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Why are the 'Fixed Stars' so FIXED?



 
 
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  #2391  
Old October 26th 07, 09:11 PM posted to sci.astro,sci.physics.relativity
Dr. Henri Wilson
external usenet poster
 
Posts: 707
Default Why are the 'Fixed Stars' so FIXED?

On Fri, 26 Oct 2007 12:46:35 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" [email protected] wrote in message
.. .
: On Fri, 26 Oct 2007 04:43:48 GMT, "Androcles"



Ok... 4Aw/c.lambda = 4Ar/lambda.(v/c)

It's as wrong as Einstein's cuckoo malformations no matter how long
its been discussed, it has no MINUS v in it, or lambda1 or lambda2.
You didn't think.


See new 'sagnac united' thread...


Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm
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  #2392  
Old October 26th 07, 09:35 PM posted to sci.astro,sci.physics.relativity
Dr. Henri Wilson
external usenet poster
 
Posts: 707
Default Why are the 'Fixed Stars' so FIXED?

On Fri, 26 Oct 2007 15:58:36 +0200, "Paul B. Andersen"
wrote:

Dr. Henri Wilson wrote:
I appreciate your efforts but it doesn't work as you say. You have ignored the
movement of the start point in the source frame. You are also ignoring the fact
that the 'intrinsic frequency' appears Doppler shifted at the detector because
the latter is moving wrt the startpoint. Even though the travel time of the
rays is the same, the number of cycles arriving at the detector differs for
each ray.
In short, you are confusing the start point frame with the source/detector
frame.
The fact is, the photon experiences one INTRINSIC cycle every wavelength
traveled.

The path lengths of the two rays are different, therefore the photons generally
end up out of phase.

Thankyou for helping me develop my model. It is all coming together nicely now.


So let's take one step at the time. So far, we have only stated what
the equation for the phase of your BaTh photon must be in the source frame.

Your talk about Doppler shift and motion relative some point in
another frame is thus utterly irrelevant.

Let's first agree on the equation describing the phase of your
BaTh photon in the source frame.
We can then take it from there later.

So read again, carefully this time:

You said:
1. 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.
The front of a BaTh photon oscillates once every absolute wavelength traveled.

This is YOUR oral description of your 'approach'.
All I do below is to express this description mathematically.
If you find an error in my math, please point out exactly what
it is, and show what the correct math should be.
Otherwise I will assume it is correct.

From your description, it follows that he 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.

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.c/l)t (in the source frame)
#
# The phase doesn't depend on x, all the photons in
# the ray have at any time the same phase.
================================================= ========

It's your model, Henri. You have now seen it expressed mathematically.
If your oral description of your 'approach' at the top is wrong,
please correct it, and I will express your changed description mathematically.

I understand that you are unable to do it, so I will have to help you.


A comprehensive reply will soon appear in the new thread 'Sagnac threads
united'


Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm
  #2393  
Old October 28th 07, 08:50 AM posted to sci.astro,sci.physics.relativity
Jerry
external usenet poster
 
Posts: 502
Default Why are the 'Fixed Stars' so FIXED?

On Oct 26, 7:58 am, "Paul B. Andersen"
wrote:
Dr. Henri Wilson wrote:

I appreciate your efforts but it doesn't work as you say. You have ignored the
movement of the start point in the source frame. You are also ignoring the fact
that the 'intrinsic frequency' appears Doppler shifted at the detector because
the latter is moving wrt the startpoint. Even though the travel time of the
rays is the same, the number of cycles arriving at the detector differs for
each ray.
In short, you are confusing the start point frame with the source/detector
frame.
The fact is, the photon experiences one INTRINSIC cycle every wavelength
traveled.

The path lengths of the two rays are different, therefore the photons generally
end up out of phase.

Thankyou for helping me develop my model. It is all coming together nicely now.


So let's take one step at the time. So far, we have only stated what
the equation for the phase of your BaTh photon must be in the source frame.

Your talk about Doppler shift and motion relative some point in
another frame is thus utterly irrelevant.

Let's first agree on the equation describing the phase of your
BaTh photon in the source frame.
We can then take it from there later.

So read again, carefully this time:

You said:
1. 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.
The front of a BaTh photon oscillates once every absolute wavelength traveled.

This is YOUR oral description of your 'approach'.
All I do below is to express this description mathematically.
If you find an error in my math, please point out exactly what
it is, and show what the correct math should be.
Otherwise I will assume it is correct.

From your description, it follows that he 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.

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.c/l)t (in the source frame)
#
# The phase doesn't depend on x, all the photons in
# the ray have at any time the same phase.
================================================== =======

It's your model, Henri. You have now seen it expressed mathematically.
If your oral description of your 'approach' at the top is wrong,
please correct it, and I will express your changed description mathematically.

I understand that you are unable to do it, so I will have to help you.


I posted a brief response in the "Sagnac Threads United" thread.

Jerry

  #2394  
Old October 28th 07, 09:13 AM posted to sci.astro,sci.physics.relativity
Jerry
external usenet poster
 
Posts: 502
Default Why are the 'Fixed Stars' so FIXED?

On Oct 28, 2:50 am, Jerry wrote:

I posted a brief response in the "Sagnac Threads United" thread.


The thread is in sci.physics.relativity where it belongs,
i.e. no cross-posting.

http://groups.google.com/group/sci.p...27208a478f7929

Jerry


  #2395  
Old October 28th 07, 12:53 PM posted to sci.astro,sci.physics.relativity
Jerry
external usenet poster
 
Posts: 502
Default Why are the 'Fixed Stars' so FIXED?

On Oct 26, 6:57 am, Jerry wrote:
On Oct 26, 6:34 am, [email protected](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
Henri Wilson's Lies

http://mysite.verizon.net/cephalobus...ri/diploma.htm
http://mysite.verizon.net/cephalobus.../deception.htm
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http://mysite.verizon.net/cephalobus...ri/history.htm
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Message copied to "Sagnac Threads United" in sci.physics.relativity

Jerry

  #2396  
Old October 29th 07, 04:27 PM posted to sci.astro,sci.physics.relativity
George Dishman[_1_]
external usenet poster
 
Posts: 2,509
Default Why are the 'Fixed Stars' so FIXED?

On 26 Oct, 10:41, [email protected](Clueless Henri Wilson) wrote:
On Fri, 26 Oct 2007 00:48:51 -0700, George Dishman wrote:
On 25 Oct, 23:20, [email protected](Clueless Henri Wilson) wrote:
On Thu, 25 Oct 2007 17:30:45 +0100, "George Dishman" wrote:
"Clueless Henri Wilson" [email protected] wrote in messagenews:[email protected] .com...


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?


Paul, I and many others have repeatedly told you
that you don't understand what a frame is but
instead of listening to the explanations you just
shout abuse. This is just another case where your
ignorance is catching you out.

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..


It is moving at that speed relative to the LAB,
not a POINT in the frame. Learn the difference
between coordinates and the physical objects
to which they relate.

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.


Actually I did comment in another reply after a quick
look.

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.


No, it is the first order effct. You should be
able to work this out Henry, you are sort of
double-counting.

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.


Very. The distance travelled is the distance
moved by a surface of given phase, so the
'intrinsic oscillation' appears when you measure
at a fixed point and is of constant phase at
a point moving at the speed of the photon.

However, even if you try to do what you want, it
isn't "every wavelength traveled", it should be
the distance travelled by a wave in the duration
of a source cycle, that is the error I have been
pointing out consistently throughout. When you
decide to listen, you will be able to correct
your maths error.

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.


Sorry Henry, you are talking nonsense. As I have
pointed out several times, a standing wave is the
combination of two travelling waves and the speed
of each is that of a point of fixed phase.

I have defined wavelength in this context.


Rubbish, the wavelength is defined as the distance
between repetitions of the periodic waveform at a
given instant.

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.


Yes but the distance moved in that time in any other
frame is not the same as the wavelength. Your web
page mistake is that you use the wavelength for the
distance moved but in the inertial frame, that is
wrong.

'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.


What you say above confirms what I said.

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.


See above, your own definition confirms it is.

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.


It is fundamental, maxwell's Equation do
not use a medium.

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


George, you are way behind..


No, you just don't have the maths ability.

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.


Nope, no medium, the speeds are relative to the observer.

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.


Obviously, and the time, distances and speeds must
all be experessed in the same coordinate scheme.

t=2piR/(c+v)

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


Oh good grief, how basic does this need to be
for you to understand?

"c" is the speed of the light, it tells you
the distance the light moves in a given time.

"v" is the speed of the splitter, it tells you
the distance the splitter moves in a given time.

"c+v" represents the sum of two speeds, it is
the sum of the distance move by the light and
the distance moved by the splitter and is also
the increase in their separation in a given time.
It is NOT the speed of either the light or the
splitter.

George

  #2397  
Old October 29th 07, 09:23 PM posted to sci.astro,sci.physics.relativity
Dr. Henri Wilson
external usenet poster
 
Posts: 707
Default Why are the 'Fixed Stars' so FIXED?

On Mon, 29 Oct 2007 09:27:38 -0700, George Dishman
wrote:

On 26 Oct, 10:41, [email protected](Clueless Henri Wilson) wrote:
On Fri, 26 Oct 2007 00:48:51 -0700, George Dishman wrote:


see my reply in "sagnac threads united"


Henri Wilson. ASTC,BSc,DSc(T)

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




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