Why are the 'Fixed Stars' so FIXED?

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

The sensible thing to do is use monochromatic light and tilt the top
miror slightly in order to produce an 'optical wedge' effect. That
produces a straight line fringe pattern rather than circles. Straight
lines are easier to count than circles and in the case of gyros, make
the direction of an acceleration easy to determine.

Henri, the image is a vertical line surrounded by fainter vertical lines.
The fringes are vertical lines because they are images of the slit.

Why would you think they are circles?

Actually Henri has a point in this case.
If a slit is used, you will obviosly get a line-pattern.
But nowadays it is more common to use a laser as a source,
which can be cosidered a point source rather than a slit.
In an interferometer the light from the source is splitted,
and the light is going along different paths to the screen.
This is equivalent to having two correlated point sources
at some distance from the screen. If these sources are on
the same optical axis, but the effective distances to them
are different, then you will get a "bulls eye" pattern on
the screen. This is often used in modern Michelson interferometers.
http://www.hartnell.cc.ca.us/physics...rferometer.pdf
Note however that the lengths of the two arms have to be slightly
different for a bulls eye pattern to form.
In a Sagnac interferometer, the two paths will be equal
(when not rotating), so this is probably not practical.
But if you skew one of the mirrors a slightly, it will have the same
effect as two point sources which are NOT on the same optical axis.
You will then get bright fringes where the distance to the sources
differ by N*lambda. These will be hyperbolas. The central line (N=0)
is straigt, N=1 will be slightly curved, N=2 more curved, etc.
http://tinyurl.com/33p3at

--
Paul

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

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

The sensible thing to do is use monochromatic light and tilt the top
miror slightly in order to produce an 'optical wedge' effect. That
produces a straight line fringe pattern rather than circles. Straight
lines are easier to count than circles and in the case of gyros, make
the direction of an acceleration easy to determine.

Henri, the image is a vertical line surrounded by fainter vertical lines.
The fringes are vertical lines because they are images of the slit.

Why would you think they are circles?

Actually Henri has a point in this case.
If a slit is used, you will obviously get a line-pattern.
But nowadays it is more common to use a laser as a source,
which can be considered a point source rather than a slit.
In an interferometer the light from the source is split,
and the light is going along different paths to the screen.
This is equivalent to having two correlated point sources
at some distance from the screen. If these sources are on
the same optical axis, but the effective distances to them
are different, then you will get a "bulls eye" pattern on
the screen. This is often used in modern Michelson interferometers.
http://www.hartnell.cc.ca.us/physics...rferometer.pdf
Note however that the lengths of the two arms have to be slightly
different for a bulls eye pattern to form.
In a Sagnac interferometer, the two paths will be equal
(when not rotating), so this is probably not practical.
But if you skew one of the mirrors a slightly, it will have the same
effect as two point sources which are NOT on the same optical axis.
You will then get bright fringes where the distance to the sources
differ by N*lambda. These will be hyperbolas. The central line (N=0)
is straight, N=1 will be slightly curved, N=2 more curved, etc.
http://tinyurl.com/33p3at

--
Paul

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

"Clueless Henri Wilson" HW@.... wrote in message
...
....
..It's very amusing listening to one clueless moron talking to another....

Yes, you and Androcles are a source of great
amusement. Keep it up chaps :-)

George

"Clueless Henri Wilson" HW@.... wrote in message
...
On Wed, 24 Oct 2007 01:10:37 -0700, George Dishman
wrote:
On 23 Oct, 22:32, HW@....(Clueless Henri Wilson) wrote:
On Tue, 23 Oct 2007 00:23:06 -0700, George Dishman
wrote:
On 21 Oct, 21:59, HW@....(Clueless Henri Wilson) wrote:
On Sun, 21 Oct 2007 12:10:14 +0100, "George Dishman"
wrote:
"Clueless Henri Wilson" HW@.... wrote in
messagenews:k31lh399msodtdq95b05hn1m8fjsg8r0s7@4a x.com...

I appreciate that....but since light MUST BE ballistic

That's your problem Henry, you lack the knowledge
of geometry to see that there is an alternative
so your religious conviction has no scientific
validity. Theere is absolutely no reason why it
"must" be ballistic, "could be" but not "must be"
and from experiment we know it isn't.

(its source is its only
speed reference) and since my approach produces the right answer, I'm
confident
that it is the correct one.

Regardless of you anser, the maths is simply
broken, you divide by the wrong number.

Incidentally, this also tends to suggest that the fringe production in a
sagnac
interferometer is something to do with the phase relationship between
INCOMING
and OUTGOING rays rather than the rejoining of the two oppositely moving
rays...I know that sounds impossible...but is it?

Yes, for two reasons. The simpler is that if
you look at the arrangement of the beam
splitter, the remaining light goes back to
the lamp but the more robust is that there
would be a path length difference of nearly
a metre (the loop length) between the originated
light and that which has bone round the loop.
That grossly exceeds the coherence length for
a filament source so there is no way to form
fringes with a detectable contrast ratio.

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.

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.

(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".

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.

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!

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.

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.

Both approaches produce the same answer.

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

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.

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.

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.

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.

George

"Paul B. Andersen" wrote in
:

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

The sensible thing to do is use monochromatic light and tilt the top
miror slightly in order to produce an 'optical wedge' effect. That
produces a straight line fringe pattern rather than circles. Straight
lines are easier to count than circles and in the case of gyros, make
the direction of an acceleration easy to determine.

Henri, the image is a vertical line surrounded by fainter vertical
lines. The fringes are vertical lines because they are images of the
slit.

Why would you think they are circles?

Actually Henri has a point in this case.

Glad to have provided a chance for Henri to see that you are fair.

If a slit is used, you will obviously get a line-pattern.

Yep. And MMX and Sagnac used slits.

But nowadays it is more common to use a laser as a source,
which can be considered a point source rather than a slit.

Yes.

In an interferometer the light from the source is split,
and the light is going along different paths to the screen.
This is equivalent to having two correlated point sources
at some distance from the screen. If these sources are on
the same optical axis, but the effective distances to them
are different, then you will get a "bulls eye" pattern on
the screen. This is often used in modern Michelson interferometers.
http://www.hartnell.cc.ca.us/physics...erferometer.pd
f Note however that the lengths of the two arms have to be slightly
different for a bulls eye pattern to form.
In a Sagnac interferometer, the two paths will be equal
(when not rotating), so this is probably not practical.
But if you skew one of the mirrors a slightly, it will have the same
effect as two point sources which are NOT on the same optical axis.
You will then get bright fringes where the distance to the sources
differ by N*lambda. These will be hyperbolas. The central line (N=0)
is straight, N=1 will be slightly curved, N=2 more curved, etc.

Hmmm. Still not a straight line, but Henri definitely gets half a point
for the 'tipped miror'.

A tip of my 'miror' to Henri.


http://tinyurl.com/33p3at






--
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 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.

If you think a photon is just a 'moving squiggly line' then you're welcome to
the idea ...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?

So your approach is NOT to count the number of wavelengths defined
by the equation sin(wt - kx). But as you can see below by my question below,
I had understood that much.

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'? :-)

No answer, Henri?'

What IS your approach which produce 'the right answer'?

And how do you define the wavelengths you are counting?


--
Paul

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

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.

--
Paul

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

Dr. Henri Wilson skrev:
On Wed, 24 Oct 2007 22:42:32 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson skrev:
YOU are treating light as though it is a classical wave that obeys the general
wave equation: A= sin[2pi(t/tor-x/lambda]
Mind your language.
This is not a wave equation, it is a particular solution to
the the wave equation:
@^2A/@t^2 = c^2*nabla^2*A
or in the 1D case:
@^2A/@t^2 = c^2*@^2A/@x^2

...traveling wave in a medium.....

This works for waves in a medium and can be literally observed in the case of
water waves.

Jerry's animation shows this....it assumes light is nothing but a 'moving sine
wave'.
My 'cycle chain' idea also describes the classical traveling wave although it
was intended to show the different path lengths.

The plain fact is, George, light does NOT behave like this.


The plain fact is that light does indeed behave like this.

You must be pretty desperate to dispute what has been
experimentally verified by innumerable experiments and
everyday practical applications for centuries.

The originator of the "modern" version (as opposed to Newton's version)
of the emission theory, Ritz, wouldn't dream of disputing that light
behave as a wave. He was no idiot!
The difference between Ritz and Einstein is that Ritz claims
that the wave equation is valid only in the rest frame of the source,
and it transforms to other frames according to the Galilean transform,
while Einstein claim that the wave equation is equally valid in
any inertial frame (from which the Lorentz transform follows).

Nobody but you have since the days of Huygen's been so stupid as to
dispute that light behaves as a wave in the macroscopic realm.
The experimental evidence simply does not allow it.

Light is still very much a mystery.

No, it isn't.
QED - the theory of light and its interaction with matter -
is the best tested theory ever. There is no aspect of light
which this theory doesn't correctly model.
(Correctly = in accordance with experimental evidence)

Wave theory cannot generally explain the behavior of light.

If you mean Maxwell's theory, that's obviously correct.
Maxwell's theory explains nothing about the quantum aspects of light.
It doesn't explain how light sources work. It doesn't explain
how detectors work. It doesn't explain how light multipliers
and lasers work.

But when it comes to optics, _which is what we now are discussing_,
and generally the transmission of EM-radiation of all wavelengths
in all kinds of media, then Maxwell's equations and Maxwell's wave
equation describe the behaviour of that radiation extremely well,
and I repeat:
No exception is ever detected _in the macroscopic realm_.

The fact that optical instruments, optical transmission systems,
and radio transmission systems work as designed prove that
EM-radiation behave as the designers assumed - that is according
to Maxwell's equations.

How many devices designed on the basis of Maxwell's equations
do you think are in operation to allow you to read this posting?
From Norway to Australia - how many satellite jumps? How many
kilometres of optical and electrical cable? How many antennae?
Oh yes, Henri. EM-radiation behave according to Maxwell's equations.

And if you think quantum effects falsifies any of that, you are wrong.
The complete and exact theory of light - QED - predicts exactly
the same as Maxwell's equations + SR in the macroscopic limit.
If it didn't, it would be falsified and wouldn't exist.

I repeat:
Nobody but you have since the days of Huygens' been so stupid as to
dispute that light behaves as a wave in the macroscopic realm.
The experimental evidence simply does not allow it.

So you are contributing with something new after all.
Your stupidity of historical proportions is hardly world shattering, though.
Very few will notice.
For which you should be glad, Mr. Ralph Rabbidge

--
Paul

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

"Dr. Henri Wilson" HW@.... wrote in message
...
: On Tue, 23 Oct 2007 23:26:46 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Tue, 23 Oct 2007 21:48:29 GMT, "Androcles"

: : wrote:
:
: :
: : The classical equation A'=Asin[2pi(t/T-x/L)] doesn't apply to light.
:
: Your constant wavelength goes the same way as uni****ation
: and h-aether. Flushed for the **** it is, dumb plagiarising sheep
shagger.
: http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
:
: 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.




: 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.


: 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.

On Thu, 25 Oct 2007 20:38:44 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
: On Tue, 23 Oct 2007 23:26:46 GMT, "Androcles"
: wrote:
:
:
: "Dr. Henri Wilson" HW@.... wrote in message
: .. .
: : On Tue, 23 Oct 2007 21:48:29 GMT, "Androcles"

: : wrote:
:
: :
: : The classical equation A'=Asin[2pi(t/T-x/L)] doesn't apply to light.
:
: Your constant wavelength goes the same way as uni****ation
: and h-aether. Flushed for the **** it is, dumb plagiarising sheep
shagger.
: http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
:
: 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.
'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.
I don't know why this kind of model has been rejected.
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.


: 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