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<font face="Helvetica, Arial, sans-serif"><span
style="mso-ansi-language:EN-US" lang="EN-US"><font size="-1">Dear
Martin, dear Adam, dear all:<br>
<br>
QM, de Broglie and his wave:<br>
<br>
It is true that de Broglie started from Special Relativity
(using some results) to develop his first considerations about
particles, so about QM. But that does of course not guarantee
that all his conclusions are correct with respect to
relativity. The de Broglie wave length is a very special case.<br>
<br>
I have shown the other day by a little example that the de
Broglie wave length as defined by himself and as used by
present physics is <i>not Lorentz-invariant</i>. We can see
(confirmed by experiments) that this wave length works
correctly if used and observed in an inertial system, in which
the double slit arrangement is at rest. But if such experiment
is observed by someone who moves with respect to this
arrangement, the results are incorrect. They can be
drastically wrong as I have shown with the numbers in my
example.<br>
<br>
How is it possible that this concept in some instances seems
correct but in others not? A plausible assumption (mentioned
earlier) seems to be that the scattering process of an
electron at such scattering device develops some local details
in which this de Broglie wave length in fact occurs; but only
there. And with this assumption we can explain both
consequences, the good and the bad one.<br>
<br>
This now has a severe consequence: As the de Broglie wave
length (and dB's considerations about it) is not
Lorentz-invariant, it cannot be valid for a free particle, as
for a free particle there is no natural reference system. And
a further, even more dramatic consequence, is that the
according part of the Schrödinger equation cannot be correct.<br>
<br>
Best regards<br>
Albrecht</font><br>
</span></font><br>
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<br>
Am 23.10.2015 um 00:50 schrieb Mark, Martin van der:<br>
<blockquote
cite="mid:3e12a835fd64467494fffec492cbfda6@AM3PR90MB0100.MGDPHG.emi.philips.com"
type="cite">
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<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Dear
Adam, I was still following up on Richards idea of the
origin of the de Broglie wavelength.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">I
was trying to make clear to you Richard and the group how
the order of events was (roughly), and how quantum mechanics
was born from special relativity and that the two theories
are in complete accordance despite a general believe that
there are unsurmountable problems with causality and that
Einstein was wrong, etc. Thank you for the article by Felix
Bloch, which confirms what I said about the origin of the
wave equation by Schroedinger, based on de Broglie’s ideas.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">I
must say that you have not yet understood the salient point
of the EPR experiment, but let me try to fix it. First of
all, before I do so, you must realize that I am not claiming
anything, I am only telling you what is the present
situation in physics and that EPR experiments show that
quantum mechanics is essentially correct.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Now
let’s go to the point that you are missing in your example
of the white and black marbles. You have forgotten that each
of the two envelopes, with a concealed marble in it, must be
delivered through the slit of a mailbox. The rule is that if
the slit is horizontal, that it selects for white/black. If
it is vertical it selects for green/red, when it is slanted
at some angle it is selecting for yellow/violet, etcetera.
While the postman is underway from sender to you, you are
allowed to alter the angle of your mailbox (perhaps you like
some particular color pair better than another pair). The
sender cannot do anything anymore about the marbles he has
put in the envelope, but in quantum mechanics it appears
that the measurement reveals colors as corresponding to the
angle of the mailboxes slit! This makes it mysterious and is
seen as a non-local action, kind of spooky indeed. It
implies that the marbles that went in the envelope at the
sender had opposite colors of all possible colors
simultaneously, somehow.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Now
I can understand that the rest of what I have written
earlier is not going to make much sense to you until you get
the above.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">I
hope this is of some help.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Very
best,
<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D">Martin<o:p></o:p></span></p>
<p class="MsoNormal"><o:p> </o:p></p>
<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 class="moz-txt-link-freetext" href="mailto:general-bounces+martin.van.der.mark=philips.com@lists.natureoflightandparticles.org">mailto:general-bounces+martin.van.der.mark=philips.com@lists.natureoflightandparticles.org</a>]<b>On
Behalf Of </b>Adam K<br>
<b>Sent:</b> donderdag 22 oktober 2015 22:19<br>
<b>To:</b> <a class="moz-txt-link-abbreviated" href="mailto:phys@a-giese.de">phys@a-giese.de</a>; Nature of Light and Particles -
General Discussion
<a class="moz-txt-link-rfc2396E" href="mailto:general@lists.natureoflightandparticles.org"><general@lists.natureoflightandparticles.org></a><br>
<b>Cc:</b> Joakim Pettersson <a class="moz-txt-link-rfc2396E" href="mailto:joakimbits@gmail.com"><joakimbits@gmail.com></a>;
Ariane Mandray <a class="moz-txt-link-rfc2396E" href="mailto:ariane.mandray@wanadoo.fr"><ariane.mandray@wanadoo.fr></a>; ARNOLD
BENN <a class="moz-txt-link-rfc2396E" href="mailto:arniebenn@mac.com"><arniebenn@mac.com></a><br>
<b>Subject:</b> Re: [General] research papers<o:p></o:p></span></p>
<p class="MsoNormal"><o:p> </o:p></p>
<div>
<p class="MsoNormal">Dear Martin,<o:p></o:p></p>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">I am not sure whether or not you were
expressing doubt as to the provenance of the wave
equation, and what Schrodinger owed to de Broglie, but in
either case you would probably enjoy reading this first
person account by Felix Bloch <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"><a moz-do-not-send="true"
href="http://www.physics.smu.edu/scalise/P5382fa15/FelixBlochPhysTodayDec1976b.pdf">http://www.physics.smu.edu/scalise/P5382fa15/FelixBlochPhysTodayDec1976b.pdf</a><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">As for EPR, it would probably make
sense to start another thread, I'll leave that to you or
anyone else who wishes to reply to decide if it's worth
doing. I am biased by my own thinking against your claim
of spooky action. To be honest, I could never bring myself
to believe that the properties of a particle are
indeterminate in the way standardly thought. The pilot
wave model makes much more sense than, and the same
predictions as, the mysterian Copenhagen, "particles are
smears and reality does not exist until I look at it" <i>niaiserie</i>. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">I could be wrong, and would welcome
correction from those with more physics experience than
me, but there seems to be an enormous conceptual leap from
the notion that A) We cannot know, even in principle,
whether 1 or 2 obtains, to B) Neither 1 nor 2 actually
obtains (they are in superposition) until actually
measured. It seems to me a grotesque error, and this is a
long story of course. Many people have weighed in. It is a
conceptual leap like that from A) The Universe exists, to
B) The Creator of the Universe does not want me to have
sex with people who are the same gender as myself. The
chasm between these propositions can only be spanned by
some enormous error. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">A positron was trapped for so long by
Dehmelt that he gave it a proper name: Priscilla. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"><i>Dehmelt says: “[t]here can be little
doubt about the identity of Priscilla during this
period, since in ultrahigh vacuum she never had a chance
to trade places with a passing antimatter twin. The
well-defined identity of this elementary particle is
something fundamentally new, which deserves to be
recognized by being given a name, as pets are give names
of persons” </i><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><i> </i><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal">I can accept that reality is nonlocal
(in a certain sense), and I have been aware of the intense
problem which is supposedly posed for relativity by Bell's
inequalities and the notion of entanglement. Supposedly
there is no actual information transmitted faster than
light, but all the same the state of one entangled
particle is altered by measuring the other one. Why is it
so hard for people to accept that there is no alteration
going on, and it is just as if I colored a marble white
and a marble black, and sent them to two different people
in the mail? The person who opens the envelope and sees a
black marble knows that the other person has the white
one, but there was no spooky action at a distance going
on. So what if there is no way of knowing,
<i>even in principle</i>, whether a particle is spin up or
spin down? Are we humans really so arrogant that by
instinct we must project ontological restrictions out of
our epistemological ones?<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">The Schrodinger equation gives us
probabilities. Quantum physics is a statistical theory.
Thus all of its predictions hold in the long time limit
(an electron in a box over sufficient cycles of the
electron's frequency), or the multiple particle limit
(diffraction experiments). This alone should be enough for
thinking individuals to realize that quantum physics is
not the end of the road. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal">Adam<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> <o:p></o:p></p>
</div>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
<div>
<p class="MsoNormal">On Thu, Oct 22, 2015 at 10:18 AM, Dr.
Albrecht Giese <<a moz-do-not-send="true"
href="mailto:genmail@a-giese.de" target="_blank">genmail@a-giese.de</a>>
wrote:<o:p></o:p></p>
<blockquote style="border:none;border-left:solid #CCCCCC
1.0pt;padding:0cm 0cm 0cm
6.0pt;margin-left:4.8pt;margin-right:0cm">
<div>
<p class="MsoNormal" style="margin-bottom:12.0pt">Hello
Richard,<br>
<br>
thank you and see my comments below.<o:p></o:p></p>
<div>
<p class="MsoNormal">Am 22.10.2015 um 00:32 schrieb
Richard Gauthier:<o:p></o:p></p>
</div>
<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal">Hello Albert (and all),<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> I think your fundamental
objection that you mentioned earlier can be
answered below.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> The left side of the big
triangle in Figure 2 in my article is a purely
mathematical unfolding of the path of the helical
trajectory, to hopefully show more clearly the
generation of de Broglie wavelengths from plane
waves emitted by the actual charged photon moving
along the helical trajectory. Nothing is actually
moving off into space along this line.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> Consider an electron moving
with velocity v horizontally along the helical
axis. Since in Figure 2 in my article, cos (theta)
= v/c , the corresponding velocity of the charged
photon along the helical path is v/ cos(theta) = c
, the speed of the charged photon, which we knew
already because the helical trajectory was defined
so that this is the case. In a short time T, the
electron has moved a distance Delectron = vT
horizontally and the photon has moved a distance
Dphoton = Delectron/cos(theta) =vT/cos(theta) = cT
along its helical trajectory.<o:p></o:p></p>
</div>
</blockquote>
<p class="MsoNormal">I agree. <o:p></o:p></p>
<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal">A plane wave front emitted from
the photon at the distance Dphoton = cT along the
photon’s helical path will intersect the base of
the big triangle (the helical axis) at the
distance along the base given by Dwavefront =
Dphoton / cos(theta) = cT/ (v/c) = T * (c^2)/v
which means the intersection point of the plane
wave with the helical axis is moving with a speed
c^2/v which is the de Broglie wave’s phase
velocity.
<o:p></o:p></p>
</div>
</blockquote>
<p class="MsoNormal">Here I disagree. If we assume the
wave front as an extended layer through the photon and
with an orientation perpendicular to the actual
direction of the photon, then the intersect point of
this layer with the axis has the same z coordinate as
the z-component of the photon's position. This is
essential. (I have built myself a little 3-d model to
see this.)<br>
<br>
When now, say at time T<sub>0</sub>, a phase maximum
of the wave front leaves the photon, then the same
phase maximum passes the intersect point on the axis
with the same z coordinate. After a while (i.e. after
the time T<sub>p</sub>=1/frequency) the next phase
maximum will exit from the photon and simultaneously
the next phase maximum will cross the axis. The new
z-value (of the photon and of the intersect point) is
now displaced from the old one by the amount delta_z =
v * T<sub>p</sub>. During this time the photon will
have moved by c * T<sub>p</sub> on its helical path.<br>
<br>
Now the spacial distance between these two phase
maxima, which is the wavelength, is: lambda<sub>photon</sub>
= c * T<sub>p</sub>, and lambda<sub>electron</sub> = v
* T<sub>p</sub>.
<br>
<br>
This is my result. Or what (which detail) is wrong?<br>
<br>
best wishes<br>
Albrecht<br>
<br>
<br>
<br>
<o:p></o:p></p>
<blockquote style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal">The length of the de Broglie
wave itself as shown previously from Figure 2 is
Ldb = Lambda-photon / cos(theta) = h/(gamma mc) /
(v/c) = h/(gamma mv). So as the electron moves
with velocity v along the z-axis, de Broglie waves
of length h/(gamma mv) produced along the z-axis
are moving with velocity c^2/v along the z-axis.
The de Broglie waves created by the circulating
charged photon will speed away from the electron
(but more will be produced) to take their place,
one de Broglie wave during each period of the
circulating charged photon (corresponding to the
moving electron). As mentioned previously, the
period of the circulating charged photon is 1/f =
1/(gamma mc^2/h) = h/(gamma mc^2/). As the
electron speeds up (v and gamma increase) the de
Broglie wavelengths h/(gamma mv) are shorter and
move more slowly, following the speed formula
c^2/v .<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<p class="MsoNormal" style="margin-bottom:12.0pt"><o:p> </o:p></p>
<div>
<p class="MsoNormal">Unpublished graphic showing the
generation of de Broglie waves from a moving
charged photon along its helical trajectory. The
corresponding moving electron is the red dot
moving to the right on the red line. The charged
photon is the blue dot moving at light speed along
the helix.The blue dot has moves a distance of one
charged photon wavelength h/(gamma mc) along the
helix from the left corner of the diagram On the
left diagonal line (representing the
mathematically unrolled helix), the blue dots
correspond to separations of 1 charged photon
h/(gamma mc) wavelength along the helical axis. In
this graphic, v/c = 0.5 so cos(theta)= 0.5 and
theta= 60 degrees. The group velocity is c^2/v =
c^2/0.5c = 2 c, the speed of the de Broglie waves
along the horizontal axis . The distances between
the intersection points on the horizontal line
each correspond to 1 de Broglie wavelength, which
in this example where v=0.5 c is h(gamma mv) = 2
x charged photon wavelength h/(gamma mc).<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> It is true that when the
electron is at rest, the wave fronts emitted by
the circulating charged photon all pass through
the center of the circular path of the charged
photon and do not intersect any helical axis,
because no helical axis is defined for a resting
electron, i.e. the pitch of the helix of the
circulating charged photon is zero. For a very
slowly moving electron, the pitch of the helix of
the circulating charged photon is very small but
non-zero, but the de Broglie wavelength is very
large, much larger than the helical pitch. Perhaps
you are confusing these two lengths — the helical
pitch of the circulating charged photon and the de
Broglie wavelength generated by the wave fronts
emitted by the circulating charged photon. The
pitch of the helix starts at zero (for v=0 of the
electron) and reaches a maximum when the speed of
the electron is c/sqrt(2) and theta = 45 degrees
(see my charged photon paper) and then the helical
pitch decreases towards zero as the speed of the
electron further increases towards the speed of
light. But the de Broglie wavelength Ldb starts
very large (when the electron is moving very
slowly) and decreases uniformly towards zero as
the speed of the electron increases, as given by
Ldb = h/gamma mv. It is the de Broglie wavelength
generated by the charged photon that has
predictive physical significance in diffraction
and double-slit experiments while the helical
pitch of the charged photon’s helical trajectory
has no current predictive physical significance
(though if experimental predictions based on the
helical pitch could be made, this could be a test
of the charged photon model).<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> I don’t have any comments
yet on your concerns about the de Broglie
wavelength that you just expressed to John W
(below).<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"> all the best,<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"> Richard<o:p></o:p></p>
</div>
<p class="MsoNormal"><o:p> </o:p></p>
<div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal">On Oct 21, 2015, at 12:42
PM, Dr. Albrecht Giese <<a
moz-do-not-send="true"
href="mailto:genmail@a-giese.de"
target="_blank"><a class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de">genmail@a-giese.de</a></a>>
wrote:<o:p></o:p></p>
</div>
<p class="MsoNormal"><o:p> </o:p></p>
<div>
<p class="MsoNormal"><span
style="font-size:10.0pt;font-family:"Helvetica",sans-serif;background:white">Dear
John W and all,<br>
<br>
about the <u>de Broglie wave</u>:<br>
<br>
There are a lot of elegant derivations for
the de Broglie wave length, that is true.
Mathematical deductions. What is about the
physics behind it?<br>
<br>
De Broglie derived this wave in his first
paper in the intention to explain, why the
internal frequency in a moving electron is
dilated, but this frequency on the other
hand has to be increased for an external
observer to reflect the increase of energy.
To get a result, he invented a "fictitious
wave" which has the phase speed c/v, where v
is the speed of the electron. And he takes
care to synchronize this wave with the
internal frequency of the electron. That
works and can be used to describe the
scattering of the electron at the double
slit. - But is this physical
understanding? De Broglie himself stated
that this solution does not fulfil the
expectation in a "complete theory". Are we
any better today?<br>
<br>
Let us envision the following situation. An
electron moves at moderate speed, say 0.1*c
(=> gamma=1.02) . An observer moves
parallel to the electron. What will the
observer see or measure? <br>
The internal frequency of the electron will
be observed by him as frequency = m<sub>0</sub>*c<sup>2</sup>/h
, because in the observer's system the
electron is at rest. The wave length of the
wave leaving the electron (e.g. in the model
of a circling photon) is now not exactly
lambda<sub>1</sub> = c/frequency , but a
little bit larger as the rulers of the
observer are a little bit contracted (by
gamma = 1.02), so this is a small effect.
What is now about the phase speed of the de
Broglie wave? For an observer at rest it
must be quite large as it is extended by the
factor c/v which is 10. For the co-moving
observer it is mathematically infinite (in
fact he will see a constant phase). This is
not explained by the time dilation (=2%), so
not compatible. And what about the de
Broglie wave length? For the co-moving
observer, who is at rest in relation to the
electron, it is lambda<sub>dB</sub> =
h/(1*m*0), which is again infinite or at
least extremely large. For the observer at
rest there is lambda<sub>dB</sub> =
h/(1.02*m*0.1c) . Also not comparable to the
co-moving observer.<br>
<br>
To summarize: these differences are not
explained by the normal SR effects. So, how
to explain these incompatible results?<br>
<br>
Now let's assume, that the electron closes
in to the double slit. Seen from the
co-moving observer, the double slit
arrangement moves towards him and the
electron. What are now the parameters which
will determine the scattering? The
(infinite) de Broglie wave length? The phase
speed which is 10*c ? Remember: For the
co-moving observer the electron does not
move. Only the double slit moves and the
screen behind the double slit will be ca. 2%
closer than in the standard case. But will
that be a real change?<br>
<br>
I do not feel that this is a situation which
in physically understood.<br>
<br>
Regards<br>
Albrecht</span><o:p></o:p></p>
<div>
<div>
<p class="MsoNormal"><span
style="font-size:9.0pt;font-family:"Helvetica",sans-serif"><br
style="text-align:start;word-spacing:0px">
<br>
</span><o:p></o:p></p>
<div>
<p class="MsoNormal"
style="background:white"><span
style="font-size:9.0pt;font-family:"Helvetica",sans-serif">Am
21.10.2015 um 16:34 schrieb John
Williamson:<o:p></o:p></span></p>
</div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt;text-align:start;word-spacing:0px">
<div>
<p class="MsoNormal"
style="background:white"><span
style="font-size:10.0pt;font-family:"Tahoma",sans-serif">Dear
all,<br>
<br>
The de Broglie wavelength is best
understood, in my view, in one of
two ways. Either read de Broglies
thesis for his derivation (if you do
not read french, Al has translated
it and it is available online).
Alternatively derive it yourself.
All you need to do is consider the
interference between a standing wave
in one (proper frame) as it
transforms to other relativistic
frames. That is standing-wave
light-in-a-box. This has been done
by may folk, many times. Martin did
it back in 1991. It is in our 1997
paper. One of the nicest
illustrations I have seen is that of
John M - circulated to all of you
earlier in this series.<br>
<br>
It is real, and quite simple.<br>
<br>
Regards, John.<o:p></o:p></span></p>
<div>
<div class="MsoNormal"
style="text-align:center;background:white"
align="center">
<hr size="2" width="100%"
align="center">
</div>
<div>
<p class="MsoNormal"
style="margin-bottom:12.0pt;background:white"><b><span
style="font-size:10.0pt;font-family:"Tahoma",sans-serif">From:</span></b><span
style="font-size:10.0pt;font-family:"Tahoma",sans-serif"> General
[<a moz-do-not-send="true"
href="mailto:general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org"
target="_blank">general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org</a>]
on behalf of Dr. Albrecht Giese
[<a moz-do-not-send="true"
href="mailto:genmail@a-giese.de"
target="_blank">genmail@a-giese.de</a>]<br>
<b>Sent:</b> Wednesday, October
21, 2015 3:14 PM<br>
<b>To:</b> Richard Gauthier<br>
<b>Cc:</b> Nature of Light and
Particles - General Discussion;
David Mathes<br>
<b>Subject:</b> Re: [General]
research papers</span><o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="margin-bottom:12.0pt;background:white">Hello
Richard,<br>
<br>
thanks for your detailed
explanation. But I have a
fundamental objection.<br>
<br>
Your figure 2 is unfortunately
(but unavoidably) 2-dimensional,
and that makes a difference to the
reality as I understand it. <br>
<br>
In your model the charged electron
moves on a helix around the axis
of the electron (or equivalently
the axis of the helix). That means
that the electron has a constant
distance to this axis. Correct?
But in the view of your figure 2
the photon seems to start on the
axis and moves away from it
forever. In this latter case the
wave front would behave as you
write it. <br>
<br>
Now, in the case of a constant
distance, the wave front as well
intersects the axis, that is true.
But this intersection point moves
along the axis at the projected
speed of the photon to this axis.
- You can consider this also in
another way. If the electron moves
during a time, say T1, in the
direction of the axis, then the
photon will during this time T1
move a longer distance, as the
length of the helical path (call
it L) is of course longer than
the length of the path of the
electron during this time (call it
Z). Now you will during the time
T1 have a number of waves (call
this N) on the helical path L. On
the other hand, the number of
waves on the length Z has also to
be N. Because otherwise after an
arbitrary time the whole situation
would diverge. As now Z is smaller
than L, the waves on the axis have
to be shorter. So, not the de
Broglie wave length. That is my
understanding. <br>
<br>
In my present view, the de Broglie
wave length has no immediate
correspondence in the physical
reality. I guess that the success
of de Broglie in using this wave
length may be understandable if we
understand in more detail, what
happens in the process of
scattering of an electron at the
double (or multiple) slits.<br>
<br>
Best wishes<br>
Albrecht<br>
<br>
<o:p></o:p></p>
<div>
<p class="MsoNormal"
style="background:white">Am
21.10.2015 um 06:28 schrieb <br>
Richard Gauthier:<o:p></o:p></p>
</div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal"
style="background:white">Hello
Albrecht,<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
Thank you for your effort to
understand the physical
process described
geometrically in my Figure 2.
You have indeed misunderstood
the Figure as you suspected.
The LEFT upper side of the big
90-degree triangle is one
wavelength h/(gamma mc) of the
charged photon, mathematically
unrolled from its two-turned
helical shape (because of the
double-loop model of the
electron) so that its full
length h/(gamma mc) along the
helical trajectory can be
easily visualized. The emitted
wave fronts described in my
article are perpendicular to
this mathematically unrolled
upper LEFT side of the
triangle (because the plane
waves emitted by the charged
photon are directed along the
direction of the helix when it
is coiled (or mathematically
uncoiled), and the plane wave
fronts are perpendicular to
this direction). The upper
RIGHT side of the big
90-degree triangle corresponds
to one of the plane wave
fronts (of constant phase
along the wave front) emitted
at one wavelength lambda =
h/(gamma mc) of the helically
circulating charged photon.
The length of the horizontal
base of the big 90-degree
triangle, defined by where
this upper RIGHT side of the
triangle (the generated plane
wave front from the charged
photon) intersects the
horizontal axis of the
helically-moving charged
photon, is the de Broglie
wavelength h/(gamma mv) of the
electron model (labeled in the
diagram). By geometry the
length (the de Broglie
wavelength) of this horizontal
base of the big right triangle
in the Figure is equal to the
top left side of the triangle
(the photon wavelength
h/(gamma mc) divided (not
multiplied) by cos(theta) =
v/c because we are calculating
the hypotenuse of the big
right triangle starting from
the upper LEFT side of this
big right triangle, which is
the adjacent side of the big
right triangle making an angle
theta with the hypotenuse. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
What you called the
projection of the charged
photon’s wavelength h/(gamma
mc) onto the horizontal axis
is actually just the distance
D that the electron has moved
with velocity v along the
x-axis in one period T of the
circulating charged photon.
That period T equals 1/f =
1/(gamma mc^2/h) = h/(gamma
mc^2). By the geometry in the
Figure, that distance D is the
adjacent side of the smaller
90-degree triangle in the left
side of the Figure, making an
angle theta with cT, the
hypotenuse of that smaller
triangle, and so D = cT cos
(theta) = cT x v/c = vT , the
distance the electron has
moved to the right with
velocity v in the time T. In
that same time T one de
Broglie wavelength has been
generated along the horizontal
axis of the circulating
charged photon. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"> I
will answer your question
about the double slit in a
separate e-mail.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
all the best,<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
Richard<o:p></o:p></p>
</div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
<div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal"
style="background:white">On
Oct 20, 2015, at 10:06 AM,
Dr. Albrecht Giese <<a
moz-do-not-send="true"
href="mailto:genmail@a-giese.de"
target="_blank"><a class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de">genmail@a-giese.de</a></a>>
wrote:<o:p></o:p></p>
</div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
<div>
<div>
<p class="MsoNormal"
style="margin-bottom:12.0pt;background:white">Hello
Richard,<br>
<br>
thank you for your
explanations. I would
like to ask further
questions and will place
them into the text
below.<o:p></o:p></p>
<div>
<p class="MsoNormal"
style="background:white">Am
19.10.2015 um 20:08
schrieb Richard
Gauthier:<o:p></o:p></p>
</div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal"
style="background:white">Hello
Albrecht,<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
Thank your for
your detailed
questions about my
electron model,
which I will answer
as best as I can. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
My approach of
using the formula
e^i(k*r-wt) =
e^i (k dot r minus
omega t) for a
plane wave emitted
by charged photons
is also used for
example in the
analysis of x-ray
diffraction from
crystals when you
have many incoming
parallel photons in
free space moving in
phase in a plane
wave. Please see for
example <span
style="font-size:10.0pt"><a
moz-do-not-send="true"
href="http://www.pa.uky.edu/%7Ekwng/phy525/lec/lecture_2.pdf"
target="_blank"><a class="moz-txt-link-freetext" href="http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf">http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf</a></a></span> .
When Max Born
studied electron
scattering using
quantum mechanics
(where he used
PHI*PHI of the
quantum wave
functions to predict
the electron
scattering
amplitudes), he also
described the
incoming electrons
as a plane wave
moving forward with
the de Broglie
wavelength towards
the target. I think
this is the general
analytical procedure
used in scattering
experiments. In my
charged photon model
the helically
circulating charged
photon,
corresponding to a
moving electron, is
emitting a plane
wave of wavelength
lambda = h/(gamma
mc) and frequency
f=(gamma mc^2)/h
along the direction
of its helical
trajectory, which
makes a forward
angle theta with the
helical axis given
by cos (theta)=v/c.
Planes of constant
phase emitted from
the charged photon
in this way
intersect the
helical axis of the
charged photon. When
a charged photon has
traveled one
relativistic
wavelength lambda =
h/(gamma mc) along
the helical axis,
the intersection
point of this wave
front with the
helical axis has
traveled (as seen
from the geometry of
Figure 2 in my
charged photon
article) a distance
lambda/cos(theta) =
lambda / (v/c) =
h/(gamma mv) i.e
the relativistic de
Broglie wavelength
along the helical
axis.<o:p></o:p></p>
</div>
</blockquote>
<p class="MsoNormal"
style="background:white">Here
I have a question with
respect to your Figure
2. The circling charged
photon is accompanied by
a wave which moves at
any moment in the
direction of the photon
on its helical path.
This wave has its normal
wavelength in the
direction along this
helical path. But if now
this wave is projected
onto the axis of the
helix, which is the axis
of the moving electron,
then the projected wave
will be shorter than the
original one. So the
equation will not be
lambda<sub>deBroglie</sub> =
lambda<sub>photon</sub> /
cos theta , but: lambda<sub>deBroglie</sub> =
lambda<sub>photon</sub> *
cos theta . The result
will not be the
(extended) de Broglie
wave but a shortened
wave. Or do I completely
misunderstand the
situation here?<br>
<br>
Or let's use another
view to the process.
Lets imagine a
scattering process of
the electron at a double
slit. This was the
experiment where the de
Broglie wavelength
turned out to be
helpful. <br>
So, when now the
electron, and that means
the cycling photon,
approaches the slits, it
will approach at a slant
angle theta at the layer
which has the slits. Now
assume the momentary
phase such that the wave
front reaches two slits
at the same time (which
means that the photon at
this moment moves
downwards or upwards,
but else straight with
respect to the azimuth).
This situation is
similar to the front
wave of a <i>single</i> normal
photon which moves
upwards or downwards by
an angle theta. There is
now no phase difference
between the right and
the left slit. Now the
question is whether this
coming-down (or -up)
will change the temporal
sequence of the phases
(say: of the maxima of
the wave). This distance
(by time or by length)
determines at which
angle the next
interference maxima to
the right or to the left
will occur behind the
slits. <br>
<br>
To my understanding the
temporal distance will
be the same distance as
of wave maxima on the
helical path of the
photon, where the latter
is lambda<sub>1</sub> =
c / frequency; frequency
= (gamma*mc<sup>2</sup>)
/ h. So, the geometric
distance of the wave
maxima passing the slits
is lambda<sub>1</sub> =
c*h / (gamma*mc<sup>2</sup>).
Also here the result is
a shortened wavelength
rather than an extended
one, so not the de
Broglie wavelength.<br>
<br>
Again my question: What
do I misunderstand?<br>
<br>
For the other topics of
your answer I
essentially agree, so I
shall stop here.<br>
<br>
Best regards<br>
Albrecht<br>
<br>
<br>
<o:p></o:p></p>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
Now as seen from
this geometry, the
slower the
electron’s velocity
v, the longer is the
electron’s de
Broglie wavelength —
also as seen from
the relativistic de
Broglie wavelength
formula Ldb =
h/(gamma mv). For a
resting electron
(v=0) the de Broglie
wavelength is
undefined in this
formula as also in
my model for v = 0.
Here, for stationary
electron, the
charged photon’s
emitted wave fronts
(for waves of
wavelength equal to
the Compton
wavelength h/mc)
intersect the axis
of the circulating
photon along its
whole length rather
than at a single
point along the
helical axis. This
condition
corresponds to the
condition where de
Broglie said
(something like)
that the electron
oscillates with the
frequency given by f
= mc^2/h for the
stationary electron,
and that the phase
of the wave of this
oscillating electron
is the same at all
points in space. But
when the electron is
moving slowly, long
de Broglie waves are
formed along the
axis of the moving
electron.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
In this basic
plane wave model
there is no
limitation on how
far to the sides of
the charged photon
the plane wave
fronts extend. In a
more detailed model
a finite
side-spreading of
the plane wave would
correspond to a
pulse of many
forward moving
electrons that is
limited in both
longitudinal and
lateral extent (here
a Fourier
description of the
wave front for a
pulse of electrons
of a particular
spatial extent would
probably come into
play), which is
beyond the present
description.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
You asked what an
observer standing
beside the resting
electron, but not in
the plane of the
charged photon's
internal circular
motion) would
observe as the
circulating charged
photon emits a plane
wave long its
trajectory. The
plane wave’s
wavelength emitted
by the circling
charged photon would
be the Compton
wavelength h/mc. So
when the charged
photon is moving
more towards (but an
an angle to) the
stationary observer,
he would observe a
wave of wavelength
h/mc (which you call
c/ny where ny is the
frequency of charged
photon’s orbital
motion) coming
towards and past
him. This is not the
de Broglie
wavelength (which is
undefined here and
is only defined on
the helical axis of
the circulating
photon for a moving
electron) but is the
Compton wavelength
h/mc of the
circulating photon
of a resting
electron. As the
charged photon moves
more away from the
observer, he would
observe a plane wave
of wavelength h/mc
moving away from him
in the direction of
the receding charged
photon. But it is
more complicated
than this, because
the observer at the
side of the
stationary electron
(circulating charged
photon) will also be
receiving all the
other plane waves
with different
phases emitted at
other angles from
the circulating
charged photon
during its whole
circular trajectory.
In fact all of these
waves from the
charged photon away
from the circular
axis or helical axis
will interfere and
may actually cancel
out or partially
cancel out (I don’t
know), leaving a net
result only along
the axis of the
electron, which if
the electron is
moving, corresponds
to the de Broglie
wavelength along
this axis. This is
hard to visualize in
3-D and this is why
I think a 3-D
computer graphic
model of this
plane-wave emitting
process for a moving
or stationary
electron would be
very helpful and
informative.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
You asked about
the electric charge
of the charged
photon and how it
affects this
process. Clearly the
plane waves emitted
by the circulating
charged photon have
to be different from
the plane waves
emitted by an
uncharged photon,
because these plane
waves generate the
quantum wave
functions PHI that
predict the
probabilities of
finding electrons or
photons respectively
in the future from
their PHI*PHI
functions. Plus the
charged photon has
to be emitting an
additional electric
field (not emitted
by a regular
uncharged photon),
for example caused
by virtual uncharged
photons as described
in QED, that
produces the
electrostatic field
of a stationary
electron or the
electro-magnetic
field around a
moving electron. <o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
I hope this helps.
Thanks again for
your excellent
questions.<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
with best
regards,<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white">
Richard<o:p></o:p></p>
</div>
<div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
<div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p
class="MsoNormal"
style="background:white">On Oct 19, 2015, at 8:13 AM, Dr. Albrecht Giese
<<a
moz-do-not-send="true"
href="mailto:genmail@a-giese.de" target="_blank"><a class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de">genmail@a-giese.de</a></a>>
wrote:<o:p></o:p></p>
</div>
<p class="MsoNormal"
style="background:white"><o:p> </o:p></p>
<div>
<div>
<p
class="MsoNormal"
style="margin-bottom:12.0pt;background:white">Richard:<br>
<br>
I am still
busy to
understand the
de Broglie
wavelength
from your
model. I think
that I
understand
your general
idea, but I
would like to
also
understand the
details. <br>
<br>
If a photon
moves straight
in the free
space, how
does the wave
look like? You
say that the
photon emits a
plane wave. If
the photon is
alone and
moves
straight, then
the wave goes
with the
photon. No
problem. And
the wave front
is in the
forward
direction.
Correct? How
far to the
sides is the
wave extended?
That may be
important in
case of the
photon in the
electron.<br>
<br>
With the
following I
refer to the
figures 1 and
2 in your
paper referred
in your
preceding
mail.<br>
<br>
In the
electron, the
photon moves
according to
your model on
a circuit. It
moves on a
helix when the
electron is in
motion. But
let take us
first the case
of the
electron at
rest, so that
the photon
moves on this
circuit. In
any moment the
plane wave
accompanied
with the
photon will
momentarily
move in the
tangential
direction of
the circuit.
But the
direction will
permanently
change to
follow the
path of the
photon on the
circuit. What
is then about
the motion of
the wave? The
front of the
wave should
follow this
circuit. Would
an observer
next to the
electron at
rest (but not
in the plane
of the
internal
motion) notice
the wave? This
can only
happen, I
think, if the
wave does not
only propagate
on a straight
path forward
but has an
extension to
the sides.
Only if this
is the case,
there will be
a wave along
the axis of
the electron.
Now an
observer next
to the
electron will
see a
modulated wave
coming from
the photon,
which will be
modulated with
the frequency
of the
rotation,
because the
photon will in
one moment be
closer to the
observer and
in the next
moment be
farer from
him. Which
wavelength
will be
noticed by the
observer? It
should be
lambda = c /
ny, where c is
the speed of
the
propagation
and ny the
frequency of
the orbital
motion. But
this lambda is
by my
understanding
not be the de
Broglie wave
length.<br>
<br>
For an
electron at
rest your
model expects
a wave with a
momentarily
similar phase
for all points
in space. How
can this
orbiting
photon cause
this? And
else, if the
electron is
not at rest
but moves at a
very small
speed, then
the situation
will not be
very different
from that of
the electron
at rest.<br>
<br>
Further: What
is the
influence of
the charge in
the photon?
There should
be a modulated
electric field
around the
electron with
a frequency ny
which follows
also from E =
h*ny, with E
the dynamical
energy of the
photon. Does
this modulated
field have any
influence to
how the
electron
interacts with
others? <br>
<br>
Some
questions,
perhaps you
can help me
for a better
understanding.<br>
<br>
With best
regards and
thanks in
advance<br>
Albrecht<br>
<br>
PS: I shall
answer you
mail from last
night
tomorrow.<br>
<br>
<o:p></o:p></p>
<div>
<p
class="MsoNormal"
style="background:white">Am 14.10.2015 um 22:32 schrieb Richard
Gauthier:<o:p></o:p></p>
</div>
<blockquote
style="margin-top:5.0pt;margin-bottom:5.0pt">
<div>
<p
class="MsoNormal"
style="background:white">Hello Albrecht,<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> I second David’s question. The last I heard
authoritatively,
from
cosmologist
Sean Carroll -
"The Particle
at the End of
the Universe”
(2012), is
that fermions
are not
affected by
the strong
nuclear force.
If they were,
I think it
would be
common
scientific
knowledge by
now. <o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white">You wrote: "<span style="background:white">I
see it as a
valuable goal
for the
further
development to
find an answer
(a </span><i>physical </i><span
style="background:white">answer!) to the question of the de Broglie
wavelength."</span><o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> My spin 1/2 charged photon model DOES give a
simple
physical
explanation
for the origin
of the de
Broglie
wavelength.
The
helically-circulating
charged photon
is proposed to
emit a plane
wave directed
along its
helical path
based on its
relativistic
wavelength
lambda =
h/(gamma mc)
and
relativistic
frequency
f=(gamma
mc^2)/h. The
wave fronts of
this plane
wave intersect
the axis of
the charged
photon’s
helical
trajectory,
which is the
path of the
electron being
modeled by the
charged
photon,
creating a de
Broglie wave
pattern of
wavelength
h/(gamma mv)
which travels
along the
charged
photon’s
helical axis
at speed
c^2/v. For a
moving
electron, the
wave fronts
emitted by the
charged photon
do not
intersect the
helical axis
perpendicularly
but at an
angle (see
Figure 2 of my
SPIE paper at <a
moz-do-not-send="true"
href="https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength"
target="_blank"><a class="moz-txt-link-freetext" href="https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength">https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength</a></a> )
that is simply
related to the
speed of the
electron being
modeled. This
physical
origin of the
electron’s de
Broglie wave
is similar to
when a series
of parallel
and
evenly-spaced
ocean waves
hits a
straight beach
at an angle
greater than
zero degrees
to the beach —
a wave pattern
is produced at
the beach that
travels in one
direction
along the
beach at a
speed faster
than the speed
of the waves
coming in from
the ocean. But
that beach
wave pattern
can't transmit
“information”
along the
beach faster
than the speed
of the ocean
waves, just as
the de Broglie
matter-wave
can’t
(according to
special
relativity)
transmit
information
faster than
light, as de
Broglie
recognized.
As far as I
know this
geometric
interpretation
for the
generation of
the
relativistic
electron's de
Broglie
wavelength,
phase
velocity, and
matter-wave
equation is
unique.<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> For a resting (v=0) electron, the de Broglie
wavelength
lambda =
h/(gamma mv)
is not defined
since one
can’t divide
by zero. It
corresponds to
the ocean wave
fronts in the
above example
hitting the
beach at a
zero degree
angle, where
no velocity of
the wave
pattern along
the beach can
be defined.<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> <span style="color:#252525;background:white">Schrödinger</span> took
de Broglie’s
matter-wave
and used it
non-relativistically
with a
potential V
to generate
the <span
style="color:#252525;background:white">Schrödinger</span> equation
and wave
mechanics,
which is
mathematically
identical in
its
predictions to
Heisenberg’s
matrix
mechanics.
Born
interpreted
Psi*Psi of
the <span
style="color:#252525;background:white">Schrödinger</span> equation
as the
probability
density for
the result of
an
experimental
measurement
and this
worked well
for
statistical
predictions.
Quantum
mechanics was
built on this
de Broglie
wave
foundation and
Born's
probabilistic
interpretation
(using Hilbert
space math.)<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> The charged photon model of the electron
might be used
to derive the <span
style="color:#252525;background:white">Schrödinger</span> equation,
considering
the electron
to be a
circulating
charged photon
that generates
the electron’s
matter-wave,
which depends
on the
electron’s
variable
kinetic energy
in a potential
field. This
needs to be
explored
further, which
I began in <a
moz-do-not-send="true"
href="https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schr%C3%B6dinger_Equation"
target="_blank"><a class="moz-txt-link-freetext" href="https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation">https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation</a></a> .
Of course, to
treat the
electron
relativistically
requires the
Dirac
equation. But
the spin 1/2
charged photon
model of the
relativistic
electron has a
number of
features of
the Dirac
electron, by
design.<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> As to why the charged photon circulates
helically
rather than
moving in a
straight line
(in the
absence of
diffraction,
etc) like an
uncharged
photon, this
could be the
effect of the
charged photon
moving in the
Higgs field,
which turns a
speed-of-light
particle with
electric
charge into a
less-than-speed-of-light
particle with
a rest mass,
which in this
case is the
electron’s
rest mass
0.511 MeV/c^2
(this value is
not predicted
by the Higgs
field theory
however.) So
the electron’s
inertia may
also be caused
by the Higgs
field. I would
not say that
an unconfined
photon has
inertia,
although it
has energy and
momentum but
no rest mass,
but opinions
differ on this
point.
“Inertia” is a
vague term and
perhaps should
be dropped— it
literally
means
"inactive,
unskilled”.<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> You said that a faster-than-light phase wave
can only be
caused by a
superposition
of waves. I’m
not sure this
is correct,
since in my
charged photon
model a single
plane wave
pattern
emitted by the
circulating
charged photon
generates the
electron’s
faster-than-light
phase wave of
speed c^2/v .
A group
velocity of an
electron model
may be
generated by a
superposition
of waves to
produce a wave
packet whose
group velocity
equals the
slower-than-light
speed of an
electron
modeled by
such an
wave-packet
approach.<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white">with best regards,<o:p></o:p></p>
</div>
<div>
<p
class="MsoNormal"
style="background:white"> Richard<o:p></o:p></p>
</div>
<p
class="MsoNormal"
style="background:white"><o:p> </o:p></p>
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