<html>
<head>
<meta content="text/html; charset=windows-1252"
http-equiv="Content-Type">
</head>
<body text="#000000" bgcolor="#FFFFFF">
<small>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<br>
</small><br>
<br>
<div class="moz-cite-prefix">Am 21.10.2015 um 16:34 schrieb John
Williamson:<br>
</div>
<blockquote
cite="mid:7DC02B7BFEAA614DA666120C8A0260C914714222@CMS08-01.campus.gla.ac.uk"
type="cite">
<meta http-equiv="Content-Type" content="text/html;
charset=windows-1252">
<style id="owaParaStyle" type="text/css">P {margin-top:0;margin-bottom:0;}</style>
<div style="direction: ltr;font-family: Tahoma;color:
#000000;font-size: 10pt;">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.<br>
<div style="font-family: Times New Roman; color: #000000;
font-size: 16px">
<hr tabindex="-1">
<div style="direction: ltr;" id="divRpF555421"><font size="2"
color="#000000" face="Tahoma"><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 Dr. Albrecht Giese [<a class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de">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<br>
</font><br>
</div>
<div>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>
<br>
<div class="moz-cite-prefix">Am 21.10.2015 um 06:28 schrieb
<br>
Richard Gauthier:<br>
</div>
<blockquote type="cite">
<div class="">Hello Albrecht,</div>
<div class=""><br class="">
</div>
<div class=""> 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. </div>
<div class=""><br class="">
</div>
<div class=""> 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. </div>
<div class=""><br class="">
</div>
<div class=""> I will answer your question about the
double slit in a separate e-mail.</div>
<div class=""><br class="">
</div>
<div class=""> all the best,</div>
<div class=""> Richard</div>
<br class="">
<div>
<blockquote type="cite" class="">
<div class="">On Oct 20, 2015, at 10:06 AM, Dr.
Albrecht Giese <<a moz-do-not-send="true"
href="mailto:genmail@a-giese.de" class=""
target="_blank">genmail@a-giese.de</a>> wrote:</div>
<br class="Apple-interchange-newline">
<div class="">
<div bgcolor="#FFFFFF" class="">Hello Richard,<br
class="">
<br class="">
thank you for your explanations. I would like to
ask further questions and will place them into the
text below.<br class="">
<br class="">
<div class="moz-cite-prefix">Am 19.10.2015 um
20:08 schrieb Richard Gauthier:<br class="">
</div>
<blockquote type="cite" class="">
<div class="">Hello Albrecht,</div>
<div class=""><br class="">
</div>
<div class=""> Thank your for your detailed
questions about my electron model, which I
will answer as best as I can. </div>
<div class=""><br class="">
</div>
<div class=""> 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 <a
moz-do-not-send="true"
href="http://www.pa.uky.edu/%7Ekwng/phy525/lec/lecture_2.pdf"
class="" target="_blank"><font class=""
size="2"><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></font></a> .
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.</div>
</blockquote>
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 class="">deBroglie</sub> =
lambda<sub class="">photon</sub> / cos theta ,
but: lambda<sub class="">deBroglie</sub> = lambda<sub
class="">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 class="">
<br class="">
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 class="">
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 class="">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 class="">
<br class="">
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
class="">1</sub> = c / frequency; frequency =
(gamma*mc<sup class="">2</sup>) / h. So, the
geometric distance of the wave maxima passing the
slits is lambda<sub class="">1</sub> = c*h /
(gamma*mc<sup class="">2</sup>). Also here the
result is a shortened wavelength rather than an
extended one, so not the de Broglie wavelength.<br
class="">
<br class="">
Again my question: What do I misunderstand?<br
class="">
<br class="">
For the other topics of your answer I essentially
agree, so I shall stop here.<br class="">
<br class="">
Best regards<br class="">
Albrecht<br class="">
<br class="">
<blockquote type="cite" class="">
<div class=""><br class="">
</div>
<div class=""> 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.</div>
<div class=""><br class="">
</div>
<div class=""> 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.</div>
<div class=""><br class="">
</div>
<div class=""> 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.</div>
<div class=""><br class="">
</div>
<div class=""> 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. </div>
<div class=""><br class="">
</div>
<div class=""> I hope this helps. Thanks
again for your excellent questions.</div>
<div class=""><br class="">
</div>
<div class=""> with best regards,</div>
<div class=""> Richard</div>
<div class=""><br class="">
</div>
<br class="">
<div class="">
<blockquote type="cite" class="">
<div class="">On Oct 19, 2015, at 8:13 AM,
Dr. Albrecht Giese <<a
moz-do-not-send="true"
href="mailto:genmail@a-giese.de"
class="" target="_blank"><a class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de">genmail@a-giese.de</a></a>>
wrote:</div>
<br class="Apple-interchange-newline">
<div class="">
<div bgcolor="#FFFFFF" class="">Richard:<br
class="">
<br class="">
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 class="">
<br class="">
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
class="">
<br class="">
With the following I refer to the
figures 1 and 2 in your paper referred
in your preceding mail.<br class="">
<br class="">
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 class="">
<br class="">
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
class="">
<br class="">
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 class="">
<br class="">
Some questions, perhaps you can help me
for a better understanding.<br class="">
<br class="">
With best regards and thanks in advance<br
class="">
Albrecht<br class="">
<br class="">
PS: I shall answer you mail from last
night tomorrow.<br class="">
<br class="">
<br class="">
<div class="moz-cite-prefix">Am
14.10.2015 um 22:32 schrieb Richard
Gauthier:<br class="">
</div>
<blockquote type="cite" class="">
<div class="">Hello Albrecht,</div>
<div class=""><br class="">
</div>
<div class=""> 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. </div>
<div class=""><br class="">
</div>
<div class="">You wrote: "<span
class=""
style="font-family:HelveticaNeue,'Helvetica
Neue',Helvetica,Arial,'Lucida
Grande',sans-serif;
font-size:16px;
background-color:rgb(255,255,255)">I
see it as a valuable goal for the
further development to find an
answer (a</span><span class=""
style="font-family:HelveticaNeue,'Helvetica
Neue',Helvetica,Arial,'Lucida
Grande',sans-serif;
font-size:16px;
background-color:rgb(255,255,255)"> </span><i
class=""
style="font-family:HelveticaNeue,'Helvetica
Neue',Helvetica,Arial,'Lucida
Grande',sans-serif;
font-size:16px">physical </i><span
class=""
style="font-family:HelveticaNeue,'Helvetica
Neue',Helvetica,Arial,'Lucida
Grande',sans-serif;
font-size:16px;
background-color:rgb(255,255,255)">answer!)
to the question of the de Broglie
wavelength."</span></div>
<div class=""> 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"
class="moz-txt-link-freetext"
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.</div>
<div class=""><br class="">
</div>
<div class=""> 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.</div>
<div class=""><br class="">
</div>
<div class=""> <span class=""
style="color:rgb(37,37,37);
line-height:22px;
background-color:rgb(255,255,255)">Schrödinger</span> took
de Broglie’s matter-wave and used
it non-relativistically with a
potential V to generate the <span
class=""
style="color:rgb(37,37,37);
line-height:22px;
background-color:rgb(255,255,255)">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 class=""
style="color:rgb(37,37,37);
line-height:22px;
background-color:rgb(255,255,255)">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.)</div>
<div class=""><br class="">
</div>
<div class=""> The charged photon
model of the electron might be used
to derive the <span class=""
style="color:rgb(37,37,37);
line-height:22px;
background-color:rgb(255,255,255)">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"
class="" target="_blank">https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation</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.</div>
<div class=""><br class="">
</div>
<div class=""> 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”.</div>
<div class=""><br class="">
</div>
<div class=""> 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.</div>
<div class=""><br class="">
</div>
<div class="">with best regards,</div>
<div class=""> Richard</div>
<br>
</blockquote>
</div>
</div>
</blockquote>
</div>
</blockquote>
</div>
</div>
</blockquote>
</div>
</blockquote>
<br>
<br>
<br>
<hr style="border:none; color:#909090;
background-color:#B0B0B0; height:1px; width:99%">
<table style="border-collapse:collapse; border:none">
<tbody>
<tr>
<td style="border:none; padding:0px 15px 0px 8px"><a
moz-do-not-send="true"
href="https://www.avast.com/antivirus"
target="_blank"><img moz-do-not-send="true"
src="http://static.avast.com/emails/avast-mail-stamp.png"
alt="Avast logo" border="0">
</a></td>
<td>
<p style="color:#3d4d5a;
font-family:"Calibri","Verdana","Arial","Helvetica";
font-size:12pt">
Diese E-Mail wurde von Avast Antivirus-Software
auf Viren geprüft. <br>
<a moz-do-not-send="true"
href="https://www.avast.com/antivirus"
target="_blank">www.avast.com</a> </p>
</td>
</tr>
</tbody>
</table>
<br>
</div>
</div>
</div>
</blockquote>
<br>
<br /><br />
<hr style='border:none; color:#909090; background-color:#B0B0B0; height: 1px; width: 99%;' />
<table style='border-collapse:collapse;border:none;'>
<tr>
<td style='border:none;padding:0px 15px 0px 8px'>
<a href="https://www.avast.com/antivirus">
<img border=0 src="http://static.avast.com/emails/avast-mail-stamp.png" alt="Avast logo" />
</a>
</td>
<td>
<p style='color:#3d4d5a; font-family:"Calibri","Verdana","Arial","Helvetica"; font-size:12pt;'>
Diese E-Mail wurde von Avast Antivirus-Software auf Viren geprüft.
<br><a href="https://www.avast.com/antivirus">www.avast.com</a>
</p>
</td>
</tr>
</table>
<br />
</body>
</html>