Re: How the Aether Physics Model came to be

From: FrediFizzx (fredifizzx_at_hotmail.com)
Date: 02/16/05 

Date: Wed, 16 Feb 2005 01:22:05 -0800

"David Thomson" <news5@volantis.org> wrote in message
news:4212bec4$1_1@127.0.0.1...
| "FrediFizzx" <fredifizzx@hotmail.com> wrote in message
| news:377s27F59j245U1@individual.net...
| > Your e^2 = 8pia e.emax^2 is simply the same as e^2 = alpha*hbar*c in cgs
| > units. Your emax is just +,- sqrt(hbar*c).
|
| ??? You really need to double check your math. My equation is in MKS
units,
| but it really doesn't matter. The equation has the same value no matter
| what system of units is used.
|
| e.emax^2 = 1.400 x 10^-37 coul^2
| sqrt(hbar*c) = 1.778 x 10^-13 kg^.5 m^1.5 / sec
|
| You're way off.

Not way off. I said your emax not emax^2. In SI units it is,

4pi*eps0*hbar*c = 3.5 x 10^-36 coul^2

My relativistic quantum charge unit is bigger than yours. And more simple
and reasonable. Your expression is simply;

eps0*hbar*c/2 = 1.3996 x 10^-37 coul^2

Who are you trying to kid with e.emax^2? 4pi*eps0*hbar*c reduces to +,- 1
in natural units. A perfect quantum charge unit. Don't you know that in
MKS or SI units the Coulomb constant, k_e, is 1/4pi*eps0. Always. But
there is no getting around,

e^2 = alpha*hbar*c in cgs units. Or,

e^2 = alpha*4pi*eps0*hbar*c in SI units.

| > Wheeler proposed back in 1957 that
| > sqrt(hbar*c) might be spacetime charge in his Quantum Geometrodynamics
| > paper.
|
| What's the reference? sqrt(hbar*c) has no meaning. It certainly does not
| resolve to charge units.

JA Wheeler, "On the Nature of Quantum Geometrodynamics," Annals Phys. 2,
604-614 (1957).

See above for charge units. Please learn how to convert to and from cgs and
other unit systems.

| > As far as I can see, any spacetime medium theory should be consistent
with
| > most all of quantum theory and relativity. Some re-interpretation is
| > definitely necessary in a few places.
|
| I know you have put a lot of thought into this. But I'm still convinced
of
| the accuracy and utility of the Aether Physics Model. I don't think you
| understand it, though, based upon your equations above.

I told you to send me a copy of the book. I will read it. You might have
some ideas that are useful.

FrediFizzx