The source of fundamental spin and charge.
ThomasL283
electron cube model has an edge length of (lambda) / 2pi where lambda is the
electrons published Compton wavelength. The cube is spinning because the
momentum of the trapped photons adds in front and back cube faces. The volume
of this spinning cube is a cylinder:
Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
It is well known that the electrical potential energy (Js) in the electron is
equal to the electron's rest mass energy times the fine structure constant (a).
Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
The spinning cube sweeps out a volume of space that allows the power density to
be calculated. Since the (E) and (H) fields have identical energy, we use the
factor of two. The power density (Pe) of the spinning cube becomes:
Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per cubic meter.
The impedance of space is the ratio between the electric (E) and magnetic (H)
field energy. Z=376.730313474969 kg m^2 s^-3 A^-2 Ohms, and allows calculating
the (E) and (H) value. Either (E) or (H) can be used to derive the fundamental
charge, so we will use just the (E in Volt per meter) rather than the (H in
Ampere per meter) as follows:
E= (Pe Z)^0.5 = 3.862591921 x 10^16 kg m s^-3 A^-1 Volt per meter.
The charge density (D) then becomes, where Eo is the permittivity of the
vacuum:
D= Eo x E = 3.42001143 x 10^5 s A m^2 Note: (D) has the dimensions of a
charge (s A) divided by the current loop areas (m^2). The spinning cubes two
current loop areas (L2) are:
L2 = (lambda squared) / 4 pi =4.68471083x10^-25 m^2
It follows that the fundamental charge (e) is simply:
e = D L2 = 1.602176462 s A which is exactly the published value.
The (Spin) angular momentum of the spinning VPP electron cube model is:
Spin = (Js lambda) / (4pi c a) = h/4pi exactly.
The above calculations are based on the 1998 CODATA adjustment of the
constants. Had we used the 1973 CODATA published adjustment, the equations
would have returned the 1973 value for (e) and (h/4pi). Had we used the 1986
adjustment, the eqautions would have returned the 1986 value for (e).
The equations, above, are general for any sized cube of spinning EM structure.
If you put these equations into your favorite math soft ware (I now use the
Mathcad 2001 professional) Try changing lambda to any value what-so-ever and
the equations still result in exactly (e) and (h/4pi), proving sub atomic
particles form factors are all spinning cubes of EM energy. This demonstrated
independence of size, explains the mystery of why the (e) and (h/4pi) are
ubiquitous in all sizes of sub atomic particles, and prove the VPP models are
correct.
These equations prove that the free neutrino cannot have a spin of (h/4pi) or
it would have mass (energy) and a fundamental charge, it does not.
The VPP model shows that the neutrino only spins when in concert with the
electron (or positron) and it is by this mechanism that the neutrino like
structure stores 782.354 keV in the neutron's structure, using the scaling
given in the VPP proton model. The model gives the mass of the proton to
within 3ppm and the mass of the neutron to within 40ppb of the CODATA, to prove
the concepts.
Compare these VPP part per million (ppm) results with the 30 year lack of
results by the quark model for the proton and neutron, and you will throw rocks
at the SM.
Regards: Tom:
Tom Lockyer (75 and retired) See "Vector Partcles 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)
John C. Polasek
>The particles are modeled as spinning cubes of EM field energy. The VPP
>electron cube model has an edge length of (lambda) / 2pi where lambda is the
>electrons published Compton wavelength. The cube is spinning because the
>momentum of the trapped photons adds in front and back cube faces. The volume
>of this spinning cube is a cylinder:
>Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
How'd you get that? It's the cube times pi/2. But, OK
>It is well known that the electrical potential energy (Js) in the electron is
>equal to the electron's rest mass energy times the fine structure constant (a).
>
>Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
>
>The spinning cube sweeps out a volume of space that allows the power density to
>be calculated. Since the (E) and (H) fields have identical energy, we use the
>factor of two. The power density (Pe) of the spinning cube becomes:
>
>Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per cubic meter.
This needs work. Your units here are watt/square meter, though the
numbers are right.
>The impedance of space is the ratio between the electric (E) and magnetic (H)
>field energy. Z=376.730313474969 kg m^2 s^-3 A^-2 Ohms, and allows calculating
>the (E) and (H) value. Either (E) or (H) can be used to derive the fundamental
>charge, so we will use just the (E in Volt per meter) rather than the (H in
>Ampere per meter) as follows:
>
>E= (Pe Z)^0.5 = 3.862591921 x 10^16 kg m s^-3 A^-1 Volt per meter.
Using Pe watt/m^2 you get E as volts x sqrt(meters).
Tom, you probably mistyped somewhere.
John C. Polasek
ThomasL283
>Date: 11/8/2001 6:07 PM Pacific Standard Time
>Message-id: <3beb397b.43979467@news-server>
>On 08 Nov 2001 22:35:08 GMT, thoma...@aol.com (ThomasL283) wrote:
>
>>The particles are modeled as spinning cubes of EM field energy. The VPP
>>electron cube model has an edge length of (lambda) / 2pi where lambda is the
>>electrons published Compton
>wavelength. The cube is spinning because the
>>momentum of the trapped photons adds in front and back cube faces. The
>volume
>>of this spinning cube is a cylinder:
>>Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
>How'd you get that? It's the cube times pi/2. But, OK
No check me if this is wrong:
The cube has an edge length of lambda / 2pi. So the radius, from the center of
a cube face, to the corner of a cube face, is (sqr2) (lambda / 4pi). The area
swept out by a cube face is then:
(((sqr2 lambda) / 4pi))^2) x pi = ((2 lambda ^2) / 16 pi^2) x pi = ((2
lambda^2)/16 pi)
Since the heigth of the cube (lambda / 2pi) is also the heigth of the spinning
cylinder, so then the volume swept out is just the product of the area of one
face times the heigth of (lambda/2pi)
((2lambda^2)/16Pi)) (lambda / 2pi) = or:
(lambda cubed) / 16 pi^2 as stated.
>snip<>>
>>Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per cubic meter.
>This needs work. Your units here are watt/square meter, though the
>numbers are right.
Yes, thanks John, it should be Watt per sqr meter.
V/m x A/m = VA/ m^2, how dumb of me.
The claims for all results are correct. Any sized cube (even as big as the
solar system) describes a spinning volume that always gives exactly (e) and
(h/4pi).
This model proves that the fractional charges postulated for the quarks are
"not even wrong". And this also makes the W and Z electroweak theory "not even
wrong" as well.
John C. Polasek
In place of 2 Js c you need to divide by a radius to get angular
velocity, as c/edge*.5. (After which the development may become less
interesting).
>The claims for all results are correct. Any sized cube (even as big as the
>solar system) describes a spinning volume that always gives exactly (e) and
>(h/4pi).
>
>This model proves that the fractional charges postulated for the quarks are
>"not even wrong". And this also makes the W and Z electroweak theory "not even
>wrong" as well.
>
>Regards: Tom:
>Tom Lockyer (75 and retired) See "Vector Partcles 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)
John C. Polasek
ThomasL283
Wrote in:
>Message-id: <3bec7fa9.40744091@news-server>
>On 09 Nov 2001 22:30:18 GMT, thoma...@aol.com (ThomasL283) wrote:
>snip<
>>>>Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per cubic meter.
>>
>>>This needs work. Your units here are watt/square meter, though the
>>>numbers are right.
>>
>>Yes, thanks John, it should be Watt per sqr meter.
>In place of 2 Js c you need to divide by a radius to get angular
>velocity, as c/edge*.5. (After which the development may become less
>interesting).
Yes, John. Note that he spinning cube does derive the spin angular momentum (
and the quantum (h) by extension, from the geometry) In fact this is the math
used to calculate the spin of ANY SIZED cube of spinning EM energy; See follow
on article to this post:
>>The claims for all results are correct. Any sized cube (even as big as the
>>solar system) describes a spinning volume that always gives exactly (e) and
>>(h/4pi).
>>This model proves that the fractional charges postulated for the quarks are
>>"not even wrong". And this also makes the W and Z electroweak theory "not
>even
>>wrong" as well.
Regards: Tom:
P.S. I repeated the original math and corrected a typo in the charge density
exponent, along with examples of results using a very large cube, and a very
small cube. In a new article that follows:
ThomasL283
electron cube model has an edge length of (lambda) / 2pi where lambda is the
electrons published Compton wavelength. The cube is spinning because the
momentum of the trapped photons adds in front and back cube faces. The volume
of this spinning cube is a cylinder:
Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
It is well known that the electrical potential energy (Js) in the electron is
equal to the electron's rest mass energy times the fine structure constant (a).
Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
The spinning cube sweeps out a volume of space that allows the power density to
be calculated. Since the (E) and (H) fields have identical energy, we use the
factor of two. The power density (Pe) of the spinning cube becomes:
Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per square meter.
The impedance of space is the ratio between the electric (E) and magnetic (H)
field energy. Z=376.730313474969 kg m^2 s^-3 A^-2 Ohms, and allows calculating
the (E) and (H) value. Either (E) or (H) can be used to derive the fundamental
charge, so we will use just the (E in Volt per meter) rather than the (H in
Ampere per meter) as follows:
E= (Pe Z)^0.5 = 3.862591921 x 10^16 kg m s^-3 A^-1 Volt per meter.
The charge density (D) then becomes, where Eo is the permittivity of the
vacuum:
D= Eo x E = 3.42001143 x 10^5 s A m^-2 Note: (D) has the dimensions of a
charge (s A) divided by the current loop areas (m^2). The spinning cubes two
current loop areas (L2) are:
L2 = (lambda squared) / 4 pi =4.68471083x10^-25 m^2
It follows that the fundamental charge (e) is simply:
e = D L2 = 1.602176462 s A which is exactly the published value.
The (Spin) angular momentum of the spinning VPP electron cube model is:
Spin = (Js lambda) / (4pi c a) = h/4pi exactly.
The equations, above, are general for any sized cube of spinning EM structure.
If you put these equations into your favorite math soft ware (I now use the
Mathcad 2001 professional) Try changing lambda to any value what-so-ever and
the equations still result in exactly (e) and (h/4pi), proving sub atomic
particles form factors are all spinning cubes of EM energy. This demonstrated
independence of size, explains the mystery of why the (e) and (h/4pi) are
ubiquitous in all sizes of sub atomic particles, and prove the VPP models are
correct.
Just for demonstration purposes, here are results from a very large to a very
small spinning cube of EM energy:
First very large:
Lambda= 1.9x10^30 m (meters).
Volume= 4.34351249126747 x 10^88 m^3
Power density=1.05316881643797 x 10^-137 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
Then I tried very small spinning cube of EM energy:
Lambda= 1.9 x 10^-30 m (meters)
Volume= 4.34351249126747 x 10^-92 m^3
Power density= 1.05316881643797 x 10^103 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
So, as I repeat, the math proves that particles that have a spin of (h/4pi)
MUST also have a fundamental charge of exactly (e).
Regards: Tom:
Jim Carr
>
>The particles are modeled as spinning cubes of EM field energy.
And the flaws in that model were pointed out to you so long
ago that they probably are not even archived on Google. Note
that, rather than justifying your continued posting of nonsense,
this shows that you should have known better by now.
>The VPP
>electron cube model has an edge length of (lambda) / 2pi where lambda is the
>electrons published Compton wavelength.
This is falsified by experiments done decades ago.
--
James Carr <j...@scri.fsu.edu> http://www.scri.fsu.edu/~jac/
SirCam Warning: read http://www.cert.org/advisories/CA-2001-22.html
e-mail info: new...@fbi.gov pyr...@ftc.gov enfor...@sec.gov
ThomasL283
Wrote in:
>Message-id: <9tesam$ovp$1...@news.fsu.edu>
>In article <20011108173508...@mb-fw.aol.com>
>thoma...@aol.com (ThomasL283) writes:
>>
>>The particles are modeled as spinning cubes of EM field energy.
> <... snip ...>
>
> And the flaws in that model were pointed out to you so long
> ago that they probably are not even archived on Google.
Nope, make no mistake about it. I have the correct model for the structure of
energy and matter.
Put these equations into your favorite math soft ware (I now use the Mathcad
2001 Professional)
Set your software to simultaneously do dimensional and numeric calculations.
Get the CODATA values for the constants and set them up in your variables list.
And then, for gods sake swallow your pride and do the math, or shut up.
The proof is the geometry of the spinning EM cube, regardless of it's size,
always gives the fundamental charge (e) and spin (1/2hbar) EXACTLY.
To wit:
The particles are modeled as spinning cubes of EM field energy.
For one example the VPP electron cube model has an edge length of (lambda) /
2pi where lambda is the electrons published Compton wavelength.
The cube is spinning because the momentum of the trapped photons adds in front
and back cube faces. No other model shows a mechanism for spin.
The volume of this spinning cube is a cylinder:
Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
It is well known that the electrical potential energy (Js) in the electron is
equal to the electron's rest mass energy times the fine structure constant (a).
Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
The spinning cube sweeps out a volume of space that allows the power density to
be calculated.
Since the (E) and (H) fields have identical energy, we use the factor of two.
The power density (Pe) of the spinning cube becomes:
Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per square meter.
The impedance of space (Z) is the ratio between the electric (E) and magnetic
(H) field energy.
Z=376.730313474969 kg m^2 s^-3 A^-2 Ohms, and allows calculating the (E) and
(H) value.
Either (E) or (H) can be used to derive the fundamental charge, so we will use
just the (E in Volt per meter) rather than the (H in Ampere per meter) as
follows:
E= (Pe Z)^0.5 = 3.862591921 x 10^16 kg m s^-3 A^-1 Volt per meter.
The charge density (D) then becomes, where Eo is the permittivity of the
vacuum:
D= Eo x E = 3.42001143 x 10^5 s A m^-2
Note: (D) has the dimensions of a charge (s A) divided by the current loop
areas (m^2). The spinning cubes two current loop areas (L2) are:
L2 = (lambda squared) / 4 pi =4.68471083x10^-25 m^2
It follows that the fundamental charge (e) is simply:
e = D L2 = 1.602176462 s A which is exactly the published value.
The (Spin) angular momentum of the spinning VPP electron cube model is:
Spin = (Js lambda) / (4pi c a) = h/4pi exactly.
The equations, above, are general for any sized cube of spinning EM structure.
Try changing lambda to any value what-so-ever and the equations still result in
exactly (e) and (h/4pi), proving sub atomic particles form factors are all
spinning cubes of EM energy.
This demonstrated independence of size, explains the mystery of why the (e)
and (h/4pi) are ubiquitous in all sizes of sub atomic particles, and prove the
VPP models are correct.
Just for demonstration purposes, here are results from a very large to a very
small spinning cube of EM energy:
First for very large cube set
Lambda= 1.9x10^30 m (meters).
Volume= 4.34351249126747 x 10^88 m^3
Power density=1.05316881643797 x 10^-137 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
Then for a very small spinning cube of EM energy, set :
Lambda= 1.9 x 10^-30 m (meters)
Volume= 4.34351249126747 x 10^-92 m^3
Power density= 1.05316881643797 x 10^103 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
>>The VPP
>>electron cube model has an edge length of (lambda) / 2pi where lambda is the
>>electrons published Compton wavelength.
> This is falsified by experiments done decades ago.
Swallow your pride and actually do the math above, then come back and admit the
math works.
The secret is the geometry of a spinning cube maintains the constants in their
proper ratios. This proves that particles model as spinning cubes of electro
(E) and magnetic (H) fields is the correct approach.
Jim Carr
thoma...@aol.com (ThomasL283) writes:
>
... in response to article <20011110134554...@mb-cj.aol.com>
written by "ThomasL283 (thoma...@aol.com)" ...
>
You are talking to yourself, Thomas, and ignoring the fact that
your model was junked a long time ago.
<... snip repost of same old junk from years ago ...>
Out of curiousity, when did you first post this, Tom?
ThomasL283
In article <20011110135908...@mb-cj.aol.com>
thoma...@aol.com (ThomasL283) writes:
>
... in response to article <20011110134554...@mb-cj.aol.com>
written by "ThomasL283 (thoma...@aol.com)" ...
>
> You are talking to yourself, Thomas, and ignoring the fact that
>your model was junked a long time ago.
<... snip repost of same old junk from years ago ...>
>Out of curiousity, when did you first post this, Tom?
Jim, the math is new. The model was not junked. A group of your cohorts
ganged up and cluttered the newsgroup with their own brand of snake oil. They
did not want or even try to see the my POV. They argued, much like you, on the
basis that they had the correct ideas, and that I was some sort of fringe
player.
The fact remains that the VPP models are mathematically consistent, and this
can not be said of the QCD and QED junk you teach.
That being said, this is a public forum and there may be others that will take
the time to do the math I asked you to do.
So here it is again, and this time don't weasel out of doing the proofs.
To wit:
The particles are modeled as spinning cubes of EM field energy. The VPP
electron cube model has an edge length of (lambda) / 2pi where lambda is the
electrons published Compton wavelength. The cube is spinning because the
momentum of the trapped photons adds in front and back cube faces. The volume
of this spinning cube is a cylinder:
Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
It is well known that the electrical potential energy (Js) in the electron is
equal to the electron's rest mass energy times the fine structure constant (a).
Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
The spinning cube sweeps out a volume of space that allows the power density to
be calculated. Since the (E) and (H) fields have identical energy, we use the
factor of two. The power density (Pe) of the spinning cube becomes:
Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per square meter.
The impedance of space is the ratio between the electric (E) and magnetic (H)
field energy. Z=376.730313474969 kg m^2 s^-3 A^-2 Ohms, and allows calculating
the (E) and (H) value. Either (E) or (H) can be used to derive the fundamental
charge, so we will use just the (E in Volt per meter) rather than the (H in
Ampere per meter) as follows:
E= (Pe Z)^0.5 = 3.862591921 x 10^16 kg m s^-3 A^-1 Volt per meter.
The charge density (D) then becomes, where Eo is the permittivity of the
vacuum:
D= Eo x E = 3.42001143 x 10^5 s A m^-2 Note: (D) has the dimensions of a
charge (s A) divided by the current loop areas (m^2). The spinning cubes two
current loop areas (L2) are:
L2 = (lambda squared) / 4 pi =4.68471083x10^-25 m^2
It follows that the fundamental charge (e) is simply:
e = D L2 = 1.602176462 s A which is exactly the published value.
The (Spin) angular momentum of the spinning VPP electron cube model is:
Spin = (Js lambda) / (4pi c a) = h/4pi exactly.
The equations, above, are general for any sized cube of spinning EM structure.
If you put these equations into your favorite math soft ware (I now use the
Mathcad 2001 professional) Try changing lambda to any value what-so-ever and
the equations still result in exactly (e) and (h/4pi), proving sub atomic
particles form factors are all spinning cubes of EM energy. This demonstrated
independence of size, explains the mystery of why the (e) and (h/4pi) are
ubiquitous in all sizes of sub atomic particles, and prove the VPP models are
correct.
Just for demonstration purposes, here are results from a very large to a very
small spinning cube of EM energy:
First very large:
Lambda= 1.9x10^30 m (meters).
Volume= 4.34351249126747 x 10^88 m^3
Power density=1.05316881643797 x 10^-137 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
Then I tried very small spinning cube of EM energy:
Lambda= 1.9 x 10^-30 m (meters)
Volume= 4.34351249126747 x 10^-92 m^3
Power density= 1.05316881643797 x 10^103 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
Regards: Tom:
Daniel McCarty
>From: thoma...@aol.com (ThomasL283)
>Date: 24-Nov-01 12:45 PM Central Standard Time
>Message-id: <20011124134556...@mb-fv.aol.com>
>
>j...@dirac.csit.fsu.edu (Jim Carr)
>Wrote in:
>Message-id: <9tn568$lg9$1...@news.fsu.edu>
>
>In article <20011110135908...@mb-cj.aol.com>
>thoma...@aol.com (ThomasL283) writes:
>>
> ... in response to article <20011110134554...@mb-cj.aol.com>
> written by "ThomasL283 (thoma...@aol.com)" ...
>>
>
>> You are talking to yourself, Thomas, and ignoring the fact that
> >your model was junked a long time ago.
>
> <... snip repost of same old junk from years ago ...>
>
> >Out of curiousity, when did you first post this, Tom?
>
>Jim, the math is new. The model was not junked. A group of your cohorts
>ganged up and cluttered the newsgroup with their own brand of snake oil.
>They
>did not want or even try to see the my POV. They argued, much like you, on
>the
>basis that they had the correct ideas, and that I was some sort of fringe
>player.
>
>The fact remains that the VPP models are mathematically consistent, and this
>can not be said of the QCD and QED junk you teach.
>
>That being said, this is a public forum and there may be others that will
>take
>the time to do the math I asked you to do.
>
>So here it is again, and this time don't weasel out of doing the proofs.
>
the charge value back out by combining those other values in the traditional
way. You proved nothing.
ThomasL283
>Date: 11/24/2001 2:50 PM Pacific Standard Time
>Message-id: <20011124175028...@mb-dd.aol.com>
>>Subject: Re: The source of fundamental spin and charge.
>>From: thoma...@aol.com (ThomasL283)
>>Date: 24-Nov-01 12:45 PM Central Standard Time
>>Message-id: <20011124134556...@mb-fv.aol.com>
>>
>>That being said, this is a public forum and there may be others that will
>>take
>>the time to do the math I asked you to do.
>>
>>So here it is again, and this time don't weasel out of doing the proofs.
>>
> You put the charge in as combinations of other standard values, then drew
>the charge value back out by combining those other values in the traditional
>way. You proved nothing.
>
You really don't see what has been done.
While it is true that the fundamental constants are all related to one another,
the secret to the equations working is the UNIQUE geometry of the spinning cube
of EM energy, maintaining the constants in their proper ratios.
Notice the two examples that are with very large and very small cubes of EM
energy.
*****************
First very large:
Lambda= 1.9x10^30 m (meters).
Gives a large
Volume= 4.34351249126747 x 10^88 m^3
Gives a small
Power density=1.05316881643797 x 10^-137 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
*******************
Then very small spinning cube of EM energy:
Lambda= 1.9 x 10^-30 m (meters)
Gives a small
Volume= 4.34351249126747 x 10^-92 m^3
And a large
Power density= 1.05316881643797 x 10^103 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
******************
The geometry of the spinning EM cube describes the volume as a cylinder.
The volume is used to solve for the electric (E) and magnetic (H) field energy
intrinsic in the spinning cube volume of the EM energy.
The two areas (L2) swept out by the cube faces give the required current loop
areas, to obtain the charge, from
e= Eo x E x L2
Where:
E=v/m electric field from the cube.
L2= two current loops from the cube.
Note: the only constant used is:
Eo= permittivity of the vacuum
Everything else is derived from the spinning cube of EM energy, and proves the
point.
Here is the math for charge and spin again, for those lurkers who may have
missed the equations.
Regards: Tom:
************************************
To wit:
The particles are modeled as spinning cubes of EM field energy. The VPP
electron cube model has an edge length of (lambda) / 2pi where lambda is the
electrons published Compton wavelength. The cube is spinning because the
momentum of the trapped photons adds in front and back cube faces. The volume
of this spinning cube is a cylinder:
Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
It is well known that the electrical potential energy (Js) in the electron is
equal to the electron's rest mass energy times the fine structure constant (a).
Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
The spinning cube sweeps out a volume of space that allows the power density to
be calculated.
Since the (E) and (H) fields have identical energy, we use the factor of two.
The power density (Pe) of the spinning cube becomes:
Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per square meter.
The impedance of space (Z) is the ratio between the electric (E) and magnetic
(H) field energy.
ThomasL283
I hope you realize that my examples of very large and very small spinning cubes
of EM energy are cases of writing checks that nature can't cash. (Writing
checks nature can't cash is something the SM garbage theories do without
apology or a second thought.)
I find that nature can only form the (lowest energy) EM cubes as those of the
electron and positron, and that the smallest possible (largest energy) EM cube
particles are the core particles of the proton and neutron. These, limits on
size, establish the boundary conditions that assures the same particles are
always formed here (or on alpha century).
These clear boundary conditions point out the SM fallacy of postulating basic
particles of higher energy, than the proton or neutron. Anything heavier
then the proton or neutron must be a composite, not a basic entity.
BTW, remember that either E (volt per meter) or H (Ampere per meter) can be
used to calculate the fundamental charge from the geometry of any size EM
energy cube what-so-ever? Here is the math comparison:
E= (Pe x Z)^0.5
and H= (Pe/Z)^0.5
Where Pe is the power density, Z=impedance of space.
Then for E, the charge density D=Eo x E
and for H, the charge density D1= H/c
Where: Eo is the permittivity of the vacuum, and c is the velocity of light.
Then from the cube geometry:
e= D x L2 Or e= D1 x L2
Where L2 is the two current loops from the spinning cube geometry, and
D and D1 then identically give us the fundamental charge:
(e=1.602176462 x 10^-19 sA)
When these vector particle physics (VPP) models were developed about 25 years
ago, the assumption was made that the nested cubes of the proton and neutron
model would each have a fundamental charge. This assumption was then supported
by the calculation of the electrical potential energy between nested cubes
adding the needed 5% to the proton mass, to give the mass of the proton model
to within 3ppm of the CODATA.
Now, it has been shown that those old VPP assumptions were correct. One more
example (of many) that the VPP models are mathematically consistent.
These VPP math proofs also point out the standard models use of 1/3e and 2/3e
fractional charges are not mathematically consistent, i.e. the garbage of
using two negative charges, the minus 1/3e (minus) a minus 1e, to produce a
fractional (plus) 2/3e positive charge in electroweak theory.
Those silly garbage ideas, of the old particle physics, cannot account for the
actual numbers 1/3e = 0.5340588 x 10^-19 sA or the actual numbers 2/3e=
1.0681176 x 10^-19 sA )
Where is smoky Joe to counter argue?
Regards: Tom:
Jim Carr
Wrote in <9tesam$ovp$1...@news.fsu.edu>
|
| In article <20011108173508...@mb-fw.aol.com>
| thoma...@aol.com (ThomasL283) writes:
| >
| >The particles are modeled as spinning cubes of EM field energy.
| <... snip ...>
|
| And the flaws in that model were pointed out to you so long
| that, rather than justifying your continued posting of nonsense,
| this shows that you should have known better by now.
|
| >electron cube model has an edge length of (lambda) / 2pi where lambda is the
| >electrons published Compton wavelength.
|
| This is falsified by experiments done decades ago.
In article <20011120235238...@mb-fh.aol.com>
thoma...@aol.com (ThomasL283) writes:
>
>Nope, make no mistake about it.
You are lying about what I have posted in the past, and you know it.
>I have the correct model for the structure of energy and matter.
Then why couldn't you ever respond to my criticisms of it?
Indeed, why do you just restate your model rather than respond
to the comment above?
>Put these equations into your favorite math soft ware (I now use the Mathcad
>2001 Professional)
Your equations do not address the questions I posed for you ages ago,
and in the article you are not responding to, when it was pointed out
that your model
>For one example the VPP electron cube model has an edge length of (lambda) /
>2pi where lambda is the electrons published Compton wavelength.
disagrees with a huge amount of data on electron size and structure
that you never address.
>Swallow your pride and actually do the math above, then come back and admit
>the math works.
You have that backwards. It is you who must show that e-e scattering
agrees with experiment by applying your model to it, given that data
already show the size is much smaller than you claim it is.
ThomasL283
Wrote in:
>Message-id: <9uf12d$sg3$1...@news.fsu.edu>
>
>| In article <20011108173508...@mb-fw.aol.com>
>| thoma...@aol.com (ThomasL283) writes:
>| >
>| >The particles are modeled as spinning cubes of EM field energy.
>| <... snip ...>
>|
>| And the flaws in that model were pointed out to you so long
>| ago that they probably are not even archived on Google. Note
>| that, rather than justifying your continued posting of nonsense,
>| this shows that you should have known better by now.
If you remember, the argument was on the fact that Maxwell's equation did not
conserve the energy when the E and H fields seem to go identically to zero. I
tried to show that Maxwell does not model the photon so cannot explain coherent
versus non coherent radiation.
I tried to get you and others to see that the trig identity (sin^2 + cos^2=1)
does straight line the energy over all time, and was the model for the photon,
to no avail.
We were all still arguing when I went into the hospital in Feb 1998. I was on
life supports and in the hospital for three months (hospital bill was 800
thousand dollars).
So you all thought you had won the argument, because I "went away".
>>Nope, make no mistake about it.
>>I have the correct model for the structure of energy and matter.
>
>
>>Put these equations into your favorite math soft ware (I now use the Mathcad
>>2001 Professional)
> Your equations do not address the questions I posed for you ages ago,
> and in the article you are not responding to, when it was pointed out
> that your model
I can't argue with the deeply ingrained (false) beliefs that the electron is a
point particle. and the positron is a point particle. And the muon is a point
particle that comes apart into three other point particles. (God what crap).
Surely you don't cling to those false ideas?
>>Swallow your pride and actually do the math above, then come back and admit
>>the math works.
> You have that backwards. It is you who must show that e-e scattering
> agrees with experiment by applying your model to it, given that data
> already show the size is much smaller than you claim it is.
Again, you are misinterpreting those scattering experiments. The e-e
scattering is elastic by definition, and it is their Coulombic field
scattering force reaction that is considered as a point in the math, the
electrons don't even touch. I've made this point (sorry for the pun) to you
numerous times.
If you do the math for the fundamental spin and charge you will see that it
PROVES the VPP model is correct.
Here it is again, and for God's sake do the math and then tell me how a WRONG
model can give the correct results.
TO WIT:
The particles are modeled as spinning cubes of EM field energy. The VPP
electron cube model has an edge length of (lambda) / 2pi where lambda is the
electrons published Compton wavelength. The cube is spinning because the
momentum of the trapped photons adds in front and back cube faces. The volume
of this spinning cube is a cylinder:
Volume = (lambda cubed) / 16 pi^2 = 9.0452224x10^-38 m^3
It is well known that the electrical potential energy (Js) in the electron is
equal to the electron's rest mass energy times the fine structure constant (a).
Js= (h c a)/ lambda = 5.97441851 x 10^-16 kg m^2 s^-2 Joule
The spinning cube sweeps out a volume of space that allows the power density to
be calculated. Since the (E) and (H) fields have identical energy, we use the
factor of two. The power density (Pe) of the spinning cube becomes:
Pe= (2 Js c)/ Volume = 3.9602909 x 10^30 kg s^-3 Watt per square meter.
The impedance of space is the ratio between the electric (E) and magnetic (H)
field energy. Z=376.730313474969 kg m^2 s^-3 A^-2 Ohms, and allows calculating
the (E) and (H) value. Either (E) or (H) can be used to derive the fundamental
charge, so we will use just the (E in Volt per meter) rather than the (H in
Ampere per meter) as follows:
E= (Pe Z)^0.5 = 3.862591921 x 10^16 kg m s^-3 A^-1 Volt per meter.
The charge density (D) then becomes, where Eo is the permittivity of the
vacuum:
D= Eo x E = 3.42001143 x 10^5 s A m^-2 Note: (D) has the dimensions of a
charge (s A) divided by the current loop areas (m^2). The spinning cubes two
current loop areas (L2) are:
L2 = (lambda squared) / 4 pi =4.68471083x10^-25 m^2
It follows that the fundamental charge (e) is simply:
e = D L2 = 1.602176462 s A which is exactly the published value.
The (Spin) angular momentum of the spinning VPP electron cube model is:
Spin = (Js lambda) / (4pi c a) = h/4pi exactly.
The equations, above, are general for any sized cube of spinning EM structure.
If you put these equations into your favorite math soft ware (I now use the
Mathcad 2001 professional) Try changing lambda to any value what-so-ever and
the equations still result in exactly (e) and (h/4pi), proving sub atomic
particles form factors are all spinning cubes of EM energy. This demonstrated
independence of size, explains the mystery of why the (e) and (h/4pi) are
ubiquitous in all sizes of sub atomic particles, and prove the VPP models are
correct.
Just for demonstration purposes, here are results from a very large to a very
small spinning cube of EM energy:
First very large:
Lambda= 1.9x10^30 m (meters).
Volume= 4.34351249126747 x 10^88 m^3
Power density=1.05316881643797 x 10^-137 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
Then I tried very small spinning cube of EM energy:
Lambda= 1.9 x 10^-30 m (meters)
Volume= 4.34351249126747 x 10^-92 m^3
Power density= 1.05316881643797 x 10^103 kg s^-3 VA/m^2
Charge (e) = 1.60217646218343 x 10^-19 s A Exactly the fundamental charge.
Spin = 5.27285798210393 x 10^-35 kg m^2 s^-1 Exactly (h/4pi)
Regards: Tom:
Larry Shultis
If I understand your electron model right, your electron would
decrease in
size as it is measured at higher and higher energies since energy
would be
transfered to it by the probing particle increasing its mass. As it
gets smaller it would still have the same charge but a greater mass
due to the spin about a smaller radius. Since your electron gets
smaller at higher energies, it would tend to look like it would limit
to a point as the probing energy increases.
Larry
ThomasL283
>Date: 12/4/2001 10:11 AM Pacific Standard Time
>Message-id: <5c32181d.01120...@posting.google.com>
>snip<>> I can't argue with the deeply ingrained (false) beliefs that the
electron
>is a
>> point particle. and the positron is a point particle. And the muon is a
>point
>> particle that comes apart into three other point particles. (God what crap)
>If I understand your electron model right, your electron would
>decrease in
>size as it is measured at higher and higher energies since energy
>would be
>transfered to it by the probing particle increasing its mass. As it
>gets smaller it would still have the same charge but a greater mass
Larry, I never gave that idea much thought. I suppose that one could consider
that particles react with their environment through their de Broglie
wavelengths (lambda/ mc) where (m) is the relativistic mass.
The main thrust of the model has been to successfully show that the photon
produced cube geometry supports every known particle's fundamental physical
constants. The models find proof by using the proton and neutron models to
model the structures of nuclei,
ThomasL283
>Date: 12/4/2001 10:10 PM Pacific Standard Time
>Message-id: <20011205011053...@mb-fw.aol.com>
Sorry Larry; I botched the de Broglie eqaution.
Should be: (lambda= h/ mv)