Hello AI
In the following, arbituary units are
suppressed for clarity.

Special Relativity relies on two apparently
paradoxical postulates,

SR1) Velocity = x/t is relative.
(no absolute velocity)
(measuring velocity requires the relativity
of two references)

SR2) Velocity "c" is invariant.
(c is relativistically invariant)

Let's introduce two mirror postulates for
Unitivity, that again appear paradoxical,

U1) Fundamental Charge "q" is relative.
(no absolute charge)
(measuring charge requires two relating
charges)

U2) Charge "q" is invariant.
(charge is relativistically invariant)

SR did not require a continuum, it is a
relational theory but we find classical
GR incorporates Newton's continuum by an
osmosis. In that way classical GR retro-
graded by re-embracing the continuum.

Using postulates U1 and U2 together with
the Principle of Covariance, a relational
GR theory evolves.

For example, solving AE's Law, G_uv = T_uv,
using postulate (U1) renders,

G_00 = E(a).E(b) = T_00 ,

where E(a) and E(b) are Electric fields of
relative charges "a" and "b".

A fundametric solution to the above yields,
(using G_00 ~ Nabla^2 g_00, Grav&Cosmo 7.1.7)

g_00 ~ (a/r)*(b/r) ~ Coulomb's Force.

Rest energy expressed in electrical terms is,

p = (a*b/r) , g_00 = p/r == mass/radius,

where the g_00 is the gravitational potential
in static GR.

Apparently then gravitational force depends
upon the "change" in Coulomb's Force.

End of Intro.
Ken S. Tucker

"Ken S. Tucker" wrote:

Hello AI
In the following, arbituary units are
suppressed for clarity.

Special Relativity relies on two apparently
paradoxical postulates,

SR1) Velocity = x/t is relative.
(no absolute velocity)
(measuring velocity requires the relativity
of two references)

A measured velocity is absolute, ...
it's _value_ is relative.

There's no paradox there. This is
a common problem in all measurements.
Take the last presidential election
results for instance.


SR2) Velocity "c" is invariant.
(c is relativistically invariant)

c is a speed, not a velocity.
Big difference.

Hi Tron

tron wrote:
"Ken S. Tucker" wrote:

Hello AI
In the following, arbituary units are
suppressed for clarity.

Special Relativity relies on two apparently
paradoxical postulates,

SR1) Velocity = x/t is relative.
(no absolute velocity)
(measuring velocity requires the relativity
of two references)

A measured velocity is absolute, ...
it's _value_ is relative.
There's no paradox there. This is
a common problem in all measurements.
Take the last presidential election
results for instance.

That makes no sense to me, I do agree comparative
speed is absolute, is that what you mean?

SR2) Velocity "c" is invariant.
(c is relativistically invariant)

c is a speed, not a velocity.
Big difference.

A constant velocity means light is not deflected.
It's another way of saying, we're in SR not GR.
Ken

RELATIONAL VS CONTINUUM RELATIVITY.

1) We can define a 2nd rank tensor,

A_a B_b = - A_b B_a

(That tensor can be proven real when the norm
of a^ab(A_a B_b) =/=0, is calculated, where a^ab
is the asymemtric fundametric tensor).

and specify

(A_a B_b);c =0 , (a symmetry and conservation)

and then find,

A_a;c B_b = - B_b;c A_a

is that asymmetric in indices "a,b,c" ?
and symmetric if A and B are exchanged?

2) Set

(A_a A_b);c =0

and find

A_a;c A_b = - A_b;c A_a

and see an asymmetry in indices "a" and "b",
but does it follow "c" and "a" or "b" are
asymmetric?

In problem (1) the vectors A and B are distinct,
so one is applying "relational" GR.

In problem (2) vector "A" can exist on a continuum,
as it is not relative to another vector so it can
apply to a point.

I find in (1) a,b,c, are antisymmetrical permutations
IFF A and B are antisymmetric.

However if A=B, as in case (2) A=A then a,b,c are
NOT asymmetic, nor are they symmetric.
They appear anti in a,b and symmetric in a,c and
b,c.

So to summarize,

When a symmetric relation exists between A and B
defined by A=B then

2.1)
A,B == A,A == symm == (ac=ca , bc = cb, ab = -ba)

= abc = acb = cba

1.1)
A,B anti == (ac=-ca, bc=-cb , ba=-ab)

= abc = cab = bca

Eq.(1.1) corresponds to

Fab,c + Fca,b + Fbc,a = 0 (Maxwell's)

ie. the photon and likely the typical boson,
which has spin 1.

OTOH, Eq. (2.1) has 1/2 spin, since the ac,ca and
the bc,cb spin =0, because they're symmetric.
Ken
(KxS1xT)