Nature abhors ''NO medium''.
brian a m stuckless
Electromagnetic energy in the form of radiation does NOT
propagate without a medium.
There is a medium everywhere measured to date.
Note the following attempts to achieve ''NO medium''.
Web quotes:
"Best was achieved by hanging a plate off the back of the
shuttle, achieving around 1000 atoms per cubic centimeter.
This would be a year or two back?"
[10^9 atoms per cubic meter]
"Best achieved on earth would be a million atoms per cc,
or a thousand times less sparse."
[10^12 atoms per cubic meter --at 10^-8 Pascal]
"These don't compare too well with naturally occuring vacuums,
out in intergalactic space there is vacuum where 10 meters on
average separate the atoms, the shuttle vacuum was 10^6
times more dense by volume."
[0.001 atoms per cubic meter --but a medium none-the-less]
There are aproximately 10^25 molar particles per cubic meter
at international Standard Temperature and Pressure (STP).
Mention the background radiation, other already-in-transit
photons, etc, and already-in-transit phonons. Yes, ''phonons''!.
A space of 10meter atoms is just as much a medium as a
space of atoms at Standard Temperature and Pressure.
One atom of water is just as wet as an ocean.
Water is to an ocean, as are trees to a forest,
--or a medium to aether.
We can't see aether for the medium.
''In-vacu'' is a misnomer, --nature abhors ''NO medium''.
Yours truly,
`````arcsign````` mar 10, 2001
Paul Lutus
news:01c0a98a$32aaa8c0$LocalHost@default...
> Nature abhors ''NO medium''.
>
> Electromagnetic energy in the form of radiation does NOT
> propagate without a medium.
In your first line, you invite educated readers to stop reading. Why not
conceal your total ignorance of physics until later?
In the remainder of your post, the utter ignorance of your position is
dramatically made clear.
--
Paul Lutus
www.arachnoid.com
Joe Rongen
news:01c0a98a$32aaa8c0$LocalHost@default...
>
> Electromagnetic energy in the form of radiation does NOT
> propagate without a medium.
Interesting, so how did you manage to
read this reply from a CRT type monitor?
Those electrons in the CRT are in a good
vacuum and controlled by the deflection coils
around the CRT's neck.
Where is your 'medium'?
-------
Regards Joe
j...@alpha.to
brian a m stuckless
Dear Paul Michael Lutus Varney,
It's time for you to give up your Yosimite Sam act, Paul.
All ''educated readers'' aren't GR-tivity-'fruitcake' buffoons.
Wasn't it YOU:
--Who measured ''NO medium'' and found no aether?
--Who could clearly distinguish one ''no feeling'' from
another, in the GR-tivity 'gedanken-equivalence
freefalling elevator principle', --concluding that it was
distinctly ''freefalling up'' ?
Please STOP trolling, Michael.
Yours truly,
`````arcsign````` mar 10, 2001
^^^^^^^^^^references follow:
Nature abhors ''NO medium''.
Electromagnetic energy in the form of radiation does NOT
propagate without a medium.
There is a medium everywhere measured to date.
^^^^^^^^^^Paul Michael Lutus Varney reference:
Re: Nature abhors ''NO medium''.
Paul Lutus <nos...@nosite.com> wrote in article
<Wiwq6.43298$lj4.1...@news6.giganews.com>...
^^^^^^^^^end of post.
Harry H Conover
: Nature abhors ''NO medium''.
:
: Electromagnetic energy in the form of radiation does NOT
: propagate without a medium.
...and your evidence for this rather remarkable conclusion is....?
Harry C.
Paul Stowe
con...@tiac.net (Harry H Conover) wrote:
>brian a m stuckless (bast...@avalon.nf.ca) wrote:
>: Nature abhors ''NO medium''.
>:
>: Electromagnetic energy in the form of radiation does NOT
>: propagate without a medium.
>
>
>....and your evidence for this rather remarkable conclusion is....?
Well for me, the "Characteristic Impedance of the Medium"...
Paul Stowe
Harry H Conover
: In article <98ee7r$7...@news-central.tiac.net>,
If not (what is it) 377-Ohms or so, is there a figure that you
would prefer? Why?
Sorry, but I don't see a connection between the characteristic
impedance of free space and a need for a propagation medium.
Harry C.
Paul Lutus
Your post is a series of outright lies, with no imaginable connection to
physics.
--
Paul Lutus
www.arachnoid.com
brian a m stuckless
Harry H Conover <con...@tiac.net> wrote
in article <98ee7r$7...@news-central.tiac.net>...
^^^^^^^^^^^^reply:
Dear Harry,
Please finish reading the post.
Electromagnetic energy in the form of radiation does NOT
(i.e. CANNOT) propagate without a medium because there is
a medium everywhere.
''NO medium'' cannot be achieved, even in deep space.
Therefore, there is no known space with ''NO medium'' in it.
No measurement of any kind can be made in any state
which has never been achieved. ''In-vacu'' is a misnomer.
Yours truly,
`````arcsign````` mar 10, 2001
^^^^^^^^^original post follows:
Nature abhors ''NO medium''.
Electromagnetic energy in the form of radiation does NOT
propagate without a medium.
There is a medium everywhere measured to date.
^^^^^^^^^end of post.
Paul Lutus
> Re: Nature abhors ''NO medium''.
>
> Harry H Conover <con...@tiac.net> wrote
> in article <98ee7r$7...@news-central.tiac.net>...
> > brian a m stuckless (bast...@avalon.nf.ca) wrote:
> > : Nature abhors ''NO medium''.
> > :
> > : Electromagnetic energy in the form of radiation does NOT
> > : propagate without a medium.
> >
> >
> > ...and your evidence for this rather remarkable conclusion is....?
> >
> > Harry C.
> >
> ^^^^^^^^^^^^reply:
>
> Dear Harry,
>
> Please finish reading the post.
>
> Electromagnetic energy in the form of radiation does NOT
> (i.e. CANNOT) propagate without a medium because there is
> a medium everywhere.
Repeating a lie doesn't make it true, nor add evidentiary weight to it. You
were asked to provide evidence, not repetition of the same tired lie.
If your reasoning deserved the name (it does not), it would be called
"circular."
Look up the word "evidence" in your Scrabble word list.
--
Paul Lutus
www.arachnoid.com
Paul Stowe
con...@tiac.net (Harry H Conover) wrote:
>Paul Stowe (pst...@ix.netcom.com) wrote:
>: In article <98ee7r$7...@news-central.tiac.net>,
>: con...@tiac.net (Harry H Conover) wrote:
>:
>: >brian a m stuckless (bast...@avalon.nf.ca) wrote:
>: >: Nature abhors ''NO medium''.
>: >:
>: >: Electromagnetic energy in the form of radiation does NOT
>: >: propagate without a medium.
>: >
>: >
>: >....and your evidence for this rather remarkable conclusion is....?
>:
>: Well for me, the "Characteristic Impedance of the Medium"...
>
>If not (what is it) 377-Ohms or so, is there a figure that you
>would prefer? Why?
>
>Sorry, but I don't see a connection between the characteristic
>impedance of free space and a need for a propagation medium.
OK, you did ask...
Consider in particular the general properties of a medium:
All known media are either fluidic (gas, liquid, or Plasma [excluding
a possible aether for now]) or a solid, and (regardless of type) have
certain traits in common. These are what are designated 'fingerprint'
characteristics. Including:
1. At some scale they must met the definition of a field
2. They exhibit the condition known as action at a distance
3. Their bulk or large-scale properties can be described by
Continuum Mechanics
All manifest the following empirical physical properties:
a. A Characteristic wave & general disturbance
propagation speed [c]
b. A Pressure or modulus [P (gas), B (liquid), Y
(solid)]
c. A Coefficient of compressibility [u = 1/kP, 1/kB,
1/kY]
d. A Mass Density [rho]
e. An Impedance [Z]
f. Temperature [T]
Now let's see how we empirically discern these one at a time.
I listed these deliberately in order with wave speed first. All known
mediums have a characteristic speed at which disturbances (such as a
sudden pressure pulse) can & will disperse (which is the root cause of
action at a distance in such systems). In a simple mono-atomic gas,
this is:
kP
c^2 = --- Where k = 3(degrees of freedom)
rho
For any liquid we see the same equation as:
kB
c^2 = --- Where k = 1(by convention is factored in B directly) & B is
rho the bulk modulus
For a solid we have two wave speeds, the longitudinal as:
kY
c^2 = --- Where k = 3{ideal} (but ranges from 2 to 4 in real solids)
rho & Y is 'Young's Modulus'
And the shearing transverse components (S_h & S_v) as:
Y
c^2 = ---
rho
Conversely the 'coefficient of compressibility' is simply the inverse of
the pressure or modulus, i.e.
1 1 1
u = -, or u = - or, u = -
P B Y
Thus the a single general propagation equation becomes:
k
c^2 = ------
u(rho)
And if we factor k into u as:
1 1 1
u = --, or u = -- or, u = --
kP kB kY
We get a general propagation speed equation of the form:
1
c^2 = ------
u(rho)
But u is now specific to both the type of medium and disturbance (as in
a solid, a transverse or longitudinal type) being quantified...
Now, if we empirically we 'measure' only c in any unitary system, we
cannot directly discern either u or rho. We know their product, but
not the components values.
Usually we get these by empirically determining the medium's density
and solving for the overall modulus. For gases however, pressure can be
directly measured and the density can be approximated. However, if we
cannot empirically measure either, there is a slight problem.
Luckily for us Maxwell's equations are also based upon a specific model
of a physical medium (see below). From this model and resulting
equations we get:
1
c^2 = ------
u(eps)
where [u] is given the name Permeability and [eps] Permitivitty. Anyone
should immediately see the similarity in form to the above. Since the
empirical evidence clearly shows that light is characteristically a
transverse wave, the form of:
Y
c^2 = ---
rho
should be applicable and u would take the form of 1/Y. One should 'expect'
therefore to probably have a component of light that would be the longitudinal
component of:
3Y
c^2 = ---
rho
varying from the transverse value of c by the Sqrt(3) [1.732...]. We will get
back to this when discussing unique empirical evidence for the aethereal medium.
That Maxwell's model fully supports his equations is recognized in standard
literature, for example:
“There is an analogy between fluid dynamics for
solenoidal fields and electrodynamics: the
vortex strength corresponds to the current
intensity and the vortex vector to the current
density. Vortices are surrounded by velocity
lines (streamlines for steady flows) just as
electric currents are by magnetic lines of force.
In these terms the flow velocity is said to be
induced by the vorticity. The formula for induced
velocity corresponds EXACTLY to the law of Biot
and Savart for the magnetic effect of an electric
current ."
Handbook of Physics",Condon & Odishaw, McGraw-Hill 1967,
Part 3, Chapter 2 (Fluid Mechanics) Section 6 Page 3-20
& 3-21]
Which is of course, what Maxwell said he was modeling and attempting to describe
and quantify with his now famous equations. Further, solenoidal fields have an
interesting feature; the vorticity suppresses longitudinal propagation,
transforming it into transverse action by gyroscopic action. At very short
ranges however, this component can be manifested. But the large-scale result is
propagation of purely transverse wave action. Kelvin showed that small linear
disturbances in such a system (which he termed a vortex sponge) result in the
characteristic behavior of light.
So we have a basis which might suggest that light could be in fact, as the
equation presents itself, propagation of wave action in a vortex sponge (or
solenoidal field). But which is which? Does permeability represent the
coefficient of compressibility or density?
Well we now look to standard Continuum Mechanics to deduce these properties:
The basic continuity equation of Continuum Mechanics is given as:
d(rho)/dt + (rho)Div v = 0
Where rho is the field density, and v is the mean velocity. If the field is
incompressible this simplifies to:
(rho)Div v = 0
Since with the incompressible assumption, there can be no change in density.
We can further simplify the equation by removing density (dividing it from both
sides) we then get:
Div v = 0
This definition requires infinite propagation speeds of any perturbations in
such incompressible systems, eliminating any possibility of wave activity.
Conversely, in compressible mediums we see that (rho)Div v equals the time rate
of change in the density d(rho)/dt. For the limit, as a volume element [s] go
to zero, we get:
s(rho)Div v = s(d(rho)/dt)
This is based on the observation that for the two terms to sum to zero, and
therefore must have opposite signs. This leads directly to:
mDiv v = dm/dt
And cannot be zero. This is an important finding, it describes a unique
characteristic of all compressible systems. The result of this is a fixed finite
propagation speed for any perturbations in the resulting continuum,
leading to the ability of the resulting field to have acoustic behavior.
In general, the physical consequences of this definition has been overlooked
due to an almost universal adoption of the 'assumption' of incompressibility, in
evaluating the general behavior of such systems. This eliminates many higher
order terms, greatly simplifying the equations, and generally doesn't introduce
significant errors in the results obtained.
It does however eliminate this property and any resulting consequences from any
such evaluations. As should be obvious, as a limit, this definition has a
unique value fixed by the density and velocity of the constitute continuum.
So, what is the above equation saying? It appears to be saying that
compressible medium will have a basic oscillation of density fluctuation
occurring continuously. Moreover, given a generally uniform density and
velocity, this fluctuation will have a distinctive frequency associated with
this activity. This is clearly demonstrated by the relationship:
Div v = d/dt
When applied to Electromagnetism where would one find this? There is a
fundamental property that has remained undefined (and given arbitrary units),
this is charge [q]. So, if we assign to charge the units [kg/sec] and assume it
is a result of the definition above, what is the result?
In Coulomb's law, the force resulting from the interaction of two charges is given
to be:
F = [1/4pi(eps)][qq/r^2]
Following our assumption we find that permitivitty [eps] must have units of
density to get a result in units of force.
This provides us with a means of mapping the EM terms into the more commonly
known medium properties.
But we did make an intuitive leap in the above, namely we assumed that the
property called charge was related to the inherent compressibility of the
medium. Let's now look at the next characteristic property of compressible
mediums, namely the “Characteristic Impedance of the Medium”. The term usually
designated [Z] is:
---
P / u
Z = - = c[rho] = / ---
c v rho
Knowing that for our discussion P is young's modulus [Y] such that u = 1/Y, thus
Y = 1/u so, we would expect that uc should yield 376.7. It does, of course.
However, the astute reader will immediately recognize that Z is actually the
inverse the Z as used in EM theory where ada [n] represents the above value.
However, the fact remains clear, the signature relationship is one to one.
Referenced below is Sec 4, Chapter 8 (Acoustics), Part III (Mechanics of
Deformable Bodies) of Condon & Odishaw's “Handbook of Physics”, McGraw-Hill 3
Edition describing this:
=======================================================================
4. Boundary Conditions. Impedance and Absorption Coefficients
Accompanying the differential equations (8.2) to (8.9) or Kirchhoff's formula
are the boundary conditions of continuity of velocity normal to the boundary and
continuity of pressure.
The ratio between the particle velocity at a point in a field and the sound
pressure is termed the specific acoustic admittance.
n*_i u*_i
---- = ---- i = 1,2,3; p = density; P = pressure
pc P
It is a vector with the same direction as u*_i. The inverse of n*/pc in a given
direction is termed the “specific acoustic impedance”.
Z = kpc = (n*/pc)^-1 = (theta – ix)pc
For a plane wave in the x direction the specific acoustic impedance or the
“characteristic impedance of the medium” is real and equals
P
Z = - = pc
u
(for air ~_- 41.5 cgs at 20'C) [51, where p = density, c = velocity of sound
(f--340 m/sec at 20'C in air). The radiation impedance of a vibrating surface
is the ratio of the pressure at the boundary and the particle velocity of the
surface.
The ratio between the pressure and normal velocity at a boundary is referred to
as the normal impedance of the boundary. In general this quantity is not known a
priori and can be determined first after the field has been found, utilizing the
boundary conditions mentioned above. The normal impedance will then in general
be a function of the angle of incidence. However, for some special material (a
"wall" with pores normal to the surface) the particle velocity is always normal
to the boundary and will depend only on the local pressure at the point under
consideration. For such a “locally reacting” or “point-reacting” boundary the
normal impedance will be independent of the sound field and can be specified in
advance as a characteristic property of the boundary. Under those conditions the
analysis of many field problems, sound waves in rooms, etc., is considerably
simplified (6]. Many materials met in practice are approximately locally
reacting, e.g., perforated porous tiles, dense porous homogeneous material,
cavity-resonator arrangements, etc.
...
======================================================================
Finally as mentioned above, the ideal relationship of a longitudinal wave speed
(which is also the RMS of a standard distribution under kinetic theory) and the
transverse wave speed is a characteristic Sqrt(3). Maxwell's vortex model would
predict that such a property should exist in nature and be observable. Further,
in Maxwell's model this phenomena should only be manifest over very short
distances. In 1993 speed measurements were made for tunneling photons in a one
micron thick glass and compared to light not involved (passing thru equivalent
air). The photon that tunneled arrived on average 1.7 time faster than the non-
tunneling cousins. See Scientific American, August 1993 “Faster than Light?”
for complete details. But, needless to say, a measured 1.7 is in agreement to
the 1.732 resulting from the idealized value.
Dr Petr Beckmann in his book “Einstein Plus Two” (pg 176-178) observes the same
1.73 factor manifesting itself as the slope of a line representing the Titus-
Bode relationship. He calls this “the mysterious number 1.73”.
Now at this juncture one must make a philosophical decision, either say well,
that's one hell of a series of coincidences or, there seems to be direct
correlation between the two. Ockham's Razor would argue for the latter, and the
former seem logically flawed...
Paul Stowe
Alan Boswell
Paul Stowe wrote:
>
>> >:
> >: Electromagnetic energy in the form of radiation does NOT
> >: propagate without a medium.
> >
> >
> >....and your evidence for this rather remarkable conclusion is....?
>
> Well for me, the "Characteristic Impedance of the Medium"...
>
that's just a shorthand phrase for "the characteristic
impedance of EM plane waves in a vacuum" i.e. in nothing.
Alan
Richard Herring
And *that* is shorthand for "self-inflicted constant because we
perversely insist on measuring electric and magnetic fields in
different units".
It doesn't tell us anything "deep" about the nature of the
medium or the lack of one.
--
Richard Herring | <richard...@baesystems.com>
brian a m stuckless
Dear Harry,
Nature abhors ''NO medium''.
There is a medium everywhere measured, to date.
The calculated medium of intergalactic deep space has atoms
which occupy a space of aproximately 10^3 m^3 / atom, each.
[0.001 atoms per cubic meter]
[= 10^-3 atoms / m^3]
The measured medium achieved by hanging a plate off the
back of the space-shuttle was aproximately 10^-9 m^3 / atom,
each.
[1,000,000,000 atoms per cubic meter]
[= 10^9 atoms / m^3]
The very least medium achieved on earth is aproximately
10^-12 m^3 / atom, each.
[1, 000,000,000,000 atoms per cubic meter]
[= 10^12 atoms / m^3]
The international standard molar medium has atoms with ~
10^-25 m^3 / atom, each.
[10,000,000,000,000,000,000,000,000 atoms per cubic meter]
[= 10^25 atoms per cubic meter]
Note that the very least measured medium achieved on earth had
10^3 times MORE atoms / m^3 than the space-shuttle test medium,
and 10^15 times MORE atoms / m^3 than deep space. --However,
that's only 10^13 times LESS atoms / m^3 than the density of the
international standard molar medium.
Therefore, with a density spread of 28 orders of magnitude, the speed
of light has only ever been measured on earth in a medium 2 orders
of magnitude closer to the international standard molar density than
it is to deep space density(which is still not a vacuum).
''In vacu'' is a gross misnomer.
These mediums also contained already-in-transit electromagnetic
energy in the form of radiation, and already-in-transit phonons.
Only atoms were counted.
Deep space with 1000 m^3 per atom is just as much a medium
as the international STP medium with 10^-25 m^3 per atom.
One atom of water is just as wet as an ocean.
Water is to an ocean, as are trees to a forest,
and trees are to a forest, as a medium is to aether.
It seems, you can't see aether for the medium.
Yours truly,
`````arcsign````` mar 12, 2001
Harry H Conover
: in article <98egg8$s...@news-central.tiac.net>
: > Paul Stowe (pst...@ix.netcom.com) wrote:
: > : In article <98ee7r$7...@news-central.tiac.net>,
: > : con...@tiac.net (Harry H Conover) wrote:
: > : >brian a m stuckless (bast...@avalon.nf.ca) wrote:
: > : >: Nature abhors ''NO medium''.
: > : >:
: > : >: Electromagnetic energy in the form of radiation does NOT
: > : >: propagate without a medium.
: > : >
: > : > ....and your evidence for this rather remarkable conclusion
: > : > is....?
: > :
: > : Well for me, the "Characteristic Impedance of the Medium"...
: >
: > If not (what is it) 377-Ohms or so, is there a figure that you
: > would prefer? Why?
: >
: > Sorry, but I don't see a connection between the characteristic
: > impedance of free space and a need for a propagation medium.
: >
: > Harry C.
: ^^^^^^^^^^^^^^^^^^reply:
: Dear Harry,
:
: Nature abhors ''NO medium''.
:
: There is a medium everywhere measured, to date.
Yes, and the most common propagation medium for electromagnetic
waves is a vacuum. It's properties too are measured and well
known. Also, the more perfect the vacuum, the more perfect are
its propagation characteristics for electromagnetic waves.
Notice how well light and other electromagnetic waves propagate
through free space (vacuum), with any contamination of the vacuum
only serving to attenuate them.
[non-relevant material snipped, including statement that confuse
contamination of the progation medium with the medium itself]
: Therefore, with a density spread of 28 orders of magnitude, the speed
: of light has only ever been measured on earth in a medium 2 orders
: of magnitude closer to the international standard molar density than
: it is to deep space density(which is still not a vacuum).
You point bein what -- that the the available media index of refraction
range is limited?
Your point escapes me, but I believe like another genetleman who
contributes to this thread, you both confuse the properties of
media that transmit matter waves with those that transmit
electromagnetic waves. Since totally different physics concepts
are involved for these two very differnt forms of wave motion, the
media characteristics for each type of waveform transmission are
entirely different.
A simple experiment will demonstrate to you that other than causing
very slight changes in the media index of refraction and attenuation
characteristics, there is essentially no difference in the propagation
of electrmagnetic waves through a rarified atmoshere or through a
near perfect vacuum.
Simply place an interferrometer (sp?) in a small vacuum system. Take
and initial set of measurements with the vacuum system pumped down
to about 0.1 Torr, a very poor vacuum with a good fraction of an
atmosphere of pressure remaining. Observe the frige pattern. Then
pump the chamber down to, say, 10^-6 Torr, about the level of vacuum
present in your TV CRT. You won't see much of a fringe shift, indicating
the the index of refraction and correspondingly the propagation velocity
of light in the media has barely changed, despite the fact the the
pressure within the chamber has undergone a pressure change of over
5-Orders of Magnitude.
This pretty well eliminates any possiblity that matter contamination
of vacuum plays any significant role in the transmission of light or
any other form of electromagnetic wave through it.
I'd also point out that once you grasp the concept and understand
electromagnetic waves, you'll also understand why the require no
material transmission media -- a vacuum suffices quite nicely,
thank you.
Hope this helps.
Harry C.
Alan Boswell
brian a m stuckless wrote:
>
>
> Deep space with 1000 m^3 per atom is just as much a medium
> as the international STP medium with 10^-25 m^3 per atom.
>
OK, but this argument applies equally if the density of atoms is one per
cubic light-year. Would that be a "medium"? What density would you
accept as indicating a vacuum? If your true vacuum requires no atoms at
all in the entire universe, it's a little exclusive I think, and the
argument is not leading anywhere.
Alan
Bilge
>
> Nature abhors ''NO medium''.
That must be why they advertise their services on late night TV
for readings via 1-900 lines.
>There is a medium everywhere measured, to date.
>
>The calculated medium of intergalactic deep space has atoms
>which occupy a space of aproximately 10^3 m^3 / atom, each.
>[0.001 atoms per cubic meter]
>[= 10^-3 atoms / m^3]
I'll take this as seriously as possible. Assuming I have a
proton target (H, the stuff in space), with a thickness of
1 nano-gram/cm^2, (which is so thin as to be ridiculous in
terms of stopping power), what does this translate to in terms
of distance for your 10^-3 atoms/m^3? Why, approximately 5 x 10^17
cm or about 16 million light years.
This doesn't constitute a medium no matter how you look at it.
In order to qualify, other particles have to have some chance of
hitting one of the constituents and the constituents have to be
close enough together to actually exhibit some sort of effect
based on interacting amoung themselves. This is the most optimistic
one may possibly be and not get struck by lightning.
[...]
>The measured medium achieved by hanging a plate off the
>back of the space-shuttle was aproximately 10^-9 m^3 / atom,
>each.
>[1,000,000,000 atoms per cubic meter]
>[= 10^9 atoms / m^3]
>
>The very least medium achieved on earth is aproximately
>10^-12 m^3 / atom, each.
>[1, 000,000,000,000 atoms per cubic meter]
>[= 10^12 atoms / m^3]
Do you ever consider the physical implications associated with
the numbers you throw out or are you one of those people that starts
punching numbers into a calculator mindlessly? Do you have the slightest
idea what the mean free path of an atom of anything happens to be
in even the most awful vacuums? Say, like 10^-4 torr?
>Note that the very least measured medium achieved on earth had
>10^3 times MORE atoms / m^3 than the space-shuttle test medium,
>and 10^15 times MORE atoms / m^3 than deep space. --However,
>that's only 10^13 times LESS atoms / m^3 than the density of the
>international standard molar medium.
Do you really believe this constitutes an argument for anything
but a sanity hearing?
[...]
>''In vacu'' is a gross misnomer.
Only if you happen to be among those that consider the ouija board an
indispensible scientific instrument which is almost as valuable as
your divining rod and magic wand.
>One atom of water is just as wet as an ocean.
There is no such thing as an atom of water.
>Water is to an ocean, as are trees to a forest,
>and trees are to a forest, as a medium is to aether.
>
>It seems, you can't see aether for the medium.
Or you've sniffed too much of it and gone from medium to
well done or completely baked.
Gordon D. Pusch
It's also worth pointing out that in the case of sound, a medium cannot
support wavelengths much shorter than the mean interparticle spacing;
below this cutoff, energy propagates via diffusion as _heat_, not waves.
If stuckless is claiming that the medium of ``deep space with 1000 m^3 per
atom'' (i.e., mean spacing ~10 meters) can support wavelengths that are
over _TWELVE ORDERS OF MAGNITUDE_ shorter, he had better provide a
mechanism for why light still behaves like a wave in it, instead of
a random diffusion process.
-- Gordon D. Pusch
perl -e '$_ = "gdpusch\@NO.xnet.SPAM.com\n"; s/NO\.//; s/SPAM\.//; print;'
Aladar Stolmar
"Gordon D. Pusch" <gdp...@NO.xnet.SPAM.com> wrote in message
news:m2vgpdu...@pusch.IntegratedGenomics.com...
Excellent remark.
The resolution is the understanding of particles and photons - including the
elements of fields or virtual photons - as collision event progressions.
With this understanding we fill-up the space with spontaneous collision events
and the spacing between them will be able to support very large frequency
waves also.
sch...@gefen.cc.biu.ac.il
: It's also worth pointing out that in the case of sound, a medium cannot
: support wavelengths much shorter than the mean interparticle spacing;
: below this cutoff, energy propagates via diffusion as _heat_, not waves.
Do you mean "mean interparticle spacing" or "mean free path"?
-----
Richard Schultz sch...@mail.biu.ac.il
Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel
Opinions expressed are mine alone, and not those of Bar-Ilan University
-----
". . .Mr Schutz [sic] acts like a functional electro-terrorist who
impeads [sic] scientific communications with his too oft-silliness."
-- Mitchell Swartz, sci.physics.fusion article <EEI1o...@world.std.com>
Richard Herring
> "Gordon D. Pusch" <gdp...@NO.xnet.SPAM.com> wrote in message
> news:m2vgpdu...@pusch.IntegratedGenomics.com...
> > Alan Boswell <alan.b...@nospam.baesystems.com> writes:
> >
> > > brian a m stuckless wrote:
> > >>
> > >> Deep space with 1000 m^3 per atom is just as much a medium
> > >> as the international STP medium with 10^-25 m^3 per atom.
> > >
> > > OK, but this argument applies equally if the density of atoms is one per
> > > cubic light-year. Would that be a "medium"? What density would you
> > > accept as indicating a vacuum? If your true vacuum requires no atoms at
> > > all in the entire universe, it's a little exclusive I think, and the
> > > argument is not leading anywhere.
> >
> > It's also worth pointing out that in the case of sound, a medium cannot
> > support wavelengths much shorter than the mean interparticle spacing;
> > below this cutoff, energy propagates via diffusion as _heat_, not waves.
> >
> > If stuckless is claiming that the medium of ``deep space with 1000 m^3 per
> > atom'' (i.e., mean spacing ~10 meters) can support wavelengths that are
> > over _TWELVE ORDERS OF MAGNITUDE_ shorter, he had better provide a
> > mechanism for why light still behaves like a wave in it, instead of
> > a random diffusion process.
> Excellent remark.
> The resolution is the understanding of particles and photons - including the
> elements of fields or virtual photons - as collision event progressions.
Except that photons don't collide with other photons.
Their dynamics couldn't be more different from those of particles
in a gas carrying sound waves.
> With this understanding we fill-up the space with spontaneous collision events
> and the spacing between them will be able to support very large frequency
> waves also.
Nonsense.
--
Richard Herring | <richard...@baesystems.com>
Gordon D. Pusch
> In sci.physics.fusion Gordon D. Pusch <gdp...@no.xnet.spam.com> wrote:
>
>: It's also worth pointing out that in the case of sound, a medium cannot
>: support wavelengths much shorter than the mean interparticle spacing;
>: below this cutoff, energy propagates via diffusion as _heat_, not waves.
>
> Do you mean "mean interparticle spacing" or "mean free path"?
Take your pick --- they both represent cutoff scales. (Think about
``aliasing'' on a spring/mass chain: You can't resolve a wavelength
shorter than the particle spacing --- they get folded back via the
Brillouin-zone construction.)
brian a m stuckless
```snip```
> electromagnetic waves,-- ```snip``` --require no
> thank you.
>
> Hope this helps.
> Harry C.
Dear Harry,
What ''vacuum'' ?
There is a medium everywhere measured, to date.
Note that the very least measured medium achieved on earth
has 10^15 times MORE atoms / m^3 than deep space(which is
still not a vacuum).
The speed of light has only ever been measured on earth in a
medium at least 15 orders of magnitude more dense than deep
space.
[1, 000,000,000,000,000 atoms / m^3 more than deep space]
[10,000,000 atoms / m^3 less than molar gas standard(~Air)]
'One atom of water is just as wet as an ocean.
Water is to an ocean, as are trees to a forest,
and trees are to a forest, as a medium is to aether.
It seems, you can't see aether for the medium.
Yours truly,
`````arcsign````` mar 13, 2001
^^^^^^^^^^^^^^^^^^^^^^^^^^^^reference follows:
Nature abhors ''NO medium'', too.
As per conservative anaylsis:
a. The calculated medium of intergalactic deep space has atoms
which occupy a space of approximately 10^3 m^3 / atom, each.
[0.001 atoms per cubic meter]
[= 10^-3 atoms / m^3]
b. The measured medium achieved by hanging a plate off the
back of the space-shuttle was approximately 10^-9 m^3 / atom,
each.
[1,000,000,000 atoms per cubic meter]
[= 10^9 atoms / m^3]
c. The very least medium achieved on earth is approximately
10^-12 m^3 / atom, each.
[1, 000,000,000,000 atoms per cubic meter]
[= 10^12 atoms / m^3]
d. The very least medium on earth --in which light has ever been
actually measured-- has approximately 10^-16 m^3 / atom, each.
[10,000,000,000,000,000 atoms per cubic meter], rangeing up to
[1,000,000,000,000,000,000,000 atoms per cubic meter]
[= a range of 10^16 ~ 10^21 atoms / m^3]
e. The international standard molar gas medium has atoms with ~
10^-25 m^3 / atom, each.
[10,000,000,000,000,000,000,000,000 atoms per cubic meter]
[= 10^25 atoms per cubic meter]
NOTE: The very least measured medium on earth in which light
has ever been actually measured had:
~10^7 times MORE atoms / m^3 than the space-shuttle test medium,
and ~10^19 times MORE atoms / m^3 than deep space. --However,
that's only ~10^6 times LESS atoms / m^3 than the density of the
international standard molar gas medium.
Therefore, with a density spread of 28 orders of magnitude, the
very least medium --in which the speed of light has ever actually
been measured-- on earth, is a medium 7 orders of magnitude closer
to the international standard molar density than it is to deep space
density(which is still not a vacuum) --clearly demonstrated as
follows:
a. Deep space:
[0.001 atoms per cubic meter]
[= 10^-3 atoms / m^3]
b. Space suttle:
[1,000,000,000 atoms per cubic meter]
[= 10^9 atoms / m^3]
c. Very least test medium achieved on earth:
[1, 000,000,000,000 atoms per cubic meter]
[= 10^12 atoms / m^3]
d. Light test range(as-measured on earth):
[10,000,000,000,000,000 atoms per cubic meter], --to:
[1,000,000,000,000,000,000,000 atoms per cubic meter]
[= a range of 10^16 ~ 10^21 atoms / m^3]
e. Molar gas standard:
[10,000,000,000,000,000,000,000,000 atoms per cubic meter]
[= 10^25 atoms per cubic meter]
These mediums also contain already-in-transit electromagnetic
energy in the form of radiation, and already-in-transit phonons.
In the beginning was the phonon.
`````arcsign````` mar 13, 2001
^^^^^^^^^^^end of post.
Aladar Stolmar
"Richard Herring" <r...@gmrc.gecm.com> wrote in message
news:98la2s$41a$8...@miranda.gmrc.gecm.com...
Pay attention! The collision events are the constructing elements of photons,
and everything else. The collision event progressions are self-supporting
systems, the collision events are collisions of elements of an underlying
subltance.
> Their dynamics couldn't be more different from those of particles
> in a gas carrying sound waves.
Their constructing elements could be understood by this analogy, using
infinite speed constructing elements, colliding with each other.
>
> > With this understanding we fill-up the space with spontaneous collision
events
> > and the spacing between them will be able to support very large frequency
> > waves also.
>
> Nonsense.
On the contrary, a very easy useful representation of everything.
>
> --
> Richard Herring | <richard...@baesystems.com>
--
Aladar
http://www2.3dresearch.com/~alistolmar/
Harry H Conover
: Harry H Conover <con...@tiac.net> wrote in article
: <98j9pb$5...@news-central.tiac.net>...
: ```snip```
: > electromagnetic waves,-- ```snip``` --require no
: > material transmission media -- a vacuum suffices quite nicely,
: > thank you.
: >
: > Hope this helps.
: > Harry C.
: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^reply:
: Dear Harry,
:
: What ''vacuum'' ?
:
: There is a medium everywhere measured, to date.
<plonk> Sound of another troll going into the byte crapper!
Harry C.
Maury Markowitz
> What ''vacuum'' ?
>
> There is a medium everywhere measured, to date.
The two are logically unrelated. Just because there are atoms in space
means nothing whatsoever in terms of whether or not they are the medium of
light. You might as well say that grass is the medium of wind, because you
often see grass blowing in the wind.
How can you demonstrate that these atoms are not the medium of light?
Trivially easy, measure energy exchange. You'll find that the atoms in space
have a nice linear effect on the transmission of light whether or not they
are interacting strongly - the implication being that the light is still
transmitting even though it's not interacting with the atoms. The
interaction with the atoms is the definition of the medium.
Of course all of this stems from your apparent misunderstanding that light
"is" a wave. It is not, it is a particle. Particles have no need for a
medium.
Maury
Alan Boswell
Richard Herring wrote:
And *that* is shorthand for "self-inflicted constant because we
> perversely insist on measuring electric and magnetic fields in
> different units".
I quite agree, except for "perverse". Don't different physical
quantites require different units, like voltage in volts and current in
amps?
Alan
Richard Herring
No. Only different dimensions.
Read Jackson's Appendix on Units and Dimensions for the full story.
Look at cgs units, for example. Either permittivity or permeability
is a dimensionless 1, and the other is 1/c^2. There's no independent
dimension of charge, so voltage and current are measured in statvolts
and statamperes or abvolts and abamperes. All are combinations
of centimetres, grams and seconds.
In fact that c^2 is only there because we insist on measuring distance
and time in different units, even though we know from SR that they
are just two aspects of a single unified space-time. Set it to 1,
and you get Gaussian units, where c, Z, epsilon_0 and mu_0 are all 1.
--
Richard Herring | <richard...@baesystems.com>
Rene Tschaggelar
It is a bit hard to grasp, I know.
Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
Joe Rongen wrote:
>
> "brian a m stuckless" <bast...@avalon.nf.ca> wrote in message
> news:01c0a98a$32aaa8c0$LocalHost@default...
> > Nature abhors ''NO medium''.
> >
> > Electromagnetic energy in the form of radiation does NOT
> > propagate without a medium.
>
> Interesting, so how did you manage to
> read this reply from a CRT type monitor?
>
> Those electrons in the CRT are in a good
> vacuum and controlled by the deflection coils
> around the CRT's neck.
>
> Where is your 'medium'?
Gordon D. Pusch
> Vaccum is not nothing, but carrier of EM fields.
> It is a bit hard to grasp, I know.
EM fields are perfectly capable of propagating through nothing.
It is a bit hard to grasp, I know.
rbw...@rmi.net
<bast...@avalon.nf.
>
>Water is to an ocean, as are trees to a forest,
>
>We can't see aether for the medium.
>
>''In-vacu'' is a misnomer, --nature abhors ''NO medium''.
>
> Yours truly,
>`````arcsign````` mar 10, 2001
>
I would like to get your opinion of a idea I have about the
Michelson-Morley experiment. We take the Galillean transformation equations
x'=x-vt
y'=y
z'=z
t'=t
Now consider the last equation, the one about time. Scientists say they
have proven that time passes slower in a moving system, etc., and use two
cesium clocks to prove their theory, a clock in an orbiting satellite
registering less time than an identical clock on earth, etc. Therefore, they
say, the Galillean transformation equations cannot be used, and we have to
use the Lorentz equations to describe transmission of light. Well, forget
about the clocks, which obviously have a difference of kinetic energy, and
consider the rotation of the earth during one orbit of the satellite. Are
scientists trying to tell us that the earth rotates fewer degrees during an
orbit as observed from the satellite?
Sorry, I don't believe it, so I am just going to use degrees of rotation of
the earth as my basis for time in these calculations and go ahead and use the
Galillean transformation equations, making the wild assumption that the earth
rotates the same number of degrees during the orbit of a satellite as it does
during the orbit of a satellite. I hope I am right about this.
Now having established t'=t, disregarding clocks altogether for this
calculation, we take two sets of coordinates, K at rest and K' in motion in
the +x direction relative to K. At the time the origins of K and K'
coincide, we send two photons, one in the +x direction and one in the -x
direction. After a time of t has elapsed, we observe from K that K' has
traveled a distance of vt. The two photons are equal distances from the
origin of K, x for the +x photon and -x for the -x photon.
Now we take the same event that Einstein used to base his theory on, the
arrival of the +x photon at a distance of x from the origin of K, and
describe events from K'. According to the Michelson-Morley experiment, the
two photons have to be equal distances from the origin of K'. Consequently,
we see that seen from K', when the +x photon reaches its position, the origin
of K' has not traveled a distance of vt from the origin of K, and the -x
photon is a proportionally shorter distance on its path to a distance of -x
from the origin of K.
If we designate the time when the +x photon reaches its position as
measured in K' as t3, then we see that t3 is a lesser time than t, explaining
why the origin of K' is at a shorter distance from the origin of K than vt.
We can express the time and distance for this event from K' by the equation
c(t3) = x - v(t3)
where c(t3) is the distance to the +x photon from the origin of K'. The same
equation can be used to calculate the time the -x photon will reach a
distance of -x from the origin of K, in this case being a longer time than t.
Notice that the time of t3 agrees with the equation t'=t because when K'
reaches a distance of vt from K, t' will equal t, just as the Galillean
transformation equations show.
This would indicate that photons, instead of being little pellets or
beebees are really pure energy which leads energy from the other frame of
reference in proportion to velocity in the direction of motion.
I arrived at this idea a few weeks ago, but since I am neither a scientist
nor well educated, I am having some difficulty communicating it to others. I
would like to get your opinion and any suggestions you might have.
Robert B. Winn
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brian a m stuckless
Alan Boswell <alan.b...@nospam.baesystems.com> wrote
in article <3AADEBD1...@nospam.baesystems.com>
> brian a m stuckless wrote:
> >
> > Deep space with 1000 m^3 per atom is just as much a medium
> > as the international STP medium with 10^-25 m^3 per atom.
This is in agreement with Maxwell's equations derivation.
> OK, but this argument applies equally if the density of atoms is
> one per cubic light-year. Would that be a "medium"?--
Where do YOU speculate that condition exists? It's
very, very, far out of the range of this discussion.
> What density would you accept as indicating a vacuum?--
NOT ten million Billion parts per cubic meter Which is
19 orders of magnitude MORE than deep space, per m^3.
You are confusing ''reduced pressure'' --or specifically,
''reduction'', with ''vacuum''.
> --If your true vacuum requires no atoms at
> all in the entire universe,--
We know that's NOT the case, Jim.
Ten million Billion parts per cubic meter = 10^16 parts / m^3
= 10.000,000,000,000,000 particles / m^3
--is a far cry from ''no atoms at all in the entire universe,--''.
''NO molar particles'' doth NOT a ''universe'' make.
You can't see the forest for the trees, Alan.
Yours truly,
`````arcsign````` mar 29, 2001
> Alan
>