Re: Basics series proposed
From: Timo Nieminen (uqtniemi_at_mailbox.uq.edu.au)
Date: 01/03/05
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Date: Mon, 3 Jan 2005 14:26:27 +1000
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.
> > > 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?
What is the momentum of a photon?
What is the angular momentum of a photon?
> 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?)?
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.
> > 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.
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?
> 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.
> > > > Do note that the usual Maxwellian non-relativistic classical
> Doppler
> > > shift
> > > > says nothing about photons. Inserting the basic semi-classical
> > > photon, the
> > > > astute reader will realise that Doppler shift of frequency is not
> due
> > > to
> > > > the change in rate of reception of photons.
> > > >
> > > > Recommendation: Learn some classical wave theory.
> > >
> > > A machine gun approaching a target causes a higher frequency of
> strikes
> > > than one which is stationary. This is because the velocity of the
> > > bullets has increased, and the wavelength (distance between them)
> > > remains the same (rate of fire and muzzle velocity ref gun).
> >
> > And yet, sound waves, which have speed independent of the motion of
> the
> > source, also show Doppler shift. Clearly, Doppler shift doesn't
> require
> > constant velocity with respect to the source, or constant wavelength.
>
> For the nth time; what has a compression front in a gas got to do with
> photons travelling through a vacuum???????????
What do machine-gun bullets have to do with photons travelling through a
vacuum?
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.
> > 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.
> > Why not try to test it experimentally?
> >
> > 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?
> 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.
> > > Now, to believe frequency change (Doppler) for light is NOT due to
> > > motion of the source, one must accept the breath-taking magic that
> the
> > > light source KNOWS that it is moving ref the target, and adjusts
> its
> > > "muzzle velocity" accordingly..........or else accept space=jello
> > > (aether)
> >
> > So, if frequency of light depends on the rate that photons are
> received at
> > (a dubious proposition, easily overturned by simple experiment, but
> since
> > it's the basis of your argument, we'll make use of it), where do the
> extra
> > photons go in gravitational redshift?
>
> They don't "go" anywhere! They have had their emission velocities
> reduced while escaping the grav field, and there are more of them in
> transit at any instant.
You have been stating repeatedly that the frequency of light depends on
the rate that photons are received at. How does the flux of photons change
when the source is stationary?
To use your favoured machine-gun analogy, a stationary machine gun firing
upwards at a stationary target at 600 rpm will hit that target at 600 rpm,
as long as the bullets reach.
So, explain gravitational redshift in terms of your theory.
Actually, it would be useful if you could quantify the effect of gravity
on photons (ie on their speed, energy, and momentum) in your theory. This
would tell us what your theory predicts the rest mass of a photon to be.
Are you brave enough to subject your theory to testing?
-- Timo
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