Thomas Lockyer's VPP Model is absolutely correct
FrediFizzx
flames (and QFT also). There is absolutely no doubt left in my mind that
the VPP electron wave geometry is the correct model for the electron (and
positron). The cosine-cosine wave *does not* violate Maxwell. The photon
wave volume is lambda cubed divided by (2*pi) squared. The photon wave is
lambda divided by 2*pi wide and the same height. The photon wave is one
wavelength long. The photon (energy) lives at 90 degrees and 270 degrees.
Plug *those* into Maxwell and you will see. So my derivation of photon wave
energy density from the Maxwell travelling wave equations is correct. It is
on the second page of the PDF and the second equation from the bottom. I
challenge anyone to prove that it violates Maxwell with the above
"conditions".
http://www.flashrock.com/upload/photonsincos.pdf
The cosine-cosine phantom wave is in. Thomas was right but missed that it
is in Maxwell. In a way he was correct. It "closes the loop" on the photon
wave and is why it can be shown that Maxwell conserves energy. The
magnitudes I was referring are the electric and magnetic field
magnitudes. That is all the photon wave cares about and is why my
derivation is correct.
Jacques, if you can't see that this is starring you right in the face, then
you need to step back for a minute and take a really good look at all the
evidence. Another thing, Thomas was absolutely correct in that the
constants and their equations come from the electron wave geometry. I was
wrong. The model is not derived from the constants and their equations.
They are what they are because of the electron wave geometry. Now I really
understand Thomas' frustration in this matter. As far as I am concerned, he
can spam the newsgroup all he wants. It is absolutely the correct classical
wave model.
If any of you lurkers out there want to get famous, figure out how this
works in the atom. My take on it is that the proton, in the case of the
hydrogen atom, lives inside of the electron wave (the proton itself is a
similar wave spinning cube). Once the electron's charge is nulled out, it
can absorb and emit photons. It does this by changing its size in
quanitized steps to keep its geometry balanced. Thomas has already shown
that the length of the cube sides does not matter for charge. So the charge
stays balanced with the proton's charge.
All particles in reality are waves. If it wan't for Aybar's description of
how a sine/cosine wave can be a point particle, I would still be scratching
my head. But no more. Thomas' discovery is truely a discovery of the
century. And maybe for all time.
More coming later.
FrediFizzx
FrediFizzx
news:Cz3I8.6950$0z2.20...@newssvr21.news.prodigy.com...
Folks, as promised here is the more.
http://www.flashrock.com/upload/maxwellphoton.pdf
The above document shows without a doubt that there is a cos-cos component
in addition to the sin-sin component in the photon wave. More evidence that
VPP is correct. And further more, it shows that the cos-cos component is
actually carrying the photon energy. Now I don't think that the wave is
longitudinal until the electron-positron pair is formed. If anyone thinks
they can show how that works, go for it. After all these years we are
finally going to have true classical wave pictures of particles. Thank you
very much, Thomas.
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> wrote in message
| news:Cz3I8.6950$0z2.20...@newssvr21.news.prodigy.com...
Whoa! Disregard the following paragraph. I had bronchitis and was taking
way too much cough syrup. The atom is still explained by QM. The electron
wave geometry would have to be a 10,000 to 100,000 times bigger.
| | If any of you lurkers out there want to get famous, figure out how this
| | works in the atom. My take on it is that the proton, in the case of the
| | hydrogen atom, lives inside of the electron wave (the proton itself is a
| | similar wave spinning cube). Once the electron's charge is nulled out,
it
| | can absorb and emit photons. It does this by changing its size in
| | quanitized steps to keep its geometry balanced. Thomas has already
shown
| | that the length of the cube sides does not matter for charge. So the
| charge
| | stays balanced with the proton's charge.
| |
| | All particles in reality are waves. If it wan't for Aybar's description
| of
| | how a sine/cosine wave can be a point particle, I would still be
| scratching
| | my head. But no more. Thomas' discovery is truely a discovery of the
| | century. And maybe for all time.
It is easy to realize that a sine or cosine wave would be a point particle.
For any instance of time, there is only one point that describes the
location in space. In the case of the photon wave, there should be two
points that describe the photon. One for the electric wave and one for the
magnetic wave. For the VPP electron wave geometry there should be at least
four points at any one instance of time and maybe it is 8 points that gives
the location in space for any instance of time.
Another thing is that it is easy to show that the photon wave has no
mechanism for motion. The motion imparted to the photon wave has to come
from whatever initiates it. This means the mechanism for photon wave motion
*has* to be in the electron (or positron), etc.
FrediFizzx
cherring17
"Fredi Fizzx" <fredi fizz...@ahahhotmail.com> wrote in message
news:ciOM8.363$w83.32...@newssvr21.news.prodigy.com...
fizzxhaha@ahahhotmail.com Fredi Fizzx
cherring17" <cherr...@home.com> wrote in message
news:gHvN8.59950$092.2...@news1.east.cox.net...
What makes you think this is the correct model? VPP model theory gives a
reason why the electron-positron pair forms at the exact energy required so
that every electron and positron are identical to each other. I don't see
how this happens with the magnetic ring theory.
Plus the VPP electron wave model geometry is derivable from two sources. A
good indication that it is correct. What is the magnetic ring model derived
from? EM waves can be point particles. VPP shows how photon waves can
latch up via their electric and magnetic forces to make charge, spin, mass,
etc.
FrediFizzx
cherring17
space to cause this. Thomas' calculations simply imply that the
properties of particles are contained within the photon, something
I have believed for years for which I posted several discussions here
last year. Since one geometry can be easily converted to another I
dare say that the same values that Thomas calculates should be
obtainable from K.M.Doshi's model when one considers the following
explanation for particle production, again, which I posted here sometime
last year (see below). Doshi has extended the math much further than I did.
If
one then follows along Thomas' computations but uses instead Doshi's
model I think the same results can be derived.
***************************************************
Previous post:
If one studies the motion of an electron in a betatron, the path can be
describe by considering the circular magnetic field the electron
created as it travels forward. The influence of the
magnetic field of the betatron is to enhance the electron's magnetic field
on one side and oppose it on the other. The result "upsets" the Poynting
vector which normally guarantees that the electron will travel in a straight
line outside the field of the betatron. This disruption of the Poynting
vector causes the electron to follow curved path.
Taken to the extreme, a photon, a wave packet, traveling close to the
strong electric spherical vector field surrounding a nucleus where the
whole packet appears
at one instance within the field (short wavelength) could be totally
destabilized and thereby turn completely in on itself. The positive field
would be repelled away from the nucleus and the negative field would be
attracted toward the nucleus. The magnetic fields would be separated equally
to support the new static electric fields thus creating the magnetic moment
of each newly formed particle, an electron and a positron, one having spin
in one direction and one having spin in the other direction realtive to the
direction of the magnetic poles. The Poynting vector turning completed in on
itself would produce the spins of each particle.
Photons are E&M waves with a leading field that is either positive-north or
negative-south and when the wavelength is very short, on the order of the
atomic scale, matter interacts with the leading field first as though it is
a particle. There is a self imposed boundary due to c and the wavelength
which define a "size" and, again, when the wavelength is on the order of the
atomic scale the photon "looks" like a particle from the front.
When the photon passes obliquely through a strong Coulomb field in the
vicinity of a nucleus (strongly curved) and since E fields add,
there will be an affect upon the Poynting vector causing it to curve. There
has to be an affect on the Poynting vector because the total E field is now
radial so the instantaneous S vector must change direction in order to
maintain orthonality. Both the positive-north and negative-south fields will
be affect independently. At some critical field strength a critical
curvature should occur causing 2 particles to form. The positive particle
will be repelled away from the Coulomb and the negative particle will be
attracted toward the Coulomb field. Each particle will gain a magnetic
dipole field due to the rotating electric field, hence, we have 2 particles
of opposite charge with their electric fields pointing out from their
respective surfaces, each has a dipole B field and a Poynting vector mapping
out a circle (rotating). All three vectors for each particle are still
orthogonal and now self-sustaining. Energy is now trapped in the form of
static fields which, since E=m*c^2, is now mass. Spins will be in the same
direction (which I had originally thought to be opposite) and will be given
by...
(1/2)[h/(2*pi)] = E_o/w_o = (m_o*c)*(wl_1/2)/pi
where E_o is the rest mass energy of either electron or positron, w_o is the
angular frequency given by 2*pi*v_o such that v_o is the photon frequency
for minimal conversion, m_o is the rest mass of an electron or positron,
wl_1/2 is the half wavelength associated with v_o and the rest have their
normally understood meanings.
The derivation is simple but one must assume that linear EM vectors can be
forced to wrap into rotating vectors. With this assumption and starting with
E = hv...
Take E/v=h and multiply by 1/2 then divide by 2*pi and knowing that for the
conversion energy v = v_o and that E/2 equals E_o, the energy mass of either
the electron or positron, one can easily show...
E_o/(2*pi*v_o) = (1/2)[h/(2*pi)] eq 1 (the right side is spin
units)
Further, since 2*pi*v_o is angular frequency w_o, we get...
E_o/w_o = (1/2)[h/(2*pi)] eq2 (simpliest form)
Which means that half the energy from the photon that can create an
electron-positron pair, roughly 1.022 Mev, divide by the angular frequency
equivalent to the linear frequency of the photon is particle spin, a
rotating vector but circular.
Going back to the left side of eq 1 and knowing that v_o = c/wl_o where wl_o
is the photon wavelength and that (wl_o)/2 = wl_1/2, half wavelength and
that
E_o = m_o*c^2...
E_o/(2*pi*v_o) = (m_o*c^2)/ {2*pi*[c/(2*wl_1/2)]} = (m_o*wl_1/2*c)/pi
hence...
(m_o*c)*(wl_1/2)/pi = (1/2)[h/(2*pi)] eq 3. (the mass and half
wavelength form)
Which means that a form of momentum related to the photon, m_o*c, times a
conversion factor from meters to radians, wl_1/2/pi, equals particle spin,
angular momentum.
Eq. 2 in terms of the fine structure constant, given by a =
[(2*pi)(e_o)^2]/(h*c) where e_o is the fundumental electric charge, we
have...
E_o/w_o = [(e_o)^2]/(2*a*c)
or
e_o = [(2*a*c*E_o)/(w_o)]^(1/2) = [(m_o)*(c^2)*(wl_o)*(a)]/(pi)
Note: wl_o = (1/2)*(Compton wavelength)
Besides a rotating system of two particles, something equivalent to a toroid
should be stable. Then there would be the electric field equivalent of a
toroid which
should also be stable with no charge, just a magnetic moment. These are
candidates for neutrinos in an analagous fashion to your postulated cubic
particles (and I emphaizes the postulated! - "Vector Particle and Nuclear
Models" by T.N. Lockyer - something not supported in free space!)
I agree with the use of Vector Particale to describe matter but not the
notion of cubic fields. The data and graphs may prove useful to analysis
but the rest is junk in my opinion.
Magnetic pole particles are postualted but have never been observed. Indeed
from one of Dirac's equations, g_o = (1/2)*[(h*c)/(2*pi*e_o)] where g_o
would have to be a fundumental magnetic charge and e_o is the fundumental
electric charge.
"Fredi Fizzx" <fredi fizz...@ahahhotmail.com> wrote in message
news:mPdP8.4736$Ge4.23...@newssvr21.news.prodigy.com...
fizzxhaha@ahahhotmail.com Fredi Fizzx
| Thomas' model postulates cubic particles. There is no boundry in
| space to cause this. Thomas' calculations simply imply that the
| properties of particles are contained within the photon, something
| I have believed for years for which I posted several discussions here
| last year. Since one geometry can be easily converted to another I
| dare say that the same values that Thomas calculates should be
| obtainable from K.M.Doshi's model when one considers the following
| explanation for particle production, again, which I posted here sometime
| last year (see below). Doshi has extended the math much further than I
did.
| If
| one then follows along Thomas' computations but uses instead Doshi's
| model I think the same results can be derived.
Well only the VPP vectors are cubic. The VPP electron wave geometry (the
waves are not shown in Thomas' models) is more than cubic. Since the cube
corners are rotating at c, and there should be some quanitized wave
components there, it maybe looks like two quanitized wave "rings" or
"doughnuts" latched together. So it will look like a cylinder of sorts with
the ends "puffed" out or kind of dumbbell shaped. Now, it could be possible
that this thing is also "tumbling" in the other direction so this would give
it a spherical look.
Yes, I remember your previous posts on the subject and they were
interesting. Doshi's model is interesting also. I will be looking at it
more extensively in the near future especially if my work on the VPP model
stalls out.
FrediFizzx
cherring17
orthogonal arrangement that exists between the E and H in the photon, we
will proceed to construct particle models in the form of vector cube
frameworks. (p.18, 1st para.)". He labors under the incorrect assumption
that orthogonal means square, hence cubic in 3D space.
Further, "The belief that subatomic particles should be spherical may be
aesthetically pleasing, but nature is under no obligation to agree with our
preconceived notions." (same page and para.). Then on p. 28, figure 4.5 he
specifically shows a cubic electron model. To this I say no. Nature gives
us curved orthogonal fields in the macroscopic world and cannot possibly
support cubed fields at the quantum foam level or just above this level, so
how does it suddenly order itself to cubes somewhere in between?
There must be structure to support cubed fields as one would find in a wave
guide (see http://www1.borg.umn.edu/waveguides.html). Since Thomas'
background is Engineering, I dare say he got his concept from EE and
waveguides.
"Fredi Fizzx" <fredi fizz...@ahahhotmail.com> wrote in message
news:IhyQ8.6483$IZ6.34...@newssvr21.news.prodigy.com...
fizzxhaha@ahahhotmail.com Fredi Fizzx
| I have Thomas' book. I know what he said and means, "To preserve the
| orthogonal arrangement that exists between the E and H in the photon, we
| will proceed to construct particle models in the form of vector cube
| frameworks. (p.18, 1st para.)". He labors under the incorrect assumption
| that orthogonal means square, hence cubic in 3D space.
|
| Further, "The belief that subatomic particles should be spherical may be
| aesthetically pleasing, but nature is under no obligation to agree with
our
| preconceived notions." (same page and para.). Then on p. 28, figure 4.5 he
| specifically shows a cubic electron model. To this I say no. Nature gives
| us curved orthogonal fields in the macroscopic world and cannot possibly
| support cubed fields at the quantum foam level or just above this level,
so
| how does it suddenly order itself to cubes somewhere in between?
What is an example of a curved orthogonal field?
| There must be structure to support cubed fields as one would find in a
wave
| guide (see http://www1.borg.umn.edu/waveguides.html). Since Thomas'
| background is Engineering, I dare say he got his concept from EE and
| waveguides.
Well I don't think that he did. I think his model is just showing the
vector forces as cubic but in reality is more complex. IOW, I am agreeing
with you that cube model is bizarre. But that his model is possibly a first
step to discovering what the "real" EM wave geometry of the electron is.
There are some features built into his ideas that are very promising. Such
as the size of the cube doesn't matter for charge value. The geometry can
indeed be as big as an atom and still have elementary charge. I just
haven't got a clue as to how it would get 10,000 to 100,000 times bigger
when it is in an atom. But then on the artilcle about Maris' electron
bubbles, the electron bubble that encloses the wavefunction is even bigger
than that. Very mysterious.
Now, your previous post has a great description of what happens with the
photon when the electron-positron pair forms. I like it very much.
FrediFizzx
larry
> I have Thomas' book. I know what he said and means, "To preserve the
> orthogonal arrangement that exists between the E and H in the photon, we
> will proceed to construct particle models in the form of vector cube
> frameworks. (p.18, 1st para.)". He labors under the incorrect assumption
> that orthogonal means square, hence cubic in 3D space.
Since the VPP model has two parallel 1/2 e current rings, it should be
possible to calculate the the repulsion between them and thus find
the necessary magnetic field strength to hold the thing together and
thus the distance between the rings for different ring sizes. Thomas'
related rates calculation says nothing (as was pointed out to him years
ago) about the length of the cylinder formed by the rotating current
rings. The VPP electron seems just on the verge of flying apart.
As for Fredi's attempt to reify "wave function" aka "wavefunction"
into a real thing, "wave function" is a mathematical abstraction about
objective reality. It describes the wave nature of a particle or group
of particles. The discussion should be on whether a real entity with
a describing wave function can separate into two real entities each
described in some manner by 1/2 of the original wave function.
larry
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:3D14BC94...@charter.net...
| cherring17 wrote:
|
| > I have Thomas' book. I know what he said and means, "To preserve the
| > orthogonal arrangement that exists between the E and H in the photon, we
| > will proceed to construct particle models in the form of vector cube
| > frameworks. (p.18, 1st para.)". He labors under the incorrect assumption
| > that orthogonal means square, hence cubic in 3D space.
|
|
| Since the VPP model has two parallel 1/2 e current rings, it should be
| possible to calculate the the repulsion between them and thus find
| the necessary magnetic field strength to hold the thing together and
| thus the distance between the rings for different ring sizes. Thomas'
| related rates calculation says nothing (as was pointed out to him years
| ago) about the length of the cylinder formed by the rotating current
| rings. The VPP electron seems just on the verge of flying apart.
Good point. However if we look at the cube geometry construction we notice
that the EH axial components between the two rings are "out of phase" so
this may complicate things as far as trying to calculate the force between
the two rings. In the VPP calculations that Thomas did, it showed that size
didn't matter for charge so maybe this size adjustment only applies to the
length of the cylinder. IOW, the as the cylinder length becomes longer, the
forces still stay balanced keeping the two rings latched together. Now the
rings can't change size because I think that would affect the magneton of
the electron. The "related rates calculation" does have some say about it
though since it is puting the E and H component lengths at lambda_C divided
by two pi. And the H component length is the length of the cylinder for the
electron.
http://members.aol.com/tnlockyer/cubedimensions.gif
Thanks for the tip. I will take a close look at this in the near future as
I am going to start an in-depth mathematical analysis of the VPP electron
wave model very soon with the help of a tutor. We will probably take a
close look at the magnetic ring theory found by Cherring also.
| As for Fredi's attempt to reify "wave function" aka "wavefunction"
|
| into a real thing, "wave function" is a mathematical abstraction about
| objective reality. It describes the wave nature of a particle or group
| of particles. The discussion should be on whether a real entity with
| a describing wave function can separate into two real entities each
| described in some manner by 1/2 of the original wave function.
Well, actually I am contending that the abstract "wave function" has basis
in reality because all particles are in reality composed of "real"
quanitized EM waves. And that the "real" wave geometry rules for particles
have not been discovered yet. Well, maybe part of them has been discovered
now. We will see.
FrediFizzx
cherring17
You might want to save the HTML of the site because the e-mail address at
the top is no longer valid and it's possible the site may go away at some
future time.
cherring17
The magnetic field around an oscillating magnet, oscillating in the
direction of the poles, creates an E field at all instantaneous points in
space that must be orthogonal to the magnetic field lines at all those same
points in space yet the magnet field curves through space from one pole to
the other.
fizzxhaha@ahahhotmail.com Fredi Fizzx
|
| "Fredi Fizzx" <fredi fizz...@ahahhotmail.com> wrote in message
| news:%z4R8.7062$TD2.38...@newssvr21.news.prodigy.com...
| > "larry" <gold...@charter.net> wrote in message
| > news:3D14BC94...@charter.net...
| > | cherring17 wrote:
| > Thanks for the tip. I will take a close look at this in the near future
| as
| > I am going to start an in-depth mathematical analysis of the VPP
electron
| > wave model very soon with the help of a tutor. We will probably take a
| > close look at the magnetic ring theory found by Cherring also.
|
| You might want to save the HTML of the site because the e-mail address at
| the top is no longer valid and it's possible the site may go away at some
| future time.
Yes, I did save it. I also converted it to a PDF file so I can email as one
file. Let me know if you would like a copy of the PDF file.
FrediFizzx
larry
> "larry" <gold...@charter.net> wrote in message
> news:3D14BC94...@charter.net...
> | cherring17 wrote:
> |
> | > I have Thomas' book. I know what he said and means, "To preserve the
> | > orthogonal arrangement that exists between the E and H in the photon, we
> | > will proceed to construct particle models in the form of vector cube
> | > frameworks. (p.18, 1st para.)". He labors under the incorrect assumption
> | > that orthogonal means square, hence cubic in 3D space.
> |
> |
> | Since the VPP model has two parallel 1/2 e current rings, it should be
> | possible to calculate the the repulsion between them and thus find
> | the necessary magnetic field strength to hold the thing together and
> | thus the distance between the rings for different ring sizes. Thomas'
> | related rates calculation says nothing (as was pointed out to him years
> | ago) about the length of the cylinder formed by the rotating current
> | rings. The VPP electron seems just on the verge of flying apart.
>
> Good point. However if we look at the cube geometry construction we notice
> that the EH axial components between the two rings are "out of phase" so
> this may complicate things as far as trying to calculate the force between
> the two rings. In the VPP calculations that Thomas did, it showed that size
> didn't matter for charge so maybe this size adjustment only applies to the
> length of the cylinder. IOW, the as the cylinder length becomes longer, the
> forces still stay balanced keeping the two rings latched together. Now the
Since the magnetic field generated by the rotating charge ring must
equal the repulsion of the two rings there is only one possible distance
at which the VPP electron will hold together. That distance will depend
upon the rotation rate. I assume that that rate is c.
Another problem is whether it can be shown that the charge distribution
of the VPP electron can somewhat simulate the charge distribution of a
central point charge and thus be in accord with Maxwell. A more
realistic non-pointlike electron model would have a rotating sphere of
charge somewhat like charges distributed throughout a rotating
dielectric sphere in the macroscopic world. I suppose that it would be
difficult to get the proper g-factor for such a sphere of charge.
Larry
fizzxhaha@ahahhotmail.com Fredi Fizzx
Yep, a very *big* problem to solve. And a very good reason why everyone
objects to the electron having any structure. Well, if the electron is a
complex quantized EM wave structure (which I believe it is), then the
concept of pure point charge needs to be abandoned. This is something that
man made up anywise to make calculations easier starting with Coulomb's law.
However, point charges blow up at zero size because of the infinities they
create. This should have been a first clue that something isn't right. In
fact, with pure mathematics, you are trying to divide by zero. You actually
never can get to absolute zero size or the the equation is meaningless. So
by Coulomb's law itself, I have to contend that pure point charges *do not
exist* ever in reality.
But your above thoughts do scrap the notion that the cylinder length could
become longer under normal conditions. On page 27 of Thomas' book, he does
show that the magnetic and electric forces are equal at Compton wavelength
divided by 2 pi. So I think this locks in the cylinder length to that
dimension. However, his calculation is a simplified view of what is going
on, so I am going to take an in-depth look at it. It may be that it all
reduces to the simplified view anywise. Thomas has the electron rings as
current rings that produce charge, etc. And I believe the cube corners are
traveling at c?
Now the current rings are what is strange to me. If the electron has
current rings then the positron would have to have charge rings? It seems
to me that the positron has to be the monopole charge associated with the
magnetic field in VPP. Or in any model that uses the photon for
construction of elementary particles? Well, it does explain why there are
no magnetic monopoles but does totally screw up other concepts. However,
relativity does show us that the magnetic field and electric field can
appear as the same force under different observation conditions. And the
VPP models are relativistic beasts since the cube corners (where the
quanitized action has to be) are traveling at c.
FrediFizzx
Michael Moroney
>Since the magnetic field generated by the rotating charge ring must
>equal the repulsion of the two rings there is only one possible distance
>at which the VPP electron will hold together. That distance will depend
>upon the rotation rate. I assume that that rate is c.
Actually his model has the corners of his cube exceeding c (sqrt(2)*c
to be exact). One of many problems with that model.
-Mike
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:Gy6pDB...@world.std.com...
Hmmm? Yes, it is actually the midpoint of the sides that are traveling at
c. Well, since this is a quanitzed EM wave model, I am not sure that this
is a problem that the corners are traveling at sgrt(2)*c if the corners are
just representing the peak of the wave. Certainly something to investigate
in more detail. I think maybe it would require a somewhat different
configuration of the vectors.
FrediFizzx
larry
Wouldn't any length less than the Planck length be sufficiently small
that a particle would appear to be point-like? Any field produced by
such a particle would appear to come from a point since a spherical
gaussian surface with diameter of the Planck length could be imagined
around which the electric field would seem to be coming radially from
the center. Zero in objective reality means non-existence or lack of
existence. Point-like is an abstraction describing a particle that has
a field that is equal in intensity omni-directionally, i.e., a central
field, and does not mean, necessarily, that the particle has zero
radius.
> create. This should have been a first clue that something isn't right. In
> fact, with pure mathematics, you are trying to divide by zero. You actually
> never can get to absolute zero size or the the equation is meaningless. So
> by Coulomb's law itself, I have to contend that pure point charges *do not
> exist* ever in reality.
>
> But your above thoughts do scrap the notion that the cylinder length could
> become longer under normal conditions. On page 27 of Thomas' book, he does
> show that the magnetic and electric forces are equal at Compton wavelength
> divided by 2 pi. So I think this locks in the cylinder length to that
> dimension. However, his calculation is a simplified view of what is going
Since a rotating ring of charge produces, just as a rotating charged
disk does, a magnetic field, and the two parallel rings of charge have
electric fields which are repulsive, the length of the cylinder
will be the length at which the attraction of the magnetic fields from
the rotating rings exactly equals the repulsion of the electric fields.
That is not the same as the equilibrium length between E and B. Thomas
assumes, as with traveling electro-magnetic waves which are made of real
photons, that the photon must have a must have an actual magnetic
component which is not just an effect due to the motion of the photon
relative to an observer.
Another problem is the motion of the VPP photon being generated by it's
own internal actions which would likely not conserve energy. It is
more likely that inertial motion is due to motion relative to the local
sum of all the gravitational fields from all the particles in the universe.
> on, so I am going to take an in-depth look at it. It may be that it all
> reduces to the simplified view anywise. Thomas has the electron rings as
> current rings that produce charge, etc. And I believe the cube corners are
> traveling at c?
>
> Now the current rings are what is strange to me. If the electron has
> current rings then the positron would have to have charge rings? It seems
> to me that the positron has to be the monopole charge associated with the
> magnetic field in VPP. Or in any model that uses the photon for
> construction of elementary particles? Well, it does explain why there are
> no magnetic monopoles but does totally screw up other concepts. However,
> relativity does show us that the magnetic field and electric field can
> appear as the same force under different observation conditions. And the
As the charged rotating disk experiment shows, the electric field is
primary and the magnetic field is a relativistic effect due to relative
motion with respect to the electric field.
Larry
fizzxhaha@ahahhotmail.com Fredi Fizzx
Probably so. But even as small as the Planck length, you get energies that
are way too big (using the elementary charge). I get about 7 millions
joules for that. For something around the size of the VPP model I get 3
times 10E-16 joules. For a charge the size of 10E-18 meters I get 10E-10
joules. So these size ranges are much more reasonable energy-wise. So
right off the bat, I would have to say that Coulomb's law needs to have a
minimum size incorporated into it for charge. All we have to do is figure
out what it is (the real size of the electron for starters). IOW, you can't
just throw a charge into the equation without specifying its size.
Coulomb's law is assuming point charges otherwise. It is not reflecting
reality. This would be the first modification necessary to Maxwell's
equations.
I have been thinking that the initial or inertial motion of the photon is
due to the "dynamo" effect of the VPP electron. Thomas doesn't think so but
the more I think about it, the more sense it makes. When the electron
absorbs and then emits a photon, what happens? Something else I am going to
take a good look at. This means that the EH axial components would have to
form during the electron-positron pair formation and then the photon does
not need to have an EH axial component. So that whole issue of getting the
VPP photon to fit Maxwell would become a non-issue.
| > on, so I am going to take an in-depth look at it. It may be that it all
| > reduces to the simplified view anywise. Thomas has the electron rings
as
| > current rings that produce charge, etc. And I believe the cube corners
are
| > traveling at c?
| >
| > Now the current rings are what is strange to me. If the electron has
| > current rings then the positron would have to have charge rings? It
seems
| > to me that the positron has to be the monopole charge associated with
the
| > magnetic field in VPP. Or in any model that uses the photon for
| > construction of elementary particles? Well, it does explain why there
are
| > no magnetic monopoles but does totally screw up other concepts.
However,
| > relativity does show us that the magnetic field and electric field can
| > appear as the same force under different observation conditions. And
the
|
|
| As the charged rotating disk experiment shows, the electric field is
| primary and the magnetic field is a relativistic effect due to relative
| motion with respect to the electric field.
| Larry
But what do you think about the positron being the monopole charge for the
magnetic field? This one makes me stutter a bit. And still has me
scratching my head.
| > VPP models are relativistic beasts since the cube corners (where the
| > quanitized action has to be) are traveling at c.
I had this wrong as the midpoints of the sides are traveling at c. The cube
corners are where the wave peaks would be. So is the "action" at the
midpoint of the EH axial sides? This is where Thomas has mass being formed
I think. Well actually it is more complex than that.
FrediFizzx
larry
[snip}
The VPP electron and positron have the attribute charge (the ability to
produce a field of virtual photons in the vacuum) at eight corners and
the VPP photon has no charge but forms all other particles. Therefore
the photon must have attributes for both negative and positive
charge when combined to form, say, electrons or positrons. Thomas seems
not to distinguish between the direction of the electric field as to
whether it is for a negative or for a positive charge but, rather, uses
a magnetic component for the positive charge. It would seem that the
photon would be modeled with a negative charge and a positive
charge and that, when combined into an electron, the positive charge in
the rotating system is seen as an affective magnetic field. In the
positron, the negative charge in the rotating system is seen as a an
affective magnetic field. But the magnetic fields cancel and the
opposite magnetic fields produced by the two rotating current rings
counter the repulsive force from the charged rings. Perhaps there is no
need for the magnetic monopole.
Larry
larry
Are my posts messed up with very long line lengths?
I had to change to Netscape because there is a bug
in Microsoft Outlook Express which causes it to take
about a minute to open and about five minutes to
close down. I have not found a cure for that problem
yet.
Could you post a Mathcad document of your calculations.
I do not have time to convert the Adobe .pcf to .mcd
format.
Thanks,
Larry
fizzxhaha@ahahhotmail.com Fredi Fizzx
I think I see what you are saying. The electron and positron are creating
the complementary electric or magnetic fields from the other action. Neat!
Let me see if I have it right. I think I had it backwards before. The
electron has two rotating rings each comprised of two EH axial components
and two E fields. But, yes, what are these E fields? Are they positive or
negative or both? If a wave, they would have to be a whole wavelength from
corner to corner with the peaks of the wave at the corners and being
"positive" at the corners. With the "negative" going part of the waves
extending toward the ring center. Or where would the "negative" part of the
wave be? Maybe not at the center? The only thing we can maybe say for sure
is that the "positive" peaks would be at the corners. Could the "negative"
going part of the wave extend toward the other ring or away from it? I am
trying to draw a picture of it and it is somewhat complex. I think if we
look at the orientation of the H wave going from ring to ring at the corners
may give us a clue. They too would have to have the peak of the wave at
each corner and be one wavelength long. Now it seems they would have their
"negative" going part of the wave oriented in toward the center of the model
putting them at a 45 degree angle if looking thru from one of the ring
sides.
So would this affect how the E wave in the ring is oriented? Man, this
thing is more complex than one might have thought. And I haven't even got
to the wave configuration of the EH axial component yet. Anywise, the
corners end up with "positive" sections of both E and H waves (and of the EH
axial waves) producing a current ring which gives a negative charge. Is
this because of Benjamin Franklin's convention of how he assigned charge and
the electron should really be "positive"? I think I missed on getting your
idea in here. Help me out.
I think I am getting an idea about where the anomalous magnetic moment comes
from. What about centrifugal force affecting this rotating beast? Causing
the rings to slightly increase their diameters.
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
| Fredi,
| Are my posts messed up with very long line lengths?
| I had to change to Netscape because there is a bug
| in Microsoft Outlook Express which causes it to take
| about a minute to open and about five minutes to
| close down. I have not found a cure for that problem
| yet.
Your recent posts have been fine.
| Could you post a Mathcad document of your calculations.
| I do not have time to convert the Adobe .pcf to .mcd
| format.
| Thanks,
| Larry
Yes I can very easily. Which calculations? I have so many now I have lost
track. The photon wave volume? The derivations of the model geometry from
the constants and their equations? The ones on Maxwell's traveling wave
equations (I have some mistakes in those that I have to fix)? I haven't
started any yet on the excruciating details of the model. But soon I will
be if I can get a starting point worked out in pictorial form. Once I get a
picture of how this thing really looks with the waves, I think it will be
easier. Thomas' picture is really just a simplified version but works well
for many things.
What version do you need them in? They are all Mathcad 2001i format
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
I guess I should have been saying wave "crests" and "troughs" in the
description above but I think you know what I was meaning.
FrediFizzx
larry
> "larry" <gold...@charter.net> wrote in message
> news:3D17EF0C...@charter.net...
> | Fredi,
> | Are my posts messed up with very long line lengths?
> | I had to change to Netscape because there is a bug
> | in Microsoft Outlook Express which causes it to take
> | about a minute to open and about five minutes to
> | close down. I have not found a cure for that problem
> | yet.
>
> Your recent posts have been fine.
I was wondering because when I bring them up of the
group, the text extends across my whole 1280 x 1024 screen.
>
> | Could you post a Mathcad document of your calculations.
> | I do not have time to convert the Adobe .pcf to .mcd
> | format.
> | Thanks,
> | Larry
>
> Yes I can very easily. Which calculations? I have so many now I have lost
> track. The photon wave volume? The derivations of the model geometry from
> the constants and their equations? The ones on Maxwell's traveling wave
> equations (I have some mistakes in those that I have to fix)? I haven't
> started any yet on the excruciating details of the model. But soon I will
> be if I can get a starting point worked out in pictorial form. Once I get a
> picture of how this thing really looks with the waves, I think it will be
> easier. Thomas' picture is really just a simplified version but works well
> for many things.
>
> What version do you need them in? They are all Mathcad 2001i format
Maybe, if possible, just your future versions. I have Mathcad 8.0
Professional.
Larry
>
> FrediFizzx
>
>
>
>
FrediFizzx
Larry, never mind about the picture I was trying to visualize above. It is
not even close. I think the only thing we can say is that the wave crests
are at the corners somehow. Say we have the crest peak of E at one of the
corners, then one of the nodes will be hanging outside of the circular ring?
Does something have to return this back to the model? And the other node
will be at the center and following thru to the E vector (crest trough) on
the opposite side of the cube face going the same rotational direction.
Kitty-corner. And then the wave continues to the next node outside of the
circular ring again. So it looks something like a fan blade? Wow! Strange
beast this thing is.
FrediFizzx
larry
My photon composed of equal bound positive and negative charges would
split, in pair production, into an electron (negative charge part) and a
positron (positive charge part) with the spin split between the two. The
problem is that they would not produce VPP particles unless the charge
could be continuously distributed as two rotating rings held together by
the magnetic force due to the rotating rings of charge. Perhaps this is
where the wave structure would be appropriate.
I do not see the photon moving by itself. All matter in inertial motion
and EM radiation in inertial motion moves due to motion relative to all
the gravitational fields encountered by the matter or radiation.
Far away from matter where the net gravitational field is zero, matter
and radiation, at some velocity v, will encounter a constant
gravitational field > 0 in the direction of motion and thus move with a
constant velocity v. Near matter, they will also encounter a non
constant field which is accelerative in the direction of the center of
mass.
I was wondering if experiments have ever been done where and electron,
say, with initial velocity v, is passed through the interior of a
charged spherical shell where the net EM field is zero. I would expect
that the electron would tend to, after removing any influence from
gravitational forces, accelerate inside a positively charged spherical
shell and decelerate inside a negatively charged spherical shell due to
encountering an excess field in the direction of motion.
Larry
>
> FrediFizzx
>
>
Jim Heckman
On 27-Jun-2002, larry <gold...@charter.net> wrote:
> I was wondering if experiments have ever been done where and electron,
> say, with initial velocity v, is passed through the interior of a
> charged spherical shell where the net EM field is zero. I would expect
> that the electron would tend to, after removing any influence from
> gravitational forces, accelerate inside a positively charged spherical
> shell and decelerate inside a negatively charged spherical shell due to
> encountering an excess field in the direction of motion.
If the charge, whether positive or negative, is uniformly distributed on
the shell, the force on the electron is zero -- for the same reason
there's no gravitational force on a particle inside a spherically
symmetric shell. It's a general feature of inverse-square forces.
--
Jim Heckman
larry
Suppose that all that existed was a large non rotating sphercal shell
and a small mass inside the sphere. If at rest with respect to the shell
it would remain at rest. If it has an initial velocity v, it will feel a
constant net gravitational field in the direction of motion and thus
continue to move in the direction of motion at velocity v, i.e., in
inertial motion.
I understand that the force would be zero on an electron at rest with
respect to the charged shell. But, would a moving electron not feel a
greater flux in the direction of motion and thus be affected by a
force? After all, inside the sphere, at all points, there are equal and
opposite electric field elements omnidirectionally just as in the
gravitational case. So it would appear that a moving charge would meet a
greater field strength in the direction of motion and feel a magnetic
field which will accelerate or decelerate the electron depending on the
charge on the sphere.
Larry
>
>
Jim Heckman
On 29-Jun-2002, larry <gold...@charter.net> wrote:
> Jim Heckman wrote:
>
> > On 27-Jun-2002, larry <gold...@charter.net> wrote:
> >
> >
> >>I was wondering if experiments have ever been done where and electron,
> >>say, with initial velocity v, is passed through the interior of a
> >>charged spherical shell where the net EM field is zero. I would expect
> >>that the electron would tend to, after removing any influence from
> >>gravitational forces, accelerate inside a positively charged spherical
> >>shell and decelerate inside a negatively charged spherical shell due to
> >>encountering an excess field in the direction of motion.
> >
> > If the charge, whether positive or negative, is uniformly distributed on
> > the shell, the force on the electron is zero -- for the same reason
> > there's no gravitational force on a particle inside a spherically
> > symmetric shell. It's a general feature of inverse-square forces.
>
> Suppose that all that existed was a large non rotating sphercal shell
> and a small mass inside the sphere. If at rest with respect to the shell
> it would remain at rest. If it has an initial velocity v, it will feel a
> constant net gravitational field
The gravitational potential is constant throughout the interior of the
sphere. There is no net force, whatever the motion of the test
particle.
> in the direction of motion and thus
> continue to move in the direction of motion at velocity v, i.e., in
> inertial motion.
Yes, it will continue in inertial motion. Hence, no force.
> I understand that the force would be zero on an electron at rest with
> respect to the charged shell. But, would a moving electron not feel a
> greater flux in the direction of motion and thus be affected by a
> force?
No. The electric potential is constant throughout the interior of the
sphere. There is no net force, whatever the motion of the test
particle.
> After all, inside the sphere, at all points, there are equal and
> opposite electric field elements omnidirectionally just as in the
> gravitational case. So it would appear that a moving charge would meet a
> greater field strength in the direction of motion
There is *no* field strength inside the sphere. E = 0 everywhere.
> and feel a magnetic
> field which will accelerate or decelerate the electron depending on the
> charge on the sphere.
If the charges on the sphere are at rest, they do not generate a
magnetic field.
--
Jim Heckman
Michael Moroney
>Suppose that all that existed was a large non rotating sphercal shell
>and a small mass inside the sphere. If at rest with respect to the shell
>it would remain at rest. If it has an initial velocity v, it will feel a
>constant net gravitational field in the direction of motion and thus
>continue to move in the direction of motion at velocity v, i.e., in
>inertial motion.
The first part of the last sentence contradicts the second part. The
first part says there is a force on the mass, while the second part
(inertial motion) says there is no force.
BTW, the first part is incorrect. (the gravitational field is everywhere
zero) The gravitational force from the mass on any part of the sphere
exactly cancels that on the opposite part of the sphere.
-Mike
larry
> larry <gold...@charter.net> writes:
>
>
>>Suppose that all that existed was a large non rotating sphercal shell
>>and a small mass inside the sphere. If at rest with respect to the shell
>>it would remain at rest. If it has an initial velocity v, it will feel a
>>constant net gravitational field in the direction of motion and thus
>>continue to move in the direction of motion at velocity v, i.e., in
>>inertial motion.
>>
>
> The first part of the last sentence contradicts the second part. The
> first part says there is a force on the mass, while the second part
> (inertial motion) says there is no force.
"Force" was an inappropriate chosen word for the affect since it implies
an acceleration. As I see it, the gravitational flux causes motion. If
the flux has a constant value, then the velocity is constant and if the
flux increases or decreases, then the motion is accelerative or
decelerative, respectively. If the field is composed of gravitons and is
seen to have a constant flux omnidirectionally, the particle will remain
at rest but if the field has a net non zero flux, then the particle will
either move with constant velocity in the direction of motion, or if the
net non zero flux is changing as the mass moves, the mass will
accelerate, decelerate, or or have a jerk component of motion.
>
> BTW, the first part is incorrect. (the gravitational field is everywhere
> zero) The gravitational force from the mass on any part of the sphere
> exactly cancels that on the opposite part of the sphere.
True, the gravitational force at any interior point is zero and a
stationary mass will be affected by a net force of zero. That does not
mean that there is no field interior to the shell. If that were so, it
would be as though the interior shields a mass outside the sphere from
the gravitational field of one hemisphere from the mass. What bothers me
is that, if there were a spherical gaussian surface around a moving mass
inside the shell, would not there be a greater gravitational flux in the
direction of motion which would be the cause of the motion? Is that not
the meaning of Wheeler's "mass there causes inertia here", (meaning that
inertia of a body is caused by the gravitational fields of all the other
matter in the universe)?
Larry
>
> -Mike
>