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

From: FrediFizzx (fredifizzx_at_hotmail.com)
Date: 09/27/04 

Date: Sun, 26 Sep 2004 20:13:35 -0700

"Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
news:ssgel05ghq71imjv7aja6qau8gkf9vuehn@4ax.com...
| On Sat, 25 Sep 2004 20:22:15 -0700, "FrediFizzx" <fredifizzx@hotmail.com>
wrote:
|
| >"Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
| >news:poobl0193du3ge0e3frn6me3rf1gk7t9ip@4ax.com...
| >|
| >| In 2001 I authored a paper (and co-authored another) in which, within
the
| >| LeSage Model of Gravity, it was determined that the linear power flux
[Q]
| >| was 1.6E+08 Watts/m^2 (kg/sec^3) and field omni-directional momentum
flux
| >| [k] was 4piQ/c, or 6.707 [kg/m-sec^2]. These were emperically related
to
| >| the model's predicted gravitational induction heating and non-orbital
| >| linear drag.
| >|
| >| Ref: "Pushing Gravity"
| >|
| >|
http://www.amazon.com/exec/obidos/tg/detail/-/0968368972/103-0163631-0051051?v=glance
| >| Pages 183-200
| >|
| >| For the gravitational heating the equation is as follows,
| >|
| >| J = 2QG/c^2 ~ 2.4E-19 (m/sec^3)
| >|
| >| Where G = Gravitational Constant
| >| c = Light Speed
| >|
| >| and thus radiated equilibrium power flux is,
| >|
| >| w = JM/R [kg/sec^3]
| >|
| >| Where M = Mass of the Body
| >| R = The Body's Spherical Radius
| >|
| >| Given that, in LeSage's model the Graviational constant is actually
| >| the product of two physical terms, the omni momentum flux k and the
| >| material attenuation coefficient [S in units of m^2/kg] we have,
| >|
| >| G = kS^2
| >|
| >| Thus the linear motion induced drag coefficient [a'] is simply,
| >|
| >| a' = -kS(v/c)
| >|
| >| In earlier works I had determined that elemental charge could be
| >| related to the fundamental property of compressibility Ref:
| >| http://www.mountainman.com.au/charge_ps.htm This maps the dimension
| >| of Coulomb in the SI system to mass/length/time dimensions of
| >| kg/sec (harmonic oscillation). This, in turn, puts the SI
| >| expressions of Permitivitty [z] and Permeability [u] as the EM
| >| field's mass density [kg/m^3] and inverse Energy density (just like
| >| all other material media). Thus, as would be expected,
| >|
| >| [Energy Density]
| >| c^2 = ----------------
| >| [Mass Density]
| >|
| >| Given E = mc^2, and 1/u = Energy Density...
| >|
| >| To furher validate the validity of the 'interpretation' (nothing
| >| is changed from SI except the definition of Coulomb, which is
| >| converted from the arbitrary C to kg/sec). We find several
| >| predictions resulting from this specific formulation.
| >|
| >| 1. The Charge to Mass Ratio is a frequency
| >| 2. All constants of nature should be reducible to specific
| >| quantum values of Momentum [P], an Interaction Length [L],
| >| and the "Root Mean Speed" c.
| >|
| >| Indeed, the is seen, Ref: http://www.mountainman.com.au/p_stowe.html
| >|
| >| My discussion today is to show, that, perhaps this method can relate
| >| the LeSagian omni momentum flux (k) determined in 2001 to the QM/EM
| >| parameters, q, h, z. Where h is Planck's Constant and q is elemental
| >| charge. Thus to match k @ ~6.7E+00 kg/m-sec^2 we need to get a fit
| >| in both magnitude and the dimensions. I've found that,
| >|
| >| 1 / q^2 \^2
| >| k = ---- | --- | = 6.744E+00 kg/m-sec^2
| >| z2pi \ 2h /
| >|
| >| Conversely,
| >|
| >| a^2
| >| k = ---- = 6.744E+00 kg/m-sec^2
| >| u2pi
| >|
| >| Where a = the Fine Structure Constant ([a]lpha)
| >|
| >| Thus the Gravitational constant G would become,
| >|
| >| 1 / Sq^2 \^2
| >| G = ---- | ---- | = 6.73E-11 m^3/kg-sec^2
| >| z2pi \ 2h /
| >|
| >| Where S (~3.1456E-06 m^2/kg) remains still undefined in terms of any
| >| other constants.
| >
| > 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.

| > 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 persee.
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.

Now the exciting part. Assuming that a single aetheron can make the
electron "vortex", 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.

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.

Now, is there a way to plug in this mass quantum into Lesagian gravity? You
can handle that part if so. ;-)

FrediFizzx

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