but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this
occuring at c), and at the destination we detect the last kicked
photon. Result is the same as if the original photon travelled. One
exception: if there was a box void of photons (but how?), light
couldn't be created in it. Just like sound cannot in space. So,
compton scattering, how does that disprove this? I'm so sorry

---------
Dear wespe:

"wespe" wrote in message
om...
ok.. this just dawned on me, I had to post this, sorry..

In electricity, electrons are not moving very fast, but the electric
current moves at almost speed of light, because that's a wave. So, why
not say, aether is a photon gas, photons are all around moving
randomly with small speeds, but it's the light (photon current) that
moves at c (and its frequency just like AC frequency) So, say, the
photons we "receive" from the sun are not the original photons, but
the photons that were already around. How would we know?

We have local light sources. All light travels at c. We have one
mechanism for creation of one photon from the energy of another,
Compton
scattering. And it does not provide a specular image.

Sorry.

David A. Smith

Dear wespe:

"wespe" wrote in message
om...
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this
occuring at c), and at the destination we detect the last kicked
photon. Result is the same as if the original photon travelled. One
exception: if there was a box void of photons (but how?), light
couldn't be created in it. Just like sound cannot in space. So,
compton scattering, how does that disprove this? I'm so sorry

You can empty a space of certain wavelengths with a Faraday cage (radio -
microwave), or concrete (visible thru UV). Yet this light can still be
created in these spaces, transmitted, and received. Yes, these spaces will
still have other wavelengths of light zooming around. No matter where in
these spaces you check, however, they are the emitted wavelength, or its
scattered reflections. This requires that there be extraordinary
communication (FTL for one) between photons, and that very intense beams of
light cannot exist (but do) since there would be no "carriers" available in
certain spaces.

Again, I state that there are no such observed photon-photon interactions
that you imagine. The only photon-photon interactions that have been
observed are matter creators. Otherwise, photons don't interact with
anything except charged particles (Compton scattering).

Now what will perhaps get you excited, is that QM has a single "real"
propagating photon be a series of virtual photons that are "called up" from
somewhere, and disappearing after they are not needed. Perhaps another
poster can supply a reason why this model is an improvement over having two
types of photons (virtual and real). The problem is that these virtual
photons are not observable, and even the path of a single photon is not
observable.

You don't need to say "sorry" each time. You should post *after* someone
else's reply... what you did is called "top posting" and is considered
impolite. Notice how I posted after yours, and have responded to what you
wrote. What we say and ask here, is archived for posterity.

David A. Smith

wespe:
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this

Photons don't interact with each other, therefore they cannot
``collide''.

Dear Bilge:

"Bilge" wrote in message
...
wespe:
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this

Photons don't interact with each other, therefore they cannot
``collide''.

Not that multi-GeV photons are really what he was talking about, but how
about LEP? These are photon-photon "collisions", no? Certainly not
support for a "bucket brigade" like wespe was imagining...

David A. Smith

N:dlzc D:aol T:com \(dlzc\):
Dear Bilge:

"Bilge" wrote in message
ue-al.net...
wespe:
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this

Photons don't interact with each other, therefore they cannot
``collide''.

Not that multi-GeV photons are really what he was talking about, but how
about LEP? These are photon-photon "collisions", no? Certainly not
support for a "bucket brigade" like wespe was imagining...


Yes, you are right. Photon-photon collisions _can_ occur, but the
process occurs only at higher orders and at a minimum, requires two
pair productions and anihilations. The feynman diagram ususally
used to illustrate photon-photon scattering is:

~ ~ There are five additional diagrams at this order
~ ~ and the total cross section is the coherent sum.
~+----+~
^ |
| v
+----+
~ ~
~ ~
~ ~

"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:xefpc.38085$iy5.28609@okepread05...
Dear wespe:

"wespe" wrote in message
om...
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this
occuring at c), and at the destination we detect the last kicked
photon. Result is the same as if the original photon travelled. One
exception: if there was a box void of photons (but how?), light
couldn't be created in it. Just like sound cannot in space. So,
compton scattering, how does that disprove this? I'm so sorry

You can empty a space of certain wavelengths with a Faraday cage (radio -
microwave), or concrete (visible thru UV). Yet this light can still be
created in these spaces, transmitted, and received. Yes, these spaces will
still have other wavelengths of light zooming around. No matter where in
these spaces you check, however, they are the emitted wavelength, or its
scattered reflections. This requires that there be extraordinary
communication (FTL for one) between photons, and that very intense beams of
light cannot exist (but do) since there would be no "carriers" available in
certain spaces.

Again, I state that there are no such observed photon-photon interactions
that you imagine. The only photon-photon interactions that have been
observed are matter creators. Otherwise, photons don't interact with
anything except charged particles (Compton scattering).

Now what will perhaps get you excited, is that QM has a single "real"
propagating photon be a series of virtual photons that are "called up" from
somewhere, and disappearing after they are not needed. Perhaps another
poster can supply a reason why this model is an improvement over having two
types of photons (virtual and real). The problem is that these virtual
photons are not observable, and even the path of a single photon is not
observable.

You don't need to say "sorry" each time. You should post *after* someone
else's reply... what you did is called "top posting" and is considered
impolite. Notice how I posted after yours, and have responded to what you
wrote. What we say and ask here, is archived for posterity.

David A. Smith

I know. I did that because I was excited and didn't want to wait for
your reply to appear on google.

ok, first, a correction: I falsely claimed a single photon cannot be
detected. Like an air molecule from a very faint and highly
directional sound could be detected, so can a photon, I guess. And, as
sound is not wind, photon is not light, so there was some confusion
due to my naming. Call them virtual/real photons or whatever, doesn't
matter. The problem is how to prove or disprove.

Since I really started to think light is very much like sound,
question is: how can we show that sound is not wind, without being
able to follow air molecules? The faraday cage you mention could help,
if I only knew how to build a faraday cage for sound, then I could say
why it is like that for light. Also the box void of photons I
mentioned, could be something like: a sphere covered with one-way
mirror and containing a light source. I imagine photons would be
pumped out but not allowed to enter, so after some time, the carrier
medium would be gone and we wouldn't see any more light from outside.
I doubt anyone has built such a thing.

Then, we talk about photon frequency, but an individual (virtual)
photon should not carry this property (like an air molecule), but
would still have some little energy. So we would detect this energy
and say it has a corresponding frequency, while in fact it wouldn't. I
don't know how to tell the difference. Also, if we think of light's
wavelike properties, is it always sinusoidal or can it have different
waveforms like sound? How can we create or detect different waveforms
of light? According to fourier transform, if light consists of
travelling photons, we would need lots of photons of harmonic
frequencies to build different waveforms, but we wouldn't need that if
light is what I imagine.

regards.

Dear wespe:

"wespe" wrote in message
om...
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in
message news:xefpc.38085$iy5.28609@okepread05...
Dear wespe:

"wespe" wrote in message
om...
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this
occuring at c), and at the destination we detect the last kicked
photon. Result is the same as if the original photon travelled. One
exception: if there was a box void of photons (but how?), light
couldn't be created in it. Just like sound cannot in space. So,
compton scattering, how does that disprove this? I'm so sorry

You can empty a space of certain wavelengths with a Faraday cage
(radio -
microwave), or concrete (visible thru UV). Yet this light can still be
created in these spaces, transmitted, and received. Yes, these spaces
will
still have other wavelengths of light zooming around. No matter where
in
these spaces you check, however, they are the emitted wavelength, or
its
scattered reflections. This requires that there be extraordinary
communication (FTL for one) between photons, and that very intense
beams of
light cannot exist (but do) since there would be no "carriers"
available in
certain spaces.

Again, I state that there are no such observed photon-photon
interactions
that you imagine. The only photon-photon interactions that have been
observed are matter creators. Otherwise, photons don't interact with
anything except charged particles (Compton scattering).

Now what will perhaps get you excited, is that QM has a single "real"
propagating photon be a series of virtual photons that are "called up"
from
somewhere, and disappearing after they are not needed. Perhaps another
poster can supply a reason why this model is an improvement over having
two
types of photons (virtual and real). The problem is that these virtual
photons are not observable, and even the path of a single photon is not
observable.

You don't need to say "sorry" each time. You should post *after*
someone
else's reply... what you did is called "top posting" and is considered
impolite. Notice how I posted after yours, and have responded to what
you
wrote. What we say and ask here, is archived for posterity.

I know. I did that because I was excited and didn't want to wait for
your reply to appear on google.

ok, first, a correction: I falsely claimed a single photon cannot be
detected. Like an air molecule from a very faint and highly
directional sound could be detected, so can a photon, I guess. And, as
sound is not wind, photon is not light, so there was some confusion
due to my naming. Call them virtual/real photons or whatever, doesn't
matter. The problem is how to prove or disprove.

Can't prove. Can only disprove. OK?

Since I really started to think light is very much like sound,
question is: how can we show that sound is not wind, without being
able to follow air molecules?

Can look at the pressure as a function of time. If the pressure
oscillates, it is sound. If it continuous (or at least a period of many
seconds) it is wind.

The faraday cage you mention could help,
if I only knew how to build a faraday cage for sound, then I could say
why it is like that for light.

Look up "anechoic chamber", for the sound analog.

Also the box void of photons I
mentioned, could be something like: a sphere covered with one-way
mirror and containing a light source. I imagine photons would be
pumped out but not allowed to enter, so after some time, the carrier
medium would be gone and we wouldn't see any more light from outside.
I doubt anyone has built such a thing.

Also not possible, since the box will be made of real matter, and will
radiate IR radiation to "announce" its temperature. But a dark room will
do for visible light.

Then, we talk about photon frequency, but an individual (virtual)
photon should not carry this property (like an air molecule), but
would still have some little energy.

Not a requirement to have "a little energy". No observer would agree on
what this "little energy" was.

So we would detect this energy
and say it has a corresponding frequency, while in fact it wouldn't.

A bolometer does this (energy measurement) as does the photoelectric
effect, and if wavelength and c will give you frequency, then a diffraction
grating can "sort out" frequencies for you.

I
don't know how to tell the difference. Also, if we think of light's
wavelike properties, is it always sinusoidal or can it have different
waveforms like sound?

This is a very broad topic. The light we encounter is always transverse,
and can be polarized in different ways. This is something that sound
cannot do.

How can we create or detect different waveforms
of light?

You cannot for a single photon. To assume waveform is to assign structure
where it is indetectable.

According to fourier transform, if light consists of
travelling photons, we would need lots of photons of harmonic
frequencies to build different waveforms, but we wouldn't need that if
light is what I imagine.

You can imitate a waveform of various shapes using amplitude modulation,
which is what the AM radio is based on. As to constructing a propagating
"square wave", we cannot do this, as nature turns it into a sine wave, and
then into hash... with dispersion. This is encountered all the time in
communcations through optical fibers.

David A. Smith

"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:luvpc.41836$iy5.29808@okepread05...
Dear wespe:


"wespe" wrote in message
om...
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in
message news:xefpc.38085$iy5.28609@okepread05...
Dear wespe:

"wespe" wrote in message
om...
but you don't understand.. Nothing has to change. All I'm saying is a
photon doesn't have to travel from one place to another at c. It can
kick the nearby photon, and that one kicks another, and so on (this
occuring at c), and at the destination we detect the last kicked
photon. Result is the same as if the original photon travelled. One
exception: if there was a box void of photons (but how?), light
couldn't be created in it. Just like sound cannot in space. So,
compton scattering, how does that disprove this? I'm so sorry

You can empty a space of certain wavelengths with a Faraday cage
(radio -
microwave), or concrete (visible thru UV). Yet this light can still be
created in these spaces, transmitted, and received. Yes, these spaces
will
still have other wavelengths of light zooming around. No matter where
in
these spaces you check, however, they are the emitted wavelength, or
its
scattered reflections. This requires that there be extraordinary
communication (FTL for one) between photons, and that very intense
beams of
light cannot exist (but do) since there would be no "carriers"
available in
certain spaces.

Again, I state that there are no such observed photon-photon
interactions
that you imagine. The only photon-photon interactions that have been
observed are matter creators. Otherwise, photons don't interact with
anything except charged particles (Compton scattering).

Now what will perhaps get you excited, is that QM has a single "real"
propagating photon be a series of virtual photons that are "called up"
from
somewhere, and disappearing after they are not needed. Perhaps another
poster can supply a reason why this model is an improvement over having
two
types of photons (virtual and real). The problem is that these virtual
photons are not observable, and even the path of a single photon is not
observable.

You don't need to say "sorry" each time. You should post *after*
someone
else's reply... what you did is called "top posting" and is considered
impolite. Notice how I posted after yours, and have responded to what
you
wrote. What we say and ask here, is archived for posterity.

I know. I did that because I was excited and didn't want to wait for
your reply to appear on google.

ok, first, a correction: I falsely claimed a single photon cannot be
detected. Like an air molecule from a very faint and highly
directional sound could be detected, so can a photon, I guess. And, as
sound is not wind, photon is not light, so there was some confusion
due to my naming. Call them virtual/real photons or whatever, doesn't
matter. The problem is how to prove or disprove.

Can't prove. Can only disprove. OK?

** prove something wrong and you disprove it, right? anyway english is
not my mother tounge. ok.


Since I really started to think light is very much like sound,
question is: how can we show that sound is not wind, without being
able to follow air molecules?

Can look at the pressure as a function of time. If the pressure
oscillates, it is sound. If it continuous (or at least a period of many
seconds) it is wind.


** but we need a method which also applies to light. we don't have a
barometer for light, so this won't do. What we have, I think, is
detectors that get triggered when sufficient energy is absorbed. They
don't give us a waveform like a barometer can for sound. Or, with
raido waves, I think we could see a real waveform for very low
frequencies, but there's always an oscillation there, so why do say
it's a continuous stream of photons that travelled from the source.

The faraday cage you mention could help,
if I only knew how to build a faraday cage for sound, then I could say
why it is like that for light.

Look up "anechoic chamber", for the sound analog.

** well it's nice but I think walls of that chamber just absorbs sound
and prevents echo. Does a faraday cage work like that? I think it
shields by reflecting certain frequencies, not absorbing (but I'm not
sure. Then, what's a faraday cage that works for visible light, a
mirror?). Anyway, it's a filter for light. So what we need is a
mechanical filter for sound waves. I forgot why I wanted that.


Also the box void of photons I
mentioned, could be something like: a sphere covered with one-way
mirror and containing a light source. I imagine photons would be
pumped out but not allowed to enter, so after some time, the carrier
medium would be gone and we wouldn't see any more light from outside.
I doubt anyone has built such a thing.

Also not possible, since the box will be made of real matter, and will
radiate IR radiation to "announce" its temperature. But a dark room will
do for visible light.

** ok, to clear that up: what I was trying to create was: a vacuum in
which light cannot propogate (of course I'm assuming there exists this
medium full of photons, which is the subject of this thread). Then,
such a box would not be able to announce its temparature to its inside
space. I don't see why it makes such a box impossible. Dark room.. I
don't see how that's relevant.


Then, we talk about photon frequency, but an individual (virtual)
photon should not carry this property (like an air molecule), but
would still have some little energy.

Not a requirement to have "a little energy". No observer would agree on
what this "little energy" was.

** I was talking about the energy that a single photon carries. Why is
that not a requirement? Observers may not agree on how much this
energy is, but as long as it's not zero, we can calculate a frequency
value for it, despite there's no real vibration for a single photon.
That is, in my model, that is, if light propogates like sound.


So we would detect this energy
and say it has a corresponding frequency, while in fact it wouldn't.

A bolometer does this (energy measurement) as does the photoelectric
effect, and if wavelength and c will give you frequency, then a diffraction
grating can "sort out" frequencies for you.


** As I said above, if bolometer was the light analog for barometer,
it would be useful here, otherwise it doesn't help. Diffraction
grating.. Yeah, that can do. But does it work with a single photon?
That is, if a single photon carries a frequency property, a
diffraction grate should be able to determine its frequency. If it
can, I'm lost.

I
don't know how to tell the difference. Also, if we think of light's
wavelike properties, is it always sinusoidal or can it have different
waveforms like sound?

This is a very broad topic. The light we encounter is always transverse,
and can be polarized in different ways. This is something that sound
cannot do.


** Light can be polarized.. Sound cannot.. yeah, that presents a
problem with my model. OK, how about this: suppose a photon is
vibrating, giving it an attribute of a certain direction, so it can be
polarized. (but I'm not saying it's vibrating at the same frequency of
light, just some vibration enough to allow polarizing) That could be
like the temperature of air molecules. Suppose there are two sound
sources, one is surrounded by cold air, one is surrounded by hot air.
If sound waves are able to carry some of this temperature as they
travel, we could polarize sound waves too. If this is correct, that
would mean: light must not be able to preserve its polarization over
very long distances. But remember, with all this, I'm assuming my
model is correct, I'm trying to imagine various things, I'm not
delusional, I can withdraw my argument easily.

How can we create or detect different waveforms
of light?

You cannot for a single photon. To assume waveform is to assign structure
where it is indetectable.


** no waveform for a single photon.. well that's certainly not against
my model.

According to fourier transform, if light consists of
travelling photons, we would need lots of photons of harmonic
frequencies to build different waveforms, but we wouldn't need that if
light is what I imagine.

You can imitate a waveform of various shapes using amplitude modulation,
which is what the AM radio is based on. As to constructing a propagating
"square wave", we cannot do this, as nature turns it into a sine wave, and
then into hash... with dispersion. This is encountered all the time in
communcations through optical fibers.


** Not clear what you mean by "imitate". Is it a real waveform or
superimposed sine waves (different photons)? Ok not squarewave, but
say, a sawtooth wave, can it propogate? Can't we just analyse this by
creating 1 Hz radio wave.

David A. Smith

** thanks for your patience with me so far

Dear wespe:

"wespe" wrote in message
m...
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in
message news:luvpc.41836$iy5.29808@okepread05...
....
ok, first, a correction: I falsely claimed a single photon cannot be
detected. Like an air molecule from a very faint and highly
directional sound could be detected, so can a photon, I guess. And,
as
sound is not wind, photon is not light, so there was some confusion
due to my naming. Call them virtual/real photons or whatever, doesn't
matter. The problem is how to prove or disprove.

Can't prove. Can only disprove. OK?

** prove something wrong and you disprove it, right? anyway english is
not my mother tounge. ok.

This is basic science. An hypothesis is formed, then experimentally
tested. If it survives the test(s) it becomes a theory. Theories can then
be disproved by experiment, but not "proved". They only survive another
day...

Since I really started to think light is very much like sound,
question is: how can we show that sound is not wind, without being
able to follow air molecules?

Can look at the pressure as a function of time. If the pressure
oscillates, it is sound. If it continuous (or at least a period of
many
seconds) it is wind.


** but we need a method which also applies to light. we don't have a
barometer for light, so this won't do. What we have, I think, is
detectors that get triggered when sufficient energy is absorbed. They
don't give us a waveform like a barometer can for sound. Or, with
raido waves, I think we could see a real waveform for very low
frequencies, but there's always an oscillation there, so why do say
it's a continuous stream of photons that travelled from the source.

The only way to detect propagating photons is to absorb some of them. In
quantum wells, it is possible to detect information about photons without
absorbing them, but wouldn't otherwise apply.

The faraday cage you mention could help,
if I only knew how to build a faraday cage for sound, then I could
say
why it is like that for light.

Look up "anechoic chamber", for the sound analog.

** well it's nice but I think walls of that chamber just absorbs sound
and prevents echo. Does a faraday cage work like that? I think it
shields by reflecting certain frequencies, not absorbing (but I'm not
sure. Then, what's a faraday cage that works for visible light, a
mirror?). Anyway, it's a filter for light. So what we need is a
mechanical filter for sound waves. I forgot why I wanted that.

A Faraday cage acts like a mirror, to protect the internals, from
experiencing the design wavelengths-of-interest, from external sources.

As to forgetting, if it is of importance, you will remember it.

Also the box void of photons I
mentioned, could be something like: a sphere covered with one-way
mirror and containing a light source. I imagine photons would be
pumped out but not allowed to enter, so after some time, the carrier
medium would be gone and we wouldn't see any more light from outside.
I doubt anyone has built such a thing.

Also not possible, since the box will be made of real matter, and will
radiate IR radiation to "announce" its temperature. But a dark room
will
do for visible light.

** ok, to clear that up: what I was trying to create was: a vacuum in
which light cannot propogate (of course I'm assuming there exists this
medium full of photons, which is the subject of this thread). Then,
such a box would not be able to announce its temparature to its inside
space. I don't see why it makes such a box impossible. Dark room.. I
don't see how that's relevant.

It is not possible to create such a box. Even if you lined it with miles
of diamond, it would still announce its temperature (as best it could) at
the characteristic bond energy of the C-C bond. What would be possible,
however, would be to create an intense concentration of light, and pass it
through a very cold, shielded space. The intense beam should be
anomalously dissipated, if there were insufficient "photon gas" carriers.

Then, we talk about photon frequency, but an individual (virtual)
photon should not carry this property (like an air molecule), but
would still have some little energy.

Not a requirement to have "a little energy". No observer would agree
on
what this "little energy" was.

** I was talking about the energy that a single photon carries. Why is
that not a requirement? Observers may not agree on how much this
energy is, but as long as it's not zero, we can calculate a frequency
value for it, despite there's no real vibration for a single photon.
That is, in my model, that is, if light propogates like sound.

"calculating" a frequency is not the same as agreeing on the frequency.
And measuring the frequency for a single photon *is not possible*, as it
violates Heisenberg's uncertainty principle. We can neither know the path
nor position for a quantum object that we know precisely the energy of.

It is well known that sound propagates as a longitudinal (pressure) wave,
and light propagates as a transverse wave. So light will not precisely
propagate like sound. At question is whether a "bucket-brigade" mechanism
is required for the propagation of light.

So we would detect this energy
and say it has a corresponding frequency, while in fact it wouldn't.

A bolometer does this (energy measurement) as does the photoelectric
effect, and if wavelength and c will give you frequency, then a
diffraction
grating can "sort out" frequencies for you.


** As I said above, if bolometer was the light analog for barometer,
it would be useful here, otherwise it doesn't help.

It is more along the lines of integrating the barometer over one pressure
cycle. Indicating transmitted energy.

Diffraction
grating.. Yeah, that can do. But does it work with a single photon?
That is, if a single photon carries a frequency property, a
diffraction grate should be able to determine its frequency. If it
can, I'm lost.

It can sort each photon, and on a *large population* of such photons, a
pattern emerges. The photoelectric effect will deliver a single electron
above a threshold energy, and this can be detected. So this might do?

I
don't know how to tell the difference. Also, if we think of light's
wavelike properties, is it always sinusoidal or can it have different
waveforms like sound?

This is a very broad topic. The light we encounter is always
transverse,
and can be polarized in different ways. This is something that sound
cannot do.


** Light can be polarized.. Sound cannot.. yeah, that presents a
problem with my model. OK, how about this: suppose a photon is
vibrating, giving it an attribute of a certain direction, so it can be
polarized. (but I'm not saying it's vibrating at the same frequency of
light, just some vibration enough to allow polarizing)

How it is vibrating is very much a function of its characterisitc
frequency. The means to polarize it depend strongly on its characterisitc
frequency. So its "vibration" and frequency are interrelated.

That could be
like the temperature of air molecules. Suppose there are two sound
sources, one is surrounded by cold air, one is surrounded by hot air.
If sound waves are able to carry some of this temperature as they
travel, we could polarize sound waves too.

No. The temperature of air is poorly transmitted by radiation. That is
why gases are used for insulation. Their only means of "emitting" heat is
by convection. It is possible in IR wavelengths to see through smoke an
flames and measure soot particle temperature.

If this is correct, that
would mean: light must not be able to preserve its polarization over
very long distances. But remember, with all this, I'm assuming my
model is correct, I'm trying to imagine various things, I'm not
delusional, I can withdraw my argument easily.

Delusion would have little to do with it. It is sometimes neccesary to try
on clothing, to judge its fit.

The problem with longitudinal waves (sound) is that they have two of three
axes that are not involved in their propagation. Transverse waves have
three axes that are involved in their propagation. Two transverse (or
normal to the direction of propagation), that the E and B field vectors can
express "into", and then the direction of propagation itself. It is likely
that the photon has no finite "expression" along its direction of motion
however. Until "it" gets there, or once "it" leaves, the Universe is none
the wiser.

How can we create or detect different waveforms
of light?

You cannot for a single photon. To assume waveform is to assign
structure
where it is indetectable.


** no waveform for a single photon.. well that's certainly not against
my model.

Agreed, however the concern was expression of other waveforms than
sinusoidal. This can only occur with waveform populations. But Maxwell's
equations (still based on populations) seem to say that even a single
photon will express a sinusoidal EM field, or at least behave as if it did.

According to fourier transform, if light consists of
travelling photons, we would need lots of photons of harmonic
frequencies to build different waveforms, but we wouldn't need that
if
light is what I imagine.

You can imitate a waveform of various shapes using amplitude
modulation,
which is what the AM radio is based on. As to constructing a
propagating
"square wave", we cannot do this, as nature turns it into a sine wave,
and
then into hash... with dispersion. This is encountered all the time in
communcations through optical fibers.


** Not clear what you mean by "imitate". Is it a real waveform or
superimposed sine waves (different photons)? Ok not squarewave, but
say, a sawtooth wave, can it propogate? Can't we just analyse this by
creating 1 Hz radio wave.

Imitate/construct/fake/produce-for-specific-detector. It is generating a
stream of photons with the necessary intensity vs. time to get your
detector to "display" the waveform of interest.

1 Hz transmission will require very large antennae. It is unlikley that
this is feasable from an Earth antennae.

David A. Smith

wespe:
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:xefpc.38085$iy5.28609@okepread05...
[...]
ok, first, a correction: I falsely claimed a single photon cannot be
detected. Like an air molecule from a very faint and highly
directional sound could be detected, so can a photon, I guess. And, as
sound is not wind, photon is not light, so there was some confusion
due to my naming. Call them virtual/real photons or whatever, doesn't
matter. The problem is how to prove or disprove.

That is simple. We have a theory which defines the properties of
what we call a photon. We make measurements and the measurements
we make agree or do not agree with the predictions of the theory.

Since I really started to think light is very much like sound,
question is: how can we show that sound is not wind, without being
able to follow air molecules?

Light is very different from sound. The ``wave'' analogy is is
a very crude way to describe light (and in my opinion, a very poor
way to describe light).
[...]

Then, we talk about photon frequency, but an individual (virtual)
photon should not carry this property (like an air molecule), but
would still have some little energy.

Photons do not have a property called ``frequency''. That is purely
a property of the relationship between the source and the observer.
A photon has two properties, it's spin and it's mass. It has a mass
of zero and a spin of 1. Those are lorentz invariants.

So we would detect this energy
and say it has a corresponding frequency, while in fact it wouldn't. I
don't know how to tell the difference. Also, if we think of light's
wavelike properties, is it always sinusoidal or can it have different
waveforms like sound?

A pulse of laser light is just such a waveform.

How can we create or detect different waveforms
of light? According to fourier transform, if light consists of
travelling photons, we would need lots of photons of harmonic
frequencies to build different waveforms, but we wouldn't need that if
light is what I imagine.

That is not the right way to think of it in terms of photons. The
photon(s) in a pulse of laser light have frequencies which are in-
determinate as defined by the uncertainty relations.