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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
cite="mid:E7357BA6-790D-4BE3-8DDD-93E324D75D3A@gmail.com"
type="cite">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<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="">genmail@a-giese.de</a>>
wrote:</div>
<br class="Apple-interchange-newline">
<div class="">
<meta content="text/html; charset=utf-8"
http-equiv="Content-Type" class="">
<div text="#000000" 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
cite="mid:10655753-DF29-4EDC-91E0-27701A9B1CC2@gmail.com"
type="cite" class="">
<meta http-equiv="Content-Type" content="text/html;
charset=utf-8" 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=""><font class="" size="2">http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf</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
cite="mid:10655753-DF29-4EDC-91E0-27701A9B1CC2@gmail.com"
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>
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<!--StartFragment--><!--EndFragment--><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="">genmail@a-giese.de</a>>
wrote:</div>
<br class="Apple-interchange-newline">
<div class="">
<meta content="text/html; charset=utf-8"
http-equiv="Content-Type" class="">
<div text="#000000" 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
cite="mid:26CF7357-2118-4391-889D-150E1F112A8C@gmail.com"
type="cite" class="">
<meta http-equiv="Content-Type"
content="text/html; charset=utf-8" 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
style="font-family: HelveticaNeue,
'Helvetica Neue', Helvetica, Arial,
'Lucida Grande', sans-serif; font-size:
16px; background-color: rgb(255, 255,
255);" class="">I see it as a valuable
goal for the further development to find
an answer (a</span><span
style="font-family: HelveticaNeue,
'Helvetica Neue', Helvetica, Arial,
'Lucida Grande', sans-serif; font-size:
16px; background-color: rgb(255, 255,
255);" class=""> </span><i class=""
style="font-family: HelveticaNeue,
'Helvetica Neue', Helvetica, Arial,
'Lucida Grande', sans-serif; font-size:
16px;">physical </i><span
style="font-family: HelveticaNeue,
'Helvetica Neue', Helvetica, Arial,
'Lucida Grande', sans-serif; font-size:
16px; background-color: rgb(255, 255,
255);" class="">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"><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 style="color: rgb(37,
37, 37); line-height: 22px;
background-color: rgb(255, 255, 255);"
class="">Schrödinger</span> took de
Broglie’s matter-wave and used it
non-relativistically with a potential V to
generate the <span style="color: rgb(37, 37,
37); line-height: 22px; background-color:
rgb(255, 255, 255);" class="">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: rgb(37, 37, 37);
line-height: 22px; background-color:
rgb(255, 255, 255);" class="">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
style="color: rgb(37, 37, 37);
line-height: 22px; background-color:
rgb(255, 255, 255);" class="">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="">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>
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