From:Franz Heymann (notfranz.heymann@btopenworld.com)
Subject:Re: To Aleksandr Timofeev on interference
Newsgroup:sci.physics
Date:2004-12-27 21:33:07 PST
"Sergey Karavashkin" <selflab@go.com> wrote in message
news:1305a9da.0412271511.67bfb958@posting.google.com...
[snip]
Dear Sasha, you have nothing to be in opposition. My opinion which you
took from my phrase
Interference in no case is a mathematical fiction, as it is
observed in nature independently of digital either analogue means of
study, and whether do we study it or not.
just says, interference is a physical phenomenon, in no case a
mathematical fiction.
What a fatuous and puerile statement
[snip]
Franz
Sergey:
Franz, you again threw a stone to your own garden! This was not me but
your colleague Duglas who stated so and whose name you snipped. And
you are going on telling absolute nonsense about statistic pattern of
photon beam. Aleksandr recalled just to the point a pearl of you:
<< Fourthly, a macroscopic radio wave is the wavefunction of an
extremely large assembly of photons, all coherently sharing the
same wave function. Detecting one photon at one of the antennae
simultaneously with the detection of another (coherent) photon at
another antenna is then possible. >> [Franz Heymann]
Will you again delete your original posts? ;-)
Well, I wrote you before, for radio waves you cannot compare correctly
the quantum size of photon and wavelength. Supposing the statistical
pattern of photon beam, you don't understand that you are even more
falling into your own trap.
First, if a large association of photons is distributed by one wave
function, some energy has to be spent on it. At the same time,
accordingly to photon theory, photon is born in one act of atom
excitation: one excitation - one photon. In accordance with
Schroedinger equation, the wave function corresponds to this
excitation. And just photon contains the whole energy of excitation.
They do not suppose any additional energy to control photons. ;-)
Second, imagining the photon well less than the wavelength, you
violate another postulate - uncharged photon. If you integrate E-field
of the wave over the length of your photon, it will not be zero and
will be time-variable. From this directly follows the discrepancy with
experiment. As is known, E-field of radio wave can be indicated both
in space and time. Also is known, it is variable both in amplitude and
direction. Judging by your conviction, the association of photons at
each moment of time has to create E-field in its interaction with the
receiver; this means, the association has to be charged. And not
simply charged but spatially sign-alternating! If so, we must not see
any light, the more from far astronomic bodies. According to your
relativistic statistics which you don't properly know but of which you
are pissing by something boiling, such unstable association of charged
bodies has to be averaged during little parts of a micro-second. ;-)
Third, from this second item one more absurd follows. You hopefully
will not argue, usually a light beam carries some spectrum. If we
locate our spectral device in free space at any distance from the
source, we will yield one and the same spectral composition of the
light beam. We can see the difference only for very far objects, due
to the Hubble shift. (And this addition already contradicts your
quantum-relativistic interpretation of this phenomenon, as in your
mind the Hubble effect is caused exceptionally by Doppler shift. But
if the source and receiver are reciprocally stationary, there has not
to be any shift). But even if you retain the affection of Hubble's red
shift, charged photons have to ruin the light beam. I will easily
prove you it. See.
If photons were charged, nothing prevents them from affecting each
other just so as they are affecting the receiver. The more, if they
were distributed in space and each wave of spectrum has its
wavelength, this would naturally form the packets where these photons
were located quite close to interact or even have an opposite
orientation! This naturally turns the spectrum into something averaged
with the lost relation between the photon energy and wave frequency.
However, we see a spectrum, not some grey background. ;-)
The fourth follows from this third. If the wave function only
distributed photon in space, it is not the function of the very
photon! Consequently, Planck equation
E = h*nu
does not relate to photon, only to some their distribution! But then,
what is the energy of the very photon, and which is its concern to the
Bohr's results? In Bohr's theory, just as in Schroedinger equation,
there are radiated photons of definite frequency and energy related to
this frequency? If all photons were equal, there does not exist the
basic equation with whose help you describe the pulse of photon etc.
This corroborates again, why discussions with you are long ago out of
my interest. People here say about such as you, wise head but fool its
master.
Sergey ;-)
.
|
