<div dir="ltr">Dear Martin,<div><br></div><div>I am not sure whether or not you were expressing doubt as to the provenance of the wave equation, and what Schrodinger owed to de Broglie, but in either case you would probably enjoy reading this first person account by Felix Bloch </div><div><br></div><div><a href="http://www.physics.smu.edu/scalise/P5382fa15/FelixBlochPhysTodayDec1976b.pdf">http://www.physics.smu.edu/scalise/P5382fa15/FelixBlochPhysTodayDec1976b.pdf</a></div><div><br></div><div>As for EPR, it would probably make sense to start another thread, I'll leave that to you or anyone else who wishes to reply to decide if it's worth doing. I am biased by my own thinking against your claim of spooky action. To be honest, I could never bring myself to believe that the properties of a particle are indeterminate in the way standardly thought. The pilot wave model makes much more sense than, and the same predictions as, the mysterian Copenhagen, "particles are smears and reality does not exist until I look at it" <i>niaiserie</i>. </div><div><br></div><div>I could be wrong, and would welcome correction from those with more physics experience than me, but there seems to be an enormous conceptual leap from the notion that A) We cannot know, even in principle, whether 1 or 2 obtains, to B) Neither 1 nor 2 actually obtains (they are in superposition) until actually measured. It seems to me a grotesque error, and this is a long story of course. Many people have weighed in. It is a conceptual leap like that from A) The Universe exists, to B) The Creator of the Universe does not want me to have sex with people who are the same gender as myself. The chasm between these propositions can only be spanned by some enormous error. </div><div><br></div><div>A positron was trapped for so long by Dehmelt that he gave it a proper name: Priscilla. </div><div><br></div><div><i>Dehmelt says: “[t]here can be little doubt
about the identity of Priscilla during this period, since in
ultrahigh vacuum she never had a chance to trade places
with a passing antimatter twin. The well-defined identity of
this elementary particle is something fundamentally new,
which deserves to be recognized by being given a name, as
pets are give names of persons” <br></i></div><div><i> </i></div><div>I can accept that reality is nonlocal (in a certain sense), and I have been aware of the intense problem which is supposedly posed for relativity by Bell's inequalities and the notion of entanglement. Supposedly there is no actual information transmitted faster than light, but all the same the state of one entangled particle is altered by measuring the other one. Why is it so hard for people to accept that there is no alteration going on, and it is just as if I colored a marble white and a marble black, and sent them to two different people in the mail? The person who opens the envelope and sees a black marble knows that the other person has the white one, but there was no spooky action at a distance going on. So what if there is no way of knowing, <i>even in principle</i>, whether a particle is spin up or spin down? Are we humans really so arrogant that by instinct we must project ontological restrictions out of our epistemological ones?</div><div><br></div><div>The Schrodinger equation gives us probabilities. Quantum physics is a statistical theory. Thus all of its predictions hold in the long time limit (an electron in a box over sufficient cycles of the electron's frequency), or the multiple particle limit (diffraction experiments). This alone should be enough for thinking individuals to realize that quantum physics is not the end of the road. </div><div><br></div><div>Adam</div><div><br></div><div> </div><div><br></div><div> </div></div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Oct 22, 2015 at 10:18 AM, Dr. Albrecht Giese <span dir="ltr"><<a href="mailto:genmail@a-giese.de" target="_blank">genmail@a-giese.de</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div text="#000000" bgcolor="#FFFFFF">
Hello Richard,<br>
<br>
thank you and see my comments below.<span class=""><br>
<br>
<div>Am 22.10.2015 um 00:32 schrieb Richard
Gauthier:<br>
</div>
<blockquote type="cite">
<div>Hello Albert (and all),</div>
<div><br>
</div>
<div> I think your fundamental objection that you
mentioned earlier can be answered below.</div>
<div><br>
</div>
<div> 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><br>
</div>
<div> 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></span>
I agree.
<span class=""><blockquote type="cite">
<div> 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></span>
Here I disagree. If we assume the wave front as an extended layer
through the photon and with an orientation perpendicular to the
actual direction of the photon, then the intersect point of this
layer with the axis has the same z coordinate as the z-component of
the photon's position. This is essential. (I have built myself a
little 3-d model to see this.)<br>
<br>
When now, say at time T<sub>0</sub>, a phase maximum of the wave
front leaves the photon, then the same phase maximum passes the
intersect point on the axis with the same z coordinate. After a
while (i.e. after the time T<sub>p</sub>=1/frequency) the next phase
maximum will exit from the photon and simultaneously the next phase
maximum will cross the axis. The new z-value (of the photon and of
the intersect point) is now displaced from the old one by the amount
delta_z = v * T<sub>p</sub>. During this time the photon will have
moved by c * T<sub>p</sub> on its helical path.<br>
<br>
Now the spacial distance between these two phase maxima, which is
the wavelength, is: lambda<sub>photon</sub> = c * T<sub>p</sub>, and
lambda<sub>electron</sub> = v * T<sub>p</sub>. <br>
<br>
This is my result. Or what (which detail) is wrong?<br>
<br>
best wishes<br>
Albrecht<br>
<br>
<br>
<blockquote type="cite"><span class="">
<div>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><br>
</div>
<br>
<fieldset></fieldset>
<br>
</span><div>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><br>
</div>
<div> 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><br>
</div>
<div> 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><br>
</div>
<div> all the best,</div>
<div> Richard</div>
<br>
<div>
<blockquote type="cite">
<div>On Oct 21, 2015, at 12:42 PM, Dr. Albrecht Giese
<<a href="mailto:genmail@a-giese.de" target="_blank">genmail@a-giese.de</a>>
wrote:</div>
<br>
<div><small style="font-family:Helvetica;font-style:normal;font-variant:normal;font-weight:normal;letter-spacing:normal;line-height:normal;text-align:start;text-indent:0px;text-transform:none;white-space:normal;word-spacing:0px;background-color:rgb(255,255,255)">Dear John
W and all,<br>
<br>
about the<span> </span><u>de Broglie wave</u>:<br>
<br>
There are a lot of elegant derivations for the de Broglie
wave length, that is true. Mathematical deductions. What
is about the physics behind it?<br>
<br>
De Broglie derived this wave in his first paper in the
intention to explain, why the internal frequency in a
moving electron is dilated, but this frequency on the
other hand has to be increased for an external observer to
reflect the increase of energy. To get a result, he
invented a "fictitious wave" which has the phase speed
c/v, where v is the speed of the electron. And he takes
care to synchronize this wave with the internal frequency
of the electron. That works and can be used to describe
the scattering of the electron at the double slit. - But
is this physical understanding? De Broglie himself stated
that this solution does not fulfil the expectation in a
"complete theory". Are we any better today?<br>
<br>
Let us envision the following situation. An electron moves
at moderate speed, say 0.1*c (=> gamma=1.02) . An
observer moves parallel to the electron. What will the
observer see or measure?<span> </span><br>
The internal frequency of the electron will be observed by
him as frequency = m<sub>0</sub>*c<sup>2</sup>/h
, because in the observer's system the electron is at
rest. The wave length of the wave leaving the electron
(e.g. in the model of a circling photon) is now not
exactly lambda<sub>1</sub><span> </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>dB</sub><span> </span>= h/(1*m*0), which
is again infinite or at least extremely large. For the
observer at rest there is lambda<sub>dB</sub><span> </span>= h/(1.02*m*0.1c)
. Also not comparable to the co-moving observer.<br>
<br>
To summarize: these differences are not explained by the
normal SR effects. So, how to explain these incompatible
results?<br>
<br>
Now let's assume, that the electron closes in to the
double slit. Seen from the co-moving observer, the double
slit arrangement moves towards him and the electron. What
are now the parameters which will determine the
scattering? The (infinite) de Broglie wave length? The
phase speed which is 10*c ? Remember: For the co-moving
observer the electron does not move. Only the double slit
moves and the screen behind the double slit will be ca. 2%
closer than in the standard case. But will that be a real
change?<br>
<br>
I do not feel that this is a situation which in physically
understood.<br>
<br>
Regards<br>
Albrecht<br>
</small><div><div class="h5"><br style="font-family:Helvetica;font-size:12px;font-style:normal;font-variant:normal;font-weight:normal;letter-spacing:normal;line-height:normal;text-align:start;text-indent:0px;text-transform:none;white-space:normal;word-spacing:0px;background-color:rgb(255,255,255)">
<br style="font-family:Helvetica;font-size:12px;font-style:normal;font-variant:normal;font-weight:normal;letter-spacing:normal;line-height:normal;text-align:start;text-indent:0px;text-transform:none;white-space:normal;word-spacing:0px;background-color:rgb(255,255,255)">
<div style="font-family:Helvetica;font-size:12px;font-style:normal;font-variant:normal;font-weight:normal;letter-spacing:normal;line-height:normal;text-align:start;text-indent:0px;text-transform:none;white-space:normal;word-spacing:0px;background-color:rgb(255,255,255)">Am 21.10.2015 um
16:34 schrieb John Williamson:<br>
</div>
<blockquote type="cite" style="font-family:Helvetica;font-size:12px;font-style:normal;font-variant:normal;font-weight:normal;letter-spacing:normal;line-height:normal;text-align:start;text-indent:0px;text-transform:none;white-space:normal;word-spacing:0px;background-color:rgb(255,255,255)">
<div style="direction:ltr;font-family:Tahoma;font-size:10pt">Dear all,<br>
<br>
The de Broglie wavelength is best understood, in my
view, in one of two ways. Either read de Broglies thesis
for his derivation (if you do not read french, Al has
translated it and it is available online). Alternatively
derive it yourself. All you need to do is consider the
interference between a standing wave in one (proper
frame) as it transforms to other relativistic frames.
That is standing-wave light-in-a-box. This has been done
by may folk, many times. Martin did it back in 1991. It
is in our 1997 paper. One of the nicest illustrations I
have seen is that of John M - circulated to all of you
earlier in this series.<br>
<br>
It is real, and quite simple.<br>
<br>
Regards, John.<br>
<div style="font-family:'Times New Roman';font-size:16px">
<hr>
<div style="direction:ltr"><font size="2" face="Tahoma"><b>From:</b><span> </span>General [<a href="mailto:general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org" target="_blank"></a><a href="mailto:general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org" target="_blank">general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org</a>]
on behalf of Dr. Albrecht Giese [<a href="mailto:genmail@a-giese.de" target="_blank"></a><a href="mailto:genmail@a-giese.de" target="_blank">genmail@a-giese.de</a>]<br>
<b>Sent:</b><span> </span>Wednesday,
October 21, 2015 3:14 PM<br>
<b>To:</b><span> </span>Richard
Gauthier<br>
<b>Cc:</b><span> </span>Nature of
Light and Particles - General Discussion; David
Mathes<br>
<b>Subject:</b><span> </span>Re:
[General] research papers<br>
</font><br>
</div>
<div>Hello Richard,<br>
<br>
thanks for your detailed explanation. But I have a
fundamental objection.<br>
<br>
Your figure 2 is unfortunately (but unavoidably)
2-dimensional, and that makes a difference to the
reality as I understand it.<span> </span><br>
<br>
In your model the charged electron moves on a helix
around the axis of the electron (or equivalently the
axis of the helix). That means that the electron has
a constant distance to this axis. Correct? But in
the view of your figure 2 the photon seems to start
on the axis and moves away from it forever. In this
latter case the wave front would behave as you write
it.<span> </span><br>
<br>
Now, in the case of a constant distance, the wave
front as well intersects the axis, that is true. But
this intersection point moves along the axis at the
projected speed of the photon to this axis. - You
can consider this also in another way. If the
electron moves during a time, say T1, in the
direction of the axis, then the photon will during
this time T1 move a longer distance, as the length
of the helical path (call it L) is of course longer
than the length of the path of the electron during
this time (call it Z). Now you will during the time
T1 have a number of waves (call this N) on the
helical path L. On the other hand, the number of
waves on the length Z has also to be N. Because
otherwise after an arbitrary time the whole
situation would diverge. As now Z is smaller than L,
the waves on the axis have to be shorter. So, not
the de Broglie wave length. That is my
understanding.<span> </span><br>
<br>
In my present view, the de Broglie wave length has
no immediate correspondence in the physical reality.
I guess that the success of de Broglie in using this
wave length may be understandable if we understand
in more detail, what happens in the process of
scattering of an electron at the double (or
multiple) slits.<br>
<br>
Best wishes<br>
Albrecht<br>
<br>
<br>
<div>Am 21.10.2015 um 06:28
schrieb<span> </span><br>
Richard Gauthier:<br>
</div>
<blockquote type="cite">
<div>Hello Albrecht,</div>
<div><br>
</div>
<div> 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><br>
</div>
<div> 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><br>
</div>
<div> I will answer your question about
the double slit in a separate e-mail.</div>
<div><br>
</div>
<div> <span> </span>all the
best,</div>
<div> <span> </span>Richard</div>
<br>
<div>
<blockquote type="cite">
<div>On Oct 20, 2015, at 10:06 AM,
Dr. Albrecht Giese <<a href="mailto:genmail@a-giese.de" target="_blank"></a><a href="mailto:genmail@a-giese.de" target="_blank">genmail@a-giese.de</a>>
wrote:</div>
<br>
<div>
<div bgcolor="#FFFFFF">Hello
Richard,<br>
<br>
thank you for your explanations. I would
like to ask further questions and will
place them into the text below.<br>
<br>
<div>Am 19.10.2015
um 20:08 schrieb Richard Gauthier:<br>
</div>
<blockquote type="cite">
<div>Hello Albrecht,</div>
<div><br>
</div>
<div> <span> </span>Thank
your for your detailed questions about
my electron model, which I will answer
as best as I can. </div>
<div><br>
</div>
<div> 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 size="2"><a href="http://www.pa.uky.edu/%7Ekwng/phy525/lec/lecture_2.pdf" target="_blank"></a><a href="http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf" target="_blank">http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf</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>deBroglie</sub><span> </span>=
lambda<sub>photon</sub><span> </span>/
cos theta , but: lambda<sub>deBroglie</sub><span> </span>=
lambda<sub>photon</sub><span> </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>
<br>
Or let's use another view to the process.
Lets imagine a scattering process of the
electron at a double slit. This was the
experiment where the de Broglie wavelength
turned out to be helpful.<span> </span><br>
So, when now the electron, and that means
the cycling photon, approaches the slits,
it will approach at a slant angle theta at
the layer which has the slits. Now assume
the momentary phase such that the wave
front reaches two slits at the same time
(which means that the photon at this
moment moves downwards or upwards, but
else straight with respect to the
azimuth). This situation is similar to the
front wave of a<span> </span><i>single</i><span> </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> </span><br>
<br>
To my understanding the temporal distance
will be the same distance as of wave
maxima on the helical path of the photon,
where the latter is lambda<sub>1</sub><span> </span>=
c / frequency; frequency = (gamma*mc<sup>2</sup>) / h. So, the geometric
distance of the wave maxima passing the
slits is lambda<sub>1</sub><span> </span>=
c*h / (gamma*mc<sup>2</sup>).
Also here the result is a shortened
wavelength rather than an extended one, so
not the de Broglie wavelength.<br>
<br>
Again my question: What do I
misunderstand?<br>
<br>
For the other topics of your answer I
essentially agree, so I shall stop here.<br>
<br>
Best regards<br>
Albrecht<br>
<br>
<blockquote type="cite">
<div><br>
</div>
<div> 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><br>
</div>
<div> 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><br>
</div>
<div> 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><br>
</div>
<div> <span> </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><br>
</div>
<div> <span> </span>I
hope this helps. Thanks again for your
excellent questions.</div>
<div><br>
</div>
<div> <span> </span>with
best regards,</div>
<div> Richard</div>
<div><br>
</div>
<br>
<div>
<blockquote type="cite">
<div>On Oct 19, 2015, at
8:13 AM, Dr. Albrecht Giese <<a href="mailto:genmail@a-giese.de" target="_blank"></a><a href="mailto:genmail@a-giese.de" target="_blank">genmail@a-giese.de</a>>
wrote:</div>
<br>
<div>
<div bgcolor="#FFFFFF">Richard:<br>
<br>
I am still busy to understand
the de Broglie wavelength from
your model. I think that I
understand your general idea,
but I would like to also
understand the details.<span> </span><br>
<br>
If a photon moves straight in
the free space, how does the
wave look like? You say that the
photon emits a plane wave. If
the photon is alone and moves
straight, then the wave goes
with the photon. No problem. And
the wave front is in the forward
direction. Correct? How far to
the sides is the wave extended?
That may be important in case of
the photon in the electron.<br>
<br>
With the following I refer to
the figures 1 and 2 in your
paper referred in your preceding
mail.<br>
<br>
In the electron, the photon
moves according to your model on
a circuit. It moves on a helix
when the electron is in motion.
But let take us first the case
of the electron at rest, so that
the photon moves on this
circuit. In any moment the plane
wave accompanied with the photon
will momentarily move in the
tangential direction of the
circuit. But the direction will
permanently change to follow the
path of the photon on the
circuit. What is then about the
motion of the wave? The front of
the wave should follow this
circuit. Would an observer next
to the electron at rest (but not
in the plane of the internal
motion) notice the wave? This
can only happen, I think, if the
wave does not only propagate on
a straight path forward but has
an extension to the sides. Only
if this is the case, there will
be a wave along the axis of the
electron. Now an observer next
to the electron will see a
modulated wave coming from the
photon, which will be modulated
with the frequency of the
rotation, because the photon
will in one moment be closer to
the observer and in the next
moment be farer from him. Which
wavelength will be noticed by
the observer? It should be
lambda = c / ny, where c is the
speed of the propagation and ny
the frequency of the orbital
motion. But this lambda is by my
understanding not be the de
Broglie wave length.<br>
<br>
For an electron at rest your
model expects a wave with a
momentarily similar phase for
all points in space. How can
this orbiting photon cause this?
And else, if the electron is not
at rest but moves at a very
small speed, then the situation
will not be very different from
that of the electron at rest.<br>
<br>
Further: What is the influence
of the charge in the photon?
There should be a modulated
electric field around the
electron with a frequency ny
which follows also from E =
h*ny, with E the dynamical
energy of the photon. Does this
modulated field have any
influence to how the electron
interacts with others?<span> </span><br>
<br>
Some questions, perhaps you can
help me for a better
understanding.<br>
<br>
With best regards and thanks in
advance<br>
Albrecht<br>
<br>
PS: I shall answer you mail from
last night tomorrow.<br>
<br>
<br>
<div>Am
14.10.2015 um 22:32 schrieb
Richard Gauthier:<br>
</div>
<blockquote type="cite">
<div>Hello Albrecht,</div>
<div><br>
</div>
<div> <span> </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><br>
</div>
<div>You wrote: "<span 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 style="font-size:16px;background-color:rgb(255,255,255)"> </span><i style="font-size:16px">physical </i><span style="font-size:16px;background-color:rgb(255,255,255)">answer!)
to the question of the de
Broglie wavelength."</span></div>
<div> <span> </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 href="https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength" target="_blank"></a><a href="https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength" target="_blank">https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength</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><br>
</div>
<div> <span> </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><br>
</div>
<div> <span 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 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 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><br>
</div>
<div> <span> </span>The
charged photon model of the
electron might be used to
derive the <span style="color:rgb(37,37,37);line-height:22px;background-color:rgb(255,255,255)">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 href="https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schr%C3%B6dinger_Equation" target="_blank"></a><a href="https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation" target="_blank">https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation</a> .
Of course, to treat the
electron relativistically
requires the Dirac equation.
But the spin 1/2 charged
photon model of the
relativistic electron has a
number of features of the
Dirac electron, by design.</div>
<div><br>
</div>
<div> <span> </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><br>
</div>
<div> <span> </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><br>
</div>
<div>with best
regards,</div>
<div> Richard</div>
<br>
</blockquote>
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