Photon is Not a Point Particle?

Can the photon differ from being a point particle within the wave
pulse to not actually existing within the wave itself? Instead, the EM
wave pulse itself would actually act as the photon particle? The
length of the pulse particle would be a fixed length given by the time
that the electron requires to fall to the lower energy level. I have
hypothesized here for simplicity that the time to fall to a lower
energy level is always a constant for any electron. Then a blue
colored EM wave pulse would have more wavelengths inside this pulse
while a red pulse would have less. Thus the wavelength would not be
determining the size of the pulse, but instead the time for the
electron to fall would be.

The electron would temporarily spiral around the nucleus (instead of
jumping) from a higher energy level to a lower? Even if it did jump
instead of spiraling, it is still going to temporarily exist between
the levels anyways. For simplicity again assume two 2D orbits. Now
rotate out of the page 90 degrees. Thus, when viewed from the side,
this spiraling around the nucleus would produce a half wave dipole
antenna producing a wave pulse. One axis of this wave would be
vertically polarized while another circularly polarized. The pulse
would decrease in amplitude from Ehigh to Elow. Using calculus
integration would verify consistency to E=hf=hc/wavelength.

Dear rraw...:

On Apr 20, 9:40 am, "
wrote:
Can the photon differ from being a point particle
within the wave pulse to not actually existing
within the wave itself?

What would cause the photon to manifest? How would a group of
photons, more or less co-located be differntiable from a "wave"?

Instead, the EM wave pulse itself would actually
act as the photon particle?

Fails photoelectric effect.

The length of the pulse particle would be a fixed
length given by the time that the electron
requires to fall to the lower energy level. I have
hypothesized here...

Waste of effort. Both particle and wave are mutually exclusive, and
both are only models made up by "flatlanders", and ppor fits to the
quantum realm.

David A. Smith

On Apr 20, 9:40*am, "
wrote:
Can the photon differ from being a point particle within the wave
pulse to not actually existing within the wave itself? Instead, the EM
wave pulse itself would actually act as the photon particle? The
length of the pulse particle would be a fixed length given by the time
that the electron requires to fall to the lower energy level. I have
hypothesized here for simplicity that the time to fall to a lower
energy level is always a constant for any electron. Then a blue
colored EM wave pulse would have more wavelengths inside this pulse
while a red pulse would have less. Thus the wavelength would not be
determining the size of the pulse, but instead the time for the
electron to fall would be.

The electron would temporarily spiral around the nucleus (instead of
jumping) from a higher energy level to a lower? Even if it did jump
instead of spiraling, it is still going to temporarily exist between
the levels anyways. For simplicity again assume two 2D orbits. Now
rotate out of the page 90 degrees. Thus, when viewed from the side,
this spiraling around the nucleus would produce a half wave dipole
antenna producing a wave pulse. One axis of this wave would be
vertically polarized while another circularly polarized. The pulse
would decrease in amplitude from Ehigh to Elow. Using calculus
integration would verify consistency to E=hf=hc/wavelength.

I favor that electrons and positrons make up photons (aka matter/
antimatter), thus our photon represents near zero mass.

~ BG

Experiments (such as the photoelectric effect) show that the photon is
definitely a point particle. The wave description of the photon
pertains to its travel, not the particle. The photon has no classical
existence between emission and impingement, so the quantum description
models its potential destination points. That's all.

Eric Flesch

Dear Eric Flesch:

"Eric Flesch" wrote in message
...
Experiments (such as the photoelectric effect) show that
the photon is definitely a point particle. The wave
description of the photon

.... and electrons, neutrons, nucleii, and molecules larger then
C-60 buckyballs...

pertains to its travel, not the particle.

It applies to the system of particle + observer + Universe. The
wave behavior arises because the butcher unwittingly has his
thumb on the scale.

The photon has no classical existence between
emission and impingement, so the quantum description
models its potential destination points. That's all.

What about diffraction (self-interference), and polarization
effects? Does the photon "change its nature" in flight?

David A. Smith

On Thu, 23 Apr 2009, "N:dlzc D:aol T: (etc)" wrote:
"Eric Flesch" wrote in message
pertains to its travel, not the particle.

It applies to the system of particle + observer + Universe.

That's a vast way of saying nothing.

The photon has no classical existence between
emission and impingement, so the quantum description
models its potential destination points. That's all.

What about diffraction (self-interference), and polarization
effects? Does the photon "change its nature" in flight?

The photon is never in flight. It moves conductively from emission to
registration. We, the observers, have to come up with the big dance
about its flight path. To the photon, it simply steps across. Time
does not exist for the photon.

cheers

Dear Eric Flesch:

"Eric Flesch" wrote in message
...
On Thu, 23 Apr 2009, "N:dlzc D:aol T: (etc)" wrote:
"Eric Flesch" wrote in message
pertains to its travel, not the particle.

It applies to the system of particle + observer +
Universe.

That's a vast way of saying nothing.

No, that is a way of saying that when you measure / observe wave
behvior, you do so in a framework involving the entire Universe,
no choices. Distance and duration do that. Particleness
involves discrete exchanges, no extent, no duration. The photon
does not change, the test does.

The photon has no classical existence between
emission and impingement, so the quantum
description models its potential destination points.
That's all.

What about diffraction (self-interference), and
polarization effects? Does the photon "change its
nature" in flight?

The photon is never in flight. It moves conductively
from emission to registration. We, the observers,
have to come up with the big dance about its flight
path. To the photon, it simply steps across. Time
does not exist for the photon.

The photon responds to changes along its path (see the above
phenomenon). What is more, the Lorentz transforms *cannot* be
applied to a frame at "c". How can you say the photon cannot
experience change (has no time)?

David A. Smith

"N:dlzc D:aol T:com (dlzc)" wrote in message
...
Dear Eric Flesch:

"Eric Flesch" wrote in message
...
On Thu, 23 Apr 2009, "N:dlzc D:aol T: (etc)" wrote:
"Eric Flesch" wrote in message
pertains to its travel, not the particle.

It applies to the system of particle + observer +
Universe.

That's a vast way of saying nothing.

No, that is a way of saying that when you measure / observe wave behvior,
you do so in a framework involving the entire Universe, no choices.
Distance and duration do that. Particleness involves discrete exchanges,
no extent, no duration. The photon does not change, the test does.

The photon has no classical existence between
emission and impingement, so the quantum
description models its potential destination points.
That's all.

What about diffraction (self-interference), and
polarization effects? Does the photon "change its
nature" in flight?

The photon is never in flight. It moves conductively
from emission to registration. We, the observers,
have to come up with the big dance about its flight
path. To the photon, it simply steps across. Time
does not exist for the photon.

The photon responds to changes along its path (see the above phenomenon).
What is more, the Lorentz transforms *cannot* be applied to a frame at
"c". How can you say the photon cannot experience change (has no time)?

David A. Smith
Even though I would agree with you, Smiffy, Eric has a point.

tau = (t-vx/c^2) / sqrt(1-v^2/c^2)

So for a photon,
tau = (t-vx/c^2) / sqrt(1-c^2/c^2)
= (t-cx)/c^2)/ sqrt(0)
= (t-cx)/c^2)/ 0
and that is division by zero which is undefined.

However,
tau = (t-vx/c^2) / sqrt(1-v^2/c^2)

Substituting x for its value,

tau = (t-v * (vt) /c^2) / sqrt(1-v^2/c^2)
= (t- tv^2/c^2) / sqrt(1-v^2/c^2)
= t(1-v^2c^2) / sqrt(1-v^2/c^2)
= t * sqrt(1-v^2/c^2)
Now I substitute v for c,
= t * sqrt(1-c^2/c^2)
tau = t * 0
= 0

Time, for the photon, does not pass (according to the Messiah Einstein).
He even says this himself.
"For velocities greater than that of light our deliberations become
meaningless; we shall, however, find in what follows, that the velocity of
light in our theory plays the part, physically, of an infinitely great
velocity."

The only way a photon can infinite velocity is if it gets here in no time at
all.

So it is not Eric that is saying the photon cannot experience change
(has no time), but the dork Einstein himself.

How can you heretically contradict your Lord and Master, the Holey
Pope of Relativity, St Rabbi Einstein and blame Eric for your ****-up?

You are a troll, aren't you?

On Fri, 24 Apr 2009, "N:dlzc D:aol T: (etc)" wrote:
"Eric Flesch" wrote in message
That's a vast way of saying nothing.

No, that is a way of saying that when you measure / observe wave
behvior, you do so in a framework involving the entire Universe,
no choices. Distance and duration do that. Particleness
involves discrete exchanges, no extent, no duration. The photon
does not change, the test does.

Yes.

The photon is never in flight. It moves conductively
from emission to registration. We, the observers,
have to come up with the big dance about its flight
path. To the photon, it simply steps across. Time
does not exist for the photon.

The photon responds to changes along its path (see the above
phenomenon). What is more, the Lorentz transforms *cannot* be
applied to a frame at "c". How can you say the photon cannot
experience change (has no time)?

All these changes influence where the photon ends up. When the photon
is emitted, it does not know where it ends, but its ending is
instantanous, to it. So neither distance nor time apply. The
Schroedinger equations map this null-dimensional scenatio into our
space-time manifold.

As John Wheeler said of the delayed choice experiment, "we cannot
speak of what the photon is doing before it has done it." The answer
is simply that the photon never does it all all. It has no classical
existence in flight -- which is to say, it isn't even there. Its path
changes, the photon never does.

"Eric Flesch" wrote in message
...
On Fri, 24 Apr 2009, "N:dlzc D:aol T: (etc)" wrote:
"Eric Flesch" wrote in message
That's a vast way of saying nothing.

No, that is a way of saying that when you measure / observe wave
behvior, you do so in a framework involving the entire Universe,
no choices. Distance and duration do that. Particleness
involves discrete exchanges, no extent, no duration. The photon
does not change, the test does.

Yes.

The photon is never in flight. It moves conductively
from emission to registration. We, the observers,
have to come up with the big dance about its flight
path. To the photon, it simply steps across. Time
does not exist for the photon.

The photon responds to changes along its path (see the above
phenomenon). What is more, the Lorentz transforms *cannot* be
applied to a frame at "c". How can you say the photon cannot
experience change (has no time)?

All these changes influence where the photon ends up. When the photon
is emitted, it does not know where it ends, but its ending is
instantanous, to it. So neither distance nor time apply. The
Schroedinger equations map this null-dimensional scenatio into our
space-time manifold.

As John Wheeler said of the delayed choice experiment, "we cannot
speak of what the photon is doing before it has done it." The answer
is simply that the photon never does it all all. It has no classical
existence in flight -- which is to say, it isn't even there. Its path
changes, the photon never does.


Yeah, Wheeler was a crank mystic, but what did he care? He made
his money out of it.