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Hello Richard (and all),<br>
<br>
thank you, Richard, for your informations. You find my answers and
comments in your text.<br>
<br>
However I see here two general problems which should be reviewed by
all. <br>
<br>
1.) The fact that the de Broglie wave regarding its definition and
its use is <i>not </i>Lorentz-invariant. So it is incompatible
with our physical understanding since 1905.<br>
<br>
2.) If the photon is seen as the ingredient of the electron, we need
a much clearer definition and understanding what the photon is and
what its effects are in detail (like the wave front emitted).
Otherwise there are too many insufficiently defined situations as
visible in the discussion further down. - And clearly we do not get
any help from quantum mechanics for this, after Heisenberg has
stated that it is completely useless to look into an elementary
particle, and the physical community has accepted this since that
time.<br>
<br>
<div class="moz-cite-prefix">Am 26.10.2015 um 00:29 schrieb Richard
Gauthier:<br>
</div>
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
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<div class="">Hello Albrecht (and all),</div>
<div class=""><br class="">
</div>
<div class=""> Thanks for your further questions. First of all,
I think your comments on the de Broglie wavelength and the
double-slit experiment as observed by a moving observer (moving
with the electron or with another speed), were quite astute. I
can’t see how an apparently stationary group of electrons, with
an undefined or near infinite de Broglie wavelength, approached
with speed v by a double slit apparatus, will form a wavy
statistical interference pattern of points on the approaching
screen on the other side of the approaching double slit
apparatus, equal to the wavy statistical interference pattern
produced by electrons moving at speed v approaching a stationary
double-slit apparatus and screen. But apparently the pattern
will be the same since this transverse wavy pattern should be
invariant with respect to the longitudinal motion of the
observer with respect to the double slit experiment (electrons
plus double-slit and screen apparatus). Perhaps someone else can
explain how this would work. The double slit pattern formed by
PHOTONS of a particular wavelength with respect to the
double-slit apparatus should not be affected in a similar way by
the motion of the observer. The same wavy transverse statistical
pattern of photon spots on the screen behind the double slits
should also occur independent of the velocity of the observer
with respect to the photons and double-slit apparatus plus
screen. Of course, in this case the observer can’t move as fast
as the approaching photons, which makes a difference between the
photon and the electron double-slit experiments, in relation to
a moving observer. <br>
</div>
</blockquote>
True, for a photon coming in straight, there is no conflict. But I
have mentioned earlier another view. Look what the circling photon
does when the electron approaches the double slit. The photon
performs a scattering at the double slit. The difference to the
photon coming straight is that the photon approaches the double slit
at a - maybe - very flat angle. This will change the angle at which
the light beam leaves the double slit at the other side. So the
diffraction pattern on the screen will not be on a straight line
perpendicular
to the direction of the slits. It will be positioned more or less on
a circuit. But the structure of the pattern, i.e. the spread of the
maxima, will be according to photon scattering and not to electron
scattering. The deflections will be different from the ones of a
photon moving straight in by a factor, which is between 1 and c/v
depending on the direction (up/down or right/left). And so different
from the de Broglie assumption which deviates by a factor of
(c^2/v^2) from the angular deflection of a photon coming in
straight.
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class="">However, the problem with the ELECTRON double-slit
experiment with a moving observer might be resolved by thinking
of the stationary electrons (as seen by an observer moving with
them) as composed of circulating charged photons, with each
electron producing a standing wave composed of Compton waves
moving in opposite directions, approached by a moving
double-slit apparatus. </div>
</blockquote>
Let's look at a collection of electrons having an opposite rotation
and so the waves move to different directions. If now an observer
moves at a moderate speed towards the double slit (so not co-moving
at the same speed as the electron) he should now see a diffraction
pattern which is smeared out as both wave contribute in a different
way. But again this will be different of what an observer at rest
will see.<br>
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class=""> Now to answer your further questions about the
charged photon model and the proposed quantum wave function
describing quantum plane waves emitted by the helically
circulating charged photon.</div>
<div class=""><br class="">
</div>
<div class="">
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); float: none;
display: inline !important;">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</span><br class=""
style="background-color: rgb(255, 255, 255);">
<span class="" style="background-color: rgb(255, 255, 255);
float: none; display: inline !important;"> 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
. </span></blockquote>
</div>
<div class=""><br class="">
</div>
<div class="">For the v=0 or near zero resting electron, where
gamma=1, the k wave number relations k(electron) =(gamma
mv)/hbar for the resting electron’s de Broglie wave and
k(photon)=(gamma mc)/hbar for the circulating photon BOTH apply.
For the resting electron, v=0 so k=0 since k(electron)
=2pi/LAMBDAdb and LAMBDAdb goes to infinity for a resting
electron since LAMBDAdb = h/mv and v -> 0. But for the
circulating charged photon in a resting electron where gamma =
1, k(photon) = (gamma mc)/hbar -> mc/hbar applies also
because here for the circulating photon, k(photon)=
2pi/LAMBDAphoton = 2pi/ComptonWavelength = 2pi/(h/mc) = mc/hbar
. So there is no contradiction here or for any other electron
speed, which is always less than c even though the circulating
charged photon’s speed is always c .</div>
</blockquote>
For the photon I do not see any problem here. But further down in
the sequence of your arguments, you explain the de Broglie
wavelength by the intercept of the wave front of the photon with the
axis. At this early point of your sequence there is nothing about
this intercept. So, if you use the de Broglie wavelength of the
electron
<i> here </i>as an argument, then you use here something which you
deduce only later. This is what I meant as circular reasoning.
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class="">
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); display: inline
!important;">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. </span></blockquote>
</div>
<div class=""><br class="">
</div>
<div class="">I am not deriving the de Broglie wavelength by
circular reasoning. Although we already know the electron’s de
Broglie wavelength formula experimentally, the electron’s de
Broglie wavelength formula is derived in my model from the
circulating charged photon’s wavelength LAMBDA(photon) =
h/(gamma mc) along its helical trajectory. This photon
wavelength Lambda(photon) along the helical trajectory is simply
derived from the proposed charged photon’s energy E= hf = gamma
mc^2 for the relativistic moving electron. The z=component of
the helically circulating charged photon’s wave vector k(photon)
value along the helical axis is k(axis) = k(photon ) x
cos(theta) which corresponds to the de Broglie wavelength
h/(gamma mv) .</div>
</blockquote>
Correct, but earlier you use the de Broglie wavelength before you
derive it. Not true?
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class="">
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); display: inline
!important;">2.) And anyway, for a photon the de Broglie
wave length is identical the wave length of the phase wave
as v=c .</span></blockquote>
</div>
<div class=""><br class="">
</div>
<div class=""> No, the wavelength h/(gamma mc) of the circulating
charged photon is not identical to the de Broglie wavelength
h/(gamma mv) of the phase wave. The speed of the circulating
charged photon is c, while the speed of the moving electron is v
(less than c) and the phase velocity of the de Broglie wave is
c^2/v (always greater than c)</div>
</blockquote>
I understand your arguments so that you first present the charged
photon with its properties. Its role in the set up of the electron
comes later and the de Broglie wavelength is derived later. So at
this point, where the photon is the subject, we can state that the
de Broglie wave of the <i>photon </i>is identical with the normal
phase wave of the <i>photon </i>which is formally given by the
relation v=c, don't we?
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class="">
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); float: none;
display: inline !important;">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 </span><i
class="" style="background-color: rgb(255, 255, 255);">plane </i><span
class="" style="background-color: rgb(255, 255, 255); float:
none; display: inline !important;">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.) </span><br
class="" style="background-color: rgb(255, 255, 255);">
<br class="" style="background-color: rgb(255, 255, 255);">
<span class="" style="background-color: rgb(255, 255, 255);
float: none; display: inline !important;">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. </span></blockquote>
</div>
<div class=""><br class="">
</div>
<div class="">Mathematical plane waves e^i(k dot r-wt) are
generally used in practical physics beam experiments all the
time to model for example a beam of photons in a laser, a beam
of electrons in an electron scattering experiment, or a beam of
other particles in a high energy particle collider experiment.
Plane wave math is a good approximation when the particles are
all of the same momentum, moving in same direction (past some
stationary point in the beam) and are basically independent of
each other in the beam. This is standard practice for
calculating scattering probability amplitudes in quantum theory.
And also in practice the plane waves in a particle beam are
probably not assumed to extend beyond the beam, where there are
anyway (by definition) no beam particles to scatter. So no
infinite propagation speeds for producing plane waves are
assumed in these practical applications of mathematical plane
waves. The same would be true in charged-photon modeling in
electron beam experiments.</div>
</blockquote>
You are right that a light beam is understood as having a plane
front perpendicular to the motion of the photons. And correctly you
mention that such beam is in experiments built by a huge number of
photons. By the superposition of the single fields it is very
plausible to assume this plane front. But careful: if we reduce the
problem to a mathematical one then we do what the Copenhagen QM does
all day. And I think that no one in our community wants this way. <br>
<br>
But questions:<br>
<br>
What is about a single photon? At radar systems I have worked with
it can be measured that in case of a beam, which is kind of
collimated by an aperture, if one goes away from that beam by more
than a wavelength, there is nothing any more (except small
contributions caused by bending). What is about the circling photon?
What size can be assumed for the photon in the electron? You refer
to the uncertainty principle. If this causes the photon to be
smeared out by an amount which is roughly related to the size of the
orbit (so the size of the electron), how well defined is the
intercept point where this wave crosses the axis? <br>
<br>
I have the impression that we are here on the border between QM (you
mention Heisenberg) and a classical understanding of this process.
If we follow quantum mechanics then the electron is a structureless
point which is surrounded by a cloud of virtual charges. This is
present understanding of main stream. But I have the impression that
nobody of us in this round wants to understand particle physics in
this way.<br>
<br>
So, let's stay with the classical understanding. Now the question
is, what does the wave front if the direction of the photon changes?
This does normally not happen in experiments, so we don't know it
from practice. And if the photon continues with speed c, then the
outer regions of the front have to move with superluminal speed. Is
this accepted? If in the radar case the beam is redirected by a wave
guide then the speed is reduced below c, technically reasoned by the
impedance of the wave guide. <br>
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""> The position of a single circulating charged photon
(i.e. electron with a fixed momentum gamma mv) could not be
located at all along its helical length, according to
Heisenberg’s uncertainty principle. But a long pulse of many
electrons in a beam of finite width could still be modeled by
circulating charged photons emanating quantum plane waves at an
angle theta to the electron beam, where cos(theta)=v/c, and
whose projected quantum wave function along the z=axis (beam
axis) for each electron is the electron's quantum plane wave
function with the de Broglie wavelength. The superluminal phase
velocity c^2/v comes in when the wave vector k of these
charged-photon quantum plane waves at angle theta to the beam,
intersects the beam direction, generating de Broglie waves along
the beam direction for each electron, moving with phase velocity
c^2/v. These phase waves (also as de Broglie described them) are
not physical waves moving superluminally (which would violate
relativity). Rather they are (for charged photons) like the
wave-like motion along a beach when successive parallel waves
hit a beach at an angle, causing a disturbance that travels up
the beach at a speed faster than the speed of the waves
themselves. It is these de Broglie phase waves which predict the
scattering of the electrons during a collision or scattering
process.</div>
<div class=""><br class="">
</div>
<div class="">
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); display: inline
!important;">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. </span></blockquote>
<br class="">
</div>
<div class="">See above about the limited-width electron beam as
corresponding to a limited width laser beam.The charged photon's
quantum plane wave cone half-angle would be 45 degrees only if v
= c/sqrt(2) and so cos (theta) = 1/sqrt(2) = 0.707 so that
theta = 45 degrees). The angle theta (your half-angle) in the
charged photon model can vary between near 90 degrees (very slow
electrons) and near 0 degrees (for highly relativistic
electrons).</div>
</blockquote>
This is misunderstanding. At this point I do not mean the angle
built by the relation of v and c, but the cone in which the wave
front leaves the photon if we have a classical understanding.
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""> In the laser (as I understand it) the many coherent
photons can also be considered as point sources each generating
plane waves (because the laser intensity can be drastically
reduced or filtered to one photon at a time without changing
photon scattering results). The charged photon plane waves will
be emitted from the circulating charged photon in a cone whose
circulating k wave vector makes a half angle of theta . This
cone-sweep at 1/2 angle theta and the intersection of these
corresponding emitted light-speed plane waves with the
longitudinal z-axis will generate waves moving superluminally
along the z=axis which will be the de Broglie waves. I think it
would be great to have a 3D animation of this, for different
values of v (and therefore different values of theta).</div>
</blockquote>
I do not believe that a laser beam can be reduced so far that only
one photon is moving inside. This photon has to stimulate the next
radiation of an atom in the gas, and this is a process of low
probability in the single case. So I expect that a laser which is
too much reduced will stop its radiation. And the coherence of the
radiation is anyway only possible if there is a sufficient density
of photons. If the laser beam is filtered on the other hand so that
only single photons are in the beam, these photons are also assumed
to build an interference pattern at a double slit. My understanding
to explain this is that a photon is extended in a similar way as the
electron is extended. Then the tip of the cone which I mentioned
would not be a sharp peak but a bit flat, and that could be
sufficient to explain the scattering observed as a bit like a plane.
<br>
<br>
But as I said before: we need a much better understanding of how the
photon is built in order to use it in your model. At present I have
even the impression that the photon which we need for this model is
more complex than the electron which it is supposed to explain.<br>
<br>
Best regards<br>
Albrecht<br>
<br>
<blockquote
cite="mid:460F80CA-A624-4BB6-B837-D873F2643782@gmail.com"
type="cite">
<div class=""><br class="">
</div>
<div class="">
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); display: inline
!important;">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.</span></blockquote>
<br class="">
</div>
<div class="">See the plane-wave reply above for real beams.</div>
<div class=""><br class="">
</div>
<div class="">with best regards,</div>
<div class=""> Richard</div>
<div class=""><br class="">
</div>
<div class=""><br class="">
</div>
<div class=""><br class="">
</div>
<div class=""><br class="">
</div>
<br class="">
<div class="">
<blockquote type="cite" class="">
<div class="">On Oct 25, 2015, at 6:21 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=""><span 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); float: none;
display: inline !important;" class="">Hello Richard,</span><br
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);" class="">
<br 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);" class="">
<span 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); float: none;
display: inline !important;" class="">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.<span class="Apple-converted-space"> </span></span><br
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);" class="">
<br 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);" class="">
<span 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); float: none;
display: inline !important;" class="">I have added some
further comments below in your text.</span><br
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);" class="">
<br 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);" 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 23.10.2015 um
22:41 schrieb Richard Gauthier:<br class="">
</div>
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite" 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);" class="">
<div class="">Hello Albrecht (and others)</div>
<div class=""><br class="">
</div>
<div class=""> <span class="Apple-converted-space"> </span>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=""> <span class="Apple-converted-space"> </span>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.<span
class="Apple-converted-space"> </span></div>
</blockquote>
<span 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); float: none;
display: inline !important;" class="">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</span><br
style="font-family: Helvetica; font-size: 12px;
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normal; letter-spacing: normal; line-height: normal;
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background-color: rgb(255, 255, 255);" class="">
<span 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); float: none;
display: inline !important;" class=""> 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
.<span class="Apple-converted-space"> </span></span><br
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;
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background-color: rgb(255, 255, 255);" class="">
</div>
</blockquote>
<div class=""><br class="">
</div>
<blockquote type="cite" class=""><span class=""
style="background-color: rgb(255, 255, 255); display: inline
!important;">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. </span></blockquote>
<blockquote type="cite" class="">
<div class=""><br 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);" class="">
<span 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); float: none;
display: inline !important;" class="">Here you try to
apply the de Broglie wave length to the circling photon
which you cannot do by two reasons:</span><br
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);" class="">
<span 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); float: none;
display: inline !important;" class="">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.<span
class="Apple-converted-space"> </span></span><br
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);" class="">
<span 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); float: none;
display: inline !important;" class="">2.) And anyway, for
a photon the de Broglie wave length is identical the wave
length of the phase wave as v=c .</span>
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite" 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);" class="">
<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>
<span 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); float: none;
display: inline !important;" class="">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<span
class="Apple-converted-space"> </span></span><i
style="font-family: Helvetica; font-size: 12px;
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="">plane<span
class="Apple-converted-space"> </span></i><span
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); float: none;
display: inline !important;" class="">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.)<span
class="Apple-converted-space"> </span></span><br
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);" class="">
<br 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);" class="">
<span 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); float: none;
display: inline !important;" class="">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.<span class="Apple-converted-space"> </span></span><br
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);" class="">
<br 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);" class="">
<span 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); float: none;
display: inline !important;" class="">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.<span
class="Apple-converted-space"> </span></span><br
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);" class="">
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite" 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);" class="">
<div class=""><br class="">
</div>
<div class=""> <span class="Apple-converted-space"> </span>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>
<span 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); float: none;
display: inline !important;" class="">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.</span>
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite" 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);" class="">
<div class=""><br class="">
</div>
<div class=""> <span class="Apple-converted-space"> </span>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.<span
class="Apple-converted-space"> </span><br class="">
</div>
</blockquote>
<span 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); float: none;
display: inline !important;" class="">For the laser
example please see above.</span><br 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);" class="">
<br 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);" class="">
<span 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); float: none;
display: inline !important;" class="">Best regards</span><br
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);" class="">
<span 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); float: none;
display: inline !important;" class="">Albrecht</span>
<blockquote
cite="mid:F1DF4916-2B0C-4571-8246-07F9B59977D3@gmail.com"
type="cite" 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);" class="">
<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=""> <span class="Apple-converted-space"> </span>Richard</div>
.<br class="">
<div class="">
<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">
<div class="">
<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="">
<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.<span
class="Apple-converted-space"> </span></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><span
class="Apple-converted-space"> </span>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><span
class="Apple-converted-space"> </span>= c * T<sub
class="">p</sub>, and lambda<sub class="">electron</sub><span
class="Apple-converted-space"> </span>= v * T<sub
class="">p</sub>.<span
class="Apple-converted-space"> </span><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="">
<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=""> <span
class="Apple-converted-space"> </span>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=""> <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 21, 2015, at 12:42 PM,
Dr. Albrecht Giese <<a
moz-do-not-send="true"
href="mailto:genmail@a-giese.de"
class=""><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=""><small class=""
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);">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 class="" 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);">
<br class="" 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);">
<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" class="" 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);">
<div class="" style="direction: ltr;
font-family: Tahoma; font-size: 10pt;">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 class="" style="font-family:
'Times New Roman'; font-size: 16px;">
<hr tabindex="-1" class="">
<div id="divRpF555421" class=""
style="direction: ltr;"><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">genmail@a-giese.de</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>
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