Re: Basics series proposed

jgreenfield_at_seol.net.au
Date: 01/04/05 

Date: 3 Jan 2005 22:50:52 -0800

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?
>
> > > > 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
>
> What is the momentum of a photon?

" " " spin rate "

>
> What is the angular momentum of a photon?

as above
>
> > 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?
(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.
>
> 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.

DHR's will be calling Mills and Boon "classics" next
>
> 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.
>
> > 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!
>
> > > > > 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

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???????????????

Jim G
c'=c+v

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