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<DIV><FONT color=#000080 size=2 face=Arial>Thanks Richard,</FONT></DIV>
<DIV><FONT color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV><FONT color=#000080 size=2 face=Arial>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><FONT color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV><FONT color=#000080 size=2 face=Arial>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><FONT color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV><FONT color=#000080 size=2 face=Arial>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 <EM>not</EM> 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><FONT color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV><FONT color=#000080 size=2 face=Arial>Best regards,</FONT></DIV>
<DIV><FONT color=#000080 size=2 face=Arial>Grahame</FONT></DIV>
<BLOCKQUOTE
style="BORDER-LEFT: #000080 2px solid; PADDING-LEFT: 5px; PADDING-RIGHT: 0px; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="FONT: 10pt arial; BACKGROUND: #e4e4e4; font-color: black"><B>From:</B>
<A title=richgauthier@gmail.com href="mailto:richgauthier@gmail.com">Richard
Gauthier</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=general@lists.natureoflightandparticles.org
href="mailto:general@lists.natureoflightandparticles.org">Nature of Light and
Particles - General Discussion</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Thursday, July 07, 2016 6:42
AM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [General] double photon
cycle, subjective v objective realities</DIV>
<DIV><BR></DIV>
<DIV>Chip and Grahame,</DIV>
<DIV> 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> Richard</DIV><BR>
<DIV>
<BLOCKQUOTE type="cite">
<DIV>On Jul 6, 2016, at 10:18 AM, Dr Grahame Blackwell <<A
href="mailto:grahame@starweave.com">grahame@starweave.com</A>>
wrote:</DIV><BR class=Apple-interchange-newline>
<DIV>
<DIV
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color=#000080 size=2 face=Arial>Yes Chip,</FONT></DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>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>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>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
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>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
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>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
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>I hope that helps make things
clearer.</FONT></DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial></FONT> </DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>Best regards,</FONT></DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2 face=Arial>Grahame</FONT></DIV>
<DIV
style="TEXT-TRANSFORM: none; BACKGROUND-COLOR: rgb(255,255,255); TEXT-INDENT: 0px; FONT: 12px Helvetica; WHITE-SPACE: normal; LETTER-SPACING: normal; WORD-SPACING: 0px; -webkit-text-stroke-width: 0px"><FONT
color=#000080 size=2
face=Arial></FONT> </DIV></DIV></BLOCKQUOTE></DIV></BLOCKQUOTE></BODY></HTML>