Re: LeSagian Gravitational Field Momentum Flux linked to EM/QM constants

From: FrediFizzx (fredifizzx_at_hotmail.com)
Date: 09/29/04 

Date: Tue, 28 Sep 2004 19:26:53 -0700

"FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:2rrsdgF1ek01jU1@uni-berlin.de...
| "Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
| news:gmbhl0lttr52mo4pv4k4kkp8dmift84ept@4ax.com...
| | On Sun, 26 Sep 2004 20:13:35 -0700, "FrediFizzx"
<fredifizzx@hotmail.com>
| wrote:
| |
| | >"Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
| | >news:ssgel05ghq71imjv7aja6qau8gkf9vuehn@4ax.com...
| |
| | [Snip...]
| |
| | >|> If we put all your units back to SI then we get,
| | >|>
| | >|> G = (1/(2pi*eps0*tesla^2))(S*(e*tesla)^2/2h)^2 ~= 6.673E-11
| m^3/kg-sec^2
| | >|>
| | >|> Where eps0 is the SI electric constant, e is electronic charge,
tesla
| | >|> is the SI unit of magnetic flux density (weber/meter^2) and h is
| Planck's
| | >|> constant. S is your term
| | >|>
| | >|> But one has to ask here: What is the *physical* significance of
| setting
| | >|> tesla = 1? It might be fine to choose the B field to be
| dimensionless,
| | >|> but you have to have a physical reason for choosing a tesla to equal
| 1.
| | >|
| | >| I didn't choose, those that set SI 'chose'. I think it has to do
with
| the
| | >| physical parameters of the materials used to set the SI system. Just
| like
| | >| I didn't choose to have 1 Coulomb = 1.602E-19 kg/sec. Had I choosen
I
| | >| would not have done so...
| | >
| | > I think you mean a couloumb = 1kg/sec. Electronic charge is equal to
| | > 1.602E-19 kg/sec.
| |
| | Yeah... I'd have set 1 Coulomb = 1.60E-19 kg/sec
| |
| | >|> A tesla is just a SI derived unit.???
| | >|
| | >| Precisely, just like a Coulomb is...
| | >|
| | >|> If the B field is dimensionless, where would it actually equal 1?
And
| | >|> why?
| | >|
| | >| Ah, why... That's the big question. I haven't bothered to go back
| over
| | >| all of the definitions of SI (example, 1 cc = 1g of H2O @ STP) to try
| to
| | >| find out. I kinda' don't care.
| | >
| | > You should care as this is a big part of your theory. Try this on and
| see
| | > if you like it. After thinking about it for a while, it seems to have
| more
| | > to do with the definition of the ampere than with B fields per see.
| | > A tesla is kg/(sec^2*amp), so if we set equal to one then an amp =
| kg/sec^2.
| | > We get the pure mechanical definition of an ampere. Now does this
| really
| | > hold up for Ampere's law? It seems like it might as this modification
| of SI
| | > gives us coul = kg/sec and amp = kg/sec^2. Which gives us a magnetic
| | > constant, k_m = sec^2*meter/kg and electric constant,
| | > k_e = 10^-7*c^2*sec^2*meter/kg or ~ 8.99*10^9*meter^3/kg.
| | > Everything is pretty clean except for the electric constant but is
| consistent.
| | > OK, so I think there is physical justification for setting a tesla =
1.
| |
| | Yup.
| |
| | > Now the exciting part. Assuming that a single aetheron can make the
| | > electron "vortex", ...
| |
| | Then it's not 'an aetheron' it is an aether entity. Vortices are
fluidic
| | structures consisting of many individual particles. But, OK I'll go
with
| | the thought...
|
| OK, thanks. You really need to try to imagine what is going on if space
and
| time are being defined at this small level. At least the space and time
| that is important to us. If a single aetheron is going faster than c in
its
| little "space" that it is defining, then macroscopically it *is* like a
| bunch of particles. Thus we can have a vortex from this situation.
|
| | > ... then it is simple to figure out the mass quantum of the
| | > aetheron. If coul = kg/sec, then we just take electronic charge and
the
| | > compton frequency of the electron and solve for mass. I am going to
use
| SI
| | > units so others can follow more easily.
| | >
| | > m_q = e*tesla/w_C = e*tesla*hbar/(m_e*c^2) ~= 1.1578E-4 eV/c^2.
| | >
| | > Where m_q is the mass quantum, e is electron charge, hbar is reduced
| Planck
| | > constant, m_e is electron mass. So we get a really tiny tiny mass
| quantum
| | > this way. Dang, it is about 10 billion times smaller than the mass of
| the
| | > electron but produces the electron! The only way to account for this
is
| | > that it is going faster than the speed of light in the electron!
Since
| | > spacetime is being defined at this level, then there is no reason to
| assume
| | > that SR is true here. And I don't think anyone has actually proven
that
| it
| | > holds at very small scales. If it is being defined at this level,
then
| we
| | > don't have to make that assumption as long as it emerges from this on
| the
| | > macroscopic scales.
| |
| | Don't confoze inertia with fundamental mass. Inertia may be a result
of
| | field configurations NOT wanting to be changed. Thus the actual 'mass'
| | (not to be confosed with matter inertia mass) could be different.
|
| If this mass quantum is true, then most of the mass we "sense" is not
| fundamental mass. But it is what you would call effective mass.
|
| | > OK, so orginally I was thinking the aetheron might be massless but I
| think
| | > you are right and it does have a tiny tiny mass. And I think the
above
| | > calculation is correct for it. Now all we need is the exact geometry
of
| the
| | > coupled vacuum oscillators (vortices) and everything should fall into
| place.
| | > I am 99.99 percent sure that two counter circulating aetherons make
the
| | > primary EM vortex cells. Since they are going faster than c, then for
| sure
| | > they are stringy-like and then cloud-like to our perspective and only
| one
| | > aetheron can make a vortex. The end product of the spin matrix of
| coupled
| | > vacuum oscillators is that massless photons go at c macroscopically.
| This
| | > should be able to also fit neatly with the Standard Model and thus a
| model
| | > for matter is already there. The only thing missing is the exact
model
| for
| | > the quantum vacuum.
| |
| |
| | The area of a ring torus is 4pi^2Rr. Two coupled would simply be
| 2(4pi^2Rr)
| | with the geometric part being 8pi^2... Then to evaluate in arbitrarily
| | defined three dimensions, we need the eigen-value Sqrt(3)... thus,
| |
| | Sqrt(3)8pi^2 would be primal to this configuration.
|
| Yes, to that configuration but is it the right one? We have to be able to
| get protons, etc. from it also.
|
| | > Now, is there a way to plug in this mass quantum into Lesagian
gravity?
| You
| | > can handle that part if so. ;-)
| |
| | So, you have a mass of the vortex @ ~2.08E-40 kg?
|
| No. That is the "bare" mass of the aetheron. If the vortex is an
electron,
| then its mass is the electron mass relative to the rest of everything
else.

Now, are there or would there be any "free" aetherons? Or are they all
bound up in the vacuum spin matrix vortices?

FrediFizzx

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