Why are the 'Fixed Stars' so FIXED?

On Fri, 26 Oct 2007 04:41:34 GMT, "Androcles"
wrote:


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


: 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.

v=wR.

What's the wavelength of this sine wave?
http://spaceflight.nasa.gov/realdata/tracking/

I don't know ahnd don't care.

The wavelength of a photon is the distance it travels in the source frame
during one cycle of its intrinsic oscillation. It move at c wrt its source so
lambda=c/nu.

Since all lenThs are absolute, wavElength is also the same in all frames...as
is the intrinsic period....BUT NOT THE OBSERVED FREQUENCY.

: : 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




Henri Wilson. ASTC,BSc,DSc(T)

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

On Oct 25, 4:45 pm, HW@....(Dr. Henri Wilson) wrote:
On Wed, 24 Oct 2007 21:47:32 -0700, Jerry
wrote:

On Oct 24, 4:14 pm, HW@....(Dr. Henri Wilson) wrote:
On Wed, 24 Oct 2007 02:13:45 -0700, Jerry
wrote:
On Oct 24, 3:10 am, George Dishman wrote:
On 23 Oct, 22:32, HW@....(Clueless Henri Wilson) wrote:

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.

I think that you have gone -way- over Henri's head with
mention of coherence length.

To Henri:

Early experimentalists such as Michelson and Morley, Sagnac
etc. used monochromatic sources only during the alignment
stages while setting up their interferometers. Actual
experimental runs were always performed with white light.
The reason for this is that white light creates a distinctive
pattern of a central bright white fringe surrounded by a
rapidly fading set of colored fringes. The advantage of this
is that the central fringe of equal path length is always
readily identifiable, whereas monochromatic light produces
uniform fringes in which it is virtually impossible to
determine the central fringe of equal path length.

The distinctive pattern of fringes formed by white light
enabled Michelson and Morley, who recorded their observations
visually, not to "get lost" while figuring out how far their
fringes were displaced from their fiducial marks. In the
Michelson and Gale experiment, which was a giant Sagnac
setup, the central fringe, in the absence of rotation, would
appear precisely midway between the two images of the slit.
This enabled them to calibrate their apparatus for zero
rotational velocity; it was thus not necessary for them to
halt the rotation of the Earth to get a zero reading, which
would have been somewhat impractical in the absence of divine
intervention (Joshua 10:12-15).

Note that I stated that the pattern of colored fringes
surrounding the central bright fringe fades rapidly. This is
because the spacing between the red fringes and the blue
fringes is different. Within a few fringe widths from the
central fringe, the colored fringes overlap until the fringe
pattern is no longer perceptible. Since each fringe represents
a half wave difference in path length to the two images of the
source slit, this means that the path lengths must be
precisely matched, otherwise it would be impossible to see any
fringes at all.

This distance to which the path lengths must be matched,
otherwise fringes are invisible, is known as the "coherence
length". The coherence length for white light is no more
than a handful of microns. Your notion that "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"
is totally ridiculous to anybody who knows anything at all
about optics.

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.

You have COMPLETELY lost the point. Earlier, you made the stupid
and asinine speculation that "fringe production in a sagnac
interferometer [has] something to do with the phase relationship
between INCOMING and OUTGOING rays rather than the rejoining of
the two oppositely moving rays."

George's point was that since white light Sagnac interferometers
are perfectly functional, your speculation is dead in the water.

You didn't read properly. I SAID IT WAS IMPOSSIBLE.

Oh, dear me. Another blatant lie.

Henri, if you want to make such statements, hoping that
the previous history of our conversation would have
scrolled off people's newsreaders, you could at least
make the effort to trim the quote chain from this message,
or perhaps falsify the quote chain, which is another
technique which you have previously employed.

You wrote, as can be seen above in this same message,
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?

In other words, you did NOT state that it was impossible.
You were stating it as a possibility to consider, even if
it sounded unlikely.

Jerry

Henri Wilson's Lies
(1)Fakes Diploma (2)Uses Deceptive Language (3)Fakes Program
(4)Intentionally Misquotes (5)Snips (6)Accuses Others of Lying
1 http://mysite.verizon.net/cephalobus...ri/diploma.htm
2 http://mysite.verizon.net/cephalobus.../deception.htm
3 http://mysite.verizon.net/cephalobus...rt_aurigae.htm
4 http://mysite.verizon.net/cephalobus...ri/history.htm
5 http://mysite.verizon.net/cephalobus...enri/snips.htm
6 http://mysite.verizon.net/cephalobus...ri/accuses.htm

Dr. Henri Wilson wrote:
On Thu, 25 Oct 2007 20:24:22 +0200, "Paul B. Andersen"
wrote:
What IS your approach which produce 'the right answer'?

And how do you define the wavelengths you are counting?

A photon has an intrinsic oscillation of an unknown nature. During the absolute
time interval defined by one period of that oscillation, an identifiable point
in the photon body moves through a 'spatial interval' at c wrt the source. The
absolute distance it moves in the source frame is its 'wavelength'. Like ALL
lengths, that wavelength is the same in all frames.

Unlike your 'fixed squiggly line' theory, the phase at the 'front' of a BaTh
photon changes as it moves.

Quite.
"the front of a BaTh photon oscillates once every absolute wavelength traveled."

"Oscillates once" can only mean that the phase increases by 2pi radians.

So according to what you say above, the phase at the front of any photon
must in the source frame fulfill the equation:
phi(t+T, x+cT) - phi(t,x) = 2pi
where T is the 'absolute time interval of one oscillation'
and cT = l is "the absolute distance it moves during T", that is the wavelength
T = l/c

If we assume that phi(t,x) is a linear function of x and t,
phi(t,x) = at + bx
we get:
(at + aT + bx + bcT)-(at + bx) = 2pi
aT+bcT = 2pi
b = (2pi+acT)/T = 2p/T + ac
Inserting T = l/c, we find:
phi(t,x) = at + (a/c + 2pi/l)x

Since the phase of any photon in a ray of photons must fulfill
this equation, it gives us the phase of the photon found at x at
the time t.

But what is a?

We know that the phase of the photon emitted from the source at x = x1
at the time t+T must be 2pi more than the photon emtted at the time t.
So:
phi(t+T,x1)-phi(t,x1) = 2pi
aT = 2pi
a = 2pi/T = 2pi.c/l (usually called the angular frequency w, of course)

So the equation becomes:
phi(t,x) = (2pi.c/l)t +((2pi.c/l)/c - 2pi/l)x = (2pi.c/l)t

So according to your BaTh:

phi(t,x) = 2pi.l/c (in the source frame)

The phase doesn't depend on x, all the photons in the ray have at
any time the same phase.

Satisfied, Henri?
If not, please show exactly where I was wrong, and show what
the correct equation for phi(t,x) should be.

--
Paul

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

On Oct 26, 4:25 am, "Paul B. Andersen"
wrote:

So the equation becomes:
phi(t,x) = (2pi.c/l)t +((2pi.c/l)/c - 2pi/l)x = (2pi.c/l)t

So according to your BaTh:

phi(t,x) = 2pi.l/c (in the source frame)

Transcription error?

phi(t,x) = (2pi.c/l)t (in the source frame)

I suppose that I'll have to modify my applet...although I'm
working on another one that's rather more serious...

The phase doesn't depend on x, all the photons in the ray have at
any time the same phase.

Satisfied, Henri?
If not, please show exactly where I was wrong, and show what
the correct equation for phi(t,x) should be.

Jerry

On Fri, 26 Oct 2007 00:48:51 -0700, George Dishman
wrote:

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.




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.

wHAT?..THAT YOU ARE ACTING DUMB?

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".

Gord, your not acting at all.....
The point is moving in the rotating frame, the light moves at c+v wrt that
point in the rotating frame....because the source is moving at v in the
nonrotating frame even though the emission point is not..



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 :-)

OK.


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.

think again George. I'm not losing it. You are Andersen are..
It's only a small second order effect anyway. It's the one involved when you
replace c^2-v^2 with c^2. Don't worry about it.


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.

Indeed they do!!!!
And their 'intrinsic oscillation' is out of phase because it goes through 1
cycle every wavelength traveled.
Simple isn't it.


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.

George, the leading edge - and indeed the whole waveform - of your 'moving
wiggles' does not change.
The leading edge of a BaTh photon goes trough a cycle for every wavelength
moved. I have defined wavelength in this context.

The absolute wavelength of light is the distance moved in the source frame
during one cycle of its 'intrinsic oscillation'. Since it moves at c relative
to the source (in the source frame), lambda = c/nu. Lambda is an absolute
length and the same in all frames.

'nu' should not be confused with the 'inferred frequency', which is the number
of wavelengths arriving per second...or nu(c+v)/c.

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.

Wrong.
see above.



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

See above, your maths is wrong.

See above. It is not.

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.

See above. It is not.

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



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.

It has always been too small to worry about.

As I said, it appears you don't have the ability
to understand the maths involved.

George, you are way behind..


They don't
apply to photon particles.

Obviously but aggregating photons must produce
Maxwell's Equations.

In a medium. Speeds must always be specified relative to something George.

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.

Travel time is distance/speed.

t=2piR/(c+v)

What does the 'c+v' represent, George?



George



Henri Wilson. ASTC,BSc,DSc(T)

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

On Fri, 26 Oct 2007 04:43:48 GMT, "Androcles"
wrote:


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

:
: : 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.

You tried that one last time...It ain't funnuy any more...

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.

So waves approach the shore because water likes to flow uphill...is that what
you're saying?


: 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.

not funny any more. ..just a tactic used by pommie ****pots to avoid having to
answer difficult questions.

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.

w is angular velocity.....= v/r
Any decent engineer should know that......but apparently not you or Dishman.




Henri Wilson. ASTC,BSc,DSc(T)

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

On Fri, 26 Oct 2007 04:53:19 GMT, "Androcles"
wrote:


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

: That's not a fringe shift...

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

Bloody dope...


: 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

Aha! Seto would be proud of you..... hahahahahaha!

YOU HAVE MODELLED THE BLOODY AETHER.

Tell me. What makes the light reflect from the mirror at three times its
incident speed(relative to the mirror)?

Hahahahaha!

: 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

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

On Thu, 25 Oct 2007 08:43:29 +0000 (UTC), bz
wrote:

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

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.

Crap. We were talking about the MMx and four mirror sagnac. Lines rather
than circles appear if the top mirror is angled slightly to form an
optical wedge. This was a popular version of the MM interferometer.


Why would you think they are circles?

No wonder you are so confused about Sagnac. Your mind keeps running
around in circles.

Don't try to discuss things you know nothing about, bob.

If you stuck to your own rule, you would post a LOT less nonsense.

What I said is correct for the original MMX and Sagnac experiments. They
both used a vertical slit and hence saw a series of vertical lines, not
circles.






Henri Wilson. ASTC,BSc,DSc(T)

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





--
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

On Wed, 24 Oct 2007 02:13:45 -0700, Jerry
wrote:

On Oct 24, 3:10 am, George Dishman wrote:
On 23 Oct, 22:32, HW@....(Clueless Henri Wilson) wrote:

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.

I think that you have gone -way- over Henri's head with
mention of coherence length.

To Henri:

Early experimentalists such as Michelson and Morley, Sagnac
etc. used monochromatic sources only during the alignment
stages while setting up their interferometers. Actual
experimental runs were always performed with white light.
The reason for this is that white light creates a distinctive
pattern of a central bright white fringe surrounded by a
rapidly fading set of colored fringes. The advantage of this
is that the central fringe of equal path length is always
readily identifiable, whereas monochromatic light produces
uniform fringes in which it is virtually impossible to
determine the central fringe of equal path length.

I know. I once made a michelson interferometer. I was quite easy to adjust.

The distinctive pattern of fringes formed by white light
enabled Michelson and Morley, who recorded their observations
visually, not to "get lost" while figuring out how far their
fringes were displaced from their fiducial marks.

The deliberate tilting of the top mirror to create an optical wedge was a later
innovation that produced almost straight line fringes. It is obviously easier
to measure the sideways displacement of a line than to estimate the shade of
fairly uniform image.

Maybe circles are still preferred in metrology.

get it yet?

In the
Michelson and Gale experiment, which was a giant Sagnac
setup, the central fringe, in the absence of rotation, would
appear precisely midway between the two images of the slit.
This enabled them to calibrate their apparatus for zero
rotational velocity; it was thus not necessary for them to
halt the rotation of the Earth to get a zero reading, which
would have been somewhat impractical in the absence of divine
intervention (Joshua 10:12-15).

Note that I stated that the pattern of colored fringes
surrounding the central bright fringe fades rapidly. This is
because the spacing between the red fringes and the blue
fringes is different. Within a few fringe widths from the
central fringe, the colored fringes overlap until the fringe
pattern is no longer perceptible. Since each fringe represents
a half wave difference in path length to the two images of the
source slit, this means that the path lengths must be
precisely matched, otherwise it would be impossible to see any
fringes at all.

This distance to which the path lengths must be matched,
otherwise fringes are invisible, is known as the "coherence
length". The coherence length for white light is no more
than a handful of microns. Your notion that "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"
is totally ridiculous to anybody who knows anything at all
about optics.

It's all irelevant anyway since light moves at c wrt its source and everything
at rest wrt the source.

Jerry



Henri Wilson. ASTC,BSc,DSc(T)

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

On Oct 26, 6:34 am, HW@....(Dr. Henri Wilson) wrote:
On Wed, 24 Oct 2007 02:13:45 -0700, Jerry
wrote:

Early experimentalists such as Michelson and Morley, Sagnac
etc. used monochromatic sources only during the alignment
stages while setting up their interferometers. Actual
experimental runs were always performed with white light.
The reason for this is that white light creates a distinctive
pattern of a central bright white fringe surrounded by a
rapidly fading set of colored fringes. The advantage of this
is that the central fringe of equal path length is always
readily identifiable, whereas monochromatic light produces
uniform fringes in which it is virtually impossible to
determine the central fringe of equal path length.

I know. I once made a michelson interferometer. I was quite
easy to adjust.

Using a monochromatic light source, yes. A white light
Michelson interferometer is rather finicky because of the
short coherence length.

The distinctive pattern of fringes formed by white light
enabled Michelson and Morley, who recorded their observations
visually, not to "get lost" while figuring out how far their
fringes were displaced from their fiducial marks.

The deliberate tilting of the top mirror to create an optical
wedge was a later innovation that produced almost straight line
fringes.

Nope. See my next comments.

It is obviously easier to measure the sideways
displacement of a line than to estimate the shade of
fairly uniform image.

Tilting doesn't work with a white light interferometer.
The interference pattern doesn't extend far enough out to get
"straight" fringes, and the fringes would be colored. The early
experimentalists used SLIT sources of light.

Obviously you are accustomed to monochromatic light and lasers.

Maybe circles are still preferred in metrology.

get it yet?

Sure. But YOU sure haven't.

In the
Michelson and Gale experiment, which was a giant Sagnac
setup, the central fringe, in the absence of rotation, would
appear precisely midway between the two images of the slit.
This enabled them to calibrate their apparatus for zero
rotational velocity; it was thus not necessary for them to
halt the rotation of the Earth to get a zero reading, which
would have been somewhat impractical in the absence of divine
intervention (Joshua 10:12-15).

Note that I stated that the pattern of colored fringes
surrounding the central bright fringe fades rapidly. This is
because the spacing between the red fringes and the blue
fringes is different. Within a few fringe widths from the
central fringe, the colored fringes overlap until the fringe
pattern is no longer perceptible. Since each fringe represents
a half wave difference in path length to the two images of the
source slit, this means that the path lengths must be
precisely matched, otherwise it would be impossible to see any
fringes at all.

This distance to which the path lengths must be matched,
otherwise fringes are invisible, is known as the "coherence
length". The coherence length for white light is no more
than a handful of microns. Your notion that "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"
is totally ridiculous to anybody who knows anything at all
about optics.

It's all irelevant anyway since light moves at c wrt its source
and everything at rest wrt the source.

Anything you don't understand is "irrelevant"?

Jerry
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