Why are the 'Fixed Stars' so FIXED?

"Henri Wilson" HW@.... wrote in message
...

....

A photomultiplier produces a flash for each photon, you should know
that. The basic physics is the photoelectric effect. An electron
ejected
by a photon creates a cascade that generates enough light on the
final phosphor to be measured.

A very sensitive PM might pick up single photons.

All PMs pick up single photons, that's their job!

Their main job is to amplify very weak light signals. A single photon
could
barely be seen above the noise.

This is the experiment done with electrons rather than
photons but if you saw a video of the photon version
it would look exactly the same:

http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm

The site seemed slow and I had to download the movie
rather than view it on-line but it's worth a look so
that you understand the appearance of what we are
discussing. The regions where most photons land are
of course the same as the locations of the fringes
predicted by Huygens' method hence K=1.

George

On 4 May 2007 01:08:57 -0700, George Dishman wrote:

On 4 May, 00:35, HW@....(Henri Wilson) wrote:
On 3 May 2007 02:58:34 -0700, George Dishman wrote:



I have George...
You refuse to listen.
There are plenty of instances where the BaTh does NOT require continued
ballistic movement.

That's why I don't listen, the theory is that light moves at c+v. If
it is to be a theory it must apply ALWAYS so the moment you
say sometimes it doesn't your description is self-contradictory.

I gave you another analogy where there is continued but
reduced movement.
You are assuming the ends of a photon are completely independent of each other.

You give lots of analogies but they don't reflect the theory, c+v.
That
equation applies to every wave in your classical wavetrain version of
a photon so the analogy is wrong.

Your problem George, is that you keep flipping backwards and forwards between
classical wave theory and BaTh.

The very fact that an individual photon can exist means that it must have some
kind of identifying property and 'structure' that distinguishes it from
'nothing'. There is absolutely no reasn to assume the ends will continue to
move in way pulsar pulses appear to.

There is no reson to assume anything Henry, but ballistic theory
says each end moves at c+v where v is the speed of the source when
the end was emitted.

No George, you just want to believe that is true.
If it were true it wouldn't match the evidence. So, since there is an
alternative theory that works - ie., mine - let's choose it instead.


Wavefronts approach a series of wires (the asterisks are the wires
seen
along their length.

_____________________________
_____________ _____________
_____________ | _____________
_____________ | _____________
_____________ v _____________
_____________________________
_____________________________
_____________________________

* * * * * * * *

The field induces current in the wires which then re-radiate. That
raiation should be isotropic in a plane perpendicular to the axis
of the wires but interferences creates 'lobes' or preferred
directions.
There is a peak of radiation at any angle (which depends on the
frequency or wavelength) where the waves arrive in phase. The
reflected wavefronts might go like this:

\ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \
\
* * * * * * * *

Changing the scale so the horizontal line is the grating, the
signal arriving vertically downwards is reflected at some angle
like this and the maximum field, say measured with a dipole
and oscilloscope as mentioned in my other post, might have
a maximum at 'A'. What you are telling me is that TDoppler
affects the waves at a macrosopic level hence the angle to
A but that if we think of that wave as bing delivered in small
bursts, each burst of classical waves (which you are calling
a photon) is affected by an amount which is smaller by the
factor K so the bursts all get deposited at location B.

|
| \A
| /\
| / \B
__|/_ \

You can't have all the "photons" arriving at B and the energy
appearing at A.

The question you want me to answer is, "how is an individual photon affected by
being in a group of photons?"

No, I'm not asking you a question, I am pointing out something
that is so obvious I shouldn't have to waste my time saying it.

It is not obvious at all. It isn't even true. The wave energy goes to A and the
intrinsic photon energy to B.
Think of a diffracted water wave. The KE of the individual molecules doesn't go
to the nodes.

"Photon" is the name we give to a packet of energy. You cannot
say that the photons will land at point B above and also that the
energy is all deposited at point A.

I don't see why not. I say the summed photon energy is not the wave energy.

I doubt if anyone can give you a decent answer to that....but like I said
before, I doubt if individual photons making up an RF signal would be
diffracted in the same way as the macroscopic wave itself.

Henry, the macroscopic wave is nothing more than the
collection of photons. You don't have some sort of wierd
composite of a wave being carried along by photons like
pall-bearers carrying a coffin.

Look, this 'wave' you keep refering to is just a graph of a varying physical
process. I'm suggesting various physical processes that might fit in with all
the known data. One is as I've described it....individual photons bunching
together and moving apart periodically as they travel. At the same time they
each have their own intrinsic 'oscillation' that may or may not be somehow
synchroinized with the former one.

when they fit the data? That's not a
very scientific attitude George.

The first rule of a scientific attitude is slf-consitency. Your
nonsense
predicts two contradictory locations for the light reflected from a
grating, that's why I dismiss it as nonsense.

You have twisted the meaning of what I am saying...

It seems you have at least finally understood the argument.

I have been quite aware of the argument all along George.

and since there is no
evidence either way, you don't have a leg to stand on.



....and it doesn't happen like that ...

Right, so the variation must be intrinsic.

No George.
I have explained.

Your 'explanation' was self-contradictory.

I will not do it again.

Promises, promises.

I can see I will have to...

You are the only one confused Henry, BaTh is very simple. It
says both luminosity and Doppler shift are due to to TDoppler,
and if K 1 then both get changed. The relationship between
the two is unaffected by K.

...but that doesn't fit the data George...

Yes it does Henry, Cepheid variations are intrinsic to the star.

Your classical wave theory doesn't apply to individual photons nor does your
'ballistic shrinkage'.

Your so-called "photons" are nothing more than bursts of classical
waves, the view Planck held in 1914, and a purely classical analysis
is entirely valid.

Nobody knows anything about the particle nature of a photon George,

Sorry Henry, QED tells us all about photons and is the most
precise and complete theory in science. You might be clueless
but that's your choice.

More useless stats...

so don't
try to make out they do. It is clearly not just a 'classical wave'. It is a
package of energy with definite properties and structure.

It is indeed.

At last we agree on something.

Wrong again henry, when will you learn. BaTh says it is TDoppler
that matters.

Of course,...

Good, try to remember that, you keep making the same error
and I'm fed up correcting it. My point regarding a Fouier analysis
stands.

The VDoppler component is generally negligible. Forget it...

Still can't handle reality eh, you have proved it is always
dominant.

You are a bigger dreamer than Andersen...


if it had that same frequency. It must do Henry, the individual wave
bursts
cannot be deflected by a grating at an angle different from the bulk
wave.

It is the wave formed by the group of photons that is diffracted.
If you are considering monochromatic light, naturally all photons will be
diffracted to the same angle.

Exactly, so the photons must have the same wavelength as the
macroscopic wave.

How long is a 1 hz photon, George?


If you want to use those for an analogy that accurately reflects the "c
+v"
equation, each car or bullet is but a point on the wave within the
long train
of waves that you call a photon. It's easier to think of them
representing
peaks of the sine wave but any other part is equally valid.

When water waves are diffracted, it is a macroscopic effect, not a molecular
one.

The difference is that we know the photons have the same
wavelength.

No we don't.

analogy would be that the macroscopic
waves have a minimum height and multiple overlapping waves
with the same phase add up to make a larger wave. Obviously
the larger amplitude wave has the same wavelength as the
smaller ones.

You seem to have a funny idea that a wave is just a squiggly line painted in
the sky. It isn't . It is a graphical representation of a physical process.
When are you going to get down to basics, George.

if you could 'diffract' the line of cars on the highway, you would get two
different angles. One being due to the 'wavelength' defined by the distance
between cars and the other but the lengths of the cars themselves.

I suppose you would get another angle from the fact that the wheels were all
the ame diameter.

Some analogies don't reflect the physics correctly, you must chose
them carefully.

Let's talk about the charged rubber cars then.

The rest is intrinsic.
That interpretation matches all you have discovered from
pulsars and EF Dra and there is no evidence to question it.

I gave you MY light curve for EF Dra. It fits an ADoppler type prediction..

No, the one that fitted was an error using half the period (I'm
inclined to
give you the benefit of the doubt and think it wasn't deliberate
fraud) but
the phase is actually 90 degrees out from ADoppler, it matches
VDoppler.

Not in my diagram. The brightness peak, which is eclipsed, occurs when the star
is on the LOS, furthest away...

No, that doesn't match what is observed. See below.

That is where maximum ADoppler occurs. Do you
not agree George?

No, it is 45 degrees out with the two merged curves.

So the BaTh wins again.

No, you are just in denial again.

snip repetition
but 90 degrees out from the orbital phase, the curves are VDoppler.

No George,

Yes Henry, the peak of the velocity curve is half way between
the eclipses, not coincident with them.

No George, have another look at:- Hide quoted text -

I hate Google's new interface !

The diagrams are

http://www.users.bigpond.com/hewn/efdrag.jpg
http://www.georgedishman.f2s.com/Henri/EF_Dra.png

If you put them together you will see the phase is now
45 degrees out, exactly half way between VDoppler and
ADoppler so they give exactly equal contributions.

Where did that idea come from?
My projected brightness peak lies right over the zero radial velocity mark.
That is exactly what the BaTh expects.

That
is a remarkable coincidence given VDoppler is asymptotic
to a constant level while ADoppler varies with distance. I
also haven't had time to consider whether the sense of the
contributions is valid but your program should be able to
test that. In fact I should check this out a bit more so I may
find more errors in this later.

You are wrong here george.

Coincidences aside, certainly you have again ruled out the
possibility of ADoppler being dominant or even significantly
larger than VDoppler.

The brightness curve is all ADoppler.

- Show quoted text -- Hide quoted text -

- Show quoted text -...

read more »

I'll reply to the rest later.

George



www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

On 4 May 2007 01:41:03 -0700, George Dishman wrote:

On 4 May, 00:35, HW@....(Henri Wilson) wrote:

...

No George, have another look at:www.users.bigpond.com/hewn/efdrag.jpg

The peak velocity curve is in phase with the peak brightness curve, which in in
phase with hte eclipses.

I had another look Henry, it is a fake again! The top is a cut-off
ellipse and
you have then drawn a number of dots along the actual curve by hand.

Of course.

Show a screen capture from your program, state the orbital parameters
and
_copy_ the curve onto a composite diagram like mine showing both
luminosity
and velocity curves with the correct relative phasing:

http://www.georgedishman.f2s.com/Henri/EF_Dra.png

You are a charlatan Henry, a plain old fraud.

George, my diagram was never supposed to be accurate. It was merely
demonstrating the basic idea.
I will make a more accurate one for you if you like.

George



www.users.bigpond.com/hewn/index.htm

Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

"Henri Wilson" HW@.... wrote in message
...
On 4 May 2007 01:08:57 -0700, George Dishman
wrote:

On 4 May, 00:35, HW@....(Henri Wilson) wrote:
On 3 May 2007 02:58:34 -0700, George Dishman
wrote:



I have George...
You refuse to listen.
There are plenty of instances where the BaTh does NOT require continued
ballistic movement.

That's why I don't listen, the theory is that light moves at c+v. If
it is to be a theory it must apply ALWAYS so the moment you
say sometimes it doesn't your description is self-contradictory.

I gave you another analogy where there is continued but
reduced movement.
You are assuming the ends of a photon are completely independent of each
other.

You give lots of analogies but they don't reflect the theory, c+v.
That
equation applies to every wave in your classical wavetrain version of
a photon so the analogy is wrong.

Your problem George, is that you keep flipping backwards and forwards
between
classical wave theory and BaTh.

BaTh IS a classical wave theory Henry.

The very fact that an individual photon can exist means that it must
have some
kind of identifying property and 'structure' that distinguishes it from
'nothing'. There is absolutely no reasn to assume the ends will continue
to
move in way pulsar pulses appear to.

There is no reson to assume anything Henry, but ballistic theory
says each end moves at c+v where v is the speed of the source when
the end was emitted.

No George, you just want to believe that is true.

I am just repeating back to you what the theory says.

If it were true it wouldn't match the evidence.

It does match but it requires a low value of speed
equalisation and it means that ADoppler is always
undetectable. That isn't surprising since in reality
it doesn't exist.

So, since there is an
alternative theory that works - ie., mine - let's choose it instead.

I'm not wasting my time on a self-contradictory set
of ideas.

Wavefronts approach a series of wires (the asterisks are the wires
seen
along their length.

_____________________________
_____________ _____________
_____________ | _____________
_____________ | _____________
_____________ v _____________
_____________________________
_____________________________
_____________________________

* * * * * * * *

The field induces current in the wires which then re-radiate. That
raiation should be isotropic in a plane perpendicular to the axis
of the wires but interferences creates 'lobes' or preferred
directions.
There is a peak of radiation at any angle (which depends on the
frequency or wavelength) where the waves arrive in phase. The
reflected wavefronts might go like this:

\ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \ \
\ \ \
\
* * * * * * * *

Changing the scale so the horizontal line is the grating, the
signal arriving vertically downwards is reflected at some angle
like this and the maximum field, say measured with a dipole
and oscilloscope as mentioned in my other post, might have
a maximum at 'A'. What you are telling me is that TDoppler
affects the waves at a macrosopic level hence the angle to
A but that if we think of that wave as bing delivered in small
bursts, each burst of classical waves (which you are calling
a photon) is affected by an amount which is smaller by the
factor K so the bursts all get deposited at location B.

|
| \A
| /\
| / \B
__|/_ \

You can't have all the "photons" arriving at B and the energy
appearing at A.

The question you want me to answer is, "how is an individual photon
affected by
being in a group of photons?"

No, I'm not asking you a question, I am pointing out something
that is so obvious I shouldn't have to waste my time saying it.

It is not obvious at all. It isn't even true. The wave energy goes to A
and the
intrinsic photon energy to B.

There is no "wave energy" Henry. Reduce the brightness
and you can see individual photons arriving. The first
goes to point A. A few seconds later the next goes to
point A and so on. Between them nothing happens. The
only hint of a wave is the fact that the location of
point A can be calculated by Huygens method but the
total energy delivered is no more than the sum of the
energy of the photons.

Think of a diffracted water wave. The KE of the individual molecules
doesn't go
to the nodes.

"Photon" is the name we give to a packet of energy. You cannot
say that the photons will land at point B above and also that the
energy is all deposited at point A.

I don't see why not. I say the summed photon energy is not the wave
energy.

Photons create flashes on PM tubes and nothing happens
between them.

I doubt if anyone can give you a decent answer to that....but like I
said
before, I doubt if individual photons making up an RF signal would be
diffracted in the same way as the macroscopic wave itself.

Henry, the macroscopic wave is nothing more than the
collection of photons. You don't have some sort of wierd
composite of a wave being carried along by photons like
pall-bearers carrying a coffin.

Look, this 'wave' you keep refering to is just a graph of a varying
physical
process.

Sure, it is a graph of the summed effect of the photons.

I'm suggesting various physical processes that might fit in with all
the known data.

Other than the most fundamental fact, the energy is
transferred as packets or "quanta" just like atoms of
an element.

snip repetition

if it had that same frequency. It must do Henry, the individual wave
bursts
cannot be deflected by a grating at an angle different from the bulk
wave.

It is the wave formed by the group of photons that is diffracted.
If you are considering monochromatic light, naturally all photons will
be
diffracted to the same angle.

Exactly, so the photons must have the same wavelength as the
macroscopic wave.

How long is a 1 hz photon, George?

With a sufficiently sensitive (hypothetical!) material
it would knock one electron out of one atom, no more,
just like optical photons or gamma rays.

If you want to use those for an analogy that accurately reflects the "c
+v"
equation, each car or bullet is but a point on the wave within the
long train
of waves that you call a photon. It's easier to think of them
representing
peaks of the sine wave but any other part is equally valid.

When water waves are diffracted, it is a macroscopic effect, not a
molecular
one.

The difference is that we know the photons have the same
wavelength.

No we don't.

Yes we do, a single photon has the same probability
distribution as a plot of the distribution of many
photons from the same source, obviously since the
photons are just the photons.

analogy would be that the macroscopic
waves have a minimum height and multiple overlapping waves
with the same phase add up to make a larger wave. Obviously
the larger amplitude wave has the same wavelength as the
smaller ones.

You seem to have a funny idea that a wave is just a squiggly line painted
in
the sky. It isn't . It is a graphical representation of a physical
process.
When are you going to get down to basics, George.

I've been teaching you them for days but you seem to have
trouble grasping them. I think it would be different if
you had actually seem photons arriving on a PM tube.

Yes Henry, the peak of the velocity curve is half way between
the eclipses, not coincident with them.

No George, have another look at:- Hide quoted text -

I hate Google's new interface !

The diagrams are

http://www.users.bigpond.com/hewn/efdrag.jpg
http://www.georgedishman.f2s.com/Henri/EF_Dra.png

If you put them together you will see the phase is now
45 degrees out, exactly half way between VDoppler and
ADoppler so they give exactly equal contributions.

Where did that idea come from?

The luminosity and velocity curves predited by BaTh
are in phase because both are due to TDoppler. The
peak of the luminosity curve would match the peak of
the velocity curve at 0.5 phase, the dip would match
the lowest negative peak of the velocity at 0.0 phase
and the zero of the predicted velocity would be half
way between those so at phases of 0.25 and 0.75.

My projected brightness peak lies right over the zero radial velocity
mark.

You have faked it coinciding with the zero, but I think
matching the velocity curve properly would put it 45
degrees out.

That is exactly what the BaTh expects.

That
is a remarkable coincidence given VDoppler is asymptotic
to a constant level while ADoppler varies with distance. I
also haven't had time to consider whether the sense of the
contributions is valid but your program should be able to
test that. In fact I should check this out a bit more so I may
find more errors in this later.

You are wrong here george.

I don't think so though it isn't obvious.

Coincidences aside, certainly you have again ruled out the
possibility of ADoppler being dominant or even significantly
larger than VDoppler.

The brightness curve is all ADoppler.

If you do it properly instead of faking it, I think you
will find it doesn't work. Remember also ADoppler creates
narrow peaks and wide minima and you have faked the
opposite. I know you can get that reversed by using
eccentricity but I think it might work the other way
round and make the problem worse on the other star.

Whatever the outcome, you need to do it properly to see
what really happens, you are nowhere near a match yet.

George

On May 4, 3:46 pm, HW@....(Henri Wilson) wrote:
On 4 May 2007 01:41:03 -0700, George Dishman wrote:

I had another look Henry, it is a fake again! The top is a
cut-off ellipse and you have then drawn a number of dots along
the actual curve by hand.

Of course.

Show a screen capture from your program, state the orbital
parameters and _copy_ the curve onto a composite diagram like
mine showing both luminosity and velocity curves with the
correct relative phasing:

http://www.georgedishman.f2s.com/Henri/EF_Dra.png

You are a charlatan Henry, a plain old fraud.

George, my diagram was never supposed to be accurate. It was
merely demonstrating the basic idea.

That's a pretty pathetic attempt to explain away your forged fit.

Jerry

"Henri Wilson" HW@.... wrote in message
...
On 4 May 2007 01:41:03 -0700, George Dishman
wrote:

On 4 May, 00:35, HW@....(Henri Wilson) wrote:

...

No George, have another look at:www.users.bigpond.com/hewn/efdrag.jpg

The peak velocity curve is in phase with the peak brightness curve,
which in in
phase with hte eclipses.

I had another look Henry, it is a fake again! The top is a cut-off
ellipse and
you have then drawn a number of dots along the actual curve by hand.

Of course.

Show a screen capture from your program, state the orbital parameters
and
_copy_ the curve onto a composite diagram like mine showing both
luminosity
and velocity curves with the correct relative phasing:

http://www.georgedishman.f2s.com/Henri/EF_Dra.png

You are a charlatan Henry, a plain old fraud.

George, my diagram was never supposed to be accurate. It was merely
demonstrating the basic idea.
I will make a more accurate one for you if you like.

Use your program. Set up orbital parameters that match
one of the velocity curves using the green curve (I know
it is supposed to be luminosity but it also matches
velocity with a different scale as we have discussed and
I guess you haven't added the "red curve" yet). The
luminosity variation will be small and the dips are due
solely to the eclipses.

Now adjust your speed equalisation distance until you get
the phase right relative to the eclipses which tell you
when the stars are on the LoS. You will find the effect
is pure VDoppler.

George

On Fri, 4 May 2007 13:58:52 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On 3 May 2007 02:58:34 -0700, George Dishman
wrote:

[continued]



For an inertial source, the length of the photon is c/N. ...but for an
accelerating source, it is something different, the variation being a
function
of a, not v.

Right, Different waves in the packet therfore get different speeds
and the usual c+v bunching factor due to acceleration applies.

Initially that may be true...but I'm suggesting any such differences are
quickly dampened out and the photon settles down to a length that reflects
its
average emission ACCELERATION.

And I agree, speed equalisation does precisely that.

Well that's settled then.

This would suggest that photon 'shrinkage' occurs only at origin
time....AND IT
IS ACCELERATION DEPENDENT.

No, that contradicts what you just said.

Well it could go on for a little time after emission.

Yes, the speed equalisation distance that you already
include in your program.

No. The intra-photonic movement settles down long before the inter-photonic
movement does.

(Note: two new Wilsonian terms)



But we don't agree that the rate within a photon is far greater than the
rate
BETWEEN photons.

The rate is fixed by your speed equalisation factor.

The inside of a photon has completely different properties from the space
between photons. Why should the two be the same?


Not entirely. The oscillations could be related to the orbiting of a
large
second body. ..after all the constancy of cepheid periods strongly
suggests
some kind of connection with an orbit.

No, Cepheid variation is less stable.

So are many orbit periods.

No orbital periods are more stable and don't show the
discontinuous phase changes of Cepheids.

There are plenty of complex orbit systems that would cause that effect.
There can also be a long term Vdoppler shift caused by a whole cepheid system
being in a long period orbit around a galactic centre or similar.

Its huffing is
analogous to orbiting eccentrically as far as radial velocity is
concerned.
The BaTh DOES however provide a perfectly sound and accurate
expanation for the
brightness variation, something no other theory can do.

Rubbish, plasma theory shows how the opacity changes and
thermodynamics, radiation pressure and ordinary dynamics
(momentum) does the rest.

Well, I haven't found paper yet where the author claim to have found a
convincing link between huffing and brightness.

You would be better to look in a textbook.

ROFL, that's always your answer Henry, if you can't
cope, bury your head.

Burn the book.

A photomultiplier produces a flash for each photon, you should know
that. The basic physics is the photoelectric effect. An electron
ejected
by a photon creates a cascade that generates enough light on the
final phosphor to be measured.

A very sensitive PM might pick up single photons.

All PMs pick up single photons, that's their job!

Their main job is to amplify very weak light signals. A single photon
could
barely be seen above the noise.

********, see these stills:

It's not ******** George. PMs were initially used to amplify very weak light
signals.
The fact that the principle can be used to detect single photons is an added
bonus.


http://ophelia.princeton.edu/~page/single_photon.html

There is no PM in this experiment.

What do you think might happen when you send an RF signal through a PM
George?


Same thing, EM is EM, but the energy per photon is going to be very
low so you would need a detector with much better sensitivity than
is currently available.

Yep.

That's OK.
Perfectly in accord with hte P.E. effect.

Of course, but it requires that the "wavelength" of a single photon
is the same as the macroscopic wave of which it is a part, hence
K=1.

Bull....

Plain bull!!!!

Required for self-consistency Henry, see the grating discussion above.

Not required at all. Explained above...

Sorry Henry, wittering about rubber cars or something
which conflicts with your own equations isn't an
"explanation".

It's a simple demonstration of the principle involved.


Yes, so? What is the BaTh equation?

I don't knw....How long does the contact last?

So there you are you see, you don't have any equation so
you don't know whether speed appears in it or not.

The FREQUENCY of wavecrest arrival is what the BaTh uses.

I'm happy at this stage just to match brightness and velocity curves.

You can match the velocity curves and they are VDoppler dominated,
but you cannot match the luminosity curves without speeds greater
than c.

www.users.bigpond.com/hewn/efdrag/jpg...... The BaTh wins....

Faked, and still 45 degrees wrong, you can't even
cheat successfully.

George, so far you have been a great help to me. So much so I will give you
quite a mention when I write all this up. It is now all coming together nicely.

I just hope your desperation is not going to cause you to make stupid
elementary errors like this.

THE BLOODY BRIGHTNESS PEAK IS EXACTLY IN PHASE WITH THE CENTRE OF THE ECLIPSE.

Which is the BaTh prediction.

It is not contradictory ...

It is contradictory, it would have the same photons
landing in two different places.

Monochromatic light is made up of many identical photons, all with intrinsic
'absolute wavelengths' of whatever the main beam exhibits.

An RF signal is made from many possibly varied photons, the intrinsic
wavelengths of which are not the same as the 'absolute wavelength' of the
signal.


Nope, the result would be an extreme broadening of spectral lines
which isn't displayed in any way.

Most is unified before it leaves the star's influence.

Try the sums. I think that's how the page on Sekerin gets
the speed equaisation distance of ~5 microns (from memory).
Certainly that would be "before it leaves the star's
influence." :-)

That's great!
It ensures that thermal molecular speeds are neutralised and that all light
leaves the star at exactly c wrt that star.
Thanks again George.


Speed equalization wasn't part of the theory he was commenting
on so he was right. AFAIK that bodge was added after he was dead
so he didn't comment on it at all.

Extinction refuted his arguments.

Extinction woluld not be required if his argument
was incorrect. He was right and Ritzian theory had
to be abandoned. Some cranks tried to add extinction
but it doesn't work.

De Sitter was wrong.. face it George.
....and no other experiment refutes the BaTh.

I would also add that he probably used
grossly inflated velocity figures, based on VDoppler instead of ADoppler.

I would also add that I have corrected you on that
stupid and uninformed statement three times now.

De Sitter was wrong.. face it George.
....and no other experiment refutes the BaTh.

George




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Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

On Fri, 4 May 2007 14:16:37 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .

...

A photomultiplier produces a flash for each photon, you should know
that. The basic physics is the photoelectric effect. An electron
ejected
by a photon creates a cascade that generates enough light on the
final phosphor to be measured.

A very sensitive PM might pick up single photons.

All PMs pick up single photons, that's their job!

Their main job is to amplify very weak light signals. A single photon
could
barely be seen above the noise.

This is the experiment done with electrons rather than
photons but if you saw a video of the photon version
it would look exactly the same:

http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm

Yes I'm familiar with that kind of result. De Broglie waves are quite amazing
really. It shows that matter and 'fields' are not very far apart in nature.

George, there is nothing here that surprises me. Single photons making up a
monochromatic beam should have the same wavelength as the beam itself. The beam
is just 'lots of them'.

The site seemed slow and I had to download the movie
rather than view it on-line but it's worth a look so
that you understand the appearance of what we are
discussing. The regions where most photons land are
of course the same as the locations of the fringes
predicted by Huygens' method hence K=1.


that's good.


George




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Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.

On May 4, 4:52 pm, HW@....(Henri Wilson) wrote:
On Fri, 4 May 2007 13:58:52 +0100, "George Dishman"
wrote:

http://ophelia.princeton.edu/~page/single_photon.html

There is no PM in this experiment.

Microchannel plates are photomultiplier arrays.

Your electronics knowledge is VERY out of date.

Jerry

On 4 May 2007 04:03:17 -0700, George Dishman wrote:

On 4 May, 10:33, HW@....(Henri Wilson) wrote:
On 4 May 2007 00:26:21 -0700, George Dishman wrote:
On 4 May, 03:36, HW@....(Henri Wilson) wrote:
On Fri, 4 May 2007 00:21:07 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message
My question was, "what intrinsic property of an individual photon produces
a
sensation of a 'frequency'?

No it wasn't, the question was what is the definition
of frequency and that is what I explained above. I have
restored what you cut trying to cover it up.

In other words, what aspect of photon structure 'oscillates'?

A photon has no structure so nothing oscillates in it.

Hahahaha!

What makes a photon different from anything else then George?

It has different intrinsic properties.

How can anything have 'intrinsic properties' (which can be measured in
3space1time) if it doesn't have a 'structure'?

Consider some entity A. It is made of entities B and C.
A has properties which come from the properties of B
and C plus some influence from the relationship between
B and C. For example the mass of A might be the sum
of the masses of B and C plus the binding energy of the
pair. As you go down the scale, eventually you come to
something fundamental which is not composed of other
things, and yet it must have some properties of its own.

I think you just enjoy arguing, George.


Location is a continuous variable. It is not possible to
calculate exactly where a photon will land given an
experimental setup, you can only calculate the probability
as a function of location. That is an intrinsic property of
all particles.

George, if a thousand bullets are fired at a target, the way they are
distributed around the bull follows an established statistical law.

Yes, and that is true even if the gun is locked into position.

However, if single ONE bullet is fired at the target, it has zero probability
of landing anywhere other than at the point where the gun was aimed. (please
don't mention wind shear)

No, it has exactly the same probability of landing at any location
as each of the thousand.

No it doesn't!!!!!! Probability is not a cause of anything. It's a result.

All those bullets that were normally distributed around the bull landed exactly
where they did for purely physical reasons.
Where the bullet will strike is precisely determined BEFORE it is fired. Even
factors like the nerve movements of the shooter and the wind movements are
precisely predetermined. There is no way anyone could produce a mathematical
model to predict the outcome but it is still theoretically possible.

Statistics is the most misinterpreted science of all....

Indeed, though your mistake above is less common than
others. The key here is that the pprobability for each bullet
is unaffected by the existence of any preceding shot.

That is not related to my statement.

It is
similar to tossing an unbiassed coin, the probability is
50:50 regardless of the outcome of preceding tosses, only
the variable is 2D real (location on the target) rather than
binary (heads or tails).

Yes I know that George.

If you drop a thousand ball bearings on the floor they will end up normally
distributed around the centre....BUT that does not alter the fact thta there
was a precise physical reason why every one came to rest right where it did.

No, the 'traveling oscillation' model is the macroscopic
equivalent for a group of photons.

That's also true....but it is a different package.

Just the aggregate,

The way I see it is that a monochromatic beam is just a large number of
identical photons with that particular 'wavelength'.

White light is a mixture.

A radio signal is a mixture in which groups of individual photons form sine
shaped 'bunches' which move along. ..somewhat like a water wave except the
photons move back and forth rather than up and down.

This has given me an idea. Do the individual photons move or remain at
basically the same location?
I'll have to make an animation of this.



It is not a theory, it is logically obvious, the energy cannot
be dumped in two different places at the same time.

George, there are two alternatives.
The energy/unit volume of an RF signal can be the sum of all the h.nu energy
of individual photons in that volume. ...or it could be something like
2pi^2.h.A^2.f^3/c...

Sure, I expect the formula to be different in BaTh, but
the argument still holds, that energy is deposited where
the photon lands, not somehwere else.

That's probably OK for monochromatic light but you can't deduce that the same
will apply to, say, RF.

You don't know if the photon that enters the PM is the same one that was
incident on the grating. One is absorbed and another emitted.

It makes no difference.

Anyway, we know the classical theory of gratings..

I don't think you do, you can't even work out whether speed
appears in the BaTh equations for a grating.

This argument is not about how gratings behave according to BaTh.

Of course it is.

The BaTh doesn't need gratings to verify it.

I don't know what the lowest frequency of individual
detected photons is. However, grating methods are
applied at RF regularly and work fine. The photons
carry the energy and the energy goes where the wave
equations say it will therefore so do the photons.

Water waves carry longitudinal energy...but the individual molecules go up and
down. Their vertical KE is NOT what is carried with the wave.

The wave energy is deposited where the waves lap the shore,
not somewhere else.

But the energy of the vertically oscillating water molecules is continuously
being dampened out and absorbed as heat in the ocean.

Nobody knows that actual role of individual photons in this process.

Yes we do, from the optical behaviour. EM is the same
whether high frequency or low and gratings work as well
at microwave as they do in the infra-red.

So they should. They are wavelength dependent.

Wavelength and/or frequency.

Since nobody has a clue what photon 'wavelength' or 'frequency' actually
signify, that is a pretty meaningless statement.

So why don't you know what they do? A grating reflects
an incident wave to a particular point on a screen along

Huygens.

Exactly, the place where the energy lands on the screen is
controlled by the intrinsic property of the individual photons,
but it is also where Huygens' method says it will land, hence
the wavelength and/or frequency of each photon must be the
same as the macroscopic wave, hence K=1.

Here's another analogy.
The cars on the highway are made of rubber and all carry a heavy positive
surface charge. What do you think happens to their lengths as they slow down
and speed up in different speed zones?

I think when the charge is taken to some destination, the car
also arrives at the same place. You can't send the car to
Boston and have the charge arrive in Cairo which is what you
are suggesting. Beyond that discussions of their length are
irrelevant, the length has no analog in the photon.

How do you know.
The concept matches the data very well.

George



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Einstein's Relativity - the greatest HOAX since jesus christ's virgin mother.