Re: Orthogonal distance

MusicRules wrote:
"Ken S. Tucker" <dynamics@xxxxxxxxxxxx> writes:

In this article,

http://www.vacuum-physics.com/KST/GR_Charge_Couple3.pdf

are distances "S" and "X" that are important in Eq.(4).

I'd like to clarify their meaning, ref to Pauli's "Theory of
Relativity"
chp 10, pg 27...as it is as good as any.

In that section, Pauli stated that he was discussing a Cartesian
coordinate system (X_1,X_2,X_3,X_4). At no point did Pauli ever
state that this was the appropriate coordinate system for Minkowski
spacetime. Pauli merely introduced the coordinate system on a
temporary basis to make a few points. Pauli then discussed the general
case, which has the coordinate system for Minkowski spacetime as a
special case. Specifically, at no point did Pauli ever state that
there existed a Cartesian coordinate system for Minkowski spacetime.

Define (see 35a),

S^2 = x_u x^u = x_0 x^0 + x_i x^i,

{i=1,2,3 , u=0,1}.

That's fair provided the "g_uv" are moderately constant,
as in the ref'd article.

We also will recognize the ISU finding of the equivalence
of time and spatial displacements to be related by,

x_0 x^0 = x_i x^i .

Clearly one can notice due respect paid to the fact
that covariant and contravariant indices characteristic
of a nonorthogonal spacetime are used to define "S".

In defining an Orthogonal distance "X", we need to
be able to use either,

X^2 = x_u x_u = x^u x^u

so that the covariant or contravariant projections have
no difference, characterististic of Orthogonal spacetime.

We'll be dealing with space and time measurements
in a g-field with a refractive index.

Define the term "g-field with a refractive index".

Ok, In 1983 the ISU, established the defintion of
the "metre" and second using Meter = c Seconds,
which I fully agree with, and that "c" is specified
to be in a vacuum, and furthermore, for a careful
theoretician, away from matter which can vary the
velocity of light, by deflection, or retardation as the
Shapiro Experiment finds.
Instead we live in g-fields and our *real* "c" is in
a field that bends and retards our reference "c".

Denote "C" the *refracted* coordinate speed of light
in a g-field and
"c" the usual vacuum speed of light where C<c,
(see Weinberg's "Grav&Cosmo" (9.2.5)).

Note that although C is the *coordinate* speed of light, if
you are moving inertially, and you measure the speed of light
with stationary rods and clocks, you still get a value of c
for the speed of light.

Also note an inertial frame is orthogononal.

To solve the above equations I'll specify Coordinate
Systems (CS) requiring the relations,

x_0 = x^i =Ct

x^0 = x_i = ct .

This is unnatural. The four coordinates are independent of one another.
If you want this to be the trajectory of a light ray, then you should
state this explicitly, rather than not bothering to do so. We can't read
your mind, you know.

The g-field makes x^0 =/= x_0 etc. because g_00 =/=1.
So the standard relation of the metre to the second using
"c" needs to modified in nonorthogonal spacetime.

While x_0 and x_i are definable in Special Relativity, they are not
automatically definable in General Relativity. As a consequence, your
variables x_0 and x_i are meaningless, since you are working in GR.

Even in the case of Special Relativity, one of two conditions must hold:

(1) x_0 and x^0 have opposite signs, and x_i and x^i have the same sign;

or

(2) x_0 and x^0 have the same sign, and x_i and x^i have opposite signs.

Your assignment of values satisfies neither of these conditions.

No. Actually the requirement is

ds^2 = dt^2 - dr^2 (for Lorentz Transform, LT)

and

S^2 = t^2 +r^2 (for RADAR ranging).

Please pay attention to derivatives!

Those give,

S^2 = 2cC t^2 = x_0 x^0 +x_i x^i

and

X^2 = (c^2+C^2) t^2

= x_0 x_0 + x_i x_i = x^0 x^0 + x^i x^i ,

= x_u x_u = x^u x^u .

You have still not proven this variable to be Lorentz invariant.

The LT pertains to relative motion, that is the derivatives
of the x_u x_u NOT the stationary lengths, those are
specified by RADAR ranging.

The last equation is true in any field where the
speed of light is refracted to "C", using that CS
spacetime based group specification

You never specified a group.

###
Let c0, r0 and t0 be determined in a perfect vacuum,
and r and t be physical coordinates when matter is
placed at the origin, then,

t^2 = (1+o) t0^2 , r^2 = (1-o) r0^2

r0 =c0 t0 , c0 =1.

t^2 + r^2 = t0^2 + r0^2 = X^2

1-o = g_00.

The coordinate speed of light "C" is

C = r/t ~ 1-o = g_00,

and is experimentally confirmed.
(Shapiro and deflection).
###

Between ### - ### is an application of,

x_0 = x^i =Ct
x^0 = x_i = ct .

which solves X^2 = x_u x_u = x^u x^u.

Please note how that all connects to reality.

Better understand that, because Eq.(4),

S^2 = X^2 + ab Eq.(4).

provides the strength of Coulomb force,
as a result of GR, it get's better....

Ken S. Tucker

.

Relevant Pages

  • Re: The Nanometre Twin
    ... Only a hypothetical light clock does this. ... if the principle of relativity ... what Newton's spacetime was. ... Choose a particular coordinate system, ...
    (sci.physics.relativity)
  • Re: Orthogonal distance
    ... Pauli stated that he was discussing a Cartesian ... state that this was the appropriate coordinate system for Minkowski ... which has the coordinate system for Minkowski spacetime as a ... Even in the case of Special Relativity, one of two conditions must hold: ...
    (sci.physics.relativity)
  • Re: Time dilation, spaghetti and black holes
    ... coordinate system", seems to be a baroque way of saying, "the topology ... In QM we assume a Euclidean spacetime and in quantum field ... theory we assume the flat spacetime of special relativity. ... In general relativity the geometry (and ...
    (rec.arts.sf.composition)
  • Re: Orthogonal distance
    ... Pauli stated that he was discussing a Cartesian ... state that this was the appropriate coordinate system for Minkowski ... which has the coordinate system for Minkowski spacetime as a ... automatically definable in General Relativity. ...
    (sci.physics.relativity)
  • Re: do a moving black holes eat space?
    ... You mean, choose a coordinate system? ... covers the whole manifold. ... locate them on the manifold which is the mathematical model of spacetime. ... Relativity is like Go: ...
    (sci.physics.relativity)
Loading