Why are the 'Fixed Stars' so FIXED?

On Fri, 26 Oct 2007 11:25:12 +0200, "Paul B. Andersen"
wrote:

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.

Yes, you consider a photon as a spinning pair of charges, whether or not they
are doesn't matter...the rotation defines the phase, as you say..
remember it's one rotation per 'wavelength' moved.

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.

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.

Einstein is as good as dead.... for the second time.


Henri Wilson. ASTC,BSc,DSc(T)

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

On Fri, 26 Oct 2007 02:57:23 -0700, Jerry
wrote:

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.

Phase varies with distance traveled only. 2pi rads per wavelength....

Paul is confused.

Jerry



Henri Wilson. ASTC,BSc,DSc(T)

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

On Fri, 26 Oct 2007 04:57:18 -0700, Jerry
wrote:

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


This is going nowhere.....



Henri Wilson. ASTC,BSc,DSc(T)

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

"Dr. Henri Wilson" HW@.... wrote in message
...
: 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.

"Negligible" -- Wilson.


:
: What's the wavelength of this sine wave?
: http://spaceflight.nasa.gov/realdata/tracking/
:
: I don't know ahnd don't care.

A real physicist would have no trouble answering that.
A pouting petulant crackpot who wants his own theory
desperately has an attack of the sour grapes over it.

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

Let me give you a hint. It's a bit less than Earth's diameter
because the Earth is turning in the Newtonian absolute and
stationary frame of reference.



: The wavelength of a photon is the distance it travels in the source frame
: during one cycle of its intrinsic oscillation.

Oh... you mean c = wavelength/wavetime... yeah, I can live with that.
Brilliant, Wilson, you've just defined speed = distance/time.
Congratulations on getting down the ****in' basics.

The wavelength of a photon is the distance it travels in the observer's
frame
during one cycle of its intrinsic oscillation also, and is therefore
velocity
dependent. It was only a trace anyway and of no physical significance.

: It move at c wrt its source so
: lambda=c/nu.

NO!!!! Ken Seto would say that!! (knee-jerk, knee-jerk)

It moves at c+v wrt the observer so
lambda1 = (c+v)/nu

lambda2 = (v-c)/nu


:
: Since all lenThs are absolute,

********, you sheep shagging moron.

"Dr. Henri Wilson" HW@.... wrote in message
...
: 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...

Yes it is, Wilson. You are STILL wide open to ridicule.


:
: 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?

http://en.wikipedia.org/wiki/Siphon
"the up-slope flow being driven only by hydrostatic pressure"

Water is level on the top, then it falls, then it slides down the slope on
the bottom. You don't think.
The question isn't whether you do or not, but whether you can or not.


:
: : Fringe Displacement is as negligible as v.
: : What's the Fringe Displacement of this sine wave, Wilson?
: : http://spaceflight.nasa.gov/realdata/tracking/
: : I say its about 45 minutes, what do you hallucinate AND
: : how do you get 4Aw/c.lambda with your theory?
: :
: : That the equation for one turn of a ring gyro. I didn't think
:
: Yes, we all know that, no need to keep repeating it.
: Wilson doesn't think.
:
: not funny any more. ..just a tactic used by pommie ****pots to avoid
having to
: answer difficult questions.

I've always answered your questions, you don't answer mine.

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

[Wilson the sheep shagging ****head]: "I don't know ahnd don't care."

It's still funny when you admit you don't think.


:
: you would know.
: : It's only been discussed here for about 5 years.
:
: It's as wrong as the cuckoo malformations no matter how long
: its been discussed, it has no v in it. You didn't think.
:
: w is angular velocity.....= v/r
: Any decent engineer should know that......but apparently not you or
Dishman.


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.

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 (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.

--
Paul

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

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.

--
Paul

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

Jerry wrote:
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?

Obviously.
Thanks.


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


--
Paul

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

"Dr. Henri Wilson" HW@.... wrote in message
...
: 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...
:
****ing ugly plagiarising *******.


"Anyway, this now fits in perfectly with my 'intrinsic oscillation
frequency' idea.
Thankyou Jerry for helping me develop my theory...." -- Wilson,
October 26, 2007 1:03 PM









:
: : 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.

****ing ugly plagiarising *******.
"Anyway, this now fits in perfectly with my 'intrinsic oscillation
frequency' idea.
Thankyou Jerry for helping me develop my theory...." -- Wilson,
October 26, 2007 1:03 PM



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

Why, so you can claim it was your idea?

: : 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) , FUPB.

On Fri, 26 Oct 2007 12:17:32 GMT, "Androcles"
wrote:


"Dr. Henri Wilson" HW@.... wrote in message
.. .
: 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.

"Negligible" -- Wilson.


:
: What's the wavelength of this sine wave?
: http://spaceflight.nasa.gov/realdata/tracking/
:
: I don't know ahnd don't care.

A real physicist would have no trouble answering that.
A pouting petulant crackpot who wants his own theory
desperately has an attack of the sour grapes over it.

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

Let me give you a hint. It's a bit less than Earth's diameter
because the Earth is turning in the Newtonian absolute and
stationary frame of reference.



: The wavelength of a photon is the distance it travels in the source frame
: during one cycle of its intrinsic oscillation.

Oh... you mean c = wavelength/wavetime... yeah, I can live with that.
Brilliant, Wilson, you've just defined speed = distance/time.
Congratulations on getting down the ****in' basics.

The wavelength of a photon is the distance it travels in the observer's
frame
during one cycle of its intrinsic oscillation also, and is therefore
velocity
dependent. It was only a trace anyway and of no physical significance.

: It move at c wrt its source so
: lambda=c/nu.

NO!!!! Ken Seto would say that!! (knee-jerk, knee-jerk)

It moves at c+v wrt the observer so
lambda1 = (c+v)/nu

lambda2 = (v-c)/nu


:
: Since all lenThs are absolute,

********, you sheep shagging moron.

See 'sagnac united' thread


Henri Wilson. ASTC,BSc,DSc(T)

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