Spin, Mass and Charge are all Related.
Tnlockyer
(exactly) a spin angular momentum of (1/2 h bar) and (exactly) a fundamental
charge (e) regardless of cube size.
The mechanism causing spin is the chance arrangement of momentum, going the
same direction, in back and front cube faces.
1/2 h bar = mass(energy) x radius arm X c
The mass(energy), radius arm product remains constant because the mass(energy)
stored in the spin angular momentum (Joule-seconds) varies inversely as the
size of the radius arm. The 'c' velocity of gyration is a constant velocity,
so the model always returns 1/2 h bar.
The fundamental charge (e) is calculated from the electric field (H) and
charge density (D) as:
D=Eo x H = e x m^-2 = (Ampere-seconds) per meter squared
Where H can be calculated from the Compton wavelength energy (J) x c divided by
the volume (v) of the cube giving a power density (P). Knowing that the
impedance of space (Z) then H= (P/Z)^0.5
It is straight forward that the charge (Ampere-seconds) can be obtained by
multiplying (D) by the current loop areas (m^2) of the spinning cube.
The charge density (D) increases, while the loop area decreases, as the size
of the cube decreases, so the model returns the same fundamental charge (e)
regardless of cube size.
I must conclude that all spin 1/2 particles MUST have mass and fundamental
charge (e) currents, without exception.
The quark cannot have spin 1/2 and a fractional charge, so the SM is most
probably a false model.
Some of the math and geometry may be reviewed on the web page.
Regards: Tom:
Tom: http://www.best.com/~lockyer
Any model should go together precisely, naturally and not need the theorist to
be constantly adding (ad hoc) adjustments to explain the model failures.
Tnlockyer
>The fundamental charge (e) is calculated from the electric field (E) and
>charge density (D) as:
>
>D=Eo x E = e x m^-2 = (Ampere-seconds) per meter squared
>
>
>Where E can be calculated from the Compton wavelength energy (J) x c divided
>by
> the volume (v) of the cube giving a power density (P). Knowing that the
>impedance of space is (Z) then E = (P x Z)^0.5
>
Use the above for electric field (V/m) calculations (sorry) should read my copy
more carefully.
Here are some actual numbers:
J=((h c a)/lambda) = 3.0639285695x10^-13 Joule.
V= (lambda^3) / (8 pi^2)= 1.341225144x106-45 m^3
P=(J c)/V = 6.483457097x10^40 VA/m^3
E=(P x Z)^0.5 = 5.079419825x10^21 V/m
D= E0 x E = 4.4974137127X10^10 (s A m^2)
Finally fundamental charge e=D((lambda^2)/2pi)= 1.60217733X10^-19 (s A)
CHECK.
This happens to be the core particle of the VPP proton, calculations.
Regards:
Tnlockyer
>
>J=((h c a)/lambda) = 3.0639285695x10^-13 Joule.
Where (a) is the fine structure constant which is the well known ratio between
rest mass energy and the EM energy of a particle.(a)= 7.29735308x 10^-3. The
(h) is the Planck constant, (h) =6.6260755x10^-34 Joule- seconds, (c) =
2.99792458x10^8 m/s, and lambda is the VPP core particles Compton wavelength =
4.73111828698697x10^-15 m
>
>V= (lambda^3) / (8 pi^2)= 1.341225144x10^-45 m^3
The (V) is the cylindrical volume of the VPP proton's spinning EM core.
>P=(J c)/V = 6.483457097x10^40 VA/m^2
The (P) is the power density at the VPP proton core.
>E=(P x Z)^0.5 = 5.079419825x10^21 V/m
The (E) is the electric field at the VPP proton core. The (Z) is the impedance
of space (Uo/Eo)^0.5 = 376.730313496243 Ohms (kg m^2 s^-3 A^-2)
where Uo=4pix10^-7 ( m kg s^-2 A^-2) Henry/meter and Eo=8.854187817x10^-12
(m^-3 kg^-1 s^4 A^2) Farad/meter of the vacuum.
>
>D= Eo x E = 4.4974137127X10^10 (s A m^-2)
The (D) is the charge density at the VPP core particle.
>Finally the fundamental charge (e) is simply the VPP core current loop areas
times the charge density
> e=D((lambda^2)/2pi)= 1.60217733X10^-19 (s A)
>CHECK.
This analysis can be applied to any spinning EM cube regardless of size. .
Regardless of size, these proposed VPP particle models always have the
quantized spin angular momentum of 1/2 h bar and a charge equal to the
fundamental charge (e). Further, the model clearly shows that rest mass is
entirely due to the energy stored in the spin angular momentum.
This is why I believe that the neutrino cannot be spinning in the free state or
it would have mass and charge. The VPP model shows that the neutrino
contributes mass to the composites by adding it's EM phasors to the electron or
positron's phasors, spinning up and producing charge conjugation with the
assemblage.
These VPP model results show promise to structure all subatomic particles and
atomic nuclei,
Regards: Tom.
Jacques Distler
tnlo...@aol.com (Tnlockyer) wrote:
>>Tom, a simple question : Do you believe that *any* of QM, even
>>non-relativistic -- not just the Standard Model -- is correct?
>
> [snip]
>
> I find that very little of QM can be applied to practical particle physics
>models.
That's rich.
QM does not apply to particle physics? Is there anything that it DOES
apply to? Or is it yet another branch of physics (like classical E&M,
perhaps?) that is not worth learning about?
JD
--
PGP public key: http://golem.ph.utexas.edu/~distler/distler.asc
Jim Heckman
tnlo...@aol.com (Tnlockyer) wrote:
>
> I find that the spinning EM cube particle model, proposed by VPP,
> always has (exactly) a spin angular momentum of (1/2 h bar)
>
>
> The mechanism causing spin is the chance arrangement of
> momentum, going the same direction, in back and front cube faces.
>
>
> Tom: http://www.best.com/~lockyer
Tom, a simple question : Do you believe that *any* of QM, even
non-relativistic -- not just the Standard Model -- is correct?
If not, I doubt I'll be wasting much more time even reading your posts...
If so, please explain how a quantum angular momentum derived from a
classical 'orbital' one, i.e., via (Position x Momentum) operators whose
commutator is (i h bar), can have a half-integral value : Elementary
Hermitian operator-eigenvalue theory shows that any object having
such an angular momentum would have a discontinuity in its Position
and Momentum wave functions.
--
~~ Jim Heckman ~~
-- "As I understand it, your actions have ensured that you will never
see Daniel again." -- Larissa, a witch-woman of the Lowlands.
-- "*Everything* is mutable." -- Destruction of the Endless
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
Tnlockyer
> tnlo...@aol.com (Tnlockyer) wrote:
>>
>> I find that the spinning EM cube particle model, proposed by VPP,
>> always has (exactly) a spin angular momentum of (1/2 h bar)
The key is any size EM cube will have a spin angular momentum of 1/2 h bar.
>> The mechanism causing spin is the chance arrangement of
>> momentum, going the same direction, in back and front cube faces.
To my knowledge, VPP is the only model that gives a mechanism for spin, and the
known spin angular momentum, from the geometry.
>Tom, a simple question : Do you believe that *any* of QM, even
>non-relativistic -- not just the Standard Model -- is correct?
>If so, please explain how a quantum angular momentum derived from a
>classical 'orbital' one, i.e., via (Position x Momentum) operators whose
>commutator is (i h bar), can have a
>half-integral value
<snip>
Jim, I view the spin of 1/2 integer as a short hand version of 1/2 h bar,
which is simply a number, 5.2728633x10^-35 J/s and is equal to Planck's
constant divide by 4pi. Look at the electron model on the web page and do the
math. The model does indeed return the experimentally known spin angular
momentum of 5.2728633x10^-35 Joule-seconds from a classical (Me x Rm x c) where
Rm is the mass radius of gyration of (lambda/4pi), given by the electron model
geometry.
The point is that any sized spinning EM cube will have a spin angular momentum
of exactly 1/2 h bar and a fundamental charge of exactly 1.60217733x10^-19
Ampere seconds.
I find that very little of QM can be applied to practical particle physics
models.
VPP is a structural model that gives all of the related fundamental physical
constants, and VPP seems to work very well in maintainly the constants in their
known ratios.
Regards: Tom:
Jim Heckman
> >
> > Tom, a simple question : Do you believe that *any* of QM, even
> > non-relativistic -- not just the Standard Model -- is correct?
> >
> > If so, please explain how a quantum angular momentum derived
> > from a classical 'orbital' one, i.e., via (Position x Momentum)
> > operators whose commutator is (i h bar), can have a half-integral
> > value
>
> <snip>
As per his usual M.O., Tom snips here the essence of my suggested line
of proof that his VPP model is not mathematically consistent with the
postulates of QM.
> Jim, I view the spin of 1/2 integer as a short hand version of 1/2 h
> bar, ... and is equal to Planck's constant divide by 4pi.
Duh!
> I find that very little of QM can be applied to practical particle physics
> models.
Thank you. You've answered my question. As per another part of *my*
post that you snipped, I now know better than to waste any more time
reading *your* posts, except perhaps for their entertainment value.
> Tom: http://www.best.com/~lockyer
> Any model should go together precisely, naturally and not need the
> theorist to be constantly adding (ad hoc) adjustments to explain the
> model failures.
Hey! Where's the part about: "Any model should be able to
explain/predict the results of all experiments that fall within its purview,
to within the measured experimental error."
Not to mention the part about: "Any model should be mathematically
consistent with other successful models in all areas where they
overlap."
In other words, your sig, while arguably *necessary*, is far from
*sufficient*. You *do* know enough math to understand the meaning of
these conditions, don't you?
Tnlockyer
It is an interesting test to set these equations up in your math software set
for both numerical and dimensional analysis, and use any randomly selected
length for lambda. I just did and with making the spinning cube of EM energy
have a lambda of one meter, the cube still has a spin angular momentum of 1/2 h
bar and a fundamental charge of exactly (e). Here are the numbers:
lambda= 1 m , J1=(h c a) / lambda , rest mass, kg= h / lambda c
Volume = 0.012665147955292 m^3
Power density Pp= (J1 c) / Volume becomes 3.43125408062621x10^-17 kg s^-3
Electric field E = (Pp / Z )^0.5 = 3.01794385240044x10^-10 A m^-1
Charge density D= Eo H s A m^2
Finally The Em cube with an edge length of (1 m) / 2pi has a fundamental charge
e = 1.60217733x10^-19 A s just as expected.
The spin angular momentum of the EM cube with an edge length of (1 m)/ 2pi is
the expected 1/2 h bar, to wit.
Spin = kg Rm c = 5.27286334562551x10^-35 kg m^2 s^-1
Where Rm is taken from the cube geometry as lambda / 4pi and kg is as
calculated above.(see the electron model on the web page for how the constants
ratio)
Regards: Tom:
ThomasL283
Here is the math, set it up in you computer to do both numerical and
dimensional analysis. (I find this approach reveals any inadvertant error in
the algebra)
The general equations are:
Js= (h c a ) / lambda = joule structure energy;
Jm=(h c) / lambda = joule rest mass energy;
Vol= ((lambda^2) / 4pi)) x (lambda / 2pi) = volume of the spinning cube;
Pd= (Js c) / Vol = power density of the spinning cube;
Z= (Uo / E0 )^0.5 = impedance of space;
Ed= (Pd Z)^0.5 = power density of the spinning cube;
Dd= (Eo Ed) = charge density (A s) per meter squared;
e = ((Dd (lambda)^2) / 2pi)) = calculated fundamental charge from any spinning
EM cube of energy;
The spin angular momentum equations are:
1/2 h bar = ((Jm) / c^2) x (lambda) / 4pi) x c = spin angular momentum from
any spinning EM cube of energy;
Where: h= 6.6260755x10^-34 m^2 kg s^-1
c =2,99792458x10^8 m s^-1
a = 7.29735308x10^-3
Eo=8.854187816x10^-12 m^-3 kg^-1 s^4 A^2
Uo=4pi x 10^-7 m kg s^-2 A^-2
lambda= m of any length cube of EM spinning energy;
These equations prove that the VPP model for basic particles always give the
spin angular momentum of 1/2 h bar and the fundamental charge (e) regardless of
spinning cube size, even if lambda is several meters long!
Good luck:
Regards: Tom:
Tnlockyer
I must apologize for not being able to explain my views, and giving the false
impression that the VPP model for energy was technically wrong. Let me try one
last time. Everyone, please calm down and consider this application of
elementary electrical theory as applied to a traveling wave of EM energy.
1) Energy in a traveling wave must be stored resonantly, or it will be
dissipated (that is, not conserved, i.e. lost.)
2) The vacuum assumes the characteristics of capacitance per meter
(Eo=8.854187817 E^-12 Farad per meter) and an inductance per meter (Uo=4pi E^-7
Henry per meter)
3) The vacuum acts like a resonant circuit for storing the energy of the
traveling wave.
4) The traveling wave has an electric field, Volts per meter (E) and a magnetic
field (H) Ampere per meter. Analogous to voltage and current in a resonant
circuit.
5) Schematically, when viewed from a stationary frame of reference, the
traveling wave appears as if the electrical influence (E) V/m is if phase with
the magnetic influence (H) A/m. Engineers call this 100 percent power factor
and it signifies complete dissipation (non conservation of energy, or lost
energy) See for example;
http://www.best.com/~lockyer/home1.htm
except substitute H for B (B= Uo H)
6) See Halliday and Resnick (1968) Vol II, p.945, Figure 38-2 waveforms of the
current and voltage in a capacitor. Notice the current is 90 degrees leading,
for zero dissipation, and notice the traveling wave violates this (does not
conserve energy).
7) See Halliday and Resnick (1968) Vol II, p.950, Figure 38-4 waveforms that
show the electric energy Ue= Uemax cos^2wt and magnetic energy Ub= Ubmax
sin^2wt in a resonant circuit.
AND NOTE THAT THEIR SUM IS A CONSTANT FOR ENERGY CONSERVATION in a resonant
circuit.
8) The apparent violation of energy conservation in the in phase V/m and A/m in
a traveling wave has to be corrected, because we know that energy of a
traveling wave must be conserved to be transported with no apparent loss
through light years of time.
9) The VPP model, for the transport and CONSERVATION of energy, deduces that
there must actually be two conjugate resonances coupled in time coincidence and
space quadrature.
We immediately see that this conserves the energy from:
S= 0.5((sin wt E) x (sin wt H)) + ((cos wt H) x (cos wt E))
Note the similarity to Halliday and Resnick p.950. Note the lateral sin-sin
(left terms) we see in the traveling wave. Note the axial cos-cos (right
terms) for the axial transport and conservation of energy, by the photon, as
required. The VPP photon (energy) has an instantaneous existence in time,
because the photon is a separate entity with an energy = Planck's constant x
frequency of oscillation. The lateral amplitude of radiation is always
limited to the wavelength of the energy. Coherent radiation, then is a group of
photons all in phase, and incoherent radiation has the photons at random phase.
In either case the contributions of individual photons preserves the waveform
to that given on the web page. This also means that the photon velocity is an
rms value, with peak velocity of 1.414c. The photon models as actually being
stationary twice each cycle as it obtains crest values.
These rules are used by VPP to construct basic particles as cubes of EM energy:
See:
http://www.best.com/~lockyer/home2.htm
Notice the electron and positron model are spinning cubes of EM energy. See the
following to find the proof that VPP is the correct model fro the structure of
energy and matter. To wit:
Regards: Tom:
Jacques Distler
tnlo...@aol.com (Tnlockyer) wrote:
>I must apologize for not being able to explain my views, and giving the false
>impression that the VPP model for energy was technically wrong.
Repetition does not make it better. See articles
<distler-2806...@192.168.0.1>,
<distler-2906...@192.168.0.1>, and
<distler-2906...@192.168.0.1>
where impressions meet reality. 'Nuff said.
Tnlockyer
>In article <19990701102613...@ng-fw1.aol.com>,
>tnlo...@aol.com (Tnlockyer) wrote:
>
>>I must apologize for not being able to explain my views, and giving the
>false
>>impression that the VPP model for energy was technically wrong.
>Repetition does not make it better.
Jacques, you still don't get it and continue to insult me. Look, the VPP
conserves energy in the resonances (see reference to Halliday and Resnick in my
post of yesterday.) Resonance is the only way nature has to conserve energy.
Now please explain the lack of mechanism in Maxwell's equations to show how to
conserve the energy. And, no, you can't have a power company re-supplying
energy for each cycle, that's cheating.
To be conserved, the energy has to be STORED someplace. The only other
alternative is dissipation, and Maxwell describes dissipation (not energy
conservation) very well, that's why Maxwell is of great importance in
Engineering, but Maxwell sure as hell doesn't tell where the energy is stored
in a traveling wave of EM energy.
And you still have not commented on the profound VPP revelation that any sized
spinning cube of EM energy has exactly a spin angular momentum of 1/2 h bar and
exactly a fundamental charge (e) REGARDLESS OF THE SIZE OF THE SPINNING CUBE
OF EM ENERGY.
VPP uses the fact that all spinning energy cubes have 1/2 h bar and (e) to
then scale to the composites as nested cubes of spinning EM energy. This VPP
scaling approach gives the mass of the proton and neutron to within a few
parts per million of the CODATA.
QCD cannot give particle mass or magnetic moment, so stop posting nonsence
against VPP, a model that does give coherently numerically correct models.
If QCD does not compute (mass, spin and charge), then you have to boot.
David Lamb
On 2 Jul 1999, Tnlockyer wrote:
>
> QCD cannot give particle mass or magnetic moment, so stop posting nonsence
> against VPP, a model that does give coherently numerically correct models.
>
As has been repeatedly pointed out - QCD can. Have you looked up the
references?
for a new one, try hep-lat/9906027, entitled
Baryon Masses from Lattice QCD: Beyond the Perturbative Chiral Regime
Was published this week.
Jacques Distler
tnlo...@aol.com (Tnlockyer) wrote:
>Jacques, you still don't get it and continue to insult me. Look, the VPP
>conserves energy in the resonances (see reference to Halliday and
>Resnick in my post of yesterday.) Resonance is the only way nature has
>to conserve energy.
No, it's the only thing you remember from College Physics. There is a
difference.
We have explained to you six different ways how energy is conserved in the
EM plane wave. I'm not going to waste anyone's time by trying a seventh
time. Pick up any standard texbook on E&M and work through the derivation
yourself.
And, no, your citations to unrelated pages in Halliday and Resnick buy you
nothing, as you claim Maxwell Theory is wrong while their book is BASED on
Maxwell theory.
>To be conserved, the energy has to be STORED someplace. The only other
>alternative is dissipation, and Maxwell describes dissipation (not energy
>conservation)
Maxwell's equations are NOT dissipative. Add that to the ever-lenthening
list of things you misunderstand about classical electrodynamics.
>And you still have not commented on the profound VPP revelation that
>any sized spinning cube of EM energy has exactly a spin angular momentum
>of 1/2 h bar and exactly a fundamental charge (e) REGARDLESS OF THE
>SIZE OF THE SPINNING CUBE OF EM ENERGY.
Oh yes I did.
I said it was WRONG.
Proof: both the neutron and the neutrino are spin 1/2,
electrically-neutral particles.
You still have not explained how your models is compatible with the decay
of a pion into an electron plus a neutrino which, in particular,
demonstrates conclusively that the neutrino has spin 1/2. (Note that the
nonexistent decay -- to electron plus three neutrinos -- that you
predicted ALSO would rule out the possibility of the neutrino being spin
0.)
> QCD cannot give particle mass or magnetic moment,
Untrue.
Others have provided you with citations to the contrary. But I don't
expect you to understand the papers they have cited until you have (at a
minimum) mastered classical E&M.
Since your model is supposed to be based on classical E&M, you show a
surprising lack of interest in mastering the subject.
Is there any branch of physics that you DO consider worth learning?
(You've already rejected quantum mechanics as being irrelevant to particle
physics.)
Jacques
Tnlockyer
>In article <19990702130516...@ng-cj1.aol.com>,
>tnlo...@aol.com (Tnlockyer) wrote:
>
>>Jacques, you still don't get it and continue to insult me. Look, the VPP
>>conserves energy in the resonances (see reference to Halliday and
>>Resnick in my post of yesterday.) Resonance is the only way nature has
>>to conserve energy.
>
>No, it's the only thing you remember from College Physics.
>We have explained to you six different ways how energy is conserved in the
>EM plane wave. I'm not going to waste anyone's time by trying.
No you did not, the only explanation is the analysis posted for the VPP photon.
Are you saying the VPP photon does not conserve energy?
>Pick up any standard texbook on E&M and work through the derivation
>yourself.
>And, no, your citations to unrelated pages in Halliday and Resnick buy you
>nothing, as you claim Maxwell Theory is wrong while their book is BASED on
>Maxwell theory.
Nope, Maxwell does not use the derivation for the energy storage in a resonant
circuit.
>>And you still have not commented on the profound VPP revelation that
>>any sized spinning cube of EM energy has exactly a spin angular momentum
>>of 1/2 h bar and exactly a fundamental charge (e) REGARDLESS OF THE
>>SIZE OF THE SPINNING CUBE
>OF EM ENERGY.
>
>Oh yes I did.
>
>I said it was WRONG.
Who are you, the voice of utimate authority? The math is their for everyone to
see.
Tnlockyer
the electron and positron are spinning by virtue of the momentum of the phasors
in their structures going in the same direction in back and front cube faces.
Now, these model structures can be deduced by anyone by simply exhausting all
possible ways to construct cubes from the VPP model for energy. (This is an
interesting puzzle to doodle on a rainy afternoon). See the web page.
http://www.best.com/~lockyer/home2.htm
Now set up in your computer a math software to do both numerical and
dimensional analysis and follow these equations that prove that any sized EM
cube of spinning energy will have a spin angular momentum of exactly ½ h bar =
5.2728633456 x 10^-35 m^2 kg s^-1 (Joule- seconds) and a fundamental charge of
exactly e = 1.60217733 x 10^-19 A s.
Use these variables:
h = 6.626075470853 x 10^-34 m^2 kg s^-1 (Planck constant)
c = 2.99792458 x 10^8 m s^-1 (Velocity of light)
a = 7.297353080929 x 10^-3 (Fine structure constant)
Eo = 8.854187817 x 10^-12 m^-3 kg^-1 s^4 A^4 (Permittivity of the vacuum)
Uo = 12.566370614 x 10^-7 m kg s^-2 A^-2 (Permeability of the vacuum)
lambda = 2.42631057x 10^12 m (Electron Compton for the electron model)
pi = 3.1415925
The general equations are:
Js = (h c a ) / lambda = Joule cube structure energy.
Jm = (h c ) / lambda = Joule cube rest mass energy
Vol = ((lambda^2) / 4pi )) x (lambda / 2pi) = Volume of the spinning cube
(cylinder )
Pd = ( Js c ) / Vol = Power density, Watts per square meter.
Z = ( Uo / Eo )^0.5 = Impedance of space.
Dd = (Eo Ed ) = Charge density, charge per meter squared
e = ((Dd (lambda^2) / 2pi ) = fundamental charge = 1.60217733 x 10^-19 (CHECK)
½ h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x 10^-35
J-s (CHECK)
Once you have the equations set up, it is interesting to experiment by
changing lambda to any length, in meters, whatsoever. The model always gives
the same spin and charge!!
If you are too lazy to key in the above equations and have Mathcad Professional
8 software, e-mail me and I will send you the file as an attachment.
Regards: Tom:
Jacques Distler
tnlo...@aol.com (Tnlockyer) wrote:
>>We have explained to you six different ways how energy is conserved
>>in the EM plane wave. I'm not going to waste anyone's time by trying.
>
>No you did not, the only explanation is the analysis posted for the VPP photon.
>
>Are you saying the VPP photon does not conserve energy?
I am saying that it VIOLATES Maxwell's equations.
If you wish to continue discussing it, please post the equations that you
wish use IN PLACE OF Maxwell's equations.
Then we will discuss whether your "VPP photon" is a solution to those equations.
Then we will discuss whether your equations or Maxwell's describe
electromagnetic phenomena in the real world.
Only once you write down those equations does the question of whether
energy is conserved by your solution become meaningful.
>>Pick up any standard texbook on E&M and work through the derivation
>>yourself.
>
>>And, no, your citations to unrelated pages in Halliday and Resnick
>>buy you nothing, as you claim Maxwell Theory is wrong while their
>>book is BASED on Maxwell theory.
>
>Nope, Maxwell does not use the derivation for the energy storage in
>a resonant circuit.
Which counts for precisely nothing.
You would get a lot further if you sat down and tried to understand ONE of
the half-dozen proofs which have been posted that the EM plane wave
solution to Maxwell's equations conserves energy.
I really DO think it would be within your abilities, were you not too
proud to make the attempt.
>>>And you still have not commented on the profound VPP revelation that
>>>any sized spinning cube of EM energy has exactly a spin angular momentum
>>>of 1/2 h bar and exactly a fundamental charge (e) REGARDLESS OF THE
>>>SIZE OF THE SPINNING CUBE
>
>>OF EM ENERGY.
>>
>>Oh yes I did.
>>
>>I said it was WRONG.
>
>Who are you, the voice of utimate authority?
You're right.
Let's revise that: Mother nature is the ultimate authority, and SHE says
it is wrong.
(I note that you snipped, rather than attempted to rebut, the elementary
empirical refutation.)
JD
James H. Panetta
>Use these variables:
>
>h = 6.626075470853 x 10^-34 m^2 kg s^-1 (Planck constant)
>c = 2.99792458 x 10^8 m s^-1 (Velocity of light)
>a = 7.297353080929 x 10^-3 (Fine structure constant)
>Eo = 8.854187817 x 10^-12 m^-3 kg^-1 s^4 A^4 (Permittivity of the vacuum)
>Uo = 12.566370614 x 10^-7 m kg s^-2 A^-2 (Permeability of the vacuum)
>lambda = 2.42631057x 10^12 m (Electron Compton for the electron model)
>pi = 3.1415925
Now, the definitions of these numbers:
a === e^2/(4pi*Eo*hc)
EoUo === 1/c^2
lambda === h/(m_e*c)
>The general equations are:
>
>Js = (h c a ) / lambda = Joule cube structure energy.
>Jm = (h c ) / lambda = Joule cube rest mass energy
Oh, this is too much.
Jm = hc/(h/m_e*c)
= h/h * m_e *c *c
= m_e * c^2
This trivially follows from the DEFINITION of the compton wavelength.
>Vol = ((lambda^2) / 4pi )) x (lambda / 2pi) = Volume of the spinning cube
>(cylinder )
>Pd = ( Js c ) / Vol = Power density, Watts per square meter.
>Z = ( Uo / Eo )^0.5 = Impedance of space.
>Dd = (Eo Ed ) = Charge density, charge per meter squared
You need to define Ed.
>e = ((Dd (lambda^2) / 2pi ) = fundamental charge = 1.60217733 x 10^-19 (CHECK)
>h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x 10^-35
>J-s (CHECK)
hbar =? (hc)/lambda/c^2 *c * lambda/4pi *c
= hc/4pi
Your units are incorrect.
You try to show a trivial derivation as something earthshaking, and
an incorrect derivation as proof.
--Jim
--
My opinions are mine...not SLAC's...not Penn's...not DOE's...mine.
(except by random, unforseeable coincidences)
pan...@slac.stanford.edu -- three sigma to the left of strange
Jacques Distler
pan...@bbr-dev02.SLAC.Stanford.EDU (James H. Panetta) wrote:
>Now, the definitions of these numbers:
>
>a === e^2/(4pi*Eo*hc)
>EoUo === 1/c^2
>lambda === h/(m_e*c)
>
>
>>The general equations are:
>>
>>Js = (h c a ) / lambda = Joule cube structure energy.
>>Jm = (h c ) / lambda = Joule cube rest mass energy
>
>Oh, this is too much.
>Jm = hc/(h/m_e*c)
> = h/h * m_e *c *c
> = m_e * c^2
>
>This trivially follows from the DEFINITION of the compton wavelength.
>
You're so MEAN, Jim.
How is Thomas going to sell any books if you keep giving away his secrets
for free?
Thomas's big trick is to "derive" the mass of a particle using only its
Compton wavelength as input.
Now you've gone and shown that this is a trivial identity following from
the DEFINITION of Compton wavelength.
What next? Are you going to tell us that those books which promise to help
us get rich quick in the Real Estate market don't deliver as advertised?
Tnlockyer
Ed = (Pd Z )^0.5 = electric field in volts per meter.
BTW, notice that the VPP automatically gives us the structures for BOTH and
electron and positron. If fact VPP gives us the worlds first model for the
positron and shows the positron's conjugate structure to the electron model's
structure. The VPP also shows the mechanism that causes spin angular momentum,
mass (energy) and fundamental charge, in the electron and positron, and that
only that type of structure can have those characteristics.
This VPP result suggests that we might now be able to model the difference
between a negative fundamental charge and a positive fundamental charge.
Notice the model designated the electron has all electric phasors on the
rotating faces, while the positron model has all magnetic phasors on the
rotating faces.
http://www.best.com/~lockyer/home2.htm
The VPP model is a very rich model, and I believe, VPP will be the model for
the next millennium.
Jacques Distler
tnlo...@aol.com (Tnlockyer) wrote:
>>pan...@bbr-dev02.SLAC.Stanford.EDU (James H. Panetta)writes:
>
>>This trivially follows from the DEFINITION of the compton wavelength.
>
>What you fail to see Jim, is that the geometry of the spinning cube is
>the VPP model's contribution.
I LOVE it when you talk out of both sides of your mouth.
On the one hand, your tout the astounding accuracy with which your model
(which uses the Compton wavelength as input) reproduces particle masses.
When it is pointed out to you that this follows trivially from the
definition of the Compton wavelength (and hence is no prediction at all),
you switch gears and tout the "beautiful" geometrical structure of your model.
When it is pointed out that the geometrical structure of your model is
theoretically-inconsistent (incompatible with the basic equations of
classical electromagnetism) and ruled out by experiment (your electrons
are at least a million times larger than the experimental upper bounds on
the size of the electron, your neutrinos are spin 0, whereas experiment
says the neutrino is spin 1/2, . . . ), you shift back to touting the
astounding accuracy with which your model reproduces particle masses . . .
Could you go back to arguing that Maxwell's equations are incompatible
with Special Relativity? I *really* enjoyed that one.
>My oldest son's son starts at MIT in the fall. My grandson is a
>straight A student in all subjects (since kindergarten) and has a 4.5 GPA.
Which means that, in one year's time, he'll know more physics than you do,
and will be able to start correcting your multitude of errors. I wish I
could be a fly on the wall when he comes home for Spring Break next March.
Jacques
Fred Diether
software,
| here is a collection of equations relating to the VPP particle models and
their
| maintaining the fundamental charge and spin angular momentum exactly
regardless
| of spinning cube of EM energy sizes, you can download.
|
| ftp://members.aol.com/tnlockyer/power.mcd
Tom,
Can you save it as a Mathcad 7 file and email it to me or put it up on your
ftp as power2.mcd? Thanks. I don't have Mathcad 8 yet.
Regards,
Fred Diether
fredi...@email.msn.com
Tnlockyer
>This trivially follows from the DEFINITION of the compton wavelength.
What you fail to see Jim, is that the geometry of the spinning cube is the VPP
model's contribution.
>>Dd = (Eo Ed ) = Charge density, charge per meter squared
>
>You need to define Ed.
Yes, see my earlier post. Ed= (Pd Z)^0.5
>>h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x
>10^-35
>>J-s (CHECK)
>hbar =? (hc)/lambda/c^2 *c * lambda/4pi *c
> = hc/4pi
>
>Your units are incorrect.
Nope, where did you get (hc) from? I showed this:
h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x
10^-35J-s (CHECK)
Jim, I put up the math program that does both numerical and dimensional
analysis in my FTP file.
To those of you who have access to Mathcad 8 Professional mathematics software,
here is a collection of equations relating to the VPP particle models and their
maintaining the fundamental charge and spin angular momentum exactly regardless
of spinning cube of EM energy sizes, you can download.
ftp://members.aol.com/tnlockyer/power.mcd
Set up a floppy in you're (A) drive and download the file by clicking on the
above address.
Then find someone who can run Mathcad 8 programs and print out the calculated
file.
BTW, small world department. Mathsoft (Mathcad vendor) is right across the
street from MIT.
My oldest son's son starts at MIT in the fall. My grandson is a straight A
student in all subjects (since kindergarten) and has a 4.5 GPA.
Regards: Tom:
James H. Panetta
>
>>This trivially follows from the DEFINITION of the compton wavelength.
>
>What you fail to see Jim, is that the geometry of the spinning cube is the VPP
>model's contribution.
Bullshit. Your "theory" is irrelevant if all it can reproduce is
trivial identities that follow from the definition of the terms
involved.
>>>Dd = (Eo Ed ) = Charge density, charge per meter squared
>>
>>You need to define Ed.
>
>Yes, see my earlier post. Ed= (Pd Z)^0.5
You hadn't done this when I posted. I'll give you the benefit of
the doubt, though.
>>>h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x
>>>10^-35>J-s (CHECK)
>
>>hbar =? (hc)/lambda/c^2 *c * lambda/4pi *c
>> = hc/4pi
>>
>>Your units are incorrect.
>
>Nope, where did you get (hc) from? I showed this:
>
> h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x
>10^-35J-s (CHECK)
You define Jm === hc/lambda. Trivial substitution:
hbar = ((hc/lambda)/c^2) x c x ((lambda) / 4pi) x c
= hc/4pi
>Jim, I put up the math program that does both numerical and dimensional
>analysis in my FTP file.
And it's *wrong*, as I just showed... Your 'numerical' analysis seems
to be flawed.
>My oldest son's son starts at MIT in the fall. My grandson is a straight A
>student in all subjects (since kindergarten) and has a 4.5 GPA.
Good for him. It doesn't mean your theory is worth anything, though.
James H. Panetta
>In article <7ljank$m5u$1...@nntp.Stanford.EDU>,
>pan...@bbr-dev02.SLAC.Stanford.EDU (James H. Panetta) wrote:
>
>>
>
>
>How is Thomas going to sell any books if you keep giving away his secrets
>for free?
Hush Jacques, we're part of the cabal, remember? (TINC)
>What next? Are you going to tell us that those books which promise to help
>us get rich quick in the Real Estate market don't deliver as advertised?
You can! Just buy this book, and do I have a *deal* for you on some
beachfront property in Omaha...
Tnlockyer
>>What you fail to see Jim, is that the geometry of the spinning cube is the
>VPP
>>model's contribution.
>
>Bullshit. Your "theory" is irrelevant if all it can reproduce is
>trivial identities that follow from the definition of the terms
>involved.
Please note that it is the cylindrical geometry of the spinning cube that
gives the power density and it is the charge density of the electrical energy
and the cubes current loop areas that derives the fundamental charge. And it
is the mass radius of lambda / 4pi from the cube that gives the spin angular
momentum.
But note: ANY lambda length can be substituted and the cube model will return
exactly 1/2 h bar and exactly (e).
When you get the file, try it. change lambda to 100 meters or 1000 meters,
makes no difference, the model still returns exactly 1/2 h bar and (e).
VPP is not trivial, I think VPP is telling us something profound.
>> h bar = ((Jm) / c^2) x (c) x ((lambda) / 4pi) x (c) = 5.2728633456 x
>>10^-35J-s (CHECK)
>
>You define Jm === hc/lambda. Trivial substitution:
I see what happened, Wordperfect reduces the 1/2 to graphics and your software
did not pick it up. Also there is a typo of an extra (c)
You are right, should be:
1/2 h bar = ((Jm) / c^2) x c x ((lambda) / 4pi)
>>Jim, I put up the math program that does both numerical and dimensional
>>analysis in my FTP file.
>
>And it's *wrong*, as I just showed... Your 'numerical' analysis seems
>to be flawed.
No, if you download the file you will see. Just this morning the equations
were tested before the file was posted. With the math program one can see very
quickly when a algebra (or typo) error has been made, the units come out with
fractional powers.
Tnlockyer
>I LOVE it when you talk out of both sides of your mouth.
>
>On the one hand, your tout the astounding accuracy with which your model
>(which uses the Compton wavelength as input) reproduces particle masses.
>
>When it is pointed out to you that this follows trivially from the
>definition of the Compton wavelength (and hence is no prediction at all),
Yes, it was deduced that the electron's Compton wavelength is lambda = h / me c
about 50 years ago.
But this is simply based on the fact that all fundamental constants are
related, so one can find all sorts of realtionships just by dimensional
analysis.
But, do you know WHY this relationship holds? VPP shows the geometric source
of the relationship.
>Could you go back to arguing that Maxwell's equations are incompatible
>with Special Relativity? I *really* enjoyed that one.
>
If you critics would download the program :
ftp://members.aol.com/tnlockyer/power.mcd
The proof that Maxwell does not conserve energy is nicely graphed out for all
to see.
I think your idea of energy conservation is not physcically correct. To
conserve energy, you have to show how (where) the energy is stored. Maxwell
does not do that, but the VPP energy model does, in the two conjugate
resonances.
Tnlockyer
T. Lockyer wrote:
>| To those of you who have access to Mathcad 8 Professional mathematics
>software,
>| here is a collection of equations relating to the VPP particle models and
>their
>| maintaining the fundamental charge and spin angular momentum exactly
>regardless
>| of spinning cube of EM energy sizes, you can download.
>|
>|
ftp://members.aol.com/tnlockyer/power.mcd
>
>Tom,
>
>Can you save it as a Mathcad 7 file and email it to me or put it up on your
>ftp as power2.mcd? Thanks. I don't have Mathcad 8 yet.
>
>Regards,
>Fred Diether
>fredi...@email.msn.com
Yes, I have Mathcad 7 on my old 486, but it will take some time. Unfortunately
Mathcad 8 will read Mathcad 7 but you can't go the other way.
Jacques Distler
tnlo...@aol.com (Tnlockyer) wrote:
>>I LOVE it when you talk out of both sides of your mouth.
>>
>>On the one hand, your tout the astounding accuracy with which your model
>>(which uses the Compton wavelength as input) reproduces particle masses.
>>
>
>>When it is pointed out to you that this follows trivially from the
>>definition of the Compton wavelength (and hence is no prediction at all),
>
>Yes, it was deduced that the electron's Compton wavelength is lambda = h / me c
>about 50 years ago.
It wasn't "deduced"; that was how the Compton wavelength was DEFINED.
>But this is simply based on the fact that all fundamental constants are
>related, so one can find all sorts of realtionships just by dimensional
>analysis.
>
>But, do you know WHY this relationship holds? VPP shows the geometric source
>of the relationship.
In other words, you have NO prediction of the particle mass.
You use the Compton wavelength as input. Since the Compton wavelength is
DEFINED as being inversely proportional to the mass (and the value quoted
in the tables is CALCULATED from the experimentally-measured mass), it is
an utter TRIVIALITY that the experimentally-measured value of the mass
POPS OUT at the end of your calculation.
>>Could you go back to arguing that Maxwell's equations are incompatible
>>with Special Relativity? I *really* enjoyed that one.
>
>If you critics would download the program :
>
>ftp://members.aol.com/tnlockyer/power.mcd
>
>The proof that Maxwell does not conserve energy is nicely graphed out for all
>to see.
Alas, the proof is wrong.
You are focussing your attention at one point in space and neglecting the
fact that the energy MOVES.
Just because the output came out of Mathcad does not make it any more correct.
>I think your idea of energy conservation is not physcically correct.
You think wrong.
Mine is the one that follows from Noether's theorem.
>To conserve energy, you have to show how (where) the energy is stored.
I computed for you where the energy goes. You did not (could not?) follow
the computation.
I told you to construct a "flipbook" of the function
sin^2 (k(x-ct))
and then, even if you cannot follow the ALGEBRA, you could at least SEE
where the energy goes.
If you can't even do that much for yourself, it is hard to help you.
ThomasL283
T. Lockyer writes:
>>Yes, it was deduced that the electron's Compton wavelength is lambda = h /
>me c
>>about 50 years ago.
>>But this is simply based on the fact that all fundamental constants are
>>related, so one can find all sorts of realtionships just by dimensional
>>analysis.
>>But, do you know WHY this relationship holds? VPP shows the geometric
>source
>>of the relationship.
>
>In other words, you have NO prediction of the particle mass.
>
>You use the Compton wavelength as input. Since the Compton wavelength is
>DEFINED as being inversely proportional to the mass (and the value quoted
Nope: I can derive the mass from any arbitrary wavelength using the geometry
of the VPP spinning cubes. To wit;
Problem: Take any arbitrary wavelength (lambda) and derive the quantum (h).
The problem is solvable by using the geometry of the VPP basic particle model
as a spinning cube of EM energy. Remember that the cube has an edge length
equal to (lambda / 2pi) and a mass radius of gyration of (lambda / 4pi).
The volume of the spinning cube is:
Vol = (lambda^2 / 4pi) (lambda / 2pi)
The magnetic field is:
H = (2pi e c) / lambda^2 Ampere per meter
The well known reactive impedance of space is:
Z = (Uo/Eo)^0.5 = 376.730313496243 Ohms
The electric field is:
E = H Z Volts per meter
The energy density of the cube becomes:
Jeh = 0.5 (Eo (E^2) + Uo (H^2)) Joule per meter cubed
We now can solve for the rest mass (related to lambda) knowing that the above
energy density times the cube volume is the energy of the EM structure, and
that the rest mass energy is obtained by dividing by the fine structure
constant (a) as follows.
Jm = (Jeh Vol) / a
Then the quantum is directly obtainable from the spin angular momentum (mass x
radius arm x velocity of gyration) This momentum is quantized by the geometry
of the spinning cube of EM energy. So:
h/4pi = (Jm / c^2) x (lambda / 4pi) x (c) which reduces to:
h = (Jm / c^2) x (lambda) x (c) which is the well known Compton relation:
I tested this with a lambda of 100000000 meters and still got the correct (h).
h = 6.62607549818106 x 10^-34 kg m^2 s^-1 (Joule seconds)
But, I don't expect you to understand or accept these results.
Regards: Tom:
Jacques Distler
thoma...@aol.com (ThomasL283) wrote:
>>In other words, you have NO prediction of the particle mass.
>>
>>You use the Compton wavelength as input. Since the Compton wavelength is
>>DEFINED as being inversely proportional to the mass (and the value quoted
>>in the tables is CALCULATED from the experimentally-measured mass), it is
>>an utter TRIVIALITY that the experimentally-measured value of the mass
>>POPS OUT at the end of your calculation.
>
>Nope: I can derive the mass from any arbitrary wavelength using the geometry
>of the VPP spinning cubes.
You ARE being obtuse, of course.
That's exactly the point! Given a particle's Compton wavelength, its mass
is strictly determined (and vice versa).
Contrary to your repeated claims, you do NOT have a computation of any
particle masses. Instead, you put them in BY HAND (in the form of the
Compton wavelength).
So much for the predictive power of your model.
> To wit;
>
>Problem: Take any arbitrary wavelength (lambda) and derive the quantum (h).
>[snip]
Yet another TRIVIAL application of the definitions of the quantities involved.
If you wrote out the definitions of the Fine Structure constant and the
Compton wavelength, you would see that this was an exercise in circular
(cubical?) reasoning. (It would help if you didn't have to rely on Mathcad
to do the algebra for you.)
Are you going to derive the speed of light, c, next?
How about deriving the Fine Structure constant for an encore?
Will this never end?
>But, I don't expect you to understand or accept these results.
What was there to accept?
Does your model actually make ANY correct predictions? (Since the mass of
the particle and the value of Planck's constant -- through its appearance
in the fine structure constant and the formula relating the Compton
wavelength to the mass -- are input parameters to your model, you can't
count them as "predictions".)
It makes mountains of incorrect ones (the size of your "electron" is more
than a million times larger than the experimental upper limit; your
"neutrino" is spin 0, whereas experiment says it is spin 1/2, . . . ) and
is inconsistent with the basic equations of classical electrodynamics (on
which your model is supposedly based).
Is there any point in continuing this discussion?
JD
Tnlockyer
>thoma...@aol.com (ThomasL283) wrote:
>>Nope: I can derive the mass from any arbitrary wavelength using the
>geometry
>>of the VPP spinning cubes.
>That's exactly the point! Given a particle's Compton wavelength, its mass
>is strictly determined (and vice versa).
Nope, you would have to insert the quantum (h) and the velocity of light (c).
VPP just proved to be able to calculate (h) from ANY wavelength whatso ever.
You OTOH, have to insert (h) as derived from experiment (that's cheating).
>Contrary to your repeated claims, you do NOT have a computation of any
>particle masses. Instead, you put them in BY HAND (in the form of the
>Compton wavelength).
You started making statements based on a mistaken premise that the result is
trivial, before reading the math presented in my previous post. The model DID
derive the mass from a wavelength, using only the cube geometry. (Look again
without being superficial this time)
It is the volume geometry, of the spinning cube of EM energy, that allows
deriving the mass from any arbitrary wavelength (lambda). And it is the cube's
mass radius of (lambda / 4pi) that then derives the quantum (h).
>>Problem: Take any arbitrary wavelength (lambda) and derive the quantum (h).
>>[snip]
>
>Yet another TRIVIAL application of the definitions of the quantities
>involved.
You still persist in believing that the results are trivial before you actually
do the math.
>>But, I don't expect you to understand or accept these results.
>
>What was there to accept?
>It makes mountains of incorrect ones (the size of your "electron" is more
>than a million times larger than the experimental upper limit;
Um, how does that reconcile with the electron's Compton wavelength of
2.42631058 x 10 ^-12 meters?
The electron's Compton wavelength is 2pi times the VPP electron cube model's
(lambda / 2pi) edges.
See:
http://www.best.com/~lockyer/home3.htm
>your
>"neutrino" is spin 0, whereas experiment says it is spin 1/2, . . . ) and
Jacques, the model clearly shows that the free neutrino does not spin BUT that
the neutrino spins in CONCERT with the electron. So when the neutrino is
free, (decoupled) it no longer affects the equation.
The model develops the mass of the composites by forming neutrinos 'insitu'.
Neutrino vectors combine with the assemblages vectors and spin up with the
assemblage to create rest mass, charge currents and magnetic moments, at each
layer.
To see how this works and to get the mass of the proton and neutron, see:
http://www.best.com/~lockyer/home4.htm
>Is there any point in continuing this discussion?
Probably not, but I must say the discussions with you have helped in developing
the mathematical arguments
that I might not have otherwise attempted.
Regards: Tom.
Jacques Distler
>>Contrary to your repeated claims, you do NOT have a computation of any
>>particle masses. Instead, you put them in BY HAND (in the form of the
>>Compton wavelength).
>
>You started making statements based on a mistaken premise that the result is
>trivial, before reading the math presented in my previous post. The model DID
>derive the mass from a wavelength, using only the cube geometry. (Look again
>without being superficial this time)
And the Compton wavelength listed in the tables of particle properties is
CALCULATED from the experimentally-determined mass.
To make a long story short, you take the experimentally-measured mass of
the particle as input, then after much grinding of gears, out pops a
number for the mass of the particle which agrees with the
experimentally-measured one.
Big #@$% deal.
>>>Problem: Take any arbitrary wavelength (lambda) and derive the quantum (h).
>>>[snip]
>>
>>Yet another TRIVIAL application of the definitions of the quantities
>>involved.
>
>You still persist in believing that the results are trivial before you actually
>do the math.
I looked at it. It IS trivial.
Jim Panetta has already explained that you calculation of the mass is just
circular reasoning, so let's cut to the "new" part ("Jm/ c^2" is the mass,
m):
> h/4pi = (Jm / c^2) x (lambda / 4pi) x (c)
But the DEFINITION of the Compton wavelength is
lambda = h/mc
So this is no more than a trivial application of the definition of Compton
wavelength.
Big #@$% deal.
>>>But, I don't expect you to understand or accept these results.
>>
>>What was there to accept?
>
>>It makes mountains of incorrect ones (the size of your "electron" is more
>>than a million times larger than the experimental upper limit;
>
>Um, how does that reconcile with the electron's Compton wavelength of
>2.42631058 x 10 ^-12 meters?
That's easy. The Compton wavelength of an object is in NO way related to
its size.
Your own Compton wavelength is smaller than your Schwarzschild radius. If
your Compton wavelength were a measure of your size, you would be a black
hole.
(Some might find that desirable, but it is nevertheless untrue.)
The experimental upper limit on the size of an electron is more than a
million times smaller than your model predicts. By the usual standards of
Science, this means that your model must be rejected.
>>your
>>"neutrino" is spin 0, whereas experiment says it is spin 1/2, . . . ) and
>
>Jacques, the model clearly shows that the free neutrino does not spin BUT that
>the neutrino spins in CONCERT with the electron. So when the neutrino is
>free, (decoupled) it no longer affects the equation.
Angular momentum is conserved.
Ergo the neutrino has spin 1/2. It does not "decouple" from the balance of
angular momenta, just as it does not "decouple" from the balance of linear
momenta. It is irrelevant that it is free; it still carries half-integral
angular momentum. Otherwise angular momentum conservation would be
violated.
If your model says the neutrino has spin 0, then your model is wrong.
>>Is there any point in continuing this discussion?
>
>Probably not, but I must say the discussions with you have helped in developing
>the mathematical arguments
>that I might not have otherwise attempted.
Mmmmm, like the one about how Maxwell's equations violate energy conservation?
Christopher Gomez
>
> >dis...@golem.ph.utexas.edu (Jacques Distler)writes:
>
> T. Lockyer writes:
>
> >>Yes, it was deduced that the electron's Compton wavelength is lambda = h /
> >me c
> >>about 50 years ago.
> >>But this is simply based on the fact that all fundamental constants are
> >>related, so one can find all sorts of realtionships just by dimensional
> >>analysis.
> >>But, do you know WHY this relationship holds? VPP shows the geometric
> >source
> >>of the relationship.
> >
>
> >
> >You use the Compton wavelength as input. Since the Compton wavelength is
> >DEFINED as being inversely proportional to the mass (and the value quoted
>
>
> Problem: Take any arbitrary wavelength (lambda) and derive the quantum (h).
>
> as a spinning cube of EM energy. Remember that the cube has an edge length
> equal to (lambda / 2pi) and a mass radius of gyration of (lambda / 4pi).
>
> The volume of the spinning cube is:
>
> Vol = (lambda^2 / 4pi) (lambda / 2pi)
>
> The magnetic field is:
>
> H = (2pi e c) / lambda^2 Ampere per meter
>
> The well known reactive impedance of space is:
>
> Z = (Uo/Eo)^0.5 = 376.730313496243 Ohms
>
> The electric field is:
>
> E = H Z Volts per meter
>
> The energy density of the cube becomes:
>
> Jeh = 0.5 (Eo (E^2) + Uo (H^2)) Joule per meter cubed
>
> We now can solve for the rest mass (related to lambda) knowing that the above
> energy density times the cube volume is the energy of the EM structure, and
> that the rest mass energy is obtained by dividing by the fine structure
> constant (a) as follows.
>
> Jm = (Jeh Vol) / a
>
> Then the quantum is directly obtainable from the spin angular momentum (mass x
> radius arm x velocity of gyration) This momentum is quantized by the geometry
> of the spinning cube of EM energy. So:
>
> h/4pi = (Jm / c^2) x (lambda / 4pi) x (c) which reduces to:
>
> h = (Jm / c^2) x (lambda) x (c) which is the well known Compton relation:
Hey are you an idiot?
If you substitute in "your" definition for Jm into "your" h equation above you will
see with the application of simple (elementary) algebra that the lambda's cancel out
and of course since the lambda's cancel out "your" h equation above is not dependent
lambda so it won't matter what value you plug in for lambda.
Here i will give you an example: Follow this calculation on a piece of paper and I am
sure you will see.
Use these variables:
h = 6.626075470853 x 10^-34 m^2 kg s^-1 (Planck constant)
c = 2.99792458 x 10^8 m s^-1 (Velocity of light)
a = 7.297353080929 x 10^-3 (Fine structure constant)
Eo = 8.854187817 x 10^-12 m^-3 kg^-1 s^4 A^4 (Permittivity of the vacuum)
Uo = 12.566370614 x 10^-7 m kg s^-2 A^-2 (Permeability of the vacuum)
lambda = 2.42631057x 10^12 m (Electron Compton for the electron model)
pi = 3.1415925
The general equations are:
Js = (h c a ) / lambda = Joule cube structure energy.
Jm = (h c ) / Number_Of_Angles_On_A_Pin_Head = Joule cube rest mass energy
Vol = ((lambda^2) / 4pi )) x (lambda / 2pi) = Volume of the spinning cube
(cylinder )
Pd = ( Js c ) / Vol = Power density, Watts per square meter.
Z = ( Uo / Eo )^0.5 = Impedance of space.
Dd = (Eo Ed ) = Charge density, charge per meter squared
e = ((Dd (lambda^2) / 2pi ) = fundamental charge = 1.60217733 x 10^-19 (CHECK)
½ h bar = ((Jm) / c^2) x (c) x (( / 4pi) x (c) = 5.2728633456 x 10^-35
J-s (CHECK)
I tested this with a Number_Of_Angles_On_A_Pin_Head of 100000000 and still got the correct (h).
h = 6.62607549818106 x 10^-34 kg m^2 s^-1 (Joule seconds)
Now see If you increase the Number_Of_Angles_On_A_Pin_Head to infinity
it won't matter because the DAMN EQUATION YOU MADE DOESN'T DEPEND ON
THE DAMN LAMBDA ANYWAYS - I hope that you can see all I did was replace
lambda with Number_Of_Angles_On_A_Pin_Head to illustrate to you that
(A/B) * (B/D) = A/D and notice the B's cancel simple algebra so the ratio
of A/D does NOT depend on B what so ever.
As for your equations you just fabricated those equations above from the
definitions established 50 years ago - I would hate to BURST YOUR BUBBLE
but you need to maybe get a tutor to understand even basic physics
especially basic ElectroMagnetism.
Your equations are Mumbo Jumbo because you are using other equations which
serve as definitions and just rearranging them to look like they are related
to you spinning cube model.
Sorry to be so harsh - but sheesh are you slow or something?
Chris
Matthew Nobes
On Thu, 8 Jul 1999, Jim Heckman wrote:
[snip yet another trashing of VPP, fun for one, fun for all]
> IMHO, the only rational response to his posts (i.e., hawkings of his
> book) is a standard, short reply summarizing the many, many
> experimental and theoretical deficiencies of the VPP "model", to give
> fair warning to newbies who might otherwise actually send him money.
> (Hmm... OTOH, such a standard, summary reply might not be able to be
> "short" after all :-) -- LOL again!)
Well, I was going to write a review, but it doesn't appear that he
want's to send his book after all. Nonetheless, I am preparing a short
postscript file, with two proofs of both energy conservation, and Lorentz
invariance in classical E&M. I'll probably throw in the incorrect
predictions, and the "lambda independant" "derivation" of h (which is
acutally independant of h) that he's been spouting off about recently.
Then I'll just post a standard reply, with a link to the PS file, every
time he hawks his book.
Unfortunatly, this sort of thing takes time, and I find it
difficult to get motivated :( But watch for it soon... The energy
conservation proofs are done, and the Lorentz invariance ones won't take
as long (I spend a good fifteen minutes a day on this, so it does take
time).
>
-------------------------------------------------------------------------------
"physical discoverers have differed from barren |Matthew Nobes
speculators, not by having no metaphysics in their |c/o Theory Group
heads, but by having good metaphysics while their |TRIUMF
adversaries had bad; and by binding their metaphysics |4004 Wesbrook Mall
to their physics, instead of keeping the two asunder" |Vancouver, B.C.
William Whewell|Canada, V6T 2A3
|
www.geocities.com/CollegePark/campus/1098 |
Chris Gomez
Hey TomL are you an idiot?
If you substitute in "your" definition for Jm into "your" h equation above you will
see with the application of simple (elementary) algebra that the lambda's cancel out
and of course since the lambda's cancel out,"your" h equation above is not dependent
on lambda so it won't matter what value you plug in for lambda, they just cancel each other.
Here I will give you an example: Follow this calculation on a piece of paper and I am
sure you will see.
Use these variables:
h = 6.626075470853 x 10^-34 m^2 kg s^-1 (Planck constant)
c = 2.99792458 x 10^8 m s^-1 (Velocity of light)
a = 7.297353080929 x 10^-3 (Fine structure constant)
Eo = 8.854187817 x 10^-12 m^-3 kg^-1 s^4 A^4 (Permittivity of the vacuum)
Uo = 12.566370614 x 10^-7 m kg s^-2 A^-2 (Permeability of the vacuum)
lambda = 2.42631057x 10^12 m (Electron Compton for the electron model)
pi = 3.1415925
The general equations are:
Js = (h c a ) / lambda = Joule cube structure energy.
Jm = (h c ) / Number_Of_Angles_On_A_Pin_Head = Joule cube rest mass energy
Vol = ((lambda^2) / 4pi )) x (lambda / 2pi) = Volume of the spinning cube
(cylinder )
Pd = ( Js c ) / Vol = Power density, Watts per square meter.
Z = ( Uo / Eo )^0.5 = Impedance of space.
Dd = (Eo Ed ) = Charge density, charge per meter squared
e = ((Dd (lambda^2) / 2pi ) = fundamental charge = 1.60217733 x 10^-19 (CHECK)
½ h bar = ((Jm) / c^2) x (c) x (( / 4pi) x (c) = 5.2728633456 x 10^-35
J-s (CHECK)
I tested this with a Number_Of_Angles_On_A_Pin_Head of 100000000 and still got the correct (h).
h = 6.62607549818106 x 10^-34 kg m^2 s^-1 (Joule seconds)
Now see If you increase the Number_Of_Angles_On_A_Pin_Head to infinity
it won't matter because the DAMN EQUATION YOU MADE DOESN'T DEPEND ON
THE DAMN LAMBDA ANYWAYS - I hope that you can see all I did was replace your
"lambda" with Number_Of_Angles_On_A_Pin_Head to illustrate to you the elementary
algebraic principle demonstrated below:
(A/B) * (B/D) = A/D and notice the B's cancel (from simple algebra) so the ratio
of A/D does NOT depend on B what so ever.
As for your equations you just fabricated those equations above from the
definitions established 50 years ago - I would hate to BURST YOUR BUBBLE
but you need to get a tutor and try to understand basic physics before you claim
to become a Theoretical Physicist.
Jim Heckman
dis...@golem.ph.utexas.edu (Jacques Distler) wrote:
>
> Your own Compton wavelength is smaller than your Schwarzschild
> radius. If your Compton wavelength were a measure of your size, you
> would be a black hole.
>
> (Some might find that desirable, but it is nevertheless untrue.)
ROTFLMAO!
Hmm... Perhaps the fact that we're all black holes will be the next
stunning prediction of the VPP "model". ;-)
Honestly, I don't know why anyone even continues to hold
conversations with Mr. Lockyer, whose theories about (meta)physics
are of about as much practical use as those of Aristotle. (What? You
want me to actually drop two rocks of different weights and measure
how long they take to fall? Why would I do that? *Obviously* heavier
objects fall faster: That's what makes them heavier!)
Mr. Lockyer's "model" is inconsistent with: (1) Virtually all experimental
results; (1) Classical EM; (2) SR; (3) QM; ... (and for all I know,
Newtonian mechanics). In addition, he has demonstrated over and over
that he has no understanding of, and no *wish* to understand, any of
these (experimentally highly successful) physical theories -- nor to
reconcile the predictions of his "model" with actual experiment.
IMHO, the only rational response to his posts (i.e., hawkings of his
book) is a standard, short reply summarizing the many, many
experimental and theoretical deficiencies of the VPP "model", to give
fair warning to newbies who might otherwise actually send him money.
(Hmm... OTOH, such a standard, summary reply might not be able to be
"short" after all :-) -- LOL again!)
> JD
>
> PGP public key: http://golem.ph.utexas.edu/~distler/distler.asc
--
~~ Jim Heckman ~~
-- "As I understand it, your actions have ensured that you will never
see Daniel again." -- Larissa, a witch-woman of the Lowlands.
-- "*Everything* is mutable." -- Destruction of the Endless
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
Matthew Nobes
On Tue, 6 Jul 1999, Christopher Gomez wrote:
[snip more VPP crap from Tom]
>
> Hey are you an idiot?
He thinks calssical electromagnetism doesn't conserve energy and
is not Lorentz invariant, despite the fact that both facts have been
demonstrated to him numerous times. So yes he is an idiot.
> If you substitute in "your" definition for Jm into "your" h equation
> above you will see with the application of simple (elementary) algebra
> that the lambda's cancel out and of course since the lambda's cancel out
> "your" h equation above is not dependent lambda so it won't matter what
> value you plug in for lambda.
Furthermore it is not even a defintion for h, since he has
something that goes like h=C/a where C is a constant and a is the fine
structure constant. But a^{-1} is proportional to h so the h's cancel out
and he's left with C'=1 where C' is another constant. I think he has
succesfully reproduced c^{2}=1/(epsilon0 mu0), but that may be giving him
too much credit.
> Here i will give you an example: Follow this calculation on a piece of
> paper and I am sure you will see.
No he won't. He'll respond with his usual nonsense about how the
geometrical structure is what is importent. Furthermore, it seems that he
is incabable of doing these calculations by hand, he has to use Mathcad 8!
[snip yet another trashing of the VPP "derivation" of h]
> I tested this with a Number_Of_Angles_On_A_Pin_Head of 100000000 and
> still got the correct (h).
>
> h = 6.62607549818106 x 10^-34 kg m^2 s^-1 (Joule seconds)
>
> Now see If you increase the Number_Of_Angles_On_A_Pin_Head to infinity
> it won't matter because the DAMN EQUATION YOU MADE DOESN'T DEPEND ON
> THE DAMN LAMBDA ANYWAYS - I hope that you can see all I did was replace
> lambda with Number_Of_Angles_On_A_Pin_Head to illustrate to you that
ROTFLMAO, that's good.
[snip more]
> Sorry to be so harsh - but sheesh are you slow or something?
He's a crank who won't let go of his precious ideas, despite the
fact that they've been shown to be theoretically wrong, and contradictory
to experiment (his "radius" for a spinning electro cube is 4 orders of
magnitude above the current (LEP) upper limit). Furthermore he predicts
that a free neutrino is a boson (now that's funny).
Jacques Distler
> Furthermore it is not even a defintion for h, since he has
>something that goes like h=C/a where C is a constant and a is the fine
>structure constant. But a^{-1} is proportional to h so the h's cancel out
>and he's left with C'=1 where C' is another constant. I think he has
>succesfully reproduced c^{2}=1/(epsilon0 mu0), but that may be giving him
>too much credit.
To be fair to TL, I think what he has is an expression of the form C/a, where
a=C/h. This indeed does give h (even without the aid of Mathcad 8), but it
does so in a fashion that follows from the DEFINITION of the quantity "a".
Specifically, he had
h = m c lambda
where lambda is the Compton wavelength. Yet another brilliant identity which
follows from the definition of lambda.
I expect him to follow up with a derivation of the fine structure constant.
But he might surprise us and instead, as Jim Heckman suggests, prove that we
are all black holes.
JD
--
Jim Carr
tnlo...@aol.com (Tnlockyer) wrote:
}
} >I LOVE it when you talk out of both sides of your mouth.
} >
} >On the one hand, your tout the astounding accuracy with which your model
} >(which uses the Compton wavelength as input) reproduces particle masses.
} >
}
} >When it is pointed out to you that this follows trivially from the
} >definition of the Compton wavelength (and hence is no prediction at all),
}
} h / me c about 50 years ago.
In article <distler-0507...@192.168.0.1>
dis...@golem.ph.utexas.edu (Jacques Distler) writes:
>
>It wasn't "deduced"; that was how the Compton wavelength was DEFINED.
Been down that road before; hope you enjoy the journey. ;-)
I'll just add that the "Compton wavelength" defines how the
wavelength associated with a momentum changes due to recoil
of a particle with mass m as a result of elastic scattering.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Jacques Distler
wrote:
>>It wasn't "deduced"; that was how the Compton wavelength was DEFINED.
>
> Been down that road before; hope you enjoy the journey. ;-)
>
> I'll just add that the "Compton wavelength" defines how the
> wavelength associated with a momentum changes due to recoil
> of a particle with mass m as a result of elastic scattering.
More succinctly, it is the wavelength of a photon whose energy is equal to
the rest-energy of the particle in question. This has, as I (vainly)
emphasized, NOTHING to do with the "size" of the particle in question.