"><b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif">Sent:</span></b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif">
Sunday, January 08, 2017 9:22 PM<o:p></o:p></span></p>
</div>
<div>
<p class="MsoNormal"><b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif">Subject:</span></b><span
style="font-size:10.0pt;font-family:"Arial",sans-serif"> Re:
[General] On particle radius<o:p></o:p></span></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<p class="MsoNormal"><span style="color:black" lang="EN-GB">Hi
Dr Graham Blackwell<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">I like the
way you clearly and succinctly write.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">Let me
explain some of the reasons why I feel the radius of the
electron decreases with velocity.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">In order to
accelerate the electron at rest, we must apply energy
(force through distance).<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black">The only way
to apply energy to the electron, when we get down to the
basis, is to add energy to its existing confined wave
structure. Plancks rule suggests that this confined
wave structure with energy added has a wavelength which
is (h c)/E. If this is the case and the momentum of this
wave remains p=E/c, then in order to be a spin ½ hbar
particle, it seems the electron must have a radius which
is r = (h c)/(4 pi E). Where E is the new total energy
with velocity throughout this paragraph.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">Then when we
calculate the mass of this particle from its confined
momentum (as Richard has pointed out) we get the
expected relativistic (total) mass of the moving
particle. m = E/(r w c) = E/c^2= E Eo Uo. Which is
exactly equivalent to m = y m. [where w = c/r (angular
frequency)].<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">This is the
only scenario I have found where all of the expected
parameters are accommodated, and I have searched
extensively for other possibilities.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">We also note
that the scattering cross-section of an electron at
relativistic velocities is very small, and agrees with
these assumptions quite well.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">In order for
the electron radius to remain the same size with
velocity I think we have to ignore things which seem
quite important, and these specific things appear to be
required in order to tie several of the pieces of the
puzzle together. It seems the picture is just not
complete unless the radius of the electron is reduced
with velocity.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">Thoughts?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:black">Chip<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:black"><o:p> </o:p></span></p>
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