Re: Spacetime asymmetrics (kst).Bell

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 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,

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

or

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

We need to determine the outer product,

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

Is that ok with you Mr. Bell?
Regards
Ken S. Tucker
.

Relevant Pages

  • Re: Tom Van Flandern and Newtonian Gravity
    ... from electroSTATICS, where all charges are stationary, so there is no ... time dependence in the charge density or any of the fields. ... published in our "Foundations of Physics" paper. ... the potentials; it is entirely a matter of convenience. ...
    (sci.physics.relativity)
  • Re: Spacetime asymmetrics (kst).Bell
    ... magnetism and electrostatics via the spacetime ... Have you derived actual equations for the elements of the metric tensor? ... potentials, while he primes the positional potentials. ... I think you are viewing things as a charge coasting in an exterior charge density. ...
    (sci.physics.relativity)
  • Re: Spacetime asymmetrics (kst).Bell
    ... magnetism and electrostatics via the spacetime ... Have you derived actual equations for the elements of the metric tensor? ... potentials, while he primes the positional potentials. ... I think you are viewing things as a charge coasting in an exterior charge density. ...
    (sci.physics.relativity)
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