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| Inertial... |
 Posted: Tue Oct 27, 2009 7:31 am |
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<tominlaguna at (no spam) yahoo.com> wrote in message
news:shpde598m1o1ksut08l49llkf9je92dvp0 at (no spam) 4ax.com...
[quote]On Sun, 25 Oct 2009 16:02:15 -0000, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
We hold the mirror "stationary" and move the source at v.
Relative to the mirror, the approach velocity is c+v.
Do we agree? Yes, we do.
Park yourself at the stationary mirror... and don't move.
Source is moving toward you at +v.
Source emits light at +c relative to itself.
Light arrives at stationary mirror at +c+v = + (c+v).
Light is reflected off stationary mirror at -c-v = - (c+v).
Light flies back toward source which is still moving at +v.
When light meets source they have a relative velocity of:
Vrel = - (c+v) - (+v) = -(c+2v).
If you stayed put, you would see while using vectors that I arrived at
the correct 2v relationship.
[/quote]
Yeup
[quote]The approach velocity of the missile is still c relative to
the moving source.
The missile hits a stationary mirror which returns a bounce at -c
[/quote]
Why does he get -c for the reflection when the light was incident at c+v?
Is it that he is using re-emission rather than an elastic collision? |
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| Androcles... |
| Posted: Tue Oct 27, 2009 7:34 am |
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<tominlaguna at (no spam) yahoo.com> wrote in message
news:shpde598m1o1ksut08l49llkf9je92dvp0 at (no spam) 4ax.com...
[quote]Park yourself at the stationary mirror... and don't move.
Source is moving toward you at +v.
Yep.
Source emits light at +c relative to itself.
Yep.
Light arrives at stationary mirror at +c+v = + (c+v).
[/quote]
Yep.
[quote]Light is reflected off stationary mirror at -c-v = - (c+v).
[/quote]
Yep.
[quote]Light flies back toward source which is still moving at +v.
[/quote]
Yep.
[quote]When light meets source they have a relative velocity of:
Vrel = - (c+v) - (+v) = -(c+2v).
[/quote]
Yep.
[quote]If you stayed put, you would see while using vectors that I arrived at
the correct 2v relationship.
[/quote]
Yep.
[quote]Let's do the frequency bit.
Two identical arrows with identical RPM and hence frequency,
one travels at c and the other at c+v, like this:
http://www.androcles01.pwp.blueyonder.co.uk/Wave/Arrows.gif
At the finish line, the faster c+v arrow has not completed the same
number of revolutions as the slower c arrow.
To do so, it has to go beyond the finish line and start its return
from further away. Since speed = distance/time, this reduces
its return speed from -(c+2v) to -(c+v). Of course it actually
reflects without going past the finishing post, but begins its return
journey with some partial revolution to make up causing the source
to detect a lesser count of revolutions as it passes.
So the return frequency is
f' = f * -(c+v)/-c and NOT f' = f (c+2v)/c
Makes no sense...
[/quote]
Park yourself at the stationary mirror... and don't move.
The *measured* frequency at the source is f' = f (c+2v)/c
The *measured* frequency at the mirror is f' = f (c+v)/c
The *measured* frequency at the missile is f' = f *c/c = f.
But let's look at some REAL data.
http://www.britastro.org/vss/
Select Light Curves
Select Visual Light Curves
Select V1493 Aql (row 4 column 3)
There is a star that suddenly jumps 6 magnitudes, begins to dim,
brightens again to 4 magnitudes above its normal state and then
gradually settles back to normal, all inside 3 months.
And this is why:
http://www.androcles01.pwp.blueyonder.co.uk/Doolin'sStar.GIF
for a constant emitter in an orbit.
Wilson says it can't happen because it doesn't agree with his theory,
so it must explode twice. |
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| Tom Roberts... |
| Posted: Tue Oct 27, 2009 8:02 am |
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Guest
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Jonah Thomas wrote:
[quote]I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
[/quote]
A light source emits periodic waves, but as I have discussed before, one
must consider a short light pulse. Consider just one wavecrest from the
source, and perform the analysis of time delay to the detector. Then
consider the next wavecrest from the source and re-do the analysis --
you will obtain EXACTLY the same time delay for each path. So the
successive waves emitted dt apart by the source will arrive dt apart at
the detector, and thus there is no frequency change.
[This of course assumes a monochromatic source. That does
not happen in practice. But typical light sources have a
coherence length longer than the difference in path
lengths, and that is sufficiently monochromatic for this
to apply.]
This is a simple symmetry of the system, called time translation [*]. A
very general principle of physics is to analyze the symmetries of a
system, as they often tell you important and useful facts about the
system with much simpler analysis (e.g. here I did not need to know
anything about the light paths except that they don't change from one
wavecrest to the next).
[*] Here time translation and rotational symmetry are
combined, as after a time translation of dt there is
also a rotation of d\phi=\omega*dt (\omega is the angular
rate of rotation). But the apparatus does not change
as it rotates.
Tom Roberts |
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| doug... |
| Posted: Tue Oct 27, 2009 8:41 am |
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Jonah Thomas wrote:
[quote]"Inertial" <relatively at (no spam) rest.com> wrote:
"Jonah Thomas" <jethomas5 at (no spam) gmail.com> wrote
"Inertial" <relatively at (no spam) rest.com> wrote:
"Jonah Thomas" <jethomas5 at (no spam) gmail.com> wrote
Darwin123 <drosen0000 at (no spam) yahoo.com> wrote:
I notice that
there are two types of detectors described. In one configuration
the> experimenter measures spatial fringe shift, and in the other
the> >> experimenter measures temporal beats. The supposed observer
in the> >> inertial frame is supposed to be measuring differences in
wavelength> for the fringe shift, and differences in frequency for
the temporal> beats.
So any theory about Sagnac should predict both effects. Theories
with constant lightspeed will inevitably predict both effects.
Only one of those is Sagnac. There are no frequency beats in
Sagnac> because there is no difference in frequency. Though similar
looking> in a diagram there are other issues afoot in the ring gyro.
Oh? So Sagnac, ring gyros, and Wang are all different?
Yes. Similar but difference
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
Then you've not been paying attention to the previous posts. If the
beams arrive at the detector with the same speed after travelling for
the same time (as per ballistic theory), then there is no Doppler
effect and no difference in frequency.
Oh! I mean in reality, not in ballistic theory. Is there a difference in
frequency? Has it been tested?
[/quote]
A difference in time give a stable fringe pattern. A difference in
frequency gives a changing fringe pattern since the two beams are
no longer coherent. So, yes, it would be seen.
[quote]
There is a theory predicting the result that uses
only speed differences, and if that theory is correct then frequency
differences would contaminate the data.
No contamination
Do you know whether anyone has tested for frequency differences in
Sagnac?
If there is no 'beat' in the interference (phase difference) and the
phase simply shifts but doesn't change over time, then there is no
frequency difference. Any frequency difference would mean the phase
difference changes over time. There is no phase change over time in
Saganc (for constant angular rotation).
It looks like you're giving a theoretical reason why it shouldn't
happen. Or are you predicting that if it did happen we would see results
that we do not see?[/quote] |
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| Darwin123... |
| Posted: Tue Oct 27, 2009 9:04 am |
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On Oct 27, 1:53 am, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote in message
news:20091027010207.305ff818.jethomas5 at (no spam) gmail.com...
"Inertial" <relativ... at (no spam) rest.com> wrote:
"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote
Darwin123 <drosen0... at (no spam) yahoo.com> wrote:
I found a link. It appears that there are two ways of measuring
"the" Sagnac effect: one with fringe position and one with time.
Here> is a link.
http://www.cleonis.nl/physics/phys256/sagnac.php
Thank you! I had seen that page but hadn't given it the attention it
deserved.
All it is is a copy/paste of the wikipedia page on the Sagnac effect
That's likely where I saw it.
The following are my selections from this link. I notice that
there are two types of detectors described. In one configuration
the> experimenter measures spatial fringe shift, and in the other the
experimenter measures temporal beats. The supposed observer in the
inertial frame is supposed to be measuring differences in
wavelength> for the fringe shift, and differences in frequency for
the temporal> beats.
So any theory about Sagnac should predict both effects. Theories
with constant lightspeed will inevitably predict both effects.
Only one of those is Sagnac. There are no frequency beats in Sagnac
because there is no difference in frequency. Though similar looking
in a diagram there are other issues afoot in the ring gyro.
Oh? So Sagnac, ring gyros, and Wang are all different?
Yes. Similar but difference
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
Then you've not been paying attention to the previous posts. If the beams
arrive at the detector with the same speed after travelling for the same
time (as per ballistic theory), then there is no Doppler effect and no
difference in frequency.
There is a theory predicting the result that uses
only speed differences, and if that theory is correct then frequency
differences would contaminate the data.
No contamination
Do you know whether anyone has tested for frequency differences in
Sagnac?
If there is no 'beat' in the interference (phase difference) and the phase
simply shifts but doesn't change over time, then there is no frequency
difference. Any frequency difference would mean the phase difference
changes over time. There is no phase change over time in Saganc (for
constant angular rotation).
[/quote]
There are two implementations of the Sagnac effect. In one, the
beats are measured. The experimenter measures a periodic intensity
variation with time.
By intensity, I mean the envelope of the optical field. Within
the envelope, the spacing between peaks does not change. In other
words there is the "carrier frequency" that is not effected by the
acceleration of the cavity. I think this carrier frequency is what
some posters are referring to when they say that there is only one
unchanging frequency in the Saganc field.
Start thinking like a radio or optical engineer. The fact that
the envelope is changing means that the time integrated spectrum of
the system is composed of two two sidebands. If one were to actually
use a high resolution grating, one would split the signal into two
sidebands. This is the effect of the "relativistic Doppler effect."
Using simple trigonometry, or Fourier transform, one can show
mathematically that a temporal modulation with time has to result in
the generation of sidebands. Claiming that the Sagnac Cavity has only
one frequency comes at a price. The price is that one has to consider
the temporal modulation in time. However, one can ignore the temporal
modulation in time by considering the signal as the superposition of
two sideband signals.
This is a manifestation of the classical uncertainty relation.
Uncertainty in time (the intensity modulation) times the uncertainty
in frequency (the distance between the two sidebands is on or about
one. There is no quantum mechanics here, so don't get too excited.
This is simply communication theory.
When we choose a TV or radio channel, we are really choosing a
carrier frequency. However, the signal at the carrier frequency has to
have a bandwidth. This bandwidth is based on the sidebands that
combined to make the signal.
When looking at almost any optical or radio cavity, one has an
initial choice of whether to analyze the signal as one standing wave
modulated in space and time or a superposition of many traveling
waves. The initial choice is arbitrary, but one has to be consistent
with ones choice for the rest of the analysis.
For myself, I choose to think about the Sagnac signal as a
superposition of two traveling waves of constant intensity moving in
opposite directions. Therefore, I think of each traveling wave as
undergoing a relativistic Doppler shift. In this view, there have too
be two different frequencies: the clockwise frequency and the
anticlockwise frequency. In turn, a relativistic Doppler shift is
comprised of a longitudinal component and a transverse component.
A relativistic Doppler shift has to be used rather than a classical
Doppler shift? The choice in that case has to do with the nature of
the reflective surfaces. In a reflective surface has movable electrons
which obey Lorentz invariant mechanics, not Galileo invariant
mechanics. Since there are no surfaces with Galilean invariant
electrons, one can not use the classical Doppler shift.
Someone else can do the analysis for a standing wave in the Saganc
cavity. Then one would have one carrier frequency controlling the
signal. However, the time modulation would have to turn out the same.
It is a matter of trigonometry, not fundamental physics. |
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| ... |
| Posted: Tue Oct 27, 2009 9:15 am |
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Guest
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On Wed, 28 Oct 2009 00:31:21 +1100, "Inertial" <relatively at (no spam) rest.com>
wrote:
[quote]tominlaguna at (no spam) yahoo.com> wrote in message
news:shpde598m1o1ksut08l49llkf9je92dvp0 at (no spam) 4ax.com...
On Sun, 25 Oct 2009 16:02:15 -0000, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
We hold the mirror "stationary" and move the source at v.
Relative to the mirror, the approach velocity is c+v.
Do we agree? Yes, we do.
Park yourself at the stationary mirror... and don't move.
Source is moving toward you at +v.
Source emits light at +c relative to itself.
Light arrives at stationary mirror at +c+v = + (c+v).
Light is reflected off stationary mirror at -c-v = - (c+v).
Light flies back toward source which is still moving at +v.
When light meets source they have a relative velocity of:
Vrel = - (c+v) - (+v) = -(c+2v).
If you stayed put, you would see while using vectors that I arrived at
the correct 2v relationship.
Yeup
The approach velocity of the missile is still c relative to
the moving source.
The missile hits a stationary mirror which returns a bounce at -c
Why does he get -c for the reflection when the light was incident at c+v?
Is it that he is using re-emission rather than an elastic collision?
[/quote]
Because he won't sit still. |
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| Androcles... |
| Posted: Tue Oct 27, 2009 9:26 am |
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<tominlaguna at (no spam) yahoo.com> wrote in message
news:3j3ee5ln86sapsruvl5f73f14o806gvsdo at (no spam) 4ax.com...
[quote]On Wed, 28 Oct 2009 00:31:21 +1100, "Inertial" <relatively at (no spam) rest.com
wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:shpde598m1o1ksut08l49llkf9je92dvp0 at (no spam) 4ax.com...
On Sun, 25 Oct 2009 16:02:15 -0000, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
We hold the mirror "stationary" and move the source at v.
Relative to the mirror, the approach velocity is c+v.
Do we agree? Yes, we do.
Park yourself at the stationary mirror... and don't move.
Source is moving toward you at +v.
Source emits light at +c relative to itself.
Light arrives at stationary mirror at +c+v = + (c+v).
Light is reflected off stationary mirror at -c-v = - (c+v).
Light flies back toward source which is still moving at +v.
When light meets source they have a relative velocity of:
Vrel = - (c+v) - (+v) = -(c+2v).
If you stayed put, you would see while using vectors that I arrived at
the correct 2v relationship.
Yeup
The approach velocity of the missile is still c relative to
the moving source.
The missile hits a stationary mirror which returns a bounce at -c
Why does he get -c for the reflection when the light was incident at c+v?
Is it that he is using re-emission rather than an elastic collision?
Because he won't sit still.
[/quote]
Ants in my pants, I'm not inertial. |
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| Darwin123... |
| Posted: Tue Oct 27, 2009 9:26 am |
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Guest
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On Oct 27, 8:06 am, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote in message
news:20091027031535.6fab69f6.jethomas5 at (no spam) gmail.com...
"Inertial" <relativ... at (no spam) rest.com> wrote:
"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote
"Inertial" <relativ... at (no spam) rest.com> wrote:
"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote
Darwin123 <drosen0... at (no spam) yahoo.com> wrote:
I notice that
there are two types of detectors described. In one configuration
the> experimenter measures spatial fringe shift, and in the other
the> >> experimenter measures temporal beats. The supposed observer
in the> >> inertial frame is supposed to be measuring differences in
wavelength> for the fringe shift, and differences in frequency for
the temporal> beats.
So any theory about Sagnac should predict both effects. Theories
with constant lightspeed will inevitably predict both effects.
Only one of those is Sagnac. There are no frequency beats in
Sagnac> because there is no difference in frequency. Though similar
looking> in a diagram there are other issues afoot in the ring gyro.
Oh? So Sagnac, ring gyros, and Wang are all different?
Yes. Similar but difference
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
Then you've not been paying attention to the previous posts. If the
beams arrive at the detector with the same speed after travelling for
the same time (as per ballistic theory), then there is no Doppler
effect and no difference in frequency.
Oh! I mean in reality, not in ballistic theory. Is there a difference in
frequency? Has it been tested?
There is a theory predicting the result that uses
only speed differences, and if that theory is correct then frequency
differences would contaminate the data.
No contamination
Do you know whether anyone has tested for frequency differences in
Sagnac?
If there is no 'beat' in the interference (phase difference) and the
phase simply shifts but doesn't change over time, then there is no
frequency difference. Any frequency difference would mean the phase
difference changes over time. There is no phase change over time in
Saganc (for constant angular rotation).
It looks like you're giving a theoretical reason why it shouldn't
happen. Or are you predicting that if it did happen we would see results
that we do not see?
If there was a difference in frequency there would be a changing phase
difference in sagnac. There isn't.
[/quote]
If one looks at the Sagac radiation as a standing wave, there
most certainly is a phase change. The intensity of the output signal
varies in time. The fringes move. This describes a standing wave with
a changing phase.
I think your mind is jumping between a standing wave
representation of the Sagnac cavity and a traveling wave
representation of the Sagnac cavity. A difference in frequency between
traveling waves results in a change in phase of the standing wave.
The standing wave has a "constant" frequency called the carrier
wave. The standing wave is the superposition of two traveling waves
each with a different sideband frequency.
There is no relativity and no quantum mechanics. This is a
description of experimental results. There is no unique Fourier series
that describes a wave. Any wave can be considered either as a sum of
standing waves or a sum of traveling waves.
I am almost sure that what you say is the one frequency of the
Sagnac cavity is the "carrier frequency", which describes the
frequency of the standing wave. However, is prefer to think about the
sideband frequencies, which are the properties of the counter
propagating traveling waves. Both descriptions are valid. |
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| Darwin123... |
| Posted: Tue Oct 27, 2009 9:32 am |
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On Oct 16, 11:39 am, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p>
wrote:
[quote]tominlag... at (no spam) yahoo.com> wrote in message
news:to2hd5pm6nvm8qdul7kecjf5do7imn1ng5 at (no spam) 4ax.com...
I missed the opportunity to comment on this subject when a thread was
started by Jonah Thomas last month. I hope to continue the discussion
from this new starting point.
Sue posted a link to a Wang & et al paper which describes their fiber
optical gyro (FOG) experiments. That paper has been superseded by:
http://arxiv.org/ftp/physics/papers/0609/0609235.pdf. This latest
paper provides a more detailed account of that work.
Figure 3 of the new Wang paper shows that when a linear section of the
FOG is moved in translation, there is a fringe shift that is
proportional to the length of that section and the speed of its
motion. Most people that I have discussed this with believe that Dr.
Wang has demonstrated that his design can detect translational motion.
I disagree. They measured the acceleration of the fiber section from
zero to some constant velocity.
The Wang paper has lead me to conclude that the "Sagnac effect" is a
phenomenon peculiar to situations when the source and/or receiver are
experiencing acceleration. There are "Sagnac devices" that can detect
that phenomenon, but they should not be confused with the phenomenon
itself. Examples of the devices are: the passive Sagnac
interferometer devices of Sagnac, Pogany, Michelson-Gale, and
Dufour-Prunier; the active Sagnac interferometer devices of
Macek-Davis, Stedmann, modern laser gyros; and finally the "one-way"
Sagnac system of devices known as GPS.
A simple analogy of the phenomenon can be understood by this example:
Assume you have a long freight car, 100 feet long. There is a dueler
located at each end with identical guns, ammo and skill. If the car
is stationary with respect to the rails or moving at a constant
velocity and both fire their guns at the same time, they both die at
the same time. But, if the train happens to accelerate forward while
the bullets are in flight, the guy at the rear of the car dies first.
The same thing would occur if the car was experiencing acceleration
throughout the gun fight. That, in my opinion, is the phenomenon of
Sagnac. Bullets are flying in two directions covering an equal
distance of 100 feet, but one arrives sooner than the other due to the
acceleration of the receiver.
Paul Anderson was describing a type of device while he thought he was
describing the effect. The generalized Sagnac effect does not deal
with enclosed areas and angular velocity; several detection devices
are based on those criteria, but the phenomenon is not exclusive to
them. Saburi in 1976 demonstrated that there was a radio signal
transit time difference east-west between two earth-stationary
receiver/transmitters. The GPS network is corrected each day to
adjust their clocks so that the one-way transmission of signals is
accurate due to the Sagnac effect. Paul also suggested the Wang
experiment was a modified Fizeau experiment. They used both hollow
fibers and solid cross-section fiber and got the same readings. Others
in the past, Pogany and Harress, investigated the use of glass prisms
in the Sagnac set-up to determine if it was a Fizeau effect, and they
concluded it was not. Post has written about this.
Tom Roberts erroneously states that the ballistic model cannot explain
Sagnac. I will acknowledge that the "re-emission" ballistic model is
denied by the Sagnac results. Tolman (1912) and Panofsky and Phillips
(1961) describe three ballistic models. Waldron (1977) describes two
of the three: the ballistic model of Ritz/Waldron and the re-emission
model. The re-emission model fails in explaining Sagnac and a host of
other experiments.
In the Ritz/Waldron model, a mirror is not a new source, and therefore
light may or may not be reflected at c with respect to it. Its speed
after reflection is based on any relative motion between the source
and the mirror. If there is no relative motion, the reflected photon
will be moving at c; if there is relative motion, v, its speed will be
c +/- v. all with respect to the mirror.
Regards,
Tom Miles
Correct. Very good analysis. One tiny flaw...
Newton's corpuscles of light model, today called photons, predates
Walter Ritz by 250 years.
The behavior of photons does not resemble the behavior of Newtons[/quote]
corpusules. Newtons corpusules do not have a frequency associated with
them. The energy in a photons is proportional to the frequency of the
wave.
The difference becomes really important when dealing with
interference effects. Newton did not predict that light could show
interference effects. Neither beats nor diffraction is consistent with
Newton's corpusules. |
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| Posted: Tue Oct 27, 2009 9:37 am |
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Guest
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On Tue, 27 Oct 2009 13:34:52 -0000, "Androcles"
<Headmaster at (no spam) Hogwarts.physics_p> wrote:
[quote]
tominlaguna at (no spam) yahoo.com> wrote in message
news:shpde598m1o1ksut08l49llkf9je92dvp0 at (no spam) 4ax.com...
Park yourself at the stationary mirror... and don't move.
Source is moving toward you at +v.
Yep.
Source emits light at +c relative to itself.
Yep.
Light arrives at stationary mirror at +c+v = + (c+v).
Yep.
Light is reflected off stationary mirror at -c-v = - (c+v).
Yep.
Light flies back toward source which is still moving at +v.
Yep.
When light meets source they have a relative velocity of:
Vrel = - (c+v) - (+v) = -(c+2v).
Yep.
If you stayed put, you would see while using vectors that I arrived at
the correct 2v relationship.
Yep.
[/quote]
That is a real concession.
[quote]
Let's do the frequency bit.
Two identical arrows with identical RPM and hence frequency,
one travels at c and the other at c+v, like this:
http://www.androcles01.pwp.blueyonder.co.uk/Wave/Arrows.gif
At the finish line, the faster c+v arrow has not completed the same
number of revolutions as the slower c arrow.
To do so, it has to go beyond the finish line and start its return
from further away. Since speed = distance/time, this reduces
its return speed from -(c+2v) to -(c+v). Of course it actually
reflects without going past the finishing post, but begins its return
journey with some partial revolution to make up causing the source
to detect a lesser count of revolutions as it passes.
So the return frequency is
f' = f * -(c+v)/-c and NOT f' = f (c+2v)/c
Makes no sense...
Park yourself at the stationary mirror... and don't move.
The *measured* frequency at the source is f' = f (c+2v)/c
[/quote]
Yippee!!!! The source is also where the measuring device is located.
[quote]The *measured* frequency at the mirror is f' = f (c+v)/c
[/quote]
Yes.
[quote]The *measured* frequency at the missile is f' = f *c/c = f.
[/quote]
We don't care; but yes.
[quote]But let's look at some REAL data.
http://www.britastro.org/vss/
Select Light Curves
Select Visual Light Curves
Select V1493 Aql (row 4 column 3)
There is a star that suddenly jumps 6 magnitudes, begins to dim,
brightens again to 4 magnitudes above its normal state and then
gradually settles back to normal, all inside 3 months.
And this is why:
http://www.androcles01.pwp.blueyonder.co.uk/Doolin'sStar.GIF
for a constant emitter in an orbit.
Wilson says it can't happen because it doesn't agree with his theory,
so it must explode twice.
[/quote] |
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| Jonah Thomas... |
| Posted: Tue Oct 27, 2009 10:00 am |
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Tom Roberts <tjroberts137 at (no spam) sbcglobal.net> wrote:
[quote]Jonah Thomas wrote:
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
A light source emits periodic waves, but as I have discussed before,
one must consider a short light pulse. Consider just one wavecrest
from the source, and perform the analysis of time delay to the
detector. Then consider the next wavecrest from the source and re-do
the analysis -- you will obtain EXACTLY the same time delay for each
path. So the successive waves emitted dt apart by the source will
arrive dt apart at the detector, and thus there is no frequency
change.
[/quote]
That makes perfect sense! Thank you!
[quote] [This of course assumes a monochromatic source. That does
not happen in practice. But typical light sources have a
coherence length longer than the difference in path
lengths, and that is sufficiently monochromatic for this
to apply.]
This is a simple symmetry of the system, called time translation [*].
A very general principle of physics is to analyze the symmetries of a
system, as they often tell you important and useful facts about the
system with much simpler analysis (e.g. here I did not need to know
anything about the light paths except that they don't change from one
wavecrest to the next).
[*] Here time translation and rotational symmetry are
combined, as after a time translation of dt there is
also a rotation of d\phi=\omega*dt (\omega is the angular
rate of rotation). But the apparatus does not change
as it rotates.
Tom Roberts[/quote] |
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| Darwin123... |
| Posted: Tue Oct 27, 2009 12:17 pm |
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On Oct 27, 3:16 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
[quote]"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:7c048a7e-bd06-425c-bed4-4c9d7a04fc56 at (no spam) b3g2000pre.googlegroups.com...
On Oct 27, 1:53 am, "Inertial" <relativ... at (no spam) rest.com> wrote:
You are a clueless babbling lunatic.
I will now prove Einstein an idiot using your assumptions in[/quote]
an even clearer manner than you have presented.
1) Standing waves made of light exist in the Michaelson-Morley
interferometer, the Sagnac cavity, and in lasers.
-They form when two counter propagating waves are
superimposed, which is the common feature in all these devices.
2) Standing waves don't move.
-The nodes remain in the same place.
3) Since the standing wave doesn't move, the velocity of the light
waves in these devices is zero.
4) Einstein said that the velocity of light in a vacuum is always c.
5) Obviously c does not equal zero
-c is on or about 300000 km/s.
6) Einstein did not see the obvious contradiction between 3, 4 and 5.
7) Therefore, Einstein was an idiot.
Why didn't you use this simple argument instead of all
that blather about c+v versus c-v? |
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| Androcles... |
| Posted: Tue Oct 27, 2009 1:16 pm |
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"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:7c048a7e-bd06-425c-bed4-4c9d7a04fc56 at (no spam) b3g2000pre.googlegroups.com...
On Oct 27, 1:53 am, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote in message
news:20091027010207.305ff818.jethomas5 at (no spam) gmail.com...
"Inertial" <relativ... at (no spam) rest.com> wrote:
"Jonah Thomas" <jethom... at (no spam) gmail.com> wrote
Darwin123 <drosen0... at (no spam) yahoo.com> wrote:
I found a link. It appears that there are two ways of measuring
"the" Sagnac effect: one with fringe position and one with time.
Here> is a link.
http://www.cleonis.nl/physics/phys256/sagnac.php
Thank you! I had seen that page but hadn't given it the attention it
deserved.
All it is is a copy/paste of the wikipedia page on the Sagnac effect
That's likely where I saw it.
The following are my selections from this link. I notice that
there are two types of detectors described. In one configuration
the> experimenter measures spatial fringe shift, and in the other the
experimenter measures temporal beats. The supposed observer in the
inertial frame is supposed to be measuring differences in
wavelength> for the fringe shift, and differences in frequency for
the temporal> beats.
So any theory about Sagnac should predict both effects. Theories
with constant lightspeed will inevitably predict both effects.
Only one of those is Sagnac. There are no frequency beats in Sagnac
because there is no difference in frequency. Though similar looking
in a diagram there are other issues afoot in the ring gyro.
Oh? So Sagnac, ring gyros, and Wang are all different?
Yes. Similar but difference
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
Then you've not been paying attention to the previous posts. If the beams
arrive at the detector with the same speed after travelling for the same
time (as per ballistic theory), then there is no Doppler effect and no
difference in frequency.
There is a theory predicting the result that uses
only speed differences, and if that theory is correct then frequency
differences would contaminate the data.
No contamination
Do you know whether anyone has tested for frequency differences in
Sagnac?
If there is no 'beat' in the interference (phase difference) and the phase
simply shifts but doesn't change over time, then there is no frequency
difference. Any frequency difference would mean the phase difference
changes over time. There is no phase change over time in Saganc (for
constant angular rotation).
[/quote]
There are two implementations of the Sagnac effect. In one, the
beats are measured. The experimenter measures a periodic intensity
variation with time.
By intensity, I mean the envelope of the optical field. Within
the envelope, the spacing between peaks does not change. In other
words there is the "carrier frequency" that is not effected by the
acceleration of the cavity. I think this carrier frequency is what
some posters are referring to when they say that there is only one
unchanging frequency in the Saganc field.
Start thinking like a radio or optical engineer. The fact that
the envelope is changing means that the time integrated spectrum of
the system is composed of two two sidebands. If one were to actually
use a high resolution grating, one would split the signal into two
sidebands. This is the effect of the "relativistic Doppler effect."
Using simple trigonometry, or Fourier transform, one can show
mathematically that a temporal modulation with time has to result in
the generation of sidebands. Claiming that the Sagnac Cavity has only
one frequency comes at a price. The price is that one has to consider
the temporal modulation in time. However, one can ignore the temporal
modulation in time by considering the signal as the superposition of
two sideband signals.
This is a manifestation of the classical uncertainty relation.
Uncertainty in time (the intensity modulation) times the uncertainty
in frequency (the distance between the two sidebands is on or about
one. There is no quantum mechanics here, so don't get too excited.
This is simply communication theory.
When we choose a TV or radio channel, we are really choosing a
carrier frequency. However, the signal at the carrier frequency has to
have a bandwidth. This bandwidth is based on the sidebands that
combined to make the signal.
When looking at almost any optical or radio cavity, one has an
initial choice of whether to analyze the signal as one standing wave
modulated in space and time or a superposition of many traveling
waves. The initial choice is arbitrary, but one has to be consistent
with ones choice for the rest of the analysis.
For myself, I choose to think about the Sagnac signal as a
superposition of two traveling waves of constant intensity moving in
opposite directions. Therefore, I think of each traveling wave as
undergoing a relativistic Doppler shift.
=============================================
You are a clueless babbling lunatic. |
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| Darwin123... |
| Posted: Tue Oct 27, 2009 1:59 pm |
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On Oct 27, 6:32 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:8726a9ec-2c0d-4d34-a1d5-cedf92ed3415 at (no spam) y23g2000yqd.googlegroups.com...
7) Therefore, Einstein was an idiot.
Why didn't you use this simple argument instead of all
that blather about c+v versus c-v?
Cute.
Does a standing wave have a wavelength?
Yes.
Does a standing wave have a frequency?
Yes.[/quote]
I know full well that frequency times wavelength is the "speed" of
a wave, even though the nodes of the wave aren't going anywhere. I
wasn't presenting this as a real argument <duhh>.
I was pointing out that that the "speed of light" isn't defined as
the actual motion of energy. The speed of light characterizes a
property of the vacuum.
Androcles sees "c+v" and "c-v" and automatically interprets this
as a speed of light. After making this identification, he accuses
Einstein of contradicting himself. Einstein said the speed of light is
c, and then he says it is c+v or c-v. What an idiot!
However, light energy isn't traveling at either "c+v" or "c-v".
The energy in an interferometer is really stored in a standing wave,
which doesn't move at all. So Androcles didn't really think carefully
as to what "speed of light" means. His arguments imply that the speed
of light is the speed of some form of energy.
The speed of a standing wave is frequency times wavelength. If
speed is calculated this way, then the speed of light is c in all the
interferometers. The frequency and wavelength changes due to Doppler
shift. But the speed of light is "c".
The "c+v" and the "c-v" are caused by changes in frequency, not
changes in "speed of light." The wavelengths changed in such a way as
to leave the speed of light as "c". |
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| Androcles... |
| Posted: Tue Oct 27, 2009 2:07 pm |
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"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:df4793bf-c6ef-443d-be9a-65462f61189b at (no spam) z4g2000prh.googlegroups.com...
On Oct 16, 11:39 am, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p>
wrote:
[quote]tominlag... at (no spam) yahoo.com> wrote in message
news:to2hd5pm6nvm8qdul7kecjf5do7imn1ng5 at (no spam) 4ax.com...
I missed the opportunity to comment on this subject when a thread was
started by Jonah Thomas last month. I hope to continue the discussion
from this new starting point.
Sue posted a link to a Wang & et al paper which describes their fiber
optical gyro (FOG) experiments. That paper has been superseded by:
http://arxiv.org/ftp/physics/papers/0609/0609235.pdf. This latest
paper provides a more detailed account of that work.
Figure 3 of the new Wang paper shows that when a linear section of the
FOG is moved in translation, there is a fringe shift that is
proportional to the length of that section and the speed of its
motion. Most people that I have discussed this with believe that Dr.
Wang has demonstrated that his design can detect translational motion.
I disagree. They measured the acceleration of the fiber section from
zero to some constant velocity.
The Wang paper has lead me to conclude that the "Sagnac effect" is a
phenomenon peculiar to situations when the source and/or receiver are
experiencing acceleration. There are "Sagnac devices" that can detect
that phenomenon, but they should not be confused with the phenomenon
itself. Examples of the devices are: the passive Sagnac
interferometer devices of Sagnac, Pogany, Michelson-Gale, and
Dufour-Prunier; the active Sagnac interferometer devices of
Macek-Davis, Stedmann, modern laser gyros; and finally the "one-way"
Sagnac system of devices known as GPS.
A simple analogy of the phenomenon can be understood by this example:
Assume you have a long freight car, 100 feet long. There is a dueler
located at each end with identical guns, ammo and skill. If the car
is stationary with respect to the rails or moving at a constant
velocity and both fire their guns at the same time, they both die at
the same time. But, if the train happens to accelerate forward while
the bullets are in flight, the guy at the rear of the car dies first.
The same thing would occur if the car was experiencing acceleration
throughout the gun fight. That, in my opinion, is the phenomenon of
Sagnac. Bullets are flying in two directions covering an equal
distance of 100 feet, but one arrives sooner than the other due to the
acceleration of the receiver.
Paul Anderson was describing a type of device while he thought he was
describing the effect. The generalized Sagnac effect does not deal
with enclosed areas and angular velocity; several detection devices
are based on those criteria, but the phenomenon is not exclusive to
them. Saburi in 1976 demonstrated that there was a radio signal
transit time difference east-west between two earth-stationary
receiver/transmitters. The GPS network is corrected each day to
adjust their clocks so that the one-way transmission of signals is
accurate due to the Sagnac effect. Paul also suggested the Wang
experiment was a modified Fizeau experiment. They used both hollow
fibers and solid cross-section fiber and got the same readings. Others
in the past, Pogany and Harress, investigated the use of glass prisms
in the Sagnac set-up to determine if it was a Fizeau effect, and they
concluded it was not. Post has written about this.
Tom Roberts erroneously states that the ballistic model cannot explain
Sagnac. I will acknowledge that the "re-emission" ballistic model is
denied by the Sagnac results. Tolman (1912) and Panofsky and Phillips
(1961) describe three ballistic models. Waldron (1977) describes two
of the three: the ballistic model of Ritz/Waldron and the re-emission
model. The re-emission model fails in explaining Sagnac and a host of
other experiments.
In the Ritz/Waldron model, a mirror is not a new source, and therefore
light may or may not be reflected at c with respect to it. Its speed
after reflection is based on any relative motion between the source
and the mirror. If there is no relative motion, the reflected photon
will be moving at c; if there is relative motion, v, its speed will be
c +/- v. all with respect to the mirror.
Regards,
Tom Miles
Correct. Very good analysis. One tiny flaw...
Newton's corpuscles of light model, today called photons, predates
Walter Ritz by 250 years.
The behavior of photons does not resemble the behavior of Newtons[/quote]
corpusules.
=================================================
Yes they do. They are particles of light and they zip along at a good
speed. Newton's corpuscles of light are today called photons, you
stupid argumentative jerk.
"It seems that Light is propagated in time, spending in its passage from
the sun to us about seven Minutes of time:" -- DEFIN. II of Opticks Or,
A Treatise of the Reflections, Refractions, Inflections and Colours of
Light - Sir Isaac Newton.
"the velocity of light in our theory plays the part, physically, of an
infinitely great velocity" --§ 4. Physical Meaning of the Equations
Obtained in Respect to Moving Rigid Bodies and Moving Clocks
-- ON THE ELECTRODYNAMICS OF MOVING BODIES By A. Einstein
Relativistic fuckin' bullshit, and you are a relativistic fuckin'
bullshitter. |
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