Evidence of the Existence of the Aether
reticher
It is the currently accepted "truth" that there is no experimental evidence of
the Aether. This seems surprising because, if the Aether made its presence any
more obvious, physicists would have toothmarks on their butts from where the
Aether had jumped up and bit them. A few examples:
Empty space, other than occupying a volume, has at least two observable
properties, its dielectric constant and its permeability. If space is "empty",
what is it that has these properties? Since, if space consisted of a volume of
"nothingness", it cannot have any properties other than volume. Since it does,
space it would seem that it does contain a rejected entity, which was called
the Aether in the 19th Century!.
The velocity of light in free space is determined by the dielectric constant
and the permeabiluty of space in accordance with accepted physical laws.
Similarly, the velocity of sound in a steel rod is determined, using an
equation analogous to that used to determine the velocity of light, by the
elasticity and density of the steel. While a modern physicist seems to assert
that the existence of the dielectric constant and the permeability of space
does not require the existence of a medium (i.e.- the Aether) having these
properties, it seems certain that he would not be foolish enough to assert
that, since the velocity of sound in a steel rod is determined only by its
elasticity and its density, he could remove the rod and retain the density and
elasiticity so that the sound could propagate. One might reasonably wonder if
something is lacking in the mentality of physicists who would tolerate such a
double intellectual standard.
Special Relativity (and the Lorentz Contraction Aether Theory) tell us that
the velocity of light is independent of the velocity of its source. Such an
independence is characteristic of a wavelike disturbance propagating thorugh a
medium, it is not characteristic of an entity propagating ballistically through
"empty" space. For example, the sound of a rifle shot travels at the velocity
of sound in air and which is independent of the velocity of the rifle. The
velocity of the bullet from the rifle, however, is additively determined by the
velocity of the rifle. If we abolish the Aether the observed independence of
the velocity of light from the velocity of its source would seem to require
that some form of "magic" must be at work since light would then have no way to
determine how fast it is supposed to travel. Since the writer finds the idea of
"magic" accounting for observations in the science of physics extremely hard to
accept, he would be greatful if someone could provide some other explanation
which does not require a medium (the Aether). (P.S.- Handwaving arguments about
spacetime don't serve the purpose.)
If one examines Dirac"s treatment of the creation of complementary particles
from high energy photons, he finds that Dirac had to invent a "sea of negative
energy pervading all of space" in order to make his mathematics work. Aside
from the fact that "negative energy", in the sense that Dirac used it, would
seem not only to be unexplained but to be impossible, how does that postulated
"sea of negative energy" differ from the Aether? Both say that "something"
exists in a space which had been arbitrarily assumed to be empty!
Current theory asserts that forces which "act at a distance" do so as a result
of "virtual particles" (e.g.- virtual photons, gravitons, gluons, etc.) which
pop into and out of existence so rapidly that quantum uncertainties prevent the
Law of Conservation of Energy from being violated. There would seem to be at
least two problems with this concept:-
The first is the problem is the number of particles involved. At any instant
of time, every particle in the Univserse gravitationally attracts.every other
particle and they must therefore be exchanging gravitons. Furthermore, these
gravitons, since they are virtual, must be extremely short lived. The number of
gravitons involved in this concept is staggering. At any instant for example,
assuming that the presently accepted value of 10^80 particles in the Universe
is reasonably correct, there must be at least 0.5*(10^160) particles existing
at one time. The size of this number is increased by the fact that they must be
short lived and therefore must be replaced continuously (gravity appears to be
continuous), suggests that a bit of skepticism would be in order.
The second problem is that of expaining how the exchange of "virtual
particles" accomplishes the production of an "attractive force". The production
of a repulsive force by the exchange of particles is obvious. Two athletes
throwing a medicine ball back and forth experience a repulsive force because of
momentum exchange between them. There does not seem to have ever been a viable
explanation as to how these particles act to produce an attractive force. The
momentum exchange requirement to produce an attractive force prohibits the
production of such a force unless there is a substrate (e.g.:- the Aether) with
which they can exchange momentum. A boomerang returns to the thrower because it
exchanges momentum with such a medium (air). If space were empty (no Aether),
such a momentum exchange cannot occur and the virtual particles could only
produce a repulsive force. The writer is aware that the mathematics associated
with the concept of "virtual partcles" allows for the production of attractive
forces. He is also aware of the fact that there are many mathematically treated
problems in which the mathematics deals with conditions outside of other
constraints related to the problem and which limit the applicability of the
mathematics. This would seem to be such a case. Allow the Aether or its
equivalent to be present as a substrate and the mathemetics of "virtual
photons" works.
There are two interpretations of Quantum Theory. One interpretation involves
parallel Universes coexisting in the same space as our own that are created
every time that a particle makes a "quantum choice". This approach not only
suffers the problem of absurdly large numbers described for gravitons, it
requires that an amount of energy equal to the energy content of our entire
Universe multipled by the number of particles in the Universe be created at
every instant of time. It also requires that the volume of space occupied by
each particle contain an amount of energy equal to the energy equivalent of
that enormous amount of energy. Since the writer accepts the concepts that
energy is conserved and two entities cannot occupy the same space at the same
time, he considers the interpretation to be rather foolish.
The other interpretation of Quantum Theory requires that quantum effects
propagate at an "infinite" velocity. Most of the acedemic hierarchy seems to
consider this interpretation to be both "spooky" and wrong since "Special
Relativity clearly shows that nothing can travel faster than light". To the
contrary, Special Relativity shows the opposite. What it shows is that an
entity represented by energy cannot travel faster than the velocity of light
and the scientific community has fallen into the trap of assuming that
everything which is observable is represented by some form of energy. The
reason that the Special Theory of Relativity imposes its restrictions on
velocity is that the Lorentz Transformation for energy, in a force-length-time
system of units is 1/(1-V^2/C^2)^0.5. This means that, at the velocity of
light, energy becomes infinite and that, at velocities above that of light,
energy becomes imaginary. The effect limits the propagation of energy, and
hence the velocity of conventional communication, to the velocity of light.
Experiments have shown that the polarization of "paired photons" is coupled so
that changing the polarization direction (the quantum number) of one causes the
polarization direction of the other to change in correspondence. Experiments
have also shown that the coupling propagates at at least 4 times the velocity
of light and perhaps at an infinite velocity. (It should be noted that changing
the polarization direction of a photon does not change its energy content so
the the coupling of polarization direction between them does not involve energy
transfer). The polarizarion direction of photon is an angle which is measured
in radians (the distance along an arc divide by the radius of that arc). The
Lorentz Transformation for Angle is therefore equal to the Lorentz
Transformation for Length divided by the Lorentz Transformation fo Length and
is therefore equal to unity and is independent of velocity. (Above and below a
velocity of C, the value of unity is obvious. At the velocity of C, the value
is nominally indeterminate but can be shown to be equal to the value of unity
by borrowing a technique from Calculus). With a value of unity for the Lorentz
Transformation for Angle, it seems reasonable to assert that Special Relativity
requires the polarization couplingof paired photons to travel at the infinite
velocity required by the second, more rational interpretation of Quantum
Theory. Assuming that this is the case, then the current concept of "spacetime"
as a single entity must be in error. An absolute time and velocity reference is
required.
If one examines both the Special Theory of Relativity (STR)and the
Lorentz-Contraction Aether Theory (LCAT), he finds that they are actually the
same theory in that they are cross derivable. (LCAT) is actually a special case
solution of (STR) and cannot be disproven without disproving (STR) as well.
LCAT asserts that the Aether exists but because information cannot propagte
faster than light (quantum theory tells us that this is not true), we cannot
find our velocity with respect to it. STR tells us that because our velocity
with respect to the Aether cannot be measured, it does not exist in the theory!
It does not precluide its existence nor its effect from being observed by other
means. Dr. Einstein asserted "remember gentlemen, we have not disproven the
existence of the Aether, we have only proven that we do not need it (for
computations)".
Unfortunately, Modern Physics seems to have been taken over by mathematical
idiot savants who lack an appreciation for the implications of mechanism and
strive to suppress the contributions of those who would dare to say "Hey, wait
a minute"
http://www.members.aol.com/reticher/site.htm
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parts, please do not raise objections that are not related to material that you
have read at the Website. This posting is merely a summary.
E-Mail reti...@aol.com
The material at the Website has been posted continuously for over 5 years. In
that time THERE HAVE BEEN NO OBJECTIVE REBUTTALS OF ANY OF THE MATERIAL
PRESENTED. There have only been hand waving arguments by individuals who have
mindlessly accepted the prevailing wisdom without questioning it. If anyone
provides a significant rebuttal that cannot be objectively answered, the
material at the Website will be withdrawn.
Onoang Blood
evidence of
> the Aether. This seems surprising because, if the Aether made its presence
any
> more obvious, physicists would have toothmarks on their butts from where
the
> Aether had jumped up and bit them. A few examples:
>
> Empty space, other than occupying a volume, has at least two observable
> properties, its dielectric constant and its permeability. If space is
"empty",
> what is it that has these properties? Since, if space consisted of a
volume of
> "nothingness", it cannot have any properties other than volume. Since it
does,
> space it would seem that it does contain a rejected entity, which was
called
> the Aether in the 19th Century!.
Not true. You are confusing mathematical quantities with physical ones.
You have not explained why is something has a dialectric constant it must
have "stuff" in it.
>
> The velocity of light in free space is determined by the dielectric
constant
> and the permeabiluty of space in accordance with accepted physical laws.
These again are just math quantities and have no physical reality. The
velocity of light is determined by the relationship of time to space in the
geometry of space.
> Similarly, the velocity of sound in a steel rod is determined, using an
> equation analogous to that used to determine the velocity of light, by the
> elasticity and density of the steel.
You are beating a very old drum. You cannot compare the two because they
are based on two different principles entirely. The "wave" of light is just
the photon quantum field. Quantum fields are in themselves not physical.
They are nothing more then mathematical entities we use to derive properties
of the physical particles. Waves in a medium, however, are actual motion of
physical stuff. Totally different.
> While a modern physicist seems to assert
> that the existence of the dielectric constant and the permeability of
space
> does not require the existence of a medium (i.e.- the Aether) having these
> properties, it seems certain that he would not be foolish enough to assert
> that, since the velocity of sound in a steel rod is determined only by its
> elasticity and its density, he could remove the rod and retain the density
and
> elasiticity so that the sound could propagate.
We couldn't. Sound is a physical wave hence requiring physical stuff.
Light is NOT a physical wave. It is a mathematical wave: a quantum wave.
Hence it does not require physical "stuff" to move.
> One might reasonably wonder if
> something is lacking in the mentality of physicists who would tolerate
such a
> double intellectual standard.
*sigh*. You just fail to understand how it works. You are comparing apples
to oranges and then saying "see! wallah they are the same!! why is everyone
but me so stupid?".
>
> Special Relativity (and the Lorentz Contraction Aether Theory) tell us
that
> the velocity of light is independent of the velocity of its source. Such
an
> independence is characteristic of a wavelike disturbance propagating
thorugh a
> medium, it is not characteristic of an entity propagating ballistically
through
> "empty" space.
Again you don't understand the theory. Since you are stupidly attached to
physical things, what would be the physical mechanism that would cause
moving aether to contract? Moving air does not contract the air in such a
manner as the speed of sound is always the same. Why? According to you
they both work on the same physical principles...
> For example, the sound of a rifle shot travels at the velocity
> of sound in air and which is independent of the velocity of the rifle.
Not true actually. Read a book (or two).
> The
> velocity of the bullet from the rifle, however, is additively determined
by the
> velocity of the rifle. If we abolish the Aether the observed independence
of
> the velocity of light from the velocity of its source would seem to
require
> that some form of "magic" must be at work since light would then have no
way to
> determine how fast it is supposed to travel.
I knows how fast to travel because it's speed is intrinsic. It comes from
the geometry of space-time.
> Since the writer finds the idea of
> "magic" accounting for observations in the science of physics extremely
hard to
> accept, he would be greatful if someone could provide some other
explanation
> which does not require a medium (the Aether). (P.S.- Handwaving arguments
about
> spacetime don't serve the purpose.)
Lorentz Aether contracts are the magic here. Space-time is at the core of
how it works. To give a full treatment here would take too long. If you
really want me to explain how ita ll works I'd be happy to help. However, I
really want to make sure you are going to try an learn it and not make
"handwaving" arguments to dismiss it. In other words, I don't want to waste
my time on you if you don't actually care to know.
>
> If one examines Dirac"s treatment of the creation of complementary
particles
> from high energy photons, he finds that Dirac had to invent a "sea of
negative
> energy pervading all of space" in order to make his mathematics work.
Yes. This was later fixed in Quantum Field Theory. Dirac' theory is not
the complete answer.
> Aside
> from the fact that "negative energy", in the sense that Dirac used it,
would
> seem not only to be unexplained but to be impossible, how does that
postulated
> "sea of negative energy" differ from the Aether? Both say that "something"
> exists in a space which had been arbitrarily assumed to be empty!
You are confusing things. As a photon (or any other particle) moves through
empty space, it sometimes encounters pairs of particles that have popped out
of the void and interacts with them. The space between two particles is
still empty. Macroscopically it is full of particle pairs, this is true.
But just as light moving through a medium (like glass etc.) is slowed, it
still moves at speed c between interactions with atoms. It is the
interactions that give it the appearance of being slowed. In essence, you
have confused microscopic interactions with macroscopic interactions.
>
> Current theory asserts that forces which "act at a distance" do so as a
result
> of "virtual particles" (e.g.- virtual photons, gravitons, gluons, etc.)
which
> pop into and out of existence so rapidly that quantum uncertainties
prevent the
> Law of Conservation of Energy from being violated. There would seem to be
at
> least two problems with this concept:-
None. Energy conservation IS violated. For a short time. Energy is only
macroscopically conserved.
>
> The first is the problem is the number of particles involved. At any
instant
> of time, every particle in the Univserse gravitationally attracts.every
other
> particle and they must therefore be exchanging gravitons. Furthermore,
these
> gravitons, since they are virtual, must be extremely short lived. The
number of
> gravitons involved in this concept is staggering. At any instant for
example,
> assuming that the presently accepted value of 10^80 particles in the
Universe
> is reasonably correct, there must be at least 0.5*(10^160) particles
existing
> at one time. The size of this number is increased by the fact that they
must be
> short lived and therefore must be replaced continuously (gravity appears
to be
> continuous), suggests that a bit of skepticism would be in order.
Nope. That is how it works. Your numbers are basically pulled from your
rear. However, even that aside, there are INFINITE number of particles in
the sea. As you look at shorter and shorter time periods, the amount of
energy conservation violation (and hence the number of particles present)
goes up. This is also true for electromagnetic forces aswell. These
particles are charged after all. The math of quantum electro dynamics (QED)
treats these cases entirely. The sum of all these infiinte number of
interactions gives finite numbers. These numbers have been tested and are
correct.
>
> The second problem is that of expaining how the exchange of "virtual
> particles" accomplishes the production of an "attractive force". The
production
> of a repulsive force by the exchange of particles is obvious. Two athletes
> throwing a medicine ball back and forth experience a repulsive force
because of
> momentum exchange between them. There does not seem to have ever been a
viable
> explanation as to how these particles act to produce an attractive force.
You are confusing a physcial picture with how the system really works. It
is not truly a ball tossing of a photon form one particle to another. The
photon isn't tossed at all. That physical picture can be quite misleading.
Feynmann invented that picture when solving for QED because he could then
write his mathematical equations in picture form which made it easier to
solve. However, this is indeed NOT how it works. Each picture actually
just represents a piece of a huge sum. This sum is what we call
"pertubation theory". In other words, its just math. Feynmann was just
solving a math problem and invented the incorrect picture to help him solve
it. I hope that clears it up...
> The
> momentum exchange requirement to produce an attractive force prohibits the
> production of such a force unless there is a substrate (e.g.:- the Aether)
with
> which they can exchange momentum. A boomerang returns to the thrower
because it
> exchanges momentum with such a medium (air). If space were empty (no
Aether),
> such a momentum exchange cannot occur and the virtual particles could only
> produce a repulsive force. The writer is aware that the mathematics
associated
> with the concept of "virtual partcles" allows for the production of
attractive
> forces. He is also aware of the fact that there are many mathematically
treated
> problems in which the mathematics deals with conditions outside of other
> constraints related to the problem and which limit the applicability of
the
> mathematics. This would seem to be such a case. Allow the Aether or its
> equivalent to be present as a substrate and the mathemetics of "virtual
> photons" works.
You don't understand it at all. Quite saying you do and then making
statements that show otherwise.
>
> There are two interpretations of Quantum Theory. One interpretation
involves
> parallel Universes coexisting in the same space as our own that are
created
> every time that a particle makes a "quantum choice". This approach not
only
> suffers the problem of absurdly large numbers described for gravitons, it
> requires that an amount of energy equal to the energy content of our
entire
> Universe multipled by the number of particles in the Universe be created
at
> every instant of time. It also requires that the volume of space occupied
by
> each particle contain an amount of energy equal to the energy equivalent
of
> that enormous amount of energy. Since the writer accepts the concepts that
> energy is conserved and two entities cannot occupy the same space at the
same
> time, he considers the interpretation to be rather foolish.
If you like. It doesn't matter anyway. This theory is not derived from
Quantum Theory at all. It is meta-physics at best.
>
> The other interpretation of Quantum Theory requires that quantum effects
> propagate at an "infinite" velocity. Most of the acedemic hierarchy seems
to
> consider this interpretation to be both "spooky" and wrong since "Special
> Relativity clearly shows that nothing can travel faster than light".
Wrong. This is just the result of people making the mistake of considering
one quantum field as a combination of two seperated fields. It is really
just people mixing up their pictures in their mind. Quantum fields are just
mathematical entities that gives us information about particles and systems
of particles (in quantum theory, they are the same thing).
> To the
> contrary, Special Relativity shows the opposite. What it shows is that an
> entity represented by energy cannot travel faster than the velocity of
light
> and the scientific community has fallen into the trap of assuming that
> everything which is observable is represented by some form of energy. The
> reason that the Special Theory of Relativity imposes its restrictions on
> velocity is that the Lorentz Transformation for energy, in a
force-length-time
> system of units is 1/(1-V^2/C^2)^0.5. This means that, at the velocity of
> light, energy becomes infinite and that, at velocities above that of
light,
> energy becomes imaginary. The effect limits the propagation of energy, and
> hence the velocity of conventional communication, to the velocity of
light.
>
> Experiments have shown that the polarization of "paired photons" is
coupled so
> that changing the polarization direction (the quantum number) of one
causes the
> polarization direction of the other to change in correspondence.
Wrong. They do not "change" one. They observe what it is. Suppose you
have two coins. You know that one is heads and one is tails. Someone takes
the two coins, has you close your eyes, and then places one in each of his
hands. You then are allowed to open your eyes and choose to look at one of
the hands. Lo and behold you see a tails coin. Hence, you know the other
hand has the heads coin. Is that magic? No. That is what they are doing
in the experiments. They know the total polarization of the two photons.
Hence, when they look at one photon and read it's polarization, they can
then deduce the polarization of the other.
Information really hasn't traveled from one photon telling the other "hey!!
they looked at me and I was spin up, so you be spin down OK?". Information
didn't travel at all. We simple revealed the state of the whole system by
looking at a piece of the system. The wavefunction of the system collapsed.
It was not that one wavefunction collapsed and "caused" the other to
collapse at an infinite speed. You have to look at the system as a whole to
understand it. Quantum Field Theory teaches that any system is not just
some single particle. It is rather ALWAYS a complete system of infinite
paticles.
Experiments
> have also shown that the coupling propagates at at least 4 times the
velocity
> of light and perhaps at an infinite velocity. (It should be noted that
changing
> the polarization direction of a photon does not change its energy content
so
> the the coupling of polarization direction between them does not involve
energy
> transfer).
Again just a misunderstanding of what the experiment truly shows.
> The polarizarion direction of photon is an angle which is measured
> in radians (the distance along an arc divide by the radius of that arc).
The
> Lorentz Transformation for Angle is therefore equal to the Lorentz
> Transformation for Length divided by the Lorentz Transformation fo Length
and
> is therefore equal to unity and is independent of velocity. (Above and
below a
> velocity of C, the value of unity is obvious. At the velocity of C, the
value
> is nominally indeterminate but can be shown to be equal to the value of
unity
> by borrowing a technique from Calculus). With a value of unity for the
Lorentz
> Transformation for Angle, it seems reasonable to assert that Special
Relativity
> requires the polarization couplingof paired photons to travel at the
infinite
> velocity required by the second, more rational interpretation of Quantum
> Theory. Assuming that this is the case, then the current concept of
"spacetime"
> as a single entity must be in error. An absolute time and velocity
reference is
> required.
Your deductions are completely wrong here. First of all, Lorentz
transformation preserve angles (Isometric) as you noted. The Lorentz
transform you refer to is: Sqrt(1- (v/c)^2). If this is 1 as you noted,
then v = 0. It does NOT = inifinity. Besides, angles are preserved in
Lorentz transforms no matter what v is and hence has NO connection to any
transmission speed.
>
> If one examines both the Special Theory of Relativity (STR)and the
> Lorentz-Contraction Aether Theory (LCAT), he finds that they are actually
the
> same theory in that they are cross derivable.
They give the same basic transforms yes. But beyond that they are VERY
different.
> (LCAT) is actually a special case
> solution of (STR) and cannot be disproven without disproving (STR) as
well.
Not true. They are not special cases of each other.
> LCAT asserts that the Aether exists but because information cannot
propagte
> faster than light (quantum theory tells us that this is not true), we
cannot
> find our velocity with respect to it. STR tells us that because our
velocity
> with respect to the Aether cannot be measured, it does not exist in the
theory!
> It does not precluide its existence nor its effect from being observed by
other
> means. Dr. Einstein asserted "remember gentlemen, we have not disproven
the
> existence of the Aether, we have only proven that we do not need it (for
> computations)".
Use Occam's Razor.
>
> Unfortunately, Modern Physics seems to have been taken over by
mathematical
> idiot savants who lack an appreciation for the implications of mechanism
and
> strive to suppress the contributions of those who would dare to say "Hey,
wait
> a minute"
Nope. You just are too stupid to learn the math (which is actually quite
simple) and have therefor dismissed it all as nonsense. It is typical of
stupid people to dismiss or fear that which they don't understand. Again,
if you really want to learn it (even if your motive is to try and change it)
then I would be happy to teach what I know. There is nothing wrong with
questioning what is considered "known". But you have to at least understand
what you are changing before you can try to change it.
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:Ebi79.22798$27.6...@twister.rdc-kc.rr.com...
| > It is the currently accepted "truth" that there is no experimental
| evidence of
| > the Aether. This seems surprising because, if the Aether made its
presence
| any
| > more obvious, physicists would have toothmarks on their butts from where
| the
| > Aether had jumped up and bit them. A few examples:
| >
| > Empty space, other than occupying a volume, has at least two observable
| > properties, its dielectric constant and its permeability. If space is
| "empty",
| > what is it that has these properties? Since, if space consisted of a
| volume of
| > "nothingness", it cannot have any properties other than volume. Since it
| does,
| > space it would seem that it does contain a rejected entity, which was
| called
| > the Aether in the 19th Century!.
|
| Not true. You are confusing mathematical quantities with physical ones.
| You have not explained why is something has a dialectric constant it must
| have "stuff" in it.
You are wasting your time if you expect reticher to reply to your message.
He only spams the newsgroup and hardly ever replies. If you want a
response, you have to email him. I guess he doesn't like to share.
| >
| > The velocity of light in free space is determined by the dielectric
| constant
| > and the permeabiluty of space in accordance with accepted physical laws.
|
| These again are just math quantities and have no physical reality. The
| velocity of light is determined by the relationship of time to space in
the
| geometry of space.
Yes, but it is a known fact that capacitance and inductance are purely
geometric. Geometric of what? Well, it would have to be space itself. So
I would have to postulate that space itself is capacitance and inductance.
The combination of the two, which are inseperable, create a space "field".
At least as far as EM is concerned. Space does have a known impedance also.
About 377 ohms. I believe it is these properties of space that *is* the
relationship of time to space in the geometry of space that you are talking
about. Time equals 2*pi*sqrt(L*C). This is what makes the speed of light
exactly what it is.
| > Similarly, the velocity of sound in a steel rod is determined, using an
| > equation analogous to that used to determine the velocity of light, by
the
| > elasticity and density of the steel.
|
| You are beating a very old drum. You cannot compare the two because they
| are based on two different principles entirely. The "wave" of light is
just
| the photon quantum field. Quantum fields are in themselves not physical.
| They are nothing more then mathematical entities we use to derive
properties
| of the physical particles. Waves in a medium, however, are actual motion
of
| physical stuff. Totally different.
Photons do make EM waves and they are waves of *real* electric and magnetic
fields. To me, electric and magnetic fields are real physical stuff.
| > While a modern physicist seems to assert
| > that the existence of the dielectric constant and the permeability of
| space
| > does not require the existence of a medium (i.e.- the Aether) having
these
| > properties, it seems certain that he would not be foolish enough to
assert
| > that, since the velocity of sound in a steel rod is determined only by
its
| > elasticity and its density, he could remove the rod and retain the
density
| and
| > elasiticity so that the sound could propagate.
|
| We couldn't. Sound is a physical wave hence requiring physical stuff.
| Light is NOT a physical wave. It is a mathematical wave: a quantum wave.
| Hence it does not require physical "stuff" to move.
Well, I would say more like light creates its own medium using the L and C
properties of space. Photons do make up EM waves which are not really just
a mathematical wave. EM waves are as real as something can be. It is the
complementary action of E making M and M making E that is part of the
medium. And I doubt very much if this could work without also the
complementary action of L and C space. So the combo action of EM with LC
creates the "medium" which allows light to happen.
FrediFizzx
Onoang Blood
> |
> | Not true. You are confusing mathematical quantities with physical ones.
> | You have not explained why is something has a dialectric constant it
must
> | have "stuff" in it.
>
> You are wasting your time if you expect reticher to reply to your message.
> He only spams the newsgroup and hardly ever replies. If you want a
> response, you have to email him. I guess he doesn't like to share.
Yes I know. But I have to try at least once.
>
> | >
> | > The velocity of light in free space is determined by the dielectric
> | constant
> | > and the permeabiluty of space in accordance with accepted physical
laws.
> |
> | These again are just math quantities and have no physical reality. The
> | velocity of light is determined by the relationship of time to space in
> the
> | geometry of space.
>
> Yes, but it is a known fact that capacitance and inductance are purely
> geometric. Geometric of what? Well, it would have to be space itself.
So
> I would have to postulate that space itself is capacitance and inductance.
> The combination of the two, which are inseperable, create a space "field".
> At least as far as EM is concerned. Space does have a known impedance
also.
> About 377 ohms. I believe it is these properties of space that *is* the
> relationship of time to space in the geometry of space that you are
talking
> about. Time equals 2*pi*sqrt(L*C). This is what makes the speed of light
> exactly what it is.
Hmmm. So what is inductance exactly? Capacitance? Are not these just
values placed in simplistic Electronics equations to help derive things like
current, voltages, and resistance? Capacitance is related to the potential
energy stored in a device (called a capacitor ;-) ) that holds unlike
charges very close but won't let them touch. Pretty simple really.
"Geometry" plays in only on the level that the physical configuration of the
charges effects the measured "capacitance". Space-time geometry is a whole
different animal. So where in that description did I talk about light? No
place. Why? Because light has nothing to do with capacitance and neither
does space time. You are attempting to put two unrelated things together
because they make a pretty picture in your mind. While that may give you
nice things to draw on paper it is highly misleading and will send you down
the wrong road.
In higher level physics you will learn that the EM fields you know and love
from electronics are all in fact the result of a single 4-vector typically
called "A". This vector field turns out to be the quantum field operator
for the photon. The speed of light only turns up for photons because they
have no rest mass. That is, c is a part of space-time itself. Photons sit
on that space-time and hence aquire it's properties. At no place in this
model is C or L given to space. And why would we want to replace a single
parameter called "c" with two parameters anyway?
>
> | > Similarly, the velocity of sound in a steel rod is determined, using
an
> | > equation analogous to that used to determine the velocity of light, by
> the
> | > elasticity and density of the steel.
> |
> | You are beating a very old drum. You cannot compare the two because
they
> | are based on two different principles entirely. The "wave" of light is
> just
> | the photon quantum field. Quantum fields are in themselves not
physical.
> | They are nothing more then mathematical entities we use to derive
> properties
> | of the physical particles. Waves in a medium, however, are actual
motion
> of
> | physical stuff. Totally different.
>
> Photons do make EM waves and they are waves of *real* electric and
magnetic
> fields. To me, electric and magnetic fields are real physical stuff.
Wrong. Electric and Magnetic fields are in fact "derived". They are not
real at all. They both are derived from the "true" field A. That is why it
is called "electromagnetism" and not "electric + magnetism". This has been
known since the days of Maxwell long before Einstein. BTW: photons do NOT
make EM waves. They have no electric charge. They ARE EM waves.
And how exactly do you conclude that EM fields are "real physical stuff"?
The A field is nothing more then the quantum field for the photon. The
particle photon is the "real physical stuff". The mathematical field A
simply tells the particle how to behave.
>
> | > While a modern physicist seems to assert
> | > that the existence of the dielectric constant and the permeability of
> | space
> | > does not require the existence of a medium (i.e.- the Aether) having
> these
> | > properties, it seems certain that he would not be foolish enough to
> assert
> | > that, since the velocity of sound in a steel rod is determined only by
> its
> | > elasticity and its density, he could remove the rod and retain the
> density
> | and
> | > elasiticity so that the sound could propagate.
> |
> | We couldn't. Sound is a physical wave hence requiring physical stuff.
> | Light is NOT a physical wave. It is a mathematical wave: a quantum
wave.
> | Hence it does not require physical "stuff" to move.
>
> Well, I would say more like light creates its own medium using the L and C
> properties of space. Photons do make up EM waves which are not really
just
> a mathematical wave.
Yes they are. Read up on Quantum Mechanics.
> EM waves are as real as something can be. It is the
> complementary action of E making M and M making E that is part of the
> medium.
No. They interplay because you have been fooled. A is the source field. E
and M are derived. There is in fact only one field for electromagnetism not
two. That field is no more "real" then the quantum field for the electron
is "real". You cannot hold the quantum field in your hand, toss it around,
etc. It is nothing more then a mathematical tool we use to derive
properties of the physical particles. You cannot deduce that simply because
there is money under pillow, that the tooth fairy is real. QM fields are
like that. They are the story, not the result. They say: "this is how you
measure a photon's momentum" etc. They are not the momentum itself. Get
it?
> And I doubt very much if this could work without also the
> complementary action of L and C space. So the combo action of EM with LC
> creates the "medium" which allows light to happen.
No. Space time does not have any physical properties like C or L. You can
go ahead and describe it that way if you wish but you are wasting your time.
C and L are measurements of properties from physical devices. They are in
fact simply fudge factors in electronics equations to help solve for
currents, resistances, and voltages. And often bad ones at that. They are
attached to physical configurations of metal plates and magnets etc, and in
more advanced books they may even show you the real source of these numbers
(in terms of EM equations). They are NOT fundemental. Only the EM
equations are. Those are the laws of physics we are concerned with. BTW,
once you know how to derive E and M from the one source field A, all of
Maxwell's field equations reduce to a single equation. It is a wave
equation for A with a gauge condition. It is the same equation (should be
no surprise) as the quantum field equation for photons.
reticher
responses, however, are, for the most part, pure gibberish. To say that light
is not a wave is absurd. Start with a radio wave, which can be observed to be a
wave and weaken without changing its polarization and eventually it is found to
be composed of individual particles. The simplest way of seening what those
particles is like is to multiply the original wave by an impulse function. One
winds up witht the photon still as a wave but which id composed of (perhaps) a
single cycle. At that level it appears to be botha particle and a wave, and it
is this wave that the dielectric constant and permeability of the "empty" space
transports.
I suggest you read the material at http://www.members.aol.com/reticher/site.htm
. If you can get past the fact that you have been taught correctly and already
know it all, you might just learn something.
reticher
To date, I have found many reasons to believe that the Aether exists, but no
one has presented an arguement to justify that it doesn't. Handwaving arguments
based upon mathematics don't count. "Figures don't lie but liars figure"
reticher
There are seveal magnets holding papers on my refrigerator which exhibit an
external magnetic field but not an electric field. The electric field on the
face of my TV set will attract the hair on my arms, but there is not associated
with any magnetic field. It is only when the electic and magnetiuc field
propated as a wave by interchanging energy that they behave in the way that you
assume fo electric and magnetic fields.
I suggest that you digest thematerial at
http://www.members.aol.com/reticher/site.htm. You might learns something,
assuming of course, that you don't already know all there is to know.
fizzxhaha@ahahhotmail.com Fredi Fizzx
| (snip)
| > |
| > | Not true. You are confusing mathematical quantities with physical
ones.
| > | You have not explained why is something has a dialectric constant it
| must
| > | have "stuff" in it.
| >
| > You are wasting your time if you expect reticher to reply to your
message.
| > He only spams the newsgroup and hardly ever replies. If you want a
| > response, you have to email him. I guess he doesn't like to share.
|
| Yes I know. But I have to try at least once.
Amazing! Mr. Reticher did reply.
Hmmm. What is the source of Z_0? The impedance of space? Yes, if you look
at inductance and capacitance they way you are looking at it, they are as
you describe. I probably should qualify a bit here. We know that the
geometry of space-time is definitely involved in how a capacitor works and
how an inductor works. So I would have to say that that this factor of
space-time geometry involving L and C is also important for EM radiation.
And it is the source of Z_0.
| In higher level physics you will learn that the EM fields you know and
love
| from electronics are all in fact the result of a single 4-vector typically
| called "A". This vector field turns out to be the quantum field operator
| for the photon. The speed of light only turns up for photons because they
| have no rest mass. That is, c is a part of space-time itself. Photons
sit
| on that space-time and hence aquire it's properties. At no place in this
| model is C or L given to space. And why would we want to replace a single
| parameter called "c" with two parameters anyway?
L and C of space *are* one parameter. They are inseperable when using for
EM radiation. And I would postulate at this point that L and C as one
parameter is involved with the single 4-vector "A". You say that c is part
of space-time itself. How does that work exactly? It doesn't make a lot of
sense. It make much more sense that space-time has a property that is what
limits c to c. And this property can be modeled in terms of L and C. Both.
You would have to use both together as a single space-time field at least as
far as EM radiation goes.
Boy, you need to go back to high school physics or something. EM waves are
composed of photons. I don't think one photon can make an EM wave according
to current theory. However, if it can, then I would like that better.
| And how exactly do you conclude that EM fields are "real physical stuff"?
| The A field is nothing more then the quantum field for the photon. The
| particle photon is the "real physical stuff". The mathematical field A
| simply tells the particle how to behave.
The same way Reticher described in his reply. I have magnets holding papers
to my frig. The hair on my arm stands up near a charged up item by electric
fields. Believe me, electric and magnetic fields are real. Just as radio
waves are real and the light we see are real EM waves. OK, so all of this
is due to photon exchange. The real photon is creating real electric and
magnetic fields and real EM waves. What Universe do you come from?
Sorry, real photons make real EM radio wave, light that we see waves, etc.
| > EM waves are as real as something can be. It is the
| > complementary action of E making M and M making E that is part of the
| > medium.
|
| No. They interplay because you have been fooled. A is the source field.
E
| and M are derived. There is in fact only one field for electromagnetism
not
| two. That field is no more "real" then the quantum field for the electron
| is "real". You cannot hold the quantum field in your hand, toss it
around,
| etc. It is nothing more then a mathematical tool we use to derive
| properties of the physical particles. You cannot deduce that simply
because
| there is money under pillow, that the tooth fairy is real. QM fields are
| like that. They are the story, not the result. They say: "this is how
you
| measure a photon's momentum" etc. They are not the momentum itself. Get
| it?
This is all fine if *you* want to be fooled. It is OK, because it works
well. And I would not discourage using the mathematical tools you are
describing. However, I suspect that there is a more *real* way to describe
what is going on with EM, space-time, etc. Instead of having to resort to
"mathematical tools".
| > And I doubt very much if this could work without also the
| > complementary action of L and C space. So the combo action of EM with
LC
| > creates the "medium" which allows light to happen.
|
| No. Space time does not have any physical properties like C or L. You
can
| go ahead and describe it that way if you wish but you are wasting your
time.
| C and L are measurements of properties from physical devices. They are in
| fact simply fudge factors in electronics equations to help solve for
| currents, resistances, and voltages. And often bad ones at that. They
are
| attached to physical configurations of metal plates and magnets etc, and
in
| more advanced books they may even show you the real source of these
numbers
| (in terms of EM equations). They are NOT fundemental. Only the EM
| equations are. Those are the laws of physics we are concerned with. BTW,
| once you know how to derive E and M from the one source field A, all of
| Maxwell's field equations reduce to a single equation. It is a wave
| equation for A with a gauge condition. It is the same equation (should be
| no surprise) as the quantum field equation for photons.
Well, I would have to beg to differ with you. I say that space-time does
have or can be modeled in terms L and C of a single field quanity. And one
fellow has used this property to unify gravity with EM.
http://www.todds.info/physics.htm
FrediFizzx
Onoang Blood
interaction.
> There are seveal magnets holding papers on my refrigerator which exhibit
an
> external magnetic field but not an electric field. The electric field on
the
> face of my TV set will attract the hair on my arms, but there is not
associated
> with any magnetic field. It is only when the electic and magnetiuc field
> propated as a wave by interchanging energy that they behave in the way
that you
> assume fo electric and magnetic fields.
There is a relationship between the two fields. It is all contained in
Maxwell's equations. Both fields can be derived from a single field A.
This has been known since Maxwell long before Einstein was even a baby.
Don't be so damned stupid. There exists configurations of the one field A
that give only electric fields. And there exists configurations of the one
field A that give only magnetic fields. Magnetic and Electric fields are
derived and not fundemental. Again all this has been known since long
before Einstein.
Onoang Blood
"reticher" <reti...@aol.com> wrote in message
news:20020817110132...@mb-ct.aol.com...
Those mathematical arguments are not handwaving. You are just too stupid to
read a mathematics book.
Onoang Blood
Your
> responses, however, are, for the most part, pure gibberish.
Like this statement you just made?
> To say that light
> is not a wave is absurd. Start with a radio wave, which can be observed to
be a
> wave and weaken without changing its polarization and eventually it is
found to
> be composed of individual particles.
No that is not how it works. Damn you are clueless. "Observed to be a
wave" is merely the observation that the photons have a quantum field. That
their properties, be it momentum energy etc, are dictated by a mathematical
wave. Observation of the photon reveals it to be a particle. That is how
photoelectricity works for God's sake.
> The simplest way of seening what those
> particles is like is to multiply the original wave by an impulse function.
One
> winds up witht the photon still as a wave but which id composed of
(perhaps) a
> single cycle.
No no no. You sinply do not understand quantum mechanics at all. I am
beginning to see that you are a complete waste of time. You not only have
no interest in listening to anyone but yourself but you immediately dismiss
everyhting but your own worthless theory because it has "math". Buy a clue.
> At that level it appears to be botha particle and a wave, and it
> is this wave that the dielectric constant and permeability of the "empty"
space
> transports.
Nope.
> I suggest you read the material at
http://www.members.aol.com/reticher/site.htm
> . If you can get past the fact that you have been taught correctly and
already
> know it all, you might just learn something.
You are the one who needs to learn something.
fizzxhaha@ahahhotmail.com Fredi Fizzx
He is not being stupid. It is you that is being stupid. You were saying
that electric and magnetic fields are not real. Bullcrap, they are real no
matter what the source is that is creating them.
FrediFizzx
Onoang Blood
"Fredi Fizzx" <fredi fizz...@ahahhotmail.com> wrote in message
news:hww79.1221$FO5.23...@newssvr21.news.prodigy.com...
You are stupid. They are not real, they are derived:
F_mn = dA_m/dx_n - dA_n/dx_m
Maxwell's equations are then:
dF_mn/dx_n = j_m
jm is the 4-vector "current". F_mn is the derived fields you seem to have a
boner for. This equation can then be rewritten in terms of the proper field
A:
d^2A_m/dx_n^2 = j_m
It's just the wave equation with j being the source.
There, now I've written all the equations out that are contained in any
decent EM book. For whatever reason, you are too stupid to go read a book.
Onoang Blood
Wrong geometry buddy. GR may effect a capaitor yes, but not in any
fundemental way. The "geometry" that effects a capaitor and inductor is
strictly related to the physical configuration of the metallic plates ,
magnets, etc. Read a book on how these two devices work.
> So I would have to say that that this factor of
> space-time geometry involving L and C is also important for EM radiation.
> And it is the source of Z_0.
What are talking about Z_0? Explain a bit more and I'll address this.
> | In higher level physics you will learn that the EM fields you know and
> love
> | from electronics are all in fact the result of a single 4-vector
typically
> | called "A". This vector field turns out to be the quantum field
operator
> | for the photon. The speed of light only turns up for photons because
they
> | have no rest mass. That is, c is a part of space-time itself. Photons
> sit
> | on that space-time and hence aquire it's properties. At no place in
this
> | model is C or L given to space. And why would we want to replace a
single
> | parameter called "c" with two parameters anyway?
>
> L and C of space *are* one parameter. They are inseperable when using for
> EM radiation. And I would postulate at this point that L and C as one
> parameter is involved with the single 4-vector "A". You say that c is
part
> of space-time itself. How does that work exactly?
Glad you asked. "c" is simply the ratio of our definition of time to our
definition of space. Is this arbitrary? Actually yes. If you redefine "c"
then you in effect redefine what a meter and a second are. "Light" are just
particles that travel at this ratio speed. Well call it "c". We have
already defined what a second and a meter are and so when we measure "c"
against those two (second and meter) we comes up with the value we have
today. In SR, the line element of space-time is given as:
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
You could perform a simple coorindate transformation:
t' = at
x' = x
y' = y
z' = z
Then your line element is:
ds^2' = c^2 dt'^2 - dx'^2 - dy'^2 - dz'^2 = (ca)^2 dt^2 - dx^2 - dy^2 - dz^2
Hence, in the primed coordinate system, c looks like the value ca. a is
arbitrary, hence the value "c" is truly arbitrary. This coorindate
transform is the same as to redefine your measure of "second".
> It doesn't make a lot of
> sense. It make much more sense that space-time has a property that is
what
> limits c to c. And this property can be modeled in terms of L and C.
Both.
> You would have to use both together as a single space-time field at least
as
> far as EM radiation goes.
"c" (in terms of SR) actually has nothing to do with EM at all. "c" is a
parameter of the space-time as defined above. You do not need some
arbitrary field of "L" and "C", together or not, to see how SR works.
You are dumb. EM waves ARE photons. They work together. They are one and
the same. One does not "create" the other. The EM wave is the quantum
field for the photon. I learned this while attending college for physics,
which was my major. Now with a PhD in physics I'm pretty sure I know what
the current theories say about this. "One photon" is actually pretty
meaningless. You would learn that if you ever got to the level of quantum
field theory.
>
> | And how exactly do you conclude that EM fields are "real physical
stuff"?
> | The A field is nothing more then the quantum field for the photon. The
> | particle photon is the "real physical stuff". The mathematical field A
> | simply tells the particle how to behave.
>
> The same way Reticher described in his reply.
Agreeing with him is dangerous ;-)
> I have magnets holding papers
> to my frig. The hair on my arm stands up near a charged up item by
electric
> fields. Believe me, electric and magnetic fields are real. Just as radio
> waves are real and the light we see are real EM waves. OK, so all of this
> is due to photon exchange. The real photon is creating real electric and
> magnetic fields and real EM waves. What Universe do you come from?
The photons have a physical effect. Yes. I know. But that does not make
the E and M fields physical. PHOTONS DO NOT create EM waves. That is wrong
period. If you won't even attempt to learn even basic Electromagnetism then
stay the hell away from this newsgroup. It has been known since Maxwell.
Before relativity was invented back when everyone still thought about
Aether. This simple fact stands on its own seperate from any quantum theory
or relativity theory. It is a basic known fact from Electromagnetism. If
you want references to some good books I can point you in the right
direction. Just stop wasting my time if you have no plans to read them.
Nope.
OK. Forget it. I tried to help but you are worthless.
Differ all you like. You are stupid and that is that.
pst...@ix.netcom.com
"Onoang Blood" <OBL...@mn.rr.com> wrote:
>
>"Fredi Fizzx" <fredi fizz...@ahahhotmail.com> wrote in message
>news:hww79.1221$FO5.23...@newssvr21.news.prodigy.com...
>> "Onoang Blood" <OBL...@mn.rr.com> wrote in message
>> news:lrw79.9948$Hf.3...@twister.kc.rr.com...
>>>> Electric and magnetic fields do not epend upon electromagnetic
>>>> interaction. There are seveal magnets holding papers on
>>>> my refrigerator which exhibit an external magnetic field but
>>>> not an electric field. The electric field on the face of my
>>>> TV set will attract the hair on my arms, but there is not
>>>> associated with any magnetic field. It is only when the
>>>> electic and magnetic field propagated as a wave by
>>>> interchanging energy that they behave in the way that you
>>>> assume fo electric and magnetic fields.
>>>
>>> There is a relationship between the two fields. It is all
>>> contained in Maxwell's equations. Both fields can be derived
>>> from a single field A. This has been known since Maxwell long
>>> before Einstein was even a baby. Don't be so damned stupid.
>>> There exists configurations of the one field A that give only
>>> electric fields. And there exists configurations of the
>>> one field A that give only magnetic fields. Magnetic and
>>> Electric fields are derived and not fundemental. Again all
>>> this has been known since long before Einstein.
>>
>>
>> He is not being stupid. It is you that is being stupid. You
>> were saying that electric and magnetic fields are not real.
>> Bullcrap, they are real no matter what the source is that is
>> creating them.
>
> You are stupid. They are not real, they are derived:
>
> F_mn = dA_m/dx_n - dA_n/dx_m
>
> Maxwell's equations are then:
>
> dF_mn/dx_n = j_m
>
> jm is the 4-vector "current". ...
Current of what?
> ... F_mn is the derived fields you seem to have a boner for. ...
Can it cause physically effects?
> This equation can then be rewritten in terms of the proper field A:
>
> d^2A_m/dx_n^2 = j_m
>
> It's just the wave equation with j being the source.
So? Your point being what?
> There, now I've written all the equations out that are contained in
> any decent EM book. ...
Let's say you parroted the equations... The 'real' question is,
what do they mean, or conversely, what physically, are they
representing?
> For whatever reason, you are too stupid to go read a book.
And you seem too stupid to actually try to reason out what others
have symbolically represented in mathematics. A math symbol or
formulation is neither real or unreal, but if what it represent is
empirically measurable, the 'what' it represents is quite 'real'...
Are the 'empirical' properties designated as an 'electric field'
and those designated as a 'magnetic field' the same? If not, then
both are distinct, and thus quite 'real'. Are they related to each
other?, sure! But both are real in the empirical observable sense.
That one can 'derive' them from the vector potential say something,
but NOT they are 'unreal'.
Paul Stowe
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:ajmu37$i84$1...@slb6.atl.mindspring.net...
Yes, thank you very much. Mr. Blood seems to have lost touch with reality
or something. Maybe too much math running across his brain. As far as I am
concerned, most things I can sense definitely fall under the definition of
real.
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
Oh brother! That is not it. Did you bother to read any of Todd's articles
at the link I provided. Or are you too stupid to read them? Why don't you
read them, then maybe we can have a more intelligent discussion?
http://www.todds.info/physics.htm
FrediFizzx
Onoang Blood
> >> He is not being stupid. It is you that is being stupid. You
> >> were saying that electric and magnetic fields are not real.
> >> Bullcrap, they are real no matter what the source is that is
> >> creating them.
> >
> > You are stupid. They are not real, they are derived:
> >
> > F_mn = dA_m/dx_n - dA_n/dx_m
> >
> > Maxwell's equations are then:
> >
> > dF_mn/dx_n = j_m
> >
> > jm is the 4-vector "current". ...
>
> Current of what?
the time component is related directly to the charge density. The three
space-like components are the actual current vector. Current in exactly the
same sense as we all know. Moving charges.
>
> > ... F_mn is the derived fields you seem to have a boner for. ...
>
> Can it cause physically effects?
No. Only the source field A causes physical effects.
> > This equation can then be rewritten in terms of the proper field A:
> >
> > d^2A_m/dx_n^2 = j_m
> >
> > It's just the wave equation with j being the source.
>
> So? Your point being what?
Read the thread.
> > There, now I've written all the equations out that are contained in
> > any decent EM book. ...
>
> Let's say you parroted the equations... The 'real' question is,
> what do they mean, or conversely, what physically, are they
> representing?
In a nutshell, they stem from the source principle of least action. Put
another way, they represent a symmetry of nature. This symmetry manifests
itself as charge conservation and the above equations.
>
> > For whatever reason, you are too stupid to go read a book.
>
> And you seem too stupid to actually try to reason out what others
> have symbolically represented in mathematics. A math symbol or
> formulation is neither real or unreal, but if what it represent is
> empirically measurable, the 'what' it represents is quite 'real'...
If you adjust your definition of "real" you can conclude anything. The
photon is "real". It is what we measure. You don't ever measure the
electric or magnetic field. Rather, you measure the interaction of the
photons. The devices used then translate that interaction into numbers we
associate to things like "voltage" and even your precious "electric" or
"magnetic" fields.
This is true even in gravity. You do not measure the gravitational field.
It is not real, just a mathematical construct. You measure the motion of
objects. You then *use* the graviational field (math) to make predictions.
> Are the 'empirical' properties designated as an 'electric field'
> and those designated as a 'magnetic field' the same? If not, then
> both are distinct, and thus quite 'real'. Are they related to each
> other?, sure! But both are real in the empirical observable sense.
No they are not. They are derived form a single source field. This is
ancient physics from Maxwell.
>
> That one can 'derive' them from the vector potential say something,
> but NOT they are 'unreal'.
They are not real because they cannot be directly measured. You don't
measure the E or M fields. You measure the photons and then *translate*
that into the values of your fields. They are convenient mathematical
constructs and nothing more.
>
> Paul Stowe
That explains your lack of understanding.
pst...@ix.netcom.com
"Onoang Blood" <OBL...@mn.rr.com> wrote:
> (snip)
>
>>>> He is not being stupid. It is you that is being stupid. You
>>>> were saying that electric and magnetic fields are not real.
>>>> Bullcrap, they are real no matter what the source is that is
>>>> creating them.
>>>
>>> You are stupid. They are not real, they are derived:
>>>
>>> F_mn = dA_m/dx_n - dA_n/dx_m
>>>
>>> Maxwell's equations are then:
>>>
>>> dF_mn/dx_n = j_m
>>>
>>> jm is the 4-vector "current". ...
>>
>> Current of what?
>
> the time component is related directly to the charge density.
> The three space-like components are the actual current vector.
> Current in exactly the same sense as we all know. Moving
> charges.
See below...
>>> ... F_mn is the derived fields you seem to have a boner for. ...
>>
>> Can it cause physically effects?
>
> No. Only the source field A causes physical effects.
That's odd, I've been told authoritatively that, quote,
"... Unfortunately for you, Maxwell never found any
physical phenomena that could be attributed to the
vector potential. In classical E&M, A is a
mathematical artifice that represents nothing but
the freedom to make a gauge transformation. I can
pointlessly quote volumes of text to support this,
too. ..."
(Posted by a respondent going by the handle "Bilge")
and...
"Apart from certain constant factors that will become
important to us later in this chapter, these
formulations do not differ greatly from typical
modern presentations of "Maxwell's equations" except
in the prominence given here to the vector A. In
modern texts A is normally not included in the
fundamental list, but is treated separately as the
"vector potential." Logically, it is not essential
as a first principle. For Maxwell, on the other hand,
'A' appears at the outset, since it represents for
him the momentum stored in the resting field,
manifested in various ways in all electromagnetic
phenomena. For moderns, it is logically unnecessary
and its inclusion among the basic premises would be
inelegant; for Maxwell, it is conceptually at the
foundation of the entire theory and it takes first
place among the elements of the science!"
(page 383, "Maxwell on the Electromagnetic Field, A Guided Study",
Thomas K. Simpson, Cambridge University Press - 1997)
Now I agree with you that the vector potential IS a representation
of a 'source field'. Thus the expressions,
__
B = \/ x A
__
E = -\/V - @A/@t
Where @ represents the 'partial' differential symbol
And, as Simpon states, "For Maxwell, on the other hand, A appears
at the outset, since it represents for him the momentum stored in
the resting field, manifested in various ways in all electromagnetic
phenomena".
But which is 'real'? Some argue A isn't 'real' as illustrated by
Bilge, and you appear to be arguing exactly the opposite.
My point was/is they are ALL 'real', with B and E being observably
separable elements. For Maxwell this was how he saw it also. I
reference Simpson's book (noted above) or see Maxwell's papers.
Specifically, "On Faraday's Lines of Force" and "On the Physical
Lines of Force".
Be advised that Maxwell's model from which the now famous equations
were developed was a fluidic Helmholtzian vortex ring lattice.
Thus was fundamentally founded upon vortex dynamics. What 'current'
meant to Maxwell was clearly enumerated in those works and was called
by Maxwell the 'electrotonic state'. The vector potential (A) is
fundamentally related to a certain vorticity in the ring lattice system.
>>> This equation can then be rewritten in terms of the proper field A:
>>>
>>> d^2A_m/dx_n^2 = j_m
>>>
>>> It's just the wave equation with j being the source.
>>
>> So? Your point being what?
>
> Read the thread.
I have ...
>>> There, now I've written all the equations out that are contained in
>>> any decent EM book. ...
>>
>> Let's say you parroted the equations... The 'real' question is,
>> what do they mean, or conversely, what physically, are they
>> representing?
>
> In a nutshell, they stem from the source principle of least
> action. Put another way, they represent a symmetry of nature.
> This symmetry manifests itself as charge conservation and the
> above equations.
These are 'motherhood' philosphical statements. The 'principle
of least action' is due to the statistical behavior of complex
systems containing n degrees of freedom . The question then
becomes, formally, 'what is action'?. The term 'symmetry of
nature' means what, physically?
>>> For whatever reason, you are too stupid to go read a book.
>>
>> And you seem too stupid to actually try to reason out what others
>> have symbolically represented in mathematics. A math symbol or
>> formulation is neither real or unreal, but if what it represent is
>> empirically measurable, the 'what' it represents is quite 'real'...
>
> If you adjust your definition of "real" you can conclude anything.
> The photon is "real". It is what we measure. You don't ever
> measure the electric or magnetic field. Rather, you measure the
> interaction of the photons. The devices used then translate that
> interaction into numbers we associate to things like "voltage" and
> even your precious "electric" or "magnetic" fields.
Very good. This IS the crux of the issue, the definition of 'real'
in an empirical sense. Now one should conclude that BOTH you and Mr.
Fizzx are right, it just a matter of interpetation of the definition
of 'real'.
> This is true even in gravity. You do not measure the gravitational
> field. It is not real, just a mathematical construct.
If it's 'not real' what is the 'mathematical construct' representing?
Also, if it isn't 'real' in the empirical sense, where does the
manifestations associated to it come from or go to?
> You measure the motion of objects. You then *use* the graviational
> field (math) to make predictions.
We have a very fundamental difference of opinion (and mental
concept) of 'real' here...
>> Are the 'empirical' properties designated as an 'electric field'
>> and those designated as a 'magnetic field' the same? If not, then
>> both are distinct, and thus quite 'real'. Are they related to each
>> other?, sure! But both are real in the empirical observable sense.
>
> No they are not. They are derived form a single source field. This
> is ancient physics from Maxwell.
OK fine, can you point me to where in Maxwell's papers I can find
this? I've read them all and I didn't see that. BTW, as far as I
can discern, neither did Simpson. In fact, the modern form of the
vector potential doesn't exist in Maxwell's work. Note that neither
does the Lorentz force equation.
>> That one can 'derive' them from the vector potential say something,
>> but NOT they are 'unreal'.
>
> They are not real because they cannot be directly measured. You
> don't measure the E or M fields. You measure the photons and then
> *translate* that into the values of your fields. They are convenient
> mathematical constructs and nothing more.
All so-called 'convenient mathematical constructs' are such only
BECAUSE they represent and quantify some empirically observed
process. Thus ALL SUCH must represent objective reality, or it isn't,
by definition, science.
Paul Stowe
Monitek
>Date: 17/08/02 19:50 GMT Daylight Time
>Message-id: <GWw79.9954$Hf.3...@twister.kc.rr.com>
Dear Onoang Blood,
s=ut + 1/2 ft^2 .......
is an equation describing the distance car will move in a given time.
Using this equation we can predict precisely where the car will be at some time
in the future given some elementary data about its current movement.
Question: Given that you can predict the movement and position of the car, what
makes the car move?
Fredi Fizzx is absolutely correct when he says that light creates it own medium
in which the waveform oscillates as the em wave progresses. The medium is the
dielectric medium of Maxwell created by polarising (separating e-p pairs from)
the vacuum as the em wave passes. A practical demonstration of vacuum
polarisation is to create an alternating charge field between two electrically
conducting plates placed in such a way as to be parallel to each other, and
watch as electromagnetic radiation emanates from the gap between the two
plates. Now, if electromagnetic radiation can be created by an alternating
charge field, then, electromagnetic radiation upon passing two similar plates
to those which created the emanating electromagnetic radiation, should recreate
an alternating charge upon the aforementioned two electrically conducting
plates placed in such a way as to be parallel to each other, surprise surprise
that is exactly what happens.
Your explanation for this effect is.................
Regards,
monitek (Arden Barker)
Monitek
>Date: 18/08/02 04:15 GMT Daylight Time
>Message-id: <CjE79.34270$mj7.6...@twister.rdc-kc.rr.com>
to Onoang Blood,
You measure the E field with a capacitor. You measure an M field with an
inductor.
my previous post in this thread covers the capacitive measurement. The
inductive measurement consists of aligning a coil with the direction of
propagation of the EM wave and measuring a potential difference across the ends
of the coil. Turning the coil through 90 degrees eliminates the potential
difference at the ends of the coil due to the em wave being examined.
Thats how total nobrainers with no understanding like Maxwell, Faraday and
Hertz deduced that light as we know it was an alternating electromagnetic
field. Maxwell required a dielectric medium to create the mathematics of emr.
The dielectric is non other than combined electron-positron pairs lying dormant
in the vacuum. It is the properties of these e-p pairs which give rise to the
permitivity and permeability of the vacuum.
I think you need less reading and more practical.
Regards,
Monitek (Arden Barker)
PS. Before anybody writes: Monitek regards the accomplishments of Maxwell,
Faraday and Hertz as some of the finest ever achieved by any scientists
anywhere, particularly when you take into account the available apparatus for
their experiments. The deductive capabilities of these guys was pure genius.
Alfred Einstead
> You are stupid. They are not real, they are derived:
>
> F_mn = dA_m/dx_n - dA_n/dx_m
They are real, A is derived from F:
F_mn = dA_m/dx_n - dA_n/dx_m
> Maxwell's equations are then:
> dF_mn/dx_n = j_m
> This equation can then be rewritten in terms of the proper field A:
>
> d^2A_m/dx_n^2 = j_m
No. The equation is
[]A_m = d_m (d^n A_n) + j_m.
d^n A_n is completely arbitrary and can be anything,
because ... A is derived, not F.
Onoang Blood
Well it is good to see some posts having a remote smell of intelligence ;-)
To read older books would not give you the whole story. Indeed the full
story I did not give as to cover such topics would be two great for the
feable minds that have posted in the group. But for you perhaps there is
hope...
A is indead not physical either. Rather it is the quantum field of the
photon. The things that is "real" is the photon. The field A is the
photon's quantum field. Quantum Fields are mathematical constructs used to
derive properties of the "real" particles. They tell us hting slike
momentum, energy, etc. In fact, as you noted, A is directl related to the
energy and momentum contained in the photon.
> (page 383, "Maxwell on the Electromagnetic Field, A Guided Study",
> Thomas K. Simpson, Cambridge University Press - 1997)
>
> Now I agree with you that the vector potential IS a representation
> of a 'source field'. Thus the expressions,
> __
> B = \/ x A
> __
> E = -\/V - @A/@t
>
> Where @ represents the 'partial' differential symbol
>
> And, as Simpon states, "For Maxwell, on the other hand, A appears
> at the outset, since it represents for him the momentum stored in
> the resting field, manifested in various ways in all electromagnetic
> phenomena".
>
> But which is 'real'? Some argue A isn't 'real' as illustrated by
> Bilge, and you appear to be arguing exactly the opposite.
Not true in the end. However Quantum Mechanics was too far out of reach for
the prior posters. Why indeed would one assume that two fields derived from
a single source field would be the "real" fields? Does not Maxwell's
equations show these two fields are derived from the single source field A?
It is this fact that shows both Electric and Magnetic phenomenon are united
at not distinct. In physics, that which is "real" are things which cannot
be derived from others. They are things that are "fundemental". In
mathematics terms, "real" are the axioms form which all theorems are
derived.
Is the atom real? Or is it the particles that make it up? In physics we
say the fundmental particles are the source. The atom is a construct of
those things. It is not fundemental itself.
>
> My point was/is they are ALL 'real', with B and E being observably
> separable elements. For Maxwell this was how he saw it also. I
> reference Simpson's book (noted above) or see Maxwell's papers.
> Specifically, "On Faraday's Lines of Force" and "On the Physical
> Lines of Force".
This is not to current view of phsyics since quantum theory or even days
after Maxwell. You are crossing the subjective meaning of "real" with that
held by physics. I do not want to argue over symantics. Indeed you may run
around and say your hand is "real" or that your computer is "real". In the
realm of every day life this is heald true. But in the realm of physics, we
know that your hand and your computer are made of fundmental pieces which
have no other parts. These pieces are the source and are what is truly
real. These fundmental particles, which in the Standard Model include the
photon, have quantum fields. For the photon this is the field A. These
fields are not physical but mere matematical constructs used to describe the
"real" particles.
> >>> There, now I've written all the equations out that are contained in
> >>> any decent EM book. ...
> >>
> >> Let's say you parroted the equations... The 'real' question is,
> >> what do they mean, or conversely, what physically, are they
> >> representing?
> >
> > In a nutshell, they stem from the source principle of least
> > action. Put another way, they represent a symmetry of nature.
> > This symmetry manifests itself as charge conservation and the
> > above equations.
>
> These are 'motherhood' philosphical statements. The 'principle
> of least action' is due to the statistical behavior of complex
> systems containing n degrees of freedom . The question then
> becomes, formally, 'what is action'?. The term 'symmetry of
> nature' means what, physically?
Yes. In quantum theory it turns out that there is no such thing as a single
particle. In fact, ALL systems must be treated statistically. They ALL
have infinite degrees of freedom. If you wish to know more about the
symmetry and were that plays you will need to read a many quantum physics
books. Start by looking up "Noether's Theorem". It shows that in any
quantum system, if a symmetry exists there must be a cooresponding conserved
quantity. EM equations stem from a symmetry of gauge invariance.
>
> >>> For whatever reason, you are too stupid to go read a book.
> >>
> >> And you seem too stupid to actually try to reason out what others
> >> have symbolically represented in mathematics. A math symbol or
> >> formulation is neither real or unreal, but if what it represent is
> >> empirically measurable, the 'what' it represents is quite 'real'...
> >
> > If you adjust your definition of "real" you can conclude anything.
> > The photon is "real". It is what we measure. You don't ever
> > measure the electric or magnetic field. Rather, you measure the
> > interaction of the photons. The devices used then translate that
> > interaction into numbers we associate to things like "voltage" and
> > even your precious "electric" or "magnetic" fields.
>
> Very good. This IS the crux of the issue, the definition of 'real'
> in an empirical sense. Now one should conclude that BOTH you and Mr.
> Fizzx are right, it just a matter of interpetation of the definition
> of 'real'.
Well I don't want to mix words. Symmantics... The prior argument occurred
only because it was clear that the poster did not even know basic EM. I
don't think he understood what A was. He clearly did not know what it later
lead too: the quantum field of the photon. I only tried to explain it and
he became stupidly hostile. He asked no questioned which shows his lack of
understanding and overall ignorance.
>
> > This is true even in gravity. You do not measure the gravitational
> > field. It is not real, just a mathematical construct.
>
> If it's 'not real' what is the 'mathematical construct' representing?
> Also, if it isn't 'real' in the empirical sense, where does the
> manifestations associated to it come from or go to?
Gravitons. The mathematical construct is the graviational quantum field.
As oyu know this is wand waving for gravity because a current theory of
quantum gravity does not yet exist ;-) You could, though, make the same
argument using space-time instead of gravitons.
>
> > You measure the motion of objects. You then *use* the graviational
> > field (math) to make predictions.
>
> We have a very fundamental difference of opinion (and mental
> concept) of 'real' here...
You attach yourself too much to pictures. It is what gets all in trouble
when they read about quantum fields and relativity. Too many try and hold
on to what they consider "common sense" and get lead to false results.
Indeed the world exists in a manner we cannot draw as pretty pictures on
paper. There is more going on then meets the eye. Fields were invented by
Newton long ago as a means to describe what was happening. He used them to
aide in prediction. At what point the mathematical construct of field
slipped into "real" I do now know. What I do know is now that quantum field
thoery is with us we have returned back from using fields as "real" things.
They are once again where they belong: mathematical constructs.
Indeed, even coorindate systems are nothing but mathematical constructs.
Hence we have the principle of relativity. We ascribe points in the space
with numbers. Why? Because it allows us to use our math to make
predictions and to describe the world. I can say where something is in
space and time in terms of numbers if I describe each point in space by a
set of numbers. This mapping of points to numbers is what a coorinate
system is all about. Why then would the physical world care about our
choice of mapping? It doesn't. And hence the laws of physics should be
written in such a fashion that our arbitrary choice of coorindate system
would not matter. The coorindate system is not *real*. It is created by us
to aid in our description of the physical stuff in space. This is the
principle of relativity: All laws of physics must be independent of
cooridnate system. That is why Einstein chose Tensor Calculus to describe
General Relativity. Tensors equations are by definition independent of
coordinate system.
In quantum mechanics such is the same. The physical things are the
particles. The quantum field is something we created to aid in our
description of the particles. Rules govern the particles and hence also
govern the fields used to describe them. These rules are really no
different then what we have had for a long time: energy and momentum
conservation etc. If you read books on quantum mechanics and follow it's
evolution in chronological order you will see this plainly.
> >> Are the 'empirical' properties designated as an 'electric field'
> >> and those designated as a 'magnetic field' the same? If not, then
> >> both are distinct, and thus quite 'real'. Are they related to each
> >> other?, sure! But both are real in the empirical observable sense.
> >
> > No they are not. They are derived form a single source field. This
> > is ancient physics from Maxwell.
>
> OK fine, can you point me to where in Maxwell's papers I can find
> this? I've read them all and I didn't see that. BTW, as far as I
> can discern, neither did Simpson. In fact, the modern form of the
> vector potential doesn't exist in Maxwell's work. Note that neither
> does the Lorentz force equation.
Maxwell would obviously not have known that his field A would end up the
quantum field of the photon.
> >> That one can 'derive' them from the vector potential say something,
> >> but NOT they are 'unreal'.
> >
> > They are not real because they cannot be directly measured. You
> > don't measure the E or M fields. You measure the photons and then
> > *translate* that into the values of your fields. They are convenient
> > mathematical constructs and nothing more.
>
> All so-called 'convenient mathematical constructs' are such only
> BECAUSE they represent and quantify some empirically observed
> process. Thus ALL SUCH must represent objective reality, or it isn't,
> by definition, science.
No. Math is not reality. Language is not thought. When you speak words
from your mouth, they are just a translation, a representation, of your
thoughts. After all, cannot you speak the same meaning but in different
languages? The thought in this example is what is "real". The words are
constructs to aid in communication. You can write programs the detect the
spoken word and print it out on your screen (Microsoft has such, but it
sucks ;-) ). Just as with EM. The photon is the base. It is what is
real. We have created the construct of E and M to aid in our description of
things and have even created devices to detect these fields. But do not be
duped by the end result. Look deeper into the process. These devices do
not detect E or M as they are not real. The physical process is simply the
exchange of photons. The devices perform translations from the detected
photon exchange into our created E and M fields.
Whether or not you agree as to how this works is not the question here. The
current position of the physics community is this (a summary):
Photons are the physical entity. That being so, we dub them "real". Hence
that is our definition of the word.
Photons have a mathematical construct used by us to determine their physical
properties. This "thing" is the quantum field for the photon commonly
denoted "A".
The E and M fields are not fundemental. Not only are they derived fields
from the single source field A, but that source field isn't real. Hence E
and M are not real.
Onoang Blood
> to Onoang Blood,
>
> You measure the E field with a capacitor. You measure an M field with an
> inductor.
This is a residual effect. We now know you are measuring photons not
fields.
> my previous post in this thread covers the capacitive measurement. The
> inductive measurement consists of aligning a coil with the direction of
> propagation of the EM wave and measuring a potential difference across the
ends
> of the coil. Turning the coil through 90 degrees eliminates the potential
> difference at the ends of the coil due to the em wave being examined.
I know how these devices work. Apparently you do not know what physics is
going on to make them work. Physics has moved well beyond the simplistic
understanding in the days of Maxwell.
> Thats how total nobrainers with no understanding like Maxwell, Faraday
and
> Hertz deduced that light as we know it was an alternating electromagnetic
> field. Maxwell required a dielectric medium to create the mathematics of
emr.
> The dielectric is non other than combined electron-positron pairs lying
dormant
> in the vacuum.
This last statement is your opinion and not that of current physics. Yes
the void is filled with pairs. But since you obviously know little of this
process you made a simple stupid error. The vacuum is filled with a LOT
more then just electrons and positions. Not only that, but the permeability
and permitivity of the vacuum is that of empty space. You know, the space
BETWEEN the particles. The presence of particles in space would only give
rise to a MACROSCOPIC permitivity and permeability. Indeed they do so in
materials which are filled with particles. But between the particles, where
the photons travel, there is nothing and in that nothing does the
permitivity and permeability of free space live.
> It is the properties of these e-p pairs which give rise to the
> permitivity and permeability of the vacuum.
No. It is a property of empty space.
> I think you need less reading and more practical.
Quite the opposite for you.
>
> Regards,
> Monitek (Arden Barker)
>
> PS. Before anybody writes: Monitek regards the accomplishments of Maxwell,
> Faraday and Hertz as some of the finest ever achieved by any scientists
> anywhere, particularly when you take into account the available apparatus
for
> their experiments. The deductive capabilities of these guys was pure
genius.
Good. Now read further for there is a lot more accomplishments in physics
since then.
Onoang Blood
>
> Dear Onoang Blood,
>
> s=ut + 1/2 ft^2 .......
>
> is an equation describing the distance car will move in a given time.
>
> Using this equation we can predict precisely where the car will be at some
time
> in the future given some elementary data about its current movement.
>
> Question: Given that you can predict the movement and position of the car,
what
> makes the car move?
Someone pushed it, maybe Fred Flintstone used his feet. Or perhaps it has
an engine inside? Physics 101: the equation above says nothing about what
makes the car move because the source of acceleration (f in your equation...
why you chose f instead of a...) is irrelevent when it comes to the results
of said acceleration.
> Fredi Fizzx is absolutely correct when he says that light creates it own
medium
> in which the waveform oscillates as the em wave progresses.
This is his (and your) opinion. Current physics says there is no medium.
Your example can be explained without one...
> The medium is the
> dielectric medium of Maxwell created by polarising (separating e-p pairs
from)
> the vacuum as the em wave passes.
See a different post. You are mixing ideas here that don't mix. Likely
because you do not fully understand the source of those pairs you describe.
For if you did, then you would know all about quantum field thoery and would
know already that EM phenomenon are the result of particles called photons,
not fields like E and M.
> A practical demonstration of vacuum
> polarisation is to create an alternating charge field between two
electrically
> conducting plates placed in such a way as to be parallel to each other,
and
> watch as electromagnetic radiation emanates from the gap between the two
> plates. Now, if electromagnetic radiation can be created by an alternating
> charge field, then, electromagnetic radiation upon passing two similar
plates
> to those which created the emanating electromagnetic radiation, should
recreate
> an alternating charge upon the aforementioned two electrically conducting
> plates placed in such a way as to be parallel to each other, surprise
surprise
> that is exactly what happens.
And why does this explaination require Aether? After all, let's not beat
around the bush as that is in effect what you are trying to describe.
Vacuum polarization in no way requires any new physics. It is easily
explained by current quantum field theory. Just do some searches on the
internet for "vacuum polarization" and "quantum field theory". This effect
is explained by photons as is ALL electromagnetic effects. There is no need
for the false field E or M. There is no need for some wave moving in a
medium. It doesn't. Photons are exchanged between charges including those
in the vacuum. That is how it gets polarized. Also do some reading from
Feynmann. Look up "Feynmann" and "QED". There is another great place to
look.
>
> Your explanation for this effect is.................
Quantum mechanics. Read up!
Onoang Blood
> > F_mn = dA_m/dx_n - dA_n/dx_m
>
> They are real, A is derived from F:
>
> F_mn = dA_m/dx_n - dA_n/dx_m
Geez. That was a stupid remark. And how do you conclude that from the
above equation? Because you ant o be argumentative?
> > Maxwell's equations are then:
> > dF_mn/dx_n = j_m
> > This equation can then be rewritten in terms of the proper field A:
> >
> > d^2A_m/dx_n^2 = j_m
>
> No. The equation is
> []A_m = d_m (d^n A_n) + j_m.
>
> d^n A_n is completely arbitrary and can be anything,
> because ... A is derived, not F.
Nope. The gauge condition is arbitrary yes. But this in no way makes A the
derived quantity. j is the source for A. F is NOT the source for A. Read
up!
Monitek
>Date: 18/08/02 18:40 GMT Daylight Time
>Message-id: <B_Q79.36113$mj7.6...@twister.rdc-kc.rr.com>
>
>(snip - please read thread...)
>> to Onoang Blood,
>>
>> You measure the E field with a capacitor. You measure an M field with an
>> inductor.
>
>This is a residual effect. We now know you are measuring photons not
>fields.
>
There are no photons only fields that is why emr has no mass.
>> my previous post in this thread covers the capacitive measurement. The
>> inductive measurement consists of aligning a coil with the direction of
>> propagation of the EM wave and measuring a potential difference across the
>ends
>> of the coil. Turning the coil through 90 degrees eliminates the potential
>> difference at the ends of the coil due to the em wave being examined.
>
>I know how these devices work. Apparently you do not know what physics is
>going on to make them work. Physics has moved well beyond the simplistic
>understanding in the days of Maxwell.
>
The physics is straight forward, a flowing current induces an opposite current
in the coil. The current flow is the e-p pairs separating.
>> Thats how total nobrainers with no understanding like Maxwell, Faraday
>and
>> Hertz deduced that light as we know it was an alternating electromagnetic
>> field. Maxwell required a dielectric medium to create the mathematics of
>emr.
>> The dielectric is non other than combined electron-positron pairs lying
>dormant
>> in the vacuum.
>
>This last statement is your opinion and not that of current physics.
I assume you mean modern physics. Let us see.
>Yes the void is filled with pairs.
Yes. So you agree there is a candidate for the dielectric medium.
>But since you obviously know little of this process you made a simple stupid
error. The >vacuum is filled with a LOT more then just electrons and
positions. Not only that, but the >permeability and permitivity of the vacuum
is that of empty space. You know, the space
>BETWEEN the particles. The presence of particles in space would only give
>rise to a MACROSCOPIC permitivity and permeability. Indeed they do so in
>materials which are filled with particles. But between the particles, where
>the photons travel, there is nothing and in that nothing does the
>permitivity and permeability of free space live.
>
No there are only e-p pairs in the vacuum there is no need for any other
particles in the vacuum so why have them? The physical size of a wave of emr
precludes the possibility of it travelling between the particles in the vacuum.
I postulate the e-p pairs are the maxwell dielectric medium.
>> It is the properties of these e-p pairs which give rise to the
>> permitivity and permeability of the vacuum.
>
>No. It is a property of empty space.
>
The impedance of the vacuum is 370 ohms, whereas the impedance of the empty
space between the particles of the vacuum is infinite.
>> I think you need less reading and more practical.
>
>Quite the opposite for you.
>
>>
>> Regards,
>> Monitek (Arden Barker)
>>
>> PS. Before anybody writes: Monitek regards the accomplishments of Maxwell,
>> Faraday and Hertz as some of the finest ever achieved by any scientists
>> anywhere, particularly when you take into account the available apparatus
>for
>> their experiments. The deductive capabilities of these guys was pure
>genius.
>
>Good. Now read further for there is a lot more accomplishments in physics
>since then.
>
Regards,
Monitek (Arden Barker)
Monitek
>Date: 18/08/02 18:56 GMT Daylight Time
>Message-id: <vdR79.36116$mj7.6...@twister.rdc-kc.rr.com>
>
>(snip)
>>
>> Dear Onoang Blood,
>>
>> s=ut + 1/2 ft^2 .......
>>
>> is an equation describing the distance car will move in a given time.
>>
>> Using this equation we can predict precisely where the car will be at some
>time
>> in the future given some elementary data about its current movement.
>>
>> Question: Given that you can predict the movement and position of the car,
>what
>> makes the car move?
>
>Someone pushed it, maybe Fred Flintstone used his feet. Or perhaps it has
>an engine inside? Physics 101: the equation above says nothing about what
>makes the car move because the source of acceleration (f in your equation...
>why you chose f instead of a...) is irrelevent when it comes to the results
>of said acceleration.
>
Yes - correct answer. The point is the maths doesnt tell us how the thing works
only what it will do next. Some of us out here in cyberspace like to know what
is going on under the bonnet (hood) so please let us muse.
>> Fredi Fizzx is absolutely correct when he says that light creates it own
>medium
>> in which the waveform oscillates as the em wave progresses.
>
>This is his (and your) opinion. Current physics says there is no medium.
>Your example can be explained without one...
>
And you said:
I quote:
>Yes the void is filled with pairs.
So if the void is filled with pairs. Doesn't it occur to you that they may do
something other than sit there quietly. The mathematics of QED may not need a
medium to travel in, ( wasnt that achieved by moving into the frame of the
light wave?) But that does not preclude a medium being responsible for the
propagation of emr, particularly if we both agree a candidate is there.
>> The medium is the
>> dielectric medium of Maxwell created by polarising (separating e-p pairs
>from)
>> the vacuum as the em wave passes.
>
>See a different post. You are mixing ideas here that don't mix. Likely
>because you do not fully understand the source of those pairs you describe.
>For if you did, then you would know all about quantum field thoery and would
>know already that EM phenomenon are the result of particles called photons,
>not fields like E and M.
>
I would love to know the source of the e-p pairs. I think they came from the
Big Bang.
Yes but one can not use the word Aether. Zero point energy, polarised vacuum,
Dirac Sea, e-p pairs, Quantum Vacuum, Quantum Foam to name but a few are all
euphemisms for the same thing the 'A' word.
As I have said before Photons dont exist. EMR is a wave motion in the Dirac
Sea. They weren't absolutely sure until fairly recently that there was anything
in the vacuum - now they are as sure as experiment will allow.
Feynmann does do a nice pictogram for the purposes of aid memoire.
>>
>> Your explanation for this effect is.................
>
>Quantum mechanics. Read up!
>
Regards,
Monitek (Arden Barker)
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:20020818052607...@mb-fh.aol.com...
| >From: "Onoang Blood" OBL...@mn.rr.com
| >Date: 18/08/02 04:15 GMT Daylight Time
| >Message-id: <CjE79.34270$mj7.6...@twister.rdc-kc.rr.com>
| to Onoang Blood,
|
| You measure the E field with a capacitor. You measure an M field with an
| inductor.
| my previous post in this thread covers the capacitive measurement. The
| inductive measurement consists of aligning a coil with the direction of
| propagation of the EM wave and measuring a potential difference across the
ends
| of the coil. Turning the coil through 90 degrees eliminates the potential
| difference at the ends of the coil due to the em wave being examined.
|
| Thats how total nobrainers with no understanding like Maxwell, Faraday
and
| Hertz deduced that light as we know it was an alternating electromagnetic
| field. Maxwell required a dielectric medium to create the mathematics of
emr.
| The dielectric is non other than combined electron-positron pairs lying
dormant
| in the vacuum. It is the properties of these e-p pairs which give rise to
the
| permitivity and permeability of the vacuum.
What you are saying above is exactly what I have been researching. A
connection between vacuum polarization and L and C of space.
| I think you need less reading and more practical.
|
| Regards,
| Monitek (Arden Barker)
|
| PS. Before anybody writes: Monitek regards the accomplishments of Maxwell,
| Faraday and Hertz as some of the finest ever achieved by any scientists
| anywhere, particularly when you take into account the available apparatus
for
| their experiments. The deductive capabilities of these guys was pure
genius.
Yes they most certainly were pure genius, they built the foundation that
Einstein and others used to finish figuring most of it out. Someone that I
would add is Tesla for his practical applications of that genius.
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
| (snip)
| Photons are the physical entity. That being so, we dub them "real".
Hence
| that is our definition of the word.
| Photons have a mathematical construct used by us to determine their
physical
| properties. This "thing" is the quantum field for the photon commonly
| denoted "A".
| The E and M fields are not fundemental. Not only are they derived fields
| from the single source field A, but that source field isn't real. Hence E
| and M are not real.
You have completely lost touch with reality. Have you ever put iron filings
on a piece of paper and then put a magnet under the paper? Believe me, the
iron filings line up with a *real* magnetic field. Have you ever rubbed a
balloon on your head and then felt your hair stand on end as you pull the
balloon away? This is a *real* electric field. Sure the source is photons,
but photons can make *real* electric fields and *real* magnetic fields. In
fact photons are the source for most all forces we define as *real* other
than the force related to gravity. Get real man!
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
Correction: And if what Todd Desiato is claiming is true, then photons are
basically the source of gravity also.
http://www.todds.info/physics.htm
FrediFizzx
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:20020817110132...@mb-ct.aol.com...
| How does space have an "L" and a "C" property if it is empty?
Simple. Space is not empty. Nor can it ever be completely empty or devoid
of EM radiation. Look out into space and you can easily see and detect that
EM radiation is everywhere. There is no place that it is not. Only the
density of it varies. Plus space has an intrinsic impedance of about 377
ohms. Where does this come from? As far as EM is concerned, it must come
from the vacuum polarization. This is most likely the source of the L and C
properties of space-time also. So space is far from what we would think as
"empty". Space is not "empty" and never will be. The vacuum polarization
must be the source of your "aether" also.
FrediFizzx
Jan Emil Larsen
"Onoang Blood" <OBL...@mn.rr.com> skrev i en meddelelse
news:Jww79.9950$Hf.3...@twister.kc.rr.com...
> You are the one who needs to learn something.
But you seem to be a lousy teacher, who hides behind meta-reality and math.
While you may well be right in the formalism, you are not in the attitude.
An expert is capable of presenting his knowlegde to a lay-man. Of cource, if
you don't really know, but only cite your books, you can't.
Monitek
>Date: 18/08/02 20:49 GMT Daylight Time
>Message-id: <eTS79.1039$RZ2.94...@newssvr14.news.prodigy.com>
I would be interested in discussing your line of thought.
Monitek (Arden Barker)
fizzxhaha@ahahhotmail.com Fredi Fizzx
Have you read any of the articles by Todd Desiato on how he is unifying
gravity and EM?
http://www.todds.info/physics.htm
Read this if you haven't done so. I think you will find it interesting.
Basically, my thoughts are that anytime you have EM radiation, you also have
inductance and capacitance happening. Where does this come from? It has to
come from space-time itself. And if the vacuum polarization is true, then
it must be connected somehow. Even though Todd mentions it, I don't think
he really explains the connection directly. Maybe you can get more out of
it. But I will make an attempt here. You have to have something that is
opposing current or the speed of light would be infinite. Yet you have to
have something (in space) that allows current to conduct or the speed of
light would be zero. Inductance is opposing and capacitance is conducting.
Working together they set the speed of light exactly to what it is. I think
the way Todd is describing it is that the polarization forces of the vacuum
are causing a "refraction" and this is where the L and C properties of space
are being manifested. Or that this refraction can be modeled like L and C.
FrediFizzx
larry
> "Onoang Blood" <OBL...@mn.rr.com> wrote in message
> news:Ebi79.22798$27.6...@twister.rdc-kc.rr.com...
> | > It is the currently accepted "truth" that there is no experimental
> | evidence of
> | > the Aether. This seems surprising because, if the Aether made its
> presence
> | any
> | > more obvious, physicists would have toothmarks on their butts from where
> | the
> | > Aether had jumped up and bit them. A few examples:
> | >
> | > Empty space, other than occupying a volume, has at least two observable
> | > properties, its dielectric constant and its permeability. If space is
> | "empty",
> | > what is it that has these properties? Since, if space consisted of a
> | volume of
> | > "nothingness", it cannot have any properties other than volume. Since it
> | does,
> | > space it would seem that it does contain a rejected entity, which was
> | called
> | > the Aether in the 19th Century!.
> |
> | Not true. You are confusing mathematical quantities with physical ones.
> | You have not explained why is something has a dialectric constant it must
> | have "stuff" in it.
>
> You are wasting your time if you expect reticher to reply to your message.
> He only spams the newsgroup and hardly ever replies. If you want a
> response, you have to email him. I guess he doesn't like to share.
>
> | >
> | > The velocity of light in free space is determined by the dielectric
> | constant
> | > and the permeabiluty of space in accordance with accepted physical laws.
> |
> | These again are just math quantities and have no physical reality. The
> | velocity of light is determined by the relationship of time to space in
> the
> | geometry of space.
>
> Yes, but it is a known fact that capacitance and inductance are purely
> geometric. Geometric of what? Well, it would have to be space itself. So
> I would have to postulate that space itself is capacitance and inductance.
It appears that many forget that a field strength of zero does not mean
that there is no EM fields in the space under consideration. E.g., a
charged spherical shell has a net field strength of zero in its
interior, but is filled with EM radiation. If it were not, a charged
test body exterior to the sphere would not be affected by a central
field. The same reasoning applies to a spherical shell of matter. The
gravitational field has a net value of zero in the interior of the
sphere, but the sphere is filled with whatever makes up a gravitational
field, which also is the cause of inertial motion. So a real physical
capacitance and inductance is not out of the question for that which
lies between material bodies. There should also be effects on bodies
moving with respect to to radiation within the spherical shells, in
the gravitational case a body should encounter a greater flux in the
direction of motion, say, at velocity v, and thus meet a constant flux
which would cause the continuation of the motion at velocity v.
Larry
> The combination of the two, which are inseperable, create a space "field".
> At least as far as EM is concerned. Space does have a known impedance also.
> About 377 ohms. I believe it is these properties of space that *is* the
> relationship of time to space in the geometry of space that you are talking
> about. Time equals 2*pi*sqrt(L*C). This is what makes the speed of light
> exactly what it is.
>
> | > Similarly, the velocity of sound in a steel rod is determined, using an
> | > equation analogous to that used to determine the velocity of light, by
> the
> | > elasticity and density of the steel.
> |
> | You are beating a very old drum. You cannot compare the two because they
> | are based on two different principles entirely. The "wave" of light is
> just
> | the photon quantum field. Quantum fields are in themselves not physical.
> | They are nothing more then mathematical entities we use to derive
> properties
> | of the physical particles. Waves in a medium, however, are actual motion
> of
> | physical stuff. Totally different.
>
> Photons do make EM waves and they are waves of *real* electric and magnetic
> fields. To me, electric and magnetic fields are real physical stuff.
>
> | > While a modern physicist seems to assert
> | > that the existence of the dielectric constant and the permeability of
> | space
> | > does not require the existence of a medium (i.e.- the Aether) having
> these
> | > properties, it seems certain that he would not be foolish enough to
> assert
> | > that, since the velocity of sound in a steel rod is determined only by
> its
> | > elasticity and its density, he could remove the rod and retain the
> density
> | and
> | > elasiticity so that the sound could propagate.
> |
> | We couldn't. Sound is a physical wave hence requiring physical stuff.
> | Light is NOT a physical wave. It is a mathematical wave: a quantum wave.
> | Hence it does not require physical "stuff" to move.
>
> Well, I would say more like light creates its own medium using the L and C
> properties of space. Photons do make up EM waves which are not really just
> a mathematical wave. EM waves are as real as something can be. It is the
> complementary action of E making M and M making E that is part of the
> medium. And I doubt very much if this could work without also the
> complementary action of L and C space. So the combo action of EM with LC
> creates the "medium" which allows light to happen.
>
> FrediFizzx
>
>
Todd Desiato
Correction, photons are the source of gravity between EM fields. In my most
recent work other fields are not considered or included. Only EM is looked
at and the forces between EM fields.
A more recent paper can be found at;
http://www.todds.info/gaugex/gtevs.pdf
Gauge Transformations of the Electromagnetic Vacuum States
The “Back Road” to Quantum Gravity
Where the gravity of the EM fields is shown to come about by a superposition
of EM fields.
Best Regards,
Todd Desiato
Todd Desiato
> >From: "Onoang Blood" OBL...@mn.rr.com
> >Date: 18/08/02 18:40 GMT Daylight Time
> >Message-id: <B_Q79.36113$mj7.6...@twister.rdc-kc.rr.com>
> >
> >(snip - please read thread...)
> >> to Onoang Blood,
> >>
> >> You measure the E field with a capacitor. You measure an M field with
an
> >> inductor.
> >
> >This is a residual effect. We now know you are measuring photons not
> >fields.
> >
>
> There are no photons only fields that is why emr has no mass.
No that is not why the EM field has no mass and is completely irrelevant.
There is more than adequate evidence of the existence of photons.
The EM field is massless because;
E^2 - p^2 = 0
for EM radiation, and because if photons had mass light would not travel at
the speed of light in a vacuum, and EM fields would not be gauge invariant.
In QFT fields are composed of the "field quanta", commonly referred to as
particles but the "classical" notion of a particle does not apply.
> >> my previous post in this thread covers the capacitive measurement. The
> >> inductive measurement consists of aligning a coil with the direction of
> >> propagation of the EM wave and measuring a potential difference across
the
> >ends
> >> of the coil. Turning the coil through 90 degrees eliminates the
potential
> >> difference at the ends of the coil due to the em wave being examined.
> >
> >I know how these devices work. Apparently you do not know what physics
is
> >going on to make them work. Physics has moved well beyond the simplistic
> >understanding in the days of Maxwell.
> >
>
> The physics is straight forward, a flowing current induces an opposite
current
> in the coil. The current flow is the e-p pairs separating.
There are no electron-positron pairs in a typical induction coil! The energy
density is not nearly high enough to create them. To create an e-p pair
requires an energy of at least,
E = 2*m*c^2, where m is the mass of the electron.
In an inductor you can speak of electrons and holes, but these holes are not
the holes in the Dirac Sea that are referred to as positrons. They are
missing electrons in the conduction band of the metal which appear to
propagate in the opposite direction as the electron current flow. There is a
difference between holes in a conduction band and holes in a vacuum.
>
> >> Thats how total nobrainers with no understanding like Maxwell, Faraday
> >and
> >> Hertz deduced that light as we know it was an alternating
electromagnetic
> >> field. Maxwell required a dielectric medium to create the mathematics
of
> >emr.
> >> The dielectric is non other than combined electron-positron pairs lying
> >dormant
> >> in the vacuum.
> >
>
> >This last statement is your opinion and not that of current physics.
>
> I assume you mean modern physics. Let us see.
>
> >Yes the void is filled with pairs.
>
> Yes. So you agree there is a candidate for the dielectric medium.
The probability of e-p pairs forming depends on the energy density of the
field, and for most vacuum states it is a very small probability. EM fields
of photons are a polarizable medium without the need for e-p pairs to be
forming. Where e-p pairs do form is very near to the charge of an electron
or in very strong electric fields. When they form there is a noticeable
change in the vacuum polarization and a measurable change in the
permittivity which becomes larger than the value in free space. So you
cannot attribute the permittivity of vacuum to be due to e-p pairs. I've
been there and done that and it didn't work out.
What gives the vacuum polarizability in free space is the presence of the
zero-point EM field. When you superimpose other EM fields onto the
zero-point field there are natural, derivable forces which act to minimize
the energy density in that region by rotating the fields to be out-of-phase
and cancel each other out. This force is easily seen to be exerted between 2
electric field vectors superimposed at a point. This superposition leads to
a probability density term that is analogous to the interference term in a
hologram. As in a hologram it is the interference pattern that holds the
information about both fields and their equations of motion. As in GR the
end result is that gravity is obtained from the tensor T_uv, which even for
a superposition of EM fields is not too hard to calculate. The
polarizability is then interpreted as geometry because that is what the
equations of motion show us about how the fields will behave.
I would say that IF you want "evidence of the Aether", then all the evidence
we have about EM fields tells us that the aether must be made up of EM
fields of photons.
Best Regards,
Todd Desiato
Jim Heckman
On 22-Aug-2002, larry <gold...@charter.net> wrote:
[...]
> It appears that many forget that a field strength of zero does not mean
> that there is no EM fields in the space under consideration. E.g., a
> charged spherical shell has a net field strength of zero in its
> interior, but is filled with EM radiation. If it were not, a charged
> test body exterior to the sphere would not be affected by a central
> field. [...]
??
--
Jim Heckman
Monitek
>Date: 22/08/02 20:10 GMT Daylight Time
>Message-id: <XGa99.2368$ja.1...@twister.socal.rr.com>
>
>
>"Monitek" <mon...@aol.com> wrote in message
>news:20020818145234...@mb-fp.aol.com...
>> >From: "Onoang Blood" OBL...@mn.rr.com
>> >Date: 18/08/02 18:40 GMT Daylight Time
>> >Message-id: <B_Q79.36113$mj7.6...@twister.rdc-kc.rr.com>
>> >
>> >(snip - please read thread...)
>> >> to Onoang Blood,
>> >>
>> >> You measure the E field with a capacitor. You measure an M field with
>an
>> >> inductor.
>> >
>> >This is a residual effect. We now know you are measuring photons not
>> >fields.
>> >
>>
>> There are no photons only fields that is why emr has no mass.
>
>No that is not why the EM field has no mass and is completely irrelevant.
>There is more than adequate evidence of the existence of photons.
>
>The EM field is massless because;
>
>E^2 - p^2 = 0
>
>for EM radiation, and because if photons had mass light would not travel at
>the speed of light in a vacuum, and EM fields would not be gauge invariant.
>In QFT fields are composed of the "field quanta", commonly referred to as
>particles but the "classical" notion of a particle does not apply.
>
Monitek:
The way I see it is this:
The vacuum is a sea of annihilated electron positron pairs. The pairs can be
partially separated in the presence of charge. As they separate lets say the
positron moves to the left and the electron moves to the right. This
constitutes a current equivalent to 2 electrons moving to the right, a moving
current creates a surrounding magnetic field. I envisage EMR as a radial
disturbance exactly the same form as a radial ripple in a pond. The partially
formed leptons separate at 90 degrees to the direction of propagation of the
wave along the direction of of the annulus of the ring form in such a way that
they form a complete circle of current. This forms an inductor, the current
flowing in the inductor creates a magnetic field round the inductor which in
turn creates an inductor outside the first inductor with the current flow
induced in the opposite direction to that of the first inductor, this in turn
creates a magnetic field in the opposite direction and induces a ring with the
circulating current the same as we had in the first loop we considered.
The particles in the sea do not move in the direction of propagation and the
speed of movement of the separated pairs has no bearing on the velocity of
propagation of the wave. The speed of propagation is governed only by the
velocity of the magnetic field. The we have a moving wave in a stationary
medium with not a photon in sight.
The concept of a photon arose out of the observed effects of light on matter -
light was able to move matter particles such as other leptons. This I consider
arises from the fact that when a wave of EMR passes the e-p pairs are separated
momentarily. When separated the charge of the electron or positron reappears
and a lepton would react to this charge.The higher the energy of the EMR the
further appart the e-p pairs would be, the more charge they would have thus the
more reaction they would have with ordinary matter. The photon is not necessary
to explain the observations.
>
>> >> my previous post in this thread covers the capacitive measurement. The
>> >> inductive measurement consists of aligning a coil with the direction of
>> >> propagation of the EM wave and measuring a potential difference across
>the
>> >ends
>> >> of the coil. Turning the coil through 90 degrees eliminates the
>potential
>> >> difference at the ends of the coil due to the em wave being examined.
>> >
>> >I know how these devices work. Apparently you do not know what physics
>is
>> >going on to make them work. Physics has moved well beyond the simplistic
>> >understanding in the days of Maxwell.
>> >
>>
>> The physics is straight forward, a flowing current induces an opposite
>current
>> in the coil. The current flow is the e-p pairs separating.
>
>There are no electron-positron pairs in a typical induction coil! The energy
>density is not nearly high enough to create them. To create an e-p pair
>requires an energy of at least,
>
It it not necessary to fully separate the pairs, and the "coil" is a loop in
free space not a wire wound one. Separated e-p pairs if lined up will create a
loop and the loop will be inductive.
>E = 2*m*c^2, where m is the mass of the electron.
>
>In an inductor you can speak of electrons and holes, but these holes are not
>the holes in the Dirac Sea that are referred to as positrons. They are
>missing electrons in the conduction band of the metal which appear to
>propagate in the opposite direction as the electron current flow. There is a
>difference between holes in a conduction band and holes in a vacuum.
>
The inductor is the loop in space, the loop in space is e-p pairs. the loop in
the lab is electron flow. When we have antimatter bateries we can test it for
positron flow!
>>
>> >> Thats how total nobrainers with no understanding like Maxwell, Faraday
>> >and
>> >> Hertz deduced that light as we know it was an alternating
>electromagnetic
>> >> field. Maxwell required a dielectric medium to create the mathematics
>of
>> >emr.
>> >> The dielectric is non other than combined electron-positron pairs lying
>> >dormant
>> >> in the vacuum.
>> >
>>
>> >This last statement is your opinion and not that of current physics.
>>
>> I assume you mean modern physics. Let us see.
>>
>> >Yes the void is filled with pairs.
>>
>> Yes. So you agree there is a candidate for the dielectric medium.
>
>The probability of e-p pairs forming depends on the energy density of the
>field, and for most vacuum states it is a very small probability. EM fields
>of photons are a polarizable medium without the need for e-p pairs to be
>forming. Where e-p pairs do form is very near to the charge of an electron
>or in very strong electric fields. When they form there is a noticeable
>change in the vacuum polarization and a measurable change in the
>permittivity which becomes larger than the value in free space. So you
>cannot attribute the permittivity of vacuum to be due to e-p pairs. I've
>been there and done that and it didn't work out.
>
Your considering e-p pairs forming as separated entities, I am considering e-p
pairs which dont quite make it as independent particles.
>What gives the vacuum polarizability in free space is the presence of the
>zero-point EM field. When you superimpose other EM fields onto the
>zero-point field there are natural, derivable forces which act to minimize
>the energy density in that region by rotating the fields to be out-of-phase
>and cancel each other out. This force is easily seen to be exerted between 2
>electric field vectors superimposed at a point. This superposition leads to
>a probability density term that is analogous to the interference term in a
>hologram. As in a hologram it is the interference pattern that holds the
>information about both fields and their equations of motion. As in GR the
>end result is that gravity is obtained from the tensor T_uv, which even for
>a superposition of EM fields is not too hard to calculate. The
>polarizability is then interpreted as geometry because that is what the
>equations of motion show us about how the fields will behave.
>
The vacuum is polarisable because it contains e-p pairs. The zero point energy
or zitterbewegung is due to passing EMR popping e-p pairs in and out of
(partial) existence. The fields will cancel or reinforce depending upon the
conditions at the point in question. What is the probability density of? In all
honesty I am not following your logic for the next bit.
>I would say that IF you want "evidence of the Aether", then all the evidence
>we have about EM fields tells us that the aether must be made up of EM
>fields of photons.
>
>Best Regards,
>Todd Desiato
>
As I have already said I can not see photons only fields.
Regards,
Monitek (Arden Barker)
Todd Desiato
I strongly suggest that you learn QFT before you try to reinvent it. There
are many subtleties that you missed.
>
> The vacuum is a sea of annihilated electron positron pairs. The pairs can
be
> partially separated in the presence of charge. As they separate lets say
the
> positron moves to the left and the electron moves to the right. This
> constitutes a current equivalent to 2 electrons moving to the right, a
moving
> current creates a surrounding magnetic field. I envisage EMR as a radial
> disturbance exactly the same form as a radial ripple in a pond. The
partially
> formed leptons separate at 90 degrees to the direction of propagation of
the
> wave along the direction of of the annulus of the ring form in such a way
that
> they form a complete circle of current. This forms an inductor, the
current
> flowing in the inductor creates a magnetic field round the inductor which
in
> turn creates an inductor outside the first inductor with the current flow
> induced in the opposite direction to that of the first inductor, this in
turn
> creates a magnetic field in the opposite direction and induces a ring with
the
> circulating current the same as we had in the first loop we considered.
Your ideas are only applicable in very strong fields with high energy
density, such as within the Compton radius of an electron. At those
distances from the charge there are e-p pairs, and they sort of behave as
you describe, they act as oscillators and exchange momentum between the
charge and the vacuum. They also screen the electric charge such that the
"bare charge" is slightly larger in value than the charge "e" which we
measure.
However EM waves are not observed to behave as you describe. The free field
wave equation that results from Maxwell's equations has particular
solutions, mass and charge are not required in those solutions. That is why
it is a "free field". These solutions have certain properties such as
frequency modes, angular momentum, polarization and momentum. We ascribe
these properties to "photons" because the energy and momentum behaves as if
there are N photons with momentum p = h/\lambda. This is true even when you
consider the classical amplitudes of the field! The energy and intensity is
proportional to the frequency.
An e-p pair and a photon do not behave the same way, no matter how you
imagine them to be. You need to understand the wave equations and solutions
of both systems before you will see this for your self. Quantum physics
takes more than just a vivid imagination.
>
> The particles in the sea do not move in the direction of propagation and
the
> speed of movement of the separated pairs has no bearing on the velocity of
> propagation of the wave. The speed of propagation is governed only by the
> velocity of the magnetic field. The we have a moving wave in a stationary
> medium with not a photon in sight.
>
> The concept of a photon arose out of the observed effects of light on
matter -
> light was able to move matter particles such as other leptons. This I
consider
> arises from the fact that when a wave of EMR passes the e-p pairs are
separated
> momentarily. When separated the charge of the electron or positron
reappears
> and a lepton would react to this charge.The higher the energy of the EMR
the
> further appart the e-p pairs would be, the more charge they would have
thus the
> more reaction they would have with ordinary matter. The photon is not
necessary
> to explain the observations.
You seem to have no trouble considering particles like electrons and
postitrons, but yet reject photons as particles. There are wave equations
for both, there are interference patterns within a superposition of both
such waves, they can both be refracted and reflected, they can both be
scattered or photographed and shown to consist of many individual point-like
interactions. The "Dirac Field" and the "Maxwell Field" have different
quanta, "electrons, positrons, etc.." versus "photons". They are both fields
of waves and particles. You are neglecting a century of research in QFT so I
suggest you do not neglect anything, and start from an understanding of what
has come before you .
Best Regards,
Todd Desiato
fizzxhaha@ahahhotmail.com Fredi Fizzx
news:v8999.1804$ja.1...@twister.socal.rr.com...
Yes, just between EM fields so far. Sorry I missed that qualification,
Todd.
FrediFizzx
Monitek
>Date: 23/08/02 01:05 GMT Daylight Time
>Message-id: <%_e99.2595$ja.2...@twister.socal.rr.com>
Your still thinking in terms of pair creating in the form of now you see me now
you dont. One has to consider the inbetween stages. My ideas are based on the
principle the e-p pairs have no mas and charge when combined and have unit mass
and charge when totally separated. When they are neither in nor out they have
properties of partial mass and fractional charge. The amount of charge and mass
is related to the distance of separation until you reach the point of complete
separation. Do they ever completely separate well even in pair creation it is
only for a short time.
>However EM waves are not observed to behave as you describe.
You will have to describe the differences to which you refere here.
> The free field
>wave equation that results from Maxwell's equations has particular
>solutions, mass and charge are not required in those solutions. That is why
>it is a "free field". These solutions have certain properties such as
>frequency modes, angular momentum, polarization and momentum. We ascribe
>these properties to "photons" because the energy and momentum behaves as if
>there are N photons with momentum p = h/\lambda. This is true even when you
>consider the classical amplitudes of the field! The energy and intensity is
>proportional to the frequency.
>
I am not questioning the predictive power of the equations. I am postulating
the underlying mechanics.
>An e-p pair and a photon do not behave the same way, no matter how you
>imagine them to be.
Again you will have to describe the differences you claim to get my agreement
on that.
> You need to understand the wave equations and solutions
>of both systems before you will see this for your self. Quantum physics
>takes more than just a vivid imagination.
>
>
I need to understand the mechanics behind the equations, whatever form the
equation takes there is a mechanism behind it. Equations are only usefull for
single mode systems. ie. there is a range of values over which the equation is
valid outside this range the results are erroneous when multiple mechanisms are
occuring concurrently.
By way of example:
I throw a ball out of my window bounce it on the road and it goes into the
window of the house opposite.
Can you produce a single equation to predict the path of the ball?
Particles like electrons and positrons are a physical reality their properties
are well known and measurable. They are considered as fundamental particles ie
indivisible into other particles yet nobody knows quite what to do with the
positron. For every electron there is a positron full stop. The "dirac field"
is electrons and positrons the maxwell field is "harmonic oscillators". When
you stretch an e-p pair out the electrostatic field between the two tends to
pull then together. As the particles separate the field between them increases
due to the charge on the particles increasing, however it behaves like the
spring of the maxwellian oscillator. I suggest that e-p pairs exactly fulfill
the maxwell model.
The individual point like interactions are the e-p pairs partially separating
and creating a charge field round themselves whilst they are apart - there is
no conflict here. Wave motions an be superimposed on separating e-p pairs the
separation just gets wider. The separation is not envisaged to to create pairs,
from what I have seen this occurs with high energy EMR coupled with the
presence of charge.
Regards,
Monitek (Arden Barker)
Monitek
>Date: 22/08/02 07:09 GMT Daylight Time
>Message-id: <oe%89.2754$wW3.13...@newssvr21.news.prodigy.com>
I will have a look.
>
>Basically, my thoughts are that anytime you have EM radiation, you also have
>inductance and capacitance happening.
Yes. EMR has charge and magnetism so that is true.
> Where does this come from? It has to
>come from space-time itself. And if the vacuum polarization is true, then
>it must be connected somehow.
Yes. Yes.
> Even though Todd mentions it, I don't think
>he really explains the connection directly.
Well he didnt exactly EXPLAIN it if a post in this thread.
> Maybe you can get more out of
>it. But I will make an attempt here. You have to have something that is
>opposing current or the speed of light would be infinite. Yet you have to
>have something (in space) that allows current to conduct or the speed of
>light would be zero. Inductance is opposing and capacitance is conducting.
Absolutley bang on the button as far as I am concerned. I find that the
magnetic field propagation governs the speed of EMR.
>Working together they set the speed of light exactly to what it is. I think
>the way Todd is describing it is that the polarization forces of the vacuum
>are causing a "refraction" and this is where the L and C properties of space
>are being manifested. Or that this refraction can be modeled like L and C.
>
>FrediFizzx
>
I can not see where refraction enters the picture.
Well thats about it for the rest of the day - I am off to Warwick Castle for a
day out.
Regards,
Monitek (Arden Barker)
Todd Desiato
> >From: "Todd Desiato" todd...@serversanddomains.com
> >Date: 23/08/02 01:05 GMT Daylight Time
> >Message-id: <%_e99.2595$ja.2...@twister.socal.rr.com>
> >
> >
> >"Monitek" <mon...@aol.com> wrote in message
> >news:20020822191936...@mb-mk.aol.com...
The problem you have is that these ideas are not solutions of the equations
of motion of a free EM field. You don't get partial mass or partial charge
in the solutions of these equations. You either have an electric charge +e
and -e, and masses m_p and m_e, or you don't. The equations of motion show
you get waves in a free field, but there is no charge and no mass. This is
all part of QED which is probably the most well tested and rigorous theory
there ever was. QED is not like QCD or QG, it has many predictions that have
been verified by experiment, including predictions that cannot be made or
verified by Classical EM alone, without quantum effects. As I said you
should really learn the current theory before you try to re-write it.
> >However EM waves are not observed to behave as you describe.
>
> You will have to describe the differences to which you refere here.
There is no charge and no mass in a free EM field in vacuum. Not even
"partial" charges or masses. If their were the equation;
E^2 - p^2 = 0
would not hold and that would have obvious consequences to the gauge
invariance and Lorentz invariance of EM fields.
> > The free field
> >wave equation that results from Maxwell's equations has particular
> >solutions, mass and charge are not required in those solutions. That is
why
> >it is a "free field". These solutions have certain properties such as
> >frequency modes, angular momentum, polarization and momentum. We ascribe
> >these properties to "photons" because the energy and momentum behaves as
if
> >there are N photons with momentum p = h/\lambda. This is true even when
you
> >consider the classical amplitudes of the field! The energy and intensity
is
> >proportional to the frequency.
> >
>
> I am not questioning the predictive power of the equations. I am
postulating
> the underlying mechanics.
The "underlying mechanics" of a free EM field in a vacuum have been well
established. The field consists of photons, which is just a neat word for
saying the field has many possible frequency modes that carry energy in
discrete quanta proportional to the frequency.
> >An e-p pair and a photon do not behave the same way, no matter how you
> >imagine them to be.
>
> Again you will have to describe the differences you claim to get my
agreement
> on that.
with c=1
E^2 - p^2 = m^2 for an electron or positron and E^2 = 4m^2 for an e-p pair
to form. You do not have partial charge or mass as a solution to either the
Maxwell or Dirac fields.
Also, photons have a velocity = c, an e-p pair will have velocity < c
because they have mass, and the energy is not infinite.
Electrons and positron are Fermions, photons are Bosons, they obey a
different type of statistics.
> > You need to understand the wave equations and solutions
> >of both systems before you will see this for your self. Quantum physics
> >takes more than just a vivid imagination.
> >
> >
>
> I need to understand the mechanics behind the equations, whatever form the
> equation takes there is a mechanism behind it. Equations are only usefull
for
> single mode systems. ie. there is a range of values over which the
equation is
> valid outside this range the results are erroneous when multiple
mechanisms are
> occuring concurrently.
Then you start with a different Lagrangian, one which combines both the
Maxwell and Dirac field, and look for the solutions to the equations of
motion of that system. You don't blindly "guess" what is going on. Again,
QED is a well tested, well defined theory and you should learn it before you
try to conceive of other theories of your own.
That statement has no basis in reality. A positron is exactly like an
electron with a positive charge. It is the right-handed version of the
electron.
> For every electron there is a positron full stop.
Not true. You can say for every negative charge there is a positive charge,
or say electric lines of flux always start and end at electric charge. But
you cannot say that for every electron there is a positron because we do not
see this. There are billions of free electrons where ever you look, but
there are not billions of free positrons. Anti-matter is no that popular
:0).
> The "dirac field"
> is electrons and positrons the maxwell field is "harmonic oscillators".
When
> you stretch an e-p pair out the electrostatic field between the two tends
to
> pull then together. As the particles separate the field between them
increases
> due to the charge on the particles increasing, however it behaves like the
> spring of the maxwellian oscillator. I suggest that e-p pairs exactly
fulfill
> the maxwell model.
The Maxwell field is "photons" just like the Dirac field is "electrons", the
same method of quantization can be used on either type of field. If you
separate two charges they "are" a harmonic oscillator too, but the force
decreases with the distance between them. The charge does not change. It's
either zero or e, it is not fractional. It has never been observed to be
fractional. Your idea in this regard is seriously flawed.
Regards,
Todd Desiato
Monitek
>Date: 23/08/02 20:24 GMT Daylight Time
>Message-id: <_Zv99.5314$ja.8...@twister.socal.rr.com>...............Snip
The Equations are wave equations they dont concern themselves with the
transport medium.
QED tells me what EMR does it does not say what it is. I would like to know
what it is. When it was shown that light had particulate properties the photon
was postulated to satisfy these requirements.There is no particle which leaves
the atom at the speed of light when an electron falls down shells to a lower
energy level. There are no particles emitted by an electron when it slows
sharply. The mistake that you are making is that the particles responsible for
the movement of a light wave have no movement in the direction of the EMR wave
only a sideways movement which I have described previously, they are not
travelling at the speed of light. These movements will be at non relativistic
speed therefore EMR is classical.
>> >However EM waves are not observed to behave as you describe.
>>
>> You will have to describe the differences to which you refere here.
>
>There is no charge and no mass in a free EM field in vacuum. Not even
>"partial" charges or masses. If their were the equation;
>
>E^2 - p^2 = 0
Charge squared - momentum squared = 0 ?
I could do with you defining the terms of your equations. Please.
>
>would not hold and that would have obvious consequences to the gauge
>invariance and Lorentz invariance of EM fields.
>
I agree that there is no mass , however I would say that there is charge
involved and this charge is measurable via a capacitor- thats how radio waves
are received. I am not particularly bothered about the consequences.
Thats because nobody has considered it when the equation was derived. Basically
when e-ps are closer than 0.11 fm the above equation does not apply. Also the
above is a wave equation for a particle momentum = mass x velocity - no mass
therefore no momentum therefore no particle.
>Also, photons have a velocity = c, an e-p pair will have velocity < c
>because they have mass, and the energy is not infinite.
>
>Electrons and positron are Fermions, photons are Bosons, they obey a
>different type of statistics.
>
As you say elsewhere photons are a mathematical construct, waves in a medium is
the reality - the medium is e-p pairs in an annihilated state in the vacuum.
>
>> > You need to understand the wave equations and solutions
>> >of both systems before you will see this for your self. Quantum physics
>> >takes more than just a vivid imagination.
>> >
>> >
>>
>> I need to understand the mechanics behind the equations, whatever form the
>> equation takes there is a mechanism behind it. Equations are only usefull
>for
>> single mode systems. ie. there is a range of values over which the
>equation is
>> valid outside this range the results are erroneous when multiple
>mechanisms are
>> occuring concurrently.
>
>Then you start with a different Lagrangian, one which combines both the
>Maxwell and Dirac field, and look for the solutions to the equations of
>motion of that system. You don't blindly "guess" what is going on. Again,
>QED is a well tested, well defined theory and you should learn it before you
>try to conceive of other theories of your own.
>
>
Exchange of a virtual photon to explain action at a distance was a guess was it
not? No virtual photon exchange has ever been observed but because the maths
works then that mechanism was assumed.
Well do tell us what part you think the positron plays in the great scheme of
the universe. I am in full agreement that the positron is a positive electron.
>> For every electron there is a positron full stop.
>
>Not true. You can say for every negative charge there is a positive charge,
>or say electric lines of flux always start and end at electric charge. But
>you cannot say that for every electron there is a positron because we do not
>see this. There are billions of free electrons where ever you look, but
>there are not billions of free positrons. Anti-matter is no that popular
>:0).
>
I can say every unit positive charge is due to a positron and have experimental
evidence to support this. As every electron has a positive charge nearby in the
form of a proton, all we have to say is that the positive charge on the proton
is due to a positron in its structure and hey presto every electron has a
positron to create a pair with. Yes a positron is antimatter and given half a
chance will annihilate with an electron. You may not like this view but it fits
the facts.
>> The "dirac field"
>> is electrons and positrons the maxwell field is "harmonic oscillators".
>When
>> you stretch an e-p pair out the electrostatic field between the two tends
>to
>> pull then together. As the particles separate the field between them
>increases
>> due to the charge on the particles increasing, however it behaves like the
>> spring of the maxwellian oscillator. I suggest that e-p pairs exactly
>fulfill
>> the maxwell model.
>
>The Maxwell field is "photons" just like the Dirac field is "electrons", the
>same method of quantization can be used on either type of field.
Because they are the same thing. E-p pairs in the vacuum.
>If you
>separate two charges they "are" a harmonic oscillator too, but the force
>decreases with the distance between them.
On a micro and macro scale you are correct but on a nanoscale you are wrong. As
I have already explained as the e-p's move closer together they overlap their
own space and state neutralising themselves. The normal attractive force due to
the charge decreases because the value of the charge decreases when they fully
overlap ie occupy the same space there is no charge therefore there is no force
of attraction between them. as they separate from this point the charge starts
to reappear and the force between them increases as the distance increases this
occurs up to the point where the electron and associated positron have unit
charge. From this point onwards the force between the particles decreases with
increasing distance.
>The charge does not change. It's
>either zero or e, it is not fractional. It has never been observed to be
>fractional. Your idea in this regard is seriously flawed.
>
>Regards,
>Todd Desiato
No fractional charge observed eh! - well somebody got a Nobel under false
pretences then:
http://www.nobel.se/physics/laureates/1998/press.html
http://www.warwick.ac.uk/~phsbm/qhe.htm
http://dept.physics.upenn.edu/~mele/qcmt/p3/project3.html
I think we can assume that they are seeing effects due to fractional charge -
note its in the vacuum!
Regards,
Monitek (Arden Barker)
Todd Desiato
> >From: "Todd Desiato" todd...@serversanddomains.com
> >Date: 23/08/02 20:24 GMT Daylight Time
> >Message-id: <_Zv99.5314$ja.8...@twister.socal.rr.com>
> >
> >
> >"Monitek" <mon...@aol.com> wrote in message
> >news:20020823044513...@mb-fh.aol.com...
> ...............Snip
> >>is only for a short time.
> >
> >The problem you have is that these ideas are not solutions of the
equations
> >of motion of a free EM field. You don't get partial mass or partial
charge
> >in the solutions of these equations. You either have an electric charge
+e
> >and -e, and masses m_p and m_e, or you don't. The equations of motion
show
> >you get waves in a free field, but there is no charge and no mass. This
is
> >all part of QED which is probably the most well tested and rigorous
theory
> >there ever was. QED is not like QCD or QG, it has many predictions that
have
> >been verified by experiment, including predictions that cannot be made or
> >verified by Classical EM alone, without quantum effects. As I said you
> >should really learn the current theory before you try to re-write it.
> >
> >
>
> The Equations are wave equations they dont concern themselves with the
> transport medium.
None is required. EM makes it's own medium. The vacuum "IS" an EM field. It
cannot be removed.
> QED tells me what EMR does it does not say what it is. I would like to
know
> what it is.
A free EM field is a density of energy comprised of frequency modes, which
are dependent on the surrounding geometry and materials, which can be
polarized, and which carry energy from point A to point B in discrete quanta
called "photons". I still don't understand why you believe in electrons but
not photons. Waves are particles. I suppose you don't agree with
wave-particle duality either?
> When it was shown that light had particulate properties the photon
> was postulated to satisfy these requirements.
No, the photon was postulated by Max Planck when attempting to explain
"black body radiation" and the difference between the predictions of the
spectral energy density by the "Rayleigh-Jeans distribution" at low
frequencies and the "Wien distribution" at high frequencies. The result was
that radiation must transport energy in discrete quanta such that the energy
E=nhf. where n is the number of photons at the frequency f. That is why we
call _h_ "Planck's constant".
Then Einstein confirmed this when he showed the photoelectric effect also
indicated that the energy of light was absorbed in discrete quanta of energy
E=nhf.
It was further confirmed by the double-slit experiment using film and short
exposures, for both electrons and photons the results are the same! For very
short exposures there is no interference pattern of waves, only dots where
individual quanta struck the film.
So unless you can reproduce the works of Planck and Einstein without the
need for E=nhf, and disprove the double-slit experiment then you must agree
that photons exist.
> There is no particle which leaves
> the atom at the speed of light when an electron falls down shells to a
lower
> energy level. There are no particles emitted by an electron when it slows
> sharply. The mistake that you are making is that the particles responsible
for
> the movement of a light wave have no movement in the direction of the EMR
wave
> only a sideways movement which I have described previously, they are not
> travelling at the speed of light. These movements will be at non
relativistic
> speed therefore EMR is classical.
You are working with 19th century physics in the 21st century. As I said
before you need to learn and understand all the related physics that has
been discovered in the last 120 years before you start making up your own
theories. Seriously, crack open a book and learn! Become an expert, then
start theorizing your own ideas. At this stage you don't even understand the
history let alone the math and experimental evidence that supports modern
physics. You are deluding yourself to think you understand physics better
than the giants that came before you.
>
> >> >However EM waves are not observed to behave as you describe.
> >>
> >> You will have to describe the differences to which you refere here.
> >
> >There is no charge and no mass in a free EM field in vacuum. Not even
> >"partial" charges or masses. If their were the equation;
> >
> >E^2 - p^2 = 0
>
> Charge squared - momentum squared = 0 ?
> I could do with you defining the terms of your equations. Please.
E is Energy, p is momentum. This is the magnitude of the relativistic
momentum 4-vector. There is a well known equation for the relativistic
energy of a particle,
E = sqrt[(mc^2)^2 + (pc)^2]
This comes from the momentum 4-vector p^u = (E,p1,p2,p3) where c=1 so we
neglect it. The magnitude of this 4-vector is;
p_u*p^u = E^2 - p^2 = m^2
where m is the invariant mass of the particle in it's rest frame.
For photons m=0, for electrons and positrons m=/=0.
> >
> >would not hold and that would have obvious consequences to the gauge
> >invariance and Lorentz invariance of EM fields.
> >
>
> I agree that there is no mass , however I would say that there is charge
> involved and this charge is measurable via a capacitor- thats how radio
waves
> are received. I am not particularly bothered about the consequences.
No, if say you had 2 parallel metal plates with a small space between them,
and then you pass an EM wave between them propagating parallel to the plates
with the electric field polarized perpendicular to the plates, then the
electric field will polarize the plates at it passes but all the charge
presnet on the plates comes from the plates not from the wave. The electric
field exerts a force on the charges in the metal to polarize the metal
plates and store energy in the capacitor. Also there are no positrons in a
capacitor, the positive charge is due to a lack of electrons to balance the
nuclear charges of the metal atoms.
.............snip
> >
> >The "underlying mechanics" of a free EM field in a vacuum have been well
> >established. The field consists of photons, which is just a neat word for
> >saying the field has many possible frequency modes that carry energy in
> >discrete quanta proportional to the frequency.
> >
> >> >An e-p pair and a photon do not behave the same way, no matter how you
> >> >imagine them to be.
> >>
> >> Again you will have to describe the differences you claim to get my
> >agreement
> >> on that.
> >
> >with c=1
> >E^2 - p^2 = m^2 for an electron or positron and E^2 = 4m^2 for an e-p
pair
> >to form. You do not have partial charge or mass as a solution to either
the
> >Maxwell or Dirac fields.
> >
>
> Thats because nobody has considered it when the equation was derived.
Nobody considered it because there was no evidence to support it!
> Basically
> when e-ps are closer than 0.11 fm the above equation does not apply.
At very short distances on the order of the Compton wavelength, the
"zitterbewegung" of an electron is precisely due to electron-positron
"exchange scattering" within that tiny space. However this does not in any
way imply that EM waves carry charge or consist of e-p pairs, or that the
vacuum medium consist of e-p pairs either. It only shows that where the
electric field is very strong the energy density is large enough to create
e-p pairs.
> Also the
> above is a wave equation for a particle momentum = mass x velocity - no
mass
> therefore no momentum therefore no particle.
Photons have no mass, but they can easily be shown to carry momentum p=hc/f.
Just get yourself a cheap glass Radiometer to prove it. The blades spin
because more momentum is absorbed by the black side of the blades than by
the shinny side. Nothing penatrates the glass but light and it is obvious
that the speed is proportional to the intensity of the light.
>
> >Also, photons have a velocity = c, an e-p pair will have velocity < c
> >because they have mass, and the energy is not infinite.
> >
> >Electrons and positron are Fermions, photons are Bosons, they obey a
> >different type of statistics.
> >
>
> As you say elsewhere photons are a mathematical construct, waves in a
medium is
> the reality - the medium is e-p pairs in an annihilated state in the
vacuum.
Waves are also comprised of particles. e-p pairs in an "annihilated state"
are photons and do not carry any charge. Again you either have an e-p pair
or you don't. There are no "partial" solutions.
>
> >
> >> > You need to understand the wave equations and solutions
> >> >of both systems before you will see this for your self. Quantum
physics
> >> >takes more than just a vivid imagination.
> >> >
> >> >
> >>
> >> I need to understand the mechanics behind the equations, whatever form
the
> >> equation takes there is a mechanism behind it. Equations are only
usefull
> >for
> >> single mode systems. ie. there is a range of values over which the
> >equation is
> >> valid outside this range the results are erroneous when multiple
> >mechanisms are
> >> occuring concurrently.
> >
> >Then you start with a different Lagrangian, one which combines both the
> >Maxwell and Dirac field, and look for the solutions to the equations of
> >motion of that system. You don't blindly "guess" what is going on. Again,
> >QED is a well tested, well defined theory and you should learn it before
you
> >try to conceive of other theories of your own.
> >
> >
>
> Exchange of a virtual photon to explain action at a distance was a guess
was it
> not?
No it was not. Scalar photons are derived as solutions when quantizing the
EM field in the Lorentz gauge.
> No virtual photon exchange has ever been observed but because the maths
> works then that mechanism was assumed.
By definition "virtual" events are not observed. However there are
predictions and experiments where virtual photons become real photons and
these have been observed. See the Unruh-Davies effect for example.
.................snip
> >> For every electron there is a positron full stop.
> >
> >Not true. You can say for every negative charge there is a positive
charge,
> >or say electric lines of flux always start and end at electric charge.
But
> >you cannot say that for every electron there is a positron because we do
not
> >see this. There are billions of free electrons where ever you look, but
> >there are not billions of free positrons. Anti-matter is no that popular
> >:0).
> >
>
> I can say every unit positive charge is due to a positron and have
experimental
> evidence to support this. As every electron has a positive charge nearby
in the
> form of a proton, all we have to say is that the positive charge on the
proton
> is due to a positron in its structure and hey presto every electron has a
> positron to create a pair with. Yes a positron is antimatter and given
half a
> chance will annihilate with an electron. You may not like this view but it
fits
> the facts.
What about other charged particles such as W's, mesons, Quarks etc., that
also carry charge? These are not positrons, some are not even Fermions!
While there are reactions where a proton can be split into a neutron +
positron + neutrino, this does not mean that a proton has a positron inside
it. It is more common the other way where a neutron decays into a proton +
electron + anti-neutrino. A proton or neutron actually is 3 quarks and a
bunch of gluons. So your idea is only valid for elementary nuclear reactions
and is simplistic compared to QCD and everything we've learned since the
1940's.
> >If you
> >separate two charges they "are" a harmonic oscillator too, but the force
> >decreases with the distance between them.
>
> On a micro and macro scale you are correct but on a nanoscale you are
wrong. As
> I have already explained as the e-p's move closer together they overlap
their
> own space and state neutralising themselves.
No they either annihilate each other and become photons, or they become
positronium, which is a short-lived bound state of an electron and a
positron.
.............snip
> >The charge does not change. It's
> >either zero or e, it is not fractional. It has never been observed to be
> >fractional. Your idea in this regard is seriously flawed.
>
> No fractional charge observed eh! - well somebody got a Nobel under false
> pretences then:
>
> http://www.nobel.se/physics/laureates/1998/press.html
> http://www.warwick.ac.uk/~phsbm/qhe.htm
> http://dept.physics.upenn.edu/~mele/qcmt/p3/project3.html
>
> I think we can assume that they are seeing effects due to fractional
charge -
> note its in the vacuum!
The Fractional Quantum Hall Effect is observed in materials subjected to
very large magnetic fields. It is the "interaction of many correlated
electrons" in the material and the strong magnetic field that create
quasi-particles with fractional charges. It does not show the existence of
e-p pairs with fractional charge. This in no way implies that e-p pairs have
fractional charges or that EM radiation or vacuum consist of e-p pars. You
are implying something which is not there, taking it out of context to
support your theory. It is apparently predictable from our existing
theories, so your idea of fractionally charged e-p pairs in the vacuum are
not required and therefore do not have any consequences on our understanding
of reality.
Best Regards,
Todd Desiato
FrediFizzx
|
| "Monitek" <mon...@aol.com> wrote in message
| news:20020825125945...@mb-cv.aol.com...
| > >From: "Todd Desiato" todd...@serversanddomains.com
| > >Date: 23/08/02 20:24 GMT Daylight Time
| > >Message-id: <_Zv99.5314$ja.8...@twister.socal.rr.com>
| > >
| > >
| > >"Monitek" <mon...@aol.com> wrote in message
| > >news:20020823044513...@mb-fh.aol.com...
| > >If you
| > >separate two charges they "are" a harmonic oscillator too, but the
force
| > >decreases with the distance between them.
| >
| > On a micro and macro scale you are correct but on a nanoscale you are
| wrong. As
| > I have already explained as the e-p's move closer together they overlap
| their
| > own space and state neutralising themselves.
|
| No they either annihilate each other and become photons, or they become
| positronium, which is a short-lived bound state of an electron and a
| positron.
Todd, I know this runs contrary to QFT but the concept is not without
precedence. How do we actually know that the e+e- pair is gone forever in
the annihilation and not just disappeared and undetectable because its
combining together is totally neutral now (including mass cancellation?)?
IOW, the gamma rays that are released are just the binding energy. And it
takes that same amount of energy to separate them again. Much like how
protons and neutrons are bound together in the nucleus.
FrediFizzx
Todd Desiato
"FrediFizzx" <fredifi...@ahahhotmail.com> wrote in message
news:6xca9.554$Pw3.44...@newssvr21.news.prodigy.com...
In physics for example indistinguishable particles are the same particle. If
you cannot detect the e-p pair then there is no e-p pair. Things which have
no observable effects make no difference to reality and can be ignored. This
comes from the fact that a hypothesis that cannot be proven OR disprove is
irrelevant because it has no observable effects. You must be able to observe
something to prove or disprove it. When you observer an e-p pair or when
you observe photons there is no doubt that is what you observed. But to say
there is some state between these 2 states that is unobservable is
irrelevant because "by this definition" it is not observable, therefore it
has no effect on anything real.
Regards,
Todd Desiato
Edward Medina
> go ahead and describe it that way if you wish but you are wasting your time.
> C and L are measurements of properties from physical devices.
I love physics, however I am not an expert, the most I have is College
level Physics II and a love for particle physics, and string theory,
etc. So please, please keep this in mind with my question.
If space is a vacum and it has no properties, it should essentially
contain nothing. However, space does contain all kinds of things. In
my world space is just like our atmosphere except diluted like a
bucket of paint in the ocean.
Yeah there is paint out there but to find it you would be very hard
pressed. similarly elements, dust, and celestial objects can be found
in space. Finding one atom of Helium floating in space would be very
hard, however they are out there.
If space is "empty" and has no properties, how does light travel?
You say that electromagnetic waves (which light is part of) are
measured properties of real particles (ie photons?).
If space is empty how does light travel then, how can
we feel the radiation of heat, how can the radio signal travel in
space, etc.
I don't have any answer only questions.
Eddy
FrediFizzx
news:f581d85d.02082...@posting.google.com...
Space is not empty and far from it. For starters, there is EM radiation
everywhere througout space. Next space has an impedance of 377 ohms. And
it can also be polarized. How does light travel? Well, you could say the
same way as a baseball travels or the moon travels around the earth if you
believe that photons are like "hard" particles. If you think of photons as
wave-like oscillations then the problem becomes more difficult. Waves
require a medium. However the photon is unique in that it creates it own
medium. I believe it is a combination of its EM properties and the
inductance-capacitance properties of space that allow it to travel at c.
FrediFizzx
Edward Medina
> wave-like oscillations then the problem becomes more difficult. Waves
> require a medium. However the photon is unique in that it creates it own
> medium. I believe it is a combination of its EM properties and the
> inductance-capacitance properties of space that allow it to travel at c.
>
> FrediFizzx
I guess that's my problem.
Eddy
reticher
It is the currently accepted "truth" that there is no experimental evidence of
the Aether. This seems surprising because, if the Aether made its presence any
more obvious, physicists would have toothmarks on their butts from where the
Aether had jumped up and bit them. A few examples:
Empty space, other than occupying a volume, has at least two observable
properties, its dielectric constant and its permeability. If space is "empty",
what is it that has these properties? Since, if space consisted of a volume of
"nothingness", only a fool would assert that it can have any properties other
than volume. Since it does, space it would seem that it does contain a rejected
entity, which was called the Aether in the 19th Century!.
The velocity of light in free space is determined by the dielectric constant
and the permeabiluty of space in accordance with accepted physical laws.
Similarly, the velocity of sound in a steel rod is determined, using an
equation analogous to that used to determine the velocity of light, by the
elasticity and density of the steel. While a modern physicist seems to assert
that the existence of the dielectric constant and the permeability of space
does not require the existence of a medium (i.e.- the Aether) having these
properties, it seems certain that he would not be foolish enough to assert
that, since the velocity of sound in a steel rod is determined only by its
elasticity and its density, he could remove the rod and retain the density and
elasiticity so that the sound could propagate. One might reasonably wonder if
something is lacking in the mentality of physicists who would tolerate such a
double intellectual standard.
Special Relativity (and the Lorentz Contraction Aether Theory) tell us that
the velocity of light is independent of the velocity of its source. Such an
independence is characteristic of a wavelike disturbance propagating thorugh a
medium, it is not characteristic of an entity propagating ballistically through
"empty" space. For example, the sound of a rifle shot travels at the velocity
of sound in air and its velocity is independent of the velocity of the rifle.
The velocity of the bullet from the rifle, however, is additively determined by
the velocity of the rifle and the velocity at which the bullet is projected. If
we abolish the Aether the observed independence of the velocity of light from
the velocity of its source would seem to require that some form of "magic" must
be at work since light would then have no way to determine how fast it is
supposed to travel. Since the writer finds the idea of "magic" accounting for
observations in the science of physics extremely hard to accept, he would be
gratetful if someone could provide some other explanation which does not
require a medium (the Aether). (P.S.- Handwaving arguments about spacetime
don't serve the purpose.)
If one examines Dirac"s treatment of the creation of complementary particles
from high energy photons, he finds that Dirac had to invent a "sea of negative
energy pervading all of space" in order to make his mathematics work. Aside
from the fact that "negative energy", in the sense that Dirac used it, would
seem not only to be unexplained but to be impossible, how does that postulated
"sea of negative energy" differ from the Aether? Both say that "something"
exists in a space which had been arbitrarily assumed to be empty!
Current theory asserts that forces which "act at a distance" do so as a result
of "virtual particles" (e.g.- virtual photons, gravitons, gluons, etc.) which
pop into and out of existence so rapidly that quantum uncertainties prevent the
Law of Conservation of Energy from being violated. There would seem to be at
least two problems with this concept:-
The first is the problem is the number of particles involved. At any instant
of time, every particle in the Univserse gravitationally attracts.every other
particle and they must therefore be exchanging gravitons. Furthermore, these
gravitons, since they are virtual, must be extremely short lived. The number of
gravitons involved in this concept is staggering. At any instant for example,
assuming that the presently accepted value of 10^80 particles in the Universe
is reasonably correct, there must be at least 0.5*(10^160) particles existing
at one time. The size of this number is increased by the fact that they must be
short lived and therefore must be replaced continuously (gravity appears to be
continuous), suggests that a bit of skepticism would be in order.
The second problem is that of expaining how the exchange of "virtual
particles" accomplishes the production of an "attractive force". The production
of a repulsive force by the exchange of particles is obvious. Two athletes
throwing a medicine ball back and forth experience a repulsive force because of
momentum exchange between them. There does not seem to have ever been a viable
explanation as to how these particles act to produce an attractive force. The
momentum exchange requirement to produce an attractive force prohibits the
production of such a force unless there is a substrate (e.g.:- the Aether) with
which they can exchange momentum. A boomerang returns to the thrower because it
exchanges momentum with such a medium (air). If space were empty (no Aether),
such a momentum exchange cannot occur and the virtual particles could only
produce a repulsive force. The writer is aware that the mathematics associated
with the concept of "virtual partcles" allows for the production of attractive
forces. He is also aware of the fact that there are many mathematically treated
problems in which the mathematics deals with conditions outside of other
constraints related to the problem and which limit the applicability of the
mathematics. This would seem to be such a case. Allow the Aether or its
equivalent to be present as a substrate and the mathemetics of "virtual
photons" works.
There are two interpretations of Quantum Theory. One interpretation involves
parallel Universes coexisting in the same space as our own that are created
every time that a particle makes a "quantum choice". This approach not only
suffers the problem of absurdly large numbers described for gravitons, it
requires that an amount of energy equal to the energy content of our entire
Universe multipled by the number of particles in the Universe be created at
every instant of time. It also requires that the volume of space occupied by
each particle contain an amount of energy equal to the energy equivalent of
that enormous amount of energy. Since the writer accepts the concepts that
energy is conserved and two entities cannot occupy the same space at the same
time, he considers the interpretation to be rather foolish.
The other interpretation of Quantum Theory requires that quantum effects
propagate at an "infinite" velocity. Most of the acedemic hierarchy seems to
consider this interpretation to be both "spooky" and wrong since "Special
Relativity clearly shows that nothing can travel faster than light". To the
contrary, Special Relativity shows the opposite. What it shows is that an
entity represented by energy cannot travel faster than the velocity of light
and the scientific community has fallen into the trap of assuming that
everything which is observable is represented by some form of energy. The
reason that the Special Theory of Relativity imposes its restrictions on
velocity is that the Lorentz Transformation for energy, in a force-length-time
system of units, is 1/(1-V^2/C^2)^0.5. This means that, at the velocity of
light, energy becomes infinite and that, at velocities above that of light,
energy becomes imaginary. The effect is a limititation on the velocity of
propagation of energy. As a result, any form of communication which involves
the endoding of information in the form of energy is limited to the velocity of
light.
Experiments have shown that the polarization of "paired photons" is coupled so
that changing the polarization direction (the quantum number) of one causes the
polarization direction of the other to change in correspondence. In addition to
demonstrating this polarization coupling , experiments performed in the '80s
have also shown that the polarization coupling propagates at at least 4 times
the velocity of light and perhaps at an infinite velocity. (It should be noted
that changing the polarization direction of a photon does not change its energy
content so the the coupling of polarization direction between them does not
involve energy transfer).
The polarizarion direction of photon is an angle which is measured in radians
(the distance along an arc divide by the radius of that arc). The Lorentz
Transformation for Angle is therefore equal to the Lorentz Transformation for
Length divided by the Lorentz Transformation fo Length and is therefore equal
to unity and is independent of velocity. (Above and below a velocity of C, the
value of unity is obvious. At the velocity of C, the value is nominally
indeterminate but can be shown to be equal to the value of unity by borrowing a
technique from Calculus). With a value of unity for the Lorentz Transformation
for Angle, it seems reasonable to assert that Special Relativity requires the
polarization couplingof paired photons to travel at the infinite velocity
required by the second, more rational interpretation of Quantum Theory.
Assuming that this is the case, then the current concept of "spacetime" as a
single entity must be in error. An absolute time and velocity reference is
required.
If one examines both the Special Theory of Relativity (STR)and the
Lorentz-Contraction Aether Theory (LCAT), he finds that they are actually the
same theory in that they are cross derivable. (LCAT) is actually a special case
solution of (STR) and cannot be disproven without disproving (STR) as well.
LCAT asserts that the Aether exists but because information cannot propagte
faster than light (quantum theory tells us that this is not true), we cannot
find our velocity with respect to it. STR tells us that because our velocity
with respect to the Aether cannot be measured, it does not exist in the theory!
It does not preclude its existence nor its effect from being observed by other
means. Dr. Einstein asserted "remember gentlemen, we have not disproven the
existence of the Aether, we have only proven that we do not need it (for
computations)".
An objection to the existence of the Aether has been raised. This objection is
that if the Aether exists, "absolute time" would also have to exist. The idea
of absolute time has, however, been abolished in physical theory.
Unfortunately, its rejection did not result from physical observation or from
logical deductions based on previously proven facts. It is based solely upon a
concensus viewpoint. In other words, that rejection is based upon an unproven
(and probably unprovable opinion) rather than upon fact. The motivation for
this opinion would seem to be an overwhelming urge on the part of physicists to
consider space and time to be aspects of the same structure, spacetime, as
viewed from different reference frames so as to make their mathematical
treatment "elegsnt". Nature doesn't care about what we would like to be true
and there is no objectively valid reason to accept the viewpoint.
It is claimed that the concept of spacetime rather than space and time as
separate entities yields a simpler solution and, according to the principle of
Ockham's Razor, must therefore be the correct one. It is true, mathematical
solutions employing the spacetime concept are much simpler than solutions
involving the Aether, but the solutions are not simpler with respect to
reality. As Dr. Einstein, who maintained a belief in "absolute time" for 25
years after the publication of Special Relativity exclaimed, the "concept of
spacetime requires an infinite number of Aethers" as indeed it does. Once one
realizes this, Ockham's Razor would lead any resonable man to the Aether and
not to the concept of "spacetime" as a single entity
Unfortunately, Modern Physics seems to have been taken over by mathematical
idiot savants who lack an appreciation for the implications of mechanism and
strive to suppress the contributions of those who would dare to say "Hey, wait
a minute"
http://www.members.aol.com/reticher/site.htm
Please make and/or back up any response with an E-mail as Newsgroups are not
monitored on a regular basis. Objective responses will be treated with the same
courtesy as they are presented. To prevent the wastage of time on both of our
parts, please do not raise objections that are not related to material that you
have read at the Website. This posting is merely a summary.
E-Mail reti...@aol.com
The material at the Website has been posted continuously for over 5 years. In
that time THERE HAVE BEEN NO OBJECTIVE REBUTTALS OF ANY OF THE MATERIAL
PRESENTED. There have only been hand waving arguments by individuals who have
mindlessly accepted the prevailing wisdom without questioning it. If anyone
provides a significant rebuttal that cannot be objectively answered, the
material at the Website will be withdrawn.
reticher
reticher
Harry Potter book that from any legitimate form of science.
Monitek Arden Barker
> The idea that the photon creates its own medium sounds more like a qute from a
> Harry Potter book that from any legitimate form of science.
Well you want an aether and when someone tells you what it is you dont
like it. The "aether" consists as a sea of combined electron positron
pairs as originally postulated by Dirac (not the bastardised version
that was published when it was decided positrons did not exist){Oh why
dont people stand their ground?} The e-p pairs can not be measured in
their combined state. In the presence of charge e-p pairs in the Dirac
Sea separate slightly(Topaz & Muon g-2). In the separated state they
have the ability to act as a conductor,and they have charge and
mass(small).
As a conductor they are moving charge, the very act of separation is
charge moving, moving charge creates a magnetic field which can via
mutual induction create an adjacent current in the opposite direction.
Thus you have the basis for the electromagnetic cycle in the "aether"
which oscilates back to zilch after the wave passes.
After the wave has passed there is no physical "aether" in the vacuum,
whilst the wave is passing the "aether has physical properties. The
emr radiates as an annulus of alternating charge and magnetic field.
Whilst the e-p's are in a state of separation they are locked in a
rigid lattice in the form of a ring whose radius is equal to the
distance from the source. Thus a light wave is created from a medium
which creates itself (from our physical point of view)as it progresses
through space.
There is no photon per se we have an oscillating wave in a dirac e-p
sea, the e-p's no not move at the speed of light they oscillate
sideways to create current flow required to induce the next ring. The
propagation of the magnetic component governs the speed of the emr as
it travels through the vacuum.
You state clearly in the endless cut and paste of the same article,
that you consider that there is an Aether. I will quote your best
line:
", if the aether made its presence any more obvious , physisists would
have toothmarks on their butts from where the aether had jumped up and
bit them."
I totally agree with this statement. However, the statement above:
"> The idea that the photon creates its own medium sounds more like a
qute from a
> Harry Potter book that from any legitimate form of science."
does not fit the experimental facts which are exactly that emr creates
its own medium as it travels.
What is your point? What do YOU think the aether is?
Regards,
Monitek (Arden Barker)
PS I am temporarily in Spain and have managed to get online whist I
amd here.
Monitek Arden Barker
But e-p pairs do make observable effects-Lamb Shift, Charge of
electron (topaz), Muon g-2 expt, slac e144(or do you realy think you
can make matter out of nothing?) are observable effects how much more
evidence do you want?
>This
> comes from the fact that a hypothesis that cannot be proven OR disprove is
> irrelevant because it has no observable effects. You must be able to observe
> something to prove or disprove it. When you observer an e-p pair or when
> you observe photons there is no doubt that is what you observed. But to say
> there is some state between these 2 states that is unobservable is
> irrelevant because "by this definition" it is not observable, therefore it
> has no effect on anything real.
>
> Regards,
> Todd Desiato
No there is a state of existence for e-p pairs in between total
annihilation and and total separation which is observed when in close
proximity to a charged particle. My view that e-p pairs are the
transmission medium of EMR would also add to the list of observable
effects.
Regards,
Monitek (Arden Barker)
Monitek Arden Barker
Monitek:
This is where I started with a post in this group- I agree Yes the EMR
makes its own medium we are not in conflict on this point.
see: message ID 20020818045607...@mb-fh.aol.com
Yes I agree the vacuum is an em field and it can not be removed, the
only extra point I make is that the “EM Field” is one of
e-p pairs. How do out say EMR creates its own medium? What is the QED
explanation for creating its own medium?
> > QED tells me what EMR does it does not say what it is. I would like to
> know
> > what it is.
>
> A free EM field is a density of energy comprised of frequency modes, which
> are dependent on the surrounding geometry and materials, which can be
> polarized, and which carry energy from point A to point B in discrete quanta
> called "photons". I still don't understand why you believe in electrons but
> not photons. Waves are particles. I suppose you don't agree with
> wave-particle duality either?
>
This took me ages to figure out, why do you speak in riddles? First
mistake is the an EM field which is dependent upon surrounding
materials can in no way be described as free . You are mixing chalk
and cheese, here you are referring to black body radiation which
behaves as though it consists of discreet packages of whole number
wavelengths and it does. Now the interesting thing which I have
observed regularly is that solids emit black body radiation, liquids
emit black body radiation but gasses don’t. Solids are
crystalline and have long range order, liquids are fluid and have
short range order but gasses have no ordering at all. My suggestion is
that the former two , namely liquid and solid emit light because the
atoms are oscillating in unison.
Whereas the atoms in a gaseous state emit black body radiation
randomly and the radiation can not be seen due to cancellation i.e.
phase difference. The elimination of phase difference in the solid
state emission of black body radiation sounds like the action of a
laser, i.e. the light waves reflect from the sides of the constraining
body and create a standing wave with a node at each opposite
reflecting surface, which by definition must be a whole number of
wavelengths. This leads to the conclusion that the only way light can
travel inside a body is in such a way as to avoid cancellation. In the
gaseous state there are no reflecting surfaces therefore no laser
action, therefore no black body radiation.
Basically in this mode the EMR is filtered to obtain the frequency
modes and I suppose this does not bode well for the EMR being created
in discrete quanta, only certain discrete quanta remain to be
observed.
What does the energy of which you speak consist?
He discrete quanta of your “photon” is due to the
constraints of the system creating the EMR in this case (black body
radiation). The EMR created by discrete jumps of electrons creates a
characteristic radiation not black body radiation.
Waves are particles??
Waves are oscillations of particles. Charged particles create waves
when they move. This is not the same thing. Electrons can be isolated
and have their charge measured, and their mass measured. “
photons “ only travel at the speed of light and can not change
their velocity, have not been “weighed” . “Photons
have particulate properties but these can be explained by the e-p
vacuum argument.
> > When it was shown that light had particulate properties the photon
> > was postulated to satisfy these requirements.
>
> No, the photon was postulated by Max Planck when attempting to explain
> "black body radiation" and the difference between the predictions of the
> spectral energy density by the "Rayleigh-Jeans distribution" at low
> frequencies and the "Wien distribution" at high frequencies. The result was
> that radiation must transport energy in discrete quanta such that the energy
> E=nhf. where n is the number of photons at the frequency f. That is why we
> call _h_ "Planck's constant".
>
Where n= a whole number wavelengths in the cavity from which the
radiation
is emanating, I think you will find. It has nothing to do with the
number of photons.
You are describing thermal radiation which is created at all
frequencies randomly. The E=nhf arises from a filtering process which
I have already discussed. The blackbody radiation is filtered to
contain discreet quanta and therefore one can not say that the EMR is
created in discrete quanta by thermal means, if this is the foundation
of the quantum theory then it has just been blown.
> Then Einstein confirmed this when he showed the photoelectric effect also
> indicated that the energy of light was absorbed in discrete quanta of energy
> E=nhf.
>
> It was further confirmed by the double-slit experiment using film and short
> exposures, for both electrons and photons the results are the same! For very
> short exposures there is no interference pattern of waves, only dots where
> individual quanta struck the film.
>
> So unless you can reproduce the works of Planck and Einstein without the
> need for E=nhf, and disprove the double-slit experiment then you must agree
> that photons exist.
>
The double slit experiment merely confirms that electrons and EMR have
wave
Properties. Have you any references to a physical experiment
confirming that single electrons create interference patterns? The
double slit experiment using electrons merely confirms that moving
charged particles create waves it was this effect that defeated
Newton’s corpuscular theory of light which is of course another
word for photon. Just because an electron has been shown to have
particulate properties one can not assume that EMR also has
particulate properties because it seems to behave like an electron.
EMR has no mass therefore there is no particle in motion. There is
also a large speed difference between the phenomena. You also have to
postulate the means of existence of a massless particle, I have
already done that in a way which explains the inductive, electrostatic
and massless properties of EMR, and how the particulate properties of
EMR arise.
> > There is no particle which leaves
> > the atom at the speed of light when an electron falls down shells to a
> lower
> > energy level. There are no particles emitted by an electron when it slows
> > sharply. The mistake that you are making is that the particles responsible
> for
> > the movement of a light wave have no movement in the direction of the EMR
> wave
> > only a sideways movement which I have described previously, they are not
> > travelling at the speed of light. These movements will be at non
> relativistic
> > speed therefore EMR is classical.
>
> You are working with 19th century physics in the 21st century. As I said
> before you need to learn and understand all the related physics that has
> been discovered in the last 120 years before you start making up your own
> theories. Seriously, crack open a book and learn! Become an expert, then
> start theorizing your own ideas. At this stage you don't even understand the
> history let alone the math and experimental evidence that supports modern
> physics. You are deluding yourself to think you understand physics better
> than the giants that came before you.
>
A good salesman can sell anything its only later that you find its
crap, usually when the salesman is long gone. I have read enough to
know that something is amiss with current theories. If you think I am
wrong or my ideas don’t hold water then fight me with facts.
> >
> > >> >However EM waves are not observed to behave as you describe.
> > >>
> > >> You will have to describe the differences to which you refere here.
> > >
> > >There is no charge and no mass in a free EM field in vacuum. Not even
> > >"partial" charges or masses. If their were the equation;
> > >
> > >E^2 - p^2 = 0
 >
Therefore E=p therefore as p=mv^2 we obtain E=mv^2 c.f. E=mc^2 so
what’s new?
However there is mass hidden in your equation. The energy of a wave is
proportional to frequency, insert a constant and you have an equality
E =hf. The equating of mc^2 and hf seems to me to be very suspect, as
you are mixing chalk with cheese. How do you justify this equation?
> > Charge squared - momentum squared = 0 ?
> > I could do with you defining the terms of your equations. Please.
>
> E is Energy, p is momentum. This is the magnitude of the relativistic
> momentum 4-vector. There is a well known equation for the relativistic
> energy of a particle,
>
> E = sqrt[(mc^2)^2 + (pc)^2]
>
> This comes from the momentum 4-vector p^u = (E,p1,p2,p3) where c=1 so we
> neglect it. The magnitude of this 4-vector is;
>
> p_u*p^u = E^2 - p^2 = m^2
>
Very good I can see you understand pythagoras.
> where m is the invariant mass of the particle in it's rest frame.
> For photons m=0, for electrons and positrons m=/=0.
>
>
> > >
> > >would not hold and that would have obvious consequences to the gauge
> > >invariance and Lorentz invariance of EM fields.
> > >
> >
> > I agree that there is no mass , however I would say that there is charge
> > involved and this charge is measurable via a capacitor- thats how radio
> waves
> > are received. I am not particularly bothered about the consequences.
>
> No, if say you had 2 parallel metal plates with a small space between them,
> and then you pass an EM wave between them propagating parallel to the plates
> with the electric field polarized perpendicular to the plates, then the
> electric field will polarize the plates at it passes but all the charge
> presnet on the plates comes from the plates not from the wave. The electric
> field exerts a force on the charges in the metal to polarize the metal
> plates and store energy in the capacitor. Also there are no positrons in a
> capacitor, the positive charge is due to a lack of electrons to balance the
> nuclear charges of the metal atoms.
>
The charge on the plates comes from the EMR wave period. The electrons
in the polarised Dirac sea cause the electrons in one plate to depart
the plate, like charges repel. The positrons in the Dirac sea cause
the electrons so departed to be attracted to the other plate. As EMR
is an oscillating field the situation reversed at the next half cycle
so we get oscillating charge in the capacitor. No there are no free
positrons in the capacitor plates, the positrons are in the Dirac sea.
> .............snip
> > >
> > >The "underlying mechanics" of a free EM field in a vacuum have been well
> > >established. The field consists of photons, which is just a neat word for
> > >saying the field has many possible frequency modes that carry energy in
> > >discrete quanta proportional to the frequency.
> > >
> > >> >An e-p pair and a photon do not behave the same way, no matter how you
> > >> >imagine them to be.
> > >>
> > >> Again you will have to describe the differences you claim to get my
> agreement
> > >> on that.
> > >
> > >with c=1
> > >E^2 - p^2 = m^2 for an electron or positron and E^2 = 4m^2 for an e-p
> pair
> > >to form. You do not have partial charge or mass as a solution to either
> the
> > >Maxwell or Dirac fields.
> > >
> >
> > Thats because nobody has considered it when the equation was derived.
>
> Nobody considered it because there was no evidence to support it!
>
Yes but now there is evidence and that changes things somewhat.
> > Basically
> > when e-ps are closer than 0.11 fm the above equation does not apply.
>
> At very short distances on the order of the Compton wavelength, the
> "zitterbewegung" of an electron is precisely due to electron-positron
> "exchange scattering" within that tiny space. However this does not in any
> way imply that EM waves carry charge or consist of e-p pairs, or that the
> vacuum medium consist of e-p pairs either. It only shows that where the
> electric field is very strong the energy density is large enough to create
> e-p pairs.
>
Yes but it clearly shows that the vacuum consists of e-p pairs and
nothing else. If you have e-p pairs they must be disturbed by the
electric portion of the EMR wave. The only conclusion is that the wave
effect of EMR is a ripple in the e-p vacuum.
> > Also the
> > above is a wave equation for a particle momentum = mass x velocity - no
> mass
> > therefore no momentum therefore no particle.
>
> Photons have no mass, but they can easily be shown to carry momentum p=hc/f.
> Just get yourself a cheap glass Radiometer to prove it. The blades spin
> because more momentum is absorbed by the black side of the blades than by
> the shinny side. Nothing penatrates the glass but light and it is obvious
> that the speed is proportional to the intensity of the light.
>
>
The speed is proportional to the intensity of the light for sure but
the motion has nothing to do with the “Momentum of the light
particle”. It also works better for infrared than ultra violet
and it doesn’t work with laser light.
I have a radiometer, plasma ball , van der Graff generator,
Newton’s cradle, a laser, a 15kw RF generator, and an 8”
Newtonian reflector to play with.
Well its time for you to learn some physics:
http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html
> >
> > >Also, photons have a velocity = c, an e-p pair will have velocity < c
> > >because they have mass, and the energy is not infinite.
> > >
> > >Electrons and positron are Fermions, photons are Bosons, they obey a
> > >different type of statistics.
> > >
> >
> > As you say elsewhere photons are a mathematical construct, waves in a
> medium is
> > the reality - the medium is e-p pairs in an annihilated state in the
> vacuum.
>
> Waves are also comprised of particles. e-p pairs in an "annihilated state"
> are photons and do not carry any charge. Again you either have an e-p pair
> or you don't. There are no "partial" solutions.
>
You say waves are also comprised of particles – yes I agree and
have shown how.
e-p pairs in an annihilated state are photons. E-p pairs in the
process of annihilation emit photons the question is do the e-p pairs
accelerate towards one another and create the “photons”
before annihilation or are the photons created after the annihilation.
I happen to believe that accelerating charge creates the
“Photons” (0.511 kev gamma), after annihilation they are
neutral particles and loose their mass and charge and are off the
frame for our physics other than minuscule effects.
As the e-p pairs approach there must be a zone of space where the two
particles overlap and partially cancel themselves. The cloud of
virtual particles surrounding a charge is one of partially separated
e-p pairs, otherwise you would not have a single electron they would
appear in groups.
 .................snip
So what’s happened has the definition changed?
Physicists postulate the missing antimatter, it is well accepted that
the positrons associated with the electrons are missing. I say they
are not missing, the positive charge on the proton is due to a
positron in its structure yes that means that “antimatter”
is part of normal matter. If you follow the annihilation path of
positive and negative particles you get charge conservation. The
positive particle has a final state of a positron and the negative
final state finishes as an electron two then annihilate and disappear.
That goes for protons too. What do you think the QCD explanation is
for 3 quarks turning into a positron- do bear in mind that the
positron is considered a fundamental particle and is not divisible.
In my view quarks are a mathematical shorthand to simplify the
mathematics. Experimental evidence for their existence is tenuous. My
view is that all the particles in the particle zoo are BuckyBalls of
positronium with either a positron, electron or both in the buckyball
to give the charge. The reaction of beta plus is a special case of
pair production where the electron is trapped inside the proton and
the positron escapes to annihilate with a valence electron. The proton
becomes a neutron and the valance electron disappears.
The proton is postulated to be 3 quarks and a bunch of gluons it is
not written in tablets of stone.
>
> > >If you
> > >separate two charges they "are" a harmonic oscillator too, but the force
> > >decreases with the distance between them.
> >
> > On a micro and macro scale you are correct but on a nanoscale you are
> wrong. As
> > I have already explained as the e-p's move closer together they overlap
> their
> > own space and state neutralising themselves.
>
> No they either annihilate each other and become photons, or they become
> positronium, which is a short-lived bound state of an electron and a
> positron.
>
> .............snip
> > >The charge does not change. It's
> > >either zero or e, it is not fractional. It has never been observed to be
> > >fractional. Your idea in this regard is seriously flawed.
> >
> > No fractional charge observed eh! - well somebody got a Nobel under false
> > pretences then:
> >
> > http://www.nobel.se/physics/laureates/1998/press.html
> > http://www.warwick.ac.uk/~phsbm/qhe.htm
> > http://dept.physics.upenn.edu/~mele/qcmt/p3/project3.html
> >
> > I think we can assume that they are seeing effects due to fractional
> charge -
> > note its in the vacuum!
>
First you say fractional charges are not observed then you accept the
existence of fractionally charged quarks. You say fractional charges
are not observed then you say they are but only in special
circumstances. Well either fractional charge is observed or it
isn’t. If it is observed then it is a physical fact requiring
explanation. All charged particles are surrounded but a cloud of
partially charged e-p pairs see topaz expt. If this is a physical fact
the one can use their properties to explain other effects. The
“quantum cloud” round all charged particles creates the
electrostatic field of polarised vacuum which forms a rigid lattice to
create electrostatic action at a distance. This cloud is also proof
positive of the existence of e-p pairs in the vacuum and that they are
all pervading.
> The Fractional Quantum Hall Effect is observed in materials subjected to
> very large magnetic fields. It is the "interaction of many correlated
> electrons" in the material and the strong magnetic field that create
> quasi-particles with fractional charges. It does not show the existence of
> e-p pairs with fractional charge. This in no way implies that e-p pairs have
> fractional charges or that EM radiation or vacuum consist of e-p pairs. You
> are implying something which is not there, taking it out of context to
> support your theory. It is apparently predictable from our existing
> theories, so your idea of fractionally charged e-p pairs in the vacuum are
> not required and therefore do not have any consequences on our understanding
> of reality.
>
> Best Regards,
> Todd Desiato
Well it sure ties up a whole load of loose ends.
Regards,
Monitek (Arden Barker)
John C. Polasek
Barker) wrote:
>"Todd Desiato" <todd...@serversanddomains.com> wrote in message news:<ZGba9.31185$_7.29...@twister.socal.rr.com>...
>> "Monitek" <mon...@aol.com> wrote in message
>> news:20020825125945...@mb-cv.aol.com...
SNIP
>> > >From: "Todd Desiato" todd...@serversanddomains.com
>> > >> >An e-p pair and a photon do not behave the same way, no matter how you
>> > >> >imagine them to be.
>> > >>
>> > >> Again you will have to describe the differences you claim to get my
>> agreement
>> > >> on that.
>> > >
>> > >with c=1
>> > >E^2 - p^2 = m^2 for an electron or positron and E^2 = 4m^2 for an e-p
>> pair
>> > >to form. You do not have partial charge or mass as a solution to either
>> the
>> > >Maxwell or Dirac fields.
>> > >
>> >
>> > Thats because nobody has considered it when the equation was derived.
>>
>> Nobody considered it because there was no evidence to support it!
>>
>
>Yes but now there is evidence and that changes things somewhat.
>
>
>> > Basically
>> > when e-ps are closer than 0.11 fm the above equation does not apply.
jp:
I'm curious? Where did this number come from?
>> At very short distances on the order of the Compton wavelength,
jp:
Short, how can the CWL be short, when it's 22,000 times the .11 fm?
the
SNIP
>
>Well it sure ties up a whole load of loose ends.
>
>Regards,
>
>Monitek (Arden Barker)
John C. Polasek
Monitek Arden Barker
Monitek:
This figure 0.11 fM arises from my buckyball model of the proton.
It is assumed the proton and neutron bonding involves a hexagonal zone
of 6 partially annihilated leptons on the ball surface of each
particle giving a bond of 12 e-p pairs. The deuterium bond energy is
known and I make the assumption that the e-p pairs in the BuckyBall
annihlate to zero to give the observed mass loss of neutron proton
bonding. For simplicity I assumed that the points of the buckyball
were evenly spaced and were on a rectangular lattice on the surface
instead of the hexagonal pattern. Knowing the binding energy was equal
to 12 pairs at some reduced charge and mass I could calculate the
residual mass of the pairs. Knowing this I could calculate the number
of pairs in the buckyball the mass of the proton being known.
Having the separation distance and the reduced charge and mass one
could then plot this on a graph. The graph has two data points the
first being when the pairs annihilate separation distance is zero and
charge/mass is zero and the second is the separation distance at
reduced charge. By assuming the effect of mass and separation distance
is linear one can draw a line through the two points and exprapolate
the separation distance when the charge equals unity ie when the e-p
is fully separated and each one is not influencing its partner.
The full details and graph are in my notes on:
http://members.aol.com/medusa1989/theatom.html
Regards,
Monitek (Arden Barker)
Todd Desiato
"Monitek Arden Barker" wrote in message
news:831585e9.02090...@posting.google.com...
> "Todd Desiato" wrote in message
news:<ZGba9.31185$_7.29...@twister.socal.rr.com>...
> > "Monitek" wrote in message
> > news:20020825125945...@mb-cv.aol.com...
.............SNIP
> > > The Equations are wave equations they dont concern themselves with the
> > > transport medium.
> >
> > None is required. EM makes it's own medium. The vacuum "IS" an EM field.
It
> > cannot be removed.
> >
>
> Monitek:
> This is where I started with a post in this group- I agree Yes the EMR
> makes its own medium we are not in conflict on this point.
> see: message ID 20020818045607...@mb-fh.aol.com
>
> Yes I agree the vacuum is an em field and it can not be removed, the
> only extra point I make is that the “EM Field” is one of
> e-p pairs. How do out say EMR creates its own medium? What is the QED
> explanation for creating its own medium?
Photons! The EM field "alone" has no e-p pairs. It is strictly Boson type
particles. You must include the Dirac field to get electrons and positrons
which are the Fermions. You will find that I have changed my standing on our
argument too. I now concur that there is enough energy density in the
zero-point EM field in the range from mc^2 to 2mc^2 to have pair creation
and annihilation on a regular basis. I also learned about the idea of being
"off mass shell", meaning that the particles do not obey the equations of
motion and are in a sense tightly bound to each other. I also learned about
"Positronium" and it's many excitations of bound states.
Also, Dr. X (Jack Sarfatti) proposed that the Fermion zero-point energy
density is required to offset the Boson zero-point energy density to give us
the "flat" cosmological constant we see today. I also found out that just
like the zero-point EM field, there is also a zero-point Dirac field of
electrons and positrons. So I have no argument with you now. I think we are
in agreement.
>
>
> > > QED tells me what EMR does it does not say what it is. I would like to
> > know
> > > what it is.
> >
> > A free EM field is a density of energy comprised of frequency modes,
which
> > are dependent on the surrounding geometry and materials, which can be
> > polarized, and which carry energy from point A to point B in discrete
quanta
> > called "photons". I still don't understand why you believe in electrons
but
> > not photons. Waves are particles. I suppose you don't agree with
> > wave-particle duality either?
> >
>
> This took me ages to figure out, why do you speak in riddles? First
> mistake is the an EM field which is dependent upon surrounding
> materials can in no way be described as free . You are mixing chalk
> and cheese, here you are referring to black body radiation which
> behaves as though it consists of discreet packages of whole number
> wavelengths and it does. Now the interesting thing which I have
> observed regularly is that solids emit black body radiation, liquids
> emit black body radiation but gasses don’t. Solids are
> crystalline and have long range order, liquids are fluid and have
> short range order but gasses have no ordering at all. My suggestion is
> that the former two , namely liquid and solid emit light because the
> atoms are oscillating in unison.
I agree, this is consistent with modeling matter as a group of harmonic
oscillators, as Planck and Einstein did.
> Basically in this mode the EMR is filtered to obtain the frequency
> modes and I suppose this does not bode well for the EMR being created
> in discrete quanta, only certain discrete quanta remain to be
> observed.
>
> What does the energy of which you speak consist?
>
> He discrete quanta of your “photon” is due to the
> constraints of the system creating the EMR in this case (black body
> radiation). The EMR created by discrete jumps of electrons creates a
> characteristic radiation not black body radiation.
This is true in any system. Even in a "free" field, quantizing the field
requires defining the boundary constraints, typically a period boundary
condition is used. However I don't think this is the case for "second
quantization" so I just want to be clear that I am talking about first
quantization and not full blown QED.
> Waves are particles??
>
> Waves are oscillations of particles. Charged particles create waves
> when they move.
How do you know that particles are not just a superposition of waves?
> This is not the same thing. Electrons can be isolated
> and have their charge measured, and their mass measured. “
> photons “ only travel at the speed of light and can not change
> their velocity, have not been “weighed” . “Photons
> have particulate properties but these can be explained by the e-p
> vacuum argument.
That depends on what you refer to as weighing a photon. The wavelength's of
light can be measured quite accurately, and the momentum of each photon is
inversely proportional to the wavelength of the light.
> > > When it was shown that light had particulate properties the photon
> > > was postulated to satisfy these requirements.
> >
> > No, the photon was postulated by Max Planck when attempting to explain
> > "black body radiation" and the difference between the predictions of the
> > spectral energy density by the "Rayleigh-Jeans distribution" at low
> > frequencies and the "Wien distribution" at high frequencies. The result
was
> > that radiation must transport energy in discrete quanta such that the
energy
> > E=nhf. where n is the number of photons at the frequency f. That is why
we
> > call _h_ "Planck's constant".
> >
>
> Where n= a whole number wavelengths in the cavity from which the
> radiation
> is emanating, I think you will find. It has nothing to do with the
> number of photons.
> You are describing thermal radiation which is created at all
> frequencies randomly. The E=nhf arises from a filtering process which
> I have already discussed. The blackbody radiation is filtered to
> contain discreet quanta and therefore one can not say that the EMR is
> created in discrete quanta by thermal means, if this is the foundation
> of the quantum theory then it has just been blown.
I agree that quantization requires that we specify the boundary conditions.
But the same thing applies to an electron. An electron has an electric field
that extends to infinity, so how can you call it a localized particle? The
electron cannot be separated from it's charge and therefore cannot be
removed from the field. You cannot measure the size of an electron. It's
mass is confined to a point, in our best attempts to measure it's size, and
the field extends to infinity. In a sense the mass of the electron is due to
its interaction with the vacuum field it is in, it reacts to it's own field
and to the reflection of it's own field. So the electron has a wavelength
caused by the jitter in it's location due to all these interactions, and is
therefore also dependent on the boundary conditions.
I prefer to think of electrons and photons in the same way. Both are
particles, both are waves, both are the quanta of their respective fields.
Both are energy and this is just a measure of the way they behave. There is
no evidence which falsifies any of this so I choose to keep all of it and
use it when necessary.
...............SNIP
> > So unless you can reproduce the works of Planck and Einstein without the
> > need for E=nhf, and disprove the double-slit experiment then you must
agree
> > that photons exist.
> >
>
> The double slit experiment merely confirms that electrons and EMR have
> wave
> Properties. Have you any references to a physical experiment
> confirming that single electrons create interference patterns?
Yes.
Akira Tonomura, "The Quantum World Unveiled by Electron Waves", pub. 1998,
World Scientific, pgs, 47 - 49.
The results shown here are undeniable! They start with 3 electrons then 4,
then 10 then you can see the interference pattern being generated point by
point. There is no doubt when you look at this. The interference starts to
look like it was done by an Artist doing pointillism on paper. I have also
seen the exact same experiment done with photons in many other text books.
The
> double slit experiment using electrons merely confirms that moving
> charged particles create waves it was this effect that defeated
> Newton’s corpuscular theory of light which is of course another
> word for photon. Just because an electron has been shown to have
> particulate properties one can not assume that EMR also has
> particulate properties because it seems to behave like an electron.
> EMR has no mass therefore there is no particle in motion. There is
> also a large speed difference between the phenomena. You also have to
> postulate the means of existence of a massless particle, I have
> already done that in a way which explains the inductive, electrostatic
> and massless properties of EMR, and how the particulate properties of
> EMR arise.
Your statements are naive and do not falsifiy the data which supports that
photons are detected, are particulate and do exist. The fact that boundary
conditions are required to make the mathematics work does not falsify
anything. The fact is that matter and boundary conditions "do exist" and so
do photons. If they are the result of those boundary conditions it makes no
difference, then are there and can be detected.
Out of time!
Regards,
Todd Desitao
reticher
David Thomson
news:20020817110830...@mb-ct.aol.com...
> Electric and magnetic fields do not epend upon electromagnetic
interaction.
> There are seveal magnets holding papers on my refrigerator which exhibit
an
> external magnetic field but not an electric field. The electric field on
the
> face of my TV set will attract the hair on my arms, but there is not
associated
> with any magnetic field. It is only when the electic and magnetiuc field
> propated as a wave by interchanging energy that they behave in the way
that you
> assume fo electric and magnetic fields.
I have a unified charge equation that supports your statement. There are
two distinct types of charges. There is the elementary charge and the
strong charge. The strong charge IS magnetism.
e^2 = 8 * pi * a * eemax^2
where e is the elementary charge, a is the fine structure constant of the
electron, and eemax is the strong charge of the electron. This past week I
have shown mathematically that strong charge is the actual charge involved
in electromagnetism. This explains why photons can have electromagnetic
properties even though they have no elementary charge.
Dave