Re: Spacetime asymmetrics (kst).Bell

Hi Mr. Bell.

On Jul 17, 10:00 am, "Steve Bell" <sb...@xxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote in messagenews:a93f3286-5748-42f4-a62a-5c3a7ebb440d@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Hi Mr. Bell.

On Jul 16, 9:59 am, "Steve Bell" <sb...@xxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote in messagenews:c75c2cb1-4b02-4c18-982b-065fd74f0e38@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

To Mr. Bell.

On Jul 15, 8:44 am, "Steve Bell" <sb...@xxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote in
messagenews:604bf8ad-ffe1-4b4b-856b-6412fae3edeb@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

To Mr. Bell, I began this new thread as we have
departed nicely from the original threads "Math
& physics" intent.

On Jul 13, 8:20 am, "Steve Bell" <sb...@xxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote in
messagenews:350ade6a-69bf-40a8-9db4-467371b7d11c@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

To Bell
...
Also, I intuitively dislike an "asymmetric" geometry as the geometry of space, I
think nature is more symmetric than that. In fact, I think orthogonal, symmetric metric tensors are the only ones
nature
"uses."

Well, I use antisymmetrical metrics to convey
magnetism and electrostatics via the spacetime
field.

Have you derived actual equations for the elements of the metric tensor?

Yes.

I take it, it's 4x4. I'd like to see how gravity and
electricity and magnetism can be unified in a 4x4 metric tensor.
Steve Bell

Yes, Mr. Bell.
The asymmetricals within the 4x4 g_uv metric are,

a_uv = k*q*F_uv ,

with k a proportionality constant, q is fundamental
charge, and F_uv is the EM Field tensor.

Usually I use a QM compatible Field tensor that looks
like,

q*F_uv = A_u B'_v - A_v B'_u , Eq(a),

with A and B being potentials in order to eliminate
potential derivatives, the A_u is unprimed to stand
for positional potential, and the B'_v is a velocity
potential like this,

A_u = A x_u/s , B'_v = B dx_v/ds

A = a/s , B = b/s.

I learned that from W. Pauli, you can see his Relativity
Eq.(246a).
I depart (mildly) from Pauli by priming the velocity
potentials, while he primes the positional potentials.

What's relevent is we have eliminated the need of
potential derivatives in Eq.(a) which enables an
improved understanding of the photoelectric effect,
wherein the Quantum Theory is experimentally proved,
by disgarding the need for a continuum, required to
do derivatives of potentials.

The upshot is our a_uv asymmetrical components are now
compatible with EM and QT, to convey electrostatic and
magnetic effects via the spacetime field.

Some fella's use virtual particles to convey those
effects, and others use 5D or more, which to me is
math linguistics, not that there is anything wrong
with that, is just that I'm ok with 3D + memory.
Regards
Ken S. Tucker
kxsxt9

I think you are viewing things as a charge (q) coasting in an exterior charge density. I think better in terms of an
orbital
scenario. Let's say a central (cg at (0,0,0,)) ball of charge is rotating and generating an exterior electromagnetic field.
Then
within this field is coasting another charged particle q. Like ground state hydrogen. Let's assume the central charge just
sits
there and spins, that is, ignore that physical fact that both are actually orbiting around their common center of mass.
Under
this
scenario, can you fill in the equation:

g_uv = ?

I'm hoping for a result as "final," say, as a GR Schw result, i.e., something like g_rr = -1/(1 - 2mG/r), the r-coordinate
coefficient of a Schw metric tensor, when the metric itself is a "dtau^2" metric, not a "ds^2" metric (but of course, your
result
will not just have gravity, but will also include electromagnetism, which is what Einstein was trying to derive during many
years
of
his life).
Steve Bell

Yes Bell, your challenge is well posted.
I'll preface: I think AE was right about GR, in
the greater scheme of things but was side tracked
by war, wifes, kids, persecution etc. as well as
aging, so I think he didn't have the respect of GR
that I come to have.

Briefly Eq.(2) here,
http://physics.trak4.com/GR_Charge_Couple.pdf
unifies electrostatic and gravitational effects,
at a primitive level, but good enough.

In terms of the charges a and b, the time-time element, g_00, of your metric equals g_00 = 1 - ab/r^2. a and b are the two
"massless" charges mentioned at the beginning of your paper. I don't see any input from the masses (gravity), but of course
you
assumed them to be massless. So how does gravity come in when they are not assumed to be massless?

The source of mass is energy specifically,
energy induced into each charge from the other
(neither charge 'alone' has any mass),
http://en.wikipedia.org/wiki/Electric_potential_energy#Two_point_charges

This would seem to say no isolated charge can posses mass.

Agreed, an isolated charge does not exist, charge
is relative. Does "isolated velocity" exist(?), well
of course not.

If by "isolated" you mean "absolute," then actually, I think it does. I believe there exits a universal-sized absolute universal
space that the matter of the universe exists in at a universal point in time. In this universal space, particles have an absolute
velocity.

Well yes, the local speed of particles can all
be regarded as moving at "c" in direction of
time, in spacetime, although velocity is relative.

I prescribe to Purcell's view, 'a charge is required
to measure a charge', and therefore relative, and the
fundamental charge is an absolute (invariant) quantity.

I wonder how we have then measured the mass of an electron. I thought we
can isolate individual electrons, can't we?

I compute a charged lepton consists of 3 charges
to provide mass induction and spin, but that only
pans out at decay such as an e- e+ => gamma's.

In your (Bell) scenario above, (I like), let's take
the central fuzz ball as CS origin, generating a
field F_uv, that we'll measure with a charge "q",
to find a Lorentz force,

I'll just think the charge q is coasting in F_uv.

f_u = q*F_uv U^v , (U^v == 4velocity).

This looks like SR electrodynamics to me, except F_u would be on the left. Does f_u make up a 3x1 vector or a 4x1 vector? Is
there
any difference in your interpretation from what SR says these things are?

SR get's awkward in these problems, it really
doesn't work, GR enables the General Covariance
expressed as tensors.
What GR does (SR does not), is to convert the
electrical potential energy stored by two points
into spacetime curvature and then the two charges
themselves gravitationally attract, that's what
GR is designed to do.

So there is never any gravity present unless at least two charges are present?

Correct, two charges are the minimum to equate to mass.

That would seem to suggest that "charge" is more
"fundamental" than "mass." I personally think they are "on equal footing" as "fundamental characteristics of matter."

Agreed, a 'charge couple' is an electrical expression
of energy and thus mass. Charge couple is more detailed
than the generic term 'mass', cuz we can EM, QT and GR
out of that.

or

f^u = q*F^uv U_v , as you perfer.

Is q* equal to q with the SR relativistic correction factor?

Sorry the "*" was a multiplying sign, I'll drop
that going forward, and merely write,
f^u = q F^uv U_v.

It was hard to tell, the 4-stuff in SR has the SR time dilation factor usually taken out as a scalar constant, which premultiples
a
4x1 vector.

I hear ya, the so-called time dilation factor, is
refined in GR as a metric component such as g_00,
which is certainly NOT a scalar.
Thanks for the heads up Bell.

I still have a question. In SR electrodynamics the "little f" is a 3x1 vector containing the spatial (x,y,z) components of the
4-force F. Is that so in your math?

We need to determine the outer product,

f^u U_u = a*F^uv U_v U_u => sums to zero.

This still looks like SR electrodynamics to me. Did you mean to write q* instead of a*?

Yes, the "q" is usually the test charge relative
to an arbituary field F^uv.
When doing discrete relativitic electomagnetics,
the convention involves two charges "a" and "b",
with either "a" or "b" being the test charge.

Regards
Ken S. Tucker

I agree that to investigate how a charge particle coasts in a exterior EM field, that exterior EM field had to be generated by
other
charged particles. But I certainly think that an isolated charged particle with mass gravitationally curves space and dilates
time
all on its own. Imo, gravity does not "need" electricity.

Ok, in the absence of what 'mass is' - as our planetary
genius's argue about that and Higg's particles - this
theoretician uses, in the meantime, an electrical
definition of mass, consistent with the ISU definitions
of units, as briefed here,
http://physics.trak4.com/

In the metrics of GR, "characteristic" distances are in the metric. E.g., in the Schw metric, a distance is computed as a function
of the central mass and the distance is used in the metric. In Kerr GR, a distance is computed as a function of the central ball's
angular momentum and is used in the Kerr metric and introduces frame dragging effects. Is there a way for you to somehow find an
equivalent amount of mass that "curves spacetime" to the same degree as your electrical definition of mass?

Yes, suppose we place a number of (+) charges in
a small volume of space, and neutralize that with
(-) charges nearby, then we get something that looks
like an atom. I'm being a bit flippant, generally
though I find I can construct a charge configuration
to construct any particles mass and spin, a computer
helps, the hard part is grinding through the reactions
such as decay modes, annilations, decay times that
would be required to verify that model.

In the absence of a so-called ISU acknowleged "Unified
Field Theory", we knocked one together using 'off-the-shelf'
components, so we didn't need to build anything new, but we
term that assembly Unitivity.

Regards
Ken S. Tucker
.

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