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

From: greywolf42 (mingstb_at_marssim-ss.com)
Date: 09/27/04 

Date: Mon, 27 Sep 2004 03:07:19 GMT

"FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:2rmqruF1cjem6U1@uni-berlin.de...
> "Paul Stowe" <ps@acompletelyjunkaddress.net> wrote in message
> news:poobl0193du3ge0e3frn6me3rf1gk7t9ip@4ax.com...

{snip prior reasoning}

> | 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

Fredi, where did you come up with the "tesla" in the above pseudo-equation?
Tesla is a unit, not a parameter. (Hint: you can't mix 'epsilon_naught'
with tesla.)

> 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.

That would be a question for you to answer. After all, what is
the physical significance of (or your purpose for) shifting from 'natural'
units to SI units? Paul has not set Tesla = 1 -- in any units. You have.

> A tesla
> is just a SI derived unit. ??? If the B field is dimensionless, where
> would it actually equal 1? And why?

Feel free to answer yourself.

> | Do I have a 'solid' model for deriving k yet, no. But I find this
> | coorelation most interesting. It would be the very first time that
> | the Gravitational constant is linked to other EM/QM terms.
>
> I doubt that it has been the first time someone has tried to link G to
> EM/QM terms.

True, but irrelevant.

>You might be on to something here, but you have got to get a grip on
> the B field being dimensionless and where it might actually equal one.

Actually, Paul has no such need. You are the only one proposing such an
arrangement. 

--
greywolf42
ubi dubium ibi libertas
{remove planet for e-mail}

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