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Science Forum Index » Physics - Research Forum » Free photons
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| Alex |
 Posted: Mon Dec 04, 2006 6:01 pm |
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I have been reading Kevin Brown's intriguing "Mathpages"
http://www.mathpages.com/rr/s9-10/9-10.htm and found his arguments
against free photons to be fairly persuasive:
"From the standpoint of quantum electrodynamics, the wave properties of
electromagnetic radiation are actually wave properties of the emitter.
All the potential sources of a photon have a certain (complex)
amplitude for photon emission, and this amplitude evolves in time as we
progress along the emitter's worldline. However, as noted above, once a
photon is emitted, its phase does not advance. In a sense, the ancients
who conceived of sight as something like a blind man's incompressible
cane, feeling distant objects, were correct, because our retinas
actually are in "direct" contact, via null intervals, with the sources
of light. The null interval plays the role of the incompressible cane,
and the wavelike properties we "feel" are really the advancing quantum
phases of the source. (We might say there is less to photons than meets
the eye.)"
There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
Alex |
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| FrediFizzx |
| Posted: Tue Dec 05, 2006 11:39 am |
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"Alex" <dralexgreen@yahoo.co.uk> wrote in message
news:1165081281.208439.169920@80g2000cwy.googlegroups.com...
Quote: I have been reading Kevin Brown's intriguing "Mathpages"
http://www.mathpages.com/rr/s9-10/9-10.htm and found his arguments
against free photons to be fairly persuasive:
"From the standpoint of quantum electrodynamics, the wave properties
of
electromagnetic radiation are actually wave properties of the emitter.
All the potential sources of a photon have a certain (complex)
amplitude for photon emission, and this amplitude evolves in time as
we
progress along the emitter's worldline. However, as noted above, once
a
photon is emitted, its phase does not advance. In a sense, the
ancients
who conceived of sight as something like a blind man's incompressible
cane, feeling distant objects, were correct, because our retinas
actually are in "direct" contact, via null intervals, with the sources
of light. The null interval plays the role of the incompressible cane,
and the wavelike properties we "feel" are really the advancing quantum
phases of the source. (We might say there is less to photons than
meets
the eye.)"
There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
Not really a killer argument but if it is valid to quantize the source
free Maxwell equations, then there has to be free photons. If not
valid, then there can't be free photons. IMHO and research, it is valid
and there are free photons. But it is probably too speculative for this
group to explain my viewpoint further.  If interested, send me an
email.
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com |
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| Oh No |
| Posted: Wed Dec 06, 2006 11:18 am |
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Thus spake Alex <dralexgreen@yahoo.co.uk>
Quote: I have been reading Kevin Brown's intriguing "Mathpages"
http://www.mathpages.com/rr/s9-10/9-10.htm and found his arguments
against free photons to be fairly persuasive:
"From the standpoint of quantum electrodynamics, the wave properties of
electromagnetic radiation are actually wave properties of the emitter.
People say such things, but I have only ever been able to think that
they have not grasped the mathematical and empirical implications of
such a statement. It is not generally made with the mathematical
precision from which one could draw conclusions, but as far as I can see
it would lead to something other than the laws of quantum theory which
we observe.
Quote: All the potential sources of a photon have a certain (complex)
amplitude for photon emission, and this amplitude evolves in time as we
progress along the emitter's worldline. However, as noted above, once a
photon is emitted, its phase does not advance.
I am not sure what he means to say phase does not advance. On the face
of it that is in flat contradiction to the form of the wave function
e^i(Et - p.x)
Quote: In a sense, the ancients
who conceived of sight as something like a blind man's incompressible
cane, feeling distant objects, were correct, because our retinas
actually are in "direct" contact, via null intervals, with the sources
of light. The null interval plays the role of the incompressible cane,
and the wavelike properties we "feel" are really the advancing quantum
phases of the source. (We might say there is less to photons than meets
the eye.)"
There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
To describe "cooperative interactions between a source and emitter"
suggests that he is apparently assuming a non-local effect of a sort
which goes fundamentally against what most physicists are prepared to
believe, and which appears to be at odds with the locality condition as
it appears at a fundamental level in quantum electrodynamics.
Clearly not all issues of interpretation in this field are resolved,
otherwise Bell's theorem would not be the source of controversy which it
is. You should be extremely suspicious of anyone presenting pat answers
without a heck of a lot of rigorous analysis with which to back it up.
The simplest conclusion to draw when you see such an argument is that
the person is talking off the top of his head and has no idea what the
true issues are.
Regards
--
Charles Francis
substitute charles for NotI to email |
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| Oz |
| Posted: Wed Dec 06, 2006 11:18 am |
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Guest
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FrediFizzx <fredifizzx@hotmail.com> writes
Quote: Not really a killer argument but if it is valid to quantize the source
free Maxwell equations, then there has to be free photons. If not
valid, then there can't be free photons. IMHO and research, it is valid
and there are free photons. But it is probably too speculative for this
group to explain my viewpoint further. If interested, send me an
email.
1) When you say 'source-free' do you also mean 'detector free'?
2) Is the converse also true.
--
Oz
This post is worth absolutely nothing and is probably fallacious. |
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| Igor Khavkine |
| Posted: Fri Dec 08, 2006 10:43 am |
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Alex wrote:
Quote: There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
The term "free photon" in your post and in the website you reference is
not terribly precise. Therefore, I can't give a precise answer to your
question. The best I can do is give an answer to a precise question
that is similar to yours.
We know (i.e. it is experimentally confirmed) that electromagnetic
fields obey Maxwell's equations:
div(F) = j and div(*F) = 0,
where div is the 4-divergence and F is the Faraday tensor (containing E
and B fields), and *F is its dual. A complete theory of matter
interacting with the EM field would also include some dynamical
equations for the matter distribution whose 4-current is j. One
question is whether F is uniquely determined by j. I think this
question is as close as I can come to whether "free photons" exist.
Mathematically, the answer is clearly No, since the non-homogeneous
Maxwell equations don't have a unique solution unless some
boundary/initial conditions are imposed on the field. To say that there
are no "free photons" would be akin to saying that there is some
preferred set of boundary/initial conditions that uniquely fix F given
a specific j.
Unfortunately, physically, this proposition is difficult to test. If we
try to fix an initial condition for F at some definite time in the
past, one could argue that this specific field configuration could have
been produced by *any* field configuration in a more distant past plust
a specifically tailored j in the intervening time. Hence, unless we
know j all the way to the infinite past (or to the Big Bang), if there
is a preferred initial condition that fixes a one-to-one relationship
between F and j, it is impossible to identify.
That's the situation classically. Fourtunately, quantum mechanics comes
to the rescue. The situation is analogous to a simple harmonic
oscillator with one degree of freedom compared to one with two degrees
of freedom. Classically, you may be wondering whether the latter system
really has two independent degrees of freedom or whether there some
preferred initial condition that gives a definite relationship between
motions in the two configuration coordinates. On the other hand, in
quantum mechanics, you can compare the energy spectrum and find out
right away. The spectrum of a single degree of freedom simple harmonic
oscillator is different from that of a two degree of freedom simple
harmonic oscillator, especially if it has two distinct fundamental
frequencies.
Similarly, Quantum Electrodynamics assumes that the EM field has
degrees of freedom that are not uniquely fixed by the distribution of
other matter. The result is a prediction of vacuum EM field
fluctuations and various vacuum polarization effects that show up in
elementary particle scattering cross sections. The final verdict comes
from experiment: yes, vacuum polarization effects do show up in
particle accelerator experiments. In my book, that's a big win for
"free photons".
Hope this helps.
Igor |
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| FrediFizzx |
| Posted: Fri Dec 08, 2006 10:44 am |
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Guest
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"Oz" <Oz@farmeroz.port995.com> wrote in message
news:jkYHhlDdYndFFwwx@farmeroz.port995.com...
Quote: FrediFizzx <fredifizzx@hotmail.com> writes
Not really a killer argument but if it is valid to quantize the source
free Maxwell equations, then there has to be free photons. If not
valid, then there can't be free photons. IMHO and research, it is
valid
and there are free photons. But it is probably too speculative for
this
group to explain my viewpoint further. If interested, send me an
email.
1) When you say 'source-free' do you also mean 'detector free'?
Yes, of course.
div E = 0
div B = 0
curl E = -&B/&t
curl B = &E/&t
With c = 1
Quote: 2) Is the converse also true.
Your question is not clear to me. Are you asking if 'detector free' it
will be 'source free'? If so, yes. Perhaps it is just better to say
the Maxwell equations for the "free" field as listed above. As Milonni
puts it; "..., i.e., the field in a region where there are no
sources..." I do think in this case that would include detectors.
FrediFizzx
Quantum Vacuum Charge papers;
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf
or postscript
http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps
http://www.arxiv.org/abs/physics/0601110
http://www.vacuum-physics.com |
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| Arnold Neumaier |
| Posted: Fri Dec 08, 2006 10:45 am |
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Alex wrote:
Quote: There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
Free particles are a convenient mathematical abstraction. In Nature,
there are no free particles, only interacting ones. This holds both
for photons and for other particles like electrons.
However, in sufficiently localized (and nearly empty) regions of
space, particles can be approximately free.
Again, this holds for both photons and other particles.
It is very convenient to approximate such states by free states.
For example, this allows to explain much of quantum mechanics
in terms of particle scattering. The S-matrix interpretation depends
crucially on the fact that the ingoing and outgoing asymptotic states
of photons, electrons, quarks, etc. are free.
Thus, in this sense, free photons exist just as much (or just as little)
as free electrons.
Arnold Neumaier |
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| Neil Bates |
| Posted: Thu Dec 14, 2006 12:28 pm |
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"Alex" <dralexgreen@yahoo.co.uk> wrote in message
news:1165081281.208439.169920@80g2000cwy.googlegroups.com...
[quote]I have been reading Kevin Brown's intriguing "Mathpages"
http://www.mathpages.com/rr/s9-10/9-10.htm and found his arguments
against free photons to be fairly persuasive:
Quote:
There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
Alex
I always thought that the "killer" argument for free photons was
*maintainence* of energy and momentum conservation (and angular momentum
sometimes.) After "emission of the photon" we want something to "hold" the E
and p and S until it is found in the absorber. It is like the argument that
electric fields must "have energy" etc., so that when we push charges
together, the energy we used is "still there". This is all based on a
realist philosophical posture of wanting things to keep track between overt
interactions. Otherwise, we wouldn't even have to believe that the fields
existed in any circumstance, we could just say that things make other things
behave in various ways - whether it was overt events like photon
transmission or just the maintainence of the force between charges, etc.
BTW, for you logical positivists: What is the operational definition of the
statement, "Things continue to exist even when not being observed."? |
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| muser |
| Posted: Mon Dec 25, 2006 9:29 am |
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Neil Bates wrote:
[quote]"Alex" <dralexgreen@yahoo.co.uk> wrote in message
news:1165081281.208439.169920@80g2000cwy.googlegroups.com...
I have been reading Kevin Brown's intriguing "Mathpages"
http://www.mathpages.com/rr/s9-10/9-10.htm and found his arguments
against free photons to be fairly persuasive:
Quote:
There are numerous papers on "free photons" but I have been unable to
find any argument to justify the assumption that photons can indeed be
free rather than cbeing ooperative interactions between a source and
emitter. Does anyone have a "killer" argument for the existence of
free photons?
Alex
I always thought that the "killer" argument for free photons was
*maintainence* of energy and momentum conservation (and angular momentum
sometimes.) After "emission of the photon" we want something to "hold" the E
and p and S until it is found in the absorber. It is like the argument that
electric fields must "have energy" etc., so that when we push charges
together, the energy we used is "still there". This is all based on a
realist philosophical posture of wanting things to keep track between overt
interactions. Otherwise, we wouldn't even have to believe that the fields
existed in any circumstance, we could just say that things make other things
behave in various ways - whether it was overt events like photon
transmission or just the maintainence of the force between charges, etc.
BTW, for you logical positivists: What is the operational definition of the
statement, "Things continue to exist even when not being observed."?
Could free photons be a candidate for dark matter. How would a free
photon behave outside the EM field (is such a thing possible?), would
it gain mass? would hawking's radiation be a likily source emitter. |
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| Igor Khavkine |
| Posted: Tue Dec 26, 2006 10:35 am |
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muser wrote:
Quote: Could free photons be a candidate for dark matter. How would a free
photon behave outside the EM field (is such a thing possible?), would
it gain mass? would hawking's radiation be a likily source emitter.
Unfortnately, your question shows some misunderstanding of the concepts
of photons, the EM field, and Hawking radiation.
At the classical level, "free photons" refers to a configuration of the
EM field that satisfies the sourceless Maxwell equations. At the
quantum level, the situation is slightly more subtle, but is a more or
less direct analog. So, to speak of photons as separate from the EM
field is impossible.
Hawking radiation, as usually meant, is only present at the event
horizon of a black hole and is only physically significant (due to its
weakness) only for extremely tiny black holes. The only link between EM
radation and Hawking radiation is that the former is part of the
latter.
Any kind of matter or radiation distribution acts as a source of
curvature, and hence gravity. The CMB is evidence that non-trivial EM
fields are present everywhere in the universe. These are of course
taken into account by cosmologists, along with all visible matter and
any other kind of matter and radiation that we have evidence for, in
models of the universe. By definition, dark matter and dark energy is
what is neccesary beyond these inputs to make cosmological models agree
with observations. Hence, the EM fields that we are aware off, cannot
account for dark matter or dark energy.
Hope this helps.
Igor |
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