"Dr. Henri Wilson" HW@.... wrote in message
...
On Thu, 10 Jan 2008 21:40:36 +1100, "Jeckyl" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
. ..
On Thu, 10 Jan 2008 10:38:17 +1100, "Jeckyl" wrote:

DOPPLER SHIFTED IN THE NONROTATING FRAME OBVIOUSLY.
THEY ARE DOPLER SHIFTED IN THAT FRAME FOR THE DURATION OF THEIR
FLIGHT.

You do realise the doppler shifting doesn't change the wave in itself ..
don't you?

Listen you halfwit, the waves you see in my rayphases demo

... are two completely different frequency waves .. your demo does not
show
doppler shift at all. Get it right.

OF COURSE THE WAVES HAVE DIFFERENT FREQUENCIES.

Not by the amount shown .. you have two waves with INITIALLY (before
doppler) different frequencies in that example. So it is not valid

On Fri, 11 Jan 2008 11:32:58 +1100, "Jeckyl" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
.. .
On Wed, 9 Jan 2008 23:43:13 +1100, "Jeckyl" wrote:



There doesn't have to be an observer, idiot.

Indeed .. this imagined observer in the inertial frame has no bearing on
what happens .. yet you ar efocused on him INSTEAD of what happens at the
detector .. You've said so yourself.

The frequencies are DOPPLER SHIFTED IN THE NONROTATING FRAME.

Totally irrelevant because the detector (where the sagnac effect is found)
is MOVING in the non-rotating frame. so if you doppler shifted from the
source to the non-rotating frame initially, you need to doppler shift back
again .. and when you do so, you get no Sagnac effect. Gees.

Of course the ****ing waves are 'doppler shifted back' in the detector frame.
They were never ****ing doppler shifted at all in that frame... you ****ing
moron.

The point you cannot get into your moronic head is that they were traveling for
the same short period of time at different frequencies IN THE NONROTATING
FRAME. DURING THAT TIME, THEY WENT OUT OF PHASE.




Henri Wilson. ASTC,BSc,DSc(T)

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

Relativists believe that a rotating Sagnac viewed in the rotating frame is identical to a nonrotating one viewed from the nonrotating frame. Hahahaha! What about all the imaginary effects?

"Dr. Henri Wilson" HW@.... wrote in message
...
On Fri, 11 Jan 2008 11:32:58 +1100, "Jeckyl" wrote:

"Dr. Henri Wilson" HW@.... wrote in message
. ..
On Wed, 9 Jan 2008 23:43:13 +1100, "Jeckyl" wrote:



There doesn't have to be an observer, idiot.

Indeed .. this imagined observer in the inertial frame has no bearing on
what happens .. yet you ar efocused on him INSTEAD of what happens at the
detector .. You've said so yourself.

The frequencies are DOPPLER SHIFTED IN THE NONROTATING FRAME.

Totally irrelevant because the detector (where the sagnac effect is found)
is MOVING in the non-rotating frame. so if you doppler shifted from the
source to the non-rotating frame initially, you need to doppler shift back
again .. and when you do so, you get no Sagnac effect. Gees.

Of course the ****ing waves are 'doppler shifted back' in the detector
frame.
They were never ****ing doppler shifted at all in that frame... you
****ing
moron.

So . glad you finally see that your analysis is wrong'

It only shows that there would be a time-varying phase shift due to
differeing frequencies for a detector that was statioanry in the inertial
frame

The point you cannot get into your moronic head is that they were
traveling for
the same short period of time at different frequencies IN THE NONROTATING
FRAME. DURING THAT TIME, THEY WENT OUT OF PHASE.

I understand it perfectly .. which is why I know it is irrelevant.

Doppler shift is an appearance .. it doesn't actually change the waves
themselves .. just how they appear to a particular observer .. that a
stationary observer would see different frequencies is completely irrelevant
to sagnac.

The waves never went out of phase wrt the detector. That is what counts.

On Jan 10, 2:27 pm, HW@....(Dr. Henri Wilson) wrote:
On Thu, 10 Jan 2008 02:58:48 -0800 (PST), Jerry

Very simply, Your explanation of Sagnac predicts effects that are
never seen.

Consider two cesium atomic clocks, one clock being regarded as
the "source" S, the other clock being the "observer" O on opposite
sides of the Earth on the Equator. As the Sun rises in the East,
clock S travels along the circumference of the Earth's orbit at
30 km/s sending timing signals at a speed of 299970 km/s towards
clock O:

O ---------------------S (c-v)

In the stationary frame of the Solar System, the radio waves
emitted by clock A will have a frequency of 9191712507 Hz.

As the Sun sets in the West, clock S is
[corrected text] oriented in the opposite direction relative to O
and sends timing signals at a speed of 300030 km/s
towards clock O:

(c+v) S --------------------- O

In the stationary frame of the Solar System, the radio waves
emitted by Clock A will have a frequency of 9193551033 Hz.

Over the 12756 km diameter of the Earth, the Sagnac effect would
cause the difference in phase between signals received in the
morning by clock O from clock S versus the signals received in
the evening to be an easily detected 77071 cycles.

Your frame shifting Sagnac analysis leads to absurd prediction

All I can gather from this is that you are completely hopeless
at designing and describing experiments.

WHAT THE HELL ARE YOU TALKING ABOUT MAN? DO THE RADIO SIGNALS
PASS RIGHT THROUGH THE EARTH, OR WHAT?

How utterly pathetic of you.

Don't you realize that your theory predicts that deviations in
light path from a straight line WILL NOT AFFECT the total phase
shift attributable to orbital motion???

For example, compare bouncing a signal from S to O up and down a
link to a communications satellite versus signals traveling in a
straight line from S to O.

/\
/ \ ----- v=30km/s
/ \
S------O

The Earth in its orbit around the Sun is moving 30 km/s to the
right.

If the distance between S and O is 10000 km, then your theory
states that the radio waves traveling in a straight line from
S to O would have a speed of 300030 km/s and would be phase
shifted by 30642 cycles relative to the situation if the Earth
were not orbiting, except for the fact that the Earth is in the
way and sending signals in a straight line from S to O is not
possible.

Instead, we bounce the beam at a 60 degree angle off the
satellite. Since cos(60 degrees) = 0.5, the speed of light up and
down the communications link is 300015 km/s over a distance of
20000 km, and again we compute 30642 cycles of phase shift
relative to the "motionless Earth" scenario.

Over a 12 hour period, your theory predicts that the cesium clock
at O receives signals from S that periodically shift in phase
by +/- 30642 cycles.

Your explanation of Sagnac behavior predicts nonsense.

Jerry
Henri Wilson's Lies
http://mysite.verizon.net/cephalobus...ri/diploma.htm
http://mysite.verizon.net/cephalobus.../deception.htm
http://mysite.verizon.net/cephalobus...rt_aurigae.htm
http://mysite.verizon.net/cephalobus...ri/history.htm
http://mysite.verizon.net/cephalobus...enri/snips.htm
http://mysite.verizon.net/cephalobus...ri/accuses.htm
http://mysite.verizon.net/cephalobus...ri/oh_dear.htm

On Thu, 10 Jan 2008 22:48:28 -0800 (PST), Jerry
wrote:

On Jan 10, 2:27 pm, HW@....(Dr. Henri Wilson) wrote:
On Thu, 10 Jan 2008 02:58:48 -0800 (PST), Jerry

Very simply, Your explanation of Sagnac predicts effects that are
never seen.

Consider two cesium atomic clocks, one clock being regarded as
the "source" S, the other clock being the "observer" O on opposite
sides of the Earth on the Equator. As the Sun rises in the East,
clock S travels along the circumference of the Earth's orbit at
30 km/s sending timing signals at a speed of 299970 km/s towards
clock O:

O ---------------------S (c-v)

In the stationary frame of the Solar System, the radio waves
emitted by clock A will have a frequency of 9191712507 Hz.

As the Sun sets in the West, clock S is
[corrected text] oriented in the opposite direction relative to O
and sends timing signals at a speed of 300030 km/s
towards clock O:

(c+v) S --------------------- O

In the stationary frame of the Solar System, the radio waves
emitted by Clock A will have a frequency of 9193551033 Hz.

Over the 12756 km diameter of the Earth, the Sagnac effect would
cause the difference in phase between signals received in the
morning by clock O from clock S versus the signals received in
the evening to be an easily detected 77071 cycles.

Your frame shifting Sagnac analysis leads to absurd prediction

All I can gather from this is that you are completely hopeless
at designing and describing experiments.

WHAT THE HELL ARE YOU TALKING ABOUT MAN? DO THE RADIO SIGNALS
PASS RIGHT THROUGH THE EARTH, OR WHAT?

How utterly pathetic of you.

Don't you realize that your theory predicts that deviations in
light path from a straight line WILL NOT AFFECT the total phase
shift attributable to orbital motion???

For example, compare bouncing a signal from S to O up and down a
link to a communications satellite versus signals traveling in a
straight line from S to O.

/\
/ \ ----- v=30km/s
/ \
S------O

The Earth in its orbit around the Sun is moving 30 km/s to the
right.

If the distance between S and O is 10000 km, then your theory
states that the radio waves traveling in a straight line from
S to O would have a speed of 300030 km/s and would be phase
shifted by 30642 cycles relative to the situation if the Earth
were not orbiting, except for the fact that the Earth is in the
way and sending signals in a straight line from S to O is not
possible.

Instead, we bounce the beam at a 60 degree angle off the
satellite.

Hahahahahaha! You really are funny.
How do you propose to do that? Drill a thirty degree hole through the Earth?
Even if the clocks are placed on 1000m towers the satellite would have to be a
hell of a long way off.

I guess you wanted to use a geostationary orbit which is about 36000 kms above
Earth. i'll let YOU work out how high the towers have to be.

Since cos(60 degrees) = 0.5, the speed of light up and
down the communications link is 300015 km/s over a distance of
20000 km, and again we compute 30642 cycles of phase shift
relative to the "motionless Earth" scenario.

Over a 12 hour period, your theory predicts that the cesium clock
at O receives signals from S that periodically shift in phase
by +/- 30642 cycles.

This is not related to what happens in a ring gyro. The signal from S to O has
to be compared with one from S to O in the opposite direction....but you
wouldn't be able to detect any fringe displacement OR movement because you
cannot change the rotation speed of the Earth very easily.... and the Earth's
orbit speed around the sun is fairly constant.
You are confusing the earth's own nonrotating frame with that of its solar
orbit. The latter is close enough to being inertial and can be ignored.

Your explanation of Sagnac behavior predicts nonsense.

You would have to send the signal right around our solar orbit to see the
effect you seem to have in mind....but even then it would only minutely
indicate the orbit's departure from circularity.

Jerry



Henri Wilson. ASTC,BSc,DSc(T)

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

Relativists believe that a rotating Sagnac viewed in the rotating frame is identical to a nonrotating one viewed from the nonrotating frame. Hahahaha! What about all the imaginary effects?

On Jan 11, 3:15 am, HW@....(Dr. Henri Wilson) wrote:
On Thu, 10 Jan 2008 22:48:28 -0800 (PST), Jerry

wrote:
On Jan 10, 2:27 pm, HW@....(Dr. Henri Wilson) wrote:
On Thu, 10 Jan 2008 02:58:48 -0800 (PST), Jerry

Very simply, Your explanation of Sagnac predicts effects that are
never seen.

Consider two cesium atomic clocks, one clock being regarded as
the "source" S, the other clock being the "observer" O on opposite
sides of the Earth on the Equator. As the Sun rises in the East,
clock S travels along the circumference of the Earth's orbit at
30 km/s sending timing signals at a speed of 299970 km/s towards
clock O:

O ---------------------S (c-v)

In the stationary frame of the Solar System, the radio waves
emitted by clock A will have a frequency of 9191712507 Hz.

As the Sun sets in the West, clock S is
[corrected text] oriented in the opposite direction relative to O
and sends timing signals at a speed of 300030 km/s
towards clock O:

(c+v) S --------------------- O

In the stationary frame of the Solar System, the radio waves
emitted by Clock A will have a frequency of 9193551033 Hz.

Over the 12756 km diameter of the Earth, the Sagnac effect would
cause the difference in phase between signals received in the
morning by clock O from clock S versus the signals received in
the evening to be an easily detected 77071 cycles.

Your frame shifting Sagnac analysis leads to absurd prediction

All I can gather from this is that you are completely hopeless
at designing and describing experiments.

WHAT THE HELL ARE YOU TALKING ABOUT MAN? DO THE RADIO SIGNALS
PASS RIGHT THROUGH THE EARTH, OR WHAT?

How utterly pathetic of you.

Don't you realize that your theory predicts that deviations in
light path from a straight line WILL NOT AFFECT the total phase
shift attributable to orbital motion???

For example, compare bouncing a signal from S to O up and down a
link to a communications satellite versus signals traveling in a
straight line from S to O.

/\
/ \ ----- v=30km/s
/ \
S------O

The Earth in its orbit around the Sun is moving 30 km/s to the
right.

If the distance between S and O is 10000 km, then your theory
states that the radio waves traveling in a straight line from
S to O would have a speed of 300030 km/s and would be phase
shifted by 30642 cycles relative to the situation if the Earth
were not orbiting, except for the fact that the Earth is in the
way and sending signals in a straight line from S to O is not
possible.

Instead, we bounce the beam at a 60 degree angle off the
satellite.

Hahahahahaha! You really are funny.
How do you propose to do that? Drill a thirty degree hole
through the Earth? Even if the clocks are placed on 1000m
towers the satellite would have to be a hell of a long way off.

Now you are REALLY pathetic, you laughing idiot.
Where are your simple trig skills?

X
/\
/ \
/ \
S------O
\ /
\ /
\/
Y

Let X be the satellite, and Y be the center of the Earth.

angle OSX = angle SOX = 60 degrees
SO = SX = OX = 10000 km
YS = YO = 6378 km
angle YSO = angle YOS = arccos(5000/6378) = 38.37 degrees
angle YOX = angle YSX = 38.37+60 = 98.37 degrees

In other words, the satellite is visible 8.37 degrees above the
horizon as seen from either S or O.

I guess you wanted to use a geostationary orbit which is about
36000 kms above Earth.

Nope. I wanted to keep my numbers VERY SIMPLE. The purpose of my
exercise was to show that, given light behaving according to your
theory, deviations of the light path from a straight line will
not affect the measured phase.

Unfortunately, even SIMPLE defeats you.

i'll let YOU work out how high the towers have to be.

Ground level.

Since cos(60 degrees) = 0.5, the speed of light up and
down the communications link is 300015 km/s over a distance of
20000 km, and again we compute 30642 cycles of phase shift
relative to the "motionless Earth" scenario.

Over a 12 hour period, your theory predicts that the cesium clock
at O receives signals from S that periodically shift in phase
by +/- 30642 cycles.

This is not related to what happens in a ring gyro. The signal
from S to O has to be compared with one from S to O in the
opposite direction....but you wouldn't be able to detect any
fringe displacement OR movement because you cannot change the
rotation speed of the Earth very easily....

Irrelevant.

and the Earth's orbit speed around the sun is fairly constant.

Because of the rotation of the Earth, it takes a mere 12 hours
for THIS arrangement of source and detector:

O ---------------------S (c-v)

to turn into THIS arrangement of source and detector:

(c+v) S --------------------- O

You are confusing the earth's own nonrotating frame with that
of its solar orbit.

No, I am considering the SUN'S nonrotating frame, and Earth's
(approximately) circular orbit around the Sun.

I am considering the Earth in orbit as being part of a giant
Sagnac apparatus.

The latter is close enough to being inertial and can be ignored.

HAHAHAHAHAHAHA!!!!

Are you saying that if an advanced civilization were to set up
a giant Sagnac experiment in orbit around a star, that they
would not detect rotation?

The word "inertial" does not appear ANYWHERE in the analysis of
the Sagnac effect. What is important is the geometric arrangement
of the optical components.

Where in your analysis do you have an equation indicating the
diminishing sensitivity of the Sagnac effect with radius?

Your explanation of Sagnac behavior predicts nonsense.

You would have to send the signal right around our solar orbit
to see the effect you seem to have in mind....

Nope. The edge of the "turntable" is traveling at 30 km/s. It
is not necessary to send a signal completely around the orbit,
given that we have available accurate atomic clocks to monitor
the phase shifts predicted by your theory, and the speeds
involved are so great.

but even then it would only minutely indicate the orbit's
departure from circularity.

Dream on, Henri... (snicker)

Jerry
Henri Wilson's Lies
http://mysite.verizon.net/cephalobus...ri/diploma.htm
http://mysite.verizon.net/cephalobus.../deception.htm
http://mysite.verizon.net/cephalobus...rt_aurigae.htm
http://mysite.verizon.net/cephalobus...ri/history.htm
http://mysite.verizon.net/cephalobus...enri/snips.htm
http://mysite.verizon.net/cephalobus...ri/accuses.htm
http://mysite.verizon.net/cephalobus...ri/oh_dear.htm