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

From: FrediFizzx (fredifizzx_at_hotmail.com)
Date: 10/03/04 

Date: Sat, 2 Oct 2004 22:44:23 -0700

"Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
news:gl6ul0tvjlft46nihi7l2hdrudj9qqj5rf@4ax.com...
| On Thu, 30 Sep 2004 00:19:49 -0700, "FrediFizzx" <fredifizzx@hotmail.com>
wrote:
|
| >"Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
| >news:9ooml0hbtsoe1qmhkbqbprle2amoj8k4mb@4ax.com...
|
| [Snip...]
|
| >|> 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.
| >|
| >| I really recommend looking at Petr Beckmann's "Einstein Plus Two",
| >| Sections 2.6 & 2.7... He has the EM 'field's reaction (inertial
| >| reactive 'mass') as
| >|
| >| uq^2
| >| m = ---- (Eq. 1 Section 2.7)
| >| 6piR
| >|
| >| Where R is the Radii measured from the center of q... What is really
| >| informative (but not for Bozos like cannot conceptualized field lines)
| >| is the illustration on page '137' (funny eh?)
| >
| > Can you scan these parts of Beckmann and email to me?
|
| On the way...

Received. Thanks much. Ok, this is for a spherical test charge of radius
R. So he is saying it takes extra force to accelerate a charged test mass
over an uncharged test sphere of equal rest mass. Now is he saying that
half of the electron's rest mass is electromagnetic? Or is he saying m_e ~=
9.1E-31kg should be doubled? I don't get what you mean by "funny eh?" for
figure on page 137.

| >| Take a rubber band. Let that represent a simple torus structure. Now,
| >| do anything to it other than cut it and you STILL have A = 4pi^2rR.
| >| Knotting the torroid doesn't change its area definition (assuming you
| >| can still know r & R in its simplest form).
| >
| > Yes, but how does this produce a proton. I am going with the assumption
| > from the standard model that all fermions have the same basic geometry.
| > Well, except for maybe neutrinos.
|
| Ah, the million dollar question... I don't yet know. That's my problem
| I haven't yet figured out the physical model of matter.

Heck, go with the standard model and try to make the vacuum geometry fit it.
That is what we are trying to do. We already have a model for matter that
way. It seems to me to just be a natural progression that way. All the
clues are there already in the standard model for a vacuum medium. It only
requires a minor re-interpretation of what gauge bosons are. Though a
possible big drawback following this path, is that it seems to require the
concept of dual space. Without dual space, matter is holes in the vacuum
equilibrium. However with dual space, virtual and "less than virtual"
particles are a real possibility instead of just being math artifacts. Then
the vacuum medium lives in the "other" space of dual space. I have been
personally equating "less than virtual" with "ultra-mundane". Less than
virtual will definitely get you ultra-mundane aetherons. ;-)

| >|>| 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.
| >|
| >| See above on EM reactive inertia (mass).
| >
| > I think I have shown you this before, but I can represent electron
| > rest mass related to vacuum charge = +,- sqrt(hbar*c) in cgs units
| > from,
| >
| > m_e = (e*sqrt(hbar*c)/w_C^2)(2pi/lambda_C)^3*(1/sqrt(alpha))
| >
| > Where w_C is electron compton frequency, lambda_C is electron compton
| > wavelength. That whole expression reduces to lamda/2pi = hbar/m_e*c,
| > the compton formula.
|
| So,
|
| m = h/lamda*c
|
| Lose the hbar & 2pi...

Sure, lose them if you want to. All the higher level physics books use hbar
so everything I have been reading is in terms of that. Just what I am used
to. The point is the compton formula may just be a shortcut expression to
the expanded version. I would think to describe mass we would want a volume
of space that it occupies. Which we have. Then taking alpha to be
geometric and going with the volume, we are left with charge^2/freq^2. And
I would suppose that w_C^2 is not necessarily electron compton frequency
squared but w_C^2 = w_1*w_2. With w_1 being a frequency associated with e
and w_2 being a frequency associated with sqrt(hbar*c). We are still in cgs
units here. This shows that the mass of an electron is entirely
electromagnetic with respect to the vacuum medium.

| > So the rest mass of an electron is electronic charge times vacuum
| > charge divided by frequency squared per a volume. OK, so we also
| > have the mass quantum expression from above,
|
| > m_e = e*tesla*hbar/(m_q*c^2) in SI units,
|
| > that produces the electron rest mass. These must be related somehow. I
| > just can't see the relation yet. ??? I guess it would be helpful to
get
| > them in the same unit system. ;-) I will work on it some more
tomorrow.
|
| OK...

I am still working on this. Nothing seems to work out quite right yet. In
your units, mass is simply charge/freq.

FrediFizzx

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