Existence of photons
Martin Green
of
energy between light and matter. In 1926 Schroedinger gave us the
differential equation for calculating the
distribution of electrical charge in matter, including an exact solution
for the case of the hydrogen atom. Using this equation, we can
mathematically test out what happens when an isolated atom is exposed to
ordinary, classical electromagnetic light. The results of this calculation
show that:
increasing the intensity of the light will not effectivley cause an
electron to be emitted
there is a critical frequency which must be achieved to cause electron
emission
above this frequency, electrons are emitted even with very low light
intensities
These are exactly the so-called paradoxes that were previously thought to
be explainable only by the existence of light particles, or "photons".
Martin Green
Jim Carr
>
>Before 1926, photons were used to explain certain facts about the exchange
>of energy between light and matter.
Perhaps you noticed my article pointing out that one of those facts
was the Compton effect?
>In 1926 Schroedinger gave us the differential equation for calculating the
>distribution of electrical charge in matter, including an exact solution
>for the case of the hydrogen atom. Using this equation, we can
>mathematically test out what happens when an isolated atom is exposed to
>ordinary, classical electromagnetic light. The results of this calculation
>show ...
nothing at all about the Compton effect, which involves a free electron
rather than a bound one. Thus Schroedinger did not remove the need for
the photon derived by Einstein in 1905. Further, everything you say
about Schroedinger was true of the old Bohr-Sommerfeld model.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Martin Green
in favor of the photon theory of light. I have two problems with this.
1. Every physics textbook I have seen cites the photo-electric effect, not
the Compton effect, as the decisive argument. If the Compton effect it is a
rare occasion. Why should I believe your word against the conventional
wisdom of countless physics professors and textbooks?
2. If the Compton effect was demonstrated in 1919, how could Einstein have
"derived" the need for photons in 1905?
Jim Carr
| nothing at all about the Compton effect, which involves a free electron
| rather than a bound one. Thus Schroedinger did not remove the need for
| the photon derived by Einstein in 1905. Further, everything you say
| about Schroedinger was true of the old Bohr-Sommerfeld model.
In article <022e82b1$12b4b100$0100a8c0@mgreen>
"Martin Green" <test...@pangea.ca> writes:
>
>You place great stock in the Compton effect as being the clinching argument
>in favor of the photon theory of light.
Yes. Read Compton's paper and you will see why. It is both available
(Phys Rev Centennial collection) and easy to read, although you will
need advanced E+M to deal with the classical estimate.
>I have two problems with this.
>
>1. Every physics textbook I have seen cites the photo-electric effect, not
>the Compton effect, as the decisive argument.
Well, they are half-wrong. That is why you came here, right? To learn?
It was decisive in 1905. It remained accepted even after the "old"
QM showed another way to look at it, for reasons that require reading
the Einstein paper. Compton shut the door. Lasers nailed it shut.
I consider it possible that you have only read popular books and
low level textbooks where a problem requiring both relativity and
quantum mechanics and advanced E+M for its proper treatment might
be omitted in favor of one that is simple and non-relativistic.
There are physics textbooks where it is discussed. As one example,
the significance of the Compton effect was described in my first-year
physics textbook (Berkeley Physics Course, vol. 4).
>If the Compton effect it is a rare occasion.
I don't know what this means. Compton scattering is not rare.
>Why should I believe your word against the conventional
>wisdom of countless physics professors and textbooks?
Look at what the countless professors and textbooks say about the
Compton effect. Maybe you should look?
>2. If the Compton effect was demonstrated in 1919, how could Einstein have
>"derived" the need for photons in 1905?
The Compton effect was demonstrated in 1923.
Einstein's derivation of the photon did not depend on experiment.
He applied the methods of statistical mechanics to the electromagnetic
field. (Magnificent idea. Also note that he did not come to quantum
from the same direction as the final theory, hence his thinking that
a mechanically deterministic theory -- like classical stat mech -- is
the real foundation of quantum mechanics.) His result agreed with the
black-body result of Planck, for a different reason (since he quantized
the field rather than the oscillators), and predicted the photoelectric
effect, which would not come from Planck's idea.
When experiment (by Millikan, in 1916) showed the same constant in both
effects, the theory stood triumphant. The Nobel followed in 1921.
Martin Green
> >
> >1. Every physics textbook I have seen cites the photo-electric effect,
not
> >the Compton effect, as the decisive argument.
>
> Well, they are half-wrong.
Why would all the books be wrong, or even half-wrong?
> QM showed another way to look at it, for reasons that require reading
> the Einstein paper. Compton shut the door. Lasers nailed it shut.
>
Quantum mechanics makes individual atoms behave as tuned harmonic
oscillators. A "gas" composed of tiny, classical antennas would generate
coherent laser light in the form of classical electromagnetic rays. That's
exactly how a laser works. There is nothing about laser light that needs to
be explained by photons.
Martin Green
Fabio Di Teodoro
On 3 Mar 1999, Martin Green wrote:
>
> > >
> > >1. Every physics textbook I have seen cites the photo-electric effect,
> not
> > >the Compton effect, as the decisive argument.
> >
> > Well, they are half-wrong.
>
> Why would all the books be wrong, or even half-wrong?
he did not mean *all* the books. He meant the books you recalled.
There is a lot of literature around...
> Quantum mechanics makes individual atoms behave as tuned harmonic
> oscillators. A "gas" composed of tiny, classical antennas would generate
> coherent laser light in the form of classical electromagnetic rays.
nice...and why should the light be coherent ? who told you that the
dipoles will oscillate in phase with each other...?
you need an optical resonator, otherwise you'll get an abat-jour, not a
laser.
Anyway this can be still handled semiclassically. But see below.
> That's
> exactly how a laser works. There is nothing about laser light that needs to
> be explained by photons.
Just as an example, note that the semiclassical picture fails miserably
when you try to model strong-field effects, such as the ones
characterizing femto-second pulsed laser sources (or even longer pulses
nicely focused on the target). Note that the intensity
delivered by these sources corresponds to electric fields acting on the
single atom that are much stronger than the internal coulombic fields. The
resulting matter states are called "dressed states", and dramatically
differ from those of the unperturbed atom. There is no way to separate
these states from the state of the field as you do in semiclassical
approximation.
You need second quantization, i.e. you need photons.
And needless to say, second quantization is necessary even for much
simpler phenomena such as spontaneous emission. And spontaneous emission
crucially enters the laser emission dynamics, although in some case (but
not always) it can be accounted for by a stochastic hamiltonian whose
effect can be reduced to a phenomenological relaxation factor.
regards
Jim Carr
| >not the Compton effect, as the decisive argument.
|
| Well, they are half-wrong.
In article <022e832f$8f170c60$0100a8c0@mgreen>
"Martin Green" <test...@pangea.ca> writes:
>
>Why would all the books be wrong, or even half-wrong?
Not all books, just yours. I gave one example of a textbook that
did treat that subject, and I also explained why many textbooks
would not wish to deal with the subject at the introductory level.
The reason those books are half wrong is the same reason many
elementary textbooks are half wrong. The reason Dr. Seuss only
used 100 words (or whatever it was) in "The Cat in the Hat".
The audience is not thought to be sophisticated enough to deal
with a detailed answer, so a simple (and historically correct)
answer is given instead. How many physics textbooks treat a
pendulum where the rod has mass, or a mass on a spring where the
spring has mass? (The latter was removed from H&R in one new
edition that I looked at, and appeared only in the problems.)
| QM showed another way to look at it, for reasons that require reading
| the Einstein paper. Compton shut the door. Lasers nailed it shut.
>Quantum mechanics makes individual atoms behave as tuned harmonic
>oscillators.
Not really. They do not have the response of harmonic oscillators.
However, that is widely used to model some effects. What you seem
to overlook (or not know, although this detail has been pointed out
in this thread) is that such a model is insufficient to explain the
Compton effect if they are interacting with a classical E+M field.
>A "gas" composed of tiny, classical antennas would generate
>coherent laser light in the form of classical electromagnetic rays.
Irrelevant to the Compton effect. However, I challenge you to point
to the section in Jackson where this problem (or the Compton effect)
can be done with classical radiation theory. I don't recall any
classical derivation of stimulated emission, but would love to know
of one.
Fabio Di Teodoro
On 4 Mar 1999, Martin Green wrote:
>
> > > Quantum mechanics makes individual atoms behave as tuned harmonic
> > > oscillators. A "gas" composed of tiny, classical antennas would
> generate
> > > coherent laser light in the form of classical electromagnetic rays.
> >
> > nice...and why should the light be coherent ? who told you that the
> > dipoles will oscillate in phase with each other...?
>
> I've done the math. Have you?
i've done the physics.
Fabio
Martin Green
> > Quantum mechanics makes individual atoms behave as tuned harmonic
> > oscillators. A "gas" composed of tiny, classical antennas would
generate
> > coherent laser light in the form of classical electromagnetic rays.
>
> nice...and why should the light be coherent ? who told you that the
> dipoles will oscillate in phase with each other...?
I've done the math. Have you?
Martin
Martin Green
Jim Carr <j...@ibms48.scri.fsu.edu> wrote in article
<7bjjku$94d$1...@news.fsu.edu>...
> | ...
> | nothing at all about the Compton effect, which involves a free electron
> | rather than a bound one. Thus Schroedinger did not remove the need for
> | the photon derived by Einstein in 1905. Further, everything you say
> | about Schroedinger was true of the old Bohr-Sommerfeld model.
>
That's crap. In the Schroedinger model, the superposition of s and p states
gives you an oscillating electric dipole which is capable of radiating
classical e-m waves. There is nothing remotely similar in the
Bohr-Sommerfeld model.
Martin Green
Jim Carr
<7bjjku$94d$1...@news.fsu.edu>...
|
| | ...
| | nothing at all about the Compton effect, which involves a free electron
| | rather than a bound one. Thus Schroedinger did not remove the need for
| | the photon derived by Einstein in 1905. Further, everything you say
| | about Schroedinger was true of the old Bohr-Sommerfeld model.
In article <022e8371$5f9cca00$0100a8c0@mgreen>
"Martin Green" <test...@pangea.ca> writes:
>
>That's crap.
Nope. Nothing you have said (all of which you snipped) has
any bearing at all on the Compton effect.
>In the Schroedinger model, the superposition of s and p states
>gives you an oscillating electric dipole which is capable of radiating
>classical e-m waves. There is nothing remotely similar in the
>Bohr-Sommerfeld model.
Similar? No, because Bohr-Sommerfeld is semi-classical. But you
did not talk about that in the statements you removed up above,
only about the quantization of the levels. And you might recall
that the photoelectric effect you were talking about concerns
the absorption, not emission, of radiation.
Now, what about the angular distribution for Compton scattering?
Martin Green
> Nothing you have said (all of which you snipped) has
> any bearing at all on the Compton effect.
>
That's right. You're the one who keeps trying to invoke the Compton effect,
which seems to prove my basic point....which is:
"You don't need "photons" to explain the photo-electric effect"
If I'm wrong, you should be able to argue me down on the basis of the
photo-electric effect, without trying to confuse the argument by bringing
in "evidence" from Compton scattering.
Martin Green
Fabio Di Teodoro
> That's right. You're the one who keeps trying to invoke the Compton effect,
> which seems to prove my basic point....which is:
>
> "You don't need "photons" to explain the photo-electric effect"
If you agree with the quantum picture that every material medium absorbs
and emits
^^^^^
radiation in the form of quanta, isn't this enough to say that radiation
is made itself of quanta ?
Indeed, do you think there can be some form of electromagnetic radiation
around which does not originate from quantized matter ?
Note that, processes such as brehmsstralung and Cerenkov radiation do
involve emission from apparently unbound charged particles (therefore
featuring a continuous spectrum) but can you really *fully* describe these
processes entirely within a classical picture ? You might happen to run
into some circular argument.
regards
Jim Carr
| any bearing at all on the Compton effect.
In article <022e83da$69517000$0100a8c0@mgreen>
"Martin Green" <test...@pangea.ca> writes:
>
>That's right. You're the one who keeps trying to invoke the Compton effect,
>which seems to prove my basic point....which is:
>
> "You don't need "photons" to explain the photo-electric effect"
Perhaps you did not read the articles you replied to? That is
a well-known fact, modulo some assumptions about 2-step processes,
as I stated. It just happens to be a detail that is ignored for
various reasons (some that I listed) in many physics books.
Perhaps you did not read the article you wrote? What you seem to
forget is you wrongly concluded that there was nothing other than
the photoelectric effect that required photons as the explanation.
In http://www.dejanews.com/getdoc.xp?AN=450421331 you wrote
"These are exactly the so-called paradoxes that were previously thought
to be explainable only by the existence of light particles, or 'photons'."
which is wrong for the reasons I stated. Your list of the "certain
facts" that were covered by
"Before 1926, photons were used to explain certain facts about the
exchange of energy between light and matter."
and could be explained by the Schroedinger approach omitted the Compton
effect, discovered in 1923, which cannot be explained that way.
Wake up and smell the coffee.
Pooua
>From: Fabio Di Teodoro <fdit...@brynmawr.edu>
>Date: 3/4/99 1:06 PM Central Standard Time
>Message-id: <Pine.SO4.4.02.990304...@ada.brynmawr.edu>
>
>
>
>
>> That's right. You're the one who keeps trying to invoke the Compton effect,
>> which seems to prove my basic point....which is:
>>
>> "You don't need "photons" to explain the photo-electric effect"
I'm sorry. I missed the earlier posts. How did you think you would explain the
photoelectric effect without photons? My understanding is that the Nobel Prize
committee found Einstein's explanation the first satisfactory one to come
along.
Richard Alexander
Richard's Electronic Kingdom
http://members.aol.com/pooua
Martin Green
>
> I'm sorry. I missed the earlier posts. How did you think you would
explain the
> photoelectric effect without photons? My understanding is that the Nobel
Prize
> committee found Einstein's explanation the first satisfactory one to come
> along.
>
>
> Richard Alexander
Einstein explained the photo-electric effect before we had a complete
theory (Schroedinger's equation) that allowed us to calculate the charge
distribution within an atom. The solution of Schroedinger's equation for
the hydrogen atom shows that it contains several stable states, and that a
mixture of any of these stable states gives a miniature antenna which
according to Maxwell's equations would radiate classical e-m waves. Not
only can it radiate, but it can absorb...but only if stimulated by the
right frequency.
Once we have this picture available to us, why do we need "photons" to
explain the photo-electric effect? And why do we continue publishing
textbooks which tell students that the photo-electric effect cannot be
explained by a wave theory of light?
Martin Green
PS PLEASE don't tell me that it's because the Compton effect proves that
there are photons. That may or may not be true, and it is totally
irrelevant to the question of why the textbooks invoke the photo-electric
effect to "prove" the existence of photons.
Martin Green
> >
> > "You don't need "photons" to explain the photo-electric effect"
>
Jim Carr responded:
> That is
> a well-known fact....it just happens to be a detail that is ignored for
> various reasons (some that I listed) in many physics books.
>
>
A well-known fact that just happens to be ignored in many physics books?
Nonsense. It is not ignored by physics books...on the contrary, it is
blatanltly denied by countless textbooks which continue to maintain against
all reason that the photo-electric effect "proves" the existence of
photons.
If it is such a well-known fact, why do so many textbooks maintain exactly
the opposite?
And it would be helpful if you would stop talking about the Compton
effect...whatever that may or may not prove, it has NOTHING to do with
whether or not the photo-electric effect is consistent with classical
electromagnetism.
Jim Carr
>
>And why do we continue publishing
>textbooks which tell students that the photo-electric effect cannot be
>explained by a wave theory of light?
I told you.
By the way, you need to realize that this is not the only way that
textbooks state the issue. Some treat it correctly.
You also brush over how you prevent an electron from picking up a
second push when the energy density is high in a classical calculation.
>PS PLEASE don't tell me that it's because the Compton effect proves that
>there are photons.
And that was not the explanation I gave you. Why the strawman?
The explanation was that the historically accurate fact that photons
were used to predict a relation later observed in the photoelectric
effect is easier to teach at an elementary level than a phenomenon
that depends on relativistic physics and a knowledge of graduate
level E+M to appreciate.
Jim Carr
Jim Carr responded:
|
| "Martin Green" <test...@pangea.ca> writes:
| >
| > "You don't need "photons" to explain the photo-electric effect"
|
| a well-known fact....it just happens to be a detail that is ignored for
| various reasons (some that I listed) in many physics books.
>
>A well-known fact that just happens to be ignored in many physics books?
Yes. For reasons I explained. Why do you ignore them?
>Nonsense. It is not ignored by physics books...on the contrary, it is
>blatanltly denied by countless textbooks which continue to maintain against
>all reason that the photo-electric effect "proves" the existence of
>photons.
Full quotation of an offending textbook passage, please.
You will recall that I told you of one freshman textbook that does
place it in its proper context.
>If it is such a well-known fact, why do so many textbooks maintain exactly
>the opposite?
I told you. Why did you snip it from your reply as if it had
not been posted? I conclude you are a troll.
>And it would be helpful if you would stop talking about the Compton
>effect...whatever that may or may not prove, it has NOTHING to do with
>whether or not the photo-electric effect is consistent with classical
>electromagnetism.
But it does tell us that classical electromagnetism is wrong.
Why have you ignored my article pointing out that the Compton Effect
falsifies your original claim? I conclude you are a troll.
me...@cars3.uchicago.edu
>
>>
>> I'm sorry. I missed the earlier posts. How did you think you would
>explain the
>> photoelectric effect without photons? My understanding is that the Nobel
>Prize
>> committee found Einstein's explanation the first satisfactory one to come
>> along.
>>
>>
>> Richard Alexander
>
>Einstein explained the photo-electric effect before we had a complete
>theory (Schroedinger's equation) that allowed us to calculate the charge
>distribution within an atom. The solution of Schroedinger's equation for
>the hydrogen atom shows that it contains several stable states, and that a
>mixture of any of these stable states gives a miniature antenna which
>according to Maxwell's equations would radiate classical e-m waves. Not
>only can it radiate, but it can absorb...but only if stimulated by the
>right frequency.
>
>explain the photo-electric effect?
Because they do explain it and explain a whole slew of additional
phenomena beyond. While it is true that Shroedinger equation with
continuous wave will give you the correct energy dependence for the
process, it'll fail to properly predict the angular distribution of
the photoelectrons. Not very relevant for emission from solid, very
relevant for photoemission in vacuum.
Your whole harping on the topic is, to put it mildly, silly. Like
many other ley poeters you've this mental image of "foundational
experiments" which establish the coorectness of scientific
expalnations and which should be revisited, over and over again. Not
so. The foundational experiments are interesting from science history
point of view but as far as actual support for theories they've long
since been replaced by way more detailed and precise data.
We know now that the measurements of light deflection by the Sun,
which were taken in 1917 as support for GR, were not precise anough to
serve as such support. Yet, it doesn't matter anymore since much more
precise measurements were performed since then. We know by now that
the calculations based on which Pluto was found were based on
erroneous data. Yet it doesn't matter since Pluto is there and we can
observe it.
If you want to question any theory or model, you should start with the
most recent, not the most ancient supporting information and see if
you can find holes in it. Arguing whether America exists based on
possible inaccuracies in Columbus' diaries is silly.
>And why do we continue publishing textbooks which tell students that
>the photo-electric effect cannot be explained by a wave theory of light?
There are textbooks beyond high school textbooks and there is physics
beyond high school physics.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
s...@microtec.net
"Martin Green" <test...@pangea.ca> wrote:
>
> >
> > I'm sorry. I missed the earlier posts. How did you think you would
> > explain the photoelectric effect without photons? My understanding is that
> > the Nobel Prize committee found Einstein's explanation the first
> > satisfactory one to come along.
> >
> >
> > Richard Alexander
>
> Einstein explained the photo-electric effect before we had a complete
> theory (Schroedinger's equation) that allowed us to calculate the charge
> distribution within an atom.
> The solution of Schroedinger's equation for
> the hydrogen atom shows that it contains several stable states, and that a
> mixture of any of these stable states gives a miniature antenna which
> according to Maxwell's equations would radiate classical e-m waves. Not
> only can it radiate, but it can absorb...but only if stimulated by the
> right frequency.
From what I understand, there is no way the hydrogen atom's electron will not
absorb energy from any photon that hits it. It will however immediately
re-spit a photon if the imparted energy is not sufficient to momentarily move
it to the next quantized upper level, or completely escape if the imparted
energy is sufficient.
> Once we have this picture available to us, why do we need "photons" to
> explain the photo-electric effect?
I really am curious as to how you would explain the photo-electric effect
without the "photon idea".
André Michaud
Service de Recherche Pédagogique http://www.microtec.net/~srp/
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Martin Green
.
Jim Carr <j...@ibms48.scri.fsu.edu> wrote in article
<7bn5pq$77n$1...@news.fsu.edu>...
>
>
>
> I told you. Why did you snip it from your reply as if it had
> not been posted? I conclude you are a troll.
>
(snip)
>
> Why have you ignored my article pointing out that the Compton Effect
> falsifies your original claim? I conclude you are a troll.
>
Yes, Jim. I am a troll.
Martin Green
Martin Green
> >> I'm sorry. I missed the earlier posts. How did you think you would
> >explain the
> >> photoelectric effect without photons? My understanding is that the
Nobel
> >Prize
> >> committee found Einstein's explanation the first satisfactory one to
come
> >> along.
> >>
> >>
> >> Richard Alexander
> >
> >Einstein explained the photo-electric effect before we had a complete
> >theory (Schroedinger's equation) that allowed us to calculate the charge
> >distribution within an atom. The solution of Schroedinger's equation for
> >the hydrogen atom shows that it contains several stable states, and that
a
> >mixture of any of these stable states gives a miniature antenna which
> >according to Maxwell's equations would radiate classical e-m waves. Not
> >only can it radiate, but it can absorb...but only if stimulated by the
> >right frequency.
> >
> >explain the photo-electric effect?
What follows is Mati Meron's "rebuttal" to my claims about the
photo-electric effect. Please notice that he does not deny the truth of my
arguments, he merely says that my interest in this question is "silly".
Martin Green
Jim Carr
>
>What follows is Mati Meron's "rebuttal" to my claims about the
>photo-electric effect. Please notice that he does not deny the truth of my
>arguments, he merely says that my interest in this question is "silly".
Please note that Meron said much more than that.
His comments were correct, because your interest in the question was,
as described in http://www.dejanews.com/getdoc.xp?AN=450421331, based
on the claim that the photoelectric effect was the only experimental
support for photons -- a false claim.
Your comments above, ignoring Mati's remarks about broader support
for photons in other areas of physics, provides additional evidence
of your sincerity -- if your confession
)Yes, Jim. I am a troll.
)
)Martin Green
a bit earlier was not enough.
Robert Moestam
I just have to enter the discussion without having retrieved the articles I
have ordered concerning the Compton effect.
I think one of the questions one might ask here is wether the e-m field is
something 'physical', with intrinsic properties, or if it is just a
description of how charged particles interact over long range. If it is to
be regarded as only a description of the long range interaction then of
course all properties of the e-m field is quantified as well. But if the
field is something 'real', then I don't see how one can argue that all
aspects of e-m field can be investigated by looking at interaction between
matter and e-m field which has certain properties, well described with the
QED.
What I like to know is if someone can give an example on when it is not
sufficient to regard the e-m field as only a description of long range
interaction between charged particles. I have to admit that the view of the
e-m field as something other than physical disturbs me, but I still don't
see how one can decouple the physical effects from the e-m field from the
interaction. What is the physical evedence of that e-m field is something of
its own?
So trolls and goblins. Please attack the provocation and please kill the
question so that my nightmares can change subject ;=)
Martin Green
Martin Green wrote:
> >
> > Einstein explained the photo-electric effect before we had a complete
> > theory (Schroedinger's equation) that allowed us to calculate the
charge
> > distribution within an atom.
> > The solution of Schroedinger's equation for
> > the hydrogen atom shows that it contains several stable states, and
that a
> > mixture of any of these stable states gives a miniature antenna which
> > according to Maxwell's equations would radiate classical e-m waves.
Not
> > only can it radiate, but it can absorb...but only if stimulated by the
> > right frequency.
>
> From what I understand, there is no way the hydrogen atom's electron will
not
> absorb energy from any photon that hits it. It will however immediately
> re-spit a photon if the imparted energy is not sufficient to momentarily
move
> it to the next quantized upper level, or completely escape if the
imparted
> energy is sufficient.
>
>
> without the "photon idea".
>
> André Michaud
>
> Service de Recherche Pédagogique http://www.microtec.net/~srp/
>
Since Jim Carr agrees with me that "it is a well-konwn fact" that you can
explain the photo-electric effect without photons, maybe he should explain
how it works. But I know he is very busy with numerous other discussions.
With regard to your points mentioned above, I'm not saying you CAN'T use
"photons" to explain it...I'm just saying you don't HAVE to. Jim Carr and
Mati Meron are annoyed with me because I won't admit that the Compton
effect proves that photons exist. I don't want to get into that argument
because I don't know enough quantum mechanics to understand whether the
Compton effect does or does not prove what Meron/Carr claim it does. I
have, on the other hand, studied enough electromagnetic theory and quantum
mechanics to be able to caclulate what happens when you stimulate a
hydrogen atom with an ordinary, classical oscillating e-field. And I get
all the "paradoxical" effects that are normally explained by "photons":
transition depends on frequency, not energy
below a minimum frequency there is no transition no matter how much energy
transition always absorbs a definite amount of energy
So I think it's wrong to teach that the photo-electric effect cannot be
explained with classical electromagnetism. It is a natural consequence of
combining Maxwell's and Schroedinger's equations...something that
Carr/Meron will not deny.
Martin Green
Martin Green
Martin Green wrote, referring to Mati Meron's "rebuttal":
> >
> >Please notice that he does not deny the truth of my
> >arguments, he merely says that my interest in this question is "silly".
>
Jim Carr replied:
> Please note that Meron said much more than that.
>
Yeah, stuff about Christopher Columbus and the orbit of Pluto. So what?
Fabio Di Teodoro
On Fri, 5 Mar 1999, Robert Moestam wrote:
> I think one of the questions one might ask here is wether the e-m field is
> something 'physical', with intrinsic properties, or if it is just a
> description of how charged particles interact over long range.
the difference between the two things is not immediately evident, unless
you allude to the existence of an em field without sources, which i do not
know what could be.
(this sounds like one of those questions such as "what's the difference
between what it is and what it does")
> If it is to
> be regarded as only a description of the long range interaction then of
> course all properties of the e-m field is quantified as well. But if the
> field is something 'real', then I don't see how one can argue that all
> aspects of e-m field can be investigated by looking at interaction between
> matter and e-m field which has certain properties, well described with the
> QED.
it's not clear what you mean here.
Quantum Electrodynamics *is* a quantum description of the electromagnetic
field which, in this framework, is, of
course, quantized (unless by "quantified" you mean something else). And
QED is one of the best theories we have nowadays,
able to make predictions on the values of certain physical observables
with stunning precision, in agreement with very accurate experimental
results.
For several problems concerning radiation-matter interactions, however, a
semiclassical approach which treats the em field roughly as Maxwell did,
is used for computational convenience, provided that the approximation is
consistent with the experiment.
> What I like to know is if someone can give an example on when it is not
> sufficient to regard the e-m field as only a description of long range
> interaction between charged particles. I have to admit that the view of the
> e-m field as something other than physical disturbs me
An interaction to me is something very physical.
> but I still don't
> see how one can decouple the physical effects from the e-m field from the
> interaction. What is the physical evedence of that e-m field is something of
> its own?
yes, it's just like one of those questions.
regards
me...@cars3.uchicago.edu
photoelectric effect using a continuous wave formalism, but not for
the momentum distribution. And that's just one little example.
As for the stuff about Columbus, it was there for a purpose. You
should read it again.
drj...@my-dejanews.com
>
> Since Jim Carr agrees with me that "it is a well-konwn fact" that you can
> explain the photo-electric effect without photons, maybe he should explain
> how it works. But I know he is very busy with numerous other discussions.
> With regard to your points mentioned above, I'm not saying you CAN'T use
> "photons" to explain it...I'm just saying you don't HAVE to. Jim Carr and
> Mati Meron are annoyed with me because I won't admit that the Compton
> effect proves that photons exist. I don't want to get into that argument
> because I don't know enough quantum mechanics to understand whether the
> Compton effect does or does not prove what Meron/Carr claim it does. I
> have, on the other hand, studied enough electromagnetic theory and quantum
> mechanics to be able to caclulate what happens when you stimulate a
> hydrogen atom with an ordinary, classical oscillating e-field. And I get
> all the "paradoxical" effects that are normally explained by "photons":
>
> transition depends on frequency, not energy
> below a minimum frequency there is no transition no matter how much
energy
> transition always absorbs a definite amount of energy
>
> So I think it's wrong to teach that the photo-electric effect cannot be
> explained with classical electromagnetism. It is a natural consequence of
> combining Maxwell's and Schroedinger's equations...something that
> Carr/Meron will not deny.
>
> Martin Green
>
I've been following this discussion, periodically, with bemused interest.
There are a couple of points in your arguement that jump out at me. First of
all, what is a photon other than an oscillating e-field? Secondly, someone
who speaks with such authority on this subject should know that the energy of
a photon is a function of frequency.
You should keep in mind that the "classical" solution is often a subset of the
QM solution. There is no reason to be surprised at "similarities" between the
classical solution and the classical solution.
s...@microtec.net
"Martin Green" <test...@pangea.ca> wrote:
>
> Martin Green wrote:
> > >
> > > Einstein explained the photo-electric effect before we had a complete
> > > theory (Schroedinger's equation) that allowed us to calculate the charge
> > > distribution within an atom.
> > > The solution of Schroedinger's equation for
> > > the hydrogen atom shows that it contains several stable states, and that a
> > > mixture of any of these stable states gives a miniature antenna which
> > > according to Maxwell's equations would radiate classical e-m waves.
> > > Not only can it radiate, but it can absorb...but only if stimulated by the
> > > right frequency.
> >
> Andre Michaud replied:
> > From what I understand, there is no way the hydrogen atom's electron will
> > not absorb energy from any photon that hits it. It will however immediately
> > re-spit a photon if the imparted energy is not sufficient to momentarily
> > move it to the next quantized upper level, or completely escape if the
> > imparted energy is sufficient.
> >
> > without the "photon idea".
> >
> Since Jim Carr agrees with me that "it is a well-konwn fact" that you can
> explain the photo-electric effect without photons, maybe he should explain
> how it works. But I know he is very busy with numerous other discussions.
> With regard to your points mentioned above, I'm not saying you CAN'T use
> "photons" to explain it...I'm just saying you don't HAVE to.
That's right. With the maths available since Schrödinger and many others, it
is true that the photo-electric effect, as well as all other aspects of the
behaviour of EM field interaction with electrons and atoms in general can be
very precisely described.
Personnally, however, I have become very conscious, maybe too much so :o], of
the difference between the mathematical _description_ that can be made and
the _real_ fundamental particles that are out there, that already existed
before we had maths to kind of measure their movements and interactions.
To me, the mathematical descriptions your are talking about are fine. But
what I am constantly trying to understand, are the real particles and their
real interactions..
> Jim Carr and
> Mati Meron are annoyed with me because I won't admit that the Compton
> effect proves that photons exist. I don't want to get into that argument
> because I don't know enough quantum mechanics to understand whether the
> Compton effect does or does not prove what Meron/Carr claim it does.
I can tell you that you presently need not dig into QM to understand the
Compton effect, but you would have to locate a reference book that dates back
at least to the sixties or even earlier, for it not to describe it through
quantum mechanics.
Very simply put, photons (if one accepts the idea, that is) kind of interact
with electrons the way billiard balls interact with each other (very bad
parallel, obviously, but the general idea is there) There is exchange of
energy at each encounter, generally speaking, the photon loosing some to the
electron. This effect has been exhaustively explored in the first half of the
century and I am certain you would find it interesting if you had the chance
and time to locate a book that describe it without the filter of quantum
mechanics.
> I have, on the other hand, studied enough electromagnetic theory and quantum
> mechanics to be able to caclulate what happens when you stimulate a
> hydrogen atom with an ordinary, classical oscillating e-field. And I get
> all the "paradoxical" effects that are normally explained by "photons":
I agree that it is quite possible. But again here, I like to visualise the
_real_ stuff behind the mathematical description.
> transition depends on frequency, not energy
Aren't both directly linked?
> below a minimum frequency there is no transition no matter how much
> energy
> transition always absorbs a definite amount of energy
Agreed, that below a given minimum, there is no change to next quantized
level, but from what I understand, if a lower energy photon connects with an
electron, even though it will _not_ result in a jump to next level, it may
result for example in a change in orientation of the orbit of the electron on
the same level. Quite a bit of energy can be expended right there without any
measurement showing the actual transfer of energy, not even an increase in
speed of the electron, except maybe the rebounding photon coming out with
less energy than when it collided with the electron (Compton effect).
> So I think it's wrong to teach that the photo-electric effect cannot be
> explained with classical electromagnetism. It is a natural consequence of
> combining Maxwell's and Schroedinger's equations...something that
> Carr/Meron will not deny.
I see your point now. Thank you for clarifying.
Fundamentally, I agree that it is good to explain to youth that there may be
more than one way to skin a cat, so to speak, even in fundamental physics.
Regards
André Michaud
Service de Recherche Pédagogique http://www.microtec.net/~srp/
-----------== Posted via Deja News, The Discussion Network ==----------
me...@cars3.uchicago.edu
>.> >
>> >Martin Green wrote, referring to Mati Meron's "rebuttal":
>> >> >
>> >> >Please notice that he does not deny the truth of my
>> >> >arguments, he merely says that my interest in this question is
>"silly"
>of Pluto. So what?
>> >
>
>Meron replied:
>>
>> As for the stuff about Columbus, it was there for a purpose. You
>> should read it again.
>>
>
>teach "correct" physics by means of an "incorrect" argument....it doesn't
>matter how we know that light is made of photons, the important thing is
>that we teach students to believe that light IS made of photons. If the
>easiest way to do this is to cobble together a convincing fable based on
>half-baked physics and sanitized history, so be it.
>
I'm not justifying anything. I'm explaining to you that this whole
thinking along the lines of "foundational experiments" is plain wrong.
That's not how physics works.
The photoelectric effect was important in the sense that it directed
attention to phenomena which had no classical explanation. It has
been explained through the introduction of photons. This, in turn,
gave rise to predictions of other phenomena which were subsequently
observed. By the time it turned out that within quantum mechanics you
can explain some (but only some) aspects of the photoelectric effect
without using photons, it didn't matter anymore since plenty of other
confirmations did exist.
Furthermore, note what I wrote, "some aspects". You can explain the
energy dependence, but not the momentum distribution. Now, having a
model which fits all I observe, why would I go back to one which fits
only part.
Now, if what you want to complain about is just the fact that
textbooks aren't as good as they could've been, I fully agree with
you. They've a problem, though,when trying to explain modern physics
concepts to high school students lacking the appropriate background to
understand most of the stuff. So, they tend to gravitate (on the high
school level) to those few things which can be explained using not
much more than algebra. And, mind you, they're not using an
*incorrect* argument in this case. The photoelectric effect does
provide support to the notion of photons. This is by no means the
full support, of course.
s...@microtec.net
drj...@my-dejanews.com wrote:
> You should keep in mind that the "classical" solution is often a subset of the
> QM solution.
Interesting.
Could you be more specific?
What classical solutions are presently subsets of the QM solution and in what
way?
Robert Moestam
Fabio Di Teodoro wrote in message ...
Is the e-m field only a description of the long range interaction between
charged particles.
>(this sounds like one of those questions such as "what's the difference
>between what it is and what it does")
Yes, you are right, got me thinking. Does an e-m field anything on its own
or is the only use of it the description of the interaction between charged
particles?
If it doesn't have 'a life of its own', then I think the question, if the
e-m field has to be quantified, is answered, since the interaction between
any particles are always quantified (well described by QED). If not, I think
you can regard the QED as a matter e-m field interaction description.
If the only meaning of photons is charged particle interaction, then please
explain the difference between a photon and a virtual photon to me. Is the
only difference that a virtual photon never leaves the system under study
and a 'real' photon interact leaves the system and sometimes interact with
our measuring equipment?
Jim Carr
"Martin Green" <test...@pangea.ca> wrote:
}
} Jim Carr and
} Mati Meron are annoyed with me because I won't admit that the Compton
} effect proves that photons exist.
Proves? Another straw man. I said that the Compton Effect cannot
be explained by the quantum-matter plus classical-electromagnetism
argument you gave. Since it was your model, I don't see why you
make the excuse below that you can't calculate in it.
} I don't want to get into that argument
} because I don't know enough quantum mechanics to understand whether the
} Compton effect does or does not prove what Meron/Carr claim it does.
In article <7bq6th$v18$1...@nnrp1.dejanews.com>
s...@microtec.net writes:
>
>I can tell you that you presently need not dig into QM to understand the
>Compton effect, but you would have to locate a reference book that dates back
>at least to the sixties or even earlier, for it not to describe it through
>quantum mechanics.
What you really need to look at is the classical prediction, which
is found in E+M textbooks because it is classical radiation theory.
However, as I told him long ago, the best thing to do is just read
Compton's paper. It is available in the Phys. Rev. Centennial
collection that many libraries have even if they do not have the
journal going back to 1923. The paper is Phys. Rev. 21, 483 (1923)
and appears on page 37 of the collection. Look at figure 7 in
particular, which shows the angular distribution compared to the
classical prediction.
Jim Carr
>
>What classical solutions are presently subsets of the QM solution and in what
>way?
The trajectory of a particle in a multi-wire detector.
See qualitative discussion in Schiff and references therein.
s...@microtec.net
j...@ibms48.scri.fsu.edu (Jim Carr) wrote:
> s...@microtec.net writes:
> >
> >I can tell you that you presently need not dig into QM to understand the
> >Compton effect, but you would have to locate a reference book that dates back
> >at least to the sixties or even earlier, for it not to describe it through
> >quantum mechanics.
>
> What you really need to look at is the classical prediction, which
> is found in E+M textbooks because it is classical radiation theory.
>
> However, as I told him long ago, the best thing to do is just read
> Compton's paper. It is available in the Phys. Rev. Centennial
> collection that many libraries have even if they do not have the
> journal going back to 1923. The paper is Phys. Rev. 21, 483 (1923)
> and appears on page 37 of the collection. Look at figure 7 in
> particular, which shows the angular distribution compared to the
> classical prediction.
If he has access to the real deal, I agree.
From what I gathered, Compton understood all aspects of the question, and
nothing more was added afterwards.
Larry Mead
: >
: > I'm sorry. I missed the earlier posts. How did you think you would
: explain the
: > photoelectric effect without photons? My understanding is that the Nobel
: Prize
: > committee found Einstein's explanation the first satisfactory one to come
: > along.
: >
: >
: > Richard Alexander
: Einstein explained the photo-electric effect before we had a complete
: theory (Schroedinger's equation) that allowed us to calculate the charge
: distribution within an atom. The solution of Schroedinger's equation for
: the hydrogen atom shows that it contains several stable states, and that a
: mixture of any of these stable states gives a miniature antenna which
: according to Maxwell's equations would radiate classical e-m waves. Not
: only can it radiate, but it can absorb...but only if stimulated by the
: right frequency.
: Once we have this picture available to us, why do we need "photons" to
: explain the photo-electric effect? And why do we continue publishing
: textbooks which tell students that the photo-electric effect cannot be
: explained by a wave theory of light?
: Martin Green
: PS PLEASE don't tell me that it's because the Compton effect proves that
: there are photons. That may or may not be true, and it is totally
: irrelevant to the question of why the textbooks invoke the photo-electric
: effect to "prove" the existence of photons.
The need to quantize electrodynamics was argued thoroughly by Jaynes,
Lamb and lots of quantum optics folks in the 70's. It turns out, there
is one and only one effect which requires this: spontaneous emission.
For the details, see "The Quantum Vacuum" by Milonni.
--
Lawrence R. Mead Ph.D. (Lawren...@usm.edu)
Eschew Obfuscation! Espouse Elucidation!
www-dept.usm.edu/~physics/mead.html
Martin Green
> The need to quantize electrodynamics was argued thoroughly by Jaynes,
> Lamb and lots of quantum optics folks in the 70's. It turns out, there
> is one and only one effect which requires this: spontaneous emission.
>
Not the Compton effect???
>
> For the details, see "The Quantum Vacuum" by Milonni.
>
>
> --
> Lawrence R. Mead Ph.D. (Lawren...@usm.edu)
> Eschew Obfuscation! Espouse Elucidation!
> www-dept.usm.edu/~physics/mead.html
>
Thanks...I will see if I can find this book at the local university
library.
I recall reading something by Einstein written in the very early days,
where he placed great significance on the fact that radiation must be
emitted with a definite recoil to the atom rather than as spherically
symmetrical waves. I could not follow the argument as to why this had to
be. Are you able to provide any insight?
Martin Green
Martin Green
> without using photons, it didn't matter anymore since plenty of other
> confirmations did exist.
Plenty? Larry Mead does not agree with you. Please note his comments
elsewhere in this thread:
"The need to quantize electrodynamics was argued thoroughly by Jaynes,
Lamb and lots of quantum optics folks in the 70's. It turns out, there
is one and only one effect which requires this: spontaneous emission.
For the details, see "The Quantum Vacuum" by Milonni.
--
Lawrence R. Mead Ph.D. (Lawren...@usm.edu)
Eschew Obfuscation! Espouse Elucidation!
www-dept.usm.edu/~physics/mead.html"
>
> Now, if what you want to complain about is just the fact that
> textbooks aren't as good as they could've been, I fully agree with
> you........ mind you, they're not using an
> *incorrect* argument in this case. The photoelectric effect does
> provide support to the notion of photons.
No, Mati, the textbooks do not merely argue that the photo-electric effect
"provides support to the notion". They argue that the photo-electric effect
CANNOT be reconciled with a wave theory of light.
Martin Green
me...@cars3.uchicago.edu
>>............ within quantum mechanics you
>> can explain some (but only some) aspects of the photoelectric effect
>> without using photons, it didn't matter anymore since plenty of other
>> confirmations did exist.
>
>Plenty? Larry Mead does not agree with you.
Then talk it over with him.
>No, Mati, the textbooks do not merely argue that the photo-electric effect
>"provides support to the notion". They argue that the photo-electric effect
>CANNOT be reconciled with a wave theory of light.
>
I don't recall seeing many textbooks claiming it but then I haven't
seen many high school textbooks lately. Certainly advanced textbooks
aren't claiming anything of the sort.
Fabio Di Teodoro
> Plenty? Larry Mead does not agree with you. Please note his comments
> elsewhere in this thread:
>
> "The need to quantize electrodynamics was argued thoroughly by Jaynes,
> Lamb and lots of quantum optics folks in the 70's.
so Feynman, Schwinger and Tomonaga had got the Nobel prize in the 1965 for
having developed something not clearly "needed"...
Is that what you mean ?
drj...@my-dejanews.com
s...@microtec.net wrote:
> In article <7bq7b1$vd6$1...@nnrp1.dejanews.com>,
> drj...@my-dejanews.com wrote:
>
> > You should keep in mind that the "classical" solution is often a subset of
the
> > QM solution.
>
> Interesting.
>
> Could you be more specific?
>
> way?
>
>
> Service de Recherche Pédagogique http://www.microtec.net/~srp/
>
> -----------== Posted via Deja News, The Discussion Network ==----------
> http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
>
Actually, I can. I was first taught this in grad school in classical
mechanics. The case I'm thinking of, however, is in my notes from a lasers
class. The classical approach only gets you so far. (you start with a simple
harmonic oscillator). You have to resort to QM to get the right answer. I
wrote down you're email address. I'll dig this crap out an pass on more
detail.
po...@aol.com
me...@cars3.uchicago.edu wrote:
> In article <022e84ce$9bd1e6a0$0100a8c0@mgreen>, "Martin Green"
<test...@pangea.ca> writes:
> >
> >Martin Green wrote, referring to Mati Meron's "rebuttal":
> >> >
> >> >Please notice that he does not deny the truth of my
> >>
> >
> >Jim Carr replied:
> >
> >> Please note that Meron said much more than that.
> >>
> >Yeah, stuff about Christopher Columbus and the orbit of Pluto. So what?
> >
> Nope, stuff about how you can account for the energy dependence of the
> photoelectric effect using a continuous wave formalism, but not for
> the momentum distribution. And that's just one little example.
So, how does the momentum distribution compare between classic theory
and Einstein's theory? While you or Carr are at it, could you explain
what you mean by "momentum distribution"? I imagine that you are talking
about the velocity at which electrons leave the atom, but I don't recall
the specifics, and the specifics would be most useful in this discussion.
Richard Alexander
Richard's Electronic Kingdom
http://members.aol.com/pooua
s...@microtec.net
drj...@my-dejanews.com wrote:
> In article <7br67a$nde$1...@nnrp1.dejanews.com>,
> s...@microtec.net wrote:
> > In article <7bq7b1$vd6$1...@nnrp1.dejanews.com>,
> > drj...@my-dejanews.com wrote:
> >
> > > You should keep in mind that the "classical" solution is often a subset of
> the
> > > QM solution.
> >
> > Interesting.
> >
> > Could you be more specific?
> >
> > What classical solutions are presently subsets of the QM solution and in
> > what way?
>
> Actually, I can. I was first taught this in grad school in classical
> mechanics. The case I'm thinking of, however, is in my notes from a lasers
> class. The classical approach only gets you so far. (you start with a simple
> harmonic oscillator). You have to resort to QM to get the right answer. I
> wrote down you're email address. I'll dig this crap out an pass on more
> detail.
Hope you do, thanks! The stuff really is of interest to me.
André Michaud
Service de Recherche Pédagogique http://www.microtec.net/~srp/
-----------== Posted via Deja News, The Discussion Network ==----------
me...@cars3.uchicago.edu
>In article <F85Do...@midway.uchicago.edu>,
> me...@cars3.uchicago.edu wrote:
>> >
>> Nope, stuff about how you can account for the energy dependence of the
>> photoelectric effect using a continuous wave formalism, but not for
>> the momentum distribution. And that's just one little example.
>
>So, how does the momentum distribution compare between classic theory
>and Einstein's theory?
It's not so much classical versus Einstein as QM with classical waves
versus QM with photons, in this case. Purely classical model will not
yield the photoelectric effect.
>While you or Carr are at it, could you explain
>what you mean by "momentum distribution"? I imagine that you are talking
>about the velocity at which electrons leave the atom, but I don't recall
>the specifics, and the specifics would be most useful in this discussion.
What I really have in mind is the angular distribution, i.e. the
distribution of the directions of the momenta (or velocities). In the
case of a classical wave this distribution is strongly peaked in the
transverse direction (since that's the energy of the electric
field). Quantized EM wave (i.e. photons) gives a broader distribution
which, for high energy photons, is strongly pushed towards forward
angles. OF course, for sufficiently high photon energies the results
approach thos of Compton scattering from a free electron.
Aleksandr Timofeev
> However, as I told him long ago, the best thing to do is just read
> Compton's paper. It is available in the Phys. Rev. Centennial
> collection that many libraries have even if they do not have the
> journal going back to 1923. The paper is Phys. Rev. 21, 483 (1923)
> and appears on page 37 of the collection. Look at figure 7 in
> particular, which shows the angular distribution compared to the
> classical prediction.
>
Whether you can cite (indicate) examples of Compton's experiments on real
free electrons.
I consider an error physical hypothesis, which considers poorly bound
electrons in substance as free electrons. The poorly bound electrons in
substance belong to a quantum system, which cannot be eliminated from a model
of experiment.
---
Aleksandr Timofeev a_n_ti...@my-dejanews.com
http://solar.cini.utk.edu/~russeds/unknown/astrochem/
Martin Green
> So, how does the momentum distribution compare between classic theory
> what you mean by "momentum distribution"? I imagine that you are talking
> about the velocity at which electrons leave the atom, but I don't recall
> the specifics, and the specifics would be most useful in this discussion.
>
I wouldn't worry too much about what Meron/Carr thinks. He seems to be
making up this stuff about angular distributions as he goes along. The
reference cited elsewhere in this thread by Larry Mead makes it pretty
clear that a wave treatment of light has no problem with either the
photo-electric effect OR the Compton effect.
Martin Green
Reference:
The need to quantize electrodynamics was argued thoroughly by Jaynes,
Lamb and lots of quantum optics folks in the 70's. It turns out, there
is one a nd only one effect which requires this: spontaneous emission.
Jim Carr
>
>The need to quantize electrodynamics was argued thoroughly by Jaynes,
>Lamb and lots of quantum optics folks in the 70's. It turns out, there
>
>For the details, see "The Quantum Vacuum" by Milonni.
It figures that it would be on loan.
Do you have a reference to an article that treats Compton
classically and points out the error in Compton's calculation
of the angular distribution?
Fabio Di Teodoro
On 8 Mar 1999, Jim Carr wrote:
> In article <7brmkl$fcb$8...@thorn.cc.usm.edu>
> lrm...@orca.st.usm.edu (Larry Mead) writes:
> >
> >The need to quantize electrodynamics was argued thoroughly by Jaynes,
> >Lamb and lots of quantum optics folks in the 70's. It turns out, there
> >is one and only one effect which requires this: spontaneous emission.
>
> Do you have a reference to an article that treats Compton
> classically and points out the error in Compton's calculation
> of the angular distribution?
AFAIK, the classical treatment (em wave on free particle)
is the Thomson scattering, which however fails at high frequencies where
the photon momentum becomes comparable or larger than mc, m being the mass
of the charged particle. Barkla saw the departure from the Thomson
scattering already in the 1909, using X-rays (even though he didn't know
their wavelength, since that became clear only with the von Laue
experiments on crystals (1912 ?))
The departure from the classic scattering, in
this regime of frequencies, manifests itself as a too large differential
cross section at large scattering angles which is not consistent with the
experiment and gets fixed by the Compton scattering model.
I do not know if this relates to your comment.
Also I am not aware of an error in the Compton's calculation. To my
best knowledge, there is rather an inaccuracy inherent in his model which
results from neglecting the spin.
If i well remember (...Jackson..?) including the electron's
magnetic moment, for example, leads to the Klein-Nishina cross section for
the Compton scattering, which especially at large scattering angles,
should be slightly larger than the spin-free cross-section. I am sure,
anyway, that this is treated in the Bjorken and Drell's book.
I guess there are also effects that arise from being the charges
not really free in these experiments but rather bound to, at least,
Coulombic potentials. Unfortunately, i don't know much about this type of
"fine tuning" of the original works.
I hope, but I am not sure, this helps somewhat.
Anyway, i still don't understand in which sense the need of quantizing
electrodynamics is said to be something that belong to the 70's.
As i already said in a previous post, i thought Feynman, Schwinger and
Tomonaga got the Nobel prize for QED in the 1965 and that, at that time,
the theory had already proved to be consistent with a certain number of
experimental facts.
best regards.
Fabio Di Teodoro
>
> On 8 Mar 1999, Jim Carr wrote:
>
> > In article <7brmkl$fcb$8...@thorn.cc.usm.edu>
> > lrm...@orca.st.usm.edu (Larry Mead) writes:
> > >
> > >The need to quantize electrodynamics was argued thoroughly by Jaynes,
> > >Lamb and lots of quantum optics folks in the 70's. It turns out, there
> > >is one and only one effect which requires this: spontaneous emission.
> >
> > Do you have a reference to an article that treats Compton
> > classically and points out the error in Compton's calculation
> > of the angular distribution?
i guess i now know what you meant, for the first.
You should take a look at
"Atoms and light" by John N. Dodd (Plenum Press, New York, 1991).
In Chapter 6, the Compton scattering is treated in an entirely classical
way, without using energy and momentum conservation, but just standard
classical em + relativistic *kinematics*, by the picture of a
circularly polarized em wave impinging upon a charged particle.
The calculation is based on deriving a steady-state solution for the
down-stream motion of the particle which is superimposed to the
constant rotation at the frequency of the passing wave.
I haven't read the analysis in detail, but my first impression is that
it is quite clever.
It is, especially in light of the comments at page 55, apparent that the
standard Compton effect, i.e. the one the Compton explained using the
notion of photon, does not actually *need* this notion.
So, according to the author, the standard (spin-free) Compton effect
cannot be invoked to argue the existence of photons.
I am sorry if some of you do not have access to this book, unfortunately
the author does not provide other references in which the calculation has
been presented.
That was very educational for me. I confess i did not know about this.
In Ch.12, p.144 the author also presents a critical treatment of the
photoelectric effect. Again, he does not only address the issue of
the "photonless" interpretation of the experiment, but also adds that
some subtle question which, at first sight, *could* lead to a photon
interpretation, actually does not.
I finally suggest the readig of Chapter 14, which contains quite of a
compelling criticism on the way this subject is _taught_.
Now i agree with Mati Meron on the fact that many other experiments that
support the photon picture and the QED exist, of course.
I never thought this was in question.
However, i must note that, at least from what i understood from this
thread, the main concern made by Martin Green was an "educational" one
and, from this point of view, i have frankly to agree with him.
Indeed, i can see in the process of *learning* about this subject, the
student risks to be exposed to some evident misconception.
Later on, she/he will (hopefully) learn more on the subject and gain a
deeper insight. But this does not justify the initial misconception.
The notion of photon is introduced quite early in the intro physics
courses. It may be true that some more accurate book does avoid to
invoke the mere, plain photoelectric effect as the evidence of the
photon's existence (although, honestly, i do know several intro-physics
books which carry this misconception).
However, at this point, i have no problem to recognize that this is the
first time i have read about a classical derivation of the Compton
effect. Honestly, i think that *really* this is not included in any
intro-level book, the ones student read while they are learning physics.
I think they all present, in a more or less open way, the standard Compton
effect as one compelling "demonstration" of the existence of photon and
not as an experiment that "supports" the photon's picture.
Please provide references, in the case my claim happens to be wrong.
Maybe the details of this argument may be of interest only to people who
study history of physics, considering that the photon picture has proven
to be successful in so many applications.
Needless to say, the very Compton effect has aspects that are not
correctly addressed by the semiclassical picture (e.g. effect of the
particle's spin) and lead, for example, to the QED Klein-Nishina
differential cross section as i said elsewhere (i hope at least this does
not have a classical derivation... :-) )
However, why should a student be told, even if it's just the beginning of
her/his studies in physics, that "the photon's existence has been
proved by the mere Compton effect" when it's not the case, in that the
Compton lends itself both to a classical derivation and "corpuscolar"
derivation ?
Is it true that the Compton effect is understood by so many students as a
"test" for the existence of photons ?
Thank you all for your contributions.
best regards
Martin Green
effort I have put you through, but I hope you found it to be an interesting
excercise.
>
> Now i agree with Mati Meron on the fact that many other experiments that
> support the photon picture and the QED exist, of course.
> I never thought this was in question.
> However, i must note that, at least from what i understood from this
> thread, the main concern made by Martin Green was an "educational" one
> and, from this point of view, i have frankly to agree with him.
>
> Indeed, i can see in the process of *learning* about this subject, the
> student risks to be exposed to some evident misconception.
> Later on, she/he will (hopefully) learn more on the subject and gain a
> deeper insight. But this does not justify the initial misconception.
>
It certainly is the educational aspect of this quesiton that concerns me
the most. On the superficial level, the student is taught an argument that
is incorrect. "Hopefully" he will later learn the truth...however, the tone
of the discussion in this thread makes it clear that it is a very rare
occasion indeed when a physicist questions his own early indoctrination
into the photon "ideology".
What should be more disturbing is what this discussion reveals about the
culture of physics education. Students are led to believe that physics is
built on an almost axiomatic approach...consider the many advanced texts
which develop QED by "postulating" the creation/annihilation operators and
"derviving" everything else. The overwhelming impression is that the
student is being given a rigorously constructed logical edifice. It is
impossible for the student to understand all the steps, so he "accepts" the
truth based on the authority of the professor. Over the course of several
years of graduate study, the distinction becomes blurred in his mind
between what he has truly "understood" and what he has merely accepted on
authority. Eventually, he can no longer tell the difference between the two
kinds of "learning".
Martin Green
me...@cars3.uchicago.edu
>
>"Atoms and light" by John N. Dodd (Plenum Press, New York, 1991).
>
>In Chapter 6, the Compton scattering is treated in an entirely classical
>way, without using energy and momentum conservation, but just standard
>classical em + relativistic *kinematics*, by the picture of a
>circularly polarized em wave impinging upon a charged particle.
>The calculation is based on deriving a steady-state solution for the
>down-stream motion of the particle which is superimposed to the
>constant rotation at the frequency of the passing wave.
>I haven't read the analysis in detail, but my first impression is that
>it is quite clever.
Does he get the correct energy for the recoiling particle and, if so,
how does this magic happen (given that classical, or even QM free
electron has a continuum of energy states available)?
>
>Now i agree with Mati Meron on the fact that many other experiments that
>support the photon picture and the QED exist, of course.
>I never thought this was in question.
>However, i must note that, at least from what i understood from this
>thread, the main concern made by Martin Green was an "educational" one
>and, from this point of view, i have frankly to agree with him.
>
Would this have been his main concern I would fully agree with him.
But it doesn't appear to me that this is his main concern. I'm ready
to be corrected, of course.
Fabio Di Teodoro
> Does he get the correct energy for the recoiling particle and, if so,
> how does this magic happen (given that classical, or even QM free
> electron has a continuum of energy states available)?
all that he gets is the relationship between the frequency of the
scattered radiation, f, and the scattering angle, x, namely the well-known
result
f = f_0 * (1 + z*(1 - cos x))^-1 (*)
where f_0 is the frequency of the input wave
and x is T*f_0 where T is the so-called Compton period for the electron
which is about 8.09 * 10^-21 s.
I am sorry i can't post the whole derivation of the result, which would
take too much time.
What he roughly does is
a) assume that a circularly polarized wave impinges upon a free charge
b) the free charge goes through a helical path as a result of the combined
effect of electric and magnetic fields
c) turn to a new reference frame that moves along the downstream
drift motion of the particle
d) Lorentz-transform the wave's frequency. Observe that in this frame the
particle simply rotates in antiphase to the electric field, thereby
emitting radiation at the transformed wave's frequency
e) Find the resulting frequency of the light observed at a given
scattering angle in the lab's frame.
hope this helps to make the procedure used in that book a little
less opaque. As i said i did not check the details, but the calculation
seems plausible to me.
There might be flaws that I do not detect, anyway.
I would appreciate if you could clarify your reference to the particle's
recoil energy distribution. I confess that I haven't read the original
Compton's paper, but from my second-hand information, which of course
might be not complete, i recall that the Compton effect is basically all
in Eq.(*) except for a number of later fine corrections which are needed
to
match successive and more accurate experiments, corrections which are QED
in nature, indeed, as is well known.
If i am not wrong, Compton in the 1923 irradiated a graphite target with
(roughly) monochromatic X-rays, which at the time were already known to be
em radiation at high frequency. He examined the scattered radiation and
saw a component at the same frequency as the input radiation, plus another
component with a frequency depending on the scattering angle given by
Eq.(*), whose existence was impossible to explain in terms of the
standard Thomson scattering.
We know that the the unmodified component arises from scattering off
tightly bound electrons, which entails the recoil of the whole atom, which
makes the Compton shift negligible.
The rest of the scattered radiation was due to scattering off almost
"free" electrons, free in the sense the field's energy is stronger than
the binding energy.
I do not remeber whether and how he measured the energy of the recoil
particles. I am sure, however, that this type of measurements were made
later on and, as you imply, were used as tests for QED interactions,
together with, for example, the quite conclusive coincidence measurements
a la Hanbury-Brown-Twiss which led to more modern knowledge of photon
counting statistics.
Anyway, always referring to what several intro-physics books at undergrad
level do, Eq.(*) is basically introduced as the Compton effect.
By the way, browsing a book that was sitting on my desk, i have found
also mention of a semiclassical derivation of the Compton effect, in the
form of Eq.(*), due to Schrodinger, which might be another nice historical
curiosity.
The reference is
E. Schrodinger, Ann. Phys. vol. 82, p. 257 (1927).
I am sorry i don't have the time to read this paper.
> Would this have been his main concern I would fully agree with him.
i think/hope it was.
Indeed, it's easy to track down a couple of works about the
(semi)classical derivation of the Compton effect. They look like
interesting curiosities, which might have some relevance from an
educational viewpoint.
But it would probably take months, instead, just to list all the striking
evidences which provide compelling and full support for QED.
regards
Jim Goodman
Why not use Pople s Gaussian70 (current version)
and do a 1000 atom section of the material?
--
Jim Goodman
sa...@bellsouth.net
http://personal.msy.bellsouth.net/~sawf/
Energy and Structure of Molecules
Aleksandr Timofeev <t...@alpha.dnttm.rssi.ru> wrote in article <7bu0iu$tkk$1...@nnrp1.dejanews.com>...
>
> Whether you can cite (indicate) examples of Compton's experiments on real
> free electrons.
>
> I consider an error physical hypothesis, which considers poorly bound
> electrons in substance as free electrons. The poorly bound electrons in
> substance belong to a quantum system, which cannot be eliminated from a model
> of experiment.
>
> ---
> Aleksandr Timofeev
Jim Goodman
If Timofeev is some what correct, the scattering
angle would be in bands corresponding to the
state differences of the atom and the system
of close atoms.
--
Jim Goodman
sa...@bellsouth.net
http://personal.msy.bellsouth.net/~sawf/
Energy and Structure of Molecules
Jim Carr <j...@ibms48.scri.fsu.edu> wrote in article <7bvg26$f0r$1...@news.fsu.edu>...
> In article <7brmkl$fcb$8...@thorn.cc.usm.edu>
> lrm...@orca.st.usm.edu (Larry Mead) writes:
> >
> >The need to quantize electrodynamics was argued thoroughly by Jaynes,
> >Lamb and lots of quantum optics folks in the 70's. It turns out, there
> >is one and only one effect which requires this: spontaneous emission.
> >
> >For the details, see "The Quantum Vacuum" by Milonni.
>
> It figures that it would be on loan.
>
> Do you have a reference to an article that treats Compton
> classically and points out the error in Compton's calculation
> of the angular distribution?
>
s...@microtec.net
"Martin Green" <test...@pangea.ca> wrote:
> Thank you for your well-researched posting. I feel a bit guilty for the
> effort I have put you through, but I hope you found it to be an interesting
> excercise.
>
> >
> > support the photon picture and the QED exist, of course.
> > I never thought this was in question.
> > However, i must note that, at least from what i understood from this
> > thread, the main concern made by Martin Green was an "educational" one
> > and, from this point of view, i have frankly to agree with him.
> >
> > student risks to be exposed to some evident misconception.
> > Later on, she/he will (hopefully) learn more on the subject and gain a
> > deeper insight. But this does not justify the initial misconception.
> >
>
> It certainly is the educational aspect of this quesiton that concerns me
> the most. On the superficial level, the student is taught an argument that
> is incorrect. "Hopefully" he will later learn the truth...however, the tone
> of the discussion in this thread makes it clear that it is a very rare
> occasion indeed when a physicist questions his own early indoctrination
> into the photon "ideology".
I wonder what led you to conclude that the concept of photons was an ideology
rather than an _understanding_ of apparently real behaviour.
What do you think Planck's constant really is?
What do you think of the fact that it allows real measurement of any photon's
energy in relation with its frequency?
André Michaud
Service de Recherche Pédagogique http://www.microtec.net/~srp/
Martin Green
> I wonder what led you to conclude that the concept of photons was an
ideology
> rather than an _understanding_ of apparently real behaviour.
>
>
is based on false arguments, you cannot say that you "understand" the
truth, even if that which you believe happens to be true.
"Ideology" was perhaps a poor choice of words to describe a belief which
one subscribes to because it conforms to a way of thinking which is
prescribed by an authority. "Groupthink" might be a better way to
characterize the widespread acceptance within the physics community of
highly flawed arguments in support of the photon theory of light.
Martin Green
s...@microtec.net
I agree that groupthink, as you put it, is widespread and sometimes appears
not to be anchored very solidly in experience. But in the case of the photon
idea, there was serious reasons for it to be considered.
In fact this all began when Wien traced the curve of temperature versus color
of emitted radiation through actual experimentation. The mathematically
derived curve that can be traced with Maxwell's equations did not match at
all.
After 10 years of research and brainsweat, Planck kind of found the smallest
common denominator to the full range of possible EM-frequencies (h) which
allowed mathematically tracing a curve that _exactly_ matched Wien's curve.
There we were, with the smallest possible quantum of free energy in the
universe, the kinetic energy that is generated by a single photon pulse,
whether it is 1 nanometer long or 500 kilometer long, you get exactly the
same amount of energy. The very heartbeat of the universe.
This is what led to the idea of photons being discrete entities, so to speak,
because to explain that amplitude would be exactly constant for all pulses on
all EM manifestations in the universe, there appeared to be no other
explanation that whatever EM pulses were the result of, what produced them
kind of had to always be of exactly the same size, which naturally lead to
the idea that they are separate entities, and not part of a common field.
So you see, there is logic there, no groupthink. If you were to dig into
Wien's work and do the math, you most probably would see the logic.
Jim Carr
Fabio Di Teodoro <fdit...@brynmawr.edu> writes:
}
} "Atoms and light" by John N. Dodd (Plenum Press, New York, 1991).
}
} In Chapter 6, the Compton scattering is treated in an entirely classical
} way, without using energy and momentum conservation, but just standard
} classical em + relativistic *kinematics*, by the picture of a
} circularly polarized em wave impinging upon a charged particle.
} The calculation is based on deriving a steady-state solution for the
} down-stream motion of the particle which is superimposed to the
} constant rotation at the frequency of the passing wave.
} I haven't read the analysis in detail, but my first impression is that
} it is quite clever.
In article <F8ALM...@midway.uchicago.edu>
me...@cars3.uchicago.edu writes:
>
>Does he get the correct energy for the recoiling particle and, if so,
>how does this magic happen (given that classical, or even QM free
>electron has a continuum of energy states available)?
Go find the book, Mati. If _our_ library has it, yours will.
Very interesting reading. It looks to me like he cheats about
where I figured he had to -- and he does not calculate the angular
distribution for the Compton effect at all -- but he does do the
photoeffect by making use of the QM wavefunctions. There is also
a section on spontaneous emission, but I did not read it yet.
I disagree with his pair of strawman arguments in 14.2, for reasons
I have posted here in the past (the same issues come up for electrons,
and no one argues that that are not particles) but they can be used
to bring the pedagogical issues into focus.
} However, i must note that, at least from what i understood from this
} thread, the main concern made by Martin Green was an "educational" one
} ...
That is what he said after his original argument that there is no
physics argument for photons because the photoeffect has another
explanation.
me...@cars3.uchicago.edu
>
>> Does he get the correct energy for the recoiling particle and, if so,
>> how does this magic happen (given that classical, or even QM free
>> electron has a continuum of energy states available)?
>
Figures. So he doesn't reproduce the Compton effect.
It is always possible, using suitable assumptions, to reproduce some
part of a complex behavior using a simplified model. But, if you need
a different model and different set of assumptions for each facet of
what you observe, that's down the epicycle path.
>is the relationship between the frequency of the
>scattered radiation, f, and the scattering angle, x, namely the well-known
>result
>
>f = f_0 * (1 + z*(1 - cos x))^-1 (*)
>
>where f_0 is the frequency of the input wave
>and x is T*f_0 where T is the so-called Compton period for the electron
>which is about 8.09 * 10^-21 s.
>
>I am sorry i can't post the whole derivation of the result, which would
>take too much time.
>What he roughly does is
>
>a) assume that a circularly polarized wave impinges upon a free charge
>
>b) the free charge goes through a helical path as a result of the combined
> effect of electric and magnetic fields
>
Not quite a helical path, but never mind.
>c) turn to a new reference frame that moves along the downstream
> drift motion of the particle
>
There is a drift and superimposed z oscillation. What's worse, the
magnitude of the drift depends on the magnitude ot the radiation
field, not just the frequency.
>d) Lorentz-transform the wave's frequency. Observe that in this frame the
> particle simply rotates in antiphase to the electric field, thereby
> emitting radiation at the transformed wave's frequency
>
>e) Find the resulting frequency of the light observed at a given
> scattering angle in the lab's frame.
>
Since a calculation gives you a field dependent drift, you'll get an
intensity dependent Doppler shift. That doesn't fit.
>hope this helps to make the procedure used in that book a little
>less opaque. As i said i did not check the details, but the calculation
>seems plausible to me.
Well, I guess I'll have to find the book since a calculation (which is
rather straightforward) gives different results. Perhaps he does a
quasi QM calculation, with a QM treatment of the electron and
classical treatment of the radiation. A purely classical calculation
disagrees with the description you gave.
>
>I would appreciate if you could clarify your reference to the particle's
>recoil energy distribution. I confess that I haven't read the original
>Compton's paper, but from my second-hand information, which of course
>might be not complete, i recall that the Compton effect is basically all
>in Eq.(*) except for a number of later fine corrections which are needed
>to
>match successive and more accurate experiments, corrections which are QED
>in nature, indeed, as is well known.
>
Since Compton's calculation is based on energy and momentum
conservation, it goes without saying that it yields the energy and
momentum of both the photon and the recoiling electron. That should
be quite obvious. Now, the measurement is much easier for photons, at
least given the means at Compton's disposal, so that's what he
concentrated on.
>
>> Would this have been his main concern I would fully agree with him.
>
>i think/hope it was.
>
You're young and new to the ways of this group:-)
me...@cars3.uchicago.edu
>In article <Pine.SO4.4.02.990308...@ada.brynmawr.edu>,
>}
>} In Chapter 6, the Compton scattering is treated in an entirely classical
>} way, without using energy and momentum conservation, but just standard
>} classical em + relativistic *kinematics*, by the picture of a
>} circularly polarized em wave impinging upon a charged particle.
>} The calculation is based on deriving a steady-state solution for the
>} down-stream motion of the particle which is superimposed to the
>} constant rotation at the frequency of the passing wave.
>} I haven't read the analysis in detail, but my first impression is that
>} it is quite clever.
>
>In article <F8ALM...@midway.uchicago.edu>
>>
>>Does he get the correct energy for the recoiling particle and, if so,
>>how does this magic happen (given that classical, or even QM free
>>electron has a continuum of energy states available)?
>
>
checking things on the back of an envelope or two. Whatever I checked
doesn't seem to add up to the claims made.
> Very interesting reading. It looks to me like he cheats about
> where I figured he had to -- and he does not calculate the angular
> distribution for the Compton effect at all
From what I've checked (based on the description I got) he couldn't
have done it indeed.
> -- but he does do the photoeffect by making use of the QM
>wavefunctions.
That I believe, but that's not new. My faithful Landau does it too.
Jim Carr
>
>Since a calculation gives you a field dependent drift, you'll get an
>intensity dependent Doppler shift. That doesn't fit.
That is where he cheats. He goes from a steady state situation
to an "interaction" representation, and it is here that he
constrains the boost by relying on experimental data. In particular,
this is how he gets a photon momentum into the answer, by saying
that an E+M field can only transfer momentum quanta as well as
energy quanta. I suspect there is a way to tie up some of those
loose ends by invoking Bohr quantization of the orbital motion of
free electrons, which he does not do. (Since later he points out
that Bohr quantization is wrong by the experimental evidence for
L=0 states, you could say that those data defeat his model and thus
substantiate the photon.) However, one thing you are forced to do
by the Compton data is to say that the field can only transfer
quanta. Since, in the end, that is what photons are in QED, what
he really has is a phenomenological picture of how experiment forces
that quantization onto fields.
>Well, I guess I'll have to find the book since a calculation (which is
>rather straightforward) gives different results. Perhaps he does a
>quasi QM calculation, with a QM treatment of the electron and
>classical treatment of the radiation.
But that is what he is doing, a semi-classical argument.
He does lots of that later on. In this section he uses the form
of the Compton scattering law to force the relationships he needs.
He could have done more with quantization of L. Also, he did not
calculate the angular distribution, but I think he could get the
right answer for it. In any case, the key step is when he uses
data to the relation between frequency and the boost speed to the
(time averaged) "rest frame" of the helically moving electron.
He lets data do the quantization for him and begs the question
of why the key variable is the frequency of the light and not
some property of the quantized material part of the problem,
which is supposed to be the only place QM enters.
Jim Carr
>
>> Very interesting reading. It looks to me like he cheats about
>> where I figured he had to -- and he does not calculate the angular
>> distribution for the Compton effect at all
>
>From what I've checked (based on the description I got) he couldn't
>have done it indeed.
He could have. At that point, he just needs to do what Compton did.
I think he also could have done it cuter at one point by invoking
the Bohr rules for a free electron, but I have not worked it out.
me...@cars3.uchicago.edu
>In article <F8Cuu...@midway.uchicago.edu>
>me...@cars3.uchicago.edu writes:
>>
>>Since a calculation gives you a field dependent drift, you'll get an
>>intensity dependent Doppler shift. That doesn't fit.
>
> That is where he cheats. He goes from a steady state situation
> to an "interaction" representation, and it is here that he
> constrains the boost by relying on experimental data. In particular,
> this is how he gets a photon momentum into the answer, by saying
> that an E+M field can only transfer momentum quanta as well as
> energy quanta. I suspect there is a way to tie up some of those
> loose ends by invoking Bohr quantization of the orbital motion of
> free electrons, which he does not do. (Since later he points out
> that Bohr quantization is wrong by the experimental evidence for
> L=0 states, you could say that those data defeat his model and thus
> substantiate the photon.) However, one thing you are forced to do
> by the Compton data is to say that the field can only transfer
> quanta. Since, in the end, that is what photons are in QED, what
> he really has is a phenomenological picture of how experiment forces
> that quantization onto fields.
>
which happens to transfer energy and momentum quanta:-) Well, "a rose
by any other name...". But, given all this, I'm not sure what the
purpose of the exercise is.
>>Well, I guess I'll have to find the book since a calculation (which is
>>rather straightforward) gives different results. Perhaps he does a
>>quasi QM calculation, with a QM treatment of the electron and
>>classical treatment of the radiation.
>
> But that is what he is doing, a semi-classical argument.
>
OK.
> He does lots of that later on. In this section he uses the form
> of the Compton scattering law to force the relationships he needs.
> He could have done more with quantization of L. Also, he did not
> calculate the angular distribution, but I think he could get the
> right answer for it. In any case, the key step is when he uses
> data to the relation between frequency and the boost speed to the
> (time averaged) "rest frame" of the helically moving electron.
>
> He lets data do the quantization for him and begs the question
> of why the key variable is the frequency of the light and not
> some property of the quantized material part of the problem,
> which is supposed to be the only place QM enters.
>
It appears to me that if what he's trying to do is to show that
photons aren't needed, he ends up being not very convincing:-) Oh,
well.
me...@cars3.uchicago.edu
>In article <F8D3F...@midway.uchicago.edu>
>me...@cars3.uchicago.edu writes:
>>
>>> Very interesting reading. It looks to me like he cheats about
>>> where I figured he had to -- and he does not calculate the angular
>>> distribution for the Compton effect at all
>>
>>From what I've checked (based on the description I got) he couldn't
>>have done it indeed.
>
> He could have. At that point, he just needs to do what Compton did.
>
want to do.
> I think he also could have done it cuter at one point by invoking
> the Bohr rules for a free electron, but I have not worked it out.
>
I run out of envelopes (and time) by the time I finished the purely
classical part.
Richard Herring
> In fact this all began when Wien traced the curve of temperature versus color
> of emitted radiation through actual experimentation. The mathematically
> derived curve that can be traced with Maxwell's equations did not match at
> all.
> After 10 years of research and brainsweat, Planck kind of found the smallest
> common denominator to the full range of possible EM-frequencies (h) which
> allowed mathematically tracing a curve that _exactly_ matched Wien's curve.
So far, so good. But I think the following sacrifices clarity to rhetoric:
> There we were, with the smallest possible quantum of free energy in the
> universe,
... which would be how many joules, exactly?
> the kinetic energy that is generated by a single photon pulse,
> whether it is 1 nanometer long or 500 kilometer long, you get exactly the
> same amount of energy.
Is this claiming that photons of the same energy can have different
"lengths" or that photons of different wavelengths have the same energy?
> The very heartbeat of the universe.
Wow. Heady stuff.
--
Richard Herring | <richard...@gecm.com>
Jim Carr
j...@ibms48.scri.fsu.edu (Jim Carr) writes:
|
| >From what I've checked (based on the description I got) he couldn't
| >have done it indeed.
|
| He could have. At that point, he just needs to do what Compton did.
In article <F8DCv...@midway.uchicago.edu>
me...@cars3.uchicago.edu writes:
>
>Oh, yes, of course. But that appears to be the one thing he doesn't
>want to do.
You misunderstand. All he needs to do is boost the angular
distribution calculated just as Compton did, which he can do
once he has the frames correct (which he has, due to the use
of data to fix the momentum transfer from the field).
s...@microtec.net
richard...@gecm.com wrote:
> In article <7c3lde$m3u$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
>
> > In fact this all began when Wien traced the curve of temperature versus
color
> > of emitted radiation through actual experimentation. The mathematically
> > derived curve that can be traced with Maxwell's equations did not match at
> > all.
>
> > After 10 years of research and brainsweat, Planck kind of found the smallest
> > common denominator to the full range of possible EM-frequencies (h) which
> > allowed mathematically tracing a curve that _exactly_ matched Wien's curve.
>
> So far, so good. But I think the following sacrifices clarity to rhetoric:
Maybe :o]
> > There we were, with the smallest possible quantum of free energy in the
> > universe,
>
> ... which would be how many joules, exactly?
I'm no good at math. Maybe you could propose a way to figure it?
> > the kinetic energy that is generated by a single photon pulse,
> > whether it is 1 nanometer long or 500 kilometer long, you get exactly the
> > same amount of energy.
>
> Is this claiming that photons of the same energy can have different
> "lengths"
No, from what I understand, it merely states the verified fact that they have
the same wavelength in space. Remember, they all go at c.
> or that photons of different wavelengths have the same energy?
No, from what I understand, it merely states the verified fact that every
wave of every photon has the same energy. Here again, remember that they all
go at c, and that the frequency (the number of individual waves per unit
time) multiplied by h gives the total energy of the photon.
> > The very heartbeat of the universe.
>
> Wow. Heady stuff.
Yeah. I must have gotten a tad carried away here 8o}
s...@microtec.net
s...@microtec.net wrote:
> In article <7c5g11$dkb$2...@miranda.gmrc.gecm.com>,
> richard...@gecm.com wrote:
> > > There we were, with the smallest possible quantum of free energy in the
> > > universe,
> >
> > ... which would be how many joules, exactly?
>
> I'm no good at math. Maybe you could propose a way to figure it?
Well, there are no math to be done after all.
Just looking at my super dooper Sharp Scientific calculator EL-506L's spec
sheet, I found the exact value smack in constant # 10 for a space-time
reference frame:
6.6260755 x 10^-34 joules
Richard Herring
> In article <7c5g11$dkb$2...@miranda.gmrc.gecm.com>,
> richard...@gecm.com wrote:
> >
> > So far, so good. But I think the following sacrifices clarity to rhetoric:
> Maybe :o]
> > > There we were, with the smallest possible quantum of free energy in the
> > > universe,
> >
> > ... which would be how many joules, exactly?
> I'm no good at math. Maybe you could propose a way to figure it?
I can't, because it is impossible ;-)
Planck's constant is a quantum of *action*, not energy.
There is no "smallest possible quantum of free energy", so far as
we know.
> > > the kinetic energy that is generated by a single photon pulse,
> > > whether it is 1 nanometer long or 500 kilometer long, you get exactly the
> > > same amount of energy.
> >
> > Is this claiming that photons of the same energy can have different
> > "lengths"
> No, from what I understand, it merely states the verified fact that they have
> the same wavelength in space. Remember, they all go at c.
> > or that photons of different wavelengths have the same energy?
> No, from what I understand, it merely states the verified fact that every
> wave of every photon has the same energy. Here again, remember that they all
> go at c, and that the frequency (the number of individual waves per unit
> time) multiplied by h gives the total energy of the photon.
Photons of the same energy have the same wavelength and frequency, yes.
Photons travel at c, yes.
I don't see any way of mapping these statements to the one
above about the energy of a "photon pulse" and its length.
--
Richard Herring | <richard...@gecm.com>
s...@microtec.net
richard...@gecm.com wrote:
> In article <7c699e$vrr$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
> > In article <7c5g11$dkb$2...@miranda.gmrc.gecm.com>,
> > richard...@gecm.com wrote:
> > > In article <7c3lde$m3u$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
> > >
> > > So far, so good. But I think the following sacrifices clarity to rhetoric:
>
> > Maybe :o]
>
> > > > There we were, with the smallest possible quantum of free energy in the
> > > > universe,
> > >
> > > ... which would be how many joules, exactly?
>
> > I'm no good at math. Maybe you could propose a way to figure it?
>
> I can't, because it is impossible ;-)
Is it really? See my reasonning below.
> Planck's constant is a quantum of *action*, not energy.
> There is no "smallest possible quantum of free energy", so far as
> we know.
I am not very clear as to what you presently mean by "quantum of action" here.
Maybe you could elaborate?
As for Planck's constant not refering to energy, I wonder what aspect of
electromagnetism would not pertain to energy in one way or another.
You asked me how many joules my so called "smallest quantum of free energy"
would amount to, so I suppose joules refer to energy.
On the other hand, my super dooper calculator spec sheet tels me that Planck's
constant is in joules/sec so I suppose we speak the same language here, except
that Planck relates that energy to a time frame.
> > > > the kinetic energy that is generated by a single photon pulse,
> > > > whether it is 1 nanometer long or 500 kilometer long, you get exactly
> > > > the same amount of energy.
> > >
> > > Is this claiming that photons of the same energy can have different
> > > "lengths"
>
> > No, from what I understand, it merely states the verified fact that they
> > have the same wavelength in space. Remember, they all go at c.
>
> > > or that photons of different wavelengths have the same energy?
>
> > No, from what I understand, it merely states the verified fact that every
> > wave of every photon has the same energy. Here again, remember that they all
> > go at c, and that the frequency (the number of individual waves per unit
> > time) multiplied by h gives the total energy of the photon.
>
> Photons of the same energy have the same wavelength and frequency, yes.
> Photons travel at c, yes.
>
> I don't see any way of mapping these statements to the one
> above about the energy of a "photon pulse" and its length.
If we accept the idea that energy is conserved, and that we have a frequency
of 1 Hz (or 1 pulse, or one wave, or whatever we chose to call it), I think
that by definition, it would be 1 lightsecond long in space. Would you agree
that we have here a quantity of energy of ...
(manhandling my SD Calculator here)
... 6.6260755 x 10^-34 joules for this one pulse. Since by definition, Hz
means "pulses per second", we can get rid of the "/sec" part, since we know
that we have exactly one pulse here.
Let's say further that we have a 2 Hz EM-manifestation, would you agree that
we now have (6.6260755 x 10^-34) x 2 joules for that EM-manifestation? What
does this tell us? We now have two pulses (or waves or whatever) which will
extend half a lightsecond long in space but carry the same energy as the
single 1 lightsedond pulse we looked at before, if I understand correctly.
And so on and so on.
So, here don't we have it? Whatever lenght of each individual pulse, each of
them always seem to carry 6.6260755 x 10^-34 joules when considered in a
space-time frame.
Or, what don't I see?
Richard Herring
> In article <7c699e$vrr$1...@nnrp1.dejanews.com>,
> s...@microtec.net wrote:
> > In article <7c5g11$dkb$2...@miranda.gmrc.gecm.com>,
> > richard...@gecm.com wrote:
> > > > There we were, with the smallest possible quantum of free energy in the
> > > > universe,
> > >
> > > ... which would be how many joules, exactly?
> >
> > I'm no good at math. Maybe you could propose a way to figure it?
> Well, there are no math to be done after all.
> Just looking at my super dooper Sharp Scientific calculator EL-506L's spec
> sheet, I found the exact value smack in constant # 10 for a space-time
> reference frame:
> 6.6260755 x 10^-34 joules
Bzzzt. Wrong. Check those units again.
--
Richard Herring | <richard...@gecm.com>
Richard Herring
> In article <7c8803$an2$1...@miranda.gmrc.gecm.com>,
> richard...@gecm.com wrote:
> > In article <7c699e$vrr$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
> > > In article <7c5g11$dkb$2...@miranda.gmrc.gecm.com>,
> > > richard...@gecm.com wrote:
> >
> > > > > There we were, with the smallest possible quantum of free energy in the
> > > > > universe,
> > > >
> > > > ... which would be how many joules, exactly?
> >
> > > I'm no good at math. Maybe you could propose a way to figure it?
> >
> Is it really? See my reasonning below.
> > Planck's constant is a quantum of *action*, not energy.
> > There is no "smallest possible quantum of free energy", so far as
> > we know.
> I am not very clear as to what you presently mean by "quantum of action" here.
> Maybe you could elaborate?
Any good science dictionary should tell you that "action" is the
name given to the integral of energy over time. It has the same
dimensions as angular momentum.
> As for Planck's constant not refering to energy, I wonder what aspect of
> electromagnetism would not pertain to energy in one way or another.
By that argument, *everything* pertains to energy, which would make
the entire concept meaningless.
> You asked me how many joules my so called "smallest quantum of free energy"
> would amount to, so I suppose joules refer to energy.
Indeed they do.
> On the other hand, my super dooper calculator spec sheet tels me that Planck's
> constant is in joules/sec
No, joules *times* seconds.
> so I suppose we speak the same language here, except
> that Planck relates that energy to a time frame.
No, he says something specific about the quantisation of action or
angular momentum.
> > Photons of the same energy have the same wavelength and frequency, yes.
> > Photons travel at c, yes.
> >
> > I don't see any way of mapping these statements to the one
> > above about the energy of a "photon pulse" and its length.
> If we accept the idea that energy is conserved, and that we have a frequency
> of 1 Hz (or 1 pulse, or one wave, or whatever we chose to call it), I think
> that by definition, it would be 1 lightsecond long in space.
I don't. For a wave to have a precise frequency, its temporal extent
has to be infinite. In photon terms, if you know a photon's energy
precisely, you cannot localise it in time. Since we know its speed,
that implies that we cannot localise it in space either.
> Would you agree
> that we have here a quantity of energy of ...
> (manhandling my SD Calculator here)
> ... 6.6260755 x 10^-34 joules for this one pulse. Since by definition, Hz
> means "pulses per second", we can get rid of the "/sec" part, since we know
> that we have exactly one pulse here.
Faulty mathematics, but the correct result. Yes, each 1 Hz photon has
that amount of energy.
> Let's say further that we have a 2 Hz EM-manifestation, would you agree that
> we now have (6.6260755 x 10^-34) x 2 joules for that EM-manifestation?
Per photon, certainly.
> What
> does this tell us? We now have two pulses (or waves or whatever) which will
> extend half a lightsecond long in space but carry the same energy as the
> single 1 lightsedond pulse we looked at before, if I understand correctly.
Yes and no. Yes, the 2Hz photons each have twice the energy of
the 1Hz photons. No, we have no information about the length of any
wavepacket of either frequency.
--
Richard Herring | <richard...@gecm.com>
s...@microtec.net
richard...@gecm.com wrote:
> In article <7c8vaj$c8e$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
snip]
> > > Planck's constant is a quantum of *action*, not energy.
> > > There is no "smallest possible quantum of free energy", so far as
> > > we know.
>
> > I am not very clear as to what you presently mean by "quantum of action"
> > here. Maybe you could elaborate?
>
> Any good science dictionary should tell you that "action" is the
> name given to the integral of energy over time. It has the same
> dimensions as angular momentum.
OK I checked. You are right. Even my _regular_ dictionary confirms.
"Action of a mobile point in motion"
"Integral of mv^2 dt"
> > As for Planck's constant not refering to energy, I wonder what aspect of
> > electromagnetism would not pertain to energy in one way or another.
>
> By that argument, *everything* pertains to energy, which would make
> the entire concept meaningless.
Agreed.
> > You asked me how many joules my so called "smallest quantum of free energy"
> > would amount to, so I suppose joules refer to energy.
>
> Indeed they do.
>
> > On the other hand, my super dooper calculator spec sheet tels me that
> > Planck's constant is in joules/sec
>
> No, joules *times* seconds.
Hmm, that had escaped me. I'll think about it. My point however was that
_time_ was in the picture, and it still is, even with your correction.
> > so I suppose we speak the same language here, except
> > that Planck relates that energy to a time frame.
>
> No, he says something specific about the quantisation of action or
> angular momentum.
I admit that this doesn't connect much with my view of the photon.
> > > Photons of the same energy have the same wavelength and frequency, yes.
> > > Photons travel at c, yes.
> > >
> > > I don't see any way of mapping these statements to the one
> > > above about the energy of a "photon pulse" and its length.
>
> > If we accept the idea that energy is conserved, and that we have a frequency
> > of 1 Hz (or 1 pulse, or one wave, or whatever we chose to call it), I think
> > that by definition, it would be 1 lightsecond long in space.
>
> I don't.
What I meant really was that it needs to move a distance of 1 lightsecond to
complete it's cycle, whatever that cycle might be.
> For a wave to have a precise frequency, its temporal extent
> has to be infinite.
Here you loose me. I fail to see what time has to do with photon energy. I
see it clearly if we speak of accumulating energy somehow, but at the photon
level, I kind of see only point-events with no relation to time except with
the moment of its existance.
> In photon terms, if you know a photon's energy
> precisely, you cannot localise it in time.
In my view, if any photon is in existance at all, at any given moment, meaning
that it has been emitted, but has not yet been absorbed, there is no way that
its location in time will not be _now_ at that moment.
> Since we know its speed, that implies that we cannot localise it in space
> either.
Wow! Now that grabs me!
It might come as a surprise to you, but my visual cortex is hooked to the
outside world through sensors that are photon driven. I can see droves of
them coming at me at this very moment, and I can definitely localise very
precisely where each of them originated and where it has been absorbed. If I
chose to, knowing its speed, I could retrospectively pick any one of them,
assess its frequency and by rather simple means of triangulation, determine
with satisfying precision where it was at any point of its trajectory as it
came towards one of my sensors, even compute the number of time it completed
its recursive cycle before it reached that point.
Or, am I delusional?
> > Would you agree
> > that we have here a quantity of energy of ...
> > (manhandling my SD Calculator here)
> > ... 6.6260755 x 10^-34 joules for this one pulse. Since by definition, Hz
> > means "pulses per second", we can get rid of the "/sec" part, since we know
> > that we have exactly one pulse here.
>
> Faulty mathematics, but the correct result. Yes, each 1 Hz photon has
> that amount of energy.
>
> > Let's say further that we have a 2 Hz EM-manifestation, would you agree that
> > we now have (6.6260755 x 10^-34) x 2 joules for that EM-manifestation?
>
> Per photon, certainly.
>
> > What
> > does this tell us? We now have two pulses (or waves or whatever) which will
> > extend half a lightsecond long in space but carry the same energy as the
> > single 1 lightsedond pulse we looked at before, if I understand correctly.
>
> Yes and no. Yes, the 2Hz photons each have twice the energy of
> the 1Hz photons. No, we have no information about the length of any
> wavepacket of either frequency.
Again, when I said _length_, I was presently referring to the distance in
space a photon has to travel before completing a cycle.
That business of "smallest possible quantum of free energy" was just really
kind of exploring E = hf at what appeared to me as its "face value". To my
shame, trolling is not beneath me.
I presently think that a 2 Hz photon at all times carries its full quantum of
twice the energy of a 1 Hz photon, without any relation to time.
Considering for example the fact that the hydrogen atom's electron circles
the proton in the vicinity of 450 000 billion times in one second (if we
accept the idea that it is actually moving around the proton, that is), it
would appear to me unrealistic to expect the electron to wait any given
period of time for the photon to kind of pile in all of it's energy. This
means in my view that all of any photon's energy is instantaneously present
as it meets said electron. This is why I have difficulty in understanding
what the time factor is doing in Planck's equation.
Kenneth L. Graham
----------
In article <7cc8tf$acb$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
>> > I am not very clear as to what you presently mean by "quantum of action"
>> > here. Maybe you could elaborate?
Action, being the integral over time of a particle's Lagrangian
(Lagrangian equals kinetic minus potential energy), is minimized for the
path a particle takes as it moves through space and time. (Any good
Classical mechanics text discusses this.) When we move to the quantum
regime, we see that a particle's action is also quantized. This means that
while it is minimized, it will come in discrete integer multiples of
Planck's constant.
>> > so I suppose we speak the same language here, except
>> > that Planck relates that energy to a time frame.
>>
>> No, he says something specific about the quantisation of action or
>> angular momentum.
The angular momentum of quantum particles (photon, electron, etc.) comes
in integer multiple of Planck's constant (omitting quarks). We then say
that angular momentum is quantized; i.e. it comes in discrete lumps. A
photon has an angular momentum of one (times Planck's constant h/2Pi). What
I am trying to say is that Planck's constant does not really relate energy
to a time frame. It does relate a photon's energy to its frequency.
>> > If we accept the idea that energy is conserved, and that we have a
frequency
>> > of 1 Hz (or 1 pulse, or one wave, or whatever we chose to call it), I think
>> > that by definition, it would be 1 lightsecond long in space.
>>
>> I don't.
>
>
>> For a wave to have a precise frequency, its temporal extent
>> has to be infinite.
>
> Here you loose me. I fail to see what time has to do with photon energy. I
> see it clearly if we speak of accumulating energy somehow, but at the photon
> level, I kind of see only point-events with no relation to time except with
> the moment of its existance.
I can sum it up like this: A wave of finite spatial extent, a pulse,
does not have a precise frequency because it can mathematically expressed as
a sum of sine waves of different frequencies. To have a precise frequency,
you must let your wave extend to infinity. This is from Fourier Theory.
>
>> In photon terms, if you know a photon's energy
>> precisely, you cannot localise it in time.
>
> In my view, if any photon is in existance at all, at any given moment, meaning
> that it has been emitted, but has not yet been absorbed, there is no way that
> its location in time will not be _now_ at that moment.
>
This comes from the energy-time uncertainty principle. It says that you
cannot simultaneously know (or measure) a photons energy while localizing it
in time. What does this mean? If we are observing some transition where a
photon is emitted, due to the wave nature of the photon, we cannot know
exactly when the photon was emitted. Imagine an atom in which an electron
has been excited to some higher energy state. Eventually the electron will
drop back to its previous energy state and a photon will be emitted. Even
though we know how much energy the photon has, we do not know precisely when
the photon was emitted. Hope this helps!
KG
s...@microtec.net
"Kenneth L. Graham" <kenneth...@dartmouth.edu> wrote:
>
> ----------
> In article <7cc8tf$acb$1...@nnrp1.dejanews.com>, s...@microtec.net wrote:
>
> >> > here. Maybe you could elaborate?
>
> (Lagrangian equals kinetic minus potential energy), is minimized for the
> path a particle takes as it moves through space and time. (Any good
> Classical mechanics text discusses this.)
In the case of photons, is kinetic energy always considered as being at
maximum, and potential energy at min., or is it considered at max potential
as it travels, but goes to max kinetic when it is absorbed, or
Compton-effected, so to speak?
> When we move to the quantum
> regime, we see that a particle's action is also quantized. This means that
> while it is minimized, it will come in discrete integer multiples of
> Planck's constant.
OK.
> >> > so I suppose we speak the same language here, except
> >> > that Planck relates that energy to a time frame.
> >>
> >> No, he says something specific about the quantisation of action or
> >> angular momentum.
>
> The angular momentum of quantum particles (photon, electron, etc.) comes
> in integer multiple of Planck's constant (omitting quarks).
Why would quarks be excluded?
> We then say
> that angular momentum is quantized; i.e. it comes in discrete lumps. A
> photon has an angular momentum of one (times Planck's constant h/2Pi). What
> I am trying to say is that Planck's constant does not really relate energy
> to a time frame. It does relate a photon's energy to its frequency.
Ok, I see. I will think about this.
> >> > If we accept the idea that energy is conserved, and that we have a
> >> > frequency of 1 Hz (or 1 pulse, or one wave, or whatever we chose to call
> >> > it), I think that by definition, it would be 1 lightsecond long in space.
> >>
> >> I don't.
> >
> >> For a wave to have a precise frequency, its temporal extent
> >> has to be infinite.
> >
> > Here you loose me. I fail to see what time has to do with photon energy. I
> > see it clearly if we speak of accumulating energy somehow, but at the photon
> > level, I kind of see only point-events with no relation to time except with
> > the moment of its existance.
>
> I can sum it up like this: A wave of finite spatial extent, a pulse,
> does not have a precise frequency because it can mathematically expressed as
> a sum of sine waves of different frequencies. To have a precise frequency,
> you must let your wave extend to infinity. This is from Fourier Theory.
Ok, I see now where Richard's infinite temporal extent comes from.
> >> In photon terms, if you know a photon's energy
> >> precisely, you cannot localise it in time.
> >
> > In my view, if any photon is in existance at all, at any given moment,
> > meaning that it has been emitted, but has not yet been absorbed, there is no
> > way that its location in time will not be _now_ at that moment.
> >
> This comes from the energy-time uncertainty principle. It says that you
> cannot simultaneously know (or measure) a photons energy while localizing it
> in time. What does this mean? If we are observing some transition where a
> photon is emitted, due to the wave nature of the photon, we cannot know
> exactly when the photon was emitted.
Interesting. Considering however that we know about a given photon only
_after_ it has been absorbed by whatever detection device we might use, we do
know precisely then what frequency it had, and where it came from. It's speed
being a given, can't we retropedal then and determine precisely all its other
carecteristics?
Conversely, I cannot conceive of being able to measure anything about a
photon as it travels, since excluding thought experiments, there is no way
for us to know of it's existence before is is either absorbed or otherwise
collides with a detection device.
> Imagine an atom in which an electron
> has been excited to some higher energy state. Eventually the electron will
> drop back to its previous energy state and a photon will be emitted. Even
> though we know how much energy the photon has, we do not know precisely when
> the photon was emitted.
I have the same questions here since once we bag it with a detection device,
it would appear to me that we could compute all of that, couldn't we?
> Hope this helps!
Yes it does. Very helpfull.