Why are the 'Fixed Stars' so FIXED?

On Mon, 22 Oct 2007 00:29:16 -0700, George Dishman
wrote:

On 21 Oct, 22:34, HW@....(Clueless Henri Wilson) wrote:
On Sun, 21 Oct 2007 12:08:07 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" HW@.... wrote in message

the part of the wave that leaves the emission point IN PHASE is shown
moving at
c+v and c-v in red and blue. You can check the phasing at any point by
pausing
the program.

How am I supposed to do that Henry? By comparison
with the squiggly line? If that is the case, it
proves you wrong, the phase at the tip should be
constant.

I still can't upload my latest vesion. The server is playing games again....

George, the front of the ray is changing phase as it moves.

Don't be silly Henry. If you measure the speed of a
wave, all you can do is choose some particular phase
value and track its motion in order to find how fast
it moves. There is no other way to define the speed
of a sine wave.

George, I'll tell you secret. A light source emits photon PARTICLES not
classical water waves.


No Henry, ballistic theory says they move away
from it at c relative to the source, c+v ahead
and c-v the other way. Your phase indicator (the
wiggly line) doesn't move at all so it doesn't
represent ballistic theory.

The teeth are not supposed to move.

The speed they move at is supposed to be c+v.

the ' teeth' are a graph.

They are a circular graph of the phase of
the ray front as it moves around the ring. You can even see it happening in
Jerry's silly program. My toothed wheel is just a graph of what Jerry has
(wrongly) shown.

Jerry's program is right, you know so little
physics, you don't even know what is being
measured.

Jery has modelled a ring gyro using sound in air...but without the air...



The phase at the tip can be arbitrary on emission
but must be constant thereafter.

..so Jerry's program is wrong, eh?

No, it is correct. She could put another button
on the screen so that the magenta dots were
placed on the waves at the source when it was
pressed and you could then choose any arbitrary
phase to track, the speed would be correct and
the result would be the same, they would arrive
at the detector simultaneously.

George, it is obvious that no amount of advice will teach you the facts.
The phase difference at the detector is [path length difference/ absolute
wavelength].

PROVED BY EXPERIMENT.


Well done Henry, you grasped it. Now move each
part at c+v (in the inertial frame) from that
point and you get Jerry's applet.

..and the phase at the tip is constantly changing....

You are clueless Henry, a point of fixed phase is
the only identifiable thing whose speed you can
measure.

George, a light source emits photon PARTICLES, not squiggly lines.

...but Jerry made the mistake of arranging for the phases to be the same when
the rays meet.

No she doesn't, she arranges that they are emitted
in phase and that they travel at the right speed,
the fact that they are in phase when they arrive
is an unavoidable consequence.

No they are not. See www.users.bigpond.com/hewn/toothwheel.exe

The phases are calculated from pathlength/lambda

Wrong, pathlength/(distance moved per cycle)

Has Androcles finally won you over?


and you find they WERE in phase when they were emitted.

The waves are shown to be moving and their phase is shown by the teeth. It
is a
graph of phase.

I knew you didn't understand phase. What you
have drawn is NOT a graph of phase, any point
of given phase moves at speed c+v one way and
c-v the other. Your phase points don't move at
all.

George you are obviously confusing photon movement with the behavior of
classical waves in a medium...You seem to believe that ligtht behaves like
sound. It doesn't.

It doesn't, but the definition of the phase
of a sine wave is the same regardless of the
the application.

No George. In BaTh, wavelength is absolute.

Until you learn what phase is your program will
remain wrong. Once the light has been emitted,
what happens to the source has no effect on it.
Each part of the wave moves at c+/-v and until
you make the sine waves move, your program does
not represent ballistic theory. You remain
clueless.

The source emits photons, not squiggly lines.
Light is not sound...

RF is a propagating sine wave. Monochromatic
light is identical but at higher frequency,
and the definition of the phase velocity of
light is the velocity of a point on the wave
of fixed phase, obviously.

....and you have actually seen a point on the wave of a photon, have you George?

The phase of the tip as it travels at c+v around the ring is given by
pathlength/lambda and is represented by the circle of teeth.

Wrong, pathlength/(distance moved per cycle)

Why don't you and Androcles get togther. You seem to have a lot in common...

Both rays travel for the same time duration ..

Therefore they must arrive in phase.

WRONG. They move at different speeds and their wavelengths are the same.

I remind you, this approach produces the right answer.

but at different speeds. The
'wavelength' is the same in both. It should be obvious even to you George, that
their phases will not be the same at the detector.

It is obvious that if point of given phase take the
same time for each then the phase difference when
they arrive must be the same as when they were
emitted. They were emitted in phase so they must
arrive in phase.

doppler ...
If you start talking frequency and time, ypou must consider doppler. Remember
the detector is moving relative to the static EMISSION POINT.

It turns out you didn't know what part of a wave
was being measured to define the speed, you are
even more clueless than I had realised.

George, the math is simple. Phase difference is:
[path length difference/lambda] mod(lambda).

It is backed by experiment...

George



Henri Wilson. ASTC,BSc,DSc(T)

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

On 22 Oct, 10:37, HW@....(Dr. Henri Wilson) wrote:
On Mon, 22 Oct 2007 00:29:16 -0700, George Dishman wrote:
On 21 Oct, 22:34, HW@....(Clueless Henri Wilson) wrote:
On Sun, 21 Oct 2007 12:08:07 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" HW@.... wrote in message
the part of the wave that leaves the emission point IN PHASE is shown
moving at
c+v and c-v in red and blue. You can check the phasing at any point by
pausing
the program.

How am I supposed to do that Henry? By comparison
with the squiggly line? If that is the case, it
proves you wrong, the phase at the tip should be
constant.

I still can't upload my latest vesion. The server is playing games again....

George, the front of the ray is changing phase as it moves.

Don't be silly Henry. If you measure the speed of a
wave, all you can do is choose some particular phase
value and track its motion in order to find how fast
it moves. There is no other way to define the speed
of a sine wave.

George, I'll tell you secret. A light source emits photon PARTICLES not
classical water waves.

Henry, I'll tell you the obvious, individual photons
suffer interference effects in Young's Slits. You
have not considered which part of a photon the speed
measures. The answer is that it is a point of given
phase.

No Henry, ballistic theory says they move away
from it at c relative to the source, c+v ahead
and c-v the other way. Your phase indicator (the
wiggly line) doesn't move at all so it doesn't
represent ballistic theory.

The teeth are not supposed to move.

The speed they move at is supposed to be c+v.

the ' teeth' are a graph.

Yes, they should move at c+v according to
ballistic theory. Your software is wrong.

They are a circular graph of the phase of
the ray front as it moves around the ring. You can even see it happening in
Jerry's silly program. My toothed wheel is just a graph of what Jerry has
(wrongly) shown.

Jerry's program is right, you know so little
physics, you don't even know what is being
measured.

Jery has modelled a ring gyro using sound in air...but without the air...

Wrong again, the speeds would be c relative
to the lab using air. Jerry's speeds are
correct, yours are zero.

The phase at the tip can be arbitrary on emission
but must be constant thereafter.

..so Jerry's program is wrong, eh?

No, it is correct. She could put another button
on the screen so that the magenta dots were
placed on the waves at the source when it was
pressed and you could then choose any arbitrary
phase to track, the speed would be correct and
the result would be the same, they would arrive
at the detector simultaneously.

George, it is obvious that no amount of advice will teach you the facts.
The phase difference at the detector is [path length difference/ absolute
wavelength].

Wrong Henry, the pathlength is the distance
_moved_ so you divide by the distance _moved_
per cycle, not the wavelength, they are not
the same.

PROVED BY EXPERIMENT.

SR is proved by experiment, your crap is
wrong by mere inspection.

Well done Henry, you grasped it. Now move each
part at c+v (in the inertial frame) from that
point and you get Jerry's applet.

..and the phase at the tip is constantly changing....

You are clueless Henry, a point of fixed phase is
the only identifiable thing whose speed you can
measure.

George, a light source emits photon PARTICLES, not squiggly lines.

You drew the squiggly line Henry, and I see
them on an oscilloscope connected to an
antenna, and that is what we measure the
speed of.

...but Jerry made the mistake of arranging for the phases to be the same when
the rays meet.

No she doesn't, she arranges that they are emitted
in phase and that they travel at the right speed,
the fact that they are in phase when they arrive
is an unavoidable consequence.

No they are not.

Yes they are Henry, it cannot be otherwise from
very simple maths.

See www.users.bigpond.com/hewn/toothwheel.exe

You can draw anything whether it is right
or wrong.

The phases are calculated from pathlength/lambda

Wrong, pathlength/(distance moved per cycle)

Has Androcles finally won you over?

He gave up on the physics and went back to his
usual childish behaviour so I just killfiled
him as before. He has even less idea of this
than you. What I say above is simply comparing
apples with apples, trivial mechanics.

and you find they WERE in phase when they were emitted.

The waves are shown to be moving and their phase is shown by the teeth. It
is a
graph of phase.

I knew you didn't understand phase. What you
have drawn is NOT a graph of phase, any point
of given phase moves at speed c+v one way and
c-v the other. Your phase points don't move at
all.

George you are obviously confusing photon movement with the behavior of
classical waves in a medium...You seem to believe that ligtht behaves like
sound. It doesn't.

It doesn't, but the definition of the phase
of a sine wave is the same regardless of the
the application.

No George. In BaTh, wavelength is absolute.

So what, the definition of phase remains the
same.

Until you learn what phase is your program will
remain wrong. Once the light has been emitted,
what happens to the source has no effect on it.
Each part of the wave moves at c+/-v and until
you make the sine waves move, your program does
not represent ballistic theory. You remain
clueless.

The source emits photons, not squiggly lines.
Light is not sound...

RF is a propagating sine wave. Monochromatic
light is identical but at higher frequency,
and the definition of the phase velocity of
light is the velocity of a point on the wave
of fixed phase, obviously.

...and you have actually seen a point on the wave of a photon, have you George?

That's what you measure in any interferometer
Henry, the correlaton between the phases.

The phase of the tip as it travels at c+v around the ring is given by
pathlength/lambda and is represented by the circle of teeth.

Wrong, pathlength/(distance moved per cycle)

Why don't you and Androcles get togther. You seem to have a lot in common...

Why don't you grow up and try to understand
what you are being taught.

Both rays travel for the same time duration ..

Therefore they must arrive in phase.

WRONG. They move at different speeds and their wavelengths are the same.

Again, so what? you cannot calculate the received phase
from just those two items. What you know is that they
are emitted in phase and take equal times so points
of equal phase arrive simultaneously. There is no way
they can fail to do so.

I remind you, this approach produces the right answer.

I remind you, so does SR and it has been
properly analysed by thousands of people
while your attempt has glaring errors.

but at different speeds. The
'wavelength' is the same in both. It should be obvious even to you George, that
their phases will not be the same at the detector.

It is obvious that if point of given phase take the
same time for each then the phase difference when
they arrive must be the same as when they were
emitted. They were emitted in phase so they must
arrive in phase.

doppler ...
If you start talking frequency and time, ypou must consider doppler. Remember
the detector is moving relative to the static EMISSION POINT.

There is no Doppler shift clueless, the path
lengths are constant.

It turns out you didn't know what part of a wave
was being measured to define the speed, you are
even more clueless than I had realised.

George, the math is simple. Phase difference is:
[path length difference/lambda] mod(lambda).

Wrong as I told you above, apples and pears.

It is backed by experiment...

No it is falsified by experiment, your arithmetic
is flawed.

George

HW@....(Dr. Henri Wilson) wrote in news:usfnh3t692asdog55jmmq5onnucfkloah8@
4ax.com:

the front of the ray is changing phase as it moves

Neither the phase nor the polarization of 'the front of the ray' change 'as
it moves'.


To a stationary point that is being passed by the ray, the phase is
constantly changing, of course but the phase of the front does not change.

Drop a stone in the water.

you will see a wave /\/\/\-- traveling outward from the disturbance.

The FRONT of that wave /\ travels -- to the right.

The phase of that front is a constant /\ Positive (in this case) going
peak.
The positive going peak travels away from the source. It does NOT change
phase as it travels.

It does NOT matter if we are talking about BaTH, or SR or even waves on a
pond, the wave front maintains the same phase.



--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

Dr. Henri Wilson skrev:
On Mon, 22 Oct 2007 00:29:16 -0700, George Dishman
wrote:

On 21 Oct, 22:34, HW@....(Clueless Henri Wilson) wrote:
On Sun, 21 Oct 2007 12:08:07 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" HW@.... wrote in message

the part of the wave that leaves the emission point IN PHASE is shown
moving at
c+v and c-v in red and blue. You can check the phasing at any point by
pausing
the program.

Henri talking about his wave and and how to find the phase at any point.

How am I supposed to do that Henry? By comparison
with the squiggly line? If that is the case, it
proves you wrong, the phase at the tip should be
constant.

I still can't upload my latest vesion. The server is playing games again....

George, the front of the ray is changing phase as it moves.

Henri still talking about his wave and the phase at its front.

Don't be silly Henry. If you measure the speed of a
wave, all you can do is choose some particular phase
value and track its motion in order to find how fast
it moves. There is no other way to define the speed
of a sine wave.

George, I'll tell you secret. A light source emits photon PARTICLES not
classical water waves.

Oooooops! Something doesn't add up! Flee! Flee!
Henri won't talk about waves any more.

This _is_ fun, isn't it? :-)


--
Paul, with the morbid sense of humour

http://home.c2i.net/pb_andersen/

On Mon, 22 Oct 2007 04:42:19 -0700, George Dishman
wrote:

On 22 Oct, 10:37, HW@....(Dr. Henri Wilson) wrote:
On Mon, 22 Oct 2007 00:29:16 -0700, George Dishman wrote:
On 21 Oct, 22:34, HW@....(Clueless Henri Wilson) wrote:

George, I'll tell you secret. A light source emits photon PARTICLES not
classical water waves.

Henry, I'll tell you the obvious, individual photons
suffer interference effects in Young's Slits. You
have not considered which part of a photon the speed
measures. The answer is that it is a point of given
phase.

It is apparent that individual photons are split in two by hte mirror in a ring
gyro too.
So what?

The teeth are not supposed to move.

The speed they move at is supposed to be c+v.

the ' teeth' are a graph.

Yes, they should move at c+v according to
ballistic theory. Your software is wrong.

George, it matters not how fast the teeth spin, the pattern AT ANY INSTANT is
the same.
Is that too hard for you, too?

They are a circular graph of the phase of
the ray front as it moves around the ring. You can even see it happening in
Jerry's silly program. My toothed wheel is just a graph of what Jerry has
(wrongly) shown.

Jerry's program is right, you know so little
physics, you don't even know what is being
measured.

Jery has modelled a ring gyro using sound in air...but without the air...

Wrong again, the speeds would be c relative
to the lab using air. Jerry's speeds are
correct, yours are zero.

I said "without the air".
You and Jerry are assuming that light is a wave traveling in a medium.
it is not.



George, it is obvious that no amount of advice will teach you the facts.
The phase difference at the detector is [path length difference/ absolute
wavelength].

Wrong Henry, the pathlength is the distance
_moved_ so you divide by the distance _moved_
per cycle, not the wavelength, they are not
the same.

PROVED BY EXPERIMENT.

SR is proved by experiment, your crap is
wrong by mere inspection.

SR gets the same right answer by stealing from BaTh.



No they are not.

Yes they are Henry, it cannot be otherwise from
very simple maths.

See www.users.bigpond.com/hewn/toothwheel.exe

You can draw anything whether it is right
or wrong.

What I have drawn is quite obvious. the rays move at c+v and c-v and meet at
the detector at the same instant. One path is longer than the other and
contains more 'wavelengths'.
END OF STORY, GEORGE.

The phases are calculated from pathlength/lambda

Wrong, pathlength/(distance moved per cycle)

Has Androcles finally won you over?

He gave up on the physics and went back to his
usual childish behaviour so I just killfiled
him as before. He has even less idea of this
than you. What I say above is simply comparing
apples with apples, trivial mechanics.

Photons don't have 'cycles'....unless you mean the intrinsic ones.


It doesn't, but the definition of the phase
of a sine wave is the same regardless of the
the application.

No George. In BaTh, wavelength is absolute.

So what, the definition of phase remains the
same.

So what?

RF is a propagating sine wave. Monochromatic
light is identical but at higher frequency,
and the definition of the phase velocity of
light is the velocity of a point on the wave
of fixed phase, obviously.

...and you have actually seen a point on the wave of a photon, have you George?

That's what you measure in any interferometer
Henry, the correlaton between the phases.

....and the behavior of a ring gyro is fully explained by BaTh, as at:
www.users.bigpond.com/hewn/ringgyro.htm

It's so simple a 5yo kid could understand it.

Therefore they must arrive in phase.

WRONG. They move at different speeds and their wavelengths are the same.

Again, so what? you cannot calculate the received phase
from just those two items.

Yes I can. see the above webpage.

What you know is that they
are emitted in phase and take equal times so points
of equal phase arrive simultaneously. There is no way
they can fail to do so.

That might apply to waves in a medium...but photons are particles....
Wavelength is absolute. Frequency is infered as the arrival rate of
'wavecrests'.

I remind you, this approach produces the right answer.

I remind you, so does SR and it has been
properly analysed by thousands of people
while your attempt has glaring errors.

SR is crap from start to finish. It has never been verified or even vaguely
supported by any experiment.

doppler ...
If you start talking frequency and time, ypou must consider doppler. Remember
the detector is moving relative to the static EMISSION POINT.

There is no Doppler shift clueless, the path
lengths are constant.

The detector moves wrt the emission point.
If you use the rotating frame as Jerry has done in its latest, the emission
point and ALL the previous 'wavecrests' move backwards around the ring.

It turns out you didn't know what part of a wave
was being measured to define the speed, you are
even more clueless than I had realised.

George, the math is simple. Phase difference is:
[path length difference/lambda] mod(lambda).

Wrong as I told you above, apples and pears.

That show how much you know about maths.

It is backed by experiment...

No it is falsified by experiment, your arithmetic
is flawed.

It's so simple a 5you kid could do it....but not a relativist.

George



Henri Wilson. ASTC,BSc,DSc(T)

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

On Mon, 22 Oct 2007 20:58:02 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
On Mon, 22 Oct 2007 00:29:16 -0700, George Dishman
wrote:

On 21 Oct, 22:34, HW@....(Clueless Henri Wilson) wrote:
On Sun, 21 Oct 2007 12:08:07 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" HW@.... wrote in message

the part of the wave that leaves the emission point IN PHASE is shown
moving at
c+v and c-v in red and blue. You can check the phasing at any point by
pausing
the program.

Henri talking about his wave and and how to find the phase at any point.

How am I supposed to do that Henry? By comparison
with the squiggly line? If that is the case, it
proves you wrong, the phase at the tip should be
constant.

I still can't upload my latest vesion. The server is playing games again....

George, the front of the ray is changing phase as it moves.

Henri still talking about his wave and the phase at its front.

Don't be silly Henry. If you measure the speed of a
wave, all you can do is choose some particular phase
value and track its motion in order to find how fast
it moves. There is no other way to define the speed
of a sine wave.

George, I'll tell you secret. A light source emits photon PARTICLES not
classical water waves.

Oooooops! Something doesn't add up! Flee! Flee!
Henri won't talk about waves any more.

This _is_ fun, isn't it? :-)

Paul, getting more worried every day.....


Henri Wilson. ASTC,BSc,DSc(T)

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

On Mon, 22 Oct 2007 11:50:56 +0000 (UTC), bz
wrote:

HW@....(Dr. Henri Wilson) wrote in news:usfnh3t692asdog55jmmq5onnucfkloah8@
4ax.com:

the front of the ray is changing phase as it moves

Neither the phase nor the polarization of 'the front of the ray' change 'as
it moves'.


To a stationary point that is being passed by the ray, the phase is
constantly changing, of course but the phase of the front does not change.

Drop a stone in the water.

you will see a wave /\/\/\-- traveling outward from the disturbance.

The FRONT of that wave /\ travels -- to the right.

The phase of that front is a constant /\ Positive (in this case) going
peak.
The positive going peak travels away from the source. It does NOT change
phase as it travels.

It does NOT matter if we are talking about BaTH, or SR or even waves on a
pond, the wave front maintains the same phase.

Light is NOT like waves on water. A source emits photon particles, not squiggly
lines.



Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message
...
: On Sun, 21 Oct 2007 09:22:34 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Sun, 21 Oct 2007 00:29:04 GMT, "Androcles"

: : wrote:
:
: : :
: : That's interesting. By dimensional analysis,
: : E = h * (distance/time) / distance.
: : E = h * 1/time
: : = h (nu)
: :
: : that's what I said you ****ing idiot.
: : 'nu' isn't an intrinsic property.
:
: Yes it is, sheep-shagger.\
:
: At least the Aussies only go for the females...not like your Engish
: counterparts.

'nu' is an intrinsic property.


:
: : It is an inferred, 'relative speed' dependent
: : one.
:
: Wrong, frequency is count per unit time and time is absolute.
:
: that's what I said, dope. NUMBER ARRIVING PER ABSOLUTE SECOND.

'nu' is an intrinsic property, number LEAVING the source per absolute
second.
The horse's arse is at the other end, that's obviously what is interesting
you.


: Speed is relative.
: This set of wheels all have the same frequency:
: http://www.androcles01.pwp.blueyonder.co.uk/inphase.gif
: This set do not:
: http://www.androcles01.pwp.blueyonder.co.uk/nophase.gif
: Explain that, sheep shagger.
:
: 'ewe shagger' please....
:
Ok, "ewe shagger" it is.
'nu' is an intrinsic property, number LEAVING the source per absolute
second.
The ewe's arse is at the other end, that's obviously what is interesting
you.
If she runs away the ewe shagger's dick will fall out and the frequency will
stop.



: : Did you claim to be a physicist, faux Dr. Wilson?
: :
: : You are an engineer.
: Yes, I am.
:
: .......and a 'ram shagger' obviously ...

Rams turn and face you. Then YOU run away.

:
: : Like Dishman, you know nothing about physics.
:
: Like Dishman and Tusseladd, you know nothing about physics.
: I'm not fooled by clockwork, you are.
: This set of wheels all have the same frequency:
: http://www.androcles01.pwp.blueyonder.co.uk/inphase.gif
:
: ...this guy goes completely bonkers after 6 pm.


Wilson still can't count, even after a week of trying.

:
: This set do not:
: http://www.androcles01.pwp.blueyonder.co.uk/nophase.gif
: Explain that, sheep shagger, an engineer could.
:
: what's that? Ram shagger?

A little test, ewe shagger, to see if you can count better than a 5-yr-old.




: : You are an engineer. Like Dishman, you know nothing about physics.
: Like Dishman and Tusseladd, you know nothing about physics.
: I'm not fooled by clockwork, you are.
: This set of wheels all have the same frequency:
: http://www.androcles01.pwp.blueyonder.co.uk/inphase.gif
: This set do not:
: http://www.androcles01.pwp.blueyonder.co.uk/nophase.gif
: Explain that, sheep shagger, an engineer could.
:
: The ****ing teeth on the three wheels all have the SAME ABSOLUTE
WAVELENGTHS,
: YOU OLD DOPE.


Aha! A little closer after a full week of Wilson trying to count.
All wheels have the same circumference.
nophase.gif -- 24 teeth.
inphase.gif -- 24, 22, 26 teeth.
The ****ing teeth on the three wheels all have DIFFERENT
wavelengths, you drunken old ewe shagger, the wavelengths are
circumference/24
circumference/22
circumference/26

: There is no 'frequency'

Being a mere physicist and ewe shagger you would not know
about these that read out frequency: http://tinyurl.com/24ekpg

The gear teeth "ARRIVING PER ABSOLUTE SECOND"
are the same as those leaving per absolute second and a
60 * RPM/24 Hz
60 * RPM/22 Hz
60 * RPM/26 Hz

All different.
The wheels are in perfect phase, locked together and turning
at the same frequency.

BTW, even a 5 yr-old knows that the wagon wheels turn backwards
in those old Western movies, but you are so easy to fool because
you are a fool.

Couldn't even count to 26 and doesn't know RPM is frequency....
my, my... calls itself a physicist. Definitely a sheep shagger.

On Mon, 22 Oct 2007 21:07:18 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
: On Sun, 21 Oct 2007 09:22:34 GMT, "Androcles"
: wrote:
:


'nu' is an intrinsic property.


:
: : It is an inferred, 'relative speed' dependent
: : one.
:
: Wrong, frequency is count per unit time and time is absolute.
:
: that's what I said, dope. NUMBER ARRIVING PER ABSOLUTE SECOND.

'nu' is an intrinsic property, number LEAVING the source per absolute
second.

: : You are an engineer. Like Dishman, you know nothing about physics.
: Like Dishman and Tusseladd, you know nothing about physics.
: I'm not fooled by clockwork, you are.
: This set of wheels all have the same frequency:
: http://www.androcles01.pwp.blueyonder.co.uk/inphase.gif
: This set do not:
: http://www.androcles01.pwp.blueyonder.co.uk/nophase.gif
: Explain that, sheep shagger, an engineer could.
:
: The ****ing teeth on the three wheels all have the SAME ABSOLUTE
WAVELENGTHS,
: YOU OLD DOPE.


Aha! A little closer after a full week of Wilson trying to count.
All wheels have the same circumference.
nophase.gif -- 24 teeth.
inphase.gif -- 24, 22, 26 teeth.
The ****ing teeth on the three wheels all have DIFFERENT
wavelengths, you drunken old ewe shagger, the wavelengths are
circumference/24
circumference/22
circumference/26

How am I or anyone else supposed to know that?

: There is no 'frequency'

Being a mere physicist and ewe shagger you would not know
about these that read out frequency: http://tinyurl.com/24ekpg


The gear teeth "ARRIVING PER ABSOLUTE SECOND"
are the same as those leaving per absolute second and a
60 * RPM/24 Hz
60 * RPM/22 Hz
60 * RPM/26 Hz

All different.
The wheels are in perfect phase, locked together and turning
at the same frequency.

That's not a ring gyro.
The wheels should have the same number of teeth, so as to represent absolute
'wavelength'.




Henri Wilson. ASTC,BSc,DSc(T)

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

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

On Mon, 22 Oct 2007 11:50:56 +0000 (UTC), bz
wrote:

HW@....(Dr. Henri Wilson) wrote in
news:usfnh3t692asdog55jmmq5onnucfkloah8@ 4ax.com:

the front of the ray is changing phase as it moves

Neither the phase nor the polarization of 'the front of the ray' change
'as it moves'.


To a stationary point that is being passed by the ray, the phase is
constantly changing, of course but the phase of the front does not
change.

Drop a stone in the water.

you will see a wave /\/\/\-- traveling outward from the disturbance.

The FRONT of that wave /\ travels -- to the right.

The phase of that front is a constant /\ Positive (in this case) going
peak.
The positive going peak travels away from the source. It does NOT change
phase as it travels.

It does NOT matter if we are talking about BaTH, or SR or even waves on
a pond, the wave front maintains the same phase.

Light is NOT like waves on water. A source emits photon particles, not
squiggly lines.

Then why are you drawing squiggly lines in your simulations?


Your statement about phase makes even less sense now. You said:
the front of the ray is changing phase as it moves

How do you justify a 'front of the ray' when talking of photons? There is a
front end and a back end to a photon. There is not a 'front of the ray'.

How do you justify saying that 'it'[whatever it happens to be photon, ray
or wave] 'is changing phase as it moves' when talking of photons?

Whether we are speaking of waves or photons; whether we are considering
BatH or SR, the 'front of the ray' can NOT possibly 'change phase as it
moves'. Under some conditions, an observer at a point, observing a passing
'ray/wave/photon' might observe a wavelike phenomina and see a variation in
phase as the 'ray/wave/photon' passes. BUT that is NOT the same as
the front of the ray is changing phase as it moves


By the way, WAVES have 'phase'. Particles do NOT have phase. If we are
speaking of phase of photons and, as you have insisted at time, photons are
not waves, then photons do not have phase to 'change as they move'.