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From: tominlaguna on 27 Oct 2009 11:37
On Tue, 27 Oct 2009 13:34:52 -0000, "Androcles"
<Headmaster(a)Hogwarts.physics_p> wrote:

>
><tominlaguna(a)yahoo.com> wrote in message
>news:shpde598m1o1ksut08l49llkf9je92dvp0(a)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.

That is a real concession.

>
>>>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

Yippee!!!! The source is also where the measuring device is located.

>The *measured* frequency at the mirror is f' = f (c+v)/c

Yes.

>The *measured* frequency at the missile is f' = f *c/c = f.

We don't care; but yes.

>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.
>
From: Jonah Thomas on 27 Oct 2009 12:00
Tom Roberts <tjroberts137(a)sbcglobal.net> wrote:
> 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.

That makes perfect sense! Thank you!

> [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
From: Darwin123 on 27 Oct 2009 15:04
On Oct 27, 1:53 am, "Inertial" <relativ...(a)rest.com> wrote:
> "Jonah Thomas" <jethom...(a)gmail.com> wrote in message
>
> news:20091027010207.305ff818.jethomas5(a)gmail.com...
>
>
>
> > "Inertial" <relativ...(a)rest.com> wrote:
> >> "Jonah Thomas" <jethom...(a)gmail.com> wrote
> >> > Darwin123 <drosen0...(a)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).

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.
From: Androcles on 27 Oct 2009 15:16

"Darwin123" <drosen0000(a)yahoo.com> wrote in message
news:7c048a7e-bd06-425c-bed4-4c9d7a04fc56(a)b3g2000pre.googlegroups.com...
On Oct 27, 1:53 am, "Inertial" <relativ...(a)rest.com> wrote:
> "Jonah Thomas" <jethom...(a)gmail.com> wrote in message
>
> news:20091027010207.305ff818.jethomas5(a)gmail.com...
>
>
>
> > "Inertial" <relativ...(a)rest.com> wrote:
> >> "Jonah Thomas" <jethom...(a)gmail.com> wrote
> >> > Darwin123 <drosen0...(a)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).

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.



From: Darwin123 on 27 Oct 2009 15:26
On Oct 27, 8:06 am, "Inertial" <relativ...(a)rest.com> wrote:
> "Jonah Thomas" <jethom...(a)gmail.com> wrote in message
>
> news:20091027031535.6fab69f6.jethomas5(a)gmail.com...
>
>
>
> > "Inertial" <relativ...(a)rest.com> wrote:
> >> "Jonah Thomas" <jethom...(a)gmail.com> wrote
> >> > "Inertial" <relativ...(a)rest.com> wrote:
> >> >> "Jonah Thomas" <jethom...(a)gmail.com> wrote
> >> >> > Darwin123 <drosen0...(a)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.

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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