Why are the 'Fixed Stars' so FIXED?

On Thu, 25 Oct 2007 20:28:53 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
On Wed, 24 Oct 2007 22:50:27 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
My approach, WHICH PRODUCES THE RIGHT ANSWER says light PARTICLES do NOT behave
according to classical traveling wave equation A'=Asin[2pi(t/T-x/L)]
..why should they?
You are indeed funny, Henri. :-)

Wasn't your approach to count the number of wavelengths defined
by the equation A'=Asin[2pi(t/T-x/L)] (in a wrong way, but anyway).
If this equation does not apply to light, what are you then counting?

What is the _wave_length of your non wave?

What IS your 'approach'? :-)

Photons are paticles, not waves.

What's your approach to knocking electrons out of metals with squiggly lines?


And this is of course the only answer I will ever get.
None.

Henri's approach 'PRODUCES THE RIGHT ANSWER, but he
don't know what his approach is, and he don't know
what a wavelength is.

I just told you.





Henri Wilson. ASTC,BSc,DSc(T)

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

On Thu, 25 Oct 2007 00:53:00 +0000 (UTC), bz
wrote:

HW@....(Dr. Henri Wilson) wrote in
:


statements.

Let me explain.

Both SR and BaTh accept that each element of the rays is emitted from a
point that is stationary in the non-otating frame. That is legitimate
physics. (neither Paul nor George will acknowledge that this emission
point MOVES in the rotating frame.....because it destroys your 'rotating
frame' argument)

in either theory, once the light has left the emitter, it doesn't matter what
the emitter does.

In one theory, (emission) the light travels at c with respect to the emitter
(whether or not the emitter is moving).
In the other, (SR) the light travels at c wrt the inertial frame of reference
which was co-moving with the emitter at the instant of emission (and wrt all
other inertial frames of reference).
In ET, the light moves at c+v wrt any inertial frames of reference, where v
is the velocity wrt the inertial frame of reference that was co-moving with
the source at the instant of emission.


SR says the speed of both rays is magically adjusted to be c wrt that
static emission point.

Nope. It isn't 'magically adjusted' under SR. The value _measured_ for the
speed by the observer is independent on the relative velocity of the
observer, so all observers 'see' the light moving at c
_in_their_frame_of_reference_, but the light doesn't care how fast the
observer is moving. The light continues to move at the same speed,
independent of the motion of source or the observers.
The frequency/wavelength _measured_ by an observer IS dependent on the
relative velocity of the observer wrt the source because the KE of the light
is depends on the relative velocity of the observer wrt the source.

Ballistic theory says the velocity _measured_ by the observer is DEPENDENT on
the relative velocity of the observer to the source as is the energy and the
frequency. This leaves the wavelength as absolute.

The case of a rotating table is very simple in ballistic theory. Light
leaving in the direction of rotation is moving faster (from the lab's
viewpoint), but has further to go because the target is going away.
Light leaving in the direction opposite the motion is moving slower but has a
shorter distance to travel because the target is coming to it.
These cancel each other out and it takes the light exactly the same distance
to travel each path. That being true, there is no phase difference in the
arriving beams.


SR calculates the travel times of the rays around
the ring and finds those times to be different because of the different
path lengths. SR says that this indicates a phase difference at the
detector. (Note, SR ignores the fact that the elements emitted
simultaneously do not arrive simultaneously)

On the contrary, SR takes that into account. That IS part of the phase
difference seen.

BaTh says that the rays move at c wrt the moving source from the
(static) emission point.

The emission point is NOT static.

the emission point OF A PARTICULAR RAY ELEMENT IS static. Even Jerry got that
right.

There IS a point in the inertial frame of
reference that is co-moving with the source at the instant of emission that
can be related to a point in the laboratory's iFoR that was coincident with
the source at the instant of emission.
I am not sure which of these you are confusing with a 'static emission
point'.

The emission point can ONLY be considered to be 'static' in the rotating
frame of reference of the source. Of course that is NOT an inertial frame of
reference.

go away and learn some physics Bob.

and
elements emitted simultaneously arrive at the detector simultaneously.

correct.

and thus there is no fringe shift predicted by BaTH.

BaTh says that the phase of arrival of each ray is simply [path length
mod (absolute wavelength)]. If the phase of one is x degrees, that of
the other is 360-x.

Both approaches produce the same answer.

Both BaTH approaches predict NO fringe shift.

see: http://www.users.bigpond.com/hewn/ringgyro.exe

and read what i said about it elsewhere.


BaTH predicts no Sagnac effect. SR is consistent with Sagnac. There is no
question 'which is more likely' as BaTH is falsified by Sagnac.


SR relies on an unproven postulate ie., MAGIC to adjust both light
speeds to be 'c'.

You mistake the fact that all observers measure light as moving at c with a
mystical idea that the observation CHANGES the speed.

since when?...
..
It doesn't.

I look at the sun at noon time. It is at zenith.
You look at the sun at the same instant of time. The sun is NOT at zenith for
you.
The sun doesn't magically adjust its position to allow you to measure it at a
different location in the sky.
The perspective of the person doing the measuring is what changes.

Light doesn't magically adjust its speed. The speed doesn't change at all.
The perspective of the person doing the measuring is what changes.

In the absence of a dielectric medium, light has only one speed
reference....its source.

It requires that the two rays move at c+v and c-v wrt
the source.

NO! The distance to the target changes by +vt and -vt so the TIME for the
trips use the c+v and c-v factors.

In the standard SR ring gyro diagram, the rays are shown to move at c/-v wrt
the source.
SR refutes itself.

BaTh uses logical physics, in assuming that wavelength' is
constant in all frames....since ALL lengths are absolute and contant in
all frames.

It is 'logical' to think a lot of wrong things when facts are selectively
ignored.



Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message
...
: On Wed, 24 Oct 2007 22:46:54 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Wed, 24 Oct 2007 22:50:27 +0200, "Paul B. Andersen"
: : wrote:
: :
: : Dr. Henri Wilson skrev:
: : My approach, WHICH PRODUCES THE RIGHT ANSWER says light PARTICLES do
: NOT behave
: : according to classical traveling wave equation A'=Asin[2pi(t/T-x/L)]
: : ..why should they?
: :
: : You are indeed funny, Henri. :-)
: :
: : I like to make people laugh....It shows they are learning new things
from
: me...
: :
: : Wasn't your approach to count the number of wavelengths defined
: : by the equation A'=Asin[2pi(t/T-x/L)] (in a wrong way, but anyway).
: :
: : Paul, I don't expect you to be able to understand the physical
: significance of
: : the traveling wave equation but to put it simply, it describes what
: happens if
: : you draw a sqiggly line on a piece of paper and move it sideways.
:
: STILL cannot spell "squiggly" on the 20th.
:
: Squiggley, sqigley, squigley, skwiglie are all wrong.

Yes, you are.


:
: : If you think a photon is just a 'moving squiggly line' then you're
welcome
: to
: : the idea ...
:
: No, Wilson, it is 'stationary squiggly line', a trace in TIME.
:
: Whatever...but the G.J.P. brigade think light is just a moving sinewave.

So they are nuts. Ask the prats what the wavelength of this sinewave
is.
http://spaceflight.nasa.gov/realdata/tracking/
Ask them what the fringe shift is, I'd say about 45 minutes... TIME.
The wave is a trace in TIME.


: : but can you explain how one particular squiggly line and not
: : another will cause electrons to be released from a metal surface when
it
: hits
: : is?
:
: The wiggly line is a trace in time showing where the field WAS.
: The photon is an electric field. When the electric field gets to the
: metal surface it pulls the electron.
:
: ..so no squiggly lines are involved??
:
: That's settled then.

lambda = (c+v)/nu.
c for the shuttle, v for the Earth, here's the ****ing wave:
http://spaceflight.nasa.gov/realdata/tracking/
It's like a contrail when a plane flies overhead, it leaves a trace.
http://www.devsite.org/lab/spaceship...1-contrail.jpg

When you have a gazillion photons all going in the same
direction yet leaving at different times you MODEL just two
of them, the c+v and the c-v photons, at the instant in TIME
that they reach the detector.
This is what you find:
http://www.androcles01.pwp.blueyonde...outofphase.gif
I can't model a gazillion photons, nobody can. You have to imagine that
two photons reached the detector at the same instant with different
speeds, so the slow photon left earlier than the fast one. That's
not hard to do.
So...
lambda1 = (c+v)/nu.
lambda2 = (c-v)/nu.
To model Sagnac with the space shuttle, send a second shuttle
around the Earth in the opposite direction. It is that easy.
Where the shuttles pass each other on opposite sides of the Earth
will produce a continual "shift" on the map. NASA won't do it
though, they can't launch two shuttles for a ****in' newsgroup,
but you could model it in your BASIC.

On Thu, 25 Oct 2007 22:10:43 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
: On Thu, 25 Oct 2007 20:38:44 GMT, "Androcles"

: : MY Emission Fact kicks YOUR BaTh down the toilet.
: :
: : ...well why don't you use it to produce the right result, ie., fringe
: : displacement = 4Aw/cL
:
: 1) You have the wrong equation, no c+v included.
: 2) see 1) above.
:
: v is negligible.

Ok, fringe shift is neglible. I see you are back in ****head mode.

I didn't think you knew what an 'equation' was.
You can use c+v if you like...although it's wrong. It makes a difference of the
order of 1 part in 10^10

: 'c' is a universal constant.




:
: : BTW, what happened to old "EFOR" Len Gaasenbeek, his selected
: : papers (that he selected) and his helical photons?
: : He dropped off the radar...too much poetry, I expect.
: : He had point but still wanted to keep c, poor old bugger.
: : You remind me of him, all ego and no listening to reason.
: :
: : Len had some good ideas but was too indoctrinated as you say.
: : My 'rotating +- charge photon model is similar to his helical wave
: concept.
:
: It could fly, actually. I have no strong objection to it, I quite like
it.
: I modelled it a long time ago, the difference is the +ve and -ve are
: replaced by N and S.
: http://www.androcles01.pwp.blueyonde.../AC/Photon.gif
: Just because there is only one shown travelling doesn't mean there are
: not two poles, as shown in the "stationary" photon in the gif.
:
: This is very possible. It could use poles, charges or both.

Still have to conserve energy, AC still works, electric motors
still work, generators still work, there is no "very possible",
it ****in' IS. I see you are back in argumentative ****head
mode. Radio is not your theory, Wilson. It was all thought
of long before you or Einstein or Tusseladd or Dishpan or
Jeery, even if none of you understand it.
http://www.om3rkp.cq.sk/articles.php?lng=en&pg=91



: I don't know why this kind of model has been rejected.

Faraday didn't reject it. Tesla didn't reject it. Marconi didn't
reject it. Only dorks and tusselader reject it.


: I know that a photon cannot be a spinning electron/positron pair because
of an
: imbalance in 'the equation' the but it COULD be just the spinning
charges...or
: magnetic poles.

No it ****ing can't. AC still works, electric motors
still work, generators still work, radio still works,
there is no "could be", it's ****in' known and energy
is conserved.
I see you are back in argumentative ****head mode.
Do you start your drinking with breakfast, Wilson?

BTW, What's this wavelength?
http://spaceflight.nasa.gov/realdata/tracking/

What would it be if the Earth turned in 12 hours instead of 24?

:
: : Have you checked this out:
: : http://www.nasa.gov/multimedia/nasatv/
: : LIVE from the shuttle... I watched the launch this afternoon.
: :
: : yeh! good
:
: I watched the docking today.




Henri Wilson. ASTC,BSc,DSc(T)

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

On Thu, 25 Oct 2007 23:05:46 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
: On Wed, 24 Oct 2007 23:41:40 GMT, "Androcles"
: wrote:

: : What's so strange about that?
: :
: : Nothing. You again cannot answer th question...
:
: What's to answer? The ****ing boats move up and gain
: energy... What's your problem?
:
: Are you not aware that enegy is transmitted longitudinally by a water
wave?

Omnidirectionally, sure, all the way to the shore and the opposite shore.
Ripples make concentric circles.

: I'm asking you why it goes one way and not the other

It does go the other way. You are nuts, Wilson.

: since the water molecules
: themselves only move up and down.

Down as far as the bottom, then they have to move sideways.
They can go up as far as they like.

Yeh, they don't move up and down, they move in a rough ellipse.


: Why does a water wave appear to be moving
: towards the shore?

The bottom is sloped there. Water runs downhill. You
are only looking at the top. If you looked at the bottom
you'd see it go away from the shore.

It DOES ...but I don't think it's due to the slope.

: There is another wave going the opposite way.

Some of us already know that. It goes the opposite way on
the American shore to its direction on the Ozzie shore.

: Just look at the pretty pictures instead:
: http://www.androcles01.pwp.blueyonde...outofphase.gif
: You can see the phase shift quite easily, the red (+ve) and blue (-ve)
: are present at the same instant in time, almost canceling each other.
:
: can you produce Fringe Displacement = 4Aw/c.lambda with your theory?
:
: If not why not...

Fringe Displacement is as negligible as v.
What's the Fringe Displacement of this sine wave, Wilson?
http://spaceflight.nasa.gov/realdata/tracking/
I say its about 45 minutes, what do you hallucinate AND
how do you get 4Aw/c.lambda with your theory?

That the equation for one turn of a ring gyro. I didn't think you would know.
It's only been discussed here for about 5 years.


Henri Wilson. ASTC,BSc,DSc(T)

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

On Thu, 25 Oct 2007 23:36:48 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
: On Wed, 24 Oct 2007 22:46:54 GMT, "Androcles"

: : the idea ...
:
: No, Wilson, it is 'stationary squiggly line', a trace in TIME.
:
: Whatever...but the G.J.P. brigade think light is just a moving sinewave.

So they are nuts. Ask the prats what the wavelength of this sinewave
is.
http://spaceflight.nasa.gov/realdata/tracking/
Ask them what the fringe shift is, I'd say about 45 minutes... TIME.
The wave is a trace in TIME.

That's not a fringe shift...

:
: ..so no squiggly lines are involved??
:
: That's settled then.

lambda = (c+v)/nu.
c for the shuttle, v for the Earth, here's the ****ing wave:
http://spaceflight.nasa.gov/realdata/tracking/
It's like a contrail when a plane flies overhead, it leaves a trace.
http://www.devsite.org/lab/spaceship...1-contrail.jpg

When you have a gazillion photons all going in the same
direction yet leaving at different times you MODEL just two
of them, the c+v and the c-v photons, at the instant in TIME
that they reach the detector.
This is what you find:
http://www.androcles01.pwp.blueyonde...outofphase.gif
I can't model a gazillion photons, nobody can. You have to imagine that
two photons reached the detector at the same instant with different
speeds, so the slow photon left earlier than the fast one. That's
not hard to do.

No, the travel times are the same.

So...
lambda1 = (c+v)/nu.
lambda2 = (c-v)/nu.
To model Sagnac with the space shuttle, send a second shuttle
around the Earth in the opposite direction. It is that easy.
Where the shuttles pass each other on opposite sides of the Earth
will produce a continual "shift" on the map. NASA won't do it
though, they can't launch two shuttles for a ****in' newsgroup,
but you could model it in your BASIC.





Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message
...
: On Thu, 25 Oct 2007 22:10:43 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Thu, 25 Oct 2007 20:38:44 GMT, "Androcles"

:
: : : MY Emission Fact kicks YOUR BaTh down the toilet.
: : :
: : : ...well why don't you use it to produce the right result, ie.,
fringe
: : : displacement = 4Aw/cL
: :
: : 1) You have the wrong equation, no c+v included.
: : 2) see 1) above.
: :
: : v is negligible.
:
: Ok, fringe shift is neglible. I see you are back in ****head mode.
:
: I didn't think

We know that. The question is whether you can or not.


: you knew what an 'equation' was.
: You can use c+v if you like...although it's wrong. It makes a difference
of the
: order of 1 part in 10^10
:
No v, no fringe shift.
What's the wavelength of this sine wave?
http://spaceflight.nasa.gov/realdata/tracking/



: : 'c' is a universal constant.
:
:
:
:
: :
: : : BTW, what happened to old "EFOR" Len Gaasenbeek, his selected
: : : papers (that he selected) and his helical photons?
: : : He dropped off the radar...too much poetry, I expect.
: : : He had point but still wanted to keep c, poor old bugger.
: : : You remind me of him, all ego and no listening to reason.
: : :
: : : Len had some good ideas but was too indoctrinated as you say.
: : : My 'rotating +- charge photon model is similar to his helical wave
: : concept.
: :
: : It could fly, actually. I have no strong objection to it, I quite like
: it.
: : I modelled it a long time ago, the difference is the +ve and -ve are
: : replaced by N and S.
: : http://www.androcles01.pwp.blueyonde.../AC/Photon.gif
: : Just because there is only one shown travelling doesn't mean there are
: : not two poles, as shown in the "stationary" photon in the gif.
: :
: : This is very possible. It could use poles, charges or both.
:
: Still have to conserve energy, AC still works, electric motors
: still work, generators still work, there is no "very possible",
: it ****in' IS. I see you are back in argumentative ****head
: mode. Radio is not your theory, Wilson. It was all thought
: of long before you or Einstein or Tusseladd or Dishpan or
: Jeery, even if none of you understand it.
: http://www.om3rkp.cq.sk/articles.php?lng=en&pg=91
:
:
:
: : I don't know why this kind of model has been rejected.
:
: Faraday didn't reject it. Tesla didn't reject it. Marconi didn't
: reject it. Only dorks and tusselader reject it.
:
:
: : I know that a photon cannot be a spinning electron/positron pair
because
: of an
: : imbalance in 'the equation' the but it COULD be just the spinning
: charges...or
: : magnetic poles.
:
: No it ****ing can't. AC still works, electric motors
: still work, generators still work, radio still works,
: there is no "could be", it's ****in' known and energy
: is conserved.
: I see you are back in argumentative ****head mode.
: Do you start your drinking with breakfast, Wilson?
:
: BTW, What's this wavelength?
: http://spaceflight.nasa.gov/realdata/tracking/
:
: What would it be if the Earth turned in 12 hours instead of 24?
:
: :
: : : Have you checked this out:
: : : http://www.nasa.gov/multimedia/nasatv/
: : : LIVE from the shuttle... I watched the launch this afternoon.
: : :
: : : yeh! good
: :
: : I watched the docking today.
:
:
:
:
: Henri Wilson. ASTC,BSc,DSc(T)
:
: www.users.bigpond.com/hewn/index.htm

"Dr. Henri Wilson" HW@.... wrote in message
...
: On Thu, 25 Oct 2007 23:05:46 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Wed, 24 Oct 2007 23:41:40 GMT, "Androcles"

: : wrote:
:
: : : What's so strange about that?
: : :
: : : Nothing. You again cannot answer th question...
: :
: : What's to answer? The ****ing boats move up and gain
: : energy... What's your problem?
: :
: : Are you not aware that enegy is transmitted longitudinally by a water
: wave?
:
: Omnidirectionally, sure, all the way to the shore and the opposite shore.
: Ripples make concentric circles.
:
: : I'm asking you why it goes one way and not the other
:
: It does go the other way. You are nuts, Wilson.
:
: : since the water molecules
: : themselves only move up and down.
:
: Down as far as the bottom, then they have to move sideways.
: They can go up as far as they like.
:
: Yeh, they don't move up and down, they move in a rough ellipse.
:
:
: : Why does a water wave appear to be moving
: : towards the shore?
:
: The bottom is sloped there. Water runs downhill. You
: are only looking at the top. If you looked at the bottom
: you'd see it go away from the shore.
:
: It DOES ...but I don't think

Yes, we all know you don't think. The question isn't
whether you do or not, but whether you can or not.
You certainly don't listen.


it's due to the slope.
:
: : There is another wave going the opposite way.
:
: Some of us already know that. It goes the opposite way on
: the American shore to its direction on the Ozzie shore.
:
: : Just look at the pretty pictures instead:
: : http://www.androcles01.pwp.blueyonde...outofphase.gif
: : You can see the phase shift quite easily, the red (+ve) and blue (-ve)
: : are present at the same instant in time, almost canceling each other.
: :
: : can you produce Fringe Displacement = 4Aw/c.lambda with your theory?
: :
: : If not why not...
:
: Fringe Displacement is as negligible as v.
: What's the Fringe Displacement of this sine wave, Wilson?
: http://spaceflight.nasa.gov/realdata/tracking/
: I say its about 45 minutes, what do you hallucinate AND
: how do you get 4Aw/c.lambda with your theory?
:
: That the equation for one turn of a ring gyro. I didn't think

Yes, we all know that, no need to keep repeating it.
Wilson doesn't think.




you would know.
: It's only been discussed here for about 5 years.

It's as wrong as the cuckoo malformations no matter how long
its been discussed, it has no v in it. You didn't think.

"Dr. Henri Wilson" HW@.... wrote in message
...
: On Thu, 25 Oct 2007 23:36:48 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Wed, 24 Oct 2007 22:46:54 GMT, "Androcles"

:
: : : the idea ...
: :
: : No, Wilson, it is 'stationary squiggly line', a trace in TIME.
: :
: : Whatever...but the G.J.P. brigade think light is just a moving
sinewave.
:
: So they are nuts. Ask the prats what the wavelength of this sinewave
: is.
: http://spaceflight.nasa.gov/realdata/tracking/
: Ask them what the fringe shift is, I'd say about 45 minutes... TIME.
: The wave is a trace in TIME.
:
: That's not a fringe shift...

Ok, it's just an ordinary shift. Proportional to v, of course. You didn't
think.



:
: :
: : ..so no squiggly lines are involved??
: :
: : That's settled then.
:
: lambda = (c+v)/nu.
: c for the shuttle, v for the Earth, here's the ****ing wave:
: http://spaceflight.nasa.gov/realdata/tracking/
: It's like a contrail when a plane flies overhead, it leaves a trace.
: http://www.devsite.org/lab/spaceship...1-contrail.jpg
:
: When you have a gazillion photons all going in the same
: direction yet leaving at different times you MODEL just two
: of them, the c+v and the c-v photons, at the instant in TIME
: that they reach the detector.
: This is what you find:
: http://www.androcles01.pwp.blueyonde...outofphase.gif
: I can't model a gazillion photons, nobody can. You have to imagine that
: two photons reached the detector at the same instant with different
: speeds, so the slow photon left earlier than the fast one. That's
: not hard to do.
:
: No, the travel times are the same.

You didn't think. You never do, though, Einstein Dingleberry.

BUTTHEAD WILSON CONFIRMS:
'we establish by definition that the "time" required by
light to travel from A to B equals the "time" it requires
to travel from B to A' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Rabbi Albert Einstein
http://www.androcles01.pwp.blueyonde...rt/tAB=tBA.gif

:
: So...
: lambda1 = (c+v)/nu.
: lambda2 = (c-v)/nu.
: To model Sagnac with the space shuttle, send a second shuttle
: around the Earth in the opposite direction. It is that easy.
: Where the shuttles pass each other on opposite sides of the Earth
: will produce a continual "shift" on the map. NASA won't do it
: though, they can't launch two shuttles for a ****in' newsgroup,
: but you could model it in your BASIC.

On 25 Oct, 23:20, HW@....(Clueless Henri Wilson) wrote:
On Thu, 25 Oct 2007 17:30:45 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" HW@.... wrote in message ...

George, let me explain.

Both SR and BaTh accept that each element of the rays is emitted from a
point
that is stationary in the non-otating frame. That is legitimate physics.

No it ismn't Henry, it just shows how clueless you
are about all this. Let me explain. A "point" is
nothing more than a set of spatial coordinates in
some coordinate system. If you define a fixed set
of values for those coordinates, the point is at
rest in that system. If you define it in the
inertial system then it doesn't move in that system
and if you define a point in the rotating system it
doesn't move in that system. The two points might
be co-located at the instant the light is emitted
but thereafter they move apart.

George, if you are going to continueto be this dumb I will find it hard not to
start calling you names....in the style of Androcles.
see, for instance:http://www.mathpages.com/rr/s2-07/2-07.htm
That is your SR explanation of a ring gyro. The start point DOES NOT MOVE.

Correct Henry, exactly as I just said. Any "point" is
just a fixed set of values in a coordinate system anjd
therefore cannot move BY DEFINITION. YOU are the one
claiming that a point moves, not me.

However, none of that needs to worry you. Consider
a single wavecrest or phase front. You can say it is
emitted at a point in the inertial frame and moves
away from it at speed c+v in that frame, or you can
say it is emitted from a point in the rotating frame
and say it moves away at speed c, and you will find
that both calculations give the same place where you
will find the phase front at some later time.

George, that is the classical theory....applicable to 'moving sine waves' like
water waves.

Yes Henry, waves such as we see on an oscilloscope and
about which ballistic theory must make predictions, so
go ahead and see what your equations say.

(neither you nor Paul will acknowledge that this emission point MOVES in
the
rotating frame.....because it destroys your 'rotating frame' argument)

Henry, you are an idiot. Just a couple of days ago I
warned you I would keep on reminding you that the proof
I showed you was in the inertial frame. All you are
doing is showing everyone you can't even work out
which frame you are working in if the speed is "c+v".

George, you don't understand frames.

Henry, you are the one trying to say that fixed points
can move, it is obvious to everyone you have no idea
what a frame is.

You still can't see that the start point
is static in the inertial frame but moving backward in the rotating frame.

Thank you for proving the point.

Not
only that, every previously emitted 'wavecrest' moves backward in proportion.

That is the physics that matters. If that happened
the speed would not be c relative to the source in
the rotating frame, it woud be c relative to your
hypothetical point, which of course is what SR says,
that's why you get the "right answer".

SR says the speed of both rays is magically adjusted to be c wrt that
static
emission point.

Wrong, that is ballistic theory. The "speed equalisation"
equations says that light is emitted at c+v but then
magically gets reduced to c.

SR says the light is emitted at c in any inertial frame
and never gets adjusted. I am ignoring refractive index
for simplicity of course, your errors are so groos, such
details can be left until later.

George my server finally fixed the problem and I was abl to upload this:

http://www.users.bigpond.com/hewn/ringgyro.exe

It may be some time until I can look at it, my wife
and I are at an exhibition for the nxt two days and
I have tickets for the Dolphins vs. the Giants at
Wembley on Sunday :-)

(It doesn't work too well on Vista....will fix)

This shows the difference between your 'moving squiggly line' theory and mine.

The dots represent 'wavelengths'. The two rays meet together at the detector in
both theories...but for different reasons.
You say the phases are the same.
I say the phases are indicated by the 'circle of white teeth'....The arrival
phase for each is (pathlength mod lambda) .
If the phase of one ray is x degrees, then that of the other is 360-x

SR calculates the travel times of the rays around the ring and
finds those times to be diffferent because of the different path lengths.
SR
says that this indicates a phase difference at the detector. (Note, SR
ignores
the fact that the elements emitted simutaneously do not arrive
simultaneously)

ROFL, Henry that's a classic: "the fact that the elements
emitted simutaneously do not arrive simultaneously" is
just another way of saying there is a phase difference at
the detector!

Again you miss the point.
SR says that the elements of the rays that reunite were NOT emitted
simultaneously. The ones that DO meet at the detector were emitted with
different phases. However, since the travel times are DIFFERENT in SR, this
phase difference cancels to some extent.

You are losing it Henry, think again. The elements are
emitted in phase and have different travel times so
arrive out of phase, there is no cancellation, the
effect you descibe is what causes the signals to be
out of phase in the first place.

I suspect this is only a second order
error so you get away with it.

It is first order, it is the cause of the Sagnac Effect.

BaTh says they were emitted simultaneously but differ in phase when they
arrive....due to an intrinsic effect.

No, apply eqn [2] of the theory, ballistic theory
says they are emitted simultaneously and have equal
travel times so arrive simultaneously.

BaTh says that the rays move at c wrt the moving source from the (static)
emission point. They move at c+v and c-v (wrt the no-rotating frame)
around the
ring. BaTh says the travel times are the same and elements emitted
simultaneously arrive at the detector simltaneously.

Correct, and since they were emitted in phase that
means they arrive in phase.

no George, that's only according to your classical wave theory. it doesn't
apply

Arrival time = emission time plus travel time
in all theories.

BaTh says that the phase
of arrival of each ray is simply [pathlength mod (absolute wavelength)].
If the
phase of one is x degrees, that of the other is 360-x.

Wrong, the phase difference is pathlength / distance_per_cycle,
your algebra is plucked out of thin air and is not correct.

George, in BaTh the 'distance per cycle' is absolute and the same in all
frames.

No, you are thinking of the wavelength which as you
say is frame invariant. The distance moved per cycle
is (c+v)/f whereas the wavelength is just c/f. That
small error is why your maths is wrong.

Both approaches produce the same answer.

No Henry, they don't, your algebra is broken.

the equations and answer is given at:http://www.users.bigpond.com/hewn/ringgyro.htm

See above, your maths is wrong.

Androcles wants to use frequency instead of wavelength and is yet to come
up
with a prediction of fringe shift in spite of all his raving.

The correct approach is to form a set of simultaneous
equations for the motion of a phase front of the light
based on the motion of the source (beam splitter) and
for the motion the detector. Solving that gives the
arival time of the phase front at the detector and the
approach allows for arbitrary variations of source speed.

the equations and answer is given at:http://www.users.bigpond.com/hewn/ringgyro.htm

See above, your maths is wrong.

A simplification of that is suitable for constant speed
where the travel time is also constant hence detection
time can be found simply as emission time plus travel
time. What you find is that the arrivial time for both
beams is the same hence the detector sees the source
waveform simply delayed by the same travel time through
both paths and is always "in phase" even for completely
arbitrary waveforms.

If you want to use wavelength or whatever as an
alternative, by all means do so, but there can only be
one answer to the question "what is the phase differnce
at the detector" and all valid approaches must give that
single answer.

So which is more likely.

SR relies on an unproven postulate,

Wrong as usual, the postulate is derived from Maxwell's
Equations each of which is experimentally confirmed, and
the one-way speed is confirmed as c experimentally by the
Sagnac experiment. You don't have the ability to understand
the maths involved.

Maxwell's equations use the absolute aether as a speed reference.

No they don't, they use the observer as the
reference, the speed of the aether relative
to the observer does not appear in the equations.
As I said, it appears you don't have the ability
to understand the maths involved.

They don't
apply to photon particles.

Obviously but aggregating photons must produce
Maxwell's Equations.

ie., MAGIC, to adjust both light speeds to
be 'c'.

There is no adjustment clueless, the light is EMITTED moving
at c in the inertial frame.

It requires that the two rays move at c+v and c-v wrt the source.

Nope, they move at c relative to the source.

It states that..... but uses c+/-v in the equations.

Nope, it uses c. It appears you don't have the ability
to understand the maths involved.

BaTh
uses logical physics, ...

No, it is purely philosophical and proven to be wrong by
every experiment we have discussed other than the MMx.
Don't waste your time repeating your dogma, physics is
about calculation and it proves you wrong every time.

BaTh has never been proven wrong.

Sagnac proves it wrong as does the Shapiro delay,
the absence of detectable effect in contact binaries,
Cepheids, pulsar timing and so on. The list is
endless but it appears you don't have the ability
to understand the maths involved in any of them.

George