Re: Basics series proposed
From: Timo Nieminen (uqtniemi_at_mailbox.uq.edu.au)
Date: 01/04/05
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Date: Tue, 4 Jan 2005 18:41:14 +1000
On Tue, 3 Jan 2005, jgreenfield@seol.net.au wrote:
>
> Timo Nieminen wrote:
> > On Mon, 2 Jan 2005, jgreenfield@seol.net.au wrote:
> >
> > > Timo Nieminen wrote:
> > > > On Mon, 2 Jan 2005, jgreenfield@seol.net.au wrote:
> > > >
> > > > > Timo Nieminen wrote:
> > > > > > On Sun, 1 Jan 2005, jgreenfield@seol.net.au wrote:
> > > > > >
> > > > > > > Astute readers will understand that Doppler changed
> frequency
> > > > > (light in
> > > > > > > vacuum) is due to more or less photons arriving at the
> receiver
> > > per
> > > > > > > TIME.
> > > > > >
> > > > > > Counterexample: Rotational frequency shift. Photon flux is
> > > unchanged.
> > > > > > Frequency is.
> > > > >
> > > > > Photon flux is energy/area/time?
> > > >
> > > > No, number of photons per area per time. Energy per area per time
> > > would be
> > > > energy flux. In rotational Doppler shift, photon flux is
> unchanged,
> > > energy
> > > > flux is changed.
> > >
> > > Pedantic? The assumption seems to be that individual photon energy
> does
> > > not change.
> >
> > Whose assumption? Yours? Why would you assume that? Your emission
> theory
> > appears to suggest an increase in energy if the velocity increases
> > (although you've consistently refused to try to quantify the change).
> > Currently accepted theory explicitly assumes it changes - change in
> > frequency means there must be a change in photon energy.
>
> You refer to doppler in air regularly as a correllation with light
> propagation; so where is the energy of sound? in the frequency or the
> AMPLITUDE?
Here, we are not discussing sound, we are discussing light. You stated
that the "assumption seems to be that the individual photon energy does
not change." I pointed out that both your theory seems to state, and
currently accepted theory does state, that the photon energy does change.
I see you have no substantive reply to the comment.
I also see a complete avoidance of discussion of relevant physics.
Why dodge? Surely, unless you're just a particular persistent troll
dangling the same bait every time, or you know your beliefs are without
foundation, you would be interesting in discussing the relevant physics.
> > > > > Highly likely that there are LESS photons which have GREATER
> energy
> > > > > (frequency) leaving flux unchanged.
> > > >
> > > > Same number, either more or less energy each, depending on
> whether or
> > > not
> > > > the frequency is increased or decreased.
> > >
> > > "Energy flux is changed" (above): so if we have the SAME number of
> > > photons/time which individually have more energy, then they are
> going
> > > FASTER with the same separations (wavelength)ie increased
> frequency.
> >
> > According to your theory. Not according to currently accepted theory.
> > Perhaps if you were to quantify your theory, then it might be
> possible to
> > test experimentally. So, to be productive, why not tell us the
> following
> > results from your theory:
> >
> > In terms of speed, wavelength, and frequency:
> >
> > What is the energy of a photon?
>
> increased if given greater linear motion (KE/Newton0
Can you provide a quantitative answer?
You appear to state that the velocity of photons of greater energy is
greater. By how much? Be quantitative, please.
If you are willing to provide a quantitative answer, then the validity of
your theory can be tested.
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
> > What is the momentum of a photon?
>
> " " " spin rate "
Define "spin rate". How is the momentum of a photon related to its "spin
rate"? Are you claiming that a photon is a rigid spinning ball?
A Newtonian rigid spinning ball doesn't gain linear momentum when its spin
rate is increased. Why are photons different?
Caan you provide a quantitative answer, please?
If you are willing to provide a quantitative answer, then the validity of
your theory can be tested.
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
> > What is the angular momentum of a photon?
>
> as above
Since your "spin rate" is undefined, this is unfortunately meaningless.
Can you provide a quantitative answer, please? Let us be specific! For a
left-circularly polarised plane wave of irradiance 1W/m^2, what is the
angular momentum flux if (a) the wavelength is 633 nm and (b) if the
wavelength is 1064 nm? What is the angular momentum per photon in each
case? Sufficient to give just the vector component in the direction of
propagation in each case, assume free space.
If you are willing to provide a quantitative answer, then the validity of
your theory can be tested.
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
> > > It is so ingrained in the DHR psyche that c NEVER alters, that you
> > > continualyy overlook the fact that there are 3 variables in
> > > c= wavelength x frequency ie you CANNOT calculate one without
> KNOWING
> > > BOTH the others, and c IS variable according to the motion of its
> > > source (vacuum)
> >
> > Since when is "vacuum" a source? How do you suggest that the motion
> of
> > vacuum be measured (isn't that an ether theory idea, anyhow?)?
>
> Infantile idiot! How childish is it to deliberately misconstrue a
> reminder?
I'm trying to discuss physics, you start with unclear writing, and then
follow up with insults. You state that "c IS variable according to the
motion of its source (vacuum)." A literal reading is that you are
reminding the reader that the source is vacuum. Don't blame other for you
inability to clearly express what you mean.
However, you are kind enough to clarify what you meant by your confused
prose:
> (vacuum) was put in to remind you that at all times I discuss light
> propagation in abscence of any other material------which you either
> knew and are deliberately deceitful to suggest otherwise, or are just
> plain stupid.
although you do manage to be particularly ill-mannered about it.
Given your clarification, perhaps you could now explain what your original
statement meant? You appear to be claiming that c = wavelength x frequency
is sufficient proof of your emission theory. Is this so? If not, does your
original claim have any meaning?
Or do you feel that insults are a sufficient response in a discussion of
physics?
> > But it's irrelevant with respect to rotational frequency shift, the
> > closing\receding velocity of the source is zero.
> >
> > If you disagree, feel free to come up with a quantitative prediction
> for
> > rotational frequency shift based on your theory.
Lack of quantitative (or indeed any) answer noted.
If you are willing to provide a quantitative answer, then the validity of
your theory can be tested.
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
> > > > Photon angular momentum is independent of frequency. This is both
> an
> > > > experimental result, and a theoretical result. Straight from
> > > classical
> > > > theory, angular momentum flux for circularly polarised light is
> > > > power/(angular frequency), which means that the angular momentum
> per
> > > > photon is independent of frequency.
> > >
> > > As R disclaims a photon having ANY mass, referring to its angular
> > > momentum with classical connotations is oxymoronic.
> >
> > You think the Maxwell equations are oxymoronic? Angular momentum of
> > electromagnetic fields is a fully classical result. If you're
> unwilling to
> > learn, it's only your loss.
>
> DHR's will be calling Mills and Boon "classics" next
And what relevance does this have to do with the fact that both experiment
and currently accepted classical theory agree on the angular momentum of
electromagnetic fields?
Your lack of relevant answer is noted. Do you feel that insults are a
sufficient response in a discussion of physics?
> > Relativity alone says nothing about photons - SR and GR are classical
> (ie
> > non-quantum) theories.
> >
> > What relativity does do is provide an answer to how massless photons
> > behave. What answer does Newtonian mechanics provide?
> >
> > From the experimentally measured ratio of energy to momentum fluxes
> of
> > light, what is the Newtonian prediction for the mass of the photon?
> >
> > What does your theory say? What is the mass of a photon?
>
> Enough, that a beam of them can support (overcome the gravity acting
> on) a massive particle. FYI that just confirms the KE of the photons,
> and conservation of energy.
There are experimental measurements of electromagnetic radiation pressure
(over 100 years old, even). What is your quantitative prediction, based on
those experimental measurements?
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
However, your response is puzzling! In Newtonian mechanics, the force
exerted by a beam of light is equal to the rate of transfer of momentum
from the beam to the object the force is exerted on! But you claim that
the KE of the photons is important. Have you re-writted Newtonian
mechanics in the course of your work?
> > > I suspect a photon acts like a bar magnet, exercising 'push-pull'
> as it
> > > approaches/passes a receiver. More spins/time=>frequency......more
> > > velocity, same spin rate=>frequency
> >
> > Go ahead, define "spin", as you mean it, precisely. Provide a
> quantitative
> > version of your above hand-waving.
> >
> > And think of a better term - "spin" is already used to mean the
> intrinsic
> > angular momentum.
>
> The Pakis could do with another "rotational" bowler, since the "spin"
> word has apparently been copyrighted by AE!
You really think that there is sufficient likelihood of confusion of
techincal terms from physics and cricket that one or the other should
change? Really?
If not, do you feel that pointless irrelevancies are a sufficient response
in a discussion of physics?
> > If you believe that a bullet is a suitable model of a photon, please,
> tell
> > us the mass of a photon. Please, give a quantitative description of
> the
> > realtionship between speed, frequency, and rate of receiving photons.
Lack of any answer noted!
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
> > > > To state that Doppler shift of light *requires* constant
> wavelength
> > > is
> > > > just plain wrong.
> > >
> > > As above, the spin rate may change, but otherwise the wavelength
> won't
> > > change- the c will
> >
> > As above, that has nothing to do with the point that examples abound
> where
> > Doppler shift does not require changing velocity and constant
> wavelength.
Lack of any answer noted!
If are are unwilling to provide a quantitative answer, then it looks like
either (a) you know your theory is wrong and would fail such quantitative
testing or (b) you have no real idea what your theory predicts, in which
case, how can you be so confident it is correct? Is your intuition really
a better guide to reality than experiment?
> > > > Consider a diffraction grating. Angle of diffraction depends on
> the
> > > > wavelength, not the frequency. So, the position of a spectral
> line
> > > should
> > > > not change as the closing velocity of the source changes,
> according
> > > to
> > > > your theory.
> > >
> > > If the wavelength changes, but NOT the frequency, c has
> altered!!!!!
> >
> > Sure, but what's that got to do with anything we've been discussing?
> > Currently accepted theory states that both the wavelength and
> frequency
> > change, and you say your theory states that the frequency changes and
> the
> > wavelength is constant.
> >
> > > ...and no, the POSITION of the Spec line doesn't alter, as ALL the
> > > light shifts.
> >
> > Absolute position. As in, at what angle relative to the plane of the
> > grating will I find a particular spectral line.
> >
> > Your theory appears to imply that the absolute position will not
> change.
> > Care to test that experimentally?
Lack of any answer noted. Do you really not want to test your theory
experimentally? Surely, since you *know* that it is correct, you shoudl be
eager for the opportunity!
> > > Say EMR ranges from A...........G, and we see C-D. If the whole
> system
> > > speeds up ref us, we now see what was fomerly B-C (blue shifted
> and
> > > same position of spec lines)
> >
> > No, you miss the whole point. The angle of diffraction with a
> diffraction
> > grating depends on the *wavelength*. If the wavelength doesn't
> change,
> > then , then the spectrum *will not move*. Measured with something
> that
> > depends of frequency, then one should observe the frequency change.
> Now,
> > wouldn't that be incontrovertible evidence supporting your theory?
> >
> > And yet, you keep on refusing to even try to design such an
> experiment,
> > preferring to bemoan the lack of cooperation from NASA.
> >
> > Well, if that's the fate you *want* for your theory, that's your
> business.
Lack of any answer noted. Do you really not want to test your theory
experimentally? Surely, since you *know* that it is correct, you shoudl be
eager for the opportunity!
> A certain radio crystal emits a certain wavelength, right?
> Now take the crystal on a moving spaceship, and the frequency alters.
> So, as the wavelength CANNOT alter, and the received frequency DOES,
> how DOES c not alter???????????????
After waffling, providing empty handwaving (to be charitable), and
insults, while evading all substantial points of physics, you provide the
above almost unrelated closing note.
Why do you say the wavelength CANNOT alter? Maxwell's original (ether)
model sure stated that the wavelength from a moving source can be
different from that from a stationary source. The equations Maxwell
obtained seem to work remarkebly well in practical engineering, and
nowadays do not assume the existence of ether, but still provide the
result that the wavelength changes. Lorenz's equivalent set of equations,
with no assumptions of ether give the same answer. Relativistic quantum
field theory says the same.
It seems that your assertion that wavelength CANNOT alter is exactly the
place where the conflict between your theory and currently accepted theory
can be cheaply and conclusively tested. Well, I provided some pointers for
you to design such experiments above, but you didn't reply. Don't you want
ot cheaply and conclusively test your theory? You prefer to whine about
NASA being uncoopreative?
Or is your assertion that wavelength cannot change such a self-evident
religious belief that it can be assumed to be true, even if contradicted
by experiment or observation?
-- Timo
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