Proton and electron charge and spin
ThomasL283
Somewhere in this news feed:
>"Vertner Vergon" <Ver...@prodigy.net> wrote:
.
> The charge on the electron and proton, though opposite, are exactly
>equal despite the huge disparity of mass.
> Can anyone tell me why?
The VPP proton model math shows that the proton has a single core particle
about the same size (0.75 E-15 m) Hofstader measured at Stanford (with
electron scattering experiments conducted in the basement of the physics
building in 1956).
see: Phy. Rev. 102, 851 1956
For the math that proves the VPP model core size is correct; Click on this
link:
http://members.aol.com/tnlockyer/protoncharge.gif
print it out, with a copy of this article, staple them together, and read at
your leisure.
The VPP electron also calculates as having a fundamental charge and spin of 1/2
h bar as proof of concepts. Click on this link.
http://members.aol.com/tnlockyer/electroncharge.gif
Those who have a copy of "Vector Particle and Nuclear Models" ISBN 0963154680
see pages 34, 35, 36 and 57, 58.
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)
Michael Moroney
-Mike
ThomasL283
Somewhere in this news feed:
>>"Vertner Vergon" <Ver...@prodigy.net> wrote in message
>> The charge on the electron and proton, though opposite, are exactly
>>equal despite the huge disparity of mass.
>> Can anyone tell me why?
>The VPP proton model math shows that the proton has a single core particle
about >the same size of (0.75 E-15 m) Hofstader measured at Stanford (with
electron >scattering experiments conducted in the basement of the physics
building in 1956).
>see: Phy. Rev. 102, 851 1956
>For the math that proves the VPP model core size is correct; Click on this
link:
http://members.aol.com/tnlockyer/protoncharge.gif
>print it out, with a copy of this article, staple them together, and read at
your leisure.
>The VPP electron also calculates as having a fundamental charge and spin of
1/2 h >bar as proof of concepts. Click on this link.
http://members.aol.com/tnlockyer/electroncharge.gif
>Those who have a copy of "Vector Particle and Nuclear Models" ISBN >0963154680
see pages 34, 35, 36 and 57, 58.
>Somewhere in this news feed:
>>"Michael Moroney" <mor...@world.std.com> Wrote:
>>Oh boy! More "x=x proofs!
Mike, you still do not understand. I'll repeat it one more time.
"What makes those equations work, for particles of disparate size and mass, is
the volume geometry of the resulting spinning cube of EM energy, and the
resulting attendant size of their two current loop areas, front and back."
These volumes and current loop areas are from the form factor of the VPP
particle models, and prove, not only that particles must have a fundamental
charge,
but that the quark particle model's postulated fractional charges are false.
I effectively answered Vergon's question.
Care to use the SM (you are so proud protective of)
to answer Vergon?
Can't? Did not think so.
–Tom:
Fredi Fizzx
> Oh boy! More "x=x" proofs!
>
> -Mike
Yep, Mike is right. Your Js term has the electron charge in it
already. Look here for the definition of the fine structure constant.
http://www.flashrock.com/upload/alpha.pdf
FrediFizzx
Fredi Fizzx
You might have answered Vergon's question but you did not prove it.
You start with e and spin and get e and spin out. Definitely x=x.
Your Js term has e and spin in it already. Look here then carefully
look at your process.
http://www.flashrock.com/upload/alpha.pdf
Don't you wonder why your values comes out exactly as the Codata
values? It is because you started your proof with the Codata values.
You are going to have to start with something other than e and spin.
Or use a different process.
FrediFizzx
Michael Moroney
>>>"Michael Moroney" <mor...@world.std.com> Wrote:
>>>Oh boy! More "x=x proofs!
>Mike, you still do not understand. I'll repeat it one more time.
It is quite obvious that it is you who doesn't understand. A "proof"
that degenerates into an "x=x" formula, once simplified and identities
(such as the definition of the Bohr Magnetron) substituted is totally
worthless at proving (or disproving) anything, and presenting them as
"proofs" of anything only shows that you haven't worked them out
completely, either due to laziness or deliberate attempts to fool people
here.
>"What makes those equations work, for particles of disparate size and mass, is
<snip>
No, what makes them "work" is they all simplify to "x=x", with all the
other terms cancelling.
>I effectively answered Vergon's question.
No, you attempted to snow him.
>Care to use the SM (you are so proud protective of)
>to answer Vergon?
>Can't? Did not think so.
The standard model doesn't have all the answers in its current form, but
it answers particle physics much better than anything else there.
-Mike
p.s. I could just as easily come up with convoluted "x=x" proofs to
"prove" the Standard Model but I won't bother.
p.p.s. The "quality" of your "proofs" is falling off rapidly. You used
to be able to make me work to see how it simplifies to "x=x", but the
"spin" "proof" in electroncharge.gif is trivially seen as an "x=x" "proof",
I only had to glance at it to see it.
Tnlockyer
Wrote in:
>Message-id: <2d9a226e.02032...@posting.google.com>
>>
>> These volumes and current loop areas are from the form factor of the VPP
>> particle models, and prove, not only that particles must have a fundamental
>> charge,
>> but that the quark particle model's postulated fractional charges are
>false.
>>
>> I effectively answered Vergon's question.
>
>You might have answered Vergon's question but you did not prove it.
>You start with e and spin and get e and spin out. Definitely x=x.
>Your Js term has e and spin in it already. Look here then carefully
>look at your process.
>
>http://www.flashrock.com/upload/alpha.pdf
>
>Don't you wonder why your values comes out exactly as the Codata
>values? It is because you started your proof with the Codata values.
>You are going to have to start with something other than e and spin.
Where do you see (e) in the starting equations? I have to calculate power and
charge density and current loop areas (unique to proton core and to electron
sized particles) to get (e). It is not as trivial as you all think.
>Or use a different process.
Fredi: Please see the links again. The process is correct, and is not a
trivial "e = e" as Mike would have you believe.
http://members.aol.com/tnlockyer/protoncharge.gif
http://members.aol.com/tnlockyer/electroncharge.gif
What you all fail to see is that the power density involves the VOLUME OF THE
SPINNING CUBE. See above links:
Note the first equation for Js is unique to the proton core ( lambda c) and
then the electron (Js) is unique to the (electron's Compton wavelength)
lambda.
and (Js) used to obtain the E field, by the well known ratio between the fine
structure constant and rest mass (energy).
The mass is from the well known realtionship (h c / lambda).
For the proton core (Js) E field energy is:
1.925121 E-12 Joule
For the electron (Js) E field energy is :
5.9744185 E-16 Joule
For the proton spinning cube core particle, then, power density (Pe) is
4.26949 E44 kg s^-3
And, for the electron spinning cube model, power density (Pe) is
3.96029 E30 kg s^-3
This gives the V/m electric field strength "unique to that size spinning cube
shaped particle".
For the proton core, electric field stength is:
4.010548 E23 V/m
For the electron, electric field strength is:
3.86259 E16 V/m
From those unique values, for that size VPP cube particle, one gets the charge
density.
For the proton core it is
3.5510145 E12 (s A m^-2)
For the electron it is
3.4200114 E5 (s A m^-2)
Then with a "unique to that sized cube particle" one can calculate the area of
the two current loop areas, for proton core and for the electron.
The proton core two current loop areas are:
4.5118836 E-32 m^2
The electron two current loop areas are:
4.6847108 E-25 m^2
One then obtains the fundamental charge from the charge density (unique to that
sized cube particle) from the dimensional analysis of
(s A) = (s A m^-2) x (m^2)
Yielding the charge for BOTH the proton core and the electron "spinning cube
particles" as:
1.60217646 E-19 (s A)
Using those disparate and unique values for mass (energy) from the proton
core and the electron,
the VPP model PROVES that any sized spinning CUBE of EM energy will always have
(e) charge and (1/2 hbar) spin angular momentum.
The math ONLY works with various sized spinning
VPP CUBE shaped particles,
so the math is not a trivial "e = e"
This math has directly shown that there CANNOT be fractional charges,
and answers Vergon's "why does the proton and electron both have (e)
fundamental charge, despite disparate masses"
>> >Those who have a copy of "Vector Particle and Nuclear Models" ISBN
>>0963154680
>> see pages 34, 35, 36 and 57, 58.
>>
Regards: Tom:
Thomas Lockyer (75 and retired) See "Vector Particles and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about, and express it in numbers,
ThomasL283
>Date: 3/24/2002 4:44 PM Pacific Standard Time
>Message-id: <GtI7EE...@world.std.com>
>>>>"Michael Moroney" <mor...@world.std.com> Wrote:
>
>>>>Oh boy! More "x=x proofs!
>
>>Mike, you still do not understand. I'll repeat it one more time.
>It is quite obvious that it is you who doesn't understand. A "proof"
>that degenerates into an "x=x" formula, once simplified and identities
Then please explain why the GEOMETRY of the ANY sized spinning cube relates
the fundamental constants in their proper ratios.
Y
ou simply do not want to see the proof as anything but trivial. You are wrong
as you can be. See my answer to Fredi: about the same time you posted.
That was the proof of disparate masses having the same charge (e).
P.S. Don't try to lecture me about the constants, as supervisor of standards, I
attended conferences on the subject, and am very well aware that the constants
are all related by a number of exact mathematical expressions. And have said
so many times on these newsgroups.
In fact, those who have my book, see Chapter 1, page 4. where the various
relations among the constants are shown.
--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,
Jacques Distler
<thoma...@aol.com> wrote:
>I effectively answered Vergon's question.
I love it when the crackpots answer each other's questions.
It's like setting two of those AI programs (like Eliza the psychologist
and Zippy the Pinhead) to talk to each other. Utter gibberish, but
extremely funny to read.
I don't know which is funnier, Verner and Thomas or
<http://cslab.anu.edu.au/~daa/coffee/pin.html>
Anyway, since it is now firmly extablished that Thomas is incapable of
eve simple dimensional analysis, it is no longer tenable to pretend
that his posts constitute "Science".
We should, therefore, appreciate them as "Art". Which makes them much
more enjoyable.
Please, Thomas, keep explaining to Vergon where he has erred. Then we'd
like you to start on Archimedes Plutonium.
JD
--
PGP public key: http://golem.ph.utexas.edu/~distler/distler.asc
larry shultis
"Tnlockyer" <tnlo...@aol.com> wrote in message
news:20020324200026...@mb-cn.aol.com...
> >Fredi...@hotmail.com (Fredi Fizzx)
> Wrote in:
> >Message-id: <2d9a226e.02032...@posting.google.com>
>
> >>
> >> These volumes and current loop areas are from the form factor of the
VPP
> >> particle models, and prove, not only that particles must have a
fundamental
> >> charge,
> >> but that the quark particle model's postulated fractional charges are
> >false.
> >>
> >> I effectively answered Vergon's question.
> >
> >You might have answered Vergon's question but you did not prove it.
> >You start with e and spin and get e and spin out. Definitely x=x.
> >Your Js term has e and spin in it already. Look here then carefully
> >look at your process.
> >
> >http://www.flashrock.com/upload/alpha.pdf
> >
> >Don't you wonder why your values comes out exactly as the Codata
> >values? It is because you started your proof with the Codata values.
> >You are going to have to start with something other than e and spin.
>
> Where do you see (e) in the starting equations? I have to calculate power
and
> charge density and current loop areas (unique to proton core and to
electron
> sized particles) to get (e). It is not as trivial as you all think.
If you put any number in for lambda sub c, you will get the same calculated
spin.
You might as well have let it equal 1 when calculating the charge but may be
correct
for some of the other values you define. The charge is independent of of
your
proton core calculation.
Larry
larry shultis
"ThomasL283" <thoma...@aol.com> wrote in message
news:20020324202358...@mb-cn.aol.com...
> >mor...@world.std.spaamtrap.com (Michael Moroney)
> >Date: 3/24/2002 4:44 PM Pacific Standard Time
> >Message-id: <GtI7EE...@world.std.com>
>
> >>>>"Michael Moroney" <mor...@world.std.com> Wrote:
> >
> >>>>Oh boy! More "x=x proofs!
> >
> >>Mike, you still do not understand. I'll repeat it one more time.
>
> >It is quite obvious that it is you who doesn't understand. A "proof"
> >that degenerates into an "x=x" formula, once simplified and identities
>
> Then please explain why the GEOMETRY of the ANY sized spinning cube
relates
> the fundamental constants in their proper ratios.
It does because the length of the side of the cube just cancels out and
could
have been left out or set to 1. What is left are just the known fundamental
constants
that were put in as far as I can see. The side length just disguises them
until you
want to show that some number is equal to a known value.
The proton and neutron models of nested spinning cubes are much better, but
you
have to be careful to actually use your proton and neutron cores in
calculations and
not just have them cancel out. If they do just cancel out of the
calculations, they do
not contribute to the calculation, similarly to some of your equations which
have a
factor of 2 in both the numerator and the denominator which makes no
contribution
to the calculated value.
Larry
Fredi Fizzx
AHAHAHAHAHA!! Very funny, JD.
But please explain the following. I can't make it go away as easily
as what Thomas is saying in this thread.
http://www.flashrock.com/upload/quad.pdf
Thomas is getting close to his VPP magnetic moments for the proton and
neutron from a different source as far as I can tell. It is starting
with the known Codata binding energy of deuteron and assumes that it
is equal to the magnetic potential energy of deuteron. And also uses
the Codata magnetic moments for the proton and neutron. He derives
the "undamped" "true" magnetic moments of the proton and neutron and
they are close to his VPP moments. Check it out if you can and see if
I am missing something.
FrediFizzx
Fredi Fizzx
> >Fredi...@hotmail.com (Fredi Fizzx)
> Wrote in:
> >Message-id: <2d9a226e.02032...@posting.google.com>
>
> >>
> >> These volumes and current loop areas are from the form factor of the VPP
> >> particle models, and prove, not only that particles must have a fundamental
> >> charge,
> >> but that the quark particle model's postulated fractional charges are
> false.
> >>
> >> I effectively answered Vergon's question.
> >
> >You might have answered Vergon's question but you did not prove it.
> >You start with e and spin and get e and spin out. Definitely x=x.
> >Your Js term has e and spin in it already. Look here then carefully
> >look at your process.
> >
> >http://www.flashrock.com/upload/alpha.pdf
> >
> >Don't you wonder why your values comes out exactly as the Codata
> >values? It is because you started your proof with the Codata values.
> >You are going to have to start with something other than e and spin.
>
> Where do you see (e) in the starting equations? I have to calculate power and
> charge density and current loop areas (unique to proton core and to electron
> sized particles) to get (e). It is not as trivial as you all think.
Thomas, Js contains alpha which contains an e squared term. And also
contains the spin factor as well. What you are saying in the middle
of your document might have some value but don't say that you are
deriving e and spin when you started with those in the first place.
If I must, I can do the exact same thing you did starting with e and
spin. You have shown something here, but you have not deduced or
derived e or spin. You need to express your idea in a different way.
FrediFizzx
Vertner Vergon
news:240320022136010584%dis...@golem.ph.utexas.edu...
> <thoma...@aol.com> wrote:
>
>
> >I effectively answered Vergon's question.
>
> I love it when the crackpots answer each other's questions.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
Vergon:
I love it when some moron janitor from the university of Texas dubs
his superiors as crackpots.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> It's like setting two of those AI programs (like Eliza the psychologist
> and Zippy the Pinhead) to talk to each other. Utter gibberish, but
> extremely funny to read.
****************************************************************************
****
Vergon:
How would Jack Distiller from cow town know that I spoke udder
gibberish when all I did was recount known facts? The charges
*are* equal -- and the masses *are* disparate. (But maybe the poor sap
doesn't know that.)
****************************************************************************
*****
> I don't know which is funnier, Verner and Thomas or
> <http://cslab.anu.edu.au/~daa/coffee/pin.html>
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Vergon:
I know what I find funny -- that lofty airs of superiority often ride
(as in this case) with sub normal intelligence.
The sad thing is they don't have enough sense to hide quietly
in a corner so no one will notice them, but instead blare a lot
of noise to attract attention.
I notice the great brain did not attempt to explain the phenomenon --
just blast on his kiddie trumpet.
But then he udoubtedly is not capable of understanding the problem
let alone solving it. Poor thing.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> Anyway, since it is now firmly extablished that Thomas is incapable of
> eve simple dimensional analysis, it is no longer tenable to pretend
> that his posts constitute "Science".
>
> We should, therefore, appreciate them as "Art". Which makes them much
> more enjoyable.
>
> Please, Thomas, keep explaining to Vergon where he has erred
****************************************************************************
*****
Vergon:
No, smart ass, YOU explain it to him. In short, put your money where
your (big) mouth is.
(In case you don't know it, you've been called out.)
****************************************************************************
******
Then we'd
> like you to start on Archimedes Plutonium.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Vergon:
You're a better subject.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Michael Moroney
>>>>>Oh boy! More "x=x proofs!
>>
>>>Mike, you still do not understand. I'll repeat it one more time.
>>It is quite obvious that it is you who doesn't understand. A "proof"
>>that degenerates into an "x=x" formula, once simplified and identities
>Then please explain why the GEOMETRY of the ANY sized spinning cube relates
>the fundamental constants in their proper ratios.
Thomas, that is a CLUE that is SCREAMING in your face trying to get you
to see that your calculations are bogus, yet you STILL don't recognize it.
The fact that ANY number works should tell you that it is TOTALLY IRRELEVANT
to the so-called proof, and in this case it is irrelevant because all the
so-called length terms cancel out. Also consider that negative lengths
also "work", as do imaginary lengths (sqrt(-1)). What the hell does that
supposed to mean?
What you have done is essentially taken an identity, multiplied it by
something irrelevant, then divided it by the same irrelevant factor to
come up with the same identity and claimed it as a proof the irrelevant
factor is relevant. I can take your very same equation, multiply it by
my cat's age in dog years, divide it by the air speed of an unladen swallow
in furlongs per fortnight, divide the result by my cat's age in dog years
and multiply it by the air speed of an unladen swallow in furlongs per
fortnight and come up with the original equation, and it doesn't matter
whether I'm talking about an African or European swallow, or if I even
have a cat. Can you see that???
>Y
>ou simply do not want to see the proof as anything but trivial. You are wrong
>as you can be. See my answer to Fredi: about the same time you posted.
No, I am simply pointing out that your equations reduce to e=e, and therefore
prove nothing whatsover.
>P.S. Don't try to lecture me about the constants, as supervisor of standards, I
>attended conferences on the subject, and am very well aware that the constants
>are all related by a number of exact mathematical expressions. And have said
>so many times on these newsgroups.
What the hell? If you know that the constants relate to each other by
mathematical expressions, you should know it is valid to substitute those
expressions for the constants and simple mathematics then takes over to
simplify those equations of yours. They all reduce to "x=x".
So you just told me that you have no excuse for your piss-poor physics,
you should know better.
-Mike
Michael Moroney
>>Fredi...@hotmail.com (Fredi Fizzx)
>Wrote in:
>>Message-id: <2d9a226e.02032...@posting.google.com>
>>You might have answered Vergon's question but you did not prove it.
>>You start with e and spin and get e and spin out. Definitely x=x.
>>Your Js term has e and spin in it already. Look here then carefully
>>look at your process.
>>
>>http://www.flashrock.com/upload/alpha.pdf
>Where do you see (e) in the starting equations? I have to calculate power and
>charge density and current loop areas (unique to proton core and to electron
>sized particles) to get (e). It is not as trivial as you all think.
He just told you! Did you even look at the link? (e) is in the definition
of alpha! Here it is again: (alpha) = e^2/(4 pi Eo hbar c)
-Mike
ThomasL283
>Date: 3/24/2002 11:06 PM Pacific Standard Time
>Message-id: <fe51764e.02032...@posting.google.com>
>tnlo...@aol.com (Tnlockyer) wrote in message
>news:<20020324200026...@mb-cn.aol.com>...
>> >>
>> >> These volumes and current loop areas are from the form factor of the VPP
>> >> particle models, and prove, not only that particles must have a
>fundamental
>> >> charge,
>> >> but that the quark particle
>> >You start with e and spin and get e and spin out. Definitely x=x.
>> >Your Js term has e and spin in it already. Look here then carefully
>> >look at your process.
>> >
>> >http://www.flashrock.com/upload/alpha.pdf
>> >
Fredi: Yes you can obtain the dimensionless fine structure constant as a ratio
of other constants. That ratio is what is used to get the Joule E field, from
both the proton core and the electron models mass (energy).
>Thomas, Js contains alpha which contains an e squared term. And also
>contains the spin factor as well. What you are saying in the middle
>of your document might have some value but don't say that you are
>deriving e and spin when you started with those in the first place.
But take alpha and try to go back the other way.
Can't do it unless you introduce every value for (e) (Eo) (hbar) and (c).
But with your metric let's see how you would get the separate volume of
disparate masses to obtain the power density?
And with your metric let's see how you would get the current loop areas for
disparate masses?
What I am doing with the math is determine the electric field strength E=(V/m)
and the magnetic field strength H=(A/m) from the known ratio in the impedance
of space. 0.5(E x H) is the power density in free space.
To do that, one has to determine the power density from (E= sqr Power
density(Pe) x impedance of vacuum (Zo).
This requires knowing the volume (Vol) of the particle, Pe = (2Jsc) / Vol,
and VPP gives the volume of any sized particle.
I made a presentation to a group at NIST about a year and a half ago, and the
.gif is slide 13 from powerpoint.
At that time in my lecture, I brought out the fact if one used the 1973 CODATA,
the math returned the 1973 value for (e). If the 1986 CODATA was used, you get
the 1986 value for (e), and if you use the 1998 CODATA (as in the .gif example)
you get the 1998 value for (e).
>You have shown something here, but you have not deduced or
>derived e or spin. You need to express your idea in a different way.
>
No, it was not my intent to "derive" (e) or "spin" only to show that the proton
core and electron both have (e) and spin values alike. Only VPP geometry can
do that.
>If I must, I can do the exact same thing you did starting with e and
>spin.
>
One cannot show the unique proton core and electron particles both have (e)
and spin, without VPP geometry.
What volume and current loop areas would you assign? Why?
Regards: Tom:
ThomasL283
>Date: 3/25/2002 8:37 AM Pacific Standard Time
>Message-id: <GtJFI...@world.std.com>
>tnlo...@aol.com (Tnlockyer) writes:
>
>>>Fredi...@hotmail.com (Fredi Fizzx)
>>Wrote in:
>>>Message-id: <2d9a226e.02032...@posting.google.com>
>
>>>You might have answered Vergon's question but you did not prove it.
>>>You start with e and spin and get e and spin out. Definitely x=x.
>>>Your Js term has e and spin in it already. Look here then carefully
>>>look at your process.
>>>
>>Where do you see (e) in the starting equations? I have to calculate power
>and
>>charge density and current loop areas (unique to proton core and to electron
>>sized particles) to get (e). It is not as trivial as you all think.
>He just told you! Did you even look at the link? (e) is in the definition
>of alpha! Here it is again: (alpha) = e^2/(4 pi Eo hbar c)
>
>-Mike
I have the same eqaution in my book, see page 4.
Mike, a dimensionless alpha can be obtain from the ratio of constants, sure. So
what?
Can you start with alpha and derive (e) and spin from disparate masses?
Not unless you use the geometry of VPP.
You need the particle volume and current loop areas.
-Tom:
ThomasL283
>Date: 3/25/2002 8:24 AM Pacific Standard Time
>Message-id: <GtJEw...@world.std.com>
>
>>Then please explain why the GEOMETRY of the ANY sized spinning cube relates
>>the fundamental constants in their proper ratios.
>
>Thomas, that is a CLUE that is SCREAMING in your face trying to get you
>to see that your calculations are bogus, yet you STILL don't recognize it.
>The fact that ANY number works should tell you that it is TOTALLY IRRELEVANT
Jesus, What you guys think is a weakness is a strength.
I have said as much in my book, see page 36.
If you use 1 meter for lambda, you get a volume of 6.3325739E-3 m^3
and a power density of 1.37250013E-16 kg s^-3
Sure as hell does not cancel out when using one meter, as Larry assumed.
Spin calculation uses the resulting "structure energy" 1.4495792E-27 kg m^2
s^-2 times (one meter) lambda,
divided by the fine structure constant, alpha, 4pi and (c).
It sure as hell doesn't cancel out as Larry assumed.
Sorry for screaming at you guys.
I know that none of you has ever seen a geometric model for particle
structures,
that relates the constants in their proper ratios,
so you have nothing to judge against.
But in fact, one could say the constants are all related BECAUSE of the
geometry of space and time embodied in the VPP models.
Regards: Tom:
Michael Moroney
>>Thomas, that is a CLUE that is SCREAMING in your face trying to get you
>>to see that your calculations are bogus, yet you STILL don't recognize it.
>>The fact that ANY number works should tell you that it is TOTALLY IRRELEVANT
>Jesus, What you guys think is a weakness is a strength.
You consider it a strength that your so-called proofs are all degenerate?
>If you use 1 meter for lambda, you get a volume of 6.3325739E-3 m^3
Quit trying to change the topic. We're talking about how lambda and the
rest of the "spinning cube" crap cancels when calculating e.
>Sure as hell does not cancel out when using one meter, as Larry assumed.
It does if you stayed on topic.
>But in fact, one could say the constants are all related BECAUSE of the
>geometry of space and time embodied in the VPP models.
But one might have to get really, really drunk first.
-Mike
Michael Moroney
>>>Where do you see (e) in the starting equations? I have to calculate power
>>and
>>>charge density and current loop areas (unique to proton core and to electron
>>>sized particles) to get (e). It is not as trivial as you all think.
>>He just told you! Did you even look at the link? (e) is in the definition
>>of alpha! Here it is again: (alpha) = e^2/(4 pi Eo hbar c)
>Mike, a dimensionless alpha can be obtain from the ratio of constants, sure. So
>what?
>Can you start with alpha and derive (e) and spin from disparate masses?
Sure. Given (alpha) = e^2/(4 pi Eo hbar c), we rearrange to get:
e^2 = (alpha)*(4 pi Eo hbar c) or e = sqrt((alpha)*(4 pi Eo hbar c)).
Both the + and - square roots are valid, thus the electron and positron
charges.
>Not unless you use the geometry of VPP.
>You need the particle volume and current loop areas.
Nope. No VPP, no spinning cubes crap, no nonsense.
-Mike
Tnlockyer
>Date: 3/25/2002 11:51 AM Pacific Standard Time
>Message-id: <GtJoH5...@world.std.com>
>
>thoma...@aol.com (ThomasL283) writes:
>
>>Jesus, What you guys think is a weakness is a strength.
>
>You consider it a strength that your so-called proofs are all degenerate?
Nope, require a volume and current loop areas, unique to any given mass.
>
>>But in fact, one could say the constants are all related BECAUSE of the
>>geometry of space and time embodied in the VPP models.
>
>But one might have to get really, really drunk first.
Here is the only geometry that will give the proper volume and current loop
areas.
http://members.aol.com/tnlockyer/cubedimensions.gif
The only way to get (e) and (1/2hbar) for various masses, is to use those
above (cubedimensions.gif) relationships, to calculate the volume and current
loop areas, as in the following links.
http://members.aol.com/tnlockyer/protoncharge.gif
http://members.aol.com/tnlockyer/electroncharge.gif
Regards: Tom:
Thomas Lockyer (75 and retired) See "Vector Particles and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about, and express it in numbers,
Michael Moroney
>>mor...@world.std.spaamtrap.com (Michael Moroney)
>>You consider it a strength that your so-called proofs are all degenerate?
>Nope, require a volume and current loop areas, unique to any given mass.
There is none so blind as he who will not see.
>The only way to get (e) and (1/2hbar) for various masses, is to use those
>above (cubedimensions.gif) relationships, to calculate the volume and current
>loop areas, as in the following links.
Nope. All the cube crap cancels out, and both "proofs" are degenerate
"x=x" "proofs".
-Mike
Monitek
>Date: 24/03/02 18:43 GMT Standard Time
>Message-id: <20020324134336...@mb-cn.aol.com>
snip--------------------------------------
>>For the math that proves the VPP model core size is correct; Click on this
>link:
>
>http://members.aol.com/tnlockyer/protoncharge.gif
>
>>print it out, with a copy of this article, staple them together, and read at
>your leisure.
from which:
Js=(h.c.alpha)/lamda Joule E field energy
extracting h:
h= (Js.lamda)/(c.alpha)-----------------(1)
spin = h/(4.pi) [I'll take your word for it]
substituting for h from (1):
spin = (Js.lamda)/(4.pi.c.alpha)
Where does the spin model geometry come in?
Why are you using the mathematics of circles and spheres to represent a cubic
model?
Do you do a discount for retired purchasers of your signed book?
What does VPP mean?
regards,
Monitek (Arden Barker)
ThomasL283
>Date: 3/26/2002 1:31 AM Pacific Standard Time
>Message-id: <20020326043120...@mb-fq.aol.com>
>
>>From: thoma...@aol.com (ThomasL283)
>>Date: 24/03/02 18:43 GMT Standard Time
>>Message-id: <20020324134336...@mb-cn.aol.com>
>
>snip--------------------------------------
>
>
>spin = (Js.lamda)/(4.pi.c.alpha)
>
>
>Where does the spin model geometry come in?
Montek, see the General EM Cube dimensions in this link.
http://members.aol.com/tnlockyer/cubedimensions.gif
Mechanics of spin angular momentum are:
1/2 hbar = me x c x radius of gyration (Rm)
The only unknown, on the right side, is the radius of gyration (Rm).
Notice the cube geometry shows that is Rm = lambda/4pi.
If you have a computer, enter the mass of the electron (me) = ( 9.10938188E-31
kg )
times the velocity of light (c) = (2.99792458E8 m s^-1)
times the electron's Compton wavelength (divided by 4pi) or
(Rm) = (2.426310215E-12 / 4pi) meters.
You should get the correct; 1/2 hbar= 5.272857982E-35 kg m^2 s^-1 (Joule
seconds)
Proving the cube geometry maintains the constants in their proper ratios.
The VPP model also can derive the Bohr magneton (see page 30) and as I have
shown, the VPP cube geometry (spinning volume and current loop areas) also
shows the cube model volume has the power density to obtain the fundamental
charge (e) (see pages 34, 35,36)
My detractors cannot believe that you must have the spinning cube cyclinder
volume of (lambda^3 / 16pi^2)
and you must have related current loops (from that same cube dimensions) as
(lambda^2 / 4pi), exclusively from the vector particle physics (VPP).
This is shown in the above (cubedimensions.gif)
The only way to get (e) and (1/2hbar)
FOR VARIOUS MASSES, is to use those
above (cubedimensions.gif) relationships,
so as to calculate the unique, to each different mass, a related volume and
current
loop areas, as was proved in the following links.
http://members.aol.com/tnlockyer/protoncharge.gif
http://members.aol.com/tnlockyer/electroncharge.gif
Regards: Tom:
PS. Amazon.com only pays me $11.23 a book copy sold and I have to eat the
shipping to their warehouse in KY. So, I am open to sell the book at a
discount, direct. I will sell at a senior rate of $14.23 including S/H. TNL
Press. 1611 Fallen Leaf, Los Altos CA USA 94024
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,
Fredi Fizzx
Thomas, here it is for you in black and white. Print out the
following link and study it carefully. You have done nothing but
prove x=x. The lamba sub c cancels out. It is all algebraic
substitution. You really do need to present your idea in a different
manner because this does not work at all.
http://www.flashrock.com/upload/xequalsx.pdf
FrediFizzx
Michael Moroney
>>Where does the spin model geometry come in?
>Mechanics of spin angular momentum are:
>1/2 hbar = me x c x radius of gyration (Rm)
>The only unknown, on the right side, is the radius of gyration (Rm).
>Notice the cube geometry shows that is Rm = lambda/4pi.
>If you have a computer, enter the mass of the electron (me) = ( 9.10938188E-31
>kg )
>times the velocity of light (c) = (2.99792458E8 m s^-1)
> times the electron's Compton wavelength (divided by 4pi) or
>(Rm) = (2.426310215E-12 / 4pi) meters.
>You should get the correct; 1/2 hbar= 5.272857982E-35 kg m^2 s^-1 (Joule
>seconds)
Yet ANOTHER "x=x" 'proof'! The Compton wavelength is _defined_ as
h/mc where m is the mass of the electron (for the electron's Compton
Wavelength) Substitute h/mc for lambda and we wind up with hbar = h/2pi.
So Tom proves here that hbar = h/2pi (I guess that is a slight improvement
from the usual 'x=x' proofs)
>PS. Amazon.com only pays me $11.23 a book copy sold and I have to eat the
>shipping to their warehouse in KY. So, I am open to sell the book at a
>discount, direct. I will sell at a senior rate of $14.23 including S/H. TNL
>Press. 1611 Fallen Leaf, Los Altos CA USA 94024
I guess Thomas is more interested in sales of his "Cooke Book" than good
physics.
ThomasL283
>Date: 3/27/2002 1:12 AM Pacific Standard Time
>Message-id: <fe51764e.02032...@posting.google.com>
>> One cannot show the unique proton core and electron particles both have
>(e)
>> and spin, without VPP geometry.
>>
>> What volume and current loop areas would you assign? Why?
>
>Thomas, here it is for you in black and white. Print out the
>following link and study it carefully. You have done nothing but
>prove x=x.
You are letting the algebra blind you to the VPP source of (Vol) and (L2).
>The lamba sub c cancels out. It is all algebraic
>substitution. You really do need to present your idea in a different
>manner because this does not work at all.
It sure does work. The VPP (Vol) and (L2) are from the VPP model, not algebra,
but geometry.
>
>http://www.flashrock.com/upload/xequalsx.pdf
>
Fredi: Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16 pi^2)
eqaution.
And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi) current loop
areas.
See the geometric derivation of Volume and front and back current loop areas,
in the link:
http://members.aol.com/tnlockyer/cubedimensions.gif
My equations are the same algebra you show in the .pdf.
Difference is I show the source of the volume (Vol) and loop areas (L2)
Untill you can show an independent source for (Vol) and the L2 area,
other than VPP, I have made my point.
Regards: Tom:
ThomasL283
>Date: 3/27/2002 8:42 AM Pacific Standard Time
>Message-id: <Gtn527...@world.std.com>
>>Mechanics of spin angular momentum are:
>
>>1/2 hbar = me x c x radius of gyration (Rm)
>
>>The only unknown, on the right side, is the radius of gyration (Rm).
>
>>Notice the cube geometry shows that is Rm = lambda/4pi.
>
>
>Yet ANOTHER "x=x" 'proof'! The Compton wavelength is _defined_ as
>h/mc where m is the mass of the electron (for the electron's Compton
>Wavelength) Substitute h/mc for lambda and we wind up with hbar = h/2pi.
But Mike, you did not know that the mass radius was (lambda/4pi) au priori.
I got (lambda /4pi) from the VPP cube geometry, not the simple childish algebra
based on dimensional analysis. So VPP geometrically maintains the constants in
their proper ratios, See>
http://members.aol.com/tnlockyer/cubedimensions.gif
>>PS. Amazon.com only pays me $11.23 a book copy sold and I have to eat the
>>shipping to their warehouse in KY. So, I am open to sell the book at a
>>discount, direct. I will sell at a senior rate of $14.23 including S/H.
>TNL
>>Press. 1611 Fallen Leaf, Los Altos CA USA 94024
>I guess Thomas is more interested in sales of his "Cooke Book" than good
>physics.
>
Monitek had asked me if I gave senior discounts.
Don't knock the book. Fredi has a copy and he sees that the model makes some
predictions that cannot be dismissed.
I would like all you detractors to get a copy of the book, and then maybe these
discussions can be more to the point.
Michael Moroney
>>Yet ANOTHER "x=x" 'proof'! The Compton wavelength is _defined_ as
>>h/mc where m is the mass of the electron (for the electron's Compton
>>Wavelength) Substitute h/mc for lambda and we wind up with hbar = h/2pi.
>But Mike, you did not know that the mass radius was (lambda/4pi) au priori.
Doing the algebra backwards pulls out the Compton wavelength. There is
no reason I can see to assume this length is any sort of "radius of
gyration" as you call it. Since as far as anyone (with actual physics
knowledge) can tell, the electron has no size thus no spin radius.
This does seem contradictory since we know the electron has angular momentum
which is usually thought of an object with size spinning, but this is one
of the current mysteries of quantum mechanics, I guess.
If there is a nonzero spin radius, we already know it to be much smaller
than the Compton wavelength.
>I got (lambda /4pi) from the VPP cube geometry, not the simple childish algebra
>based on dimensional analysis.
First, you now call an established branch of mathematics, which you use,
as "childish" when use of it shows your proofs all worthless? And the
same for dimensional analysis that until recently you were whining that
you were defending? That itself is rather childish, at best.
>Don't knock the book. Fredi has a copy and he sees that the model makes some
>predictions that cannot be dismissed.
I've personally dismissed several of your claims that you claimed could
not be dismissed. I'm sure there are just another 100 or so pages of "x=x"
'Cooke Book' proofs in it.
>I would like all you detractors to get a copy of the book, and then maybe these
>discussions can be more to the point.
No thanks. Cat litter is much cheaper in the supermarket, and doesn't need
to be shredded first.
-Mike
larry shultis
> >fredi...@hotmail.com (Fredi Fizzx)
> >Date: 3/27/2002 1:12 AM Pacific Standard Time
> >Message-id: <fe51764e.02032...@posting.google.com>
>
> >> One cannot show the unique proton core and electron particles both
have
> >(e)
> >> and spin, without VPP geometry.
> >>
> >> What volume and current loop areas would you assign? Why?
> >
>
> >Thomas, here it is for you in black and white. Print out the
> >following link and study it carefully. You have done nothing but
> >prove x=x.
>
> You are letting the algebra blind you to the VPP source of (Vol) and (L2).
>
> >The lamba sub c cancels out. It is all algebraic
> >substitution. You really do need to present your idea in a different
> >manner because this does not work at all.
>
> It sure does work. The VPP (Vol) and (L2) are from the VPP model, not
algebra,
> but geometry.
Thomas, it is irrelevant where they came from. The do not take part in the
calculations for charge and spin. Any length would do. Now, if you can
take your proton core geometry and, without introducing either e or h either
explicitly or implicitly through some derived constant,
get the fundamental charge and the spin, then you have something.
>
> >
> >http://www.flashrock.com/upload/xequalsx.pdf
> >
>
> Fredi: Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16 pi^2)
> eqaution.
>
> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi) current
loop
> areas.
He is just using what you have on the web page and showing how
Vol and L2 just cancel out without making any contribution to the charge or
spin.
>
> See the geometric derivation of Volume and front and back current loop
areas,
> in the link:
>
> http://members.aol.com/tnlockyer/cubedimensions.gif
>
> My equations are the same algebra you show in the .pdf.
>
> Difference is I show the source of the volume (Vol) and loop areas (L2)
That is fine, but you must NOT introduce explicitly of implicitly what you
want to calculate.
>
> Untill you can show an independent source for (Vol) and the L2 area,
You have to point out some experiment that will substantiate that those are
the actual Vol and L2. Otherwise they do not model the particle. But even
if those are correct, you cannot introduce e and/or h to derive the charge
and the spin.
Larry
Fredi Fizzx
> >fredi...@hotmail.com (Fredi Fizzx)
> >Date: 3/27/2002 1:12 AM Pacific Standard Time
> >Message-id: <fe51764e.02032...@posting.google.com>
>
> >> One cannot show the unique proton core and electron particles both have
> (e)
> >> and spin, without VPP geometry.
> >>
> >> What volume and current loop areas would you assign? Why?
> >
>
> >Thomas, here it is for you in black and white. Print out the
> >following link and study it carefully. You have done nothing but
> >prove x=x.
>
> You are letting the algebra blind you to the VPP source of (Vol) and (L2).
>
> >The lamba sub c cancels out. It is all algebraic
> >substitution. You really do need to present your idea in a different
> >manner because this does not work at all.
>
> It sure does work. The VPP (Vol) and (L2) are from the VPP model, not algebra,
> but geometry.
>
> >
> >http://www.flashrock.com/upload/xequalsx.pdf
> >
>
>
> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi) current loop
> areas.
This in not my algebra; it is all yours with my proper substitutions.
Note in the equation at the bottom of the first page that the lamba
sub c cancels out. That means it could be any length at all (as many
here have pointed out) which means it is irrelevant.
> See the geometric derivation of Volume and front and back current loop areas,
> in the link:
>
> http://members.aol.com/tnlockyer/cubedimensions.gif
>
> My equations are the same algebra you show in the .pdf.
>
> Difference is I show the source of the volume (Vol) and loop areas (L2)
>
> Untill you can show an independent source for (Vol) and the L2 area,
Your Vol and L2 area are based on lamba sub c length which is
irrelevant so they are irrelevant. I really think you need to give up
on this idea that any length will produce elementary charge and spin.
As you can plainly see in my xequalsx.pdf, if you study it carefully,
that length is totally irrelevant. Please go back to plan "A". By
your own first book, you had a different explanation of the proton's
elementary charge anywise that I thought was better (partons with
alternating charges). Your current explanation of disparate masses
same charge does not hold up under scrutiny.
The reason partons with alternating charges was better, is that it
could maybe account for what they are seeing experimentally. The
partons would account for a certain percentage momentum with positive
charge and a certain percentage momentum with negative charge.
FrediFizzx
Tnlockyer
>Date: 3/27/2002 8:58 PM Pacific Standard Time
>Message-id: <ua58psf...@corp.supernews.com>
>
>> It sure does work. The VPP (Vol) and (L2) are from the VPP model, not
>algebra,
>> but geometry.
>
>Thomas, it is irrelevant where they came from. The do not take part in the
>calculations for charge and spin. Any length would do.
Larry. There has to be a direct relationship between Vol and L2, in the
calculations. The ratio Vol/L2 differs at different lengths. See these two
examples and calculate the ratios yourself.
http://members.aol.com/tnlockyer/protoncharge.gif
And:
http://members.aol.com/tnlockyer/electroncharge.gif
What you guys were doing is simply verifying that the dimensional analysis
algebra works out.
Sure the algebra will prove that the units cancel out (as they must) leaving
the charge. (This only means the calculations were set up correctly)
The kicker is we obtain different NUMBERS for the power and charge density,
for each different length inserted, and still we obtain the charge from the
new charge density and new L2. See above links.
The fact that any SIZED geometric cube has a charge (e) and spin (1/2hbar) is
absolutely proved (quantatively) by the related (different) field strengths and
power density from the volume and current loop areas associated with any given
length.
>Now, if you can
>take your proton core geometry and, without introducing either e or h either
>explicitly or implicitly through some derived constant,
>get the fundamental charge and the spin, then you have something.
Never work. One has to introduce some (SI) metric units to set the metric
scale. Otherwise it's furlongs and fortnights.
Here is a group of math relating the (SI) metric constants from my (NIST)
lecture slides. See also page 4 in the book.
http://members.aol.com/tnlockyer/constants.gif
Note, by tranposing the algebra, one can obtain any single unknown. WHY? The
constants are all related by the geometry of space and time, such as embodied
in the VPP model geometry.
Only geometry (like a circle always has the same (ratio) value of pi) can
maintain the constants in their appointed ratios.
Regards: Tom:
Thomas Lockyer (75 and retired) See "Vector Particles and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about, and express it in numbers,
Tnlockyer
>Date: 3/29/2002 12:12 AM Pacific Standard Time
>Message-id: <fe51764e.0203...@posting.google.com>
>thoma...@aol.com (ThomasL283) wrote in message
>news:<20020327140415...@mb-fh.aol.com>...
>> Fred, Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16 pi^2)
>> eqaution.
>>
>> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi) current
>loop
>> areas.
>
>This in not my algebra; it is all yours with my proper substitutions.
>Note in the equation at the bottom of the first page that the lamba
>sub c cancels out. That means it could be any length at all (as many
>here have pointed out) which means it is irrelevant.
God, I cannot understand you fellows not seeing that the Power density and
charge density DO DEPEND ON THE LENGTH INSERTED, and the resulting (unique)
volume and current loops must be as specified from the VPP geometry, to obtain
(e) and (1/2hbar).
Of course, if the dimensional analysis algebra did not cancel out the units,
the equations would not work.
But, geometry is not algebra.
Geometric particles with geometric relationships have never been used in
particle physics,
so you fellows seem to have trouble understanding how geometric models can
work, with the algebra.
I think we will just have to agree to disagree, and go on.
Y.Porat
> >fredi...@hotmail.com (Fredi Fizzx)
> >Date: 3/29/2002 12:12 AM Pacific Standard Time
> >Message-id: <fe51764e.0203...@posting.google.com>
>
> >thoma...@aol.com (ThomasL283) wrote in message
> >news:<20020327140415...@mb-fh.aol.com>...
>
> >> Fred, Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16 pi^2)
> >> eqaution.
> >>
> >> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi) current
> loop
> >> areas.
> >
>
> >This in not my algebra; it is all yours with my proper substitutions.
> >Note in the equation at the bottom of the first page that the lamba
> >sub c cancels out. That means it could be any length at all (as many
> >here have pointed out) which means it is irrelevant.
>
> God, I cannot understand you fellows not seeing that the Power density and
> charge density DO DEPEND ON THE LENGTH INSERTED, and the resulting (unique)
> volume and current loops must be as specified from the VPP geometry, to obtain
> (e) and (1/2hbar).
>
> Of course, if the dimensional analysis algebra did not cancel out the units,
> the equations would not work.
>
> But, geometry is not algebra.
>
> Geometric particles with geometric relationships have never been used in
> particle physics,
Geometric relationship have never been used in particle
physics ????? !!!!!!
common Thomas you seem to gorget and ignor Y.Porat
who could give you some private lessons about it
some decency fellow whan't hurt you! it will only
make you more credible
all the best
Y.Porat
---------
larry shultis
"Tnlockyer" <tnlo...@aol.com> wrote in message
news:20020329180610...@mb-mh.aol.com...
> >"larry shultis" gold...@charter.net
> >Date: 3/27/2002 8:58 PM Pacific Standard Time
> >Message-id: <ua58psf...@corp.supernews.com>
> >
>
> >> It sure does work. The VPP (Vol) and (L2) are from the VPP model, not
> >algebra,
> >> but geometry.
> >
>
> >Thomas, it is irrelevant where they came from. The do not take part in
the
> >calculations for charge and spin. Any length would do.
>
> Larry. There has to be a direct relationship between Vol and L2, in the
> calculations. The ratio Vol/L2 differs at different lengths. See these
two
> examples and calculate the ratios yourself.
>
> http://members.aol.com/tnlockyer/protoncharge.gif
>
> And:
>
> http://members.aol.com/tnlockyer/electroncharge.gif
>
> What you guys were doing is simply verifying that the dimensional analysis
> algebra works out.
>
> Sure the algebra will prove that the units cancel out (as they must)
leaving
> the charge. (This only means the calculations were set up correctly)
No, all you did to get, e.g., charge, was to do the algebra with the
equations. It does not matter if you conceal the charge e in two numbers
and reveal it again by cancelling common terms from the numerator and
the denominator. The common term does not make a contribution to the
charge. As I have said before, if you can get the charge without using
it explicitly or implicitly in you calculations, then you can say that you
have calculted it from your model.
>
> The kicker is we obtain different NUMBERS for the power and charge
density,
> for each different length inserted, and still we obtain the charge from
the
> new charge density and new L2. See above links.
That is fine and if you have a means to check those with an experiment, then
the model will be in accord with reality. For example, your calculations of
the proton mass from the model geometry may be valid if it gives a more
accurate mass with your so called normal power series than does QED
with its power series. Did you check that by inserting the QED value in
your equation and checking it with the experimental value?
>
> The fact that any SIZED geometric cube has a charge (e) and spin (1/2hbar)
is
They do because the length of a side is irrelevant in the calculations.
I.e., charge
and spin do not depend upon the size of the particle in your model.
> absolutely proved (quantatively) by the related (different) field
strengths and
> power density from the volume and current loop areas associated with any
given
> length.
>
> >Now, if you can
> >take your proton core geometry and, without introducing either e or h
either
> >explicitly or implicitly through some derived constant,
> >get the fundamental charge and the spin, then you have something.
>
> Never work. One has to introduce some (SI) metric units to set the metric
> scale. Otherwise it's furlongs and fortnights.
You set the metric scale with your geometry. Are you saying that you must
introduce the charge and spin inorder calculate the charge and spin with
your
model? That is fudging the results of your calculations.
>
> Here is a group of math relating the (SI) metric constants from my (NIST)
> lecture slides. See also page 4 in the book.
>
> http://members.aol.com/tnlockyer/constants.gif
And note how you sneak in h and e with those constants' definitions.
>
> Note, by tranposing the algebra, one can obtain any single unknown. WHY?
The
> constants are all related by the geometry of space and time, such as
embodied
> in the VPP model geometry.
Just where does the geometry spit out those constants. Unless you hang them
on
your geometry, you are unable to calculate them from the model.
>
> Only geometry (like a circle always has the same (ratio) value of pi) can
> maintain the constants in their appointed ratios.
That, you have not shown. There are perhaps many ways to do that. For
example,
the algebraic relationships between the constants maintain the proper
relationships
between the constants without any geometric model and that is all you are
showing
when you calculate charge and spin since the model geometry just cancels out
for
those constants.
Larry
larry shultis
> >fredi...@hotmail.com (Fredi Fizzx)
> >Date: 3/29/2002 12:12 AM Pacific Standard Time
> >Message-id: <fe51764e.0203...@posting.google.com>
>
> >thoma...@aol.com (ThomasL283) wrote in message
> >news:<20020327140415...@mb-fh.aol.com>...
>
> >> Fred, Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16 pi^2)
> >> eqaution.
> >>
> >> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi)
current
> >loop
> >> areas.
> >
>
> >This in not my algebra; it is all yours with my proper substitutions.
> >Note in the equation at the bottom of the first page that the lamba
> >sub c cancels out. That means it could be any length at all (as many
> >here have pointed out) which means it is irrelevant.
>
> God, I cannot understand you fellows not seeing that the Power density and
> charge density DO DEPEND ON THE LENGTH INSERTED, and the resulting
(unique)
> volume and current loops must be as specified from the VPP geometry, to
obtain
> (e) and (1/2hbar).
>
> Of course, if the dimensional analysis algebra did not cancel out the
units,
> the equations would not work.
It is not dimensional analysis. The actual values of your volume and current
loops
cancel out, not just the dimensions since they are in both the numerator and
the
denominator.
>
> But, geometry is not algebra.
>
> Geometric particles with geometric relationships have never been used in
> particle physics,
That, most likely is not true. Geometry and the more general topology has
been
used.
Larry
Tnlockyer
>Date: 3/30/2002 12:11 AM Pacific Standard Time
>Message-id: <uaasrqj...@corp.supernews.com>
>
>"Tnlockyer" <tnlo...@aol.com> wrote in message
>news:20020329180610...@mb-mh.aol.com...
>> >"larry shultis"
>> Larry. There has to be a direct relationship between Vol and L2, in the
>> calculations. The ratio Vol/L2 differs at different lengths. See these
>two
>> examples and calculate the ratios yourself.
>>
>> http://members.aol.com/tnlockyer/protoncharge.gif
>>
>> And:
>>
>> http://members.aol.com/tnlockyer/electroncharge.gif
>> What you guys were doing is simply verifying that the dimensional analysis
>> algebra works out.
>>
>> Sure the algebra will prove that the units cancel out (as they must)
>leaving
>> the charge. (This only means the calculations were set up correctly)
>
>As I have said before, if you can get the charge without using
>it explicitly or implicitly in you calculations, then you can say that you
>have calculted it from your model.
OK Larry. Fredi showed in his algebra that the fine structure constant (alpha)
had (e^2) in the equation, and then he proceeded, by agebra, to show the VPP
model volume and current loop geometry got the (e = e).
I got to thinking that there are two definitions for (alpha).
The one Fredi used was based on (alpha)= electrical potential energy / rest
mass energy.
Here is the equation for (alpha) based on the magnetic potential energy / rest
mass energy.
Both equations get the same dimensionless number, but magnetic potential
energy does not show the (e^2) and cannot directly derive (e = e).
alpha = (Uo UB UB) / (lambda^3 pi me c^2)
See the following link for Mathcad results:
http://members.aol.com/tnlockyer/magfinecons.gif
See if that magfinecons.gif equation works with Fredi's algebra.
Tnlockyer
>Date: 3/30/2002 12:20 AM Pacific Standard Time
>Message-id: <uaatbln...@corp.supernews.com>
>"Tnlockyer" <tnlo...@aol.com> wrote in message
>news:20020329182548...@mb-mh.aol.com...
>> >fredi...@hotmail.com (Fredi Fizzx)
>> >Date: 3/29/2002 12:12 AM Pacific Standard Time
>> >Message-id: <fe51764e.0203290012.97233e5@posti
>>
>> >thoma...@aol.com (ThomasL283) wrote in message
>>
>> >> Fred, Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16 pi^2)
>> >> eqaution.
>> >>
>> >> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi)
>current
>> >loop
>> >> areas.
>> >This in not my algebra; it is all yours with my proper substitutions.
>> >Note in the equation at the bottom of the first page that the lamba
>> >sub c cancels out. That means it could be any length at all (as many
>> >here have pointed out) which means it is irrelevant.
No it is not irrelevent, that was the point.
Any sized spinning (mass) cube gives the same charge and spin.
>> God, I cannot understand you fellows not seeing that the Power density and
>> charge density DO DEPEND ON THE LENGTH INSERTED, and the resulting
>(unique)
>> volume and current loops must be as specified from the VPP geometry, to
>obtain
>> (e) and (1/2hbar).
>>
>> Of course, if the dimensional analysis algebra did not cancel out the
>units,
>> the equations would not work.
>
>It is not dimensional analysis. The actual values of your volume and current
>loops
>cancel out, not just the dimensions since they are in both the numerator and
>the
>denominator.
Are you saying the VPP model does not work to give (e) and (1/2 hbar)?
The Vol = (lambda^3 / 16 pi^2) came from the VPP topology.
And the current loop areas L2= (lambda^2 / 4pi) came from the VPP topology.
I did not pick them out of the air at random, they came from the geometry.
And only those Vol and L2 relationships will cancel out making the particles
all have (e) and (1/2 hbar) as was originally proved.
see:
http://members.aol.com/tnlockyer/magfinecons.gif
For alpha without (e^2)
This magfinecons.gif equation wont work in Fredi's algebra.
It is a mistake to take a dimensionless number like alpha and start stuffing
it with numbers that get the same dimensionless ratio.
Notice NO (e^2) is showing in the magnetic potential energy derived fine
structure constant.
The VPP model works as advertized.
Michael Moroney
>>As I have said before, if you can get the charge without using
>>it explicitly or implicitly in you calculations, then you can say that you
>>have calculted it from your model.
>OK Larry. Fredi showed in his algebra that the fine structure constant (alpha)
>had (e^2) in the equation, and then he proceeded, by agebra, to show the VPP
>model volume and current loop geometry got the (e = e).
>Here is the equation for (alpha) based on the magnetic potential energy / rest
>mass energy.
> Both equations get the same dimensionless number, but magnetic potential
>energy does not show the (e^2) and cannot directly derive (e = e).
Nope. The Bohr Magnetron (UB) is _defined_ as eh/(4 pi Me), where e is
the electron's charge, and the squaring of UB gets e^2 in the numerator.
Substitute other identities (such as def. of Compton wavelength) and you
get the original definition of alpha back.
-Mike
ThomasL283
>Date: 3/30/2002 9:29 AM Pacific Standard Time
>Message-id: <Gtsr9n...@world.std.com>
>
>
>tnlo...@aol.com (Tnlockyer) writes:
>>Here is the equation for (alpha) based on the magnetic potential energy /
>rest
>>mass energy.
>
>> Both equations get the same dimensionless number, but magnetic potential
>>energy does not show the (e^2) and cannot directly derive (e = e).
>
>Nope. The Bohr Magnetron (UB) is _defined_ as eh/(4 pi Me), where e is
>the electron's charge, and the squaring of UB gets e^2 in the numerator.
>Substitute other identities (such as def. of Compton wavelength) and you
>get the original definition of alpha back.
Well sure, but Fredi did not go to the basic units, he used Eo etc.
THe point is that the fundamental constants yield the fine structure constant,
using either the electrical potential or magnetic potential energy, in ratio
with the mass (energy). These both give (alpha) because the VPP electron
model shows equal electric and magnetic energy.
(See page 34, 35 36 in the book before you shred it into kitty litter)
BTW You wont find the derivation for magnetic potential energy any other place
other than in my worthless book.
Regards: Tom:
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,
larry shultis
> >"larry shultis" gold...@charter.net
> >Date: 3/30/2002 12:20 AM Pacific Standard Time
> >Message-id: <uaatbln...@corp.supernews.com>
>
> >"Tnlockyer" <tnlo...@aol.com> wrote in message
> >news:20020329182548...@mb-mh.aol.com...
> >> >fredi...@hotmail.com (Fredi Fizzx)
> >> >Date: 3/29/2002 12:12 AM Pacific Standard Time
> >> >Message-id: <fe51764e.0203290012.97233e5@posti
>
> >>
> >> >thoma...@aol.com (ThomasL283) wrote in message
>
> >>
> >> >> Fred, Your algebra borrowed my VPP geometric (Vol)=(lambda^2 / 16
pi^2)
> >> >> eqaution.
> >> >>
> >> >> And your algebra borrowed my VPP geometric (L2)=(lambda^2 / 4pi)
> >current
> >> >loop
> >> >> areas.
>
> >> >This in not my algebra; it is all yours with my proper substitutions.
> >> >Note in the equation at the bottom of the first page that the lamba
> >> >sub c cancels out. That means it could be any length at all (as many
> >> >here have pointed out) which means it is irrelevant.
>
> No it is not irrelevent, that was the point.
> Any sized spinning (mass) cube gives the same charge and spin.
"that was the point"? Hardly. Your caption was about calculating the
charge and spin from the proton core geometry and not that you were showing
that the charge and spin were independent of the proton core geometry.
>
> >> God, I cannot understand you fellows not seeing that the Power density
and
> >> charge density DO DEPEND ON THE LENGTH INSERTED, and the resulting
> >(unique)
> >> volume and current loops must be as specified from the VPP geometry, to
> >obtain
> >> (e) and (1/2hbar).
> >>
> >> Of course, if the dimensional analysis algebra did not cancel out the
> >units,
> >> the equations would not work.
>
> >
> >It is not dimensional analysis. The actual values of your volume and
current
> >loops
> >cancel out, not just the dimensions since they are in both the numerator
and
> >the
> >denominator.
>
> Are you saying the VPP model does not work to give (e) and (1/2 hbar)?
>
> The Vol = (lambda^3 / 16 pi^2) came from the VPP topology.
>
> And the current loop areas L2= (lambda^2 / 4pi) came from the VPP
topology.
>
> I did not pick them out of the air at random, they came from the geometry.
I was saying that your claim that just dimensions were being canceled as in
dimensional analysis was wrong. If, as you do, you convert to the values of
the
constants and do the algebra, the values of L2 and Vol occur in both the
numerator and the denominator and thus cancel out, making no contribution to
the charge or spin. I am only saying that you are not calculating the charge
and
the spin from those constants. I might also say that the fact that they do
not
make a contribution to the charge and the spin, that you cannot conclude, if
you
do so, that the edge length can be of any size. I don't think you say that
since
it appears that you need to start from the Compton wavelength to get your
proton and neutron masses.
>
> And only those Vol and L2 relationships will cancel out making the
particles
> all have (e) and (1/2 hbar) as was originally proved.
>
> see:
>
> http://members.aol.com/tnlockyer/magfinecons.gif
>
> For alpha without (e^2)
>
> This magfinecons.gif equation wont work in Fredi's algebra.
>
> It is a mistake to take a dimensionless number like alpha and start
stuffing
> it with numbers that get the same dimensionless ratio.
So you do not believe that alpha has the definition given to it in terms of
other
fundamental constants? It has its structure built of fundamental constants
with
their units because that is what is necessary to balance the equation in
the
derivation of the fine-structure energy equation.
Larry
Fredi Fizzx
> > mor...@world.std.com (Michael Moroney)
> >Date: 3/30/2002 9:29 AM Pacific Standard Time
> >Message-id: <Gtsr9n...@world.std.com>
> >
>
> >
> >tnlo...@aol.com (Tnlockyer) writes:
>
> >>Here is the equation for (alpha) based on the magnetic potential energy /
> rest
> >>mass energy.
>
> >> Both equations get the same dimensionless number, but magnetic potential
> >>energy does not show the (e^2) and cannot directly derive (e = e).
>
> >
> >Nope. The Bohr Magnetron (UB) is _defined_ as eh/(4 pi Me), where e is
> >the electron's charge, and the squaring of UB gets e^2 in the numerator.
> >Substitute other identities (such as def. of Compton wavelength) and you
> >get the original definition of alpha back.
>
> Well sure, but Fredi did not go to the basic units, he used Eo etc.
>
> THe point is that the fundamental constants yield the fine structure constant,
> using either the electrical potential or magnetic potential energy, in ratio
> with the mass (energy). These both give (alpha) because the VPP electron
> model shows equal electric and magnetic energy.
> (See page 34, 35 36 in the book before you shred it into kitty litter)
>
> BTW You wont find the derivation for magnetic potential energy any other place
> other than in my worthless book.
Yes, that is how you should have stated your original argument. That
it is possible to derive the magnetic potential energy from the
constants. Not that you are deriving the fine structure constant from
magnetic potential energy. You always do things in a way that are
confusing and backward. Here it is in black and white.
http://www.flashrock.com/upload/alphaequalsalpha.pdf
If you start with the fine structure constant and other constants, you
can derive your magnetic potential energy. So what does this mean?
Take it from here.
FrediFizzx
Fredi Fizzx
> >Date: 3/30/2002 9:29 AM Pacific Standard Time
> >Message-id: <Gtsr9n...@world.std.com>
> >tnlo...@aol.com (Tnlockyer) writes:
>
> >>Here is the equation for (alpha) based on the magnetic potential energy /
> rest
> >>mass energy.
>
> >> Both equations get the same dimensionless number, but magnetic potential
> >>energy does not show the (e^2) and cannot directly derive (e = e).
> >
> >Nope. The Bohr Magnetron (UB) is _defined_ as eh/(4 pi Me), where e is
> >the electron's charge, and the squaring of UB gets e^2 in the numerator.
> >Substitute other identities (such as def. of Compton wavelength) and you
> >get the original definition of alpha back.
>
> Well sure, but Fredi did not go to the basic units, he used Eo etc.
>
> THe point is that the fundamental constants yield the fine structure constant,
> using either the electrical potential or magnetic potential energy, in ratio
> with the mass (energy). These both give (alpha) because the VPP electron
> model shows equal electric and magnetic energy.
> (See page 34, 35 36 in the book before you shred it into kitty litter)
>
> BTW You wont find the derivation for magnetic potential energy any other place
> other than in my worthless book.
Thomas, here is a mag pot energy derivation from other than your book.
http://www.flashrock.com/upload/potenergy.pdf
I am deriving both magnetic and electric potential energy from the
known constants and their known relationships without using anything
from your models. It is really a simple algebra exercise to do this.
So your derivation of mag and elect pot energy is definitey *not*
unique to your models. And "my" derivation also shows that mag and
elect pot energy are equal.
FrediFizzx
Fredi Fizzx
Why do you want to give up on this? Work it backwards. e and 1/hbar
support your model, not the other way around. All we are saying here
is that it is meaningless for you to say that your model gets and
supports e and 1/2hbar when you are putting them in the first place in
your equations. You have two things going for you (or against you).
Your calculation of the proton "core" wavelength and splitting into
two loop areas. Use those to show us something without starting with
equations that already have e and spin in it. And be careful when you
are selecting constants. Some are measured and some are defined from
measured constants. Alpha, Bohr magneton and Compton wavelength are
defined from other "measured" constants. They contain other
constants. You can easily look on the NIST website for this.
Your Slide 13 really doesn't show us much. Geometry or not.
FrediFizzx
Tnlockyer
>Date: 3/31/2002 12:25 PM Pacific Standard Time
>Message-id: <fe51764e.02033...@posting.google.com>
>Thomas, here is a mag pot energy derivation from other than your book.
>
>http://www.flashrock.com/upload/potenergy.pdf
>
>I am deriving both magnetic and electric potential energy from the
>known constants and their known relationships without using anything
>from your models. It is really a simple algebra exercise to do this.
>So your derivation of mag and elect pot energy is definitey *not*
>unique to your models. And "my" derivation also shows that mag and
>elect pot energy are equal.
Good, I had to use the fact that they HAD to be equal. So we are on the same
page.
Stop the presses:
Fredi, Mike Larry, et al.
It just occurred to me what we are forgetting in our arguments.
Sure x = x
but when x is the product of two numbers, there are an infinite set of numbers
whose product is that particular (x).
That is the case with the fundamental charge (e)=(Ampere x time)
And in the case of Planck constant (h) = (energy x time)
And time= lambda / c , in both cases.
So lambda can have any value, what-so-ever and will always yield the (e) and
(1/2Hbar) but:
Only if one uses the VPP topology,
http://members.aol.com/tnlockyer/cubedimensions.gif
Tnlockyer
>Date: 3/24/2002 10:51 AM Pacific Standard Time
>Message-id: <fe51764e.02032...@posting.google.com>
>
>
>mor...@world.std.com (Michael Moroney) wrote in message
>news:<GtHIu1...@world.std.com>...
>> Oh boy! More "x=x" proofs!
>>
>> -Mike
>
>Yep, Mike is right. Your Js term has the electron charge in it
>already. Look here for the definition of the fine structure constant.
Stop the presses:
Fredi, Mike Larry, et al.
It just occurred to me what we are forgetting in our arguments.
Sure x = x
but when x is the product of two numbers, there are an infinite set of numbers
whose product is that particular (x).
That is the case with the fundamental charge (e)=(Ampere x time)
And in the case of Planck constant (h) = (energy x time)
And time= lambda / c , in both cases.
So lambda can have any value, what-so-ever and will always yield the (e) and
(1/2Hbar) if one uses the VPP topology.
Michael Moroney
>> fredi...@hotmail.com (Fredi Fizzx)
>>> Oh boy! More "x=x" proofs!
>>Yep, Mike is right. Your Js term has the electron charge in it
>>already. Look here for the definition of the fine structure constant.
>Stop the presses:
>Fredi, Mike Larry, et al.
>It just occurred to me what we are forgetting in our arguments.
>Sure x = x
>but when x is the product of two numbers, there are an infinite set of numbers
>whose product is that particular (x).
I don't know what the heck you are trying to say here, other than somehow
trying to justify all your "x=x" 'proofs' as somehow valid./
>but when x is the product of two numbers, there are an infinite set of numbers
>whose product is that particular (x).
>That is the case with the fundamental charge (e)=(Ampere x time)
That is a poor choice, since it is more sensible to regard charge as a
fundamental unit Q, and make current Q/time.
Perhaps you need to see why "x=x" proof aren't useful. I could easily
write the equation for the electrostatic force between two charges with
the following variables:
q1 = charge 1 (coulombs)
q2 = charge 2 (coulombs)
Ss = my shoe size (m)
r = distance between charges (m)
Eo = permittivity of free space
F = force (newtons)
F = (q1*Ss)*(q2*Ss)/(4 pi Eo (Ss*r)^2)
which proves the electrostatic force depends on my shoe size, and in fact
works for any shoe size so the electrostatic force would work correctly
when I was a kid and growing.
-Mike
larry shultis
> > fredi...@hotmail.com (Fredi Fizzx)
> >Date: 3/24/2002 10:51 AM Pacific Standard Time
> >Message-id: <fe51764e.02032...@posting.google.com>
> >
>
> >
> >mor...@world.std.com (Michael Moroney) wrote in message
> >news:<GtHIu1...@world.std.com>...
> >> Oh boy! More "x=x" proofs!
> >>
> >> -Mike
> >
>
> >Yep, Mike is right. Your Js term has the electron charge in it
> >already. Look here for the definition of the fine structure constant.
>
>
> Stop the presses:
>
> Fredi, Mike Larry, et al.
> It just occurred to me what we are forgetting in our arguments.
>
> Sure x = x
> but when x is the product of two numbers, there are an infinite set of
numbers
> whose product is that particular (x).
>
> That is the case with the fundamental charge (e)=(Ampere x time)
Again you are having trouble telling the difference between units and
magnitudes. The fundamental charge "e" is e=1.602E-19 As where
"As" is 1ampere * second. You were talking about the fundamental
charge, were you not? You must deal with the magnitude of the unit, in
this case the fundamental charge "e", and not pretend that it has no
conversion factor when comparing it to one ampere-second.
Then, if you had put in the conversion factor, with e=1.602E-19 As,
would you be able to say, when speaking about the fundamental
charge, that when you have e / s =1.602E-19 A that e and s can be
an infinite number of different values, remembering that A= one ampere?
There is only one value of s.
>
> And in the case of Planck constant (h) = (energy x time)
Same confusion of dimensions and units.
>
> And time= lambda / c , in both cases.
A little bit stuck in understanding what algebra is? So you are saying
that lambda = time * c and lambda can be any value because the speed
of light and time can be any values, that there is no
particular size for your particles because time can have any value?
If that is so how can you say that you have calculated the sizes of
the proton, its core particle, or the neutron. Your argument would
let them be any sizes including approaching point-like.
This attempt will not circumvent the mathematical fact that the proton
core geometry did not take part in your calculations of elementary charge
and spin from the proton core geometry.
larry shultis
"larry shultis" <gold...@charter.net> wrote in message
news:uahah0q...@corp.supernews.com...
>
> "Tnlockyer" <tnlo...@aol.com> wrote in message
> news:20020401053323...@mb-cj.aol.com...
> > > fredi...@hotmail.com (Fredi Fizzx)
> > >Date: 3/24/2002 10:51 AM Pacific Standard Time
> > >Message-id: <fe51764e.02032...@posting.google.com>
> > >
> >
> > >
> > >mor...@world.std.com (Michael Moroney) wrote in message
> > >news:<GtHIu1...@world.std.com>...
> > >> Oh boy! More "x=x" proofs!
> > >>
> > >> -Mike
> > >
> >
> > >Yep, Mike is right. Your Js term has the electron charge in it
> > >already. Look here for the definition of the fine structure constant.
> >
> >
> > Stop the presses:
> >
> > Fredi, Mike Larry, et al.
> > It just occurred to me what we are forgetting in our arguments.
> >
> > Sure x = x
> > but when x is the product of two numbers, there are an infinite set of
> numbers
> > whose product is that particular (x).
> >
> > That is the case with the fundamental charge (e)=(Ampere x time)
>
> Again you are having trouble telling the difference between units and
That should have been trouble telling the difference between units and
dimensions.
Larry
ThomasL283
>Date: 4/1/2002 10:42 AM Pacific Standard Time
>Message-id: <uahah0q...@corp.supernews.com>
>"Tnlockyer" <tnlo...@aol.com> wrote in message
>> Sure x = x
>> but when x is the product of two numbers, there are an infinite set of
>numbers
>> whose product is that particular (x).
>>
>> That is the case with the fundamental charge (e)=(Ampere x time)
>
>Again you are having trouble telling the difference between units and
>magnitudes. The fundamental charge "e" is e=1.602E-19 As
Not true, Larry. Your ideas are incorrect.
What you fellows are forgetting is that the value of (e) does not change.
So, one can get the (e) value from the product af any A and s that will result
in 1.602E-19 A s
>with e=1.602E-19 As,
>would you be able to say, when speaking about the fundamental
>charge, that when you have e / s =1.602E-19 A that e and s can be
>an infinite number of different values, remembering that A= one ampere?
>There is only one value of s.
>
No that is not what is meant. The current and time can take on any values
whose product results in the fundamental constant (e).
See for example the results in:
http://members.aol.com/tnlockyer/VPPchargespin.gif
Notice that the electrons magnetic field strength gives us a circulating
current is 19.79633 Ampere and it takes 8.09329971 second to transverse the VPP
electron loops.
Giving the fundamental stored charge (e) as the Ampere times second product in
the loops.
Notice that I use no fine structure constant in the charge calculations.
Just the topology of the model and the Bohr magneton, to deduce the electric
and magnetic field strengths.
An second example of getting the fundamental charge, from the electric field
strength, is also given.
The spin angular momentum is also shown as derived from the product of the
electric and magnetic field strengths, and the VPP volume topology.
I believe that the fundamental constants are all related BECAUSE of the
geometry of the particles. The VPP model geometry supports ALL of the
electron's constants, EXACTLY.
See the geometry in:
http://members.aol.com/tnlockyer/cubedimensions.gif
It really is fun to see how the constants match the geometry, as demonstrated
in the VPPchargespin.gif link.
If you fellows should try to falsify the VPP model, by failing to get some
related constant.
I have tried for 25 years, and VPP has never disappointed my math checks.
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,
ThomasL283
>
>>Thomas, here is a mag pot energy derivation from other than your book.
>>
>>http://www.flashrock.com/upload/potenergy.pdf
>>
I got a good look at it and you used the equation from my book, verbatum, good.
I was pleased to see that you verified my modification of the equation from
Halliday and Resnik as correct.
My copyright was in 2000, so I am not worried that anyone is going to scoop
me, unless they can find a reference before that date.
BTW. Check out this link.
http://members.aol.com/tnlockyer/VPPchargespin.gif
I use the magnetic and electric field srengths of the electron to calculate the
fundamental charge and spin angular momentum.
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,
larry shultis
Why do you insist on using the symbols "A" and "s" which are unit names
for variable names. It just causes a lot of confusion.
For example, your equation s=8.09...E-21 s where the left hand s is a
variable
name and the right hand s is a unit name. Things do not work the same in
equations the same as they do in replacement operations in programming
languages. There you have a variable s, say, and it has a certain value. The
calculation is then done and the calculated value is then assigned to s as
its
new value. That is not an equation but rather a replacement operation.
Since the definition of the Bohr magneton contains both e and h, it would
appear that you have introduced both the charge and the spin into your
calculations.
Larry
larry shultis
> >
> >> fredi...@hotmail.com (Fredi Fizzx)
> >>Date: 3/31/2002 12:25 PM Pacific Standard Time
>
> >
> >>Thomas, here is a mag pot energy derivation from other than your book.
> >>
> >>http://www.flashrock.com/upload/potenergy.pdf
> >>
>
> I got a good look at it and you used the equation from my book, verbatum,
good.
>
>
> I was pleased to see that you verified my modification of the equation
from
> Halliday and Resnik as correct.
>
> My copyright was in 2000, so I am not worried that anyone is going to
scoop
> me, unless they can find a reference before that date.
>
> BTW. Check out this link.
>
> http://members.aol.com/tnlockyer/VPPchargespin.gif
>
> I use the magnetic and electric field srengths of the electron to
calculate the
> fundamental charge and spin angular momentum.
But you sneak them in with the Bohr magnaton.
Larry
Fredi Fizzx
> >
> >> fredi...@hotmail.com (Fredi Fizzx)
> >>Date: 3/31/2002 12:25 PM Pacific Standard Time
>
>
> >>Thomas, here is a mag pot energy derivation from other than your book.
> >>
> >>http://www.flashrock.com/upload/potenergy.pdf
> >>
>
> I got a good look at it and you used the equation from my book, verbatum, good.
>
>
> I was pleased to see that you verified my modification of the equation from
> Halliday and Resnik as correct.
>
> My copyright was in 2000, so I am not worried that anyone is going to scoop
> me, unless they can find a reference before that date.
Here is the third one of electron rest mass energy times alpha with no
wavelength.
http://www.flashrock.com/upload/mecalpha.pdf
What the hell does this mean? You were calling it magnetic pot energy
and electric pot energy. What do you call this one? And what is the
significance of electron rest mass energy times alpha? Notice that I
derived it *without* setting electric and mag pot energy equal to each
other. All three equations are equal anywise. I think you were
probably "scooped" about fifty years ago. There is nothing special
about it. Any high school algebra student could take the fundamental
constants and their known relationships and derive these equations.
> BTW. Check out this link.
>
> http://members.aol.com/tnlockyer/VPPchargespin.gif
>
> I use the magnetic and electric field srengths of the electron to calculate the
> fundamental charge and spin angular momentum.
Thomas, you are still making the same mistake. You are not predicting
anything. You insert charge and spin and then extract them at the
end. Look carefully at what you are doing. Besides being a number,
the Bohr magneton is also derived from an equation that has charge and
spin in it already. Look at the mecalpha.pdf. It is right there in
black and white. Instead of using numbers for alpha, Compton
wavelength and the Bohr magneton, use the equations.
FrediFizzx
Michael Moroney
>BTW. Check out this link.
>http://members.aol.com/tnlockyer/VPPchargespin.gif
>I use the magnetic and electric field srengths of the electron to calculate the
>fundamental charge and spin angular momentum.
Thomas, that's just more "x=x" proofs. And this one is not even creative,
since I already pointed out your "trick" of sneaking in the charge and
spin via the Bohr Magnetron before, and you do it again here.
-Mike
Vertner Vergon
"Michael Moroney" <mor...@world.std.spaamtrap.com> wrote in message
news:Gu029L...@world.std.com...
Vergon:
Mike, you are all wet and Thomas is correct.
For your information the magnetic moment (charge) is to the spin angular
momentum
as the electric charge (e.s.u.) is to the modular momentum (m_e c)
This direct proportionality justifies saying that spin angular momentum
creates magnetic
moment -- and polar expansion and contraction creates electric charge.
The ratio is one of time. Magnetic moment and electric charge are forces.
F x t = p. Angular momentum and modular momentum are, of course momenta.
I would show that graphically here but it gets all screwed up on NG ASCII.
I have it in an attachment.
If anyone wants to see it (and more) graphically they can request me to
email them
the attachment.
The "and more" is the explanation why the charges on the electron and proton
are equal
(though opposite). That revelation reinforces the initial concept.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Vertner Vergon
"Fredi Fizzx" <fredi...@hotmail.com> wrote in message
news:fe51764e.02040...@posting.google.com...
Vergon:
You guys are whistling in the wind.
Ask for my attachment and get the right answer.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
ThomasL283
>Date: 4/2/2002 8:25 PM Pacific Standard Time
>Message-id: <ual12hl...@corp.supernews.com>
>"ThomasL283" <thoma...@aol.com> wrote in message
>> No that is not what is meant. The current and time can take on any
>values
>> whose product results in the fundamental constant (e).
>>
>> See for example the results in:
>> http://members.aol.com/tnlockyer/VPPchargespin.gif
>
>
>Since the definition of the Bohr magneton contains both e and h, it would
>appear that you have introduced both the charge and the spin into your
>calculations.
>Larry
No, Larry, look again. The Bohr has dimensions of (A m^2).
The charge (e) has been dissected by the derivation algebra, leaving only a
current (A) times a loop area. This defines the magnetic moment. There is no
(h) involved.
That's why I used the new approach in (VPPchargespin.gif) so you guys could
not bring up that old false argument.
The VPP electron model's spinning volume (Vol) and current loop (Areas) are
used to retrieve the magnetic field strength, to get the Ampere from the Bohr.
(UB)
I could have gone directly to:
A=UB / (Area of VPP electron loops) to get the current. Try it.
But I wanted to have both the magnetic H and the electric E field strengths
for later calculating the (well known) power density
Jeh = 0.5(E x H)
and thus to obtain the electron's rest mass energy (Jm) from the VPP electron
Volume, and then to finally obtain the spin angular momentum, as (Jm x time) /
4pi.
>>
>> Notice that the electrons magnetic field strength gives us a circulating
>> current is 19.79633 Ampere and it
takes 8.09329971E-21 second to trans verse the VPP electron loops.
>> Giving the fundamental stored charge (e) as the Ampere times second
>product in
>> the loops.
>> I believe that the fundamental constants are all related BECAUSE of the
>> geometry of the particles.
The VPP model geometry supports ALL of the
>> electron's constants, EXACTLY.
>>
>> See the geometry in:
>>
>> http://members.aol.com/tnlockyer/cubedimensions.gif
>>
You fellows should try to falsify the VPP model, by failing to get some
related constant, from the geometry, if you can.
If you have trouble, I will try to help you out.
ThomasL283
>Date: 4/2/2002 11:26 PM Pacific Standard Time
>Message-id: <fe51764e.02040...@posting.google.com>
>
>
>thoma...@aol.com (ThomasL283) wrote in message
>> I was pleased to see that you verified my modification of the equation
>from
>> Halliday and Resnik as correct.
>>
>> My copyright was in 2000, so I am not worried that anyone is going to
>scoop
>> me, unless they can find a reference
>before that date.
>
>Here is the third one of electron rest mass energy times alpha with no
>wavelength.
>
>http://www.flashrock.com/upload/mecalpha.pdf
>What the hell does this mean? You were calling it magnetic pot energy
>and electric pot energy. What do you call this one?
> And what is the
>significance of electron rest mass energy times alpha?
Fredi: The alpha times the rest mass energy retrieves the energy in the E or H
field that accompanies the charged particle. To get the field strength from the
energy requires first obtaining the power density (Pe) using VPP geometry.
Then, E = (Pe x Zo)^0.5 and H = (Pe/Zo)^0.5
(calculation of power density requires knowing the Vol (volume) of the VPP
particle) (see page 35 in the book) .
>> BTW. Check out this link.
>>
>> http://members.aol.com/tnlockyer/VPPchargespin.gif
>>
>> I use the magnetic and electric field srengths of the electron to calculate
>the
>> fundamental charge and spin angular momentum.
>Thomas, you are still making the same mistake. You are not predicting
>anything. You insert charge and spin and then extract them at the
>end. Look carefully at what you are doing. Besides being a number,
>the Bohr magneton is also derived from an equation that has charge and
>spin in it already.
No I am not mistaken.
You guys are looking at the derivation algebra for the Bohr.
The resulting Bohr has the dimensions of (A m^2) so the charge (e) has been
lost in the dissection. And the spin is nowhere to be found in the Bohr
algebra :
Bohr = 1/2 (e c (lambda / 2 pi))
The Bohr ends up with dimensions of (A m^2) period.
You still have not realized that the geometry of the VPP electron maintains the
constants in their proper ratios.
http://members.aol.com/tnlockyer/VPPchargespin.gif
I get the circulating current A = 19.79633175 Ampere.
I could have just as well divided the Bohr by the VPP current loop areas. (A
m^2) / (m^2) = A
A = UB/ Area (try it)
Where the Area = (lambda^2) / 4pi from the VPP electron model geometry.
Where the hell is (e) in all of this? To get (e) you need the proper seconds
(s) in time. I use s = lambda / c and this (s) = 8.09329971536 E-19 s is then
multiplied by the previous determined unique 19.79633175 A .
This calcualtion gives the fundamental charge as 1.6021764516 E-19 (A s)
The time can also be derived from the geometry of the VPP electron loop areas.
(Note, The velocity of the current nodes are (sqr2 c), because the mass
centroids have a velocity of (c))
The electron's rest mass energy can be obtained from the power density of the
electron. Jeh = 0.5 (E x H)
Note E and H are given in the book, page 35.
But the same values are obtained from the VPP electron geometry H = UB/Vol and
E = (UB Zo) / Vol
where Vol is the volume of the VPP electron model geometry. (as shown in
VPPchargespin.gif)
The electron's mass energy obtained is; Jm = 8.18710413148E-14 Joule.
Getting this correct (Jm) value proves the VPP real geometric particle model
approach is correct.
Jm = (Jeh x Vol)/ c a
Where Vol = (lambda^3) / 16 pi^2 is from the VPP electron model geometry.
The VPP model geometry makes the electron a real particle that occupies space
and is physically spinning, on account of the vector directions in front and
back cube faces. See:
http://members.aol.com/tnlockyer/cubedimensions.gif
The VPP model is three dimensional
apart from the usual (non geometric, two dimensional) dimensional analysis.
ThomasL283
>Date: 4/3/2002 8:10 AM Pacific Standard Time
>Message-id: <Gu029L...@world.std.com
>
>thoma...@aol.com (ThomasL283) writes:
You are wrong. The Bohr has the dimensions of (A m^2) so the charge has been
dissected in the derivation algebra.
Bohr = 1/2 (e c (lambda/2pi))
Where is the spin you claim?
The current A in the electron is 19.79633 Ampere.
The current in the proton core is 1.015236633 E4 Ampere.
Why? because the circulation time is 8.0932997E-21 seconds in the electron
and 1.57813105E-23 seconds for the proton core.
And both the electron and the proton core particles yield the same product of
1.602176462E-19 (A s)
What you all are not realizing is that the PRODUCT of (Ampere x seconds) is a
constant.
So there are an infinite number of Ampere and seconds that yield the same
product of 1.602176462E-19 (A s)
When x is the product of two numbers the (x=x) argument does not hold up,
because x can be composed of infinite sets of number combinations, whose
product is a the required constant.
-Tom:
Michael Moroney
>>mor...@world.std.spaamtrap.com (Michael Moroney)
>>Date: 4/3/2002 8:10 AM Pacific Standard Time
>>Message-id: <Gu029L...@world.std.com
>>Thomas, that's just more "x=x" proofs. And this one is not even creative,
>>since I already pointed out your "trick" of sneaking in the charge and
>>spin via the Bohr Magnetron before, and you do it again here.
>>
>>-Mike
>You are wrong. The Bohr has the dimensions of (A m^2) so the charge has been
>dissected in the derivation algebra.
What? The Bohr Magnetron is ***defined*** as (e h)/(4 pi me). That's
defined, not even something measured. Do you know what that little "e" is?
>Bohr = 1/2 (e c (lambda/2pi))
>Where is the spin you claim?
There you go trying to hide variables again. Lambda here must be the
Compton wavelength, which is, once again, ***defined*** as h/(me c).
Making the substitution we get back the original definition, (e h)/(4 pi me)
The spin of the electron is in units of hbar, which is h/2pi.
And secondly, by writing the equation that way, you *knew* that the Bohr
Magnetron was defined in terms of e, so you have *no excuse* for not knowing
it had the electron charge built into it.
-Mike
larry shultis
> >"larry shultis" gold...@charter.net
> >Date: 4/2/2002 8:25 PM Pacific Standard Time
> >Message-id: <ual12hl...@corp.supernews.com>
>
> >"ThomasL283" <thoma...@aol.com> wrote in message
>
> >> No that is not what is meant. The current and time can take on any
> >values
> >> whose product results in the fundamental constant (e).
> >>
> >> See for example the results in:
>
> >> http://members.aol.com/tnlockyer/VPPchargespin.gif
> >
>
> >
> >Since the definition of the Bohr magneton contains both e and h, it would
> >appear that you have introduced both the charge and the spin into your
> >calculations.
> >Larry
>
> No, Larry, look again. The Bohr has dimensions of (A m^2).
Look again Thomas. The Bohr magneton = e * h bar / (2 * electron mass).
There are e and h staring out at you. Pretending that you are not just
extracting them from the Bohr magneton is wrong. How do you get
the magnitudes of your constants without using their definitions? So
you just disguise the e and h in the numbers and then get rid of the
contributions of some constants leaving e or h.
Larry
Fredi Fizzx
> > fredi...@hotmail.com (Fredi Fizzx)
> >Date: 4/2/2002 11:26 PM Pacific Standard Time
> >Message-id: <fe51764e.02040...@posting.google.com>
>> >thoma...@aol.com (ThomasL283) wrote in message
>
> >anything. You insert charge and spin and then extract them at the
> >end. Look carefully at what you are doing. Besides being a number,
> >the Bohr magneton is also derived from an equation that has charge and
> >spin in it already.
>
> No I am not mistaken.
> You guys are looking at the derivation algebra for the Bohr.
> The resulting Bohr has the dimensions of (A m^2) so the charge (e) has been
> lost in the dissection. And the spin is nowhere to be found in the Bohr
> algebra :
>
> Bohr = 1/2 (e c (lambda / 2 pi))
Where did you get the above from?
> The Bohr ends up with dimensions of (A m^2) period.
Wrong. Take a look at this:
http://www.flashrock.com/upload/bohrmag2.pdf
Read it and weep. We can easily use what the Bohr magneton is equal
to and insert it into your equation. All you are showing is that e=e
and spin equals spin.
FrediFizzx
Tnlockyer
>Date: 4/3/2002 8:05 PM Pacific Standard Time
>Message-id: <uank7so...@corp.supernews.com>
>
>"ThomasL283" <thoma...@aol.com> wrote in message
>> >Since the definition of the Bohr magneton contains both e and h, it would
>> >appear that you have introduced both the charge and the spin into your
>> >calculations.
>> >Larry
>>
>> No, Larry, look again. The Bohr has dimensions of (A m^2).
>
>Look again Thomas. The Bohr magneton = e * h bar / (2 * electron mass).
No. I define the Bohr as UB=1/2 (ec(lambda/2pi))
What you and others are doing is simply verifying that the constants are all
related to each other. But of course, and I say that in so many words in my
books. (VPP page 2, and Vector Particles and Nuclear models, page 4).
What you apparently fail to see is my point that the GEOMETRY of the PHYSICAL
VPP model shows exactly the required mathematical relationships as those
relating the constants.
See a new thread started on the subject.
And for god's sake will you all admit the VPP geometry WORKS to keep the
constants in their proper ratios?
At some point one has to introduce at least one constant to set the
International System (SI) metric. It is no sin, it is a necessity.
Regards. Tom:
Thomas Lockyer (75 and retired) See "Vector Particles and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about, and express it in numbers,
Michael Moroney
>>Look again Thomas. The Bohr magneton = e * h bar / (2 * electron mass).
>No. I define the Bohr as UB=1/2 (ec(lambda/2pi))
Nobody cares what you want to redefine the Bohr Magneton as. It already
has a definition, which Larry gave. At least your redefinition isn't
incorrect.
>What you and others are doing is simply verifying that the constants are all
>related to each other.
Actually it is you who is doing that, with your "x=x" proof and
redefinitions.
>And for god's sake will you all admit the VPP geometry WORKS to keep the
>constants in their proper ratios?
Why? All I see it being good for is to generate "x=x" proofs and waste
electrons in this newsgroup.
>At some point one has to introduce at least one constant to set the
>International System (SI) metric. It is no sin, it is a necessity.
But if you introduce a constant, manipulate the math and then extract
the same constant, you have proved nothing.
-Mike
Tnlockyer
>Date: 4/5/2002 12:23 PM Pacific Standard Time
>Message-id:
>>What you and others are doing is simply verifying that the constants are all
>>related to each other.
>
>Actually it is you who is doing that, with your "x=x" proof and
>redefinitions.
>
Ok, lets see you give me the circulating amperes from the Bohr.
Can't do it unless you use the VPP geometry two current loop Areas, that's my
point. A = UB/ Area
>>And for god's sake will you all admit the VPP geometry WORKS to keep the
>>constants in their proper ratios?
>
>Why? All I see it being good for is to generate "x=x" proofs and waste
>electrons in this newsgroup.
Sorry mike, the VPP model is much more than you want to admit. See:
http://members.aol.com/tnlockyer/cubedimensions.gif
The positron and electron models each show the source of 1/2hbar spin,
vectorially. Notice the vector directions going in the same direction in
front and back cube faces.
Also look at the model for the positron and electron in detail.
The spin involves H vectors in the positron spinning faces,
and involves E vectors in the electron spinning faces.
VPP is the first model that suggests a reason for positive and negative charge
currents.
Look, I can't take credit for making these structures, they were given to me
by simply combining the photon in all possible ways.
http://members.aol.com/tnlockyer/cfives.gif
Since then, the models have taken on a life of their own.
I am still trying to see if the model fails to support the constants.. Has not
yet!
I know the constants are all related, but can YOU tell me why?
I believe the constants are all related because of the (unique) cube (time
and space) geometry, shown in the VPP model.
Regards: Tom:
larry shultis
> >"larry shultis" gold...@charter.net
> >Date: 4/3/2002 8:05 PM Pacific Standard Time
> >Message-id: <uank7so...@corp.supernews.com>
> >
>
> >"ThomasL283" <thoma...@aol.com> wrote in message
>
> >> >Since the definition of the Bohr magneton contains both e and h, it
would
> >> >appear that you have introduced both the charge and the spin into your
> >> >calculations.
> >> >Larry
> >>
>
> >> No, Larry, look again. The Bohr has dimensions of (A m^2).
> >
>
> >Look again Thomas. The Bohr magneton = e * h bar / (2 * electron mass).
>
> No. I define the Bohr as UB=1/2 (ec(lambda/2pi))
>
> What you and others are doing is simply verifying that the constants are
all
> related to each other. But of course, and I say that in so many words in
my
> books. (VPP page 2, and Vector Particles and Nuclear models, page 4).
>
> What you apparently fail to see is my point that the GEOMETRY of the
PHYSICAL
> VPP model shows exactly the required mathematical relationships as those
> relating the constants.
>
> See a new thread started on the subject.
>
> And for god's sake will you all admit the VPP geometry WORKS to keep the
> constants in their proper ratios?
The constants are in proper ratios because your model just uses the proper
ratios of constants beginning with the Compton wave length. There is nothing
in the model that places them in the wrong ratios, i.e., you can spit out
the
various constants by algebraic operations because they were put into the
model. The model does not calculate the constants from any fundamental
fact of reality.
Larry
Jacques Distler
<tnlo...@aol.com> wrote:
>Since then, the models have taken on a life of their own.
Aha! An AI program, rather than an actual sentient human being has been
generating all these Usenet posts.
Now it all makes sense.
> I am still trying to see if the model fails to support the constants.. Has not
>yet!
>
>I know the constants are all related, but can YOU tell me why?
He has, numerous times.
They are related by virtue of their *definitions*.
The Compton wavelength (to pick one) is NOT an experimentally measured
quantity. It is *defined* in terms of the mass of the particle.
Now that we know you are an AI program, it is clear why no amount of
patient explanation of *why* what you are saying is gibberish has the
slightest impact on you.
In that spirit:
Virtual reality isnšt what it used to be.
Mike, this is getting out of hand.
I want a mega-meal in a mega-mall.
The numerical part modulates the basic units part.
Adopt my lifestyle or Išll have to press charges.
I want my VPP pressed and folded.
Išm afraid! I need something in a heavy cream sauce.
I refuse to put a bastardized charge in my variables list.
I can silence Joan Rivers with a single slice of Kraft cheese.
The null is between the electric and magnetic forces.
Zombies rule Belgium.
For gods sake don't let the Nazi book burners deter you!
Frivolity is a stern taskmaster.
The VPP models for the positron and electron were
given to us blindly.
Išm Zippy the Pinhead and Išm totally committed to the
festive mode.
The eV served its purpose in the old funky sliderule days.
Could you repeat that in Portuguese?
It is not as trivial as you all think.
I just felt a paradigm shift.
The VPP model PROVES that any sized spinning CUBE will fit in a
bowling alley.
My boxer shorts just went on a rampage through a Long Island
bowling alley.
I bowl therefore I am.
But, geometry is not algebra.
Feelings are cascading over me.
It just occurred to me what we are forgetting in our arguments.
Glazed donuts are the building blocks of the universe.
Only if one uses the VPP topology!
Gimme a glazed!
Stop the presses:
Nobody brings small problems into a laundromat.
And for god's sake will you all admit the VPP geometry WORKS?
Consciousness is vastly overrated.
I can't take credit for making these structures.
I hope my sensitive female side is wearing sensible leather pumps.
Reality distorts my sense of television.
America. I love it, I hate it, I love it, I hate it. When do
I collect unemployment?
I just forgot my Social Security number.
The VPP model works as advertized.
JD
--
PGP public key: http://golem.ph.utexas.edu/~distler/distler.asc
Fredi Fizzx
You have it backwards again. The constants support your model (which
is OK). Your model does not support the constants since you put them
in and then extract them out. If you agree that the Bohr Magneton is
equal to (e*h)/(4*pi*me) (which anyone knows it is), then anyone can
freely substitute (e*h)/(4*pi*me) for uB anytime uB is shown. Here it
is for you one more time. I will keep posting this until you get it.
http://www.flashrock.com/upload/vppprovesnada.pdf
FrediFizzx
Michael Moroney
>>>What you and others are doing is simply verifying that the constants are all
>>>related to each other.
>>
>>Actually it is you who is doing that, with your "x=x" proof and
>>redefinitions.
>>
>Ok, lets see you give me the circulating amperes from the Bohr.
Since the radius of the electron is immeasurably close to zero, its
area is also immeasurably small (possibly zero). So "current" is
a non-starter here.
>http://members.aol.com/tnlockyer/cubedimensions.gif
You keep posting this url as if it's proof of something. So far it
appears to be gibberish.
>I know the constants are all related, but can YOU tell me why?
Yes. Simple algebra.
-Mike