Coulomb=>Gravity in GR, (kst).
Ken S. Tucker
>Coulombs force
>F= Q1*Q2/r^2
>Remarkably one can assume,
>F(attractive) - F(repulsive) = F(gravity)
>This effect is not formally recognized but is a
>consequence of GR. If you want I can provide
>more details.
>Ken S. Tucker
Using basic terminology, I'll show in GR, *gravity*
is a residual effect of electrostatic force, similiar
to in SR, *magnetism* is a residual electrostatic
force.
The ususal Coulomb force is given by,
F = q1q2/r^2 (1)
with q1 and q2 being charges, and "r" is a Newtonian
distance that is not affected by any fields.
In GR the distance separating the charges is affected
by the field, so let's re-write (1) as
f = q1q2/s^2 (2)
where "s" is the distance in GR.
The *electrical potential energy* "e" of the system
is given classically by,
e = q1q2/r which is,
e(R) = (q1=q2, Repulsive) is +field energy,
e(A) = - e(R), (q1=-q2, Attractive) is -field energy.
From Einstein's Law relating curvature to energy
density, (G_uv = kT_uv) we must expect the
field energy's above to affect the curvature differently.
This can accomplished quite simply by setting,
s^2 = r^2 + q1q2 (3)
((the theoretical justification for this assumption
is a bit lengthy, so I'll just produce the results)).
Doing a bit of algebra, we sub Eq(3) into Eq(2),
f= q1q2/s^2 = q1q2/(r^2 + q1q2)
f ~= q1q2/r^2 - (q1q2/r)^2 /r^2
The1st RHS term is the ordinary Coulomb "F",
the 2nd RHS term is a residual attractive force,
(negative) independent of "F" itself repelling or
attracting and is the residual gravitational field.
By setting e = mc^2 and c=1, the 2nd term
becomes,
F(g) = - m*m /r^2, (Newton's Gravity)
therefore Coulomb force in GR is
f = F - F(g)
The F(g) term arises from the +q1q2 term in Eq(3).
Even though "r' is the same for repelling and
attractive charges, we can see at a glance that
a pair of Repelling charges have a greater separation
distance than Attracting charges ie,
s(R) > s(A).
In GR the explanation is, the curvature G_uv and the
energy density kT_uv in a repulsive field is greater
than in an attractive field, hence the difference.
Regards
Ken S. Tucker
PS: Among other sources I get the basic analysis for
the above from Kaluza's 5D theory. Please note the
Pythagorian theorem in Eq(3) ie, s^2 = r^2 + q1q2.
Given that r^2 is Newtonian and orthogonal, then
it's implied that q (either + or -) is orthogonal to r^2.
But "q" is NOT a dimension, it has a finite invariant
length (fundamental charge) treated as orthogonal to
Newton's 3D quantity "r", and "q" is usually called a
"compactified dimension".
KST
Michael Varney
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:2202379a.04032...@posting.google.com...
> From thread "Gravity force" I suggest,
Hey jackass... rather than obfuscating the threads... why not post under
them.
Ken S. Tucker
> "Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
> news:2202379a.04032...@posting.google.com...
> > From thread "Gravity force" I suggest,
>> Mr tucker I want more of your great wisdom
Thank you Dr.Varney...you are so kind...
Studying the research of Kaluza and Einstein on the
unification of EM and gravity there appears something
in common, naturally this is simplified, but it appears a
vector associated with *fundamental charge* "q" is
orthogonal to x,y,z.t is common to both theories.
Well, is that reasonable?
Recall that perpendicular lengths are unaffected by
motion, such as y=y' when motion is parallel to x
and x', so when "q" is converted to a length, the
length "q" must remain invariant independant of any
motion in spacetime, so it must follow to describe the
length "q" as orthogonal to x,y,z,t, as Kaluza did with
his additional compactified dimension "q" orthogonal
to x,y,z,t.
There is a sense Kaluza was *boot-strapping* a 5th
compactified dimension into spacetime, to the extent
of providing a metric like g_55. That 5th dimension,
would have quite different rules than and 4D spacetime
dimension, and is difficult to understand.
Einstein was dismissive of the 5D approach, on the
basis of his experience and intuition for unifying EM
and gravity. He realized only 6 more spacetime
equations were needed and these could be found in
the usual metric g_uv if g_uv was nonsymmetric.
The Electric Fields should be found in, g_i4
(i=1,2,3) and as a result of motion, magnetic fields
should be found in g12, g23, g31
Unfortunately Einstein assumed the Christoffel
connections to be non-symmetrical, and I think
that was a fatal assumption. ((Subsequently see
how antisymmetrical metrics null in symmetrical
connections, in my post,
"Einstein's Non-Symmetric Field Theory")).
I find no reasons to make the connection
non-symmetrical, even if the metric is.
However we are left with the problem of defining
"q" in the context of the antisymmetrical metric.
To do this look at the very basic Electric Field,
E = q/r^2
and see E* r^2=q.
There is a problem expressing what was done
in a covariant expression like,
2q = F^uv x_u x_v
because that would apparently sum to zero,
unless we employ relativity and rewrite that as,
2q = F^uv A_u B_v (4)
where we now relate 2 charges "a" and "b" by,
A_i = - B_i and A_4 = B_4 (5)
where A_u is the position of charge "A"
relative to charge "B" in spacetime.
Notice charge "a" is in the relatively opposite
direction to charge "b" spatially, but the time
components are equally directed.
The sophisication of the treatment will find Eq(4)
in accord with the relativistic Eq(5), then,
q1q2 = q*(1/2) F^uv A_u B_v
where q1 = a and q2 =b.
In juxtaposition, to Kaluza's treatment of the
fundamental charge "q", as some sort of absolute
Einstein treates "q" as related to another "q" so
only q1q2 is physical.
Hence, Einstiens's advanced GR accepts the
need for F(uv) within the metric, to inculde "q".
Ken S. Tucker
Tom Roberts
> Using basic terminology, I'll show in GR, [... much nonsense]
You do no such thing.
Why don't you stop trying to "show something in GR", and just admit that
you don't know the first thing about GR, or the second? Your time would
be MUCH better spent actually STUDYING GR rather than trying to "show
things in GR" that simply aren't related to GR at all, and seem to be
false (e.g. gravitation is utterly unrelated to Coulomb forces, in both
classical physics and in GR).
The "first thing" about GR is diffeomorphism invariance;
the "second thing" is local Lorentz invariance. Neither of
those are reflected in your post in any way.
Tom Roberts tjro...@lucent.com
Ken S. Tucker
>Ken S. Tucker wrote:
>> Using basic terminology, I'll show in GR,
>Why don't you stop trying to "show something in GR", and just admit that
>you don't know the first thing about GR, or the second?
The first thing is easier than the second. The second
thing requires understanding the basic interface between
GR and electrodynamics in accord with the EFE's, see
" g_00 vs p_00, and the source of gravitation?",
for the confirmation.
>Your time would be MUCH better spent actually STUDYING GR
Of course, that's where my OP comes from.
Take your own advice and read it.
>rather than trying to "show
>things in GR" that simply aren't related to GR at all, and seem to be
>false (e.g. gravitation is utterly unrelated to Coulomb forces, in both
>classical physics and in GR).
That's wrong Tom, check your EFE's (G_uv = k*T_uv)
and note electrostatic field products in the RHS. Multiply
them by a pair of invariant scalars (aka charges) to
actualize and energize a charge couple and you'll arrive
at the rationalization for my OP.
> The "first thing" about GR is diffeomorphism invariance;
> the "second thing" is local Lorentz invariance. Neither of
> those are reflected in your post in any way.
Of course not, the post was meant to be straight forward
algebra, as a conceptual back-of-envelope introduction.
If you want, you can take s^2 = r^2 + qQ and calculus,
s ds = r dr.
The RHS is easily integratible, therefore the LHS is too.
Additionally, using "/\" as delta (as in a finite increment)
another relation is,
s /\s = r /\r
and is useful in subsuming the QT into the field theory,
and vis-versa, as it permits /\ (delta) increments. That's
important if you intend to merge GR and QT.
I have found, with the appropriate definition of the metric,
in accord with the above,
s^2 = g^uv x_u x_v
ds^2 = g_uv dx^u dx^v
are compatible for an isolated charge couple.
However, if "reductionism" holds for charge couples,
then the condition "isolated" is not necessary, and thus
these last equations hold macroscopically.
>Tom Roberts tjro...@lucent.com
Hmm <shrug> read and learn (:-))...Regards
Ken S. Tucker