Bohr modeled the hydrogen electron on the wrong side of the cusp?
ThomasL283
It is known that the Bohr electron orbital model for hydrogen is not
physically correct.
( The orbital angular momentum of the ground state of hydrogen is zero.)
And, by the rules of physics, an orbiting electron should radiate energy and
spiral into the proton. Bohr tried to balance the electric attraction by the
angular momentum, but in truth, it is the magnetic moments that oppose the
electron proton charge attraction, in the near field.
One can model the hydrogen atom using the VPP electron and VPP proton models.
To wit:
Remember the undergrad physics exercise you did on calculating the electric
field some distance from a "ring of charge" that required the cosine of theta =
R/sqr (R^2 + x^2) for the analysis?
Remember it was then a student exercise to show when separation x >>>R ring
radius, that the ring charge looks like a point charge in the far field?
See for example, Halliday and Resnick, Physics part II, page 673, ring of
charge.
Bohr assumed that the electron and proton where point charges, which put his
hydrogen model electron (wrongly) in the far field. The electron and proton
magnetic moments were (wrongly) omitted , but these magnetic moments are
essential to combine with the electric energy, for creating the experimental
(-13.5984 eV) binding energy photon.
Bohr did not have the luxury of a model for the electron's structure, and
(wrongly) assumed that the electron was a point particle. The electron's
charge proves to be in the form of two charge current rings.
Only in the near field can the essential magnetic energy null with the
electric energy to create an exothermic photon of (-13.5984 eV).
The ring charge geometry creates a cusp in the electric energy between the
electron's charge loop and the virtual point charge of the proton's small
charged (Hofstadter) core. Print out the following GIF example:
http://www.members.aol.com/tnlockyer/VPPchem.gif
If you have math software, note that the electrical potential energy equation
for (eV) has two separations that return (13.5984 eV).
These two separations are the Bohr radius (5.29 E-11 meter) in the far field,
and the VPP (1.20310032 E-15 meter) in the near field, but only the near field
magnetic energy nulls. Note that these results tend to verify the VPP ring
charge structure for the electron and positron. Print out the example.
http://www.members.aol.com/tnlockyer/VPPring.gif
Analysis of electron to proton binding is similar to the successful VPP
modeling and calculation of the (-13.5984 eV) binding energy between nucleons.
Bohr's model has no way to create the photon at the null, which requires both
the electric and the magnetic energy to be equal. The exothermic photon energy
(deficit) is required to hold (bind) the electron, until (+13.5984 eV)
ionization energy is added.
Regards: Tom:
Tom Lockyer (75 and retired) See "Vector Particle and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about and express it in numbers,
you know something about it." Lord Kelvin (1824-1907)
ThomasL283
>Date: 8/11/2002 12:55 PM Pacific Daylight Time
Wrote in:
>Message-id: <fe51764e.02081...@posting.google.com>
>znip<
>> >However, what is the big X length you have defined for the magnetic energy
>> >equation?
>>
>> Fredi: That is (roughly) the distance to the center of the electron or
>about
>> (lambda_e/2). Seems like the magnetic force acts from the centroid.
>>
>
>One thing I am not following is how far away is the electron from the
>proton? If the electric and magnetic forces are equal at x, where
>does that put the electron?
>
>BTW, for those interested, the Ring of Charge in the 3rd Ed. of
>"Physics" by Halliday and Resnick is on page 588.
Fredi: The ring of charge calculations with the (corrected) equation for
(eV(x)) (see the new thread started) gives us a separation between the proton's
core and electron of 1.20310032 E-15 meter.
Remember the VPP proton core size of 0.7529 E-15 meter is in agreement with
Hofstadter's proton-electron scattering measurement at Stanford in 1956.
The electron's spin axis is aligned with the proton spin axis and both are
spinning in the same direction (much like the deuteron model). This means the
charges attract and the magnetic moments oppose, in the near field.
The near end charge ring calculates as the a very close 1.2 E-15 meter.
Notice the (eV(x)) equation uses just (-0.5e) on the theory that the charge of
the other ring is negative, and so is guarded, making a zero charge field
within in the electron's near field structure.
The magnetic field, however, extends through the electron as in a Helmholtz
coil. See:
http://www.members.aol.com/tnlockyer/VPPring.gif
Michael Moroney
>Remember the VPP proton core size of 0.7529 E-15 meter is in agreement with
>Hofstadter's proton-electron scattering measurement at Stanford in 1956.
1956? Got anything less, err, ancient?
Also, explain how the VPP claims the electron is at least 4 orders of
magnitude larger than experimental data measures it?
> The electron's spin axis is aligned with the proton spin axis and both are
>spinning in the same direction (much like the deuteron model). This means the
>charges attract and the magnetic moments oppose, in the near field.
What about the other state? (spins anti-aligned)? Hydrogen atoms exist
in both states.
-Mike
ThomasL283
>Date: 8/13/2002 6:45 AM Pacific Daylight Time
Wrote in:
>Message-id: <H0sBK9...@world.std.com>
>
>thoma...@aol.com (ThomasL283) writes:
>
>>Remember the VPP proton core size of 0.7529 E-15 meter is in agreement with
>>Hofstadter's proton-electron scattering measurement at Stanford in 1956.
>
>1956? Got anything less, err, ancient?
Mike there is later data, but I like Hofstader.
Hofstader was a prof to one of my sons (circa 1972) and my son said he was the
best teacher he ever had.
To me that was the mark of the man, his Nobel not withstanding.
>Also, explain how the VPP claims the electron is at least 4 orders of
>magnitude larger than experimental data measures it?
Mike, I have never seen such data.
I do know that the electron is a hearty beast. The electron does not seem to
break up even at several GeV.
The VPP model shows two different electron structures, one for the positron and
one for the electron.
The positron has H vectors where the electron has E vectors, (and reverse) in a
conjugate structure.
The structures are obtained (blindly) by simply combining the Poynting vector
(VPP energy model) in all possible ways.
This VPP model also automatically gives us the electron and muon neutrino pair
structures. See:
http://www.members.aol.com/tnlockyer/cfives.gif
The VPP proton model then is formed by scaling leptons. See:
http://www.members.aol.com/tnlockyer/protess.gif
The small active VPP proton model (scaled) core has the Hofstader 0.7529 E-15
meter dimensions noted above.
This VPP proton model scaled core size also predicts a magnetic moment
that is essentially the same value (independently) obtained by the quadratic
analysis of the deuteron's binding energy.
http://www.members.aol.com/tnlockyer/quadratic.gif
.>> The electron's spin axis is aligned with the proton spin axis and both are
>>spinning in the same direction (much like the deuteron model). This means
>the
>>charges attract and the magnetic moments oppose, in the near field
>
>What about the other state? (spins anti-aligned)? Hydrogen atoms exist
>in both states.
I never heard of that, but if such electron states exist in 1H, I would
suppose they are short lived. References?
Michael Moroney
>>mor...@world.std.spaamtrap.com (Michael Moroney)
>>Date: 8/13/2002 6:45 AM Pacific Daylight Time
>Wrote in:
>>Message-id: <H0sBK9...@world.std.com>
>>
>>thoma...@aol.com (ThomasL283) writes:
>>
>>>Remember the VPP proton core size of 0.7529 E-15 meter is in agreement with
>>>Hofstadter's proton-electron scattering measurement at Stanford in 1956.
>>1956? Got anything less, err, ancient?
>Mike there is later data, but I like Hofstader.
The point is, there is nearly a half century of technology between then
and now, measurements are much more accurate now. Who cares whether you
like him.
>>Also, explain how the VPP claims the electron is at least 4 orders of
>>magnitude larger than experimental data measures it?
>Mike, I have never seen such data.
Sheesh! References have been posted time and again as refutations of VPP.
Why don't you follow some of the links references?
>This VPP model also automatically gives us the electron and muon neutrino pair
>structures. See:
What about the tau and tau neutrino?
What about the electron antineutrino?
>>> The electron's spin axis is aligned with the proton spin axis and both are
>>>spinning in the same direction (much like the deuteron model). This means
>>the
>>>charges attract and the magnetic moments oppose, in the near field
>>
>>What about the other state? (spins anti-aligned)? Hydrogen atoms exist
>>in both states.
>I never heard of that, but if such electron states exist in 1H, I would
>suppose they are short lived. References?
Actually the anti-aligned state is the more stable, lower-energy state.
Google for "Hydrogen spin-flip transition", hyperfine structure, 21 cm
hydrogen line [this is the photon emitted when a hydrogen atom goes
from the parallel to the anti-parallel state]
http://instruct1.cit.cornell.edu/courses/astro101/lec06.htm
-Mike
ThomasL283
>Date: 8/15/2002 8:58 AM Pacific Daylight Time
Wrote in:
>Message-id: <H0w71s...@world.std.com>
>>>thoma...@aol.com (ThomasL283) writes:
>>>
>>>>Remember the VPP proton core size of 0.7529 E-15 meter is in agreement
>with
>>>>Hofstadter's proton-electron scattering measurement at Stanford in 1956.
>
>
>>>1956? Got anything less, err, ancient?
>
>>Mike there is later data, but I like Hofstadter.
>Who cares whether you
>like him.
Me. Cause his data seems to agree with the predictions of VPP.
Anyway his form factor gives an rms size. And this has been improved in later
efforts.
>>>Also, explain how the VPP claims the electron is at least 4 orders of
>>>magnitude larger than experimental data measures it?
>
>>Mike, I have never seen such data.
>
>Why don't you follow some of the links references?
Experiments are one thing, conclusions are another. As I showed with the
charge rings, in the far field the ring looks like a point.
If you insist on a "point" electron. What about the positron? Or muon?
>>This VPP model also automatically gives us the electron and muon neutrino
>pair
>>structures. See:
http://www.members.aol.com/tnlockyer/cfives.gif
>What about the tau and tau neutrino?
The tau is a composite particle that comes apart into mesons.
If you insist that there is a tau neutrino, then tell me how is it constructed?
>What about the electron antineutrino?
The VPP electron neutrino is it's own anti-particle, by simple rotation about
any axis 180 degrees.
VPP verifies that Majorani was right. And no, double beta decay does not
exist as proof.
The structures of the VPP proton and neutron verify that the electron neutrino
is its own antiparticle, because their vectors in the scaled layers sum
correctly.
And yes, these scaled structures are same VPP proton and neutron models that
give (n-1H) and a magnetic moment that agrees with the quadratic deduced value,
from the deuteron nucleous binding energy.
http://www.members.aol.com/tnlockyer/quadratic.gif
>>>> The electron's spin axis is aligned with the proton spin axis and both
>are
>>>>spinning in the same direction (much like the deuteron model). This means
>>>the
>>>>charges attract and the magnetic moments oppose, in the near field
>
>>>What about the other state? (spins anti-aligned)? Hydrogen atoms exist
>>>in both states.
>>I never heard of that, but if such electron states exist in 1H, I would
>>suppose they are short lived. References?
>Actually the anti-aligned state is the more stable, lower-energy state.
>Google for "Hydrogen spin-flip transition", hyperfine structure, 21 cm
>hydrogen line [this is the photon emitted when a hydrogen atom goes
>from the parallel to the anti-parallel state]
>http://instruct1.cit.cornell.edu/courses/astro101/lec06.htm
Nice pointer, thanks. But I don't think the spin flip would be more stable.
I always thought that the 21.1 cm (ultra low frequency) was splittings from
the spectrum near the ionization potential.
Learn something new every day.
Michael Moroney
>>>>
>>>>>Remember the VPP proton core size of 0.7529 E-15 meter is in agreement
>>with
>>>>>Hofstadter's proton-electron scattering measurement at Stanford in 1956.
>>
>>
>>>>1956? Got anything less, err, ancient?
>>
>>>Mike there is later data, but I like Hofstadter.
>>Who cares whether you
>>like him.
>Me. Cause his data seems to agree with the predictions of VPP.
Yet there is nearly 50 years of data that now contradicts it. (Probably
Hofstadter is not "wrong", but the margin of error was simply much larger
than it is now)
>Anyway his form factor gives an rms size. And this has been improved in later
>efforts.
Yup. The maximim size is now known to be much smaller.
>>>>Also, explain how the VPP claims the electron is at least 4 orders of
>>>>magnitude larger than experimental data measures it?
>>
>>>Mike, I have never seen such data.
>>Why don't you follow some of the links references?
>Experiments are one thing, conclusions are another. As I showed with the
>charge rings, in the far field the ring looks like a point.
WHat the heck is that supposed to mean? If your conclusion disagrees with
the experiment, throw out the experiment?
>If you insist on a "point" electron. What about the positron? Or muon?
As they are all fundamental particles, they should be point-like. I'd
expect the positron, in particular, to be identical with the electron
regarding this. The muon for the most part acts like a heavy electron,
behaving the same except for its mass (and that it decays).
>>>This VPP model also automatically gives us the electron and muon neutrino
>>pair
>>>structures. See:
>http://www.members.aol.com/tnlockyer/cfives.gif
I can't make heads or tails out of that.
>>What about the tau and tau neutrino?
>The tau is a composite particle that comes apart into mesons.
>If you insist that there is a tau neutrino, then tell me how is it constructed?
What do you mean how is it constructed? Both the tau and tau neutrino are
fundamental particles and I'd expect them to be pointlike.
>>What about the electron antineutrino?
>The VPP electron neutrino is it's own anti-particle, by simple rotation about
>any axis 180 degrees.
How do you explain how some processes happen when one source of neutrinos
is used but not a second, and another, reverse process happens when the
second source is used but not the first? Especially when the first and
second sources are themselves opposite processes (beta- decay and electron
capture for example)
> VPP verifies that Majorani was right. And no, double beta decay does not
>exist as proof.
Are you saying double beta decay doesn't exist? It's been observed.
>>>I never heard of that, but if such electron states exist in 1H, I would
>>>suppose they are short lived. References?
>>Actually the anti-aligned state is the more stable, lower-energy state.
>>Google for "Hydrogen spin-flip transition", hyperfine structure, 21 cm
>>hydrogen line [this is the photon emitted when a hydrogen atom goes
>>from the parallel to the anti-parallel state]
>>http://instruct1.cit.cornell.edu/courses/astro101/lec06.htm
>Nice pointer, thanks. But I don't think the spin flip would be more stable.
Kind of makes sense that it would. Wouldn't you expect a state where the
north and south poles of two magnets anti-aligned to be a lower energy
state than with them aligned?
-Mike
ThomasL283
>Date: 8/17/2002 1:50 PM Pacific Daylight Time
Wrote in:
>Message-id: <H109vn...@world.std.com>
>
>thoma...@aol.com (ThomasL283) writes:
Mike writes:
>>>>>Also, explain how the VPP claims the electron is at least 4 orders of
>>>>>magnitude larger than experimental data measures it?
>>>
>>Experiments are one thing, conclusions are another.
>
>WHat the heck is that supposed to mean? If your conclusion disagrees with
>the experiment, throw out the experiment?
No. In the case of "point" particles, I believe a particle sees other
particles through their deBroglie "pilot" waves.
Electron microscopes get their magnification by the wavelength of the electron
acceleration energy.
>>>>This VPP model also automatically gives us the electron and muon neutrino
>>>pair
>>>>structures. See:
>
>>http://www.members.aol.com/tnlockyer/cfives.gif
>I can't make heads or tails out of that.
>
Yes, it is somewhat daunting. I doodled for a couple of years, until putting
the neutrino model inside of the electron or positron model. See:
http://www.members.aol.com/tnlockyer/chargeconj.gif
>>>What about the tau and tau neutrino?
>>If you insist that there is a tau neutrino, then tell me how is it
>constructed?
>
>What do you mean how is it constructed? Both the tau and tau neutrino are
>fundamental particles and I'd expect them to be pointlike.
See above GIF. Particles must have unique structures, to give them their
unique characteristics. Only three pair of leptons structures are possible.
>>>What about the electron antineutrino?
>
>>The VPP electron neutrino is it's own anti-particle, by simple rotation
>about
>>any axis 180 degrees.
>> VPP verifies that Majorana was right. And no, double beta decay does not
>>exist as proof.
>
>Are you saying double beta decay doesn't exist? It's been observed.
>
I know several groups have been looking for double beta decay to prove that the
electron type neutrino is its own anti-particle.
VPP already has indicated that the electron type neutrino is it's own
anti-particle.
More than that, VPP shows how it is possible for a particle to be its own
anti-particle.
The electron type neutrino becomes its own anti-particle by a simple 180 degree
rotation about any axis, to present a conjugate to it's mate.
>>>Google for "Hydrogen spin-flip transition", hyperfine structure, 21 cm
>>>hydrogen line [this is the photon emitted when a hydrogen atom goes
>>>from the parallel to the anti-parallel state]
>>>http://instruct1.cit.cornell.edu/courses/astro101/lec06.htm
>>Nice pointer, thanks. But I don't think the spin flip would be more stable.
>
>Kind of makes sense that it would. >Wouldn't you expect a state where the
>north and south poles of two magnets anti-aligned to be a lower energy
>state than with them aligned?
>
Yes, Mike, but the energy difference is so small (~5.8 E-6 eV).
Further, in working with nuclei bindings, I find that nature prefers two
particles sharing the same spin axis, in tandem.
For example the spin of deuterium (proton and neutron) is one (never zero).
With more than two nucleons, some spins can cancel.
For example tritium (spin one half) and helium 4 (spin zero) show that some
spins can be anti-parallel when >2 nucleons are present.
Michael Moroney
>>>>>>Also, explain how the VPP claims the electron is at least 4 orders of
>>>>>>magnitude larger than experimental data measures it?
>>>>
>>>Experiments are one thing, conclusions are another.
>>
>>WHat the heck is that supposed to mean? If your conclusion disagrees with
>>the experiment, throw out the experiment?
>No. In the case of "point" particles, I believe a particle sees other
>particles through their deBroglie "pilot" waves.
That doesn't make sense. Besides, the deBroglie wavelength doesn't meet
the "pointlike" requirement.
>>>http://www.members.aol.com/tnlockyer/cfives.gif
>>I can't make heads or tails out of that.
>Yes, it is somewhat daunting. I doodled for a couple of years, until putting
>the neutrino model inside of the electron or positron model. See:
>http://www.members.aol.com/tnlockyer/chargeconj.gif
I still can't figure out what you are trying to say.
>>>>What about the tau and tau neutrino?
>>>If you insist that there is a tau neutrino, then tell me how is it
>>constructed?
>>
>>What do you mean how is it constructed? Both the tau and tau neutrino are
>>fundamental particles and I'd expect them to be pointlike.
>See above GIF. Particles must have unique structures, to give them their
>unique characteristics. Only three pair of leptons structures are possible.
You have a lot of work to do. So far your theory explains only 5 of the 12
known leptons/antileptons, and if it claims that's all there is it is
already obvious it is wrong.
>>>>What about the electron antineutrino?
>>
>>>The VPP electron neutrino is it's own anti-particle, by simple rotation
>>about
>>>any axis 180 degrees.
>>> VPP verifies that Majorana was right. And no, double beta decay does not
>>>exist as proof.
You skipped my question why neutrinos from one kind of source participate
in a certain type of reaction yet neutrinos from another type of source
don't participate. For example, v + D --> p + p + e- (D is a deuteron)
only works for neutrinos from EC or B+ decays, yet not at all from B-
decay neutrinos. The standard model explains this just fine.
>>Kind of makes sense that it would. >Wouldn't you expect a state where the
>>north and south poles of two magnets anti-aligned to be a lower energy
>>state than with them aligned?
>>
>Yes, Mike, but the energy difference is so small (~5.8 E-6 eV).
The electron and proton are fairly far apart, so the magnetic interaction
is small.
>Further, in working with nuclei bindings, I find that nature prefers two
>particles sharing the same spin axis, in tandem.
>For example the spin of deuterium (proton and neutron) is one (never zero).
That's but one example.
>With more than two nucleons, some spins can cancel.
They MUST cancel else they'd violate the Pauli exclusion pronciple.
>For example tritium (spin one half) and helium 4 (spin zero) show that some
>spins can be anti-parallel when >2 nucleons are present.
The like particles MUST have opposite spins.
-Mike
ThomasL283
>
>thoma...@aol.com (ThomasL283) writes:
>>Yes, it is somewhat daunting. I doodled for a couple of years, until
>putting
>>the neutrino model inside of the electron or positron model. See:
>
>>http://www.members.aol.com/tnlockyer/chargeconj.gif
>
>I still can't figure out what you are trying to say.
The vector models combine by adding their unique vector structures together for
a resultant virtual conjugate structure. The (chargeconj.gif) above
demonstrates the process.
This charge conjugation is thought to create electrical potential energy
between composite model (scaled) layers .
The postulated electrical potential energy between layers, in the scaled
structure of the proton and neutron models, sums to a whopping 5 percent
additional model mass. (Giving the model mass to within 3ppm of the CODATA
recomendation).
>>>>>What about the tau and tau neutrino?
>
>>>>If you insist that there is a tau neutrino, then tell me how is it
>>>constructed?
>>>
>>>What do you mean how is it constructed? Both the tau and tau neutrino are
>>>fundamental particles and I'd expect them to be pointlike.
>>Particles must have unique structures, to give them their
>>unique characteristics. Only three pair of leptons structures are possible.
>You have a lot of work to do. So far your theory explains only 5 of the 12
>known leptons/antileptons, and if it claims that's all there is it is
>already obvious it is wrong.
Mike, This model gives six basic leptons.
Electron-positron, electron type neutrino pair, and the muon neutrino pair.
The muon is modeled as an electron (or positron) and two neutrinos.
The tau is a composite with various branching ratios.
The tau neutrino structure cannot exist, despite recent claims to have possibly
"detected" the tau neutrino.
I would accept the tau neutrino if there was some good reason for it besides a
gleam in the theorists eyes.
The VPP model tells me how the neutrinos are constructed, and how they combine
to form muons and all other composite particles . Those who have the book can
see the arguments in chapter 10.
>>>>The VPP electron neutrino is it's own anti-particle, by simple rotation
>>>about
>>>>any axis 180 degrees.
>>>> VPP verifies that Majorana was right.
>You skipped my question why neutrinos from one kind of source participate
>in a certain type of reaction yet neutrinos from another type of source
>don't participate. For example, v + D --> p + p + e- (D is a deuteron)
>only works for neutrinos from EC or B+ decays, yet not at all from B-
>decay neutrinos.
Mike, that v + d---> p + p = e- reaction will only happen in certain unstable
nuclei, not with a lone deuteron.
The kinematics are as in:
http://www.members.aol.com/tnlockyer/Betadecay.gif
The energy for creating a neutron, from an electron capture (EC) by a proton,
comes from the new atomic energy created in the reaction. The nucleons, in an
unstable nucleus, rearrange into new (lower energy state) patterns, creating
energy for the reaction.
Michael Moroney
>>I still can't figure out what you are trying to say.
>The vector models combine by adding their unique vector structures together for
>a resultant virtual conjugate structure. The (chargeconj.gif) above
>demonstrates the process.
<snip>
Sorry, still makes no sense.
>>You have a lot of work to do. So far your theory explains only 5 of the 12
>>known leptons/antileptons, and if it claims that's all there is it is
>>already obvious it is wrong.
>Mike, This model gives six basic leptons.
>Electron-positron, electron type neutrino pair, and the muon neutrino pair.
So are there five or six? One of your gifs shows five.
>The muon is modeled as an electron (or positron) and two neutrinos.
"Is modeled". Only by yourself. Substantial evidence shows it to be
fundamental.
>The tau is a composite with various branching ratios.
Again, no evidence for this, and if it was composite why are there so
many decay modes for it?
>The tau neutrino structure cannot exist, despite recent claims to have possibly
>"detected" the tau neutrino.
Bury your head in the sand if you want, but the tau neutrino won't go
away. It was detected in 2000, by the reaction vT+N -> T- -> u-.
>I would accept the tau neutrino if there was some good reason for it besides a
>gleam in the theorists eyes.
I think filling in a space in a table of leptons is an awfully good
reason. Kind of like Mendeleev predicting the properties of eka-silicon
when he produced his periodic table. Besides, the physics laws required
it.
>>You skipped my question why neutrinos from one kind of source participate
>>in a certain type of reaction yet neutrinos from another type of source
>>don't participate. For example, v + D --> p + p + e- (D is a deuteron)
>>only works for neutrinos from EC or B+ decays, yet not at all from B-
>>decay neutrinos.
>Mike, that v + d---> p + p = e- reaction will only happen in certain unstable
>nuclei, not with a lone deuteron.
No, I was not talking about beta decay. This is a reaction, not a decay.
The neutrino reacts with and converts a neutron into a proton and breaks
up the deuteron. (Of course it must have a certain amount of energy
to do so) See: http://www.sno.phy.queensu.ca/sno/sno2.html#interactions
(see under charged current reaction)
Another type of reaction is in (normal) water, v(e)bar + p -> n + e+.
Again the antineutrino needs a certain energy to allow the energy to go
forward (Mn+Me-Mp)
But the neutrinos of the first type won't participate in the second
reaction and vice versa.
(Yet another reaction is v(u)bar + p -> n + u+. There is a detector in
Antarctica that uses the ice cap itself as the detector. (specifically the
hydrogen atoms of its water))
>The kinematics are as in:
>http://www.members.aol.com/tnlockyer/Betadecay.gif
This was shot down in flames _long_ ago.
-Mike
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:H19t00...@world.std.com...
| thoma...@aol.com (ThomasL283) writes:
|
| >>I still can't figure out what you are trying to say.
|
| >The vector models combine by adding their unique vector structures
together for
| >a resultant virtual conjugate structure. The (chargeconj.gif) above
| >demonstrates the process.
|
| <snip>
|
| Sorry, still makes no sense.
Tom, I never quite understood your thinking on this completely either.
Maybe you can explain it better?
| >>You have a lot of work to do. So far your theory explains only 5 of the
12
| >>known leptons/antileptons, and if it claims that's all there is it is
| >>already obvious it is wrong.
|
| >Mike, This model gives six basic leptons.
| >Electron-positron, electron type neutrino pair, and the muon neutrino
pair.
|
| So are there five or six? One of your gifs shows five.
|
| >The muon is modeled as an electron (or positron) and two neutrinos.
|
| "Is modeled". Only by yourself. Substantial evidence shows it to be
| fundamental.
I am not completely satisfied that the muon is a fundamental particle
either. How do we know for sure that it is not a composite when it decays
so easily?
| >The tau is a composite with various branching ratios.
|
| Again, no evidence for this, and if it was composite why are there so
| many decay modes for it?
My question would be what is it a composite of in VPP? Since about 64
percent of the time it decays into hadrons. The rest mostly muons,
electrons, neutrinos.
| >The tau neutrino structure cannot exist, despite recent claims to have
possibly
| >"detected" the tau neutrino.
|
| Bury your head in the sand if you want, but the tau neutrino won't go
| away. It was detected in 2000, by the reaction vT+N -> T- -> u-.
|
| >I would accept the tau neutrino if there was some good reason for it
besides a
| >gleam in the theorists eyes.
Is it possible for composite neutrino structures to exist in VPP?
| I think filling in a space in a table of leptons is an awfully good
| reason. Kind of like Mendeleev predicting the properties of eka-silicon
| when he produced his periodic table. Besides, the physics laws required
| it.
I am not so sure that that space should actually be there if the muon and
tau are really composites.
FrediFizzx
Michael Moroney
>I am not completely satisfied that the muon is a fundamental particle
>either. How do we know for sure that it is not a composite when it decays
>so easily?
Your mistake is thinking of these as if they were macroscopic objects.
You think of the muon as an egg, break an egg and you have a yolk and an
egg white so the yolk and white were always in there to begin with.
Subatomic particles don't work like that. They can react with each other
or decay in any combination as long as certain conserved quantities (such
as charge) are conserved through the reaction, and energy is conserved.
A muon can decay because there is a combination (electron, mu neutrino,
electron antineutrino) that exists that fulfills these requirements. An
electron cannot decay because there is no combination of lighter particles
that conserves charge (no lighter charged particle at all) Same with the
proton, the only reason it is stable is that there is no lighter particle
with a nonzero baryon number (conserved quantity)
As for the muon being fundamental, consider that if you calculate the
"Bohr Magneton" for the muon (use Mu instead of Me in calculation) you get
a value very close to, but not exactly, the muon's magnetic moment, just
like what happens with the electron. In other words, the muon's magnetic
moment is smaller than the electron's by a ratio very close to the mass
ratio. Not what you'd expect if there was an electron in there. If you
calculate the difference ratio, you get a muon magnetic moment anomoly
much like the electron's anomoly, and in fact they are close in value (not
exactly the same). Muonic atoms have been created (replace electron(s)
with muon(s)) and they behave just like electrons if you substitute the
muon's mass in the various equations. In other words, the muon behaves
just like a heavy electron, not a composite.
>| >The tau is a composite with various branching ratios.
>|
>| Again, no evidence for this, and if it was composite why are there so
>| many decay modes for it?
>My question would be what is it a composite of in VPP? Since about 64
>percent of the time it decays into hadrons. The rest mostly muons,
>electrons, neutrinos.
How could it possibly be a composite if it doesn't break down into
combinations of the same things each time?
>| I think filling in a space in a table of leptons is an awfully good
>| reason. Kind of like Mendeleev predicting the properties of eka-silicon
>| when he produced his periodic table. Besides, the physics laws required
>| it.
>I am not so sure that that space should actually be there if the muon and
>tau are really composites.
Even if we accept Thomas's flawed chart, there is an empty space for the
muon if we match up neutrinos and non neutrinos in columns and
electron/muon in rows like the standard model chart, so he "predicts" the
muon although his mystery cube diagrams apparently don't allow for it.
-Mike
ThomasL283
>Date: 8/23/2002 10:08 AM Pacific Daylight Time
Wrote in:
>Message-id: <H1B3M8...@world.std.com>
>"Fredi Fizzx" <fredi fizz...@ahahhotmail.com> writes:
>
>>I am not completely satisfied that the muon is a fundamental particle
>>either. How do we know for sure that it is not a composite when it decays
>>so easily?
>Your mistake is thinking of these as if they were macroscopic objects.
>You think of the muon as an egg, break an egg and you have a yolk and an
>egg white so the yolk and white were always in there to begin with.
>A muon can decay because there is a combination (electron, mu neutrino,
>electron antineutrino) that exists that fulfills these requirements.
Where do they exist? You either have to create them at the time of decay, or
more simply, consider that they exisr in the composite structure to begin with.
Either way, Mike, the result is the same.
Trouble is with your method nature has to create particles in singles, and we
know nature requires doublets.
OTOH. If the composite particle is made up of it's decay particles. The
doublets were made when the particle was formed.
>As for the muon being fundamental, consider that if you calculate the
>"Bohr Magneton" for the muon (use Mu instead of Me in calculation) you get
>a value very close to, but not exactly, the muon's magnetic moment, just
>like what happens with the electron.
Exactly, the VPP muon is a collapsed (small crossection) electron (or positron)
by the action of combining the muon neutrino structures.
See Chapter 9.
>
> >| >The tau is a composite with various branching ratios.
>>|
>>| Again, no evidence for this, and if it was composite why are there so
>>| many decay modes for it?
Fredi asks:
>>My question would be what is it a composite of in VPP? Since about 64
>>percent of the time it decays into hadrons. The rest mostly muons,
>>electrons, neutrinos.
Fredi:
The tau is composed of its decay particles.
>How could it possibly be a composite if it doesn't break down into
>combinations of the same things each time?
Mike, you are not looking at the FINAL decay particles. Pions decay into a
neutrino and muon. The muon in turn decays into an electron and two neutrinos.
You will find that you end up with the same number of electrons and neutrinos
regardless of the branching type.
>>| I think filling in a space in a table of leptons is an awfully good
>>| reason. Kind of like Mendeleev predicting the properties of eka-silicon
>>| when he produced his periodic table. Besides, the physics laws required
>>| it.
Fredi said:
>>I am not so sure that that space should actually be there if the muon and
>>tau are really composites.
>
>Even if we accept Thomas's flawed chart, there is an empty space for the
>muon if we match up neutrinos and non neutrinos in columns and
>electron/muon in rows like the standard model chart, so he "predicts" the
>muon although his mystery cube diagrams apparently don't allow for it.
>
Mike the cube diagrams do allow the composite muon. Chapter 9 shows the VPP
pion and muon models.
Turns out if you add the two muon type neutrino vectors together with an
electrons vectors the combination forms a virtual electron (muon) without
charge conjugation (See Figure 9.1) SEE also also:
http://www.members.aol.com/tnlockyer/muon.gif
Michsael Moroney
>>Your mistake is thinking of these as if they were macroscopic objects.
>>You think of the muon as an egg, break an egg and you have a yolk and an
>>egg white so the yolk and white were always in there to begin with.
>>A muon can decay because there is a combination (electron, mu neutrino,
>>electron antineutrino) that exists that fulfills these requirements.
>Where do they exist? You either have to create them at the time of decay, or
>more simply, consider that they exisr in the composite structure to begin with.
No, being created at the time of decay is not the same as previously
existing. Before decay, decay products simply don't exist. After they
do.
>Either way, Mike, the result is the same.
Nope.
>Trouble is with your method nature has to create particles in singles, and we
>know nature requires doublets.
Not quite true. You often see pair production from gammas because of the
requirement that quantum constants must be conserved, and you start with
most (charge, lepton number etc.) being zero so the most likely scenario
is X and Anti-X. But start with, say, a neutron, and you don't get pair
production. (You sort of do with the electron & antineutrino to conserve
lepton number)
The muon decay is almost the same, the product muon neutrino conserves
the muon quantum number, the electron conserves the charge, the electron
antineutrino cancels out the electron's lepton number, so the product
quantum numbers are all the same as the original.
>>As for the muon being fundamental, consider that if you calculate the
>>"Bohr Magneton" for the muon (use Mu instead of Me in calculation) you get
>>a value very close to, but not exactly, the muon's magnetic moment, just
>>like what happens with the electron.
>Exactly, the VPP muon is a collapsed (small crossection) electron (or positron)
>by the action of combining the muon neutrino structures.
Gibberish. If there was an electron in there you'd see it's magnetic
moment, not that from the muon's mass. Or, I suppose you say it's a
coincidence that the total magnetic moment just happens to match its mass.
>The tau is composed of its decay particles.
>>How could it possibly be a composite if it doesn't break down into
>>combinations of the same things each time?
>Mike, you are not looking at the FINAL decay particles. Pions decay into a
>neutrino and muon. The muon in turn decays into an electron and two neutrinos.
>You will find that you end up with the same number of electrons and neutrinos
>regardless of the branching type.
Nope. The final number of electrons, neutrinos etc. vary, depending on
the decay modes.
>>Even if we accept Thomas's flawed chart, there is an empty space for the
>>muon if we match up neutrinos and non neutrinos in columns and
>>electron/muon in rows like the standard model chart, so he "predicts" the
>>muon although his mystery cube diagrams apparently don't allow for it.
>>
> Mike the cube diagrams do allow the composite muon. Chapter 9 shows the VPP
>pion and muon models.
But it should be a fundamental particle, as the others are fundamental.
>Turns out if you add the two muon type neutrino vectors together with an
>electrons vectors the combination forms a virtual electron (muon) without
>charge conjugation (See Figure 9.1) SEE also also:
>http://www.members.aol.com/tnlockyer/muon.gif
More undecipherable nonsense. (Exception: you seem to think the muon is
an electron + muon neutrino +_ muon antineutrino. This has been
disproven)
-Mike
Michael Moroney
>>Your mistake is thinking of these as if they were macroscopic objects.
>>You think of the muon as an egg, break an egg and you have a yolk and an
>>egg white so the yolk and white were always in there to begin with.
>>A muon can decay because there is a combination (electron, mu neutrino,
>>electron antineutrino) that exists that fulfills these requirements.
>Where do they exist? You either have to create them at the time of decay, or
>more simply, consider that they exisr in the composite structure to begin with.
No, being created at the time of decay is not the same as previously
existing. Before decay, decay products simply don't exist. After they
do.
>Either way, Mike, the result is the same.
Nope.
>Trouble is with your method nature has to create particles in singles, and we
>know nature requires doublets.
Not quite true. You often see pair production from gammas because of the
requirement that quantum constants must be conserved, and you start with
most (charge, lepton number etc.) being zero so the most likely scenario
is X and Anti-X. But start with, say, a neutron, and you don't get pair
production. (You sort of do with the electron & antineutrino to conserve
lepton number)
The muon decay is almost the same, the product muon neutrino conserves
the muon quantum number, the electron conserves the charge, the electron
antineutrino cancels out the electron's lepton number, so the product
quantum numbers are all the same as the original.
>>As for the muon being fundamental, consider that if you calculate the
>>"Bohr Magneton" for the muon (use Mu instead of Me in calculation) you get
>>a value very close to, but not exactly, the muon's magnetic moment, just
>>like what happens with the electron.
>Exactly, the VPP muon is a collapsed (small crossection) electron (or positron)
>by the action of combining the muon neutrino structures.
Gibberish. If there was an electron in there you'd see it's magnetic
moment, not that from the muon's mass. Or, I suppose you say it's a
coincidence that the total magnetic moment just happens to match its mass.
>The tau is composed of its decay particles.
>>How could it possibly be a composite if it doesn't break down into
>>combinations of the same things each time?
>Mike, you are not looking at the FINAL decay particles. Pions decay into a
>neutrino and muon. The muon in turn decays into an electron and two neutrinos.
>You will find that you end up with the same number of electrons and neutrinos
>regardless of the branching type.
Nope. The final number of electrons, neutrinos etc. vary, depending on
the decay modes.
Compare, for example, the simplest decay (tau- --> vt + ve(bar) + e-) to a
more complicated decay (tau- --> vt + pi+ + pi- + pi+ + pi- + pi-) and all
the decay products of all those pi mesons.
>>Even if we accept Thomas's flawed chart, there is an empty space for the
>>muon if we match up neutrinos and non neutrinos in columns and
>>electron/muon in rows like the standard model chart, so he "predicts" the
>>muon although his mystery cube diagrams apparently don't allow for it.
>>
> Mike the cube diagrams do allow the composite muon. Chapter 9 shows the VPP
>pion and muon models.
But it should be a fundamental particle, as the others are fundamental.
>Turns out if you add the two muon type neutrino vectors together with an
>electrons vectors the combination forms a virtual electron (muon) without
>charge conjugation (See Figure 9.1) SEE also also:
>http://www.members.aol.com/tnlockyer/muon.gif
More undecipherable nonsense. (Exception: you seem to think the muon is
FrediFizzx
news:H1CyA2...@world.std.com...
| thoma...@aol.com (ThomasL283) writes:
|
| >>Your mistake is thinking of these as if they were macroscopic objects.
| >>You think of the muon as an egg, break an egg and you have a yolk and an
| >>egg white so the yolk and white were always in there to begin with.
|
| >>A muon can decay because there is a combination (electron, mu neutrino,
| >>electron antineutrino) that exists that fulfills these requirements.
|
| >Where do they exist? You either have to create them at the time of
decay, or
with.
|
| No, being created at the time of decay is not the same as previously
| existing. Before decay, decay products simply don't exist. After they
| do.
|
| >Either way, Mike, the result is the same.
|
| Nope.
Why aren't the result the same?
| >Trouble is with your method nature has to create particles in singles,
and we
| >know nature requires doublets.
|
| Not quite true. You often see pair production from gammas because of the
| requirement that quantum constants must be conserved, and you start with
| most (charge, lepton number etc.) being zero so the most likely scenario
| is X and Anti-X. But start with, say, a neutron, and you don't get pair
| production. (You sort of do with the electron & antineutrino to conserve
| lepton number)
|
| The muon decay is almost the same, the product muon neutrino conserves
| the muon quantum number, the electron conserves the charge, the electron
| antineutrino cancels out the electron's lepton number, so the product
| quantum numbers are all the same as the original.
This brings up a question I have had about VPP for Tom. The standard model
has the muon decaying into an electron, a muon neutrino and an electron
neutrino but yet VPP has it an electron, a muon neutrino and a muon
anti-neutrino. Tom, shouldn't this last one be an electron anti-neutrino?
Is it possible in VPP for this to happen? How could this possibly be
reconciled with the standard model if not an electron anti-neutrino?
| >>As for the muon being fundamental, consider that if you calculate the
| >>"Bohr Magneton" for the muon (use Mu instead of Me in calculation) you
get
| >>a value very close to, but not exactly, the muon's magnetic moment, just
| >>like what happens with the electron.
|
| >Exactly, the VPP muon is a collapsed (small crossection) electron (or
positron)
| >by the action of combining the muon neutrino structures.
|
| Gibberish. If there was an electron in there you'd see it's magnetic
| moment, not that from the muon's mass. Or, I suppose you say it's a
| coincidence that the total magnetic moment just happens to match its mass.
No, it is not a coincidence. I believe the wave geometry between the VPP
electron and the VPP muon neutrinos is interferring and collapsing (or maybe
changing is a better word) the active "points". This is making the two
current rings responsible for the muon magneton smaller thus making the
magneton smaller than the electron's. Tom has shown this diagrammatically
in his book but has never shown any math for it. I sure would like to see
the math behind the electron's "collapse" into a muon. I think the math can
be done thru wave interference mechanics. And this also goes back to Tom's
argument that size doesn't matter for charge e but does matter for
magnetons. The muon is another good example of that. So it is not
impossible for the muon to be a composite particle constructed of an
electron and muon neutrinos if we consider that the VPP electron and VPP
muon neutrinos are wave structures. They *have* to be to be able to explain
this. Same thing with the VPP proton.
| >The tau is composed of its decay particles.
|
| >>How could it possibly be a composite if it doesn't break down into
| >>combinations of the same things each time?
|
| >Mike, you are not looking at the FINAL decay particles. Pions decay into
a
| >neutrino and muon. The muon in turn decays into an electron and two
neutrinos.
|
| >You will find that you end up with the same number of electrons and
neutrinos
| >regardless of the branching type.
|
| Nope. The final number of electrons, neutrinos etc. vary, depending on
| the decay modes.
|
| >>Even if we accept Thomas's flawed chart, there is an empty space for the
| >>muon if we match up neutrinos and non neutrinos in columns and
| >>electron/muon in rows like the standard model chart, so he "predicts"
the
| >>muon although his mystery cube diagrams apparently don't allow for it.
| >>
|
| > Mike the cube diagrams do allow the composite muon. Chapter 9 shows
the VPP
| >pion and muon models.
|
| But it should be a fundamental particle, as the others are fundamental.
Doesn't the muon spontaneously decay? As well as the tau? You would think
that fundamental particles would not do this. Yes, yes, I know all about
how they can do this via QFT so you don't need to mention that. This is one
of the strange parts that is hard to accept.
| >Turns out if you add the two muon type neutrino vectors together with an
| >electrons vectors the combination forms a virtual electron (muon)
without
| >charge conjugation (See Figure 9.1) SEE also also:
|
| >http://www.members.aol.com/tnlockyer/muon.gif
|
| More undecipherable nonsense. (Exception: you seem to think the muon is
| an electron + muon neutrino +_ muon antineutrino. This has been
| disproven)
Well, it is not totally undecipherable if you have the book. Again, I would
like to see the math for it and reconcile why the quantum numbers are being
violated here with VPP. Tom, are you sure that a composite muon can't be
constructed with an electron anti-neutrino?
FrediFizzx
Michael Moroney
>| >Where do they exist? You either have to create them at the time of
>decay, or
>| >more simply, consider that they exist in the composite structure to begin
>with.
>|
>| No, being created at the time of decay is not the same as previously
>| existing. Before decay, decay products simply don't exist. After they
>| do.
>|
>| >Either way, Mike, the result is the same.
>|
>| Nope.
>Why aren't the result the same?
Because the muon and other particles wouldn't behave as a fundamental
particle if it was a composite.
OK, maybe the end results are the same, but the starting points are not.
>| Gibberish. If there was an electron in there you'd see it's magnetic
>| moment, not that from the muon's mass. Or, I suppose you say it's a
>| coincidence that the total magnetic moment just happens to match its mass.
>No, it is not a coincidence. I believe the wave geometry between the VPP
>electron and the VPP muon neutrinos is interferring and collapsing (or maybe
>changing is a better word) the active "points". This is making the two
>current rings responsible for the muon magneton smaller thus making the
<snip>
More gibberish, this makes zero sense. Also I find it odd that you seem
so eager to defend your hero when you questioned him on that very point
just a moment ago.
>| But it should be a fundamental particle, as the others are fundamental.
>Doesn't the muon spontaneously decay? As well as the tau? You would think
>that fundamental particles would not do this.
The problem is you are thinking of these 'particles' as particles,
physical solid little balls of matter (or eggs). They are more
mathematical objects than anything else. Also the 'fundamental' being
eternal fails, as follows: Doesn't a free neutron spontaneously decay into
a proton, electron and antineutrino? Then it must be composed of those
objects, I guess. Now, don't certain protons (as in F-18) spontaneously
into a neutron, positron and neutrino? I guess the proton is composed of
a positron, neutrino and neutron (which is composed of an electron,
antineutrino and a proton -- OOPS!!)
And don't bring up Thomas' extremely broken B+ decay physics, please.
-Mike
FrediFizzx
news:H1DJxJ...@world.std.com...
Wave interference makes no sense? What makes zero sense to me is how a
particle that maybe has zero dimension according to the SM can have a
magnetic moment at all. The only way to explain it is by a rotating wave
structure that has "active points". Any time you try to measure or detect
this particle with a photon, you will only see these zero dimension "active
points". However, since it is rotating at c, it sets up its electric and
magnetic force fields which are basically sustained or always there and are
not at a particular point but are spread out in space. I was merely
wondering if it is impossible in VPP to actually have the electron
anti-neurtrino in the muon structure. If so, then the quantum number for
the muon must be the same as the electron and is wrong in the SM. Same
would go for the tau.
| >| But it should be a fundamental particle, as the others are fundamental.
|
| >Doesn't the muon spontaneously decay? As well as the tau? You would
think
| >that fundamental particles would not do this.
|
| The problem is you are thinking of these 'particles' as particles,
| physical solid little balls of matter (or eggs). They are more
| mathematical objects than anything else. Also the 'fundamental' being
| eternal fails, as follows: Doesn't a free neutron spontaneously decay into
| a proton, electron and antineutrino? Then it must be composed of those
| objects, I guess. Now, don't certain protons (as in F-18) spontaneously
| into a neutron, positron and neutrino? I guess the proton is composed of
| a positron, neutrino and neutron (which is composed of an electron,
| antineutrino and a proton -- OOPS!!)
| And don't bring up Thomas' extremely broken B+ decay physics, please.
No, I don't think of these as "hard" solid-like particles at all, Mike. You
know better than that. I think of them as "soft" wave structures. The wave
structure of the electron can actually pass right thru the wave structure of
another electron, proton, muon, etc. in accordance to what their E and M
forces will allow. The only thing "hard" about them is the E and M forces
that they generate. IOW, what Tom has discovered here is the start of the
*real* wave mechanics of the electron, etc. I just wish I was sharper to
get it all figured out faster.
I don't think your analogy with the neutron and proton is quite the same.
They are known to be composite particles. I don't think I would say the
proton is "decaying" into a neutron. This can only happen in an atom or
nucleus. Never when the proton is free. But free muons and taus will
spontaneously decay. As do all other composite particles except the proton.
So that is really the big mystery. What makes it the *only* stable
composite particle when free?
I didn't bring up B+ decay, you did. But I actually don't think it is
broken all that bad.
FrediFizzx
ThomasL283
>Date: 8/24/2002 10:08 AM Pacific Daylight Time
Wrote in:
>Message-id: <H1DAr2...@world.std.com>
>thoma...@aol.com (ThomasL283) writes:
>
>>>A muon can decay because there is a combination (electron, mu neutrino,
>>>electron antineutrino) that exists that fulfills these requirements.
>
Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
neutrino) is a wrong theory, made to explain never seeing mu neutrino pair
anihilation..
The VPP reason that we never see anihilation between muon type neutrino and
anti-muon type neutrino, is that (counter to present belief) neutrinos
apparently can't move together when free.
(neutrinos have NO charge).
Also, the VPP model clearly shows that it takes the vectors or a pair of muon
type neutrinos summing with the electron's vectors, to create the muon.
http://www.members.aol.com/tnlockyer/muon.gif
This is the basis for VPP naming the one pair of automatically produced
neutrino models as the muon type neutrinos.
http://www.members.aol.com/tnlockyer/cfives.gif
(Note the electron type neutrino is it's own anti particle by rotation of 180
degrees about any axis, making three pair of leptons. PERIOD)
The VPP models have two many coincidences to be wrong.
Remember the VPP model gives the proton and neutron mass, within 3ppm, by
combining and scaling electrons and electron type neutrinos.
And the neutron model gives the neutron decay electron and neutrino particles,
and their mass contribution directly from model scaling.
(Compare this VPP result to the SM electroweak theory results. No contest!)
ThomasL283
>Date: 8/25/2002 12:09 AM Pacific Daylight Time
Wrote in:
>Message-id: <Ho%99.404$0l5.38...@newssvr14.news.prodigy.com>
>snip<> I was merely
>wondering if it is impossible in VPP to actually have the electron
>anti-neurtrino in the muon structure. If so, then the quantum number for
>the muon must be the same as the electron and is wrong in the SM. Same
>would go for the tau.
>
Fredi: As I wrote to Mike, the muon standard model with mixed pair of
neutrinos is wrong. The reason for the theory was to explain why we never see
the muon's neutrinos anihilating each other.
That was the reason for postulating a muon neutrino. Turns out there is a
muon type neutrino, but Brookhaven (1962) did not detect it with their
experiment. (See page 30, 31, 32 in the first VPP book.).
Actually the reason we never see the neutrino anihilate with it's mate is that
neutrinos are NOT CHARGED. Thus there is no attraction.
Mike said:
>Also the 'fundamental' being
>| eternal fails, as follows: Doesn't a free neutron spontaneously decay into
>| a proton, electron and antineutrino? Then it must be composed of those
>| objects, I guess. Now, don't certain protons (as in F-18) spontaneously
>| into a neutron, positron and neutrino? I guess the proton is composed of
>| a positron, neutrino and neutron
>(which is composed of an electron,
>| antineutrino and a proton -- OOPS!!)
>| And don't bring up Thomas' extremely broken B+ decay physics, please.
Come on Mike, the positron comes from pair production, within certain unstable
nuclei. The electron is captured, changing the proton into a neutron, if
fact, read the literature, (EC) and (B+decay) are competing processes.
Mike, I thought you and Jim C. understood that. I even gave you the math. You
never addressed the math. Here it is again. And it proves that the (EC) solar
reaction does NOT produce neutrinos.
http://www.members.aol.com/tnlockyer/Betadecay.gif
>This can only happen in an atom or
>nucleus. Never when the proton is free. >But free muons and taus will
>spontaneously decay. As do all other composite particles except the proton.
>So that is really the big mystery. What makes it the *only* stable
>composite particle when free?
The reason VPP gives for the stable proton is the development of electrostatic
forces between conjugating layers.
The reason VPP gives for the muon decay is the lack of electrostatic forces to
hold the muon neutrinos in the structure.
In fact the muon neutrinos account for all decaying particles (except the
decaying neutron with has an extra electron and neutrino).
When the neutron is held in nuclei, the energy of it's decay is not sufficient
to replace the binding energy, making the neutron stable against decay in
stable nuclei.
For example, the deuteron (proton and neutron) requires ~2.2MeV whilst the
decay of the neutron would only produce about 0.78MeV to 1.53MeV. Not enough!
FrediFizzx
news:20020825134339...@mb-ca.aol.com...
| >mor...@world.std.spaamtrap.com (Michael Moroney)
| >Date: 8/24/2002 10:08 AM Pacific Daylight Time
| Wrote in:
| >Message-id: <H1DAr2...@world.std.com>
|
| >thoma...@aol.com (ThomasL283) writes:
|
| >
| >>>A muon can decay because there is a combination (electron, mu neutrino,
| >>>electron antineutrino) that exists that fulfills these requirements.
| >
|
| Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
| neutrino) is a wrong theory, made to explain never seeing mu neutrino
pair
| anihilation..
I am not quite following your logic here. I thought that the use of a mixed
doublet was to balance out the quantum numbers. However, if the muon is
indeed a composite, then it shouldn't have its own quantum number to start
with? That the muon has its own quantum number does seem to be somewhat of
an "ad hoc" assumption in the SM.
| The VPP reason that we never see anihilation between muon type neutrino
and
| anti-muon type neutrino, is that (counter to present belief) neutrinos
| apparently can't move together when free.
| (neutrinos have NO charge).
|
| Also, the VPP model clearly shows that it takes the vectors or a pair of
muon
| type neutrinos summing with the electron's vectors, to create the muon.
|
| http://www.members.aol.com/tnlockyer/muon.gif
|
| This is the basis for VPP naming the one pair of automatically produced
| neutrino models as the muon type neutrinos.
Tom, I think you need to explain this better or more fully. The VPP muon
structure is something that I have never quite fully understood from your
pictograms. Let's take it vector by vector if we have to.
FrediFizzx
Michael Moroney
>>>>A muon can decay because there is a combination (electron, mu neutrino,
>>>>electron antineutrino) that exists that fulfills these requirements.
>>
>Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
>neutrino) is a wrong theory, made to explain never seeing mu neutrino pair
>anihilation..
Where did you dream that up from? The neutrino types are the way they are
to conserve lepton numbers, and have been confirmed by cosmic ray
experiments (muon decays). The standard model doesn't have a problem with
neutrinos annihilating with their antiparticle, but since they can only
react via the weak force (and gravity I guess) the cross section is so
absurdly small it would be impossible to detect. Also if they don't have
significant kinetic energy there's not much for them to annihilate to,
other than another pair of neutrinos. (dunno if they can produce photons)
>The VPP reason that we never see anihilation between muon type neutrino and
>anti-muon type neutrino, is that (counter to present belief) neutrinos
>apparently can't move together when free.
Well being fermions, two of them cannot have exactly the same state.
>Also, the VPP model clearly shows that it takes the vectors or a pair of muon
>type neutrinos summing with the electron's vectors, to create the muon.
>http://www.members.aol.com/tnlockyer/muon.gif
Violates lepton number conservation, other than that that image is
indecipherable. Also why hasn't this "exotic" been detected?
>This is the basis for VPP naming the one pair of automatically produced
>neutrino models as the muon type neutrinos.
Explain why cosmic ray experiments disagree.
>(Note the electron type neutrino is it's own anti particle by rotation of 180
>degrees about any axis, making three pair of leptons. PERIOD)
Explain why B- (anti)neutrinos take place in different reactions than B+
neutrinos.
>The VPP models have two many coincidences to be wrong.
I corrected your sentence here.
The VPP models have two many flaws to be correct. You're welcome.
-Mike
Michael Moroney
> >Also the 'fundamental' being
>>| eternal fails, as follows: Doesn't a free neutron spontaneously decay into
>>| a proton, electron and antineutrino? Then it must be composed of those
>>| objects, I guess. Now, don't certain protons (as in F-18) spontaneously
>>| into a neutron, positron and neutrino? I guess the proton is composed of
>>| a positron, neutrino and neutron
>>(which is composed of an electron,
>>| antineutrino and a proton -- OOPS!!)
>>| And don't bring up Thomas' extremely broken B+ decay physics, please.
>Come on Mike, the positron comes from pair production, within certain unstable
>nuclei. The electron is captured, changing the proton into a neutron, if
>fact, read the literature, (EC) and (B+decay) are competing processes.
I asked Fredi not to bring up that junk, and you go ahead and do it.
>Mike, I thought you and Jim C. understood that. I even gave you the math. You
>never addressed the math. Here it is again.
It is you who never addressed any of these numerous flaws:
Violates conservation of momentum
Violates conservation of angular momentum (spin)
Violates conservation of energy/mass (atom decays by EC, emits nothing,
where did the mass/energy go?! With beta+ decay of Co58 you postulate
creation of a e+e- pair yet there's not enough energy to do so!)
Violates conservation of lepton number
Violates time-reversal symmetry (EC is time reversed beta decay)
Violates a few other things I can't remember at the moment, go back and
check with google
And of course the biggie, you compare apples to oranges (the BE of Cu64 is
not the same as the BE of Ni64, they differ by the mass difference between
the proton and neutron, and when this reveals itself in the math you try
to use it as energy somehow. You should use your logic to model the decay
of a free neutron, maybe that will help you.
Please address all these flaws.
>http://www.members.aol.com/tnlockyer/Betadecay.gif
You still list energy in volts. Fix it.
>>This can only happen in an atom or
>>nucleus. Never when the proton is free. >But free muons and taus will
>>spontaneously decay. As do all other composite particles except the proton.
>>So that is really the big mystery. What makes it the *only* stable
>>composite particle when free?
The proton and neutron are both fundamental, at least as long as you stay
above the quark level. A *very* simple rule is: Any particle X can decay
to another set of particles as long as: 1) All of several fundamental
conserved constants (such as charge and spin) are conserved and 2) The
total mass of the products is less than the mass of X (the difference
becoming kinetic energy). It works for a free neutron as well as a bound
proton in F-18 (its mass contribution exceeds that of the corresponding
bound neutron in O-18)
-Mike
Michael Moroney
>| More gibberish, this makes zero sense. Also I find it odd that you seem
>| so eager to defend your hero when you questioned him on that very point
>| just a moment ago.
>Wave interference makes no sense?
What you wrote makes no sense. Gibberish.
> What makes zero sense to me is how a
>particle that maybe has zero dimension according to the SM can have a
>magnetic moment at all.
Why? The Bohr Magneton equation doesn't have length or radius as one of
its parameters. Despite your protests, you're still trying to visualize
all this as a little (rotating) solid ball.
>points". However, since it is rotating at c,
Correction: It rotates at sqrt(2)*c. No explanation of this faster-than-light
travel was ever offered.
>No, I don't think of these as "hard" solid-like particles at all, Mike. You
>know better than that. I think of them as "soft" wave structures. The wave
You say so, yet you still cling to the physical object description that
if A decays into B and C, A must be made of B and C.
>I don't think your analogy with the neutron and proton is quite the same.
>They are known to be composite particles.
Nope. (Unless you're talking of the quarks they're made of)
> I don't think I would say the
>proton is "decaying" into a neutron.
Why not?
> This can only happen in an atom or
>nucleus. Never when the proton is free.
So what? A He-4 nucleus never decays, and it has two neutrons, does this
mean neutrons don't decay? Nope. That is the reverse argument.
> But free muons and taus will
>spontaneously decay. As do all other composite particles except the proton.
>So that is really the big mystery. What makes it the *only* stable
>composite particle when free?
The proton is exactly as fundamental as the neutron, or pi+ or xi-.
Why does the tau and muon decay? I think the answer is the same as the
answer to the question why does a dog lick its balls.
Maybe if you look as a decay as a reaction between the particle and
virtual particle pairs. A muon reacts with a (virtual) muon antineutrino
to produce a virtual W-, which immediately decays into an electron and
electron antineutrino. The muon neutrino is the partner of the muon
antineutrino, its partner is gone so it becomes real.
>| And don't bring up Thomas' extremely broken B+ decay physics, please.
>I didn't bring up B+ decay, you did. But I actually don't think it is
>broken all that bad.
Come on! I've never seen so much bad physics crammed into one single
document! Besides I brought up B+ decay to show the absurdity of the
(If A decays into B and C, A must be composed of B and C) argument.
-Mike
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| More gibberish, this makes zero sense. Also I find it odd that you
seem
| >| so eager to defend your hero when you questioned him on that very point
| >| just a moment ago.
|
| >Wave interference makes no sense?
|
| What you wrote makes no sense. Gibberish.
Sorry you are too blind to see something so simple. It is a thing of total
beauty.
| > What makes zero sense to me is how a
| >particle that maybe has zero dimension according to the SM can have a
| >magnetic moment at all.
|
| Why? The Bohr Magneton equation doesn't have length or radius as one of
| its parameters. Despite your protests, you're still trying to visualize
| all this as a little (rotating) solid ball.
Wrong. h bar certainly does have units of length in it. The units for the
Bohr magneton is amps*meters^2. OK, you tell me. How exactly do you
propose to get a magnetic moment with no dimension of length? It is
physically impossible. Once again, there is nothing "solid" about what I am
visualizing. It is a rotating wave structure. And is a cylinder shape when
rotating; not a ball or sphere. And it is not so little as to have zero
dimension. There is absolutely no such thing as a point or point-like
particle. This is not difficult to provide a good example of why. Which I
already have and you have already commented on.
| >points". However, since it is rotating at c,
|
| Correction: It rotates at sqrt(2)*c. No explanation of this
faster-than-light
| travel was ever offered.
Yes there was. You just missed it. There is no mass involved at any
instantaneous point for one. It is just a wave representation. It is only
the cube corners that are rotating at the sqrt(2)*c. Since this thing is
just a wave structure, it shouldn't be a problem.
| >No, I don't think of these as "hard" solid-like particles at all, Mike.
You
| >know better than that. I think of them as "soft" wave structures. The
wave
|
| You say so, yet you still cling to the physical object description that
| if A decays into B and C, A must be made of B and C.
So so what? I don't see the connection you are making to what I said about
them being wave structures?
| >I don't think your analogy with the neutron and proton is quite the same.
| >They are known to be composite particles.
| Nope. (Unless you're talking of the quarks they're made of)
Do you think maybe next time you could hack a message up more so that
nothing makes any sense at all? Is it too hard to keep your train of
thought thru more than a couple of sentences? Please reply with paragraphs
to paragraphs.
| > I don't think I would say the
| >proton is "decaying" into a neutron.
|
| Why not?
I would say that it is changing in to a neutron. Not decaying into a
neutron. Decaying is the wrong word. Protons don't decay.
| > This can only happen in an atom or
| >nucleus. Never when the proton is free.
|
| So what? A He-4 nucleus never decays, and it has two neutrons, does this
| mean neutrons don't decay? Nope. That is the reverse argument.
This is not an argument at all much less a reverse argument.
| > But free muons and taus will
| >spontaneously decay. As do all other composite particles except the
proton.
| >So that is really the big mystery. What makes it the *only* stable
| >composite particle when free?
|
| The proton is exactly as fundamental as the neutron, or pi+ or xi-.
When did the proton become a fundamental particle? Not even the SM has it
as that. Or the neutron or pi+ or xi-.
| Why does the tau and muon decay? I think the answer is the same as the
| answer to the question why does a dog lick its balls.
Or chicken pot pie, that is why. Most everything decays to get to a lower
possible energy level. That concept will never change.
| Maybe if you look as a decay as a reaction between the particle and
| virtual particle pairs. A muon reacts with a (virtual) muon antineutrino
| to produce a virtual W-, which immediately decays into an electron and
| electron antineutrino. The muon neutrino is the partner of the muon
| antineutrino, its partner is gone so it becomes real.
Or the quantum number of the muon should be the same as an electron and it
decays to an electron, muon neutrino and muon antineutrio. Everything is
still conserved in this scenario also. What is the proof that a muon should
have its own quantum number? Answer me that. It is an "ad hoc" assumption
as far as I can tell.
FrediFizzx
ThomasL283
Somewhere in this news feed:
>mor...@world.std.spaamtrap.com (Michael Moroney)
Wrote:
>>>| objects, I guess. Now, don't certain protons (as in F-18) spontaneously
>>>| into a neutron, positron and neutrino? I guess the proton is composed of
>>>| a positron, neutrino and neutron
>>>(which is composed of an electron,
>>>| antineutrino and a proton --
>OOPS!!)
>>>| And don't bring up Thomas' extremely broken B+ decay physics, please.
>
>>Come on Mike, the positron comes from pair production, within certain
>unstable
>>nuclei. The electron is captured, changing the proton into a neutron, if
>>fact, read the literature, (EC) and (B+decay) are competing processes.
>
>>Mike, I thought you and Jim C. understood that. I even gave you the math.
>You
>>never addressed the math. Here it is again.
>
>It is you who never addressed any of these numerous flaws:
>Violates conservation of energy/mass (atom decays by EC, emits nothing,
> where did the mass/energy go?
You did not read the math. The proton must absorb the exrtra mass energy to
become a neutron.
The light weight proton cannot decay into a positron and neutrino, to make a
heavy weight neutron.
It is childish to think you can make something heavier (neutron) by removing
positron mass from a lighter (proton). Besides, to create a positron, you must
also create an electron.
(EC) and (B+) create the same daughter nucleus, except (B+) requires there be
an extra ~2 me c^2 energy to create an electron and positron pair. The
positron is emitted and the electron is captured.
>With beta+ decay of Cu64 (sic) you postulate
> creation of a e+e- pair yet there's not enough energy to do so!)
Yes there is Mike. BechgNi64 = 2.457439 MeV. You only need about 1.02 MeV to
create the e-e+ pair.
http://www.members.aol.com/tnlockyer/Betadecay.gif
>Violates time-reversal symmetry (EC is time reversed beta decay)
On the contrary, (EC) verifies time reversal symmetry. That was one of my
orginal arguments, see Google.
>And of course the biggie, you compare apples to oranges (the BE of Cu64 is
>not the same as the BE of Ni64, they differ by the mass difference between
>the proton and neutron, and when this reveals itself in the math you try
>to use it as energy somehow.
Mike, look at the math again. Remember that BE (binding energy) is LOST
energy.
In the variables list of the (Betadecay.gif) the atomic mass energy is
obtained by subtracting the published binding energy (Ni64be) from the sum of
all proton, neutron and electron masses for the (Ni64) atom.
>You still list energy in volts. Fix it.
Yes, that is non conventional, but I find it convenient to use "equivalent
volts" which is the English acronym for (eV).
By using equivalent volts, equations can be set up exclusively in the basic
units of MKSA.
With exclusively using basic units, here is no conversion to secondary units
until multiplying (equivalent volts) by a normal (1.602176462E-19 As) to get
Joules.
Just good physics (or good engineering) in my view!
Michael Moroney
>| >Wave interference makes no sense?
>|
>| What you wrote makes no sense. Gibberish.
>Sorry you are too blind to see something so simple. It is a thing of total
>beauty.
Actually, it's quite ugly, as it misleads potential future physicists or
those who just want to learn about particle physics.
>| > What makes zero sense to me is how a
>| >particle that maybe has zero dimension according to the SM can have a
>| >magnetic moment at all.
>|
>| Why? The Bohr Magneton equation doesn't have length or radius as one of
>| its parameters. Despite your protests, you're still trying to visualize
>| all this as a little (rotating) solid ball.
>Wrong. h bar certainly does have units of length in it.
No, the Bohr magneton equation doesn't have any radius term in it. If
you need to be reminded, it is e*hbar/(2*Me). Any radius will work,
including zero, since it doesn't factor in the equation at all.
The units of hbar is irrelevant. All the various physical constants in the
MKSA system contain some or all of the four fundamental dimensional units,
most will have length in there somewhere.
>The units for the Bohr magneton is amps*meters^2.
So what?
> OK, you tell me. How exactly do you
>propose to get a magnetic moment with no dimension of length?
Again, despite your protests, you're still stuck in the mindset of a little
physical spinning object, moving a little charge in a circle to create
a little electromagnet.
Remember Maxwell's equations. Magnetic fields can be created without
moving charges.
>There is absolutely no such thing as a point or point-like
>particle.
"Point-like" simply means it behaves as a point to the best of our
measurement abilities. With current technology, the electron *is*
point-like, it behaves exactly as a point does to the best of our
measuring abilities. Maybe some day we will find the electron does
in fact have a nonzero radius. But the fact remains the electron is
already *known* to be much smaller than the vpp "model".
>| >points". However, since it is rotating at c,
>|
>| Correction: It rotates at sqrt(2)*c. No explanation of this
>faster-than-light
>| travel was ever offered.
>Yes there was. You just missed it. There is no mass involved at any
>instantaneous point for one. It is just a wave representation. It is only
>the cube corners that are rotating at the sqrt(2)*c. Since this thing is
>just a wave structure, it shouldn't be a problem.
Yes it is a problem. Nothing, not matter, nor waves, nor even information
can travel at greater than c. Period.
>| You say so, yet you still cling to the physical object description that
>| if A decays into B and C, A must be made of B and C.
>So so what? I don't see the connection you are making to what I said about
>them being wave structures?
You think of them like physical objects, that is, if A decays into B and
C, A must be made of B and C. How can I make this any clearer.
>| >I don't think your analogy with the neutron and proton is quite the same.
>| >They are known to be composite particles.
>| Nope. (Unless you're talking of the quarks they're made of)
>Do you think maybe next time you could hack a message up more so that
>nothing makes any sense at all? Is it too hard to keep your train of
>thought thru more than a couple of sentences? Please reply with paragraphs
>to paragraphs.
The only mistake is a stray "|" that got inserted before my reply, and
maybe quoting the first sentence is unnecessary. I'll repeat below:
: They [proton and neutron] are known to be composite particles.
Nope. (Unless you're talking of the quarks they're made of)
>| > I don't think I would say the
>| >proton is "decaying" into a neutron.
>|
>| Why not?
>I would say that it is changing in to a neutron. Not decaying into a
>neutron. Decaying is the wrong word. Protons don't decay.
Physicists use "decay" to mean a single object (nucleus, particle)
spontaneously changing (without any external provocation) into something
else at a lower energy state. You can say the F-18 nucleus as a whole
decays, or a proton within it does.
>| > This can only happen in an atom or
>| >nucleus. Never when the proton is free.
>|
>| So what? A He-4 nucleus never decays, and it has two neutrons, does this
>| mean neutrons don't decay? Nope. That is the reverse argument.
>This is not an argument at all much less a reverse argument.
The reverse argument is a bound neutron in a He-4 nucleus lasts forever,
therefore a free neutron lasts forever. This is the reverse (swapping
bound and free, and proton for neutron) argument.
>| The proton is exactly as fundamental as the neutron, or pi+ or xi-.
>When did the proton become a fundamental particle? Not even the SM has it
>as that. Or the neutron or pi+ or xi-.
So you now believe in the quark model? OK.
>| Why does the tau and muon decay? I think the answer is the same as the
>| answer to the question why does a dog lick its balls.
>Or chicken pot pie, that is why. Most everything decays to get to a lower
>possible energy level. That concept will never change.
Yup. If it can decay to a lower energy level, it will. As long as certain
conserved quantum variables are conserved.
>| Maybe if you look as a decay as a reaction between the particle and
>| virtual particle pairs. A muon reacts with a (virtual) muon antineutrino
>| to produce a virtual W-, which immediately decays into an electron and
>| electron antineutrino. The muon neutrino is the partner of the muon
>| antineutrino, its partner is gone so it becomes real.
>Or the quantum number of the muon should be the same as an electron and it
>decays to an electron, muon neutrino and muon antineutrio. Everything is
>still conserved in this scenario also. What is the proof that a muon should
>have its own quantum number? Answer me that. It is an "ad hoc" assumption
>as far as I can tell.
Because if that were true, we would see decays of muon -> electron + electron
neutrino + electron antineutrino (because it could) and muon -> electron +
electron neutrino + muon antineutrino, and and muon -> electron + muon
neutrino + electron antineutrino, and probably modes involving the tau
neutrino. We *only* see muon -> electron + muon neutrino + electron
antineutrino, and none of the other combinations. In particular we *don't*
see muon -> electron + muon neutrino + muon antineutrino.
-Mike
ThomasL283
>Date: 8/26/2002 11:53 AM Pacific Daylight Time
Wrote:
>>>>>A muon can decay because there is a combination (electron, mu neutrino,
>>>>>electron antineutrino) that exists that fulfills these requirements.
>>>
>
>>Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
>>neutrino) is a wrong theory, made to explain never seeing mu neutrino pair
>>anihilation..
>Where did you dream that up from?
The original idea that there was a muon type neutrino used that same argument
(circa 1962).
Got their grant money to try to detect the muon neutrino, on the basis that (ve
+ v_e) never was seen to annihilate into photons.
>>Also, the VPP model clearly shows that it takes the vectors or a pair of
>muon
>>type neutrinos summing with the electron's vectors, to create the muon.
>
>>http://www.members.aol.com/tnlockyer/muon.gif
>
>Violates lepton number conservation, other than that that image is
>indecipherable.
No, it does not violate anything, Mike.
And yes, the combining of vectors of electron and muon neutrinos is hard to
decipher.
The model requires being able to visualize spacial relationships. I have to
draw pictures, and then sometimes mess up.
> Also why hasn't this "exotic" been detected?
>
The exotic (as VPP shows it) occurs by combining one muon neutrino with the
electron.
The result was labeled "exotic" because the particle seemed to be strange and
different, with only one rotating face, and the rest of the vectors being of
the same kind.
But notice the exotic gets well in a hurry when a conjugate muon neutrino
model vectors are added, giving virtual electron (or positron) model
structures, as the muon models.
>>This is the basis for VPP naming the one pair of automatically produced
>>neutrino models as the muon type neutrinos.
>Explain why cosmic ray experiments disagree.
Explain what you mean?
>>(Note the electron type neutrino is it's own anti particle by rotation of
>180
>>degrees about any axis, making three pair of leptons. PERIOD)
>Explain why B- (anti)neutrinos take place in different reactions than B+
>neutrinos.
There are no neutrinos from (B+) decays, so your question is moot.
>>The VPP models have two many coincidences to be wrong.
>
>I corrected your sentence here.
>
>The VPP models have two many flaws to be correct. You're welcome.
Thanks but no thanks. ;-{)
Michael Moroney
>>Violates conservation of energy/mass (atom decays by EC, emits nothing,
>> where did the mass/energy go?
>You did not read the math. The proton must absorb the exrtra mass energy to
>become a neutron.
Not what I asked. Cu-64 (mass 63.929767u) decays by EC into Ni-64 (mass
63.927969u) You claim no neutrino is emitted. Where did the mass
difference (0.001798u or 1.675 MeV energy) go?
Don't try to repeat your arguments how EC decays work, treat this as a
"black box" question. I have a black box with a Cu-64 atom in it. It
decays by EC into Ni-64. Nothing leaves the box. It now has less
mass. Why?
>The light weight proton cannot decay into a positron and neutrino, to make a
>heavy weight neutron.
The light weight proton decays by B+ and neutrino into an even lighter
weight neutron. Remember, the neutron is bound and has released binding
energy so it does not have the same mass as a free neutron. (same for the
proton, but less so)
> It is childish to think ...
It is childish to try to discuss physics by stating a statement about
physics is childish. Means you can't come up with a valid counterargument,
I guess.
>>With beta+ decay of Cu64 (sic) you postulate
>> creation of a e+e- pair yet there's not enough energy to do so!)
Don't change what I wrote. I wrote Co-58, not Cu-64. (But I misremembered
the isotope, I meant Ni-59 (B+) --> Co-59) Not enough energy to create
an e+e- pair in that decay. In fact, very little left over after creating
the positron. See Google.
>>Violates time-reversal symmetry (EC is time reversed beta decay)
>On the contrary, (EC) verifies time reversal symmetry. That was one of my
>orginal arguments, see Google.
Nope. You don't have a neutrino in EC decay. Remember, in B- decay, an
antineutrino is emitted. For time reversal we'd need either an antineutrino
to also be absorbed or, equivalently, a neutrino emitted.
>>And of course the biggie, you compare apples to oranges (the BE of Cu64 is
>>not the same as the BE of Ni64, they differ by the mass difference between
>>the proton and neutron, and when this reveals itself in the math you try
>>to use it as energy somehow.
>Mike, look at the math again. Remember that BE (binding energy) is LOST
>energy.
I'm not talking about the math. I'm talking about this: Ni-64 + its BE =
28 protons+electrons + 36 neutrons. But Cu-64 + its BE = 29 protons+electrons
and 35 neutrons. Different starting points, thus apples and oranges.
These differ in mass by (n-1H). See the Decay Calculator at:
http://nucleardata.nuclear.lu.se/Database/masses/ to see how that page
corrects for the mass difference.
Without that mass difference, all your calculations will be "off" by 0.782 MeV
which you incorrectly claim is energy.
Also, what about all the other flaws (no conservation of spin, etc.) in
your "theory"?
>>You still list energy in volts. Fix it.
>Yes, that is non conventional, but I find it convenient to use "equivalent
>volts" which is the English acronym for (eV).
You list energy in dimensionality kg m^2 s^-3 A. Energy has dimensionality
kg m^2 s^-2. Fix it.
-Mike
Michael Moroney
>>>Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
>>>neutrino) is a wrong theory, made to explain never seeing mu neutrino pair
>>>anihilation..
>>Where did you dream that up from?
>The original idea that there was a muon type neutrino used that same argument
>(circa 1962).
>Got their grant money to try to detect the muon neutrino, on the basis that (ve
>+ v_e) never was seen to annihilate into photons.
Nope. The 1962 muon neutrino experiment was to start with a stream of
pions decaying into muons+neutrinos, deflect the muons, and examine the
reactions of the neutrinos on a target. If electron neutrinos were the
same as muon neutrinos, they'd see equal numbers of electrons and muons
created. If they're different, they'd see only muons created. They saw
only muons.
>>Violates lepton number conservation, other than that that image is
>>indecipherable.
>No, it does not violate anything, Mike.
As I stated, lepton number parameters are not conserved. Explain why not.
>> Also why hasn't this "exotic" been detected?
>>
>The exotic (as VPP shows it) occurs by combining one muon neutrino with the
>electron.
<snip>
Gibberish.
>>>This is the basis for VPP naming the one pair of automatically produced
>>>neutrino models as the muon type neutrinos.
>>Explain why cosmic ray experiments disagree.
>Explain what you mean?
Examination of the type of neutrinos from muons decaying in the atmosphere
show both electron and muon types being generated, in the ratio predicted
by the u- -> vu + ve(bar) + e- decay
>>Explain why B- (anti)neutrinos take place in different reactions than B+
>>neutrinos.
>There are no neutrinos from (B+) decays, so your question is moot.
OK, explain why experiments show that neutrinos from different sources
have different kinds of reactions. Electron neutrinos from, say,
nuclear power plants (loaded with B- decaying junk) never participate in
the reaction X + Cl-37 --> Ar-37 + e- yet some neutrinos from other sources
do. These aren't muon neutrinos, either (they'd produce muons, not
electrons)
-Mike
ThomasL283
>Date: 8/27/2002 3:15 PM Pacific Daylight Time
Wrote:
>>>Violates conservation of energy/mass (atom decays by EC, emits nothing,
>>> where did the mass/energy go?
>thoma...@aol.com (ThomasL283) writes:
>>You did not read the math. The proton must absorb the exrtra mass energy to
>>become a neutron.
>
>Not what I asked. Cu-64 (mass 63.929767u) decays by EC into Ni-64 (mass
>63.927969u) You claim no neutrino is emitted. Where did the mass
>difference (0.001798u or 1.675 MeV energy) go?
It is expelled and the 0.7823 MeV is absorbed in the proton formation of the
neutron, along with the (captured) electron.
The energy leaves the reaction as a photon. (atomic energy).
http://www.members.aol.com/tnlockyer/Betadecay.gif
>>The light weight proton cannot decay into a positron and neutrino, to make a
>>heavy weight neutron.
>
>The light weight proton decays by B+ and neutrino into an even lighter
>weight neutron. Remember, the neutron is bound and has released binding
>energy so it does not have the same mass as a free neutron. (same for the
>proton, but less so)
Yes, The binding energy is actually greater for the neutron, in nuclei.
But the VPP model shows that the difference when (as in this case the proton
and neutron share the same spin axis, Bpn-Bnn=0.3325087 MeV. (See table 13.5
page 98).
Not enough to make up the 0.78 MeV (n-1H) neutrino expulsion energy.
>>>With beta+ decay of Cu64 (sic) you postulate
>>> creation of a e+e- pair yet there's not enough energy to do so!)
>
>Don't change what I wrote. I wrote Co-58, not Cu-64.
Sorry, I saw Co64 and thought you had a typo.
>(But I misremembered
>the isotope, I meant Ni-59 (B+) --> Co-59) Not enough energy to create
>an e+e- pair in that decay. In fact, very little left over after creating
>the positron. See Google.
>
Mike, notice the literature is exclusively (EC). Why? B+ and EC are competing
processes, and they both produce identical daughters.
What you see (Ni-59 -->Co59) is an exclusive (EC) process, according to the
energy you quote.
>>>Violates time-reversal symmetry (EC is time reversed beta decay)
>
>>On the contrary, (EC) verifies time reversal symmetry. That was one of my
>>orginal arguments, see Google.
>Nope. You don't have a neutrino in EC decay. Remember, in B- decay, an
>antineutrino is emitted. For time reversal we'd need either an antineutrino
>to also be absorbed or, equivalently, a neutrino emitted.
Remember that VPP verifies the Majorana theory that the electron type neutrino
is it's own anti-particle. (And you said they recently saw double beta decay
that also tents to prove the Majorana theory ).
Those old books mislabeled neutrinos in their ignorance. VPP science moves on.
>>>And of course the biggie, you compare apples to oranges (the BE of Cu64 is
>>>not the same as the BE of Ni64, they differ by the mass difference between
>>>the proton and neutron, and when this reveals itself in the math you try
>>>to use it as energy somehow.
>>Mike, look at the math again. Remember that BE (binding energy) is LOST
>>energy.
>I'm not talking about the math. I'm talking about this: Ni-64 + its BE =
>28 protons+electrons + 36 neutrons. But Cu-64 + its BE = 29
>protons+electrons
>and 35 neutrons.
Make that is MINUS the binding energy to get the atomic mass (energy). See
again:
http://www.members.aol.com/tnlockyer/Betadecay.gif
>Without that mass difference, all your calculations will be "off" by 0.782
>MeV
>which you incorrectly claim is energy.
>
Huh? See the above Betadecay.gif link.
The (n-1H) is accounted for EXACTLY when looking at the kinematics of (EC) and
(B+) decays, and is the energy absorbed, along with the captured electron, to
create a neutron from the proton, in both (EC) and (B+). Time reversal
symmetry strikes again. :-)
ThomasL283
>Date: 8/27/2002 3:49 PM Pacific Daylight Time
Wrote in:
>Message-id: <H1Iy2...@world.std.com>
>thoma...@aol.com (ThomasL283) writes:
>
>>>>Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
>>>>neutrino) is a wrong theory, made to explain never seeing mu neutrino
>pair
>>>>anihilation..
>>>Where did you dream that up from?
>
>>The original idea that there was a muon type neutrino used that same
>argument
>>(circa 1962).
>>Got their grant money to try to detect the muon neutrino, on the basis that
>(ve
>>+ v_e) never was seen to annihilate
>into photons.
>Nope. The 1962 muon neutrino experiment was to start with a stream of
>pions decaying into muons+neutrinos, deflect the muons, and examine the
>reactions of the neutrinos on a target.
So far so good:
>If electron neutrinos were the
>same as muon neutrinos, they'd see equal numbers of electrons and muons
>created.
Who said electron and muon neutrinos were the same?
Read the original paper. They shielded with battleship steel between their
accelerator, but their detector was hanging out in the cosmic ray stream.
They saw a lot of cosmic muons going horizontal (or nearly so). Dirty
experiment. They were right that there are muon type neutrinos, but thay sure
as hell did not see them.
Remember, the VPP is a model that actually gives us the structures for the
neutrinos types. Clearly the electron and muon type neutrinos are different
and (contrary to latest theory) neutrinos CANNOT oscillate between one type and
another.
http://www.members.aol.com/tnlockyer/cfives.gif
>>>Violates lepton number conservation, other than that that image is
>>>indecipherable.
>>No, it does not violate anything, Mike.
>
>As I stated, lepton number parameters are not conserved. Explain why not.
No, it is you who raised the question. Explain why you think so.
>>>Explain why cosmic ray experiments disagree.
>
>>Explain what you mean?
>Examination of the type of neutrinos from muons decaying in the atmosphere
>show both electron and muon types being generated, in the ratio predicted
>by the u- -> vu + ve(bar) + e- decay
Nope. The literature is dead wrong. Remember, the VPP model gives us the
neutrino structures.
One can model the muon, with two conjugate muon type neutros.
But, using one electron and one muon type neutrino makes non sense when
combining the e + ve + vu vectors.
Only e + vu + v_u makes a muon structure.
>>>Explain why B- (anti)neutrinos take place in different reactions than B+
>>>neutrinos.
>
>>There are no neutrinos from (B+) decays, so your question is moot.
>
>Electron neutrinos from, say,
>nuclear power plants (loaded with B- decaying junk) never participate in
>the reaction X + Cl-37 --> Ar-37 + e- yet some neutrinos from other sources
>do.
Easy, I know for a certainty that the above neutrino reaction cannot take
place.
It was theoretical from the start. Neutrinos cannot be the primary cause
for transmutation reactions, regardless of what has been written in
ignorance.
FrediFizzx
| >mor...@world.std.spaamtrap.com (Michael Moroney)
| >Date: 8/27/2002 3:49 PM Pacific Daylight Time
| Wrote in:
| >Message-id: <H1Iy2...@world.std.com>
|
| >thoma...@aol.com (ThomasL283) writes:
| >
| Remember, the VPP is a model that actually gives us the structures for
the
| neutrinos types. Clearly the electron and muon type neutrinos are
different
| and (contrary to latest theory) neutrinos CANNOT oscillate between one
type and
| another.
Tom, I have been pondering what they are seeing with this neutrino
oscillation business, and I am wondering if neutrinos that have been ejected
from the VPP proton or from a muon, etc. might behave like this? You say
that neutrinos can't move, but I think you mean that they can't move when a
photon interacts with them (they absorb or eject the photon depending on the
situation). However, a neutrino that is a decay product would be moving,
spinning, etc. due to its being coupled with an electron, etc. It couldn't
just stop on its own as that would be against Newton's law. So say we have
a bunch of these decay product neutrinos travelling around. You probably
would get some interactions once in a while when they do bump into protons,
etc. However, since they would still be spinning, etc. and they might have
absorbed energy or be at different energy levels, it will seem like they are
doing this oscillation thing because they will have a detectable mass
because of their decoupled spinning. And because of the way VPP neutrinos
work, the mass would not be consistent. So I think that what the
experiments are seeing with this oscillation business can be explained with
the VPP models. Most of the neutrinos around are probably in fact these
decoupled decay remnants.
One question that comes to mind, is what happens when one of these moving
VPP neutrinos absorbs a photon? Does that help to slow down its decoupled
spin and motion? It seems like to me that the VPP neutrino in this case
would have variable mass! If truely at rest and not moving or spinning,
etc. it would have no detectable mass. But the fact that they probably do
have some left over decoupled motion, it will seem to have variable amounts
of mass. So the VPP neutrino gets smaller by absorbing a photon. I would
imagine that this action would have to be quantized somehow also. Now since
the proton is stable there really shouldn't be many electron neutrinos
around compared to muon type neutrinos. I am not sure if you mention it in
your book, but can muon type neutrinos absorb photons? And what happens
when a proton is busted up and electron type neutrinos are ejected? The
freed up neutrinos should immediately emit their extra energy via photons.
Now we have lots of problems to figure out in this scenario dealing with
conservation. Or do we? I guess all the photons are released in the
fireball of the collision that busted up the proton to start with. So the
question is does the releasing of the photons that these neutrinos in the
proton had absorbed make the VPP neutrino "spin down"?
FrediFizzx
Michael Moroney
>>Not what I asked. Cu-64 (mass 63.929767u) decays by EC into Ni-64 (mass
>>63.927969u) You claim no neutrino is emitted. Where did the mass
>>difference (0.001798u or 1.675 MeV energy) go?
>It is expelled and the 0.7823 MeV is absorbed in the proton formation of the
>neutron, along with the (captured) electron.
If 0.7823 MeV is absorbed, why is the mass difference between Cu-64 and
Ni-64 1.675 MeV and not 0.8927 MeV?
>The energy leaves the reaction as a photon. (atomic energy).
First, that violates conservation of angular momentum.
Second, that violates conservation of lepton number.
Third, why have we _never_ seen photons of either 0.8927 Mev or 1.675 MeV?
Too many problems with VPP.
>Yes, The binding energy is actually greater for the neutron, in nuclei.
Meaning less mass.
>Not enough to make up the 0.78 MeV (n-1H) neutrino expulsion energy.
Huh? The neutrino is (nearly) massless, the minimal expulsion energy is
at most a few eVs.
>Mike, notice the literature is exclusively (EC). Why? B+ and EC are competing
>processes, and they both produce identical daughters.
>What you see (Ni-59 -->Co59) is an exclusive (EC) process, according to the
>energy you quote.
Not according to the energy. B+ is rare but exists.
What I found last time I looked the B+ decay was rare but existed. I can't
find the reference now, so I'll repeat my question for I-128 (B+) -> Te-128.
The decay energy is 1.252 MeV. See its decay mode at:
http://atom.kaeri.re.kr/cgi-bin/decay?I-128+EC .
I-128 has a mass 127.905805u Its decay produces Te-128 (127.904461u),
an extra shell electron (0.0005486u), a positron (0.0005486u), which means
the leftover mass/energy is 127.905805-127.904461-2*0.0005486 = 0.000247u
which is 0.230 MeV. There's not enough energy to form an e+e- pair,
it takes 1.022 MeV to form an e+e- pair, yet there's only 0.796 MeV.
Conclusion: No e+e- pair is formed, as it is impossible. Only the positron
(and a neutrino) is formed.
> >>>Violates time-reversal symmetry (EC is time reversed beta decay)
>>
>>>On the contrary, (EC) verifies time reversal symmetry. That was one of my
>>>orginal arguments, see Google.
>>Nope. You don't have a neutrino in EC decay. Remember, in B- decay, an
>>antineutrino is emitted. For time reversal we'd need either an antineutrino
>>to also be absorbed or, equivalently, a neutrino emitted.
>Remember that VPP verifies the Majorana theory that the electron type neutrino
>is it's own anti-particle. (And you said they recently saw double beta decay
>that also tents to prove the Majorana theory ).
No, neutrinoless double beta decay has not been seen. (If the neutrino is
its own antiparticle, neutrinoless double beta is possible. If not, it is
not possible. It hasn't been seen yet.)
Anyway, let's assume for a minute the electron neutrino is its own
antiparticle. EC decay would be time-reversed B- decay, meaning either
the nucleus absorbs both an electron and a neutrino, or it can absorb an
electron and emit the antiparticle of the neutrino, a neutrino. Your
description of EC decay fails.
>Those old books mislabeled neutrinos in their ignorance. VPP science moves on.
Moving far away from the realm of physics in to fantasy.
>>>Mike, look at the math again. Remember that BE (binding energy) is LOST
>>>energy.
>>I'm not talking about the math. I'm talking about this: Ni-64 + its BE =
>>28 protons+electrons + 36 neutrons. But Cu-64 + its BE = 29
>>protons+electrons
>>and 35 neutrons.
>Make that is MINUS the binding energy to get the atomic mass (energy). See
>again:
I am describing a theoretical process where you take a Ni-28 atom, add
energy equal to its binding energy and break it into its components.
>>Without that mass difference, all your calculations will be "off" by 0.782
>>MeV
>>which you incorrectly claim is energy.
>>
>Huh? See the above Betadecay.gif link.
Apples and oranges.
>The (n-1H) is accounted for EXACTLY when looking at the kinematics of (EC) and
>(B+) decays, and is the energy absorbed, along with the captured electron, to
No, it falls out of the mass difference between 28 protons+electrons and
36 neutrons, and 29 protons+electrons and 35 neutrons. Add up their masses
and subtract them from each other and you'll come up with 0.782 MeV.
You fail to compensate for it, it falls out of the equations and screws up
your results. Somehow you think it's some form of energy.
Now how about those other violations?
-Mike
Michael Moroney
>>>>>Actually Mike, the use of a mixed doublet, (i.e. mu neutrino and electron
>>>>>neutrino) is a wrong theory, made to explain never seeing mu neutrino
>>pair
>>>>>anihilation..
>>>>Where did you dream that up from?
>>
>>>The original idea that there was a muon type neutrino used that same
>>argument
>>>(circa 1962).
>>>Got their grant money to try to detect the muon neutrino, on the basis that
>>(ve
>>>+ v_e) never was seen to annihilate
>>into photons.
>>Nope. The 1962 muon neutrino experiment was to start with a stream of
>>pions decaying into muons+neutrinos, deflect the muons, and examine the
>>reactions of the neutrinos on a target.
>So far so good:
>>If electron neutrinos were the
>>same as muon neutrinos, they'd see equal numbers of electrons and muons
>>created.
>Who said electron and muon neutrinos were the same?
That was the purpose of the experiment, not anything with neutrino
annihilation.
>Read the original paper. They shielded with battleship steel between their
>accelerator, but their detector was hanging out in the cosmic ray stream.
>They saw a lot of cosmic muons going horizontal (or nearly so). Dirty
>experiment. They were right that there are muon type neutrinos, but thay sure
>as hell did not see them.
They saw the expected background and significant number of muons created
from neutrinos (well above background) The muons could not penetrate that
battleship armor, your statement about cosmic rays is a red herring.
>>>>Violates lepton number conservation, other than that that image is
>>>>indecipherable.
>>>No, it does not violate anything, Mike.
>>
>>As I stated, lepton number parameters are not conserved. Explain why not.
>No, it is you who raised the question. Explain why you think so.
Extraordinary claims require extraordinary proof, you need to explain why
lepton number parameters are not conserved.
Anyway, in every reaction, the sum (electrons-positrons+electron neutrinos-
electron antineutrinos) remains the same. Conservation of electron neutrino
number, works exactly like conservation of charge. In addition, the sum
(muons-antimuons+mu neutrinos-mu antineutrinos) remains the same. Repeat
for tau. Consider u- -> e- + vu + ve(bar). On the left side muon number
is one and electron number is zero. On the right side, muon number is
still 1 and electron number is still zero. (e-ve(bar))
>>>>Explain why cosmic ray experiments disagree.
>>
>>>Explain what you mean?
>>Examination of the type of neutrinos from muons decaying in the atmosphere
>>show both electron and muon types being generated, in the ratio predicted
>>by the u- -> vu + ve(bar) + e- decay
>Nope. The literature is dead wrong. Remember, the VPP model gives us the
>neutrino structures.
Oh, you are going to bury your head in the sand and scream "ALL THE
EXPERIMENTS ARE WRONG! I'M RIGHT!!). All those scientific papers aren't
going to go away, they collectively prove VPP false.
> But, using one electron and one muon type neutrino makes non sense when
>combining the e + ve + vu vectors.
Why not? This combination conserves e and u lepton numbers, and besides,
is what has been observed.
>Only e + vu + v_u makes a muon structure.
Experiments disagree.
>>Electron neutrinos from, say,
>>nuclear power plants (loaded with B- decaying junk) never participate in
>>the reaction X + Cl-37 --> Ar-37 + e- yet some neutrinos from other sources
>>do.
>Easy, I know for a certainty that the above neutrino reaction cannot take
>place.
But it *does* take place. There's a huge tank in Homestake Mine filled
with carbon tet that's been observing neutrinos by this reaction, for the
last 35 years. Where are the Ar-37 atoms that have been showing up in it
coming from? How about the ones that operate by Ga->Ge transmutations,
d->2p reactions and H->n + e+ reactions?
> It was theoretical from the start. Neutrinos cannot be the primary cause
>for transmutation reactions, regardless of what has been written in
>ignorance.
How do you think we observe them?
-Mike
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| >Wave interference makes no sense?
| >|
| >| What you wrote makes no sense. Gibberish.
|
| >Sorry you are too blind to see something so simple. It is a thing of
total
| >beauty.
|
| Actually, it's quite ugly, as it misleads potential future physicists or
| those who just want to learn about particle physics.
Actually I think that it does not mislead anyone. Anyone that is far enough
along in particle physics should not be mislead by this and recognize that
it is a possible alternative explanation. For me, it has actually been a
great learning tool as I expect that much of VPP will indeed be reconciled
with experimental evidence and the interpretation of that evidence
eventually. So I have learned much from studying both. Let's just say for
me it "pushed the learning curve". And that is most definitely a thing of
beauty.
| >| > What makes zero sense to me is how a
| >| >particle that maybe has zero dimension according to the SM can have a
| >| >magnetic moment at all.
| >|
| >| Why? The Bohr Magneton equation doesn't have length or radius as one
of
| >| its parameters. Despite your protests, you're still trying to
visualize
| >| all this as a little (rotating) solid ball.
|
| >Wrong. h bar certainly does have units of length in it.
|
| No, the Bohr magneton equation doesn't have any radius term in it. If
| you need to be reminded, it is e*hbar/(2*Me). Any radius will work,
| including zero, since it doesn't factor in the equation at all.
| The units of hbar is irrelevant. All the various physical constants in the
| MKSA system contain some or all of the four fundamental dimensional units,
| most will have length in there somewhere.
You are wrong about this. Any radius will not work. And this is one of the
biggest faults of QM. The lack of properly explaining magnetic moments.
Bunch of hand waving if you ask me. Psuedo "waving" that is.
| >The units for the Bohr magneton is amps*meters^2.
|
| So what?
Uhhh, I swear the units say meters^2. Magnetic moments have to be amps
times *physical area*.
| > OK, you tell me. How exactly do you
| >propose to get a magnetic moment with no dimension of length?
|
| Again, despite your protests, you're still stuck in the mindset of a
little
| physical spinning object, moving a little charge in a circle to create
| a little electromagnet.
That is not correct. There is current flowing and that is creating the
charge overall. And is also creating the magnetic moment. There is no
"little charge" moving in a circle. You have the picture totally wrong.
| Remember Maxwell's equations. Magnetic fields can be created without
| moving charges.
Of course. That is exactly what I described above. There are no moving
charges. Only currents. This makes the electron have an overall charge.
When the electron moves as a whole, then you have moving charge.
| >There is absolutely no such thing as a point or point-like
| >particle.
|
| "Point-like" simply means it behaves as a point to the best of our
| measurement abilities. With current technology, the electron *is*
| point-like, it behaves exactly as a point does to the best of our
| measuring abilities. Maybe some day we will find the electron does
| in fact have a nonzero radius. But the fact remains the electron is
| already *known* to be much smaller than the vpp "model".
Not necessarily true. It all has to do with the act of measurement of the
electron. Their is no absolute proof about your *known*.
| >| >points". However, since it is rotating at c,
| >|
| >| Correction: It rotates at sqrt(2)*c. No explanation of this
| >faster-than-light
| >| travel was ever offered.
|
| >Yes there was. You just missed it. There is no mass involved at any
| >instantaneous point for one. It is just a wave representation. It is
only
| >the cube corners that are rotating at the sqrt(2)*c. Since this thing is
| >just a wave structure, it shouldn't be a problem.
|
| Yes it is a problem. Nothing, not matter, nor waves, nor even information
| can travel at greater than c. Period.
I don't think that is exactly right. What substance is it that is actually
travelling faster than c in the VPP model? It really is no substance at
all.
| >| You say so, yet you still cling to the physical object description that
| >| if A decays into B and C, A must be made of B and C.
|
| >So so what? I don't see the connection you are making to what I said
about
| >them being wave structures?
|
| You think of them like physical objects, that is, if A decays into B and
| C, A must be made of B and C. How can I make this any clearer.
|
| >| >I don't think your analogy with the neutron and proton is quite the
same.
| >| >They are known to be composite particles.
| >| Nope. (Unless you're talking of the quarks they're made of)
|
| : They [proton and neutron] are known to be composite particles.
|
| Nope. (Unless you're talking of the quarks they're made of)
|
| >| > I don't think I would say the
| >| >proton is "decaying" into a neutron.
| >|
| >| Why not?
|
| >I would say that it is changing in to a neutron. Not decaying into a
| >neutron. Decaying is the wrong word. Protons don't decay.
|
| Physicists use "decay" to mean a single object (nucleus, particle)
| spontaneously changing (without any external provocation) into something
| else at a lower energy state. You can say the F-18 nucleus as a whole
| decays, or a proton within it does.
Fine. Whatever turns you on. We both know that changing is a better term.
| >| > This can only happen in an atom or
| >| >nucleus. Never when the proton is free.
| >|
| >| So what? A He-4 nucleus never decays, and it has two neutrons, does
this
| >| mean neutrons don't decay? Nope. That is the reverse argument.
|
| >This is not an argument at all much less a reverse argument.
|
| The reverse argument is a bound neutron in a He-4 nucleus lasts forever,
| therefore a free neutron lasts forever. This is the reverse (swapping
| bound and free, and proton for neutron) argument.
Why should this be a valid argument necessarily? We already know why free
neutrons decay.
| >| The proton is exactly as fundamental as the neutron, or pi+ or xi-.
|
| >When did the proton become a fundamental particle? Not even the SM has
it
| >as that. Or the neutron or pi+ or xi-.
|
| So you now believe in the quark model? OK.
No, I believe in the parton model. The quark model to me is most likely a
simplification of what is really going on. The proton probably has way more
than three partons or even six partons.
| >| Why does the tau and muon decay? I think the answer is the same as the
| >| answer to the question why does a dog lick its balls.
|
| >Or chicken pot pie, that is why. Most everything decays to get to a
lower
| >possible energy level. That concept will never change.
|
| Yup. If it can decay to a lower energy level, it will. As long as certain
| conserved quantum variables are conserved.
Agreed. We just don't agree on what the exact quantum variables should
actually be.
| >| Maybe if you look as a decay as a reaction between the particle and
| >| virtual particle pairs. A muon reacts with a (virtual) muon
antineutrino
| >| to produce a virtual W-, which immediately decays into an electron and
| >| electron antineutrino. The muon neutrino is the partner of the muon
| >| antineutrino, its partner is gone so it becomes real.
|
| >Or the quantum number of the muon should be the same as an electron and
it
| >decays to an electron, muon neutrino and muon antineutrio. Everything is
| >still conserved in this scenario also. What is the proof that a muon
should
| >have its own quantum number? Answer me that. It is an "ad hoc"
assumption
| >as far as I can tell.
|
| Because if that were true, we would see decays of muon -> electron +
electron
| neutrino + electron antineutrino (because it could) and muon -> electron +
| electron neutrino + muon antineutrino, and and muon -> electron + muon
| neutrino + electron antineutrino, and probably modes involving the tau
| neutrino. We *only* see muon -> electron + muon neutrino + electron
| antineutrino, and none of the other combinations. In particular we
*don't*
| see muon -> electron + muon neutrino + muon antineutrino.
How could we tell about the last one? If the muon were a composite of an
electron, muon neutrino and muon antineutrino and its quantum number was the
same as the electron then we wouldn't necessarily see all the other
reactions you mention. It couldn't decay into electron neutrinos because
they aren't there to start with.
FrediFizzx
Michael Moroney
>| Actually, it's quite ugly, as it misleads potential future physicists or
>| those who just want to learn about particle physics.
>Actually I think that it does not mislead anyone. Anyone that is far enough
>along in particle physics should not be mislead by this and recognize that
it is CRAP.
I'm worries by those less far along.
>| >Wrong. h bar certainly does have units of length in it.
>|
>| No, the Bohr magneton equation doesn't have any radius term in it. If
>| you need to be reminded, it is e*hbar/(2*Me). Any radius will work,
>| including zero, since it doesn't factor in the equation at all.
>| The units of hbar is irrelevant. All the various physical constants in the
>| MKSA system contain some or all of the four fundamental dimensional units,
>| most will have length in there somewhere.
>You are wrong about this. Any radius will not work. And this is one of the
Compute the Bohr for an electron with a radius of 1 meter. Compute the
Bohr for an electron with a radius of the diameter of the universe.
Compute it with a diameter of the Planck length. Compute for a radius of 0.
Since the radius isn't even a factor, you always get the same answer, and
any radius will work.
>| >The units for the Bohr magneton is amps*meters^2.
>|
>| So what?
>Uhhh, I swear the units say meters^2. Magnetic moments have to be amps
>times *physical area*.
Only for when there is an actual current flowing around a physical area.
As I mentioned, a magnetic field can be created with no current involved.
>| Again, despite your protests, you're still stuck in the mindset of a
>little
>| physical spinning object, moving a little charge in a circle to create
>| a little electromagnet.
>That is not correct. There is current flowing and that is creating the
>charge overall.
Current *is* flowing charge (I=dQ/dt). It does not create charge, it
is the flow of charge.
> And is also creating the magnetic moment. There is no
>"little charge" moving in a circle. You have the picture totally wrong.
Current *is* moving charge. Again, I=dQ/dt.
>| Remember Maxwell's equations. Magnetic fields can be created without
>| moving charges.
>Of course. That is exactly what I described above. There are no moving
>charges. Only currents.
For the third time, that is a contradiction. Anyway, the point, which
you missed, is that a magnetic field can be created with no current.
See Maxwell's third equation.
>Not necessarily true. It all has to do with the act of measurement of the
>electron. Their is no absolute proof about your *known*.
Many experiments have been run to find the electron's radius. All show
it to be smaller than their margin of error, no matter how small that is.
The most "recent" ones ([probably any within the last 20 years) show it
smaller than the VPP radius.
>| >| >points". However, since it is rotating at c,
>| >| Correction: It rotates at sqrt(2)*c. No explanation of this
>| >faster-than-light
>| >| travel was ever offered.
>| Yes it is a problem. Nothing, not matter, nor waves, nor even information
>| can travel at greater than c. Period.
>I don't think that is exactly right. What substance is it that is actually
>travelling faster than c in the VPP model? It really is no substance at
>all.
No, it's not just limited to "substances". It includes the various
fields (B E whatever) of the cubes, any field, any information, anything.
In fact if we lop off the corners of the cubes (to eliminate anything
travelling > c, we're left with NOTHING, except points where the original
vectors are tangent to the v=c circle.
>| >I would say that it is changing in to a neutron. Not decaying into a
>| >neutron. Decaying is the wrong word. Protons don't decay.
>|
>| Physicists use "decay" to mean a single object (nucleus, particle)
>| spontaneously changing (without any external provocation) into something
>| else at a lower energy state. You can say the F-18 nucleus as a whole
>| decays, or a proton within it does.
>Fine. Whatever turns you on. We both know that changing is a better term.
I'm just showing the term how physicicts use it. If you are trying to say
a "decaying" neutron isn't rotting, with maggots eating it, OK, but sheesh!
>| >| So what? A He-4 nucleus never decays, and it has two neutrons, does
>this
>| >| mean neutrons don't decay? Nope. That is the reverse argument.
>|
>| >This is not an argument at all much less a reverse argument.
>|
>| The reverse argument is a bound neutron in a He-4 nucleus lasts forever,
>| therefore a free neutron lasts forever. This is the reverse (swapping
>| bound and free, and proton for neutron) argument.
>Why should this be a valid argument necessarily? We already know why free
>neutrons decay.
The only reason a free neutron decays is it is more massive than the proton.
If it just so happened a proton was more massive the free neutron would
be stable. Just like any other isotope pair with the same A and Z
differs by 1. (proton = H with A=1, Z=1, free neutron is equivalent to a
nucleus with A=1, Z=0)
>| Yup. If it can decay to a lower energy level, it will. As long as certain
>| conserved quantum variables are conserved.
>Agreed. We just don't agree on what the exact quantum variables should
>actually be.
We just know the three lepton numbers are always conserved, just like
charge is always conserved. We don't know why, just that it is. We
also don't know why charge is conserved.
>| >Or the quantum number of the muon should be the same as an electron and
>it
>| >decays to an electron, muon neutrino and muon antineutrio. Everything is
>| >still conserved in this scenario also. What is the proof that a muon
>should
>| >have its own quantum number? Answer me that. It is an "ad hoc"
>assumption
>| >as far as I can tell.
>|
>| Because if that were true, we would see decays of muon -> electron +
>electron
>| neutrino + electron antineutrino (because it could) and muon -> electron +
>| electron neutrino + muon antineutrino, and and muon -> electron + muon
>| neutrino + electron antineutrino, and probably modes involving the tau
>| neutrino. We *only* see muon -> electron + muon neutrino + electron
>| antineutrino, and none of the other combinations. In particular we
>*don't*
>| see muon -> electron + muon neutrino + muon antineutrino.
>How could we tell about the last one? If the muon were a composite of an
>electron, muon neutrino and muon antineutrino and its quantum number was the
>same as the electron then we wouldn't necessarily see all the other
>reactions you mention.
Wrong. If it can do so, it *will* do so (sometimes). If it doesn't,
it's because there is some law forbidding it.
> It couldn't decay into electron neutrinos because
>they aren't there to start with.
Once again, you're still stuck with the physical object idea (If A
decays into B and C, A must be made of B and C.) Don't whine that you
aren't, it's obvious you are as long as you believe in that statement.
-Mike
Jim Panetta
>>mor...@world.std.spaamtrap.com (Michael Moroney)
>>Nope. The 1962 muon neutrino experiment was to start with a stream of
>>pions decaying into muons+neutrinos, deflect the muons, and examine the
>>reactions of the neutrinos on a target.
>
>So far so good:
>
>>If electron neutrinos were the
>>same as muon neutrinos, they'd see equal numbers of electrons and muons
>>created.
>
>Who said electron and muon neutrinos were the same?
>
>Read the original paper. They shielded with battleship steel between their
>accelerator, but their detector was hanging out in the cosmic ray stream.
>
>They saw a lot of cosmic muons going horizontal (or nearly so). Dirty
>experiment. They were right that there are muon type neutrinos, but thay sure
>as hell did not see them.
We know that the signal is from the beam of muons because they turned
the beam off and the signal went away. That's how the background from
cosmics was measured.
>Remember, the VPP is a model that actually gives us the structures
>for the neutrinos types. Clearly the electron and muon type
>neutrinos are different and (contrary to latest theory) neutrinos
>CANNOT oscillate between one type and another.
The latest *data* says that they DO.
>>Electron neutrinos from, say,
>>nuclear power plants (loaded with B- decaying junk) never participate in
>>the reaction X + Cl-37 --> Ar-37 + e- yet some neutrinos from other sources
>>do.
>
>Easy, I know for a certainty that the above neutrino reaction cannot take
>place.
Sorry, Ray Davis of U-Penn (emeritus) has about 40 years of data
showing that the CL-37 reaction does occur.
--Jim
--
My opinions are mine...not SLAC's...not Penn's...not DOE's...mine.
(except by random, unforseeable coincidences)
pan...@slac.stanford.edu -- Save the whales! Free the mallocs!
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| Actually, it's quite ugly, as it misleads potential future physicists
or
| >| those who just want to learn about particle physics.
|
| >Actually I think that it does not mislead anyone. Anyone that is far
enough
| >along in particle physics should not be mislead by this and recognize
that
|
| it is CRAP.
| I'm worries by those less far along.
Well, I wouldn't worry too much. Eventually those less far along will be
able to make up their own mind if they really care about knowing the truth.
At least they will have something to compare to. I almost get the sense
that some of you feel threatened by Tom's VPP. It is not as wacky as some
other ideas I have seen proposed here. It gets a lot of things right that
just *can't* be totally coincidental. Now to just work out the rest of it.
The VPP electron *is* a freakin' thing of beauty. I think it all goes back
to Schrodinger's wavefunction and HUP. If Schrodinger hadn't come up with
his "wavefunction" someone might have discovered the *real* wave structure
of the electron before Tom did and we would probably be much further along
in physics. And do particles really behave in accordance with HUP? Well,
of course, this question is kind of like the one about a tree falling in the
woods and making a sound if no one is there to hear it. I say it does make
a sound anywise. And HUP is only *our* limitations of observation because
we have nothing else to use besides photons. Particles do not obey or
follow or care about HUP. The real reason particle motion is so mushed out
and we have to use probablities to try to describe it is because of that
particle's wave structure. All particles are rotating wave structures. How
do you localize a wave structure? You can't. It has an extension in space
and if you try to measure it at a "point" of space, you will fail to
localize it in its entirety. And this non-localized wave structure will
(and does) generate EM forces that are also non-localized (in a "cloud") but
spread out over a certain amount of space. IOW, the source of the electric
or magnetic force is not ever from a point.
Point sources are something that should have been tossed out of physics long
ago. If the electron were about 10^-20 meters in size, it would have an
electric potential energy of about 72 GeV. Ridiculous!! If the electron is
about 10^-13 meter in size, we get an electric potential energy of about 7.2
KeV. Much more reasonable. An electron the size of about 10^-18 meters
(which you say it has to be smaller than), we get an electric potential
energy of about 720 MeV. More that its rest mass energy!! This makes no
sense. In fact, the VPP electron size makes its electric potential energy
equal to its rest mass energy times the fine structure constant. Exactly!
A freakin' thing of beauty.
| >| >Wrong. h bar certainly does have units of length in it.
| >|
| >| No, the Bohr magneton equation doesn't have any radius term in it. If
| >| you need to be reminded, it is e*hbar/(2*Me). Any radius will work,
| >| including zero, since it doesn't factor in the equation at all.
| >| The units of hbar is irrelevant. All the various physical constants in
the
| >| MKSA system contain some or all of the four fundamental dimensional
units,
| >| most will have length in there somewhere.
|
| >You are wrong about this. Any radius will not work. And this is one of
the
|
| Compute the Bohr for an electron with a radius of 1 meter. Compute the
| Bohr for an electron with a radius of the diameter of the universe.
| Compute it with a diameter of the Planck length. Compute for a radius of
0.
| Since the radius isn't even a factor, you always get the same answer, and
| any radius will work.
The Bohr for one meter would be 2.4*10^-11 amps*meter^2.
The Bohr for the universe is ridiculous but can be calculated.
The Bohr for the Planck length is also ridiculous but is 3.88*10^-46
amps*meter^2.
The Bohr for a radius of zero is zero.
So I don't know what you are talking about. Any radius does work but gives
different answers. The only electron model that does give the correct value
without manipulation is the VPP model. I wonder why? Maybe because it is
the right freakin' size?!!! Another thing of beauty!
| >| >The units for the Bohr magneton is amps*meters^2.
| >|
| >| So what?
|
| >Uhhh, I swear the units say meters^2. Magnetic moments have to be amps
| >times *physical area*.
|
| Only for when there is an actual current flowing around a physical area.
| As I mentioned, a magnetic field can be created with no current involved.
The VPP model *does* have a current flowing around an area.
| >| Again, despite your protests, you're still stuck in the mindset of a
| >little
| >| physical spinning object, moving a little charge in a circle to create
| >| a little electromagnet.
|
| >That is not correct. There is current flowing and that is creating the
| >charge overall.
|
| Current *is* flowing charge (I=dQ/dt). It does not create charge, it
| is the flow of charge.
|
| > And is also creating the magnetic moment. There is no
| >"little charge" moving in a circle. You have the picture totally wrong.
|
| Current *is* moving charge. Again, I=dQ/dt.
Or a current can be caused from a magnetic field. Remember Bilge's question
about finding the two currents in a system of a steel plate with a magnet
holding it?
"You have a permanent magnet which you use to [hold] a piece of metal.
Find the agent responsible for doing the [holding] and find 2 currents.
Hints: (1) Magnetic fields don't do work, so the magnetic field
can't do the [holding];
(2) There is a current in the permanent magnet and it induces
a current in the metal, but no charge is actually transported;
Somewhat more challenging: Find the physical origin of the currents."
Did you ever figure this out? I am still working on it but not real hard.
This is probably a good time though.
| >| Remember Maxwell's equations. Magnetic fields can be created without
| >| moving charges.
|
| >Of course. That is exactly what I described above. There are no moving
| >charges. Only currents.
|
| For the third time, that is a contradiction. Anyway, the point, which
| you missed, is that a magnetic field can be created with no current.
| See Maxwell's third equation.
It is not a contradiction. See above. I think you are contradicting
yourself.
| >Not necessarily true. It all has to do with the act of measurement of
the
| >electron. Their is no absolute proof about your *known*.
|
| Many experiments have been run to find the electron's radius. All show
| it to be smaller than their margin of error, no matter how small that is.
| The most "recent" ones ([probably any within the last 20 years) show it
| smaller than the VPP radius.
Impossible for it to be 10^-18 meters or smaller. See above. The electric
potential energy for it gets ridiculous. So there is definitely something
wrong with your *known*.
| >| >| >points". However, since it is rotating at c,
|
| >| >| Correction: It rotates at sqrt(2)*c. No explanation of this
| >| >faster-than-light
| >| >| travel was ever offered.
|
| >| Yes it is a problem. Nothing, not matter, nor waves, nor even
information
| >| can travel at greater than c. Period.
|
| >I don't think that is exactly right. What substance is it that is
actually
| >travelling faster than c in the VPP model? It really is no substance at
| >all.
|
| No, it's not just limited to "substances". It includes the various
| fields (B E whatever) of the cubes, any field, any information, anything.
| In fact if we lop off the corners of the cubes (to eliminate anything
| travelling > c, we're left with NOTHING, except points where the original
| vectors are tangent to the v=c circle.
Well then physics has a big problem. The phase velocity of EM radiation in
the vacuum is c. But the velocity of the wave going from full crest peak to
full trough peak is faster than c at the zero crossover point as the wave
travels past any point in space. Like Tom said, c is an rms velocity. If
the wave is propagating in the x direction, what is the wave velocity in the
y direction when it is at a peak? What is the wave velocity in the y
direction when it is midway between its "positive" going peak and "negative"
going peak? As the wave passes a point in space.
<snipped the rest; let's concentrate on the above until Tom answers some of
the questions I posed to him about the muon neutrinos>
FrediFizzx
ThomasL283
>Date: 8/31/2002 11:06 PM Pacific Daylight Time
Wrote in:
>Message-id: <aksapi$cq4$1...@usenet.Stanford.EDU
Somewhere in the newsfeed, ThomasL283 said:
>>Who said electron and muon neutrinos were the same?
>>
>>Read the original paper. They shielded with battleship steel between their
>>accelerator, but their detector was hanging out in the cosmic ray stream.
>>
>>They saw a lot of cosmic muons going horizontal (or nearly so). Dirty
>>experiment. They were right that there are muon type neutrinos, but thay
>sure
>>as hell did not see them.
>
>We know that the signal is from the beam of muons because they turned
>the beam off and the signal went away. That's how the background from
>cosmics was measured.
Jim: Look at the original paper. Phys. Rev. Letters, Volume 9, Number 1, July
1, 1962.
They "calibrated" the spark gap detector by INCREASING beam energy, so that
muons (more) got through the shield.
As for Cosmic ray false events, in the text they say in so many words,
Quote:
Muons from cosmic rays can and do simulate neutrino events.........Thus 1 in 90
cosmic ray events is neutrino like......Cerenkov gating and the short AGS pulse
effect a factor of (about) 10^6 since the circuits are on for only 3.5
microseconds. .....400 cosmic ray tracks which is consistent with observation.
Among these there should be 5+-1 cosmic ray induced "events".
Unquote.
>>Remember, the VPP is a model that actually gives us the structures
>>for the neutrinos types. Clearly the electron and muon type
>>neutrinos are different and (contrary to latest theory) neutrinos
>>CANNOT oscillate between one type and another. See>
http://www.members.aol.com/tnlockyer/cfives.gif
The VPP model shows only the electron and muon type neutrino structures.
No way can neutrinos change from one type to the other.
>The latest *data* says that they DO.
Jim, they use the fact they don't see enough "events" as somehow proving the
neutrinos oscillate, from one type to another.
I believe the neutrino events are non-existent, based on the fact that the
solar (EC) and (B+) decays don't produce the neutrinos they claim to "not see".
The electron and muon type neutrinos are all that nature has (or needs).
>>>Electron neutrinos from, say,
>>>nuclear power plants (loaded with B- decaying junk) never participate in
>>>the reaction X + Cl-37 --> Ar-37 + e- yet some neutrinos from other sources
>>>do.
>>Easy, I know for a certainty that the above neutrino reaction cannot take
>>place.
>Sorry, Ray Davis of U-Penn (emeritus) has about 40 years of data
>showing that the CL-37 reaction does occur.
Yes, but only if one believes you can detect events by sweeping 31/2
radioactive argon atoms out of 100,000 gallons of tetrachloroethylene, every
four days? The lack of results became the "solar neutrino problem".
Jim Panetta
>>We know that the signal is from the beam of muons because they turned
>>the beam off and the signal went away. That's how the background from
>>cosmics was measured.
>
>Jim: Look at the original paper. Phys. Rev. Letters, Volume 9, Number 1, July
>1, 1962.
> They "calibrated" the spark gap detector by INCREASING beam energy, so that
>muons (more) got through the shield.
I guarantee they took data with the beams off. If only in testing.
If the events were cosmic ray induced, they would have seen it then.
>As for Cosmic ray false events, in the text they say in so many words,
>
>Quote:
>Muons from cosmic rays can and do simulate neutrino events.........Thus
>1 in 90 cosmic ray events is neutrino like......Cerenkov gating and
>the short AGS pulse effect a factor of (about) 10^6 since the circuits
>are on for only 3.5 microseconds. .....400 cosmic ray tracks which is
>consistent with observation.
>Among these there should be 5+-1 cosmic ray induced "events".
>Unquote.
So? They saw *lots* of events, only when the beam was on. Much more
than cosmic rays could cause. Cosmic rays are constant at about
1/cm^2/sec. In our detector, we see a few hertz passing our cuts.
>>The latest *data* says that they DO.
>
>Jim, they use the fact they don't see enough "events" as somehow proving the
>neutrinos oscillate, from one type to another.
They use the data on CC, NC, and elastic events to determine the
non-nu_e factor. The total flux measured in the neutral current
reaction is consistent with solar models. The flux in the CC
reaction isn't. The only neutrinos that can participate in CC
interactions are nu_e. All can participate in NC. Thus there
is a significant non-nu_e component in the neutrino spectrum.
>I believe the neutrino events are non-existent, based on the fact
>that the solar (EC) and (B+) decays don't produce the neutrinos they
>claim to "not see".
I'm sorry you believe this, but the data says that you are
wrong.
>>>>Electron neutrinos from, say, nuclear power plants (loaded with B-
>>>>decaying junk) never participate in the reaction X + Cl-37 -->
>>>>Ar-37 + e- yet some neutrinos from other sources do.
>
>>>Easy, I know for a certainty that the above neutrino reaction cannot take
>>>place.
>
>>Sorry, Ray Davis of U-Penn (emeritus) has about 40 years of data
>>showing that the CL-37 reaction does occur.
>
>Yes, but only if one believes you can detect events by sweeping 31/2
>radioactive argon atoms out of 100,000 gallons of tetrachloroethylene, every
>four days? The lack of results became the "solar neutrino problem".
It's fairly easy, I've done it myself in smaller volumes with
different materials. A standard method for determining the
radioactive content of various oils is to bubble helium through it and
count the Radon atoms. It's the same process Ray pioneered.
I've read your model. You don't make any good predictions,
and what you do make is flawed by major misunderstandings on
the nature of E&M, Special Relativity, and Newtonian Mechanics.
Michael Moroney
>http://www.members.aol.com/tnlockyer/cfives.gif
Why do you keep posting that as proof of anything? It is too cryptic, and
is meaningless to everyone except yourself.
>The VPP model shows only the electron and muon type neutrino structures.
> No way can neutrinos change from one type to the other.
No proof VPP is correct has yet been posted. Plenty to refute it.
>>The latest *data* says that they DO.
> I believe the neutrino events are non-existent, based on the fact that the
>solar (EC) and (B+) decays don't produce the neutrinos they claim to "not see".
But we do see them. Just not as many as predicted.
>>>>Electron neutrinos from, say,
>>>>nuclear power plants (loaded with B- decaying junk) never participate in
>>>>the reaction X + Cl-37 --> Ar-37 + e- yet some neutrinos from other sources
>>>>do.
>>>Easy, I know for a certainty that the above neutrino reaction cannot take
>>>place.
>>Sorry, Ray Davis of U-Penn (emeritus) has about 40 years of data
>>showing that the CL-37 reaction does occur.
>Yes, but only if one believes you can detect events by sweeping 31/2
>radioactive argon atoms out of 100,000 gallons of tetrachloroethylene, every
>four days? The lack of results became the "solar neutrino problem".
If there was no transmutation, we'd see _zero_ Ar-37 atoms.
You just contradicted yourself when you said above we don't see B+ cosmic
ray events.
-Mike
Michael Moroney
>Point sources are something that should have been tossed out of physics long
>ago. If the electron were about 10^-20 meters in size, it would have an
>electric potential energy of about 72 GeV. Ridiculous!! If the electron is
Wrong. The electric PE would have that value _if and only if_ the
electron were composed of numerous tiny charges, with their sum total
equal to the electron's charge, all crammed into that radius against their
mutual repulsion. Unfortunately for that argument, we already know the
electron's charge is quantized, it is not composed of smaller charges.
I previously mentioned this to you. (see also the "classical electron
radius")
>| >You are wrong about this. Any radius will not work. And this is one of
>the
>|
>| Compute the Bohr for an electron with a radius of 1 meter. Compute the
>| Bohr for an electron with a radius of the diameter of the universe.
>| Compute it with a diameter of the Planck length. Compute for a radius of
>0.
>| Since the radius isn't even a factor, you always get the same answer, and
>| any radius will work.
>The Bohr for one meter would be 2.4*10^-11 amps*meter^2.
>The Bohr for the universe is ridiculous but can be calculated.
>The Bohr for the Planck length is also ridiculous but is 3.88*10^-46
>amps*meter^2.
>The Bohr for a radius of zero is zero.
I'd like to see your math. Remember, the _definition_ of the Bohr is
e*h/(4*pi*me), or equivalently, e*hbar/(2*me). Since there is no Re
term in there, its value is irrelevant.
>| >| >The units for the Bohr magneton is amps*meters^2.
>| >|
>| >| So what?
>|
>| >Uhhh, I swear the units say meters^2. Magnetic moments have to be amps
>| >times *physical area*.
>|
>| Only for when there is an actual current flowing around a physical area.
>| As I mentioned, a magnetic field can be created with no current involved.
>The VPP model *does* have a current flowing around an area.
I said you don't need a current to get a magnetic field. This means
a current is unnecessary.
>| Current *is* moving charge. Again, I=dQ/dt.
>Or a current can be caused from a magnetic field.
True but irrelevant.
> Remember Bilge's question
>about finding the two currents in a system of a steel plate with a magnet
>holding it?
> "You have a permanent magnet which you use to [hold] a piece of metal.
>Find the agent responsible for doing the [holding] and find 2 currents.
> Hints: (1) Magnetic fields don't do work, so the magnetic field
> can't do the [holding];
> (2) There is a current in the permanent magnet and it induces
> a current in the metal, but no charge is actually transported;
>Somewhat more challenging: Find the physical origin of the currents."
>Did you ever figure this out? I am still working on it but not real hard.
>This is probably a good time though.
This is simple. The orbital motion of d shell electrons in Fe produce a
magnetic field. I guess you could say there is a net current from those
electrons. In unmagnetized Fe the orientation is random so the magnetization
cancels, but in a permanent magnet or magnetized iron they are aligned.
>It is not a contradiction. See above. I think you are contradicting
>yourself.
See what? Current is the flow of charge, period. No charge, no current.
>| Many experiments have been run to find the electron's radius. All show
>| it to be smaller than their margin of error, no matter how small that is.
>| The most "recent" ones ([probably any within the last 20 years) show it
>| smaller than the VPP radius.
>Impossible for it to be 10^-18 meters or smaller. See above. The electric
>potential energy for it gets ridiculous. So there is definitely something
>wrong with your *known*.
I already corrected this mistake of yours.
>| No, it's not just limited to "substances". It includes the various
>| fields (B E whatever) of the cubes, any field, any information, anything.
>| In fact if we lop off the corners of the cubes (to eliminate anything
>| travelling > c, we're left with NOTHING, except points where the original
>| vectors are tangent to the v=c circle.
>Well then physics has a big problem. The phase velocity of EM radiation in
>the vacuum is c. But the velocity of the wave going from full crest peak to
>full trough peak is faster than c at the zero crossover point as the wave
>travels past any point in space.
WTF is this? The wave moves at a constant c.
> Like Tom said, c is an rms velocity.
^^^^^^^^^^^^^
Your first clue of the bogusness of the statement.
>the wave is propagating in the x direction, what is the wave velocity in the
>y direction when it is at a peak? What is the wave velocity in the y
>direction when it is midway between its "positive" going peak and "negative"
>going peak?
Zero. (you just said it propogates in the x direction)
-Mike
PC
> "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
>
> >Point sources are something that should have been tossed out of physics long
> >ago. If the electron were about 10^-20 meters in size, it would have an
> >electric potential energy of about 72 GeV. Ridiculous!! If the electron is
Yep, pure Goofy invention. Always distributions.
§:-)
ThomasL283
>Date: 8/30/2002 1:56 PM Pacific Daylight Time
Wrote in:
>Message-id: <H1oCu3...@world.std.com>
>snip<
>
>thoma...@aol.com (ThomasL283) writes:
>
>>It is expelled and the 0.7823 MeV is absorbed in the proton formation of the
>>neutron, along with the (captured) electron.
>
>If 0.7823 MeV is absorbed, why is the mass difference between Cu-64 and
>Ni-64 1.675 MeV and not 0.8927 MeV?
>
Mike as shown in:
http://members.aol.com/tnlockyer/Betadecay.gif
Internally the rearrangement of nucleons creates new energy of 2.457439 MeV.
Of this -1.67510606 MeV is expelled and the 0.78233293 MeV is absorbed along
with a shell electron.
This lost electron makes a neutral atom because the charge changes by one
unit, when the proton changes into a neutron.
The expelled -1.675106 MeV is the mass difference you quote for the daughter,
Ni64.
>Huh? The neutrino is (nearly) massless, the minimal expulsion energy is
>at most a few eVs.
Exactly, that is why there are no neutrinos from (EC) and (B+) decays, contrary
to popular belief.
Sadly the old timers did not have the advantage of a structure for the nuclei
that worked.
They made some assumptions that the positron somehow changed the proton into a
neutron. That reaction
(proton--->neutron + positron + neutrino)
would violate the conservation of energy.
>>What you see (Ni-59 -->Co59) is an exclusive (EC) process, according to the
>>energy you quote.
>
>Not according to the energy. B+ is rare but exists.
No, that must be a mistake, Mike. I don't know where that positron showed up
from , in your example, but it cannot be from the decay kinematics.
You also have to loose a shell electron, for the neutral atom, when proton
changes into a neutron.
If the shell electron is not absorbed, the electron has to come from pair
production,
and the positron and shell electron are then expelled.
ThomasL283
>Date: 8/30/2002 12:26 AM Pacific Daylight Time
Wrote in:
>Message-id: <07Fb9.63$4M4.13...@newssvr14.news.prodigy.com>
>| >thoma...@aol.com (ThomasL283) writes:
>| >
><snip>
>| Remember, the VPP is a model that actually gives us the structures for
>the
>| neutrinos types. Clearly the electron and muon type neutrinos are
>different
>| and (contrary to latest theory) neutrinos CANNOT oscillate between one
>type and
>| another.
>
>Tom, I have been pondering what they are seeing with this neutrino
>oscillation business, and I am wondering if neutrinos that have been ejected
>from the VPP proton or from a muon, etc. might behave like this?
Fredi: The VPP proton does not eject neutrinos (I think you meant VPP neutrons,
which do expell neutrinos).
I have gone to great lengths to show that both (EC) and (B+) do NOT expell
neutrinos, at proton to neutron formation.
But note:
The neutrino oscillations were postulated to explain the "solar neutrino
problem"
that is the complete lack of neutrino detection.
But lets get practical. Look at the VPP structures for the neutrinos.
http://www.members.aol.com/tnlockyer/cfives
Do you have any idea how (or why) one type of neutrino would change into
another?
VPP does things automatically, and I see no automatic reason for neutrino
oscillation.
(This thread is getting off topic)
ThomasL283
Make that:
http://www.members.aol.com/tnlockyer/cfives.gif
Left off the gif on the previous message.
FrediFizzx
| >"FrediFizzx" fredifi...@ahahhotmail.com
| >Date: 8/30/2002 12:26 AM Pacific Daylight Time
| Wrote in:
| >Message-id: <07Fb9.63$4M4.13...@newssvr14.news.prodigy.com>
|
| >| >thoma...@aol.com (ThomasL283) writes:
| >| >
| ><snip>
| >| Remember, the VPP is a model that actually gives us the structures for
| >the
| >| neutrinos types. Clearly the electron and muon type neutrinos are
| >different
| >| and (contrary to latest theory) neutrinos CANNOT oscillate between one
| >type and
| >| another.
|
| >
| >Tom, I have been pondering what they are seeing with this neutrino
| >oscillation business, and I am wondering if neutrinos that have been
ejected
| >from the VPP proton or from a muon, etc. might behave like this?
|
| Fredi: The VPP proton does not eject neutrinos (I think you meant VPP
neutrons,
| which do expell neutrinos).
OK then VPP neutron neutrinos. What happens to this ejected neutrino? It
was spinning in conjunction with the spin of the VPP neutron and having been
ejected, it will have both spin and linear motion until "something" acts to
stop it. We can't go against Newton's law here.
| I have gone to great lengths to show that both (EC) and (B+) do NOT
expell
| neutrinos, at proton to neutron formation.
Yes, I know. I was refering to destroyed protons in a collision and what
you think happens with the "parton" neutrinos in that case? For one, do
they keep their stored energy or is it all released in the collision
fireball? Do they keep spinning or does the release of their energy stop
them? Etc? There is a lot of questions to answer in this scenario and I
was just wondering what your ideas might be about it.
| But note:
| The neutrino oscillations were postulated to explain the "solar neutrino
| problem"
| that is the complete lack of neutrino detection.
|
| But lets get practical. Look at the VPP structures for the neutrinos.
|
| http://www.members.aol.com/tnlockyer/cfives
|
| Do you have any idea how (or why) one type of neutrino would change into
| another?
I am not proposing that one neutrino is changing into another. I am
proposing that the neutron *ejected* neutrinos do have some spin and motion.
This "left over" ejected spin would maybe give them some detectable mass.
This detectable mass may be variable. Can this account for what they are
seeing experimentally? After all, the VPP neutrinos are very very different
from the ones in the SM.
| VPP does things automatically, and I see no automatic reason for neutrino
| oscillation.
|
| (This thread is getting off topic)
Yes, this thread has been way off topic from the start. I agree that we
should get back to the hydrogen atom.
FrediFizzx
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >Point sources are something that should have been tossed out of physics
long
| >ago. If the electron were about 10^-20 meters in size, it would have an
| >electric potential energy of about 72 GeV. Ridiculous!! If the electron
is
|
| Wrong. The electric PE would have that value _if and only if_ the
| electron were composed of numerous tiny charges, with their sum total
| equal to the electron's charge, all crammed into that radius against their
| mutual repulsion. Unfortunately for that argument, we already know the
| electron's charge is quantized, it is not composed of smaller charges.
| I previously mentioned this to you. (see also the "classical electron
| radius")
Are you saying that q^2/(8*pi*eps0*R) is not valid for a single electronic
charge? If so, why not? This is the equation for the energy of a continous
charge distribution in a sphere of radius R. I think the electron qualifies
for this. How do you explain this away? And what is the cutoff point? How
many electrons do you have to have to make the equation valid? Two? Three?
0ne hundred? And don't tell me this is the work required to assemble a
bunch of point charges. In the case of the electron, it is the work to make
the elementary charge. Elementary electric potential energy.
| >| >You are wrong about this. Any radius will not work. And this is one
of
| >the
| >|
| >| Compute the Bohr for an electron with a radius of 1 meter. Compute the
| >| Bohr for an electron with a radius of the diameter of the universe.
| >| Compute it with a diameter of the Planck length. Compute for a radius
of
| >0.
| >| Since the radius isn't even a factor, you always get the same answer,
and
| >| any radius will work.
|
| >The Bohr for one meter would be 2.4*10^-11 amps*meter^2.
| >The Bohr for the universe is ridiculous but can be calculated.
| >The Bohr for the Planck length is also ridiculous but is 3.88*10^-46
| >amps*meter^2.
| >The Bohr for a radius of zero is zero.
|
| I'd like to see your math. Remember, the _definition_ of the Bohr is
| e*h/(4*pi*me), or equivalently, e*hbar/(2*me). Since there is no Re
| term in there, its value is irrelevant.
Or equivalently the Bohr is equal to e*c*lambda_C/4*pi. But the length I
used is lamba_C/2*pi the rationalized Compton wavelength. This allows to
find the actual area the magneton needs for its creation. In the case of
the VPP electron, it has two magneton loop areas. This can only come from
the model and nothing else. IOW, if you try to apply this to the classical
electron radius modeled as a sphere, it doesn't work.
| >| >| >The units for the Bohr magneton is amps*meters^2.
| >| >|
| >| >| So what?
| >|
| >| >Uhhh, I swear the units say meters^2. Magnetic moments have to be
amps
| >| >times *physical area*.
| >|
| >| Only for when there is an actual current flowing around a physical
area.
| >| As I mentioned, a magnetic field can be created with no current
involved.
|
| >The VPP model *does* have a current flowing around an area.
|
| I said you don't need a current to get a magnetic field. This means
| a current is unnecessary.
|
| >| Current *is* moving charge. Again, I=dQ/dt.
|
| >Or a current can be caused from a magnetic field.
|
| True but irrelevant.
Not irrelevant because that is exactly what is going on here. This is the
*source* of your charge, Q, which happens to be e.
| > Remember Bilge's question
| >about finding the two currents in a system of a steel plate with a magnet
| >holding it?
|
| > "You have a permanent magnet which you use to [hold] a piece of metal.
| >Find the agent responsible for doing the [holding] and find 2 currents.
|
| > Hints: (1) Magnetic fields don't do work, so the magnetic field
| > can't do the [holding];
| > (2) There is a current in the permanent magnet and it induces
| > a current in the metal, but no charge is actually
transported;
|
| >Somewhat more challenging: Find the physical origin of the currents."
|
| >Did you ever figure this out? I am still working on it but not real
hard.
| >This is probably a good time though.
|
| This is simple. The orbital motion of d shell electrons in Fe produce a
| magnetic field. I guess you could say there is a net current from those
| electrons. In unmagnetized Fe the orientation is random so the
magnetization
| cancels, but in a permanent magnet or magnetized iron they are aligned.
The question that you did not answer is how does the permanent magnet
current induce a current in the metal when no actual charge is transported?
| >It is not a contradiction. See above. I think you are contradicting
| >yourself.
|
| See what? Current is the flow of charge, period. No charge, no current.
That is not true since we are dealing with the *source* of elementary charge
e in the
first place.
| >| Many experiments have been run to find the electron's radius. All show
| >| it to be smaller than their margin of error, no matter how small that
is.
| >| The most "recent" ones ([probably any within the last 20 years) show it
| >| smaller than the VPP radius.
|
| >Impossible for it to be 10^-18 meters or smaller. See above. The
electric
| >potential energy for it gets ridiculous. So there is definitely
something
| >wrong with your *known*.
|
| I already corrected this mistake of yours.
No you didn't. Please explain in more detail above. You have not convinced
me yet.
| >| No, it's not just limited to "substances". It includes the various
| >| fields (B E whatever) of the cubes, any field, any information,
anything.
| >| In fact if we lop off the corners of the cubes (to eliminate anything
| >| travelling > c, we're left with NOTHING, except points where the
original
| >| vectors are tangent to the v=c circle.
|
| >Well then physics has a big problem. The phase velocity of EM radiation
in
| >the vacuum is c. But the velocity of the wave going from full crest peak
to
| >full trough peak is faster than c at the zero crossover point as the wave
| >travels past any point in space.
|
| WTF is this? The wave moves at a constant c.
Yes the wave moves in the x direction at a constant phase velocity of c
(well according to standard principles, but not to Tom). I am talking about
the velocity the wave has in the y direction orthogonal to x as it passes a
point in space.
| > Like Tom said, c is an rms velocity.
| >the wave is propagating in the x direction, what is the wave velocity in
the
| >y direction when it is at a peak? What is the wave velocity in the y
| >direction when it is midway between its "positive" going peak and
"negative"
| >going peak? As the wave passes a point in space.
|
| Zero. (you just said it propogates in the x direction)
The wave is propagating in the x direction but relative to an observation
point, it also has motion in the y orthogonal direction. I also said
passing a point in space that you snipped out for some reason and I put back
in. The wave velocity in the y direction when the wave is at its peaks is
zero. What is the velocity in the y direction relative to an observation
point when it is crossing the midway point between the "positive" peak and
the negative "peak"?
FrediFizzx
Michael Moroney
>| Wrong. The electric PE would have that value _if and only if_ the
>| electron were composed of numerous tiny charges, with their sum total
>| equal to the electron's charge, all crammed into that radius against their
>| mutual repulsion. Unfortunately for that argument, we already know the
>| electron's charge is quantized, it is not composed of smaller charges.
>| I previously mentioned this to you. (see also the "classical electron
>| radius")
>Are you saying that q^2/(8*pi*eps0*R) is not valid for a single electronic
>charge? If so, why not?
Yes! Because, as I just said, the charge cannot be divided into smaller
charges that can repel each other.
> This is the equation for the energy of a continous
>charge distribution in a sphere of radius R.
This is also classical physics. It will work correctly for a little
charged metal sphere with a very large number of excess electrons, but
not for an individual electron's composition.
> I think the electron qualifies
>for this. How do you explain this away?
You cannot divide its charge up.
> And what is the cutoff point? How
>many electrons do you have to have to make the equation valid? Two? Three?
>0ne hundred?
Kind of like asking how many water molecules does it take to create a
drop of water.
With two charges you have one repelling pair. With 3 charges you have
three pairs, 4 charges 6 pairs etc. It's a sum of the form n(n-1)/2,
for a larger number it can be done as an integral that derives that formula.
> And don't tell me this is the work required to assemble a
>bunch of point charges.
Sorry, but it was you who brought up potential energy as a function of
radius.
In the case of the electron, it is the work to make
>| >| any radius will work.
>|
>| >The Bohr for one meter would be 2.4*10^-11 amps*meter^2.
>| >The Bohr for the universe is ridiculous but can be calculated.
>| >The Bohr for the Planck length is also ridiculous but is 3.88*10^-46
>| >amps*meter^2.
>| >The Bohr for a radius of zero is zero.
>|
>| I'd like to see your math. Remember, the _definition_ of the Bohr is
>| e*h/(4*pi*me), or equivalently, e*hbar/(2*me). Since there is no Re
>| term in there, its value is irrelevant.
>Or equivalently the Bohr is equal to e*c*lambda_C/4*pi.
WRONG. You have been misled by Tom's faulty physics. I explicitly
mentioned the actual definition of the Bohr figuring this is where you
screwed up, and you *still* went off and used an incorrect definition.
Let's look a little more why it's wrong. Lambda_c is h/(c*me). Now
you go off and substitute various lengths for (h/(c*me)). Explain
this? h is a fundamental constant, as is c, and the electron's mass is
also a constant. So which of these constants are you consudering to
be a variable and WHY?
>| >The VPP model *does* have a current flowing around an area.
>|
>| I said you don't need a current to get a magnetic field. This means
>| a current is unnecessary.
>|
>| >| Current *is* moving charge. Again, I=dQ/dt.
>|
>| >Or a current can be caused from a magnetic field.
>|
>| True but irrelevant.
>Not irrelevant because that is exactly what is going on here.
No, there must be charges that can move for there to be a current.
Consider a coil with an alternating current in free space, There is
noting for it to induce a current in (other than itself) so it does
not induce a current.
> This is the *source* of your charge, Q, which happens to be e.
Which one of Maxwell's equations is _that_?
>| > Remember Bilge's question
>| >about finding the two currents in a system of a steel plate with a magnet
>| >holding it?
>|
>| > "You have a permanent magnet which you use to [hold] a piece of metal.
>| >Find the agent responsible for doing the [holding] and find 2 currents.
>|
>| > Hints: (1) Magnetic fields don't do work, so the magnetic field
>| > can't do the [holding];
>| > (2) There is a current in the permanent magnet and it induces
>| > a current in the metal, but no charge is actually
>transported;
>|
>| >Somewhat more challenging: Find the physical origin of the currents."
>|
>| >Did you ever figure this out? I am still working on it but not real
>hard.
>| >This is probably a good time though.
>|
>| This is simple. The orbital motion of d shell electrons in Fe produce a
>| magnetic field. I guess you could say there is a net current from those
>| electrons. In unmagnetized Fe the orientation is random so the
>magnetization
>| cancels, but in a permanent magnet or magnetized iron they are aligned.
>The question that you did not answer is how does the permanent magnet
>current induce a current in the metal when no actual charge is transported?
Two possible answers, depending on how you look at it. 1 is that the
current is always there, in the form of the iron's orbital electrons.
The other magnet just aligns them. 2 is the motion of the magnet as it
approaches the iron induces this pseudo-current.
>| >It is not a contradiction. See above. I think you are contradicting
>| >yourself.
>|
>| See what? Current is the flow of charge, period. No charge, no current.
>That is not true since we are dealing with the *source* of elementary charge
>e in the
>first place.
I have to repeat. How did you get that crap from Maxwell's Equations?
Again, look up the terms current and charge.
>| >Impossible for it to be 10^-18 meters or smaller. See above. The
>electric
>| >potential energy for it gets ridiculous. So there is definitely
>something
>| >wrong with your *known*.
>|
>| I already corrected this mistake of yours.
>No you didn't. Please explain in more detail above. You have not convinced
>me yet.
Yes I did. As I pointed out the electron cannot be created by assembling
smaller charges, there are no such things. The electron is fundamental.
>| >the vacuum is c. But the velocity of the wave going from full crest peak
>to
>| >full trough peak is faster than c at the zero crossover point as the wave
>| >travels past any point in space.
>|
>| WTF is this? The wave moves at a constant c.
>Yes the wave moves in the x direction at a constant phase velocity of c
>(well according to standard principles, but not to Tom). I am talking about
>the velocity the wave has in the y direction orthogonal to x as it passes a
>point in space.
Huh?
>The wave is propagating in the x direction but relative to an observation
>point, it also has motion in the y orthogonal direction. I also said
>passing a point in space that you snipped out for some reason and I put back
>in. The wave velocity in the y direction when the wave is at its peaks is
>zero. What is the velocity in the y direction relative to an observation
>point when it is crossing the midway point between the "positive" peak and
>the negative "peak"?
All I can think of is you are thinking of the path of a little surfer
dude "riding" the wave of a photon. Sorry, but the photon in this
case moves strictly in the x direction. No motion whatesoever in the Y
direction.
-Mike
ThomasL283
>>>pan...@moa.slac.stanford.edu (Jim Panetta) said:
>>>We know that the signal is from the beam of muons because they turned
>>>the beam off and the signal went away. That's how the background from
>>>cosmics was measured.
>>
>>Jim: Look at the original paper. Phys. Rev. Letters, Volume 9, Number 1,
>July
>>1, 1962.
>> They "calibrated" the spark gap detector by INCREASING beam energy, so that
>>muons (more) got through the shield.
>
>I guarantee they took data with the beams off. If only in testing.
>If the events were cosmic ray induced, they would have seen it then.
>than cosmic rays could cause. Cosmic rays are constant at about
>1/cm^2/sec. In our detector, we see a few hertz passing our cuts.
Exactly. But my point is that they were looking for muons. And they got some
from cosmic rays going horizontal (or nearly so). The "true" muons they
claimed were from neutrino events could just as well been strays from the
accelerator. After all they got muons in their detector from the beam by
increasing beam energy, during calibration runs. Besides, I know for a
certainty that a muon decay neutrino cannot create a new muon.
>>
>>Jim, they use the fact they don't see enough "events" as somehow proving the
>>neutrinos oscillate, from one type to another.
>
>They use the data on CC, NC, and elastic events to determine the
>non-nu_e factor. The total flux measured in the neutral current
>reaction is consistent with solar models. The flux in the CC
>reaction isn't. The only neutrinos that can participate in CC
>interactions are nu_e. All can participate in NC. Thus there
>is a significant non-nu_e component in the neutrino spectrum.
>
Unfortunately, the theoretical charged current reaction, neutral current
reaction and electron elastic scattering by neutrinos cannot happen.
Jim since VPP shows how neutrinos, electrons, protons, neutrons, muons and
pions are constructued, I know for a certainty that
neutrinos will never change a neutron into a proton and then the neutrino
change into an electron. (CC theory)
and neutrinos will never break apart the deuteron into a proton and neutron.
(NC theory).
>>I believe the neutrino events are non-existent, based on the fact
>>that the solar (EC) and (B+) decays don't produce the neutrinos they
>>claim to "not see".
>
>I'm sorry you believe this, but the data says that you are
>wrong.
Jim, if anyone claims to have seen that (EC) and (B+) create free neutrinos,
it must be a mistake. Kinematically it is impossible. See:
http://www.members.aol.com/tnlockyer/Betadecay.gif
Claiming neutrinos from (EC) or (B+) decay violates the conservation of energy.
>I've read your model. You don't make any good predictions,
>and what you do make is flawed by major misunderstandings on
>the nature of E&M, Special Relativity, and Newtonian Mechanics.
>
Jim, I don't believe you have seriously considered VPP.
Make no mistake about it, VPP, by definition, is the correct model for the
structures of energy and matter. The VPP proton and neutron structures allow
one to unify the strong and electromagnetic force, by calculating the binding
energy between individual nucleons, in nuclei.. (Never before possible). And
recently VPP has indicated that atoms using VPP electrons and VPP protons can
be modeled and is the topic of this thread.
Regards: Tom:
Jim Panetta
>
>>I guarantee they took data with the beams off. If only in testing.
>>If the events were cosmic ray induced, they would have seen it then.
>> They saw *lots* of events, only when the beam was on. Much more
>>than cosmic rays could cause. Cosmic rays are constant at about
>>1/cm^2/sec. In our detector, we see a few hertz passing our cuts.
>
>Exactly. But my point is that they were looking for muons. And they got some
>from cosmic rays going horizontal (or nearly so). The "true" muons they
>claimed were from neutrino events could just as well been strays from the
>accelerator. After all they got muons in their detector from the beam by
>increasing beam energy, during calibration runs. Besides, I know for a
>certainty that a muon decay neutrino cannot create a new muon.
WRONG. The chance for a real muon of the energy they were
beaming to reach all the way through the absorber is *far*
less than the number they saw. By *several* orders of magnitude.
More recent experiments use incredible baselines to examine the muon
neutrino in detail. No way in *hell* are you going to get beam
related muons punching through on order a kilometer of rock. And
then, just in case, you put a frigging veto detector in front of the
main volume. (CERN-PS-181, CHARM Collaboration)
>>>
>>>Jim, they use the fact they don't see enough "events" as somehow proving the
>>>neutrinos oscillate, from one type to another.
>>
>>They use the data on CC, NC, and elastic events to determine the
>>non-nu_e factor. The total flux measured in the neutral current
>>reaction is consistent with solar models. The flux in the CC
>>reaction isn't. The only neutrinos that can participate in CC
>>interactions are nu_e. All can participate in NC. Thus there
>>is a significant non-nu_e component in the neutrino spectrum.
>>
>
>Unfortunately, the theoretical charged current reaction, neutral current
>reaction and electron elastic scattering by neutrinos cannot happen.
Sorry, they *do*. This is DATA.
>Jim since VPP shows how neutrinos, electrons, protons, neutrons,
>muons and pions are constructued, I know for a certainty that
>neutrinos will never change a neutron into a proton and then the
>neutrino change into an electron. (CC theory) and neutrinos will
>never break apart the deuteron into a proton and neutron. (NC
>theory).
Sorry, it *does*. We see the neutron decays.
>>>I believe the neutrino events are non-existent, based on the fact
>>>that the solar (EC) and (B+) decays don't produce the neutrinos they
>>>claim to "not see".
>>
>>I'm sorry you believe this, but the data says that you are
>>wrong.
>
>Jim, if anyone claims to have seen that (EC) and (B+) create free
>neutrinos, it must be a mistake. Kinematically it is impossible.
>See: http://www.members.aol.com/tnlockyer/Betadecay.gif. Claiming
>neutrinos from (EC) or (B+) decay violates the conservation of
>energy.
How, exactly? I see in my calculations that energy, momentum,
and angular momentum are conserved.
>>I've read your model. You don't make any good predictions,
>>and what you do make is flawed by major misunderstandings on
>>the nature of E&M, Special Relativity, and Newtonian Mechanics.
>>
>Jim, I don't believe you have seriously considered VPP.
Why should I with so many mathematical errors?
>Make no mistake about it, VPP, by definition, is the correct model
>for the structures of energy and matter. The VPP proton and neutron
>structures allow one to unify the strong and electromagnetic force,
>by calculating the binding energy between individual nucleons, in
>nuclei.. (Never before possible). And recently VPP has indicated that
>atoms using VPP electrons and VPP protons can be modeled and is the
>topic of this thread.
By DEFINITION?!? That has all the logical veracity of the following:
"I so declare by swearing on this Genuine Buddha-with-a-clock-in-
its-stomach that the moon is made of compressed Wheatabix."
>Regards: Tom:
>
>Tom Lockyer (75 and retired) See "Vector Particle and Nuclear Models"
>0963154680 at http://www.amazon.com
>"When you can measure what you are speaking about and express it in numbers,
>you know something about it." Lord Kelvin (1824-1907)
Woohoo! Lord Kelvin -- Five Crackpot points!
(http://math.ucr.edu/home/baez/crackpot.html)
Michael Moroney
>>I guarantee they took data with the beams off. If only in testing.
>>If the events were cosmic ray induced, they would have seen it then.
>> They saw *lots* of events, only when the beam was on. Much more
>>than cosmic rays could cause. Cosmic rays are constant at about
>>1/cm^2/sec. In our detector, we see a few hertz passing our cuts.
>increasing beam energy, during calibration runs. Besides, I know for a
>certainty that a muon decay neutrino cannot create a new muon.
You _know_ for a certainty? Why? (What makes it a certainty)
>Unfortunately, the theoretical charged current reaction, neutral current
>reaction and electron elastic scattering by neutrinos cannot happen.
It's not theoretical, it's been seen.
>pions are constructued, I know for a certainty that
> neutrinos will never change a neutron into a proton and then the neutrino
>change into an electron. (CC theory)
Again you "know for a certainty"? Why? Because VPP says so I guess.
Did you ever stop and consider that VPP is just plain _wrong_? It's
very likely, given the myriad of faults it has.
>and neutrinos will never break apart the deuteron into a proton and neutron.
>(NC theory).
Bury your head in the sand and yell "It's not happening! It can't happen!!"
Meanwhile science moves forward, using the data to improve its theories.
>Jim, if anyone claims to have seen that (EC) and (B+) create free neutrinos,
>it must be a mistake. Kinematically it is impossible. See:
Kinematically the neutrino is _required_.
>Claiming neutrinos from (EC) or (B+) decay violates the conservation of energy.
Umm, no, the neutrino is required to conserve energy (and momentum, and
spin)
>Jim, I don't believe you have seriously considered VPP.
Most likely he looked at it, saw it loaded with faults an found it
worthless.
>Make no mistake about it, VPP, by definition, is the correct model for the
>structures of energy and matter.
By definition? What are you doing, trying to get a high score on the crackpot
index? Any set of beliefs that define itself to be the truth is either
religion or insane ravings. Anyone can declare that protons are made of
chocolate and electrons are vanilla and state by definition this is true.
-Mike
ThomasL283
>Somewhere in the newsfeed, ThomasL283 said:
>>
>>pan...@moa.slac.stanford.edu (Jim Panetta) said:
>>>I guarantee they took data with the beams off. If only in testing.
>>>If the events were cosmic ray induced, they would have seen it then.
>>> They saw *lots* of events, only when the beam was on. Much more
>>>than cosmic rays could cause. Cosmic rays are constant at about
>>>1/cm^2/sec. In our detector, we see a few hertz passing our cuts.
>>
>>Exactly. But my point is that they were looking for muons. And they got
>some
>>from cosmic rays going horizontal (or nearly so). The "true" muons they
>>claimed were from neutrino events could just as well been strays from the
>>accelerator. After all they got muons in their detector from the beam by
>>increasing beam energy, during calibration runs. Besides, I know for a
>>certainty that a muon decay neutrino cannot create a new muon.
>
>WRONG. The chance for a real muon of the energy they were
>beaming to reach all the way through the absorber is *far*
>less than the number they saw. By *several* orders of magnitude.
>
>More recent experiments use incredible baselines to examine the muon
>neutrino in detail. No way in *hell* are you going to get beam
>related muons punching through on order a kilometer of rock. And
>then, just in case, you put a frigging veto detector in front of the
>main volume. (CERN-PS-181, CHARM Collaboration)
Heroic efforts to get the conclusions accepted. The fact that one has to have
"veto" devices, to discount strays, points out the tentative nature of the
experiment.
It all boils down to simply believing what you want to believe.
There is fraud in the medical sciences, but no one wants to believe there
could be fraud in the physical sciences.
>>>>
>>>>Jim, they use the fact they don't see enough "events" as somehow proving
>the
>>>>neutrinos oscillate, from one type to another.
>>>
>>>They use the data on CC, NC, and elastic events to determine the
>>>non-nu_e factor. The total flux measured in the neutral current
>>>reaction is consistent with solar models. The flux in the CC
>>>reaction isn't. The only neutrinos that can participate in CC
>>>interactions are nu_e. All can participate in NC. Thus there
>>>is a significant non-nu_e component in the neutrino spectrum.
>>>
>>
>>Unfortunately, the theoretical charged current reaction, neutral current
>>reaction and electron elastic scattering by neutrinos cannot happen.
>
>Sorry, they *do*. This is DATA.
It depends on what you are willing to accept as data.. Out of several million
"events" SNO allowed about 1200 as neutrino events. They had contaminated the
detector with radioactive sources to "calibrate" the detectors, then had to go
to elaborate schemes to purify the water.
>>Jim since VPP shows how neutrinos, electrons, protons, neutrons,
>>muons and pions are constructed, I know for a certainty that
>>neutrinos will never change a neutron into a proton and then the
>>neutrino change into an electron. (CC theory) and neutrinos will
>>never break apart the deuteron into a proton and neutron. (NC
>>theory).
>
>Sorry, it *does*. We see the neutron decays.
Yes, a free neutron decays, but a nuclear neutron cannot be changed into a
proton by some theoretical neutrino after it coming 98 million miles and going
through 6800 feet of rock.
Do you really believe the far fetched idea that the neutrino can then "exchange
a W boson" changing the neutron, of the deuteron, into a proton and changing
itself into an electron?
If you do, any comments you might make on VPP validity are suspect.
>>
>>Jim, if anyone claims to have seen that (EC) and (B+) create free
>>neutrinos, it must be a mistake. Kinematically it is impossible.
>>See: http://www.members.aol.com/tnlockyer/Betadecay.gif. Claiming
>>neutrinos from (EC) or (B+) decay violates the conservation of
>>energy.
>
>How, exactly? I see in my calculations that energy, momentum,
>and angular momentum are conserved.
Yes, everyone simply uses rest mass energy of parent and daughter, and it does
show conservation of energy. But, what has been missing from all previous
calculations has been the changes in binding energy. In the example given:
http://www.members.aol.com/tnlockyer/Betadecay.gif
Look at the change in binding energy for (EC) or (B+) decay of Cu64.
It is 2.457439 MeV. The difference in atomic mass energy Cu64 to Ni64 is just
-Q2= -1.675106064 MeV. The balance (2.457439Mev-1.675106064 MeV = 0.78233293
MeV) goes into the daughters structure.
Since 0.78233293 MeV is (n-1H) exactly, the proton has absorbed this energy
(along with a shell or pair produced electron) to become a neutron.
For 40 years, everyone has ignored this binding energy change detail of
interpretation for beta decay, hence the error that has led to SNO neutrino
experiment on false pretenses.
So, SNO is going on what amounts to a fools errand. The neutrinos they claim
to detect simply DON'T exist.
>>>I've read your model. You don't make any good predictions,
>>>and what you do make is flawed by major misunderstandings on
>>>the nature of E&M, Special Relativity, and Newtonian Mechanics.
>>>
>
>>Jim, I don't believe you have seriously considered VPP.
>
>Why should I with so many mathematical errors?
Suspicions confirmed.
>>Make no mistake about it, VPP, by definition, is the correct model
>>for the structures of energy and matter. The VPP proton and neutron
>>structures allow one to unify the strong and electromagnetic force,
>>by calculating the binding energy between individual nucleons, in
>>nuclei.. (Never before possible). And recently VPP has indicated that
>>atoms using VPP electrons and VPP protons can be modeled and is the
>>topic of this thread.
>
>By DEFINITION?!? That has all the logical veracity of the following:
>"I so declare by swearing on this Genuine Buddha-with-a-clock-in-
>its-stomach that the moon is made of compressed Wheatabix."
I guess you have not followed the threads. The electron model is correct by
definition because it supports and predicts ALL related fundamental physical
constants, EXACTLY. And the proton and neutron models give the CODATA values
for the mass, and the neutron gives the decay electron and neutrino products,
directly from the rules of the model and their gives known decay mass.
Want to compare VPP results with the SM results after their 40 years of trying?
Did not think so.
Tom.
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| Wrong. The electric PE would have that value _if and only if_ the
| >| electron were composed of numerous tiny charges, with their sum total
| >| equal to the electron's charge, all crammed into that radius against
their
| >| mutual repulsion. Unfortunately for that argument, we already know the
| >| electron's charge is quantized, it is not composed of smaller charges.
| >| I previously mentioned this to you. (see also the "classical electron
| >| radius")
|
| >Are you saying that q^2/(8*pi*eps0*R) is not valid for a single
electronic
| >charge? If so, why not?
|
| Yes! Because, as I just said, the charge cannot be divided into smaller
| charges that can repel each other.
|
| > This is the equation for the energy of a continous
| >charge distribution in a sphere of radius R.
|
| This is also classical physics. It will work correctly for a little
| charged metal sphere with a very large number of excess electrons, but
| not for an individual electron's composition.
Charge is current times time in SI units. In MKS current is charge per
time. Which one is right? I say that they are both right. charge =
current*time is the same as saying current = charge/time. Is charge
producing the magneton or is the magneton producing the charge. Well, in
the case of the electron, it would seem to me that the magneton is producing
the elementary charge since to produce current thus the magneton, the
elementary charge would have to be rotating in a loop around an area. It
seems more likely to me that a chargeless electric field (and we know that
these can exist in conjunction with a chargeless magnetic field) is rotating
causing a current to make the magneton and that this is making the
elementary charge overall. I suppose you could say either scenario is
possible. However, they both require *real* physical rotation. But the
first one doesn't explain how elementary charge is created from no charge at
all while the second one does. And also explains the magnetic moment. Two
birds in the hand is better than one in the bush.
The Bohr = e*h/(4*pi*me). You can equally say that e = Bohr*4*pi*me/h or
that e = Bohr*4*pi/(lambda_C*c). I particularly like this last one. It
gets c involved in elementary charge. Maybe this is the true picture of
elementary charge.
And why is it that when Dirac solved the relativistic wave equation for
motion of a free electron, he came up with plus or minus c as the speed of
the electron? Simple, the electron is always moving at c because of its
rotation. IOW, there is no such thing as rest mass. Ever. Another problem
solved by VPP. Now to just figure out that blasted problem it created with
the cube corners moving faster than c. Well, maybe they don't move faster
than c. But this is a translational problem of going from linear motion to
rotational motion. Maybe this thing is a really strange beast and the
inside of the structure is moving at the same speed as the outside. I will
have to take a better look at all the math and see what is possible.
FrediFizzx
Michael Moroney
>| This is also classical physics. It will work correctly for a little
>| charged metal sphere with a very large number of excess electrons, but
>| not for an individual electron's composition.
>Charge is current times time in SI units. In MKS current is charge per
>time. Which one is right? I say that they are both right. charge =
Well yeah, there are many ways to come up with a set of fundamental units,
however since there is no way to have a current without charge, it seems
to me that charge is more "fundamental". But regardless, non sequitur.
>current*time is the same as saying current = charge/time. Is charge
>producing the magneton or is the magneton producing the charge.
The magneton is an approximation of the electron's magnetic moment.
> Well, in
>the case of the electron, it would seem to me that the magneton is producing
>the elementary charge since to produce current thus the magneton, the
>elementary charge would have to be rotating in a loop around an area. It
>seems more likely to me that a chargeless electric field (and we know that
>these can exist in conjunction with a chargeless magnetic field) is rotating
>causing a current to make the magneton and that this is making the
>elementary charge overall.
Gibberish. Review Maxwell's equations again. While you can create an
electric field with a (changing) magnetic field, you cannot create charge
from one.
>The Bohr = e*h/(4*pi*me). You can equally say that e = Bohr*4*pi*me/h or
>that e = Bohr*4*pi/(lambda_C*c).
No you can't. The Bohr is _defined_ in the terms of e, trying to define
e in the terms of the Bohr (defined in terms of e) results in an "x=x"
'proof'. Or more specifically, an "e=e" 'proof'. Worthless.
>solved by VPP. Now to just figure out that blasted problem it created with
>the cube corners moving faster than c.
Sounds like a big flaw to me.
> Well, maybe they don't move faster
>than c. But this is a translational problem of going from linear motion to
>rotational motion. Maybe this thing is a really strange beast and the
>inside of the structure is moving at the same speed as the outside. I will
>have to take a better look at all the math and see what is possible.
Sounds like "solving" the problem by generating gibberish.
-Mike
larry
> "Michael Moroney" <mor...@world.std.spaamtrap.com> wrote in message
> news:H1u1p0...@world.std.com...
> | "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
> |
> | >| Wrong. The electric PE would have that value _if and only if_ the
> | >| electron were composed of numerous tiny charges, with their sum total
> | >| equal to the electron's charge, all crammed into that radius against
> their
> | >| mutual repulsion. Unfortunately for that argument, we already know the
> | >| electron's charge is quantized, it is not composed of smaller charges.
> | >| I previously mentioned this to you. (see also the "classical electron
> | >| radius")
> |
> | >Are you saying that q^2/(8*pi*eps0*R) is not valid for a single
> electronic
> | >charge? If so, why not?
> |
> | Yes! Because, as I just said, the charge cannot be divided into smaller
> | charges that can repel each other.
> |
> | > This is the equation for the energy of a continous
> | >charge distribution in a sphere of radius R.
> |
> | This is also classical physics. It will work correctly for a little
> | charged metal sphere with a very large number of excess electrons, but
> | not for an individual electron's composition.
>
> Charge is current times time in SI units. In MKS current is charge per
> time. Which one is right? I say that they are both right. charge =
Time is a concept dealing with motion. Current is a concept dealing with
the motion of something. It is a relationship as is time. That something
moving is charge, so many charges per second through some reference
point. Thus charge is fundamental. The magneton is a relationship, not
something fundamental. It cannot be a cause of charge. Charge can be a
cause, as it is when in motion, of a magnetic field as shown by Oersted
(charge moving in a wire) and Rowland (charged rotating disk).
Relationships such as time or magneton cannot be a cause.
Larry
FrediFizzx
news:3D76E1A0...@charter.net...
That is all true *except* when dealing with the source of elementary charge
in the first place. And my point is that it is a rotating EM field (with no
charges present) that is making a current, the magnetic moment, and overall
elementary charge. They all exist together in perfect harmony. It is the
perfect dynamo and perpetual motion machine. Since this is involving
elementary charge in the first place, there are no smaller charges to make
the current. Sure we can just say that charge just is. But we can also
show and think of a possible mechanisn on how it is produced. This is one
possible way. This mechanism is not impossible. How is the magneton
produced otherwise? If you think the elementary charge is producing it, it
has to be physically rotating to produce a loop area current.
The biggest problem that I have with it, is that it implies that the H field
is responsible for positive charge since it is the primary field in Tom's
cube structure for the positron. This would make the positron the monopole
charge for the amps/meter field. Well, maybe that is why we can't find any
magnetic monopoles (we have found them if if it is the positron). Sure does
screw everything else up though. So how else might we get positive and
negative charge out of this beast? With the "current" field being the
primary rotating field, how would that necessarily make positive electric
charge? Yeah, this is the part I am having trouble with.
FrediFizzx
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| This is also classical physics. It will work correctly for a little
| >| charged metal sphere with a very large number of excess electrons, but
| >| not for an individual electron's composition.
|
| >Charge is current times time in SI units. In MKS current is charge per
| >time. Which one is right? I say that they are both right. charge =
|
| Well yeah, there are many ways to come up with a set of fundamental units,
| however since there is no way to have a current without charge, it seems
| to me that charge is more "fundamental". But regardless, non sequitur.
There is a way to have current without charge and the way the VPP electron
works is the only way. But in a way you are right because the current is
making the elementary charge anywise.
| >current*time is the same as saying current = charge/time. Is charge
| >producing the magneton or is the magneton producing the charge.
|
| The magneton is an approximation of the electron's magnetic moment.
|
| > Well, in
| >the case of the electron, it would seem to me that the magneton is
producing
| >the elementary charge since to produce current thus the magneton, the
| >elementary charge would have to be rotating in a loop around an area. It
| >seems more likely to me that a chargeless electric field (and we know
that
| >these can exist in conjunction with a chargeless magnetic field) is
rotating
| >causing a current to make the magneton and that this is making the
| >elementary charge overall.
|
| Gibberish. Review Maxwell's equations again. While you can create an
| electric field with a (changing) magnetic field, you cannot create charge
| from one.
Well I think a rotating EM field can produce a current, magnetic moment, and
elementary charge on the level that we are looking at. We are down to the
quantum here. Do Maxwell's equations really describe what is going on at
this level?
| >The Bohr = e*h/(4*pi*me). You can equally say that e = Bohr*4*pi*me/h
or
| >that e = Bohr*4*pi/(lambda_C*c).
|
| No you can't. The Bohr is _defined_ in the terms of e, trying to define
| e in the terms of the Bohr (defined in terms of e) results in an "x=x"
| 'proof'. Or more specifically, an "e=e" 'proof'. Worthless.
Wrong. The Bohr magneton can be defined in terms of a physical model that
shows a loop area current and you can plug that into e =
Bohr*4*pi/(lambda_C*c). Tom has shown this quite a few times.
| >solved by VPP. Now to just figure out that blasted problem it created
with
| >the cube corners moving faster than c.
|
| Sounds like a big flaw to me.
|
| > Well, maybe they don't move faster
| >than c. But this is a translational problem of going from linear motion
to
| >rotational motion. Maybe this thing is a really strange beast and the
| >inside of the structure is moving at the same speed as the outside. I
will
| >have to take a better look at all the math and see what is possible.
Well, we will see. I noticed you had no comments about Dirac's finding that
the electron motion is always at plus or minus c. How would you explain
that?
FrediFizzx
Michael Moroney
>|
>| >Charge is current times time in SI units. In MKS current is charge per
>| >time. Which one is right? I say that they are both right. charge =
Well if charge is current times time, and you think current is fundamental,
we have a problem here. If after 1 second (current*time) the electron has
a charge of e, after 2 seconds we have a charge of 2e, after a day,
86,400e, and after a while we'll have Freddi's Electron that Ate the
Universe.
>There is a way to have current without charge and the way the VPP electron
>works is the only way.
Sorry, I see no way whatsoever. Maxwell's Equations doesn't allow for it.
>| Gibberish. Review Maxwell's equations again. While you can create an
>| electric field with a (changing) magnetic field, you cannot create charge
>| from one.
>Well I think a rotating EM field can produce a current, magnetic moment, and
>elementary charge on the level that we are looking at.
Why? A rotating EM field can only induce a current (by providing forces
on charged particles), it can't create a current directly. Take a
transformer, remove its secondary, place it in empty space and apply AC to
its primary. What current does it induce? None! You removed the charges
the forces can act on when you removed the secondary.
> We are down to the
>quantum here. Do Maxwell's equations really describe what is going on at
>this level?
Maxwell's Equations work just fine for the photon. So now I see VPP is now
having to create new fundamental relationships out of thin air.
>| >The Bohr = e*h/(4*pi*me). You can equally say that e = Bohr*4*pi*me/h
>or
>| >that e = Bohr*4*pi/(lambda_C*c).
>|
>| No you can't. The Bohr is _defined_ in the terms of e, trying to define
>| e in the terms of the Bohr (defined in terms of e) results in an "x=x"
>| 'proof'. Or more specifically, an "e=e" 'proof'. Worthless.
>Wrong. The Bohr magneton can be defined in terms of a physical model that
>shows a loop area current and you can plug that into e =
>Bohr*4*pi/(lambda_C*c).
No. Once again the Bohr is _defined_ as e*h/(4*pi*me). It is a relationship
of more fundamental constants, the electron's mass and charge.
If we take your equation e=Bohr*4*pi/(lambda_C*c) we can rewrite as
e=[e*h/(4*pi*me)]*4*pi/(lambda_C*c) accurately (substituting the definition)
and we get e=e*h/(me*lambda_C*c) and by substituting the definition of
lambda_C (h/me*c) and simplifying we get e=e. A useless "x=x" proof.
Remember, the electron's actual magnetic moment differs from the Bohr
prediction.
What's worse is when I asked you to calculate the Bohr with differing radii
for the electron, you apparently took an incorrect definition of the Bohr
(in terms of lambda_C) and substituted different values for lambda_C.
Since lambda_C is *defined* as h/me*c, what justification can you give for
this? If lambda_C is h/me*c, and you felt free to substitute for it,
what changed? h, me or c? As far as I know they are all fundamental
constants.
> Tom has shown this quite a few times.
No, Tom has used this as a source of several useless "x=x" proofs.
(btw, when you first started replying to Tom, you were good at spotting
his "x=x" proofs and exposing them. Now you are suckered in by them.
What went wrong?)
>Well, we will see. I noticed you had no comments about Dirac's finding that
>the electron motion is always at plus or minus c. How would you explain
>that?
Not sure what you are talking about. All I know that electron and positron
speeds are routinely measured and are always less than c (sometimes
only slightly less).
-Mike
Michael Moroney
>| Time is a concept dealing with motion. Current is a concept dealing with
>| the motion of something. It is a relationship as is time. That something
>| moving is charge, so many charges per second through some reference
>| point. Thus charge is fundamental. The magneton is a relationship, not
>| something fundamental. It cannot be a cause of charge. Charge can be a
>| cause, as it is when in motion, of a magnetic field as shown by Oersted
>| (charge moving in a wire) and Rowland (charged rotating disk).
>| Relationships such as time or magneton cannot be a cause.
>| Larry
>That is all true *except* when dealing with the source of elementary charge
>in the first place.
Oh, now you need "special" rules.
>the current. Sure we can just say that charge just is. But we can also
But what are the sources of the electric and magnetic fields you say
creates charge. Oh, I see, they just are. So you haven't accomplished
anything.
>This is one
>possible way. This mechanism is not impossible.
According to the laws of physics as they are known it is impossible.
>The biggest problem that I have with it, is that it implies that the H field
>is responsible for positive charge since it is the primary field in Tom's
>cube structure for the positron. This would make the positron the monopole
>charge for the amps/meter field. Well, maybe that is why we can't find any
>magnetic monopoles (we have found them if if it is the positron). Sure does
>screw everything else up though. So how else might we get positive and
>negative charge out of this beast? With the "current" field being the
>primary rotating field, how would that necessarily make positive electric
>charge? Yeah, this is the part I am having trouble with.
Give it up. If the positron was a magnetic monopole, it would be
accelerated towards a magnetic pole. However, positron tracks in a bubble
chamber show a moving positron curves at a right angle to a magnetic field,
exactly the same as an electron except in the opposite direction.
-Mike
ThomasL283
From: thoma...@aol.com (ThomasL283)
Date: 9/2/2002 11:02 AM Pacific Daylight Time
Message-id: <20020902140249...@mb-ce.aol.com>
>sp?
>From: "FrediFizzx" fredifi...@ahahhotmail.com
>Date: 8/25/2002 6:45 PM Pacific Daylight Time
>Message-id: <YKfa9.1342$zx4.47...@newssvr21.news.prodigy.com>
>| >
>| >"ThomasL283" <thoma...@aol.com> wrote in
>|
>| >| I just can't imagine anything but certainty.
>| >
>| >Yes, and that certainty should be able to be calculated now that the real
>| >electron (and proton) geometry is known.
>|
>| >| I am looking for two null positions close together, or even a molecular
>| >| resonance between two or more hydrogen atoms.
>| >|
>| >| Back to the drawing board.
>| >
>OK, I think I might have a lead on the size of the atom problem. If we
>consider an ionized H_2 molecule in the ground state, that is one with just
>one electron, experiment has the center of mass of the two protons at about
>106 picometers apart (approx. 10^-10 meters). If we plug this into the
>electric potential energy equation
>e^2/(4*pi*eps_0*R) we get about 13.58 eV
>which is close to the ionization energy for H. So maybe this explains
>atomic distances? Now, what is happening with the electron? Is it bouncing
>back and forth between the two
Fredi: I don't know, have not gotten that far in the theory. FWIW, I favor
tight bonds (like springs).
BTW. I got Noggles book "Physical Chemistry using Mathcad" uses Mathcad 7, but
is still useful.
Looks interesting. Now if I can just somehow (understand) and use the Hermite
polynomial in the analysis.
Also here is the math that shows there are the Bohr and VPP null positions,
using the Mathcad solve engine.
http://www.members.aol.com/tnlockyer/twobond.gif
One solution is the Bohr radius, and the other solution is the VPP electron
null distance needed to balance the magnetic moment forces shown in the
revised:
http://www.members.aol.com/tnlockyer/VPPchem.gif
Regards, Tom:
ThomasL283
It is known that the Bohr electron orbital model for hydrogen is not
physically correct.
( The orbital angular momentum of the ground state of hydrogen is zero.)
And, by the rules of physics, an orbiting electron should radiate energy and
spiral into the proton. Bohr tried to balance the electric attraction by the
angular momentum, but in truth, it is the magnetic moments that oppose the
electron proton charge attraction, in the near field.
One can model the hydrogen atom using the VPP electron and VPP proton models.
To wit:
Remember the undergrad physics exercise you did on calculating the electric
field some distance from a "ring of charge" that required the cosine of theta =
R/sqr (R^2 + x^2) for the analysis?
Remember it was then a student exercise to show when separation x >>>R ring
radius, that the ring charge looks like a point charge in the far field?
See for example, Halliday and Resnick, Physics part II, page 673, ring of
charge.
Bohr assumed that the electron and proton where point charges, which put his
hydrogen model electron (wrongly) in the far field. The electron and proton
magnetic moments were (wrongly) omitted , but these magnetic moments are
essential to combine with the electric energy, for creating the experimental
(-13.5984 eV) binding energy photon.
Bohr did not have the luxury of a model for the electron's structure, and
(wrongly) assumed that the electron was a point particle. The electron's
charge proves to be in the form of two charge current rings.
Only in the near field can the essential magnetic energy null with the
electric energy to create an exothermic photon of (-13.5984 eV).
The ring charge geometry creates a cusp in the electric energy between the
electron's charge loop and the virtual point charge of the proton's small
charged (Hofstadter) core. Print out the following GIF example:
http://www.members.aol.com/tnlockyer/VPPchem.gif
If you have math software, note that the electrical potential energy equation
for (eV) has two separations that return (13.5984 eV).
These two separations are the Bohr radius (5.29 E-11 meter) in the far field,
and the VPP (1.20310032 E-15 meter) in the near field, but only the near field
magnetic energy nulls. Note that these results tend to verify the VPP ring
charge structure for the electron and positron. Print out the example.
http://www.members.aol.com/tnlockyer/VPPring.gif
Analysis of electron to proton binding is similar to the successful VPP
modeling and calculation of the (-13.5984 eV) binding energy between nucleons.
Bohr's model has no way to create the photon at the null, which requires both
the electric and the magnetic energy to be equal. The exothermic photon energy
(deficit) is required to hold (bind) the electron, until (+13.5984 eV)
ionization energy is added.
Regards: Tom:
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| Time is a concept dealing with motion. Current is a concept dealing
with
| >| the motion of something. It is a relationship as is time. That
something
| >| moving is charge, so many charges per second through some reference
| >| point. Thus charge is fundamental. The magneton is a relationship, not
| >| something fundamental. It cannot be a cause of charge. Charge can be a
| >| cause, as it is when in motion, of a magnetic field as shown by Oersted
| >| (charge moving in a wire) and Rowland (charged rotating disk).
| >| Relationships such as time or magneton cannot be a cause.
| >| Larry
|
| >That is all true *except* when dealing with the source of elementary
charge
| >in the first place.
|
| Oh, now you need "special" rules.
Isn't that what QM is? Special rules. Of course it is.
| >the current. Sure we can just say that charge just is. But we can also
|
| But what are the sources of the electric and magnetic fields you say
| creates charge. Oh, I see, they just are. So you haven't accomplished
| anything.
EM radiation. It is the thing that just is. Everything is made from EM
radiation.
| >This is one
| >possible way. This mechanism is not impossible.
|
| According to the laws of physics as they are known it is impossible.
Not when you apply "special rules" like QM does. QM *is* "special rules".
| >The biggest problem that I have with it, is that it implies that the H
field
| >is responsible for positive charge since it is the primary field in Tom's
| >cube structure for the positron. This would make the positron the
monopole
| >charge for the amps/meter field. Well, maybe that is why we can't find
any
| >magnetic monopoles (we have found them if if it is the positron). Sure
does
| >screw everything else up though. So how else might we get positive and
| >negative charge out of this beast? With the "current" field being the
| >primary rotating field, how would that necessarily make positive electric
| >charge? Yeah, this is the part I am having trouble with.
|
| Give it up. If the positron was a magnetic monopole, it would be
| accelerated towards a magnetic pole. However, positron tracks in a bubble
| chamber show a moving positron curves at a right angle to a magnetic
field,
| exactly the same as an electron except in the opposite direction.
I didn't say it was magnetic. I said it would be the *monopole* charge
associated with the H "current" field. It still has a positive electric
charge. Like I said, to me, this is the biggest problem concerning VPP. I
can't figure out how to reconcile this. I don't see how Tom is getting
negative and positive charge any other way. Unless there are actually more
vector combinations possible.
FrediFizzx
larry
> It is known that the Bohr electron orbital model for hydrogen is not
> physically correct.
> ( The orbital angular momentum of the ground state of hydrogen is zero.)
> And, by the rules of physics, an orbiting electron should radiate energy and
> spiral into the proton. Bohr tried to balance the electric attraction by the
> angular momentum, but in truth, it is the magnetic moments that oppose the
> electron proton charge attraction, in the near field.
>
> One can model the hydrogen atom using the VPP electron and VPP proton models.
> To wit:
>
> Remember the undergrad physics exercise you did on calculating the electric
> field some distance from a "ring of charge" that required the cosine of theta =
> R/sqr (R^2 + x^2) for the analysis?
> Remember it was then a student exercise to show when separation x >>>R ring
> radius, that the ring charge looks like a point charge in the far field?
What does it look like when not oriented on the perpendicular through
the center of the ring? That is a possible problem with a charge-ring
electron, since it would not act as a central field, i.e., as a point
source of an electric field, with the field flux equal omnidirectionally.
Larry
larry
Kind of a current of photons producing a field of virtual photons(E
field) which is the identity of a charge and since it appears as
rotating, a magnetic field is seen relative to the field of virtual
photons emitted from the ring? Can a photon on a circular path
produce virtual photons omnidirectionally? Can a number of photons
obtain circular paths other than in an extremely intense gravitational
field or within some material body? Is there something wrong with the
idea that a photon can be composed of electric and magnetic fields which
just remain in some localized spatial volume?
> perfect dynamo and perpetual motion machine. Since this is involving
> elementary charge in the first place, there are no smaller charges to make
> the current. Sure we can just say that charge just is. But we can also
Then why refer to a charge loop of 1/2 e? Somehow the VPP electron must
divide e into smaller units of charge, since the electron defines e
and VPP tries to model charge as two rotating current rings which are
the producers of the E field, which is the identity of charge.
> show and think of a possible mechanisn on how it is produced. This is one
> possible way. This mechanism is not impossible. How is the magneton
The magneton is a relationship and not something produced. The magnetic
moment is produced because of the motion of charge relative to some
other body. If an electron is really a rotating body, then a relative
magnetic field would be described as a magnetic moment relative to
another body.
> produced otherwise? If you think the elementary charge is producing it, it
> has to be physically rotating to produce a loop area current.
>
> The biggest problem that I have with it, is that it implies that the H field
Is the H field just due to relative motion? In the Rowland rotating
charged disk experiment, suppose a magnet is rotating along with the
disk above some fixed point on the disk. Does the magnet feel an H
field? No, because the H field is an effect due to relative motion with
respect to the E field.
Larry
FrediFizzx
Good thought provoking questions. I don't know the answers. Except maybe
for the circular path. See below.
|
| > perfect dynamo and perpetual motion machine. Since this is involving
| > elementary charge in the first place, there are no smaller charges to
make
| > the current. Sure we can just say that charge just is. But we can also
|
|
| Then why refer to a charge loop of 1/2 e? Somehow the VPP electron must
| divide e into smaller units of charge, since the electron defines e
| and VPP tries to model charge as two rotating current rings which are
| the producers of the E field, which is the identity of charge.
I don't really think you can look at it as a 1/2 e charge loop. That was a
simplification that Tom was showing to illustrate his point. Sure there are
two current rings but they can't exist without each other. It is what
allows the whole structure to be a little quantum dynamo. The whole
structure allows the EM wave to follow a circular path. This is how it does
it without a strong gravitational field like you mention above.
| > show and think of a possible mechanisn on how it is produced. This is
one
| > possible way. This mechanism is not impossible. How is the magneton
|
|
| The magneton is a relationship and not something produced. The magnetic
| moment is produced because of the motion of charge relative to some
| other body. If an electron is really a rotating body, then a relative
| magnetic field would be described as a magnetic moment relative to
| another body.
Yes, sorry. I meant magnetic moment. I don't quite get your last
statement.
| > produced otherwise? If you think the elementary charge is producing it,
it
| > has to be physically rotating to produce a loop area current.
| >
| > The biggest problem that I have with it, is that it implies that the H
field
|
|
| Is the H field just due to relative motion? In the Rowland rotating
| charged disk experiment, suppose a magnet is rotating along with the
| disk above some fixed point on the disk. Does the magnet feel an H
| field? No, because the H field is an effect due to relative motion with
| respect to the E field.
| Larry
I think you might be right considering relativistic motion. One book I have
successfully describes the magnetic force as purely a relativistic effect.
| > is responsible for positive charge since it is the primary field in
Tom's
| > cube structure for the positron. This would make the positron the
monopole
| > charge for the amps/meter field. Well, maybe that is why we can't find
any
| > magnetic monopoles (we have found them if if it is the positron). Sure
does
| > screw everything else up though. So how else might we get positive and
| > negative charge out of this beast? With the "current" field being the
| > primary rotating field, how would that necessarily make positive
electric
| > charge? Yeah, this is the part I am having trouble with.
So does that get us anywhere in regards to this problem?
FrediFizzx
ThomasL283
>Date: 9/6/2002 12:41 AM Pacific Daylight Time
Wrote in:
>Message-id: <k_Yd9.819$3z5.67...@newssvr21.news.prodigy.com>
>snip<
>|"larry" <gold...@charter.net> wrote in message
>snip<
>| Then why refer to a charge loop of 1/2 e? Somehow the VPP electron must
>| divide e into smaller units of charge, since the electron defines e
>| and VPP tries to model charge as two rotating current rings which are
>| the producers of the E field, which is the identity of charge.
>I don't really think you can look at it as a 1/2 e charge loop. That was a
>simplification that Tom was showing to illustrate his point. Sure there are
>two current rings but they can't exist without each other. It is what
>allows the whole structure to be a little quantum dynamo.
Yes, that is what the VPP model is apparently saying.
Note that this is born out in the hydrogen electron-proton bond length
calculations using the ring 1/2e charge of the VPP electron model.
http://www.members.aol.com/tnlockyer/twobond.gif
But note. (Q) in above GIF, the VPP electron model's near ring of (0.5e)
charge seems to be the effective value, for bonding. In this near field,
apparently the outer ring is effectively "guarded", from the protons much
smaller charged core, and thus only the one charge ring nearest seems to be
effective.
(notice this VPP equation also returns the Bohr radius in the far field, as
one of the solutions.)
What would be nice is if the electric and magnetic force equations can be
improved and combined to obtain bond length, directly, without having to use
13.5984 eV as a given. That would be final proof of the success or failure of
the VPP hydrogen model.
I am reasonably sure the electric energy equation algebra is correct, but I am
not too happy with the magnetic energy eqaution.
Analysis of the coupling between magnetic moments is not a easy problem because
of the relative physical sizes of the moments.
I don't expect ready answers, but please think about it and try to some
calculations, if you can. (Mathcad to the rescue. ;-) ).
FrediFizzx
At close range, sure the two rings would become important individually.
What is doing the "guarding"? Since only one half negative e is cancelling
out one half positive e what is the remaining one half negative and positive
charges doing? Are they forming a return circuit of some kind? So we end
up with an extended field curved around this whole structure? Hmmm. Kind
of like two bar magnets held apart slightly with a north and south pole
together only this forms two electric dipole sysems. The magnetic moments
are what is keeping the plus and minus "poles" apart ever so slightly.
Plus, to me, this whole action should be distorting the shape of the
electron and proton. Maybe a bunch as far as the electron is concerned;
maybe not as much for the proton core. Difficult to calculate this. I am
still going to try and work on what happens if the VPP proton is in the
center of the VPP electron. But I am liking the two electric dipole pair
concept now that I am thinking about it. Because the proton being in the
center of the electron would be more like electron capture. But maybe it
still works out because a neutrino is needed for capture all the way.
| What would be nice is if the electric and magnetic force equations can be
| improved and combined to obtain bond length, directly, without having to
use
| 13.5984 eV as a given. That would be final proof of the success or
failure of
| the VPP hydrogen model.
Yeah, what is the deal with Mathcad saying there is no solution when I try
to combine them? When I can see that there is a solution in your graphs.
| I am reasonably sure the electric energy equation algebra is correct, but
I am
| not too happy with the magnetic energy eqaution.
What exactly aren't you happy with in the mag equation?
| Analysis of the coupling between magnetic moments is not a easy problem
because
| of the relative physical sizes of the moments.
None of this is very easy at all. Its looking to me like the full
mathematical treatment is about as complex as the Schrodinger wave equation
solutions for the motion of an electron in the H atom.
FrediFizzx
ThomasL283
>Date: 9/7/2002 12:25 AM Pacific Daylight Time
Wroye in:
>Message-id: <gRhe9.1808$UY3.91...@newssvr21.news.prodigy.com>
>
>"ThomasL283" <thoma...@aol.com> wrote in message
>snip<
>|
>| Note that this is born out in the hydrogen electron-proton bond length
>| calculations using the ring 1/2e charge of the VPP electron model.
>|
>| http://www.members.aol.com/tnlockyer/twobond.gif
>|
>snip<
>| apparently the outer ring is effectively "guarded", from the protons much
>| smaller charged core, and thus only the one charge ring nearest seems to
>be
>| effective. (notice this VPP equation also returns the Bohr radius in the
>far field, as
>| one of the solutions.)
>At close range, sure the two rings would become important individually.
>What is doing the "guarding"?
Fredi: The use of guard rings is well known to be effective in constructing
standard capacitiors, to prevent fringing.
Seems when two conductors are at the same potential, (like the two VPP charge
rings) no current can flow between them.
This may be what the model math is showing.
The electric field of the outer ring is (apparently) effectively kept (guarded)
from the being seen by the VPP proton models charged core.
>| What would be nice is if the electric and magnetic force equations can be
>| improved and combined to obtain bond length, directly, without having to
>use
>| 13.5984 eV as a given. That would be final proof of the success or
>failure of
>| the VPP hydrogen model.
>Yeah, what is the deal with Mathcad saying there is no solution when I try
>to combine them? When I can see that there is a solution in your graphs.
You are better at algebra than I am. I failed (so far) to combine the two
equations. (This research is a young man's game. ;-( )
>| I am reasonably sure the electric energy equation algebra is correct, but
>I am
>| not too happy with the magnetic energy eqaution.
>
>What exactly aren't you happy with in the mag equation?
Well the "fudge factor" is not exactly at the centroid. .
Apparently it is not that simple.
Unlike the charges of the rings, the magnetic force requires BOTH current
rings creates the magnetic flux, the VPP electron acting much like from a two
turn DC inductor.
>| Analysis of the coupling between magnetic moments is not a easy problem
>because
>| of the relative physical sizes of the moments.
>None of this is very easy at all. Its looking to me like the full
>mathematical treatment is about as complex as the Schrodinger wave equation
>solutions for the motion of an electron in the H atom.
Perhaps the (magnetic moments) of the VPP electron and proton current rings
can be analyized as two "non concentric" coaxial inductors.
The Neumann formulae for mutual inductance only works with two mutually coupled
current rings, and is no help.
Terman, RADIO ENGINEERING (1937) (pages 794,795) shows some math for the
mutual inductance (hence coupling factor) that uses two cosine relationships,
one from each end, of the larger inductor, to the smaller inductors midpoint.
They do some subtraction of cosines and other stuff I can't quite figure out,
and come up with the mutual inductance, from the geometry.
See also one current loop analysis in Halliday and Resnick, Physics, Part II
page 862.
Still studying hopefully.
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >|
| >| >Charge is current times time in SI units. In MKS current is charge
per
| >| >time. Which one is right? I say that they are both right. charge =
|
| Well if charge is current times time, and you think current is
fundamental,
| we have a problem here. If after 1 second (current*time) the electron has
| a charge of e, after 2 seconds we have a charge of 2e, after a day,
| 86,400e, and after a while we'll have Freddi's Electron that Ate the
| Universe.
That is not right. e is a constant, time is a constant and current is a
constant in the equation. The current is sustaining e. Stop the current;
no more e.
| >There is a way to have current without charge and the way the VPP
electron
| >works is the only way.
|
| Sorry, I see no way whatsoever. Maxwell's Equations doesn't allow for it.
You have to stop thinking Maxwell here when dealing with quantum level
fundamentals. Just as you ask me to stop thinking of the electron as
rotating solid ball, you need to think in terms of what generates the
fundamental electronic charge in the first place. Is Maxwell going to be
helpful in this situation? I don't think so.
| >| Gibberish. Review Maxwell's equations again. While you can create an
| >| electric field with a (changing) magnetic field, you cannot create
charge
| >| from one.
|
| >Well I think a rotating EM field can produce a current, magnetic moment,
and
| >elementary charge on the level that we are looking at.
|
| Why? A rotating EM field can only induce a current (by providing forces
| on charged particles), it can't create a current directly. Take a
| transformer, remove its secondary, place it in empty space and apply AC to
| its primary. What current does it induce? None! You removed the charges
| the forces can act on when you removed the secondary.
|
| > We are down to the
| >quantum here. Do Maxwell's equations really describe what is going on at
| >this level?
|
| Maxwell's Equations work just fine for the photon. So now I see VPP is
now
| having to create new fundamental relationships out of thin air.
Please describe a photon using only Maxwell's equations. For that matter,
describe an electron using Maxwell's equations. They can't be fully
described only using Maxwell. You have to bring in QED or some other idea.
| >| >The Bohr = e*h/(4*pi*me). You can equally say that e = Bohr*4*pi*me/h
| >or
| >| >that e = Bohr*4*pi/(lambda_C*c).
| >|
| >| No you can't. The Bohr is _defined_ in the terms of e, trying to
define
| >| e in the terms of the Bohr (defined in terms of e) results in an "x=x"
| >| 'proof'. Or more specifically, an "e=e" 'proof'. Worthless.
|
| >Wrong. The Bohr magneton can be defined in terms of a physical model
that
| >shows a loop area current and you can plug that into e =
| >Bohr*4*pi/(lambda_C*c).
|
| No. Once again the Bohr is _defined_ as e*h/(4*pi*me). It is a
relationship
| of more fundamental constants, the electron's mass and charge.
| If we take your equation e=Bohr*4*pi/(lambda_C*c) we can rewrite as
| e=[e*h/(4*pi*me)]*4*pi/(lambda_C*c) accurately (substituting the
definition)
| and we get e=e*h/(me*lambda_C*c) and by substituting the definition of
| lambda_C (h/me*c) and simplifying we get e=e. A useless "x=x" proof.
| Remember, the electron's actual magnetic moment differs from the Bohr
| prediction.
I did not say to insert the standard definition of the Bohr. I said to
insert the loop area current from a model (which is defined from the model's
geometry and rotation parameters). Of course you will get e=e if you insert
the standard definition. Anyone can see that. The point being here that
the model geometry is consistent with and supports the Bohr. No other
physical model that I have seen yet does this. I guess it gets into sort of
a chicken or the egg kind of thing. Which came first? The Bohr equation or
the model? If the VPP model is indeed correct, then it came first and the
Bohr equation came from it. The Bohr magneton should be defined as total
current area, lambda_C^2/4*pi, times the current which is
sqrt(2*alpha*eps0*h*c)*c/lambda_C. Of course we still get the chicken or
egg thing since alpha has e in it. I totally see your point but I hope you
can see that the model is supporting all of the constants and equations. So
what is really needed here is a different equation for the current that I
have above that does not reduce back to having an e in it. But the problem
is that since we are defining the generation of e in the first place, there
is really nothing that defines it. And if you have current, you will always
have e and if you have e in motion you will have current. Well, one
difference here is the motion. The kind of motion. In the case of e for
the VPP model, it is not in motion per se if we consider the electron at
rest. Of course its internal motion is not at rest but its linear motion
is. It is the whole structure that is generating e, mass, the Bohr etc.
You definitely can't visualize this thing as a solid rotating ball. Nothing
works if you do that. It is actually very similar to visualizing it as a
Schrodinger wavefunction. But with more actual form.
| What's worse is when I asked you to calculate the Bohr with differing
radii
| for the electron, you apparently took an incorrect definition of the Bohr
| (in terms of lambda_C) and substituted different values for lambda_C.
| Since lambda_C is *defined* as h/me*c, what justification can you give for
| this? If lambda_C is h/me*c, and you felt free to substitute for it,
| what changed? h, me or c? As far as I know they are all fundamental
| constants.
Like I said before, It is no longer a Bohr magneton when doing that of
course. But still would/could be a magnetic moment for the described
system.
| > Tom has shown this quite a few times.
|
| No, Tom has used this as a source of several useless "x=x" proofs.
|
| (btw, when you first started replying to Tom, you were good at spotting
| his "x=x" proofs and exposing them. Now you are suckered in by them.
| What went wrong?)
Nothing went wrong. I simple saw that it is just as equally valid to say
that the model is generating the electron constants and equations. The VPP
model does in fact support all of them. The only one that I have seen so
far. I totally understand the circular argument thing. But I don't know
how to resolve that yet if it in fact can be resolved due to the fact that
current and charge are somewhat complementary. But the model geometry parts
do support all the fundamental constants and equations. It is a totally
chicken or egg thing.
| >Well, we will see. I noticed you had no comments about Dirac's finding
that
| >the electron motion is always at plus or minus c. How would you explain
| >that?
|
| Not sure what you are talking about. All I know that electron and
positron
| speeds are routinely measured and are always less than c (sometimes
| only slightly less).
When Dirac solved his relativistic wave equation for the motion of a free
electron for the x component of velocity, he found that the speed of the
electron was always plus or minus c. Of course, as you mention above, the
speeds are measured below c. What can be the explanation for this?
According to the following linked article, Dirac explained it as his
theoretical result was at any instantaneous moment in time. Experimental
measurements are averaged over a period of time.
http://arxiv.org/PS_cache/physics/pdf/0206/0206061.pdf
To me, it could mean that the electron's motion is always c even when it is
at rest because of its rotation. If you take the x component of its angular
velocity at any instant of time, it will equal plus or minus c.
FrediFizzx
Todd Desiato
"FrediFizzx" <fredifi...@ahahhotmail.com> wrote in message
news:_g9h9.276$at6.9...@newssvr21.news.prodigy.com...
> "Michael Moroney" <mor...@world.std.spaamtrap.com> wrote in message
> news:H1z10y...@world.std.com...
> | "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
> |
> | >|
> | >| >Charge is current times time in SI units. In MKS current is charge
> per
> | >| >time. Which one is right? I say that they are both right. charge
=
> |
> | Well if charge is current times time, and you think current is
> fundamental,
> | we have a problem here. If after 1 second (current*time) the electron
has
> | a charge of e, after 2 seconds we have a charge of 2e, after a day,
> | 86,400e, and after a while we'll have Freddi's Electron that Ate the
> | Universe.
>
> That is not right. e is a constant, time is a constant and current is a
> constant in the equation. The current is sustaining e. Stop the current;
> no more e.
To be quite accurate, current is the derivative of the charge with respect
to time, or the accumulated charge is the integral of the current over time.
This does not deny the fact that individual charges have a constant value e.
The value of e is a very small number, so even 1 Ampere of current is like
10^19 electrons per second. Any individuality is totaly obscured by the huge
number of electrons. There are some semi-conductor junctions where they can
control the current to 1 electron at a time.
> | >There is a way to have current without charge and the way the VPP
> electron
> | >works is the only way.
> |
> | Sorry, I see no way whatsoever. Maxwell's Equations doesn't allow for
it.
>
> You have to stop thinking Maxwell here when dealing with quantum level
> fundamentals. Just as you ask me to stop thinking of the electron as
> rotating solid ball, you need to think in terms of what generates the
> fundamental electronic charge in the first place. Is Maxwell going to be
> helpful in this situation? I don't think so.
Nothing generates the fundamental charge. Charge is just another way of
saying energy, a displacement from zero. In physical reality there is no way
to separate these things, so in fact it is one object;
charge = e
angular momentum = hbar/2
magnetic flux = hbar / 2e
This must be thought of as 1-object which spans from the location of the
charge to infinity. The field of an electron cannot be considered an
independent entity any more than the charge can be considered apart from
it's field. You can think of the spin of a magnetic flux line as creating e,
or the spin of e creating the flux, but it doesn't matter how you think of
it, in reality it is only 1 object. One cannot exist without the other.
> | >| Gibberish. Review Maxwell's equations again. While you can create
an
> | >| electric field with a (changing) magnetic field, you cannot create
> charge
> | >| from one.
Actually you can, at high enough energy density you create e-p pairs. It's
easier in the presence of matter, but it can be done in a vacuum too.
> | >Well I think a rotating EM field can produce a current, magnetic
moment,
> and
> | >elementary charge on the level that we are looking at.
> |
> | Why? A rotating EM field can only induce a current (by providing forces
> | on charged particles), it can't create a current directly. Take a
> | transformer, remove its secondary, place it in empty space and apply AC
to
> | its primary. What current does it induce? None! You removed the charges
> | the forces can act on when you removed the secondary.
When you energize the primary of a transformer without a secondary coil you
not only generate an electric and a magnetic field, you also generate a
magnetic vector potential "A", which is physically a circulation of photons
in space, i.e., stored energy "circulating" around the coil. It is current
of photons.
.....big snip
> | >Well, we will see. I noticed you had no comments about Dirac's finding
> that
> | >the electron motion is always at plus or minus c. How would you
explain
> | >that?
> |
> | Not sure what you are talking about. All I know that electron and
> positron
> | speeds are routinely measured and are always less than c (sometimes
> | only slightly less).
>
> When Dirac solved his relativistic wave equation for the motion of a free
> electron for the x component of velocity, he found that the speed of the
> electron was always plus or minus c. Of course, as you mention above, the
> speeds are measured below c. What can be the explanation for this?
> According to the following linked article, Dirac explained it as his
> theoretical result was at any instantaneous moment in time. Experimental
> measurements are averaged over a period of time.
>
> http://arxiv.org/PS_cache/physics/pdf/0206/0206061.pdf
>
> To me, it could mean that the electron's motion is always c even when it
is
> at rest because of its rotation. If you take the x component of its
angular
> velocity at any instant of time, it will equal plus or minus c.
>
Fred the paper you cited above is not a good reference. The author has many
of his own personal misconceptions and apparently prejudces about
Relativity. For example he equates;
gamma*mc^2 = mc^2 + E_k
but this is NOT an equality, it is a first order approximation. He uses it
as an equality and shows that it is wrong, which is true because it is only
an approximation, but he never mentions that. I gave up on the paper after I
saw what he was getting at. Wave mechanics and the Dirac equation are in
complete agreement. The motion you refer to above, where the electrons
instantaneous velocity is +/- c is not a "rotation" but you can think of it
that way if it were in a magnetic field. It is actually described by a
jitter in the instantaneous position of the electron. In a magnetic field
the average motion is a rotation which is the z component of angular
momentum. As a free electron it is bounced around by it's coupling to the
local vacuum field. The interaction of the electron and its surrounding
environment is a complicated issue, not easily explained by words. In the
most basic sense an electron is NEVER at rest, only it's average position
may be represented as some statistical distribution about a point in
space-time. To think that an electron could be at rest would deny the fact
that the vacuum is not empty and that the electron were somehow all alone in
the Universe. In fact this never happens. The uncertainty in the position is
on the order of magnitude of the Compton wavelength.
Here is a link to the actual derivation.
http://www.todds.info/Dirac/9_11.gif
http://www.todds.info/Dirac/Dirac.html
Unfortunately you need to download the jpg file and zoom it to read it. The
fact is that the derivation of the equations of motion from a Lagrangian
equation are well tested and well understood. The answer of a velocity
"operator" whose eigenvalues are +/- c must be correct. It is just a matter
then of interpretation. This is also a clue as to why it is thought that the
electron gets its a large portion of it's mass from its interaction with the
vacuum.
Regards,
Todd Desiato
larry
>
> Nothing went wrong. I simple saw that it is just as equally valid to say
> that the model is generating the electron constants and equations. The VPP
It might be good to inject a little objective reality in to this.
A model is conceptual and "generates" implies some kind of action which
a model is not capable of doing. "electron constants and equations"
describe the identity of the electron, which is the acting body. There
is no "chicken or egg thing" involved. The model describes something
about objective reality, nothing more or less. The Bohr magneton is a
relationship between constants, a number, and cannot be a cause of the
identities of the bodies involved in its definition. Current is defined
in terms of motion of charge and thus cannot be a cause of charge.
Larry
FrediFizzx
| FrediFizzx wrote:
|
|
| >
| > Nothing went wrong. I simple saw that it is just as equally valid to
say
| > that the model is generating the electron constants and equations. The
VPP
|
|
| It might be good to inject a little objective reality in to this.
| A model is conceptual and "generates" implies some kind of action which
| a model is not capable of doing. "electron constants and equations"
| describe the identity of the electron, which is the acting body. There
| is no "chicken or egg thing" involved. The model describes something
| about objective reality, nothing more or less. The Bohr magneton is a
| relationship between constants, a number, and cannot be a cause of the
| identities of the bodies involved in its definition. Current is defined
| in terms of motion of charge and thus cannot be a cause of charge.
| Larry
Well, of course I meant the model as a physical reality. *If* the model is
the way an actual electron is, then this modeled electron is in fact
*generating* charge, mass, a magnetic moment, etc. That is what I meant.
However, the main point is that the VPP electron *geometry* does support all
the electron constants and equations as well as they support it. You have
to admit that at least that the VPP model does get that right. If it didn't
do that, I wouldn't even be bothering with it.
The *standard* definition of current is in terms of motion of charge. But
what about the definition of *elementary* charge itself? What is that? Why
is that missing? I say that you have to define it in terms of a rotating E
field that had no charge to start with but as soon as it gets rotational
motion thus making a current, you get the elementary charge. The problem I
am having with this now is how to get plus and minus out of it. But a
fellow just posted a link to an interesting paper on QM as Electrodynamics
of Curvilinear Waves. However, he only offers up plus and minus loosely as
the E field pointing in and the E field pointing out on the ring and shows
no math for it. Plus I don't think his torus structure is stable.
FrediFizzx
Fredi Fizzx
All true except possibly when dealing with the fundamental creation of
charge in the first place. About 19 amps of current circulating in
the split second for one rotation equals the elementary charge. If
you think of elementary charges making this current then you are going
to get messed up. The current is always the same. It is not more
electrons flowing as time goes by. It is a constant caused by the
rotation of a negative E field.
> > | >There is a way to have current without charge and the way the VPP
> electron
> > | >works is the only way.
> > |
> > | Sorry, I see no way whatsoever. Maxwell's Equations doesn't allow for
> it.
> >
> > You have to stop thinking Maxwell here when dealing with quantum level
> > fundamentals. Just as you ask me to stop thinking of the electron as
> > rotating solid ball, you need to think in terms of what generates the
> > fundamental electronic charge in the first place. Is Maxwell going to be
> > helpful in this situation? I don't think so.
>
> Nothing generates the fundamental charge. Charge is just another way of
> saying energy, a displacement from zero. In physical reality there is no way
> to separate these things, so in fact it is one object;
>
> charge = e
> angular momentum = hbar/2
> magnetic flux = hbar / 2e
>
> This must be thought of as 1-object which spans from the location of the
> charge to infinity. The field of an electron cannot be considered an
> independent entity any more than the charge can be considered apart from
> it's field. You can think of the spin of a magnetic flux line as creating e,
> or the spin of e creating the flux, but it doesn't matter how you think of
> it, in reality it is only 1 object. One cannot exist without the other.
Yes, I agree. They definitely go together. I think that is what I
was trying to say. Just as the Bohr magneton can be defined in terms
of e, e should be able to be defined in terms of it. I do realize
though, that it requires a different way to define the Bohr magneton
for its current part. I did just get some new clues on how to maybe
make it work. I just have to figure out the plus and minus business.
OK, thanks. I did actually buy the Milonni book so I have it already
but had not gotten that far reading it yet. Yes, I also thought that
the article in the link I posted was not correct about its main theme
but I didn't think it was not telling the truth about what Dirac found
about the x component of the velocity of the electron. Well, which
interpretation is correct? Are these electrons really getting knocked
around at the speed of c by the vacuum? Or is the x component of
their angular velocity equal to c? The latter just seems more
probable to me.
FrediFizzx
Todd Desiato
"Fredi Fizzx" <fredi...@hotmail.com> wrote in message
news:fe51764e.02091...@posting.google.com...
> "Todd Desiato" <todd...@serversanddomains.com> wrote in message
news:<QVsh9.6857$V7.18...@twister.socal.rr.com>...
> > "FrediFizzx" <fredifi...@ahahhotmail.com> wrote in message
snip
I don't see how you can do this. If I integrate 19 amps over 1 second I get
19 Coulombs, not 1.602*10^-19 Coulombs.
> > Nothing generates the fundamental charge. Charge is just another way of
> > saying energy, a displacement from zero. In physical reality there is no
way
> > to separate these things, so in fact it is one object;
> >
> > charge = e
> > angular momentum = hbar/2
> > magnetic flux = hbar / 2e
> >
> > This must be thought of as 1-object which spans from the location of the
> > charge to infinity. The field of an electron cannot be considered an
> > independent entity any more than the charge can be considered apart from
> > it's field. You can think of the spin of a magnetic flux line as
creating e,
> > or the spin of e creating the flux, but it doesn't matter how you think
of
> > it, in reality it is only 1 object. One cannot exist without the other.
>
> Yes, I agree. They definitely go together. I think that is what I
> was trying to say. Just as the Bohr magneton can be defined in terms
> of e, e should be able to be defined in terms of it. I do realize
> though, that it requires a different way to define the Bohr magneton
> for its current part. I did just get some new clues on how to maybe
> make it work. I just have to figure out the plus and minus business.
It's just constant juggling. Use the Right Hand Rule for dB/dt = -curlE.
It's called a homo-polar motor or generator. A spinning magnet makes an
electric field. If you have North up and oriented along the axis of
rotation, If it spins counter-clockwise you get a positive E, radial from
the axis, (Right Hand Rule) If it spins clockwise you get a negative E
field. (Left Hand Rule).
snip
The latter is more probable in a steady magnetic field. Then it is most
probable that the electron is spinning with angular momentum hbar/2. If
there is no B field with which to orient the motion and polarize the
environment, then there is no most probable orientation of Spin. It is
basically random motion, or jitter. AKA "Zitterbewegung".
Best Regards,
Todd Desiato
FrediFizzx
Its not integrate 19 amps over 1 second. Its integrate 19 amps over
something like 10^-20 second (I don't have the exact figure with me). IOW,
charge is not amps*sec but is current*time with time being a certain time
period. Not necessarily a second long. I guess it should be delta time.
Thanks, got it.
Now I am a little confused. It seems like you are saying that it is in fact
always the x component of angular velocity but since the orientation is
random, it looks like jitter at c?
FrediFizzx
Todd Desiato
"FrediFizzx" <fredifi...@hahahotmail.com> wrote in message
news:aTLh9.226$il3.24...@newssvr14.news.prodigy.com...
> "Todd Desiato" <todd...@serversanddomains.com> wrote in message
> news:KCKh9.38254$U7.14...@twister.socal.rr.com...
> |
> | "Fredi Fizzx" <fredi...@hotmail.com> wrote in message
> | > OK, thanks. I did actually buy the Milonni book so I have it already
> | > but had not gotten that far reading it yet. Yes, I also thought that
> | > the article in the link I posted was not correct about its main theme
> | > but I didn't think it was not telling the truth about what Dirac found
> | > about the x component of the velocity of the electron. Well, which
> | > interpretation is correct? Are these electrons really getting knocked
> | > around at the speed of c by the vacuum? Or is the x component of
> | > their angular velocity equal to c? The latter just seems more
> | > probable to me.
> |
> | The latter is more probable in a steady magnetic field. Then it is most
> | probable that the electron is spinning with angular momentum hbar/2. If
> | there is no B field with which to orient the motion and polarize the
> | environment, then there is no most probable orientation of Spin. It is
> | basically random motion, or jitter. AKA "Zitterbewegung".
>
> Now I am a little confused. It seems like you are saying that it is in
fact
> always the x component of angular velocity but since the orientation is
> random, it looks like jitter at c?
>
Typically we call it the "z" component of angular momentum, but yes that is
what I'm saying. If there is no external magnetic field to orient it, it's
just jitter. I like to think of it as a handshake between the electron and
its local environment. You give me a photon, I give you a photon and we keep
exchanging in such a way that the average position and momentum remains
constant, but a function of the field at that location as the local
environment.
Best Regards,
Todd Desiato
mor...@world.std.spaamtrap
>> > | Well if charge is current times time, and you think current is
>> fundamental,
>> > | we have a problem here. If after 1 second (current*time) the electron
>> has
>> > | a charge of e, after 2 seconds we have a charge of 2e, after a day,
>> > | 86,400e, and after a while we'll have Freddi's Electron that Ate the
>> > | Universe.
>> >
>> > That is not right. e is a constant, time is a constant and current is a
>> > constant in the equation. The current is sustaining e. Stop the current;
>> > no more e.
So where does the charge that causes the current go? Like in normal
matter the moving e's, when stopped, neutralize with the never-moving
nuclei?
>All true except possibly when dealing with the fundamental creation of
>charge in the first place. About 19 amps of current circulating in
>the split second for one rotation equals the elementary charge. If
>you think of elementary charges making this current then you are going
>to get messed up. The current is always the same. It is not more
>electrons flowing as time goes by. It is a constant caused by the
>rotation of a negative E field.
Actually this is bass-ackwards. The math works out if you start with
a charge of e running around a circle with the ~10^-20 period, that single
charge repeating itself ~10^20 times per second does produce a current
equal to ~19A. (This, of course, neglects details such as the poor little
charge has to move faster than c to do this)
>> Nothing generates the fundamental charge. Charge is just another way of
>> saying energy, a displacement from zero. In physical reality there is no way
>> to separate these things, so in fact it is one object;
>>
>> charge = e
>> angular momentum = hbar/2
>> magnetic flux = hbar / 2e
No. A muon (or [anti]proton) has the same charge e but differing values
of magnetic flux. Other particles also have charge e and differing values
of angular momentum. (The W- has zero angular momentum)
>> it's field. You can think of the spin of a magnetic flux line as creating e,
That's not in Maxwell's Equations.
>> > | >| Gibberish. Review Maxwell's equations again. While you can create
>> an
>> > | >| electric field with a (changing) magnetic field, you cannot create
>> charge
>> > | >| from one.
>>
>> Actually you can, at high enough energy density you create e-p pairs. It's
>> easier in the presence of matter, but it can be done in a vacuum too.
Yes, energy can create matter in the form of, among other things, e+e-
pairs. But there is no way to create a lone charge from a magnetic field,
it's simply not one of Maxwell's Equations. Besides, this would violate
one of the most fundamental concepts of modern physics, the conservation
of charge.
(Simple experiment: Take a superconducting ring and measure its charge.
Induce different currents in it and measure the change in its charge.)
>> > | Why? A rotating EM field can only induce a current (by providing forces
>> > | on charged particles), it can't create a current directly. Take a
>> > | transformer, remove its secondary, place it in empty space and apply AC
>> to
>> > | its primary. What current does it induce? None! You removed the charges
>> > | the forces can act on when you removed the secondary.
>>
>> When you energize the primary of a transformer without a secondary coil you
>> not only generate an electric and a magnetic field, you also generate a
>> magnetic vector potential "A", which is physically a circulation of photons
>> in space, i.e., stored energy "circulating" around the coil. It is current
>> of photons.
A "current" of photons is not an electrical current. Photons have no charge.
There will be electric and magnetic fields associated with the transformer
primary, but no current.
-Mike
Todd Desiato
<mor...@world.std.spaamtrap>; <com (Michael Moroney)> wrote in message
news:H2Lq5y...@world.std.com...
I was not talking about composite particles like protons or W's, and a muon
is an excited state of an electron. It has more energy and therefor more
flux. The electron or positron is the lowest energy state of the Fermion's
in a Dirac field.
>
> >> it's field. You can think of the spin of a magnetic flux line as
creating e,
>
> That's not in Maxwell's Equations.
curlE = -dB/dt?
Show me an electron that does not have any magnetic flux or angular
momentum. The 3 quantities I listed above apply to an electron, which is all
I was referring to. Please don't try to force fit what I say to conform to
your own ideas.
> >> > | >| Gibberish. Review Maxwell's equations again. While you can
create
> >> an
> >> > | >| electric field with a (changing) magnetic field, you cannot
create
> >> charge
> >> > | >| from one.
> >>
> >> Actually you can, at high enough energy density you create e-p pairs.
It's
> >> easier in the presence of matter, but it can be done in a vacuum too.
>
> Yes, energy can create matter in the form of, among other things, e+e-
> pairs. But there is no way to create a lone charge from a magnetic field,
> it's simply not one of Maxwell's Equations. Besides, this would violate
> one of the most fundamental concepts of modern physics, the conservation
> of charge.
Agreed, I never said you could "create" an electron alone. They only come in
pairs.
>
> (Simple experiment: Take a superconducting ring and measure its charge.
> Induce different currents in it and measure the change in its charge.)
>
> >> > | Why? A rotating EM field can only induce a current (by providing
forces
> >> > | on charged particles), it can't create a current directly. Take a
> >> > | transformer, remove its secondary, place it in empty space and
apply AC
> >> to
> >> > | its primary. What current does it induce? None! You removed the
charges
> >> > | the forces can act on when you removed the secondary.
> >>
> >> When you energize the primary of a transformer without a secondary coil
you
> >> not only generate an electric and a magnetic field, you also generate a
> >> magnetic vector potential "A", which is physically a circulation of
photons
> >> in space, i.e., stored energy "circulating" around the coil. It is
current
> >> of photons.
>
> A "current" of photons is not an electrical current. Photons have no
charge.
> There will be electric and magnetic fields associated with the transformer
> primary, but no current.
>
Right, but if you drop 1 electron in that field of photons you have a
current, but this is analgous to droping a leaf in a calm river to see how
strong the current is. The "flow" (rather than use the word current) is
there even if there are no electrons to move. Also you neglected the eddy
currents induced in the core. :0)
Best Regards,
Todd Desiato
Michael Moroney
>| >There is a way to have current without charge and the way the VPP
>electron
>| >works is the only way.
>|
>| Sorry, I see no way whatsoever. Maxwell's Equations doesn't allow for it.
>You have to stop thinking Maxwell here when dealing with quantum level
>fundamentals. Just as you ask me to stop thinking of the electron as
>rotating solid ball, you need to think in terms of what generates the
>fundamental electronic charge in the first place. Is Maxwell going to be
>helpful in this situation? I don't think so.
The beauty of Maxwell's Equations is that, while they're classical, they
work at the quantum level as well.
>| Maxwell's Equations work just fine for the photon. So now I see VPP is
>now
>| having to create new fundamental relationships out of thin air.
>Please describe a photon using only Maxwell's equations. For that matter,
>describe an electron using Maxwell's equations. They can't be fully
>described only using Maxwell. You have to bring in QED or some other idea.
The point is, Maxwell's Equations work fine for the photon and the electron
even if they don't have any quantum characteristics. You go off and claim
for no reason (except to make your 'theory' work) that Maxwell's Equations
don't work any more.
>| >| No you can't. The Bohr is _defined_ in the terms of e, trying to
>define
>| >| e in the terms of the Bohr (defined in terms of e) results in an "x=x"
>| >| 'proof'. Or more specifically, an "e=e" 'proof'. Worthless.
...
>I did not say to insert the standard definition of the Bohr. I said to
>insert the loop area current from a model (which is defined from the model's
>geometry and rotation parameters).
Then you can't call it the Bohr Magneton, it is defined in terms of e,
not the other way around.
Of course you will get e=e if you insert
>the standard definition. Anyone can see that. The point being here that
>the model geometry is consistent with and supports the Bohr.
It "supports" the Bohr by being an "e=e" equation. You just rearranged
things and claim that you somehow discovered something.
>a chicken or the egg kind of thing. Which came first? The Bohr equation or
>the model?
Obviously the Bohr, since you use it to create your model.
>is really nothing that defines it. And if you have current, you will always
>have e and if you have e in motion you will have current.
Since current is defined as charge in motion, of course this sentence is
true. This sentence is almost the word equivalent of a simple "x=x" proof.
One BIG difference between what you are trying to do and the early days of
quantum mechanics is that there is no photoelectric effect or ultraviolet
catastrophe that can't be explained by the old rules but can by the new
rules. Another, of course, is showing that the old rules still work under
the new ones. (Thomas keeps ignoring me when I point out how his theories
violate a whole bunch of rules, or simply screams it doesn't violate any
rules (with no supporting evidence, of course)).
Explain conservation of charge to me if charge can be created by currents
(if the currents aren't themselves the charge in motion).
-Mike
Michael Moroney
>> No. A muon (or [anti]proton) has the same charge e but differing values
>> of magnetic flux. Other particles also have charge e and differing values
>> of angular momentum. (The W- has zero angular momentum)
>I was not talking about composite particles like protons or W's, and a muon
>is an excited state of an electron. It has more energy and therefor more
>flux. The electron or positron is the lowest energy state of the Fermion's
>in a Dirac field.
They are not composite. I won't go through that again.
>> >> it's field. You can think of the spin of a magnetic flux line as
>creating e,
>>
>> That's not in Maxwell's Equations.
>curlE = -dB/dt?
Nope. That creates an electric field, not a charge. (I interpreted the "e"
after the word "creating" as the electron's charge)
>Show me an electron that does not have any magnetic flux or angular
>momentum. The 3 quantities I listed above apply to an electron, which is all
>I was referring to. Please don't try to force fit what I say to conform to
>your own ideas.
Since a pi- has a spin of zero, I'd expect a zero magnetic moment. Since
you seem to think this is an excited electron or something, this would be
disproof.
>> A "current" of photons is not an electrical current. Photons have no
>charge.
>> There will be electric and magnetic fields associated with the transformer
>> primary, but no current.
>>
>Right, but if you drop 1 electron in that field of photons you have a
>current, but this is analgous to droping a leaf in a calm river to see how
>strong the current is.
Nope, you need the electron to even have the current.
The river analogy is incorrect. It should be something like pouring water
into a dry riverbed and measuring the flow rate to see what the slope of
the riverbed is.
(The electric field corresponds to the gravitational field, but without
a charge/mass (respectively) to act on, nothing happens other than any
dE/dt effect of the electric field)
> The "flow" (rather than use the word current) is
>there even if there are no electrons to move.
There is no electric current. That is the point. There is an electric
field which translates into a force on any charges present, but without
charges (which must be free to move) there is no current. You can argue
that the electric field is mediated by photons (and the gravitational
field by gravitrons) and they "flow" somehow but there is still no current.
> Also you neglected the eddy
>currents induced in the core. :0)
:-)
-Mike
FrediFizzx
Seems like you are skirting me asking you to fully describe a photon or
electron using only Maxwell's equations. Maybe I was wrong though. Maybe
Maxwell still can describe what is going on here. I think Todd said that
accumulated charge was the integral of current times dt. What is wrong with
this applied to elementary charge? The current is a constant here so you
are going to get charge = current*time. The time is the time interval for
each rotation. Also a constant.
| >| >| No you can't. The Bohr is _defined_ in the terms of e, trying to
| >define
| >| >| e in the terms of the Bohr (defined in terms of e) results in an
"x=x"
| >| >| 'proof'. Or more specifically, an "e=e" 'proof'. Worthless.
| ...
| >I did not say to insert the standard definition of the Bohr. I said to
| >insert the loop area current from a model (which is defined from the
model's
| >geometry and rotation parameters).
|
| Then you can't call it the Bohr Magneton, it is defined in terms of e,
| not the other way around.
|
| Of course you will get e=e if you insert
| >the standard definition. Anyone can see that. The point being here that
| >the model geometry is consistent with and supports the Bohr.
|
| It "supports" the Bohr by being an "e=e" equation. You just rearranged
| things and claim that you somehow discovered something.
|
| >a chicken or the egg kind of thing. Which came first? The Bohr equation
or
| >the model?
|
| Obviously the Bohr, since you use it to create your model.
Correction: it is not my model. No, Tom created the model from his calculus
of related rates which set the maximum dynamic Poynting vector, the max E
vector, and the max H vector all at equal physical lengths. Or something
like that. This is quite amazing in itself as it possibly explains *why*
the electron and positron form at the exact energy that they do. They can
only form when these three things are exactly equal. And they are equal at
the exact rest mass energy required. There is no trick about this. But we
do know that how Tom gets there does have some problems. Which I think can
eventually be resolved. Then we find out after this that this construction
also supports the Bohr magneton plus more. And besides that, you can derive
the model geometry parts from the Bohr equation and other related equations.
The main feature being that the model geometry is *two* current rings
latched together. This is what makes the model unique. Of course we will
get an e=e type of scenario from this because it is all self supporting. So
the big challenge is to try to break the circular arguments by showing how a
rotating E field can create current which creates the elementary charge. I
am not sure though that it is possible to really break the circular argument
because once you have e, you can say it is creating the current. Definitely
a chicken or the egg scenario in the case of elementary charge. Like Todd
said, they always are going to go together in this case and really have to
be considered as one thing.
| >is really nothing that defines it. And if you have current, you will
always
| >have e and if you have e in motion you will have current.
|
| Since current is defined as charge in motion, of course this sentence is
| true. This sentence is almost the word equivalent of a simple "x=x" proof.
Yes, I totally agree. When dealing with elementary charge, it is going to
end up like that. But what you are missing is how the model stands in the
middle of all this and supports the complementary theme of it all. What
really breaks this cycle is the creation of the e+e- pair. That is what is
really lacking here. The mechanics and the math to support that.
| One BIG difference between what you are trying to do and the early days of
| quantum mechanics is that there is no photoelectric effect or ultraviolet
| catastrophe that can't be explained by the old rules but can by the new
| rules. Another, of course, is showing that the old rules still work under
| the new ones. (Thomas keeps ignoring me when I point out how his theories
| violate a whole bunch of rules, or simply screams it doesn't violate any
| rules (with no supporting evidence, of course)).
|
| Explain conservation of charge to me if charge can be created by currents
| (if the currents aren't themselves the charge in motion).
Take a look at the article from the recent post "Quantum mechanics as
electrodynamics of curvilinear waves". Alexander shows a possible mechanism
of how it can happen. I think if we take Alexander's ideas and apply them
to the VPP model, we will really have something more concrete. The problem
I see with his torus model is that it is not as stable as the VPP model
(basically two tori latched together) and that he is showing his B and C
(electron and positron) wrong. I think his B and C rings should have one or
two nodes happening (where the H and E fields should go back down to
relative zero from a max). He shows the H and E fields as being constant
after the division. However, according to what Todd posted earlier, he is
showing the E field pointing in on one ring and out on another with the
correct rotations. This should produce plus and minus elementary charges
from rotating EM fields. The conservation of charge is because they always
have to be created in pairs thus always conserving charge.
One good feature of the torus model it that it is easier to see how this can
happen where as I haven't figured out how Tom is getting plus and minus
charges from the VPP models. I think there might be some kind of mistake
since Tom is showing the positron with the H fields in the direction of
rotation. This seems to imply to me that the positron's plus electric
charge is the monopole charge of the magnetic field. However, the VPP model
is much more complex in the way that the fields are aligned. A very strange
beast.
FrediFizzx
FrediFizzx
news:H2Lq5y...@world.std.com...
|
| >> > | Well if charge is current times time, and you think current is
| >> fundamental,
| >> > | we have a problem here. If after 1 second (current*time) the
electron
| >> has
| >> > | a charge of e, after 2 seconds we have a charge of 2e, after a day,
| >> > | 86,400e, and after a while we'll have Freddi's Electron that Ate
the
| >> > | Universe.
| >> >
| >> > That is not right. e is a constant, time is a constant and current
is a
| >> > constant in the equation. The current is sustaining e. Stop the
current;
| >> > no more e.
|
| So where does the charge that causes the current go? Like in normal
| matter the moving e's, when stopped, neutralize with the never-moving
| nuclei?
You can't picture it like that when you are dealing with a single elementary
charge. In the case of the VPP model, e is not moving internally. There is
only an EM field that is moving creating a current. It is this
current/moving EM field that is creating e overall.
| >All true except possibly when dealing with the fundamental creation of
| >charge in the first place. About 19 amps of current circulating in
| >the split second for one rotation equals the elementary charge. If
| >you think of elementary charges making this current then you are going
| >to get messed up. The current is always the same. It is not more
| >electrons flowing as time goes by. It is a constant caused by the
| >rotation of a negative E field.
|
| Actually this is bass-ackwards. The math works out if you start with
| a charge of e running around a circle with the ~10^-20 period, that single
| charge repeating itself ~10^20 times per second does produce a current
| equal to ~19A. (This, of course, neglects details such as the poor little
| charge has to move faster than c to do this)
No its not bass-ackwards. You have to stop thinking that only moving e's
can create a current. If we could take an electron at rest (linear
motion-wise) do you think the e is moving around to create the magnetic
moment? No its not. The e charge is overall. The only way to explain it
in a more classical fashion is how the VPP model works. There has to be
internal currents. If we try to measure e up real close, do we see it
wobbling around? I don't think so.
| >> Nothing generates the fundamental charge. Charge is just another way of
| >> saying energy, a displacement from zero. In physical reality there is
no way
| >> to separate these things, so in fact it is one object;
| >>
| >> charge = e
| >> angular momentum = hbar/2
| >> magnetic flux = hbar / 2e
|
| No. A muon (or [anti]proton) has the same charge e but differing values
| of magnetic flux. Other particles also have charge e and differing values
| of angular momentum. (The W- has zero angular momentum)
This is supported by VPP. Elementary charge is not dependant on size but
magnetic moment is.
| >> it's field. You can think of the spin of a magnetic flux line as
creating e,
|
| That's not in Maxwell's Equations.
|
| >> > | >| Gibberish. Review Maxwell's equations again. While you can
create
| >> an
| >> > | >| electric field with a (changing) magnetic field, you cannot
create
| >> charge
| >> > | >| from one.
| >>
| >> Actually you can, at high enough energy density you create e-p pairs.
It's
| >> easier in the presence of matter, but it can be done in a vacuum too.
|
| Yes, energy can create matter in the form of, among other things, e+e-
| pairs. But there is no way to create a lone charge from a magnetic field,
| it's simply not one of Maxwell's Equations. Besides, this would violate
| one of the most fundamental concepts of modern physics, the conservation
| of charge.
Well, no one ever said that we were trying to make a *lone* charge from a
magnetic field. Of course you can't do that. We all agree that charges
have to be created in complementary pairs.
| (Simple experiment: Take a superconducting ring and measure its charge.
| Induce different currents in it and measure the change in its charge.)
OK, what does this show us?
FrediFizzx
larry
> "larry" <gold...@charter.net> wrote in message
> news:3D86B9F...@charter.net...
> | FrediFizzx wrote:
> |
> |
> | >
> | > Nothing went wrong. I simple saw that it is just as equally valid to
> say
> | > that the model is generating the electron constants and equations. The
> VPP
> |
> |
> | It might be good to inject a little objective reality in to this.
> | A model is conceptual and "generates" implies some kind of action which
> | a model is not capable of doing. "electron constants and equations"
> | describe the identity of the electron, which is the acting body. There
> | is no "chicken or egg thing" involved. The model describes something
> | about objective reality, nothing more or less. The Bohr magneton is a
> | relationship between constants, a number, and cannot be a cause of the
> | identities of the bodies involved in its definition. Current is defined
> | in terms of motion of charge and thus cannot be a cause of charge.
> | Larry
>
> Well, of course I meant the model as a physical reality. *If* the model is
> the way an actual electron is, then this modeled electron is in fact
> *generating* charge, mass, a magnetic moment, etc. That is what I meant.
If you mean that the real existing electron has the attributes charge,
mass, a magnetic moment, etc., then ok, but "generating" is true only of
some of those attributes. Is there a process that generates charge? Or
is there a process generating the relationship of mass that the electron
has with respect to all the rest of the universe? Magnetic moment could
be said to be generated since it has to do with motion with respect to
electric fields.
> However, the main point is that the VPP electron *geometry* does support all
> the electron constants and equations as well as they support it. You have
> to admit that at least that the VPP model does get that right. If it didn't
> do that, I wouldn't even be bothering with it.
>
> The *standard* definition of current is in terms of motion of charge. But
> what about the definition of *elementary* charge itself? What is that? Why
> is that missing? I say that you have to define it in terms of a rotating E
> field that had no charge to start with but as soon as it gets rotational
> motion thus making a current, you get the elementary charge. The problem I
What is the source of the electric field between bodies due to a
rotating localized electric field which you describe as charge? Does it
just act on the vacuum to produce electric field waves propagating in
all directions?
Larry
Michael Moroney
>| >Please describe a photon using only Maxwell's equations. For that
>matter,
>| >describe an electron using Maxwell's equations. They can't be fully
>| >described only using Maxwell. You have to bring in QED or some other
>idea.
>|
>| The point is, Maxwell's Equations work fine for the photon and the
>electron
>| even if they don't have any quantum characteristics. You go off and claim
>| for no reason (except to make your 'theory' work) that Maxwell's Equations
>| don't work any more.
>Seems like you are skirting me asking you to fully describe a photon or
>electron using only Maxwell's equations. Maybe I was wrong though.
Maxwell's equations, by themselves, are insufficient to explain the
electron, or any quantum behavior for that matter. However, they still
hold at the quantum level (unlike Newton's laws that hold true for
real-life but not for relativistic motions) and that's the beauty of them.
Despite being 100% classical they work at the quantum level without
modification.
>Maxwell still can describe what is going on here. I think Todd said that
>accumulated charge was the integral of current times dt. What is wrong with
>this applied to elementary charge?
Yes, but you have to use the equation correctly. To calculate the charge
passing a point over a period of time, integrate over the time period
the current times dt. Note that this measures charges, not creates
anything. Do you think an ammeter measures the creation of electrons or
something? No, it measures their flow.
> The current is a constant here so you
>are going to get charge = current*time. The time is the time interval for
>each rotation. Also a constant.
So this works out to be a nice Physics 101 problem. "The rotation of a
charged ring produces a current of 19 amps. The ring rotates at a rate of
10^20 times per second. What is the charge on the ring?" Go through the
math, and substitute the exact VPP figures for the 19A and the 10^-20 period)
and you get the charge of e. What a wonderful "x=x" proof, VPP shows the
charge on an electron is e!
>| >a chicken or the egg kind of thing. Which came first? The Bohr equation
>or
>| >the model?
>|
>| Obviously the Bohr, since you use it to create your model.
>Correction: it is not my model. No, Tom created the model from his calculus
<snip gibberish>
OK it's Tom's model. The point is he incorporates the Bohr to (try to)
make it work, it is obvious the Bohr came first.
>| One BIG difference between what you are trying to do and the early days of
>| quantum mechanics is that there is no photoelectric effect or ultraviolet
>| catastrophe that can't be explained by the old rules but can by the new
>| rules. Another, of course, is showing that the old rules still work under
Of course what I mean is something like the solar neutrino deficit that
requires something new (neutrino masses/oscillations) to explain it.
>| Explain conservation of charge to me if charge can be created by currents
>| (if the currents aren't themselves the charge in motion).
>Take a look at the article from the recent post "Quantum mechanics as
>electrodynamics of curvilinear waves". Alexander shows a possible mechanism
No VPP answer to conservation of charge? You refer to someone else's ideas
as possible support but no direct answer?
-Mike
Michael Moroney
>| So where does the charge that causes the current go? Like in normal
>| matter the moving e's, when stopped, neutralize with the never-moving
>| nuclei?
>You can't picture it like that when you are dealing with a single elementary
>charge. In the case of the VPP model, e is not moving internally. There is
>only an EM field that is moving creating a current.
Once again, a moving EM field doesn't create a current. It induces forces
on charges, and if the charges move in response *that* creates a current.
> It is this
>current/moving EM field that is creating e overall.
For the bazillionth time, none of Mazwell's Equations state anything like
that.
>| Actually this is bass-ackwards. The math works out if you start with
>| a charge of e running around a circle with the ~10^-20 period, that single
>| charge repeating itself ~10^20 times per second does produce a current
>| equal to ~19A. (This, of course, neglects details such as the poor little
>| charge has to move faster than c to do this)
>No its not bass-ackwards.
Yes it is. Like I said, if you take a charge of e, make it run in a
circle with a period of ~10^-20 seconds, the current created by it works
out to be ~19A. Reversing the equation _calculates_ the charge necessary
to fulfill the requirements, but stating it _creates_ a charge is as silly
as saying the earth's gravitational field created the earth.
> You have to stop thinking that only moving e's
>can create a current.
Why? The only known source of current is moving charges (not necessarily
electrons).
> If we could take an electron at rest (linear
>motion-wise) do you think the e is moving around to create the magnetic
>moment? No its not.
The Bohr relation shows the magnetic moment is created from the charge,
quantum angular momentum and mass. No actual electron motion is required.
Also remember, Maxwell's Equations show that magnetic moments can be
created without currents.
>| Yes, energy can create matter in the form of, among other things, e+e-
>| pairs. But there is no way to create a lone charge from a magnetic field,
>| it's simply not one of Maxwell's Equations. Besides, this would violate
>| one of the most fundamental concepts of modern physics, the conservation
>| of charge.
>Well, no one ever said that we were trying to make a *lone* charge from a
>magnetic field. Of course you can't do that. We all agree that charges
>have to be created in complementary pairs.
Say what? You keep saying that a current in a circle creates the
electron's charge. This is creating the electron's charge from a non
conserved quantity.
>| (Simple experiment: Take a superconducting ring and measure its charge.
>| Induce different currents in it and measure the change in its charge.)
>OK, what does this show us?
Again, you keep claiming a current in a loop creates charge. This is an
experiment to (dis)prove that.
-Mike
ThomasL283
>Date: 9/18/2002 11:23 AM Pacific Daylight Time
wrote in:
>Message-id: <H2nCEt...@world.std.com>
>"FrediFizzx" <fredifi...@ahahhotmail.com> writes:
>snip<
>>Correction: it is not my model. No, Tom created the model from his calculus
>OK it's Tom's model. The point is he incorporates the Bohr to (try to)
>make it work, it is obvious the Bohr came first.
Mike, the point is that the model came before it was discovered that YES, the
geometry does give the Bohr and all other related electron fundamental physical
constants.
Getting back to the subject of my thread:
The binding energy holding the electron to the proton in hydrogen is shown in
the literature as a negative 13.5984 eV. Negative because the energy has been
expelled when binding took place.
What has been missing in analysis of the ionization binding energy has been the
magnetic moments of the electron and proton.
Magnetic moments are important to create the expelled photon, because both the
electric potential energy and magnetic potential energy MUST be exactly equal,
to create a photon.
It was shown in:
http://www.members.aol.com/tnlockyer/twobond.gif
that the electric potential energy has two solutions, the Bohr radius and VPP
bond length at 1.408252E-15 m for the published 13.5984 eV ionization energy.
However, only the near field can null with the magnetic moments, so Bohr's
radius, in the far field, is not usable.
The VPP electron shows two current rings separated by the electron's
rationalized Compton wavelength, and the VPP proton active core particle shows
two current rings separated by 7.52981533E-16 m.
Since the VPP proton core is so much smaller that the electron model, the
following analysis ignores the finite VPP proton core size with little error:
http://www.members.aol.com/tnlockyer/epbind.gif
In this model the electron and proton are spinning in the same direction.
Therefore, the electron's negative charge and the proton's positive charge
attract, while their magnetic moments repel.
Notice the equation IO is set up so the electric and magnetic forces are
subtract for the NULL.
We then adjust (x) so that the equation is as close to zero as required,
then solve for the ionization energy in convenient units of equivalent volts.
(I know you hate this equivalent volt short cut, but it does keep the basic
units intact, and is numerically equivalent.)
The magnetic potential energy is based on the mutual inductance between the
current loops, and is the sum of the left and right side current rings
potential energy.
The model gives the ionization energy to within about 10 percent of 13.5984 eV,
not very good, but at least the VPP hydrogen model can now be justified on good
engineering principles. (No fudge factors like big (X) in):
http://members.aol.com/tnlockyer/VPPchem.gif
Regards: Tom:
P.S. Mike, why don't you get a copy of the book, "Vector Particle and Nuclear
Models" like Freddie did, and then you might get to like VPP.
See the following sig for book ordering info.
FrediFizzx
Tom, I have been meaning to ask you about this but keep forgetting. Why is
it that you think when the electric and magnetic potential energy is equal
does it creates a photon? I am not fully understanding your thinking on
this.
FrediFizzx
Michael Moroney
>P.S. Mike, why don't you get a copy of the book, "Vector Particle and Nuclear
>Models" like Freddie did, and then you might get to like VPP.
I prefer Charmin. It's much softer, and you can get 6 500 sheet rolls for
a couple bucks (instead of only 160 sheets for $24.95)
:-)
-Mike
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| >Please describe a photon using only Maxwell's equations. For that
| >matter,
| >| >describe an electron using Maxwell's equations. They can't be fully
| >| >described only using Maxwell. You have to bring in QED or some other
| >idea.
| >|
| >| The point is, Maxwell's Equations work fine for the photon and the
| >electron
| >| even if they don't have any quantum characteristics. You go off and
claim
| >| for no reason (except to make your 'theory' work) that Maxwell's
Equations
| >| don't work any more.
|
| >Seems like you are skirting me asking you to fully describe a photon or
| >electron using only Maxwell's equations. Maybe I was wrong though.
|
| Maxwell's equations, by themselves, are insufficient to explain the
| electron, or any quantum behavior for that matter. However, they still
| hold at the quantum level (unlike Newton's laws that hold true for
| real-life but not for relativistic motions) and that's the beauty of them.
| Despite being 100% classical they work at the quantum level without
| modification.
Well, good. Then I should be able to work out the detailed math for the
hypothetical *generation* of the elementary quantum of charge from the
current rings that are being produced by a *uniquely* formed rotating EM
field. I will give it my best shot.
| >Maxwell still can describe what is going on here. I think Todd said that
| >accumulated charge was the integral of current times dt. What is wrong
with
| >this applied to elementary charge?
|
| Yes, but you have to use the equation correctly. To calculate the charge
| passing a point over a period of time, integrate over the time period
| the current times dt. Note that this measures charges, not creates
| anything. Do you think an ammeter measures the creation of electrons or
| something? No, it measures their flow.
But elementary charge would not be passing over a point in this case. It is
being generated by an area or actually a volume of space. In the case of a
single elementary charge, I think it would be creating it. You have to stop
thinking a bunch of charges and think of only one charge here and what is
happening.
| > The current is a constant here so you
| >are going to get charge = current*time. The time is the time interval
for
| >each rotation. Also a constant.
|
| So this works out to be a nice Physics 101 problem. "The rotation of a
| charged ring produces a current of 19 amps. The ring rotates at a rate of
| 10^20 times per second. What is the charge on the ring?" Go through the
| math, and substitute the exact VPP figures for the 19A and the 10^-20
period)
| and you get the charge of e. What a wonderful "x=x" proof, VPP shows the
| charge on an electron is e!
Where did the 19 amps come from if it is producing an elementary charge
overall? You could say that e is circulating around this ring but it is
not. e is being produced overall by the ring current. I think between what
Todd said and what Alexander is showing, I can work out the math for this
and show it. I will not have time to do it until the weekend though.
| >| >a chicken or the egg kind of thing. Which came first? The Bohr
equation
| >or
| >| >the model?
| >|
| >| Obviously the Bohr, since you use it to create your model.
|
| >Correction: it is not my model. No, Tom created the model from his
calculus
| <snip gibberish>
|
| OK it's Tom's model. The point is he incorporates the Bohr to (try to)
| make it work, it is obvious the Bohr came first.
Huh? He didn't incorporate the Bohr at all. The Bohr comes out of the
geometry naturally. And you can generate the model parts from the Bohr.
The model can actually be generated by two entirely different methods.
Calculus of related rates and from the electron constants and related
equations.
| >| Explain conservation of charge to me if charge can be created by
currents
| >| (if the currents aren't themselves the charge in motion).
|
| >Take a look at the article from the recent post "Quantum mechanics as
| >electrodynamics of curvilinear waves". Alexander shows a possible
mechanism
|
| No VPP answer to conservation of charge? You refer to someone else's ideas
| as possible support but no direct answer?
It is the same thing. Conservation of charge is that the e+e- are always
created in pairs. Alexander is just showing more detail on the possible
mechanics of how it can happen. This certainly would be similar for how the
VPP electron-positron pair is created. The charge is conserved at the time
of pair creation. What Tom has been lacking is showing the detailed
mechanics of the division and how the divided EM field creates separate plus
and minus charges. There is no conflict here with conservation of charge.
It does get conserved.
FrediFizzx
Michael Moroney
>| Maxwell's equations, by themselves, are insufficient to explain the
>| electron, or any quantum behavior for that matter. However, they still
>| hold at the quantum level (unlike Newton's laws that hold true for
>| real-life but not for relativistic motions) and that's the beauty of them.
>| Despite being 100% classical they work at the quantum level without
>| modification.
>Well, good. Then I should be able to work out the detailed math for the
>hypothetical *generation* of the elementary quantum of charge from the
>current rings that are being produced by a *uniquely* formed rotating EM
>field. I will give it my best shot.
Sorry, but there is no help for that hopeless quest from Maxwell's
Equations. It'll take a "let's make up a fact" new 'law' of physics
that you or Tom pull from some bodily orifice.
>| Yes, but you have to use the equation correctly. To calculate the charge
>| passing a point over a period of time, integrate over the time period
>| the current times dt. Note that this measures charges, not creates
>| anything. Do you think an ammeter measures the creation of electrons or
>| something? No, it measures their flow.
>But elementary charge would not be passing over a point in this case. It is
>being generated by an area or actually a volume of space. In the case of a
>single elementary charge, I think it would be creating it.
Once again, you are misusing that law of physics. You cannot create
charge, just act on it. Nobody knows why charge exists. It just does.
> You have to stop
>thinking a bunch of charges and think of only one charge here and what is
>happening.
I see nothing wrong with a single charge running around in a circle 10^20
times a second producing a current of ~19A. Electrons are all identical,
it doesn't matter whether you see one a zillion times or a zillion once.
Both produce the same current.
(actually there is something wrong with a charge running around in a
circle 10^20 times/second. It would radiate energy like mad. But I
digress...)
>| So this works out to be a nice Physics 101 problem. "The rotation of a
>| charged ring produces a current of 19 amps. The ring rotates at a rate of
>| 10^20 times per second. What is the charge on the ring?" Go through the
>| math, and substitute the exact VPP figures for the 19A and the 10^-20
>period)
>| and you get the charge of e. What a wonderful "x=x" proof, VPP shows the
>| charge on an electron is e!
>Where did the 19 amps come from if it is producing an elementary charge
>overall?
Solve the problem yourself, I will not do your homework for you! :-)
A charge in motion produces a current of 19A, what is the charge?
> You could say that e is circulating around this ring but it is
>not.
The only way to get the current is to move a charge.
> e is being produced overall by the ring current.
You keep saying this yet you have NO PROOF, other than a misused (bass-
ackwards) version of the current law. You need to review the electromag
chapter of a Physics 101 textbook.
>| >| >a chicken or the egg kind of thing. Which came first? The Bohr
>equation
>| >or
>| >| >the model?
>| >|
>| >| Obviously the Bohr, since you use it to create your model.
>|
>| >Correction: it is not my model. No, Tom created the model from his
>calculus
>| <snip gibberish>
>|
>| OK it's Tom's model. The point is he incorporates the Bohr to (try to)
>| make it work, it is obvious the Bohr came first.
>Huh? He didn't incorporate the Bohr at all. The Bohr comes out of the
>geometry naturally.
No he uses the Bohr (or a hacked-up version of the Bohr) to derive his
formulas. Of course many of them are "x=x" proofs so the Bohr falls out
of some of them again.
>| No VPP answer to conservation of charge? You refer to someone else's ideas
>| as possible support but no direct answer?
>It is the same thing. Conservation of charge is that the e+e- are always
>created in pairs. Alexander is just showing more detail on the possible
Yet you still claim running a current in a circle creates a charge. Let's
put a switch in there. Switch closed=current=charge. Switch open=no
current=no charge. Charge is no longer conserved?
Also, why would a current in a loop create a negative charge? If you
reverse the current do you now get a positive charge? But what about if
you flip the ring over (180 degree rotation) and now the current is
running back in its original direction. Is it still positive or negative?
Why? Your symmetry is all f***ed up.
-Mike
FrediFizzx
| "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
|
| >| Maxwell's equations, by themselves, are insufficient to explain the
| >| electron, or any quantum behavior for that matter. However, they still
| >| hold at the quantum level (unlike Newton's laws that hold true for
| >| real-life but not for relativistic motions) and that's the beauty of
them.
| >| Despite being 100% classical they work at the quantum level without
| >| modification.
|
| >Well, good. Then I should be able to work out the detailed math for the
| >hypothetical *generation* of the elementary quantum of charge from the
| >current rings that are being produced by a *uniquely* formed rotating EM
| >field. I will give it my best shot.
|
| Sorry, but there is no help for that hopeless quest from Maxwell's
| Equations.
OK, forget the current. Seems like we don't even need it. All I need is an
electric field and it doesn't even have to be moving. In cgs it is easy to
show that charge = E_0*r^2. So all I need to do is to figure out E_0 and r
and we are home free. In SI units it would be e = 4*pi*eps0*E_0*r^2. I
think that r would be Compton wavelength divided by 4*pi and Tom had
calculated that E_0 would be about 3.86259*10^16 volts per meter for the VPP
electron. Plug those in and you get about 1.602*10^-19 Amp*seconds for the
elementary charge. So now all that is left to do is to show how we can get
E_0 from the division of the photon's electric field. It seems that alpha
is the conversion ratio to go from photon electric energy to electron
electric energy so I already know what you are going to say about it. This
must be what is happening at division time. The conversion involves most of
the photon's energy being converted to mass so the electric energy is
reduced by alpha. Man, that is a big reduction!
So is this what is happening? So it seems to me that the total mass of the
electron is electric energy plus magnetic energy plus the remaining which
must be in the spin angular momentum. Hmmm? I think Tom maybe made a
slight mistake on his mass of the electron calculations. The energy left
over after subtracting the electric and magnetic energy has to equal the
spin angular momentum energy. This would make the radius slightly smaller.
Rats! We needed it to be slightly bigger.
| >| Yes, but you have to use the equation correctly. To calculate the
charge
| >| passing a point over a period of time, integrate over the time period
| >| the current times dt. Note that this measures charges, not creates
| >| anything. Do you think an ammeter measures the creation of electrons
or
| >| something? No, it measures their flow.
|
| >But elementary charge would not be passing over a point in this case. It
is
| >being generated by an area or actually a volume of space. In the case of
a
| >single elementary charge, I think it would be creating it.
|
| Once again, you are misusing that law of physics. You cannot create
| charge, just act on it. Nobody knows why charge exists. It just does.
Baloney, charges can be created by pair production easily. The easy
explanation is that the photon has both plus and minus charges in it and
pair production allows it to be divided out. Alexander showed a possible
mechanism.
FrediFizzx
ThomasL283
>Date: 9/18/2002 9:05 PM Pacific Daylight Time
Wrote in:
>Message-id: <H2o3Dy...@world.std.com>
>thoma...@aol.com (ThomasL283) writes:
>
>>P.S. Mike, why don't you get a copy of the book, "Vector Particle and
>Nuclear
>>Models" like Freddie did, and then you might get to like VPP.
>I prefer Charmin.
Mike, lead, follow or get out of the way.
ThomasL283
Date: 9/18/2002 6:06 PM Pacific Daylight Time
Wrote in: Message-id: <cq9i9.290$nx1.17...@newssvr21.news.prodigy.com>
"ThomasL283" <thoma...@aol.com> wrote in message
news:20020918175113...@mb-ca.aol.com...
| >mor...@world.std.spaamtrap.com (Michael Moroney)
| >Date: 9/18/2002 11:23 AM Pacific Daylight Time
| wrote in:
| >Message-id: <H2nCEt...@world.std.com>
|
| Getting back to the subject of my thread:
|
>| The binding energy holding the electron to the proton in hydrogen is shown
in
>| the literature as a negative 13.5984 eV. Negative because the energy has
been
>| expelled when binding took place.
>|
>| What has been missing in analysis of the ionization binding energy has
been the
>| magnetic moments of the electron and proton.
|
>| Magnetic moments are important to create the expelled photon, because both
the
>| electric potential energy and magnetic potential energy MUST be exactly
equal,
>| to create a photon.
>Tom, I have been meaning to ask you about this but keep forgetting. Why is
>it that you think when the electric and magnetic potential energy is equal
>does it creates a photon? I am not fully understanding your thinking on
>this.
Fredi, remember when working with the binding energy of the deuteron? (see
Chapter 12).
The capture of a neutron by the proton releases a single photon of 2.224573
MeV, creating the binding energy mass defect.
Every photon in the universe has equal electric and magnetic field strengths.
So, there is one (and only one binding length) where both electric and magnetic
forces are equal (null), and this creates the condition for creating a photon.
The energy deficite holds the nucleons at bay, until replaced.
The VPP binding energy math does give the binding energy for the deuteron and
more complex nuclei to within a few tenths of a percent (or better) of their
measured values.
The deuteron binding, as the electric and magnetic forces change at different
rates, until capture, (null) is graphed out on the cover of the book. (For
Mikes enlightenment here it is again.)
http://www.members.aol.com/tnlockyer/pngraph.gif
VPP unifies the strong and electromagnetic forces, every physicists dream.
This same approach is used in an effort to calculate the 13.5984 eV binding
energy given for the hydrogen atom.
http://www.members.aol.com/tnlockyer/epbind.gif
While the 10 percent error in the calculations is somewhat disappointing, the
fact that value is not further away, from the accepted value, is heartening.
The correct approach seems to be the mutual inductance for calculating the
magnetic potential energy.
If you do the math, one finds that the mutual inductance is double what one
calculates going from electron current rings to the proton core current rings.
Why? Apparently both electron and proton are independent DC current sources,
rather than transformer induced, one from the other. This math is also
contained in the two different (and independent) magnetic moments.
One can get similar results using the calculated double mutual inductance or
(4.09854401854498E-24 Henry) for right current loop, to proton current loops,
and (7.83067846018106E-25 Henry) for the left current loop, to the proton
current loops.
The VPP circulating electron current (19.8192886259104 A) and circulating VPP
proton current loop currents (2.03282393770294E4 A) are muliplied and the
results added for total Joules. Then divide Joules by (e= 1.602176462E-19 As)
to get binding energy in (eV) volts.
Regards: Tom:
ThomasL283
As is: VPP circulating electron current (19.8192886259104 A)
Should be: (19.7963317601877 A)
(Darn bifocals!)
Michael Moroney
>| Sorry, but there is no help for that hopeless quest from Maxwell's
>| Equations.
>OK, forget the current. Seems like we don't even need it. All I need is an
>electric field and it doesn't even have to be moving. In cgs it is easy to
>show that charge = E_0*r^2. So all I need to do is to figure out E_0 and r
>and we are home free. In SI units it would be e = 4*pi*eps0*E_0*r^2. I
>think that r would be Compton wavelength divided by 4*pi and Tom had
>calculated that E_0 would be about 3.86259*10^16 volts per meter for the VPP
>electron. Plug those in and you get about 1.602*10^-19 Amp*seconds for the
>elementary charge.
Looks like Freddi has found yet another way to run an equation backwards!
>| Once again, you are misusing that law of physics. You cannot create
>| charge, just act on it. Nobody knows why charge exists. It just does.
>Baloney, charges can be created by pair production easily.
Non sequitur. We know charge is conserved, and part of the conservation
rules is annihilation and pair production. Yet we still don't know why
charge exists, or more specifically, why charge has the properties it
does (creates forces on other charges, reacts to the presence of other
charges, obeys Maxwell's equations etc.) It's just the way of the
universe. At least magnetic effects can be explained as relativistic
effects on electric effects but that still begs the questions why electric
effects exist in the first place. Why the forces of an electron on a
proton follow the same rules whether they are separated by the diameter of
a galaxy or the electron is actually in the process of travelling through
the proton (high energy scattering experiments).
-Mike
Michael Moroney
>>mor...@world.std.spaamtraaaap.com (Michael Moroney)
>>Date: 9/18/2002 9:05 PM Pacific Daylight Time
>Wrote in:
>>Message-id: <H2o3Dy...@world.std.com>
>>thoma...@aol.com (ThomasL283) writes:
>>
>>>P.S. Mike, why don't you get a copy of the book, "Vector Particle and
>>Nuclear
>>>Models" like Freddie did, and then you might get to like VPP.
>>I prefer Charmin.
>Mike, lead, follow or get out of the way.
The problem with that limited set of choices is: 1) I will not follow
incorrect physics; 2) "get out of your way" means bad physics goes
unchallenged; and 3) If I try to lead I know you will not follow.
Fredi Fizzx
> "FrediFizzx" <fredifi...@ahahhotmail.com> writes:
>
> >| Sorry, but there is no help for that hopeless quest from Maxwell's
> >| Equations.
>
> >OK, forget the current. Seems like we don't even need it. All I need is an
> >electric field and it doesn't even have to be moving. In cgs it is easy to
> >show that charge = E_0*r^2. So all I need to do is to figure out E_0 and r
> >and we are home free. In SI units it would be e = 4*pi*eps0*E_0*r^2. I
> >think that r would be Compton wavelength divided by 4*pi and Tom had
> >calculated that E_0 would be about 3.86259*10^16 volts per meter for the VPP
> >electron. Plug those in and you get about 1.602*10^-19 Amp*seconds for the
> >elementary charge.
>
> Looks like Freddi has found yet another way to run an equation backwards!
How can you run an equality backwards? An equality is just that. An
equality. They can be run anyway that algegra allows to transpose
them. There is nothing backwards about the equation q = E_0*r^2. You
just refuse to admit that it can also apply to the single elementary
charge. Guess what. It can and does apply. So the whole trick to
getting an electric field with no original charge is by division of
the photon's electric field like Alexander is showing. Mindful that
it requires the postulate that Maxwell's equations describe a photon
and the photon is exactly one wavelength long. The division of the
photon's EM field will always produce an equal plus and minus
elementary charge thus conserving charge.
Now I have been thinking that since alpha is not really a fixed
constant (changes as we get closer to the bare electron) that this may
shed some light on the problem of the energy not adding up quite
right. So what is the true value of alpha and elementary charge of
the bare electron with no vacuum polarization effects? I think that
the Topaz experiment had alpha at about 1/128 so I will work on some
ideas and see if I can come up with a figure that is close to that.
One of the ideas being that it is the ratio between the conversion of
the photon's electric energy and the electron's electric energy.
Again postulating that Maxwell can describe a photon and it is exactly
one wavelength long. That should mean that 1/4 of a pair creating
photon's energy times *true* alpha should equal the electron's
electric energy. Then we should be able to calculate the electron's
bare charge from that.
> >| Once again, you are misusing that law of physics. You cannot create
> >| charge, just act on it. Nobody knows why charge exists. It just does.
>
> >Baloney, charges can be created by pair production easily.
>
> Non sequitur. We know charge is conserved, and part of the conservation
> rules is annihilation and pair production. Yet we still don't know why
> charge exists, or more specifically, why charge has the properties it
> does (creates forces on other charges, reacts to the presence of other
> charges, obeys Maxwell's equations etc.) It's just the way of the
> universe. At least magnetic effects can be explained as relativistic
> effects on electric effects but that still begs the questions why electric
> effects exist in the first place. Why the forces of an electron on a
> proton follow the same rules whether they are separated by the diameter of
> a galaxy or the electron is actually in the process of travelling through
> the proton (high energy scattering experiments).
Sure we don't know why electric effects exist but that doesn't prevent
us from exploring and understanding more about how they might come
about. I don't know if any of this stuff is in fact the truth but it
is fun playing around with it to see if there is a way to make it all
work and make sense. Plus I have actually learned quite a bit more
about the standard theories and others whilst doing it. And, in fact,
you have been quite helpful in these reqards because I am always
looking stuff up and studying to answer your questions or to reply to
your comments in a way that might make sense.
FrediFizzx