Dear Sergey Karavashkin:

"Sergey Karavashkin" wrote in message
om...
Oh, you don't appreciate good attitude, David. I have read attentively
your last post. You said many different things on me, but this is to
your account.

As I see, you put the question rigidly, requiring from me the
photoeffect with wave physics. However, you didn't understand, I can
explain it only to one who has a necessary amount of knowledge.
Unfortunately, you still don't demonstrate such knowledge.

There are lots of folks here that *do* have the wherewithal to hear all
that you say. Try explaining it, as you suggest.

Yes, I can explain photoeffect with the help of resonance phenomena in
EM wave interaction with the electrons of substance. In metal they are
the electrons of Fermi-gas, in semiconductors and dielectrics -
orbital electrons. I briefly said you of it before. If you want to
hear more, I'm pleased. But before I would like to make certain that
you know resonance systems enough. Aren't you against? You wrote,

Lets have it.

[David]
I took a class in "resonance systems". That is why I know that you
have
not even looked at what resonance is. I CAN calculate certain
limited sets
of exact solutions as the need arises.

[Sergey]
Okay. Since you put the question point-blank and state that you know
in resonance systems what I don't, and also

I know some resonant systems. I say you describe behaviours that are
counter to my knowledge and experience. Enlighten the group.

[David]
There is no difficulty in expressing the formulas for resonance. And
the
behaviour is well known in lots of different types of systems.

[Sergey]
I will not bother you with complicated systems. Please go to page 42
of our paper "OSCILLATION PATTERN FEATURES IN MISMATCHED FINITE
ELECTRIC LADDER FILTERS"

http://angelfire.lycos.com/la3/selft...42/load42.html

Circuit resonances. Has absolutely nothing to do with the photoelectric
effect.

....
and see formulas (23) - (25). This is the exact analytical solution
for a heterogeneous line shown in Fig. 4a in the same page. You can
make sure, these solutions are exact. It is sufficient for it to
compare the diagrams in Fig. 6, page 44

http://angelfire.lycos.com/la3/selft...44/load44.html

plotted with these formulas, with the experimental diagrams in Fig.
10, page 46

circuit resonance. Has absolutely nothing to do with the photoelectric
effect.

http://angelfire.lycos.com/la3/selft...46/load46.html

The calculation of this mechanical elastic line is surely simple,
takes several pages and a trifle of time. Please do show me, how I
made it. This will make me sure that when I begin telling you, I will
not see glassy eyes. ;-)

No glassy eyes required to see the magician is trying misdirection again.

After this we can advance into wave physics with a great speed, and
you will soon see, wave physics is not so simple as you used to think
outwardly, and photoeffect doesn't limit its scope.

Sergey.

P.S. I didn't want to touch other issues in this post, but couldn't
resist a temptation. ;-)

1. In your post you stated unambiguously that photon has a "zero"
size:

[David]
The size of the photon (as determined by experiment) is "zero", and
has nothing to do with the distance it travels before it achieves the
same
E&B orientation again.

At odd moment, could you explain me: if it has such size as you are
saying, it must be smaller than a period of wave (at least for radio
waves). It moves with the light velocity - it means, with the same
velocity as vector E varies in space. Hence, I have natural questions:

As I have said both to you and to Alexsandr, the concept of width is very
fuzzy. Width in one sense has to do with establishing how far away
something has to be to have zero effect on a particle "path". Width in a
more conventional sense has to do with establishing how far from a path
something has to be to have a definite measurable effect. I see only
English has such duality of definition.

a) the integral field of photon will be in this case non-zero, and
what about uncharged photon?

What integral and what field? The photon has no charge, correct.

b) no changes in time can occur within photon, as with it the field
variation registered by the receiver would be more either less than
the light velocity (so-called group velocity which is formed when
within some system there exists a subsystem with the time-variable
phase shift); then of what changes of E&B are you saying?

Why do you say time does not pass for the photon? It is not massive.
Lorentz transforms do not apply to the c frame. Time does obviously seem
to pass for the photon.

c) If the wave period consists of multitude photons, how photons do
correlate with each other, keeping a strong sequence for many
thousands and millions kilometres, especially when there propagates
not a monochromatic wave but a packet?

"Period" is as descriptive for single photons, as it is for a host of
similar photons.

2. As to the beyond-cutoff luminescence of substance you have written
the following:

[David]
Or systems that express temperature, which is relative motion of the
individual emitters and absorbers. This is simply saying that
quantum
mechanics is right. What is not right about what you have said is
that it
has anything to do with resonant behaviour. Absorption of a photon
does
*not* occur for electrons in orbitals that are just under a photon
energy,
and produce a near-zero KE electron, they produce a conduction
electron
with all the energy. This is NOT resonant behaviour.

a) please show me the regularity of energy of secondary quanta in
Planck formula with respect to temperature; ;-)

*Which* Planck formula? I find many attributable to him.

b) of which relative motion of individual emitters and absorbers are
you saying for a solid state of luminophor which so much distorts the
pattern of emission?

Only having answered these questions, you may judge, how much right is
QM.

QM is not the issue. Your claim that wave theory describes the
photoelectric effect is.

3. As to your following claim:

[David]
There are no behaviours that wave theory describes,
that particle theory cannot.

I guess, you are speaking here of the photon theory as a part of
particle theory. If so, I would like simply to cite Niels Bohr:

In the view of quantum theory, the discussed hypothesis cannot be
nonetheless considered as a satisfying solution. As is known, just
this hypothesis brings insurmountable difficulties in explanation of
interference phenomena - the main means in studying the radiation
properties [see: H.A. Lorentz. Phys. Zs., 1910, *11*, 349]. In any
case we can ascertain that the underlying statement of the hypothesis
of light quanta basically excludes the possibility to comprehend the
concept of frequency nu playing the main part in this theory. So the
hypothesis of light quanta is invalid to give general pattern of
processes which might include the entire amount of phenomena
considered in quantum theory applications

[Niels Bohr. On the quantum theory application to the structure of
atom. 1. The main postulates of quantum theory. Chapter 3. On formal
nature of quantum theory. Item 1. Hypothesis of light quanta].

This organically supplements what I usually say you and you stubbornly
don't hear. We all can take offence, the more when have such necessity
to avoid answering inconvenient questions. This, David, is too
trivial. ;-)

Trivial and "satisfying". And can be said for any particle. So wave
theory must extend to particles of all sorts. Are you up to that as well
as describing the photoelectric effect? Of course you were were referring
to QM, and not waves...

4. And one premature note on your dear photoeffect and your statement
that

[David]
Wave theory does not describe the photoelectric effect, and
particle theory does.

[Sergey]
Please take any book on photoeffect and open where the spectral
characteristics for metals are shown. You will see that the quantum
output dependence on frequency is not so much like a direct line as
Planck equation predicts. These curves are gently sloping near the
'red' boundary and increasing as the square of difference of
frequencies. After it you see an abrupt rise. And this rise relates to
the frequency band at which the material stops effectively reflecting
EM waves. The further the more. The curve reaches its maximum, then
the photoeffect abruptly falls. Began the material again reflecting EM
waves? Yes, but not so much abruptly as the quantum output falls.
Well, now please answer, David, my very simple question. As a result
of what there appears the maximum of absorption of EM waves by the
surface of metal?

And is completely beside the point for the photoelectric effect, since the
"work function" describes the threshold, and "resonance" fails to describe
the rest of the behaviour.

In semiconductors the quantum output has some other pattern. In the
area of red boundary you see an abrupt raise and saturation and almost
smooth plateau. Here also the surface absorbs in the area of plateau?
;-)

Of course, these are far from all questions, but if you can answer
specifically at least to these - I will be grateful. ;-)

Again, you have little regard fro what I have written to you in all
earnestness. Like a good con man, you flash little lights, and wave your
hands in other places to disguise the fact you would change the topic of
discussion.

The topic is "boundary conditions" for photons-as-waves, and you are not
even close to describing them.

David A. Smith