Re: Quantum Gravity Via Expansion-Contraction 56.0: Quantum and GR Equations Don't Require Quantum/GR Theories


OsherD wrote:


Why isn't a test of whether c is finite or not regarded as a test of GR
at present (most physicists regard it as a test of SR, Special
Relativity)? This is because infinite c is not even considered by
most physicists in view of SR. But if c were infinite, then unless it
is agreed that the equations of GR operate with c coded as 1 for
infinity and Likewise Maxwell's equations, the equations would fail.
In addition, the spacetime would not be locally Galilean (roughly
speaking, SR), in which case the Principle of Equivalence would fail.


Osher Doctorow

A full coverage horizon is something you could never become aware of
directly. Full coverage means you are within and merged with its essentially
dimensionless coverage. You become aware of it indirectly via a line horizon
always distantly constant to all positions and all velocities. No matter
your position that line of horizon will be there the same distance from you
the same as when you are in any other relative position whatsoever. No
matter your velocity that line of horizon will be there the same distance
from you the same as when you are at any other relative velocity whatsoever.
Position independent due to full coverage. Velocity independent due to full
coverage. '1'.

But the line of horizon isn't '1'. The constant distance to the line of
horizon will be a metric quantity different from '1'. A full coverage, or
infinite, horizon is dimensionless '1'. But it "collapses" to a constant
horizon, in this case 'c'. If you want to identify its full coverage, c = 1.
If you want to identify its collapsed line, c = 300,000 kps. The second,
because of the first, then becomes infinite expansion / contraction
(inflationary / deflationary) balloon quality to any object in motion.
Quanta dynamically though, anything so ballooned will simply be the thing
mirrored into countless many of the same thing observed by countless many
simultaneous observers to be occupying countless many different spaces and
times. One thing in one space and time to one observer, but altogether not
the same space nor the same time to all observers. Not the same position,
nor the same velocity, to all observers. It won't be observed to be smeared
in space and time, position and velocity, mass and energy, all over any
local universe.

What is an inertial frame of Relativity depends upon vertical level of
horizontal surface. Inertial frames are not one level absolute. A dining
room table is not an inertial frame to anything so decelerated in velocity
as to gain relativity to a particle at its particle level. This of course
may be a deceleration in velocity from zero velocity (relative to Einstein's
railroad track). But the particle might not then be a particle anymore nor
the particle level accessed be anything like a particle level anymore. It
may be a deceleration to simply some other universe not a whole lot
different from the universe decelerated from.

The Milky Way galaxy is an inertial frame to anything so accelerated in
velocity as to gain relativity to its whole as a much more solid and much
more singular unity of space and time level rather than its usual much more
dispersed near vacuity. It would take a lot more increase before expansion
is such as to have gained relativity to a universe not a whole lot different
from this universe accelerated in velocity from, but to which this entire
previously relative universe is but a point particle of it or less, an
infinitesimal point then entirely lost to the expanded (the inflated).

GLB


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