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Hello Richard,<br>
<br>
thanks for your detailed explanation. I think that it becomes more
and more visible, how difficult it is to visualize such a
3-dimensional process. <br>
<br>
I have added some further comments below in your text.<br>
<br>
<div class="moz-cite-prefix">Am 23.10.2015 um 22:41 schrieb Richard
Gauthier:<br>
</div>
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite">
<meta http-equiv="Content-Type" content="text/html;
charset=windows-1252">
<div class="">Hello Albrecht (and others)</div>
<div class=""><br class="">
</div>
<div class=""> Thank for your further comments. You arguments are
correct, according to how I previously explained the plane waves
emitted by the charged photon along its helical axis. I realized
that I misinterpreted and therefore poorly explained my own
proposed quantum plane wave function describing quantum waves
coming from the circulating charged photon. The left side of
Figure 2 is NOT merely the mathematically unwrapped helical
trajectory of the charged photon. It is instead (or in addition)
one of many “rays” of quantum plane waves emitted continuously
from the circulating charged photon. </div>
<div class=""><br class="">
</div>
<div class=""> The circulating charged photon’s proposed quantum
plane wave function Ae^i(k dot r - wt) , where k = (gamma
mv)/hbar and w = (gamma mc^2)/hbar are the wave vector and the
angular frequency of the circulating charged photon, describes
quantum plane waves emitted from the circulating charged photon
in the direction that the charged photon is moving at any point
in time. </div>
</blockquote>
The relation k = (gamma mv)/hbar cannot be applicable here, if I
understand correctly that v is the speed of the electron. If the
electron is at rest, then v=0 and so<br>
k=0. But for a photon k=0 is not possible. It is in permanent
motion and has energy, which you describe with w = (gamma
mc^2)/hbar . <br>
<br>
Here you try to apply the de Broglie wave length to the circling
photon which you cannot do by two reasons:<br>
1.) Your intention is to derive the de Broglie wave length. But you
cannot do this by using the validity of the de Broglie wave length
as a precondition. That would be circular reasoning. <br>
2.) And anyway, for a photon the de Broglie wave length is identical
the wave length of the phase wave as v=c .
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite">
<div class="">While emitting these quantum plane waves, the
charged photon curves away on its helical trajectory, continuing
to emit newer quantum plane waves at its own frequency and
wavelength. But the quantum plane waves previously emitted by
the charged photon continue in a straight line direction tangent
to the helical trajectory at the point along the trajectory
where they were emitted. Those quantum plane waves emitted from
the circulating charged photon at one location move out into
space at light-speed away from the charged photon, as indicated
by the left side of the big triangle in my Figure 2, and in the
recently posted figure showing 4 wave fronts. Their quantum
plane wave fronts DO intersect the charged photon’s helical axis
further along the axis to the right, as shown in the two
figures, creating de Broglie wavelengths along the helical axis.
And these de Broglie wavelengths DO travel away to the right
along the helical axis at the phase velocity c^2/v because their
speed is (from the geometry shown in Figure 2) Vphase = speed of
charged photon / cos(theta) = c/cos(theta) = c/(v/c) = c^2/v
.</div>
</blockquote>
Your Figures 2 and 4 assume that the circling photon emits a plane
wave. Is that possible? It means that the wave which just leaves the
photon with a certain phase is immediately spread out to all sides
until infinity. Otherwise it is not a <i>plane </i>wave. But if it
does so, it means an infinite propagation speed to all directions
perpendicular to the speed vector of the photon. This is in my
understanding in strong conflict with relativity. (And it means also
that for an observer in a system moving relative to this system
there can be a violation of causality. He can observe that a part of
the plane may exist at a certain phase even before this phase is
emitted from the photon.) <br>
<br>
If you assume such kind of plane wave then your considerations about
the wave on the axis caused by the sequence of intercept points are
applicable. But again: a plane wave of this kind violates
causality. <br>
<br>
You mention further down as a visualization the case of a laser
moving along the helix. A laser emits indeed a sort of a plane wave,
however in a limited region given by the diameter of the laser's
body. This plane wave is the result of a superposition of a huge
number of photons oscillating forth and back in the laser. In
contrast to this the photon in your model is a point source. If it
emits waves then those are restricted to the speed of light. So they
will leave the photon as a cone with a half angle of 45 degrees. (In
acoustics this is called Mach's cone.) If we start now to follow
this process using this way of propagation, we have to look how the
cone touches the axis. The motion of these intercept points on the
axis seems to be non-linear, and as further phases follow, there
will be an overlay of such phases. - Do you think it is worth to
follow this? I would like to first check whether we find an
agreement at this point. <br>
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class=""> All these emitted quantum plane waves from the
charged photon, described above by Ae^i(k dot r - wt) ,
intersect the helical axis, as described by the derived
relativistic de Broglie matter-wave function A^i(Kdb z -wt)
where Kdb is the wave number corresponding to the de Broglie
wavelength Ldb = h/(gamma mv). So Kdb =2pi /Ldb = (gamma
mv)/hbar , and w =(gamma mc^2)/hbar the angular frequency of
the charged photon, corresponding to f=(gamma mc^2)/h as
before.</div>
</blockquote>
I understand that these considerations follow again the assumption
of a "plane" wave which I do not believe to be possible as explained
above. So I shall wait for your response to that.
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class=""> This process can roughly be compared to a broad
plane-wave beam emitted from a laser while the laser moves along
a helical trajectory, directing its beam in new directions as
the laser moves along its helical path. The parallel waves
fronts from the laser intersect the axis and generate one de
Broglie-like wavelength along the axis for each photon
wavelength coming from the laser. <br>
</div>
</blockquote>
For the laser example please see above.<br>
<br>
Best regards<br>
Albrecht
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class=""> Does this new explanation answer your fundamental
objection?</div>
<div class=""><br class="">
</div>
<div class="">with best regards,</div>
<div class=""> Richard</div>
.<br class="">
<div>
<blockquote type="cite" class="">
<div class="">On Oct 22, 2015, at 10:18 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">
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<div text="#000000" bgcolor="#FFFFFF" class=""> Hello
Richard,<br class="">
<br class="">
thank you and see my comments below.<br class="">
<br class="">
<div class="moz-cite-prefix">Am 22.10.2015 um 00:32
schrieb Richard Gauthier:<br class="">
</div>
<blockquote
cite="mid:3BF40319-FF10-493F-8966-13FF1FC5FFCE@gmail.com"
type="cite" class="">
<meta http-equiv="Content-Type" content="text/html;
charset=windows-1252" class="">
<div class="">Hello Albert (and all),</div>
<div class=""><br class="">
</div>
<div class=""> I think your fundamental objection that
you mentioned earlier can be answered below.</div>
<div class=""><br class="">
</div>
<div class=""> 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.</div>
<div class=""><br class="">
</div>
<div class=""> 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.</div>
</blockquote>
I agree.
<blockquote
cite="mid:3BF40319-FF10-493F-8966-13FF1FC5FFCE@gmail.com"
type="cite" class="">
<div class=""> 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. </div>
</blockquote>
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 class="">
<br class="">
When now, say at time T<sub class="">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 class="">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 class="">p</sub>.
During this time the photon will have moved by c * T<sub
class="">p</sub> on its helical path.<br class="">
<br class="">
Now the spacial distance between these two phase maxima,
which is the wavelength, is: lambda<sub class="">photon</sub>
= c * T<sub class="">p</sub>, and lambda<sub class="">electron</sub>
= v * T<sub class="">p</sub>. <br class="">
<br class="">
This is my result. Or what (which detail) is wrong?<br
class="">
<br class="">
best wishes<br class="">
Albrecht<br class="">
<br class="">
<br class="">
<blockquote
cite="mid:3BF40319-FF10-493F-8966-13FF1FC5FFCE@gmail.com"
type="cite" class="">
<div class="">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 .</div>
<div class=""><br class="">
</div>
<br class="">
<fieldset class="mimeAttachmentHeader"></fieldset>
<br class="">
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<div class="">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).</div>
<div class=""><br class="">
</div>
<div class=""> 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).</div>
<div class=""><br class="">
</div>
<div class=""> I don’t have any comments yet on your
concerns about the de Broglie wavelength that you just
expressed to John W (below).</div>
<div class=""><br class="">
</div>
<div class=""> all the best,</div>
<div class=""> Richard</div>
<br class="">
<div class="">
<blockquote type="cite" class="">
<div class="">On Oct 21, 2015, at 12:42 PM, 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=""><small style="font-family: Helvetica;
font-style: normal; font-variant: normal;
font-weight: normal; letter-spacing: normal;
line-height: normal; orphans: auto; text-align:
start; text-indent: 0px; text-transform: none;
white-space: normal; widows: auto; word-spacing:
0px; -webkit-text-stroke-width: 0px;
background-color: rgb(255, 255, 255);" class="">Dear
John W and all,<br class="">
<br class="">
about the<span class="Apple-converted-space"> </span><u
class="">de Broglie wave</u>:<br class="">
<br class="">
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 class="">
<br class="">
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 class="">
<br class="">
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?<span class="Apple-converted-space"> </span><br
class="">
The internal frequency of the electron will be
observed by him as frequency = m<sub class="">0</sub>*c<sup
class="">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 class="">1</sub><span
class="Apple-converted-space"> </span>=
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 class="">dB</sub><span
class="Apple-converted-space"> </span>=
h/(1*m*0), which is again infinite or at least
extremely large. For the observer at rest there
is lambda<sub class="">dB</sub><span
class="Apple-converted-space"> </span>=
h/(1.02*m*0.1c) . Also not comparable to the
co-moving observer.<br class="">
<br class="">
To summarize: these differences are not
explained by the normal SR effects. So, how to
explain these incompatible results?<br class="">
<br class="">
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 class="">
<br class="">
I do not feel that this is a situation which in
physically understood.<br class="">
<br class="">
Regards<br class="">
Albrecht<br class="">
</small><br style="font-family: Helvetica;
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<br style="font-family: Helvetica; font-size:
12px; font-style: normal; font-variant: normal;
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start; text-indent: 0px; text-transform: none;
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0px; -webkit-text-stroke-width: 0px;
background-color: rgb(255, 255, 255);" class="">
<div class="moz-cite-prefix" style="font-family:
Helvetica; font-size: 12px; font-style: normal;
font-variant: normal; font-weight: normal;
letter-spacing: normal; line-height: normal;
orphans: auto; text-align: start; text-indent:
0px; text-transform: none; white-space: normal;
widows: auto; word-spacing: 0px;
-webkit-text-stroke-width: 0px;
background-color: rgb(255, 255, 255);">Am
21.10.2015 um 16:34 schrieb John Williamson:<br
class="">
</div>
<blockquote
cite="mid:7DC02B7BFEAA614DA666120C8A0260C914714222@CMS08-01.campus.gla.ac.uk"
type="cite" style="font-family: Helvetica;
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<div style="direction: ltr; font-family: Tahoma;
font-size: 10pt;" class="">Dear all,<br
class="">
<br class="">
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 class="">
<br class="">
It is real, and quite simple.<br class="">
<br class="">
Regards, John.<br class="">
<div style="font-family: 'Times New Roman';
font-size: 16px;" class="">
<hr tabindex="-1" class="">
<div id="divRpF555421" style="direction:
ltr;" class=""><font class="" size="2"
face="Tahoma"><b class="">From:</b><span
class="Apple-converted-space"> </span>General
[<a moz-do-not-send="true"
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
moz-do-not-send="true"
class="moz-txt-link-abbreviated"
href="mailto:genmail@a-giese.de"><a class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de">genmail@a-giese.de</a></a>]<br
class="">
<b class="">Sent:</b><span
class="Apple-converted-space"> </span>Wednesday,
October 21, 2015 3:14 PM<br class="">
<b class="">To:</b><span
class="Apple-converted-space"> </span>Richard
Gauthier<br class="">
<b class="">Cc:</b><span
class="Apple-converted-space"> </span>Nature
of Light and Particles - General
Discussion; David Mathes<br class="">
<b class="">Subject:</b><span
class="Apple-converted-space"> </span>Re:
[General] research papers<br class="">
</font><br class="">
</div>
<div class="">Hello Richard,<br class="">
<br class="">
thanks for your detailed explanation. But
I have a fundamental objection.<br
class="">
<br class="">
Your figure 2 is unfortunately (but
unavoidably) 2-dimensional, and that makes
a difference to the reality as I
understand it.<span
class="Apple-converted-space"> </span><br
class="">
<br class="">
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.<span
class="Apple-converted-space"> </span><br
class="">
<br class="">
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.<span
class="Apple-converted-space"> </span><br
class="">
<br class="">
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
class="">
<br class="">
Best wishes<br class="">
Albrecht<br class="">
<br class="">
<br class="">
<div class="moz-cite-prefix">Am 21.10.2015
um 06:28 schrieb<span
class="Apple-converted-space"> </span><br
class="">
Richard Gauthier:<br class="">
</div>
<blockquote type="cite" class="">
<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=""> <span
class="Apple-converted-space"> </span>all
the best,</div>
<div class=""> <span
class="Apple-converted-space"> </span>Richard</div>
<br class="">
<div class="">
<blockquote type="cite" class="">
<div class="">On Oct 20, 2015, at
10:06 AM, Dr. Albrecht Giese <<a
moz-do-not-send="true"
class="moz-txt-link-abbreviated"
href="mailto:genmail@a-giese.de"><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="">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=""> <span
class="Apple-converted-space"> </span>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 <font class=""
size="2"><a
moz-do-not-send="true"
class="moz-txt-link-freetext"
href="http://www.pa.uky.edu/%7Ekwng/phy525/lec/lecture_2.pdf"><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></font> .
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><span
class="Apple-converted-space"> </span>=
lambda<sub class="">photon</sub><span
class="Apple-converted-space"> </span>/
cos theta , but: lambda<sub
class="">deBroglie</sub><span
class="Apple-converted-space"> </span>=
lambda<sub class="">photon</sub><span
class="Apple-converted-space"> </span>*
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.<span
class="Apple-converted-space"> </span><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<span
class="Apple-converted-space"> </span><i
class="">single</i><span
class="Apple-converted-space"> </span>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.<span
class="Apple-converted-space"> </span><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><span
class="Apple-converted-space"> </span>=
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><span
class="Apple-converted-space"> </span>=
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=""> <span
class="Apple-converted-space"> </span>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=""> <span
class="Apple-converted-space"> </span>I
hope this helps. Thanks
again for your excellent
questions.</div>
<div class=""><br class="">
</div>
<div class=""> <span
class="Apple-converted-space"> </span>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"
class="moz-txt-link-abbreviated" href="mailto:genmail@a-giese.de"><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.<span
class="Apple-converted-space"> </span><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?<span
class="Apple-converted-space"> </span><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=""> <span
class="Apple-converted-space"> </span>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-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-size:
16px;
background-color:
rgb(255, 255,
255);"> </span><i
class=""
style="font-size:
16px;">physical </i><span
class=""
style="font-size:
16px;
background-color:
rgb(255, 255,
255);">answer!)
to the question
of the de
Broglie
wavelength."</span></div>
<div class=""> <span
class="Apple-converted-space"> </span>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=""> <span
class="Apple-converted-space"> </span>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=""> <span
class="Apple-converted-space"> </span>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" class="moz-txt-link-freetext"
href="https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schr%C3%B6dinger_Equation"><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.</div>
<div class=""><br
class="">
</div>
<div class=""> <span
class="Apple-converted-space"> </span>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=""> <span
class="Apple-converted-space"> </span>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 class="">
</blockquote>
</div>
</div>
</blockquote>
</div>
</blockquote>
</div>
</div>
</blockquote>
</div>
</blockquote>
<br class="">
<br class="">
<br class="">
<hr style="border: none; color: rgb(144,
144, 144); background-color: rgb(176,
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