Re: "Evidence of the Existence of the Aether"

On Oct 2, 10:27 pm, mpc755 <mpc...@xxxxxxxxx> wrote:
On Oct 2, glird wrote:
On Oct 2, mpc755 wrote:
On Oct 2, glird wrote:
On Sep 30, 1:02 pm, mpc wrote:
On Sep 30, 12:44 pm, glird wrote:
<<<<Since a displacement wave is a sorce-density
imbalance, if it is absorbed the zero level of sorce-pressure will
thereby have changed, whereupon so will the density.  BUT the increase
in density is due to an inflow or outflow of matter IN ALL DIRECTIONS,
from or into the given reaction atom. Therefore, some of the aether-
matter "associated with" the wave system can be absorbed even though
no material particle was part of that wave. >>>>

<<< I like your descriptions, however, I do not see how your description of a photon excludes it from being a physical particle of aether traveling through the aether. >>>

<< Having an entirely different metaphysical theory than others ("metaphysics" is the subject that treats the nature of the underlying realities that Physics measures), I "know" that a particle of aether is a lastingly discrete bit of matter that exists and travels as a unit, displacing the material through which it moves.
 A "photon" is Einstein's word for Plank's quantum of energy (in
which a "quantum" is a quantity and "energy" is the ability to do
work; work = Fd = mda, in which F is a quantity of "force" - which is
a net
pressure independent of the area of its application, m is the weight
(a force) of a "mass" (a quantity of matter), d is the distance the
mass moves, and a denotes acceleration.
 Once emitted, the energy doesn't travel as a unit, it spreads in all
directions, summing with the zero level per successive point; and
moving at c through a vacuum (a "vacuum" is a material-filled space
with no particles present in it). When the energy enters a given atom
it has a velocity of 1/f; where f denotes the frequency wrt that atom.
 If the resulting pattern of the displacement wave fits that of the
given atom, a whole quantum of energy will be absorbed. If not, some
of the energy will be absorbed and the rest of the wave system will
continue on with a bit less e remaining; wherefore its color will be
red-shifted when it is fully absorbed by another atom later on. >>

< When an electron transitions to a lower orbital and emits a photon, I do not see how you can 'know' the photon is not a physical particle of aether. >

A photon is a quantity of energy that is emitted and absorbed by an atom. As Plank said, though, while in transit it is NOT a discrete thing at all; therefore is not a particle of "aether" (which denotes the continuous aspect of a matter-filled field, whether or not particles are present in it)..
A particle of matter traveling through a material-
filled field is resisted by the displaced aether; thus - unlike a
light wave - slows down as it progresses.>>

< Resistance yes. Friction no. If there was friction, there would be no momentum of anything through the aether. >

Insofar as light-waves are concerned, there is resistance but no friction. (The amount of resistance is a function of the density of the traversed material.) Insofar as a particle is concerned, however, there is resistance by the material in the vacuum portion of the traversed field AND there is friction wrt any particles present in the local aether - where (as said above) the word "aether" (i.e, "ether") denotes the continuous aspect of a matter-filled field, whether or not particles are present in it.
Therefore a particle of matter - which is a mass that does have
weight WILL have a momentum of mv, where m denotes the weight of the
given mass and v is
its local velocity. >>

< You say, "Insofar as a particle is concerned,...there is friction...".>

As I said, "there is resistance by the material in the vacuum
portion of the traversed field AND there is friction wrt any particles
present".

< And you also say, "a particle of matter...WILL have a momentum". >

"when moving through a particle-filled aether"

< Either the Earth's interaction with the aether is causing friction or there is momentum. >

In a vacuum, the volume of the "empty space" is far greater than the
volume of the embedded particles; where "empty space" is actually
completely filled with the compressible continuous form of the same
matter out of which particles are made. as Earth moves through this
vacuum there is resistance by the continuous matter and there is
friction by the particles in it.
Insofar as momentum, though, the earth - "which is a mass that does
have weight" "WILL have a momentum of mv, where m denotes the weight
of the given mass and v is its local velocity".
Please pardon me for omitting a bit I thought of but decided wasn't
needed, i.e. - where "local velocity" is its speed relative to
anything the given mass might happen to hit.
Btw, whereas "mass" denotes a quantity of matter, whether particulate
or not; the m in most of our equations denotes m_w, the WEIGHT (a
force, NOT a quantity of matter) of a given mass. Hence, mv denotes
the weight of a given mass times its velocity relative to an object it
happens to hit; and ma denotes the weight of a given mass times its
acceleration; and mc^2 denotes the WEIGHT (NOT the matter per se!)
times (the speed of light in vacuo)^2.

Because non-particulate matter doesn't respond to gravity the way
particles do, a gram of weight is not a specific quantity of matter.
THAT'S why I say that a gram is the wrong unit of measure for mass.

glird
.

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