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From: tominlaguna on 19 Oct 2009 05:40
On Sun, 18 Oct 2009 23:30:19 -0500, Tom Roberts
<tjroberts137(a)sbcglobal.net> wrote:

>tominlaguna(a)yahoo.com wrote:
>> 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 That paper has been superseded by: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.
>
>Yes. This is all fully consistent with SR.
>
>
>> I disagree. They measured the acceleration of the fiber section from
>> zero to some constant velocity.
>
>Cannot be so. The signal is larger when a longer section of fiber is in
>the moving section or when the fiber makes more turns around the whole
>apparatus. That does not change the acceleration, but it does change the
>\integral v.dl of Wang et al.
>
> Their formula is only approximately correct in SR.
> They ignored the effect due to the differing propagation
> speeds in a moving fiber. For each of their apparatuses
> this is vastly smaller than the effect they describe.
>
>Moreover, imagine not turning on the source until the section of fiber
>on the conveyor is moving -- the phase shift will still be there
>(relative to zero velocity of the conveyor).
>
>
>> 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.
>
>Hmmm. Wang et al show otherwise. In their apparatus of Fig 3 neither the
>source nor detector accelerate, yet there is a signal.
>
> If one took a flexible fiber and coiled it up around a
> rotatable disk (< 1 full rotation), and placed both source
> and detector on the table nearby, one essentially has a
> Sagnac interferometer in which neither source not detector
> moves or accelerates. Their formula predicts a signal when
> the disk is rotated, even though neither the source nor
> detector are accelerated. This is consistent with the
> prediction of SR for this physical situation (again
> neglecting the different propagation speeds).
>
>This is all easily explained in SR: in all cases the light propagating
>in the two directions takes different paths IN THE INERTIAL FRAME, and
>the time difference and phase shift are proportional to the difference
>in path lengths in that frame.
>
>
>> The generalized Sagnac effect does not deal
>> with enclosed areas and angular velocity;
>
>Right. Wang et al give a formula that depends on \integral v.dl. And
>they show how it includes the usual Sagnac formula as a special case.
>Only for a rigidly rotating apparatus does "angular velocity" make
>sense, and only for that case does the enclosed area actually apply; for
>all other cases of the generalized apparatuses they consider it is
>\integral v.dl that matters, not "enclosed areas and angular velocity".
>
> For their original device, that is v times the perimeter,
> which is how I phrased it in other posts in this newsgroup.
>
>
>> Saburi in 1976 demonstrated that there was a radio signal
>> transit time difference east-west between two earth-stationary
>> receiver/transmitters.
>
>Certainly. This confirms the prediction of SR.
>
>
>> 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.
>
>This is not quite stated correctly. The GPS satellites are updated daily
>to account for their clocks' drift, variations in their orbits, and some
>other minor variations (e.g. those due to sun and moon). There is no
>possible way to "correct" for the Sagnac effect, it must be computed for
>each measurement, based on the actual satellites used (it depends on the
>geometrical relationship between GPS receiver and the satellite).
>
>
>> 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.
>
>My statement was not "erroneous". It applies to all ballistic models in
>which Snell's law holds IN THE REST FRAME OF THE MIRROR. Ballistic
>models that do not obey that are refuted by literally zillions of
>observations and experiments, and those that do obey it are refuted by
>Sagnac and all of Wang et al's observations.

Tom. Very thought provoking comments. I will have to give them more
consideration; I may have been too hasty in my conclusions about the
nature of Sagnac / Wang.

But for the moment I would like to focus on the issue of Snell's law
and how it plays into your thesis. When I think of Snell's law, I am
thinking of refraction not reflection, unless it is total internal
reflection. Nonetheless, I am not aware of any way to differentiate
between SRT and Ballistic theory when there is no relative motion
between the source and mirrors or refracting medium. I will look
forward to your comments.

OFF TOPIC: BTW, I thought your paper on the Dayton Miller data was
quite excellent.

>> 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.
>
>You have oversimplified, and in the process you omitted important
>caveats (e.g. you implicitly assume the light was traveling at c before
>hitting the mirror, and it hits the mirror perpendicularly; these are
>not necessarily so). But that model obeys Snell's law in the rest frame
>of the mirror, and is refuted.
>
>
>Tom Roberts
From: tominlaguna on 19 Oct 2009 05:42
On Sun, 18 Oct 2009 11:26:16 -0400, Jonah Thomas <jethomas5(a)gmail.com>
wrote:

>tominlaguna(a)yahoo.com wrote:
>> Jonah Thomas <jethomas5(a)gmail.com> wrote:
>> >tominlaguna(a)yahoo.com wrote:
>> >> Jonah Thomas <jethomas5(a)gmail.com> wrote:
>> >> >tominlaguna(a)yahoo.com wrote:
>
>> >> >> 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.
>
>> >> >Here is why I think there is something else going on too.
>> >> > ________________
>> >> >/ \
>> >> >\____x___________/
>> >> >
>> >> >Here is the simple form of the Wang experiment, with the
>> >> >emitter-detector x traveling around the loop. When x is on the
>> >linear> >section traveling at constant speed they get the same phase
>> >shift> >that they do when it is going around the rollers.
>> >> > ___________________________________
>> >> >/ \
>> >> >\____x______________________________/
>> >> >
>> >> >When they change the length and change nothing else, they get a
>> >> >change in phase shift proportional to the length. How has this
>> >> >changed the acceleration proportional to the phase shift?
>
>> >> In the diagrams you have presented, it appears you are describing
>> >an> experiment shown in the earlier paper cited by Sue. Please look
>> >at> the paper I referenced; specifically Fig. 3. For that test,
>> >neither> the source nor the receiver was moving; it was only a
>> >straight section> of fiber that was translated. So you have: 4 meter
>> >(or whatever the> exact size was) loop of fiber, stationary source,
>> >stationary receiver,> yet a Sagnac signal generated when you move a
>> >section of the loop.> That seems to defy logic since light traveling
>> >in opposing directions> still has to cover the 4 meter distance each
>> >way while traveling at c;> same speed in each direction, same
>> >distance to travel, but different> arrival times. There is obviously
>> >a change in the optical path length> since the physical path length
>> >remains unchanged. I contend that is> produced by acceleration;
>> >similar to the way the bullet path length is> changed in the dueling
>> >analogy.
>> >
>> >I can easily believe that you are talking about something that was
>> >produced by a change in acceleration.
>>
>> No, I was talking about a change in velocity.
>
>OK. But do you think the linear movement was accelerating when they
>measured it?
>
>> >But the effect that I pointed out from the first paper does not
>> >appear to me to have different acceleration, and yet they got a phase
>> >difference. So I think there is something other than acceleration
>> >going on to get the Wang effect. Or possibly there is a hidden
>> >acceleration that I haven't noticed. Maybe somehow if you use rollers
>> >with the same radius rotating at the same speed, it puts a bigger
>> >acceleration on the fiber if the fiber is a longer length?
>>
>> Perhaps we first need to discuss Sagnac Effect and Sagnac Instruments.
>
>No, I don't think so. If Wang's analysis of his data is right, the
>Sagnac effect is a special case of the Wang effect. If he's wrong, where
>did he go wrong? What's the better explanation for his results?
>
>> >> When I first thought about the reported results, I understood the
>> >> source of the fringe shift to be a change in the enclosed area.
>> >But> upon looking at the diagram and seeing the linearity of the
>> >plots, I> concluded that area change was not a factor, otherwise the
>> >data plot> would not be linear for the various tested speeds.
>> >
>> >Agreed.
>
>So, the enclosed area matters for traditional Sagnac *because* it
>correlates with the variables that do matter in that special case. There
>is no motion except angular rotation around some center. If the center
>is outside of the path the light takes then some of that path is moving
>against the velocity and the area is a measure of the actual angular
>speed. If the center is inside the path then of course the area is a
>measure of the angular speed. In either case Wang's formula with the dot
>product ought to work.
>
>> >> Translational speed by itself is not a factor just as it is not a
>> >> factor in the dueling analogy, where both shooters would die at the
>> >> same time when the train was moving at constant speed.
>> >
>> >Translational speed looks like a factor in the one I mentioned, once
>> >you accept that area is not a factor.
>>
>> The Wang setup described in each paper is a passive design. It only
>> can record "changes" in velocity. I know it looks like a velocity
>> meter, but it is only recording changes in velocity of the conduit. It
>> is much like the speedometer on your car: you look at the gauge and it
>> is reading 20 mph. I look at the gauge and I conclude it has recorded
>> your change in velocity from 0 to 20 mph. It remains pinned at 20 mph
>> until you accelerate or decelerate.
>
>?? If you're going at 20 mph and the speedometer says 20 mph, it looks
>to me like the speedometer is measuring velocity.
>
>If you had something that measured acceleration so that when you
>accelerate from 0 to 20 it goes up as the acceleration goes up, and down
>to zero as you reach 20 and stop accelerating, that would be something
>that measured acceleration. And if you did something like integrate the
>signal (like "store" an electric current in a capacitor) then you would
>have a velocity meter that is pinned at a particular velocity until it
>gets a new signal.
>
>But this appears to be showing an interference effect with no particular
>memory (although you don't know the actual velocity until you have a
>baseline to compare it to). If you have the thing moving at constant
>speed before you turn on the laser, won't you get the same interference
>pattern that you get if it's on during the acceleration?

Jonah. You raise some interesting questions and comments. Give me a
little tiime for thoughtful reply...
From: Androcles on 19 Oct 2009 06:02

<tominlaguna(a)yahoo.com> wrote in message
news:b1dod5pr4rd39bo06vrlme11btv3rlb0b0(a)4ax.com...
> On Sun, 18 Oct 2009 11:26:16 -0400, Jonah Thomas <jethomas5(a)gmail.com>
> wrote:
>
>>tominlaguna(a)yahoo.com wrote:
>>> Jonah Thomas <jethomas5(a)gmail.com> wrote:
>>> >tominlaguna(a)yahoo.com wrote:
>>> >> Jonah Thomas <jethomas5(a)gmail.com> wrote:
>>> >> >tominlaguna(a)yahoo.com wrote:
>>
>>> >> >> 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.
>>
>>> >> >Here is why I think there is something else going on too.
>>> >> > ________________
>>> >> >/ \
>>> >> >\____x___________/
>>> >> >
>>> >> >Here is the simple form of the Wang experiment, with the
>>> >> >emitter-detector x traveling around the loop. When x is on the
>>> >linear> >section traveling at constant speed they get the same phase
>>> >shift> >that they do when it is going around the rollers.
>>> >> > ___________________________________
>>> >> >/ \
>>> >> >\____x______________________________/
>>> >> >
>>> >> >When they change the length and change nothing else, they get a
>>> >> >change in phase shift proportional to the length. How has this
>>> >> >changed the acceleration proportional to the phase shift?
>>
>>> >> In the diagrams you have presented, it appears you are describing
>>> >an> experiment shown in the earlier paper cited by Sue. Please look
>>> >at> the paper I referenced; specifically Fig. 3. For that test,
>>> >neither> the source nor the receiver was moving; it was only a
>>> >straight section> of fiber that was translated. So you have: 4 meter
>>> >(or whatever the> exact size was) loop of fiber, stationary source,
>>> >stationary receiver,> yet a Sagnac signal generated when you move a
>>> >section of the loop.> That seems to defy logic since light traveling
>>> >in opposing directions> still has to cover the 4 meter distance each
>>> >way while traveling at c;> same speed in each direction, same
>>> >distance to travel, but different> arrival times. There is obviously
>>> >a change in the optical path length> since the physical path length
>>> >remains unchanged. I contend that is> produced by acceleration;
>>> >similar to the way the bullet path length is> changed in the dueling
>>> >analogy.
>>> >
>>> >I can easily believe that you are talking about something that was
>>> >produced by a change in acceleration.
>>>
>>> No, I was talking about a change in velocity.
>>
>>OK. But do you think the linear movement was accelerating when they
>>measured it?
>>
>>> >But the effect that I pointed out from the first paper does not
>>> >appear to me to have different acceleration, and yet they got a phase
>>> >difference. So I think there is something other than acceleration
>>> >going on to get the Wang effect. Or possibly there is a hidden
>>> >acceleration that I haven't noticed. Maybe somehow if you use rollers
>>> >with the same radius rotating at the same speed, it puts a bigger
>>> >acceleration on the fiber if the fiber is a longer length?
>>>
>>> Perhaps we first need to discuss Sagnac Effect and Sagnac Instruments.
>>
>>No, I don't think so. If Wang's analysis of his data is right, the
>>Sagnac effect is a special case of the Wang effect. If he's wrong, where
>>did he go wrong? What's the better explanation for his results?
>>
>>> >> When I first thought about the reported results, I understood the
>>> >> source of the fringe shift to be a change in the enclosed area.
>>> >But> upon looking at the diagram and seeing the linearity of the
>>> >plots, I> concluded that area change was not a factor, otherwise the
>>> >data plot> would not be linear for the various tested speeds.
>>> >
>>> >Agreed.
>>
>>So, the enclosed area matters for traditional Sagnac *because* it
>>correlates with the variables that do matter in that special case. There
>>is no motion except angular rotation around some center. If the center
>>is outside of the path the light takes then some of that path is moving
>>against the velocity and the area is a measure of the actual angular
>>speed. If the center is inside the path then of course the area is a
>>measure of the angular speed. In either case Wang's formula with the dot
>>product ought to work.
>>
>>> >> Translational speed by itself is not a factor just as it is not a
>>> >> factor in the dueling analogy, where both shooters would die at the
>>> >> same time when the train was moving at constant speed.
>>> >
>>> >Translational speed looks like a factor in the one I mentioned, once
>>> >you accept that area is not a factor.
>>>
>>> The Wang setup described in each paper is a passive design. It only
>>> can record "changes" in velocity. I know it looks like a velocity
>>> meter, but it is only recording changes in velocity of the conduit. It
>>> is much like the speedometer on your car: you look at the gauge and it
>>> is reading 20 mph. I look at the gauge and I conclude it has recorded
>>> your change in velocity from 0 to 20 mph. It remains pinned at 20 mph
>>> until you accelerate or decelerate.
>>
>>?? If you're going at 20 mph and the speedometer says 20 mph, it looks
>>to me like the speedometer is measuring velocity.
>>
>>If you had something that measured acceleration so that when you
>>accelerate from 0 to 20 it goes up as the acceleration goes up, and down
>>to zero as you reach 20 and stop accelerating, that would be something
>>that measured acceleration. And if you did something like integrate the
>>signal (like "store" an electric current in a capacitor) then you would
>>have a velocity meter that is pinned at a particular velocity until it
>>gets a new signal.
>>
>>But this appears to be showing an interference effect with no particular
>>memory (although you don't know the actual velocity until you have a
>>baseline to compare it to). If you have the thing moving at constant
>>speed before you turn on the laser, won't you get the same interference
>>pattern that you get if it's on during the acceleration?
>
> Jonah. You raise some interesting questions and comments. Give me a
> little tiime for thoughtful reply...

Better a slow correct thoughtful reply than a hasty wrong one.



From: tominlaguna on 19 Oct 2009 06:52
On Mon, 19 Oct 2009 19:02:28 +1100, "Inertial" <relatively(a)rest.com>
wrote:

>
><tominlaguna(a)yahoo.com> wrote in message
>news:e63od5t7bl229oahgcrjfud7ts7f1i113f(a)4ax.com...
>> On Sun, 18 Oct 2009 18:58:18 +0100, "Androcles"
>> <Headmaster(a)Hogwarts.physics_p> wrote:
>>
>>>
>>><tominlaguna(a)yahoo.com> wrote in message
>>>news:lt7md512qegmkrme8hh6h7icq47u302ht4(a)4ax.com...
>>>> On Sat, 17 Oct 2009 12:21:32 +0100, "Androcles"
>>>> <Headmaster(a)Hogwarts.physics_p> wrote:
>[snip]
>>>>>Both contain the same blunder, namely, there are two angles alpha
>>>>>and -alpha not one, as shown here,
>>>>> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/SagnacIdiocy.htm
>
>There are lots of angles, alpha and -alpha, made by the beams relative to
>the start/end point in the rotating frame .. which is what Androcles
>animation in that page shows. Those angles both end up as zero when the
>beams meet at the detector.
>
>Of course, this is not the alpha described in the article. Androcles is
>confused. The article describes alpha as the amount the start/end has
>rotated from its initial position, in the non-rotating frame. Androcles
>doesn't understand that, as is obvious from his irrelevant animation and
>nonsensical questions in red on that page.

I agree with most all that you have said regarding the analysis of the
Mathpages diagram. I don't understand the Androcles animation.

>In emission theory, there is only one alpha value (ioe one angle through
>whic h the start/end point has rotated in the non-rotating frame) when the
>beams arrive back at the start/end location. Because the rays meet
>simultaneously, there is no phase shift.

I do disagree with your comments about the emission theory. You
appear to be claiming that photons are travelling at c +/- omega*R. If
they were, you would be correct there would be no phase difference.
But, they only travel at c in each direction. There is no mechanism
for them to pick up or lose an omega*R component.

>In SR, the beams do not arrive simultaneously, so there are two alpha values
>.. one alpha value when the first arrives, and a slightly larger alpha value
>when the second other arrives.
>
>
>>>> Sorry, I don't see your reasoning.
>
>Noone does

Disagree. I think he is on-track most of the time.

>>>> The end point has moved for both
>>>> beams to the 1 o'clock position
>
>Androcles is rotating the clock.
>
>>>Both beams have started from and returned to the 12 o'clock position,
>
>When your clock rotates, yes.
>
>>>and the start position is now at the 11 o'clock position.
>
>More precisely .. where the start position WAS in the non-rotating frame is
>now at the 11 o'clock position in the rotating frame.
>
>>> The position
>>>of the start is history and not relevant to the simultaneous meeting of
>>>beams.
>
>Yeup .. and it is that simultaneous meeting which refutes emission theory.
>SR does not have a simultaneous meeting, because the light travels two
>different length paths at the same speed, and so they arrive at the detector
>at different times, giving a phase shift. So far Androcles has failed to
>grasp that.

His animation appears to have +/- omega*R in each ray. That is not
how I view the emission theory.

>>> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/ring.gif
>
>And that animation shows how the light in an emission theory arrives at the
>detectors at the same time and in phase, hence refuting emission theory.
>There's no relevant meaning for those two alpha values, nor for the
>animation to pause where it does .. other than Androcles own confusion about
>what is happening.
>
>>>The end point hasn't moved at all, it is still at the 12 o'clock position.
>
>The end point in the rotating frame is ALWAYS at the 12 o'clock position in
>the rotating frame .. and the start point in the rotating frame is ALSO
>ALWAYS at the 12 o'clock position in the rotating frame. The start and end
>points in the rotating frame remain at the same point in the rotating frame
>all the time.
>
>However, the start and end points in the non-rotating frame (two different
>fixed point in that frame) DO move in the rotating frame. The start point
>in the non-rotating frame moves from the 12 o'clock to 11 o'clock position
>in the rotating frame, and the end point in the non-rotating frame moves
>from the 2 o'clock to the 12 o'clock in the rotating frame.
>
>It all depends on where you mark your start and end points .. on the
>rotating frame, or on the non-rotating lab frame .. and in which frame you
>measure their positions over time.
>
>>>What part of that reasoning do you not understand? It is clear enough
>>>in the gif I drew!
>
>Its clear that Androcles is confused.
>
>> Sorry, I still don't get it... The start and end points remain
>> together when the device rotates.
>
>Yes. In the rotating device frame. In the lab frame they both rotate
>together

Agree.

>> Rays should be shown leaving the
>> start place and returning to the now displaced, start place.
>
>Yes .. at the same time according to emission theory .. so no phase shift.

Disagree. That only applies to the re-emission theory.
From: Inertial on 19 Oct 2009 08:03

<tominlaguna(a)yahoo.com> wrote in message
news:sqfod5hhbbd0o9bg90letn9p4gc8303ffe(a)4ax.com...
> On Mon, 19 Oct 2009 19:02:28 +1100, "Inertial" <relatively(a)rest.com>
> wrote:
>
>>
>><tominlaguna(a)yahoo.com> wrote in message
>>news:e63od5t7bl229oahgcrjfud7ts7f1i113f(a)4ax.com...
>>> On Sun, 18 Oct 2009 18:58:18 +0100, "Androcles"
>>> <Headmaster(a)Hogwarts.physics_p> wrote:
>>>
>>>>
>>>><tominlaguna(a)yahoo.com> wrote in message
>>>>news:lt7md512qegmkrme8hh6h7icq47u302ht4(a)4ax.com...
>>>>> On Sat, 17 Oct 2009 12:21:32 +0100, "Androcles"
>>>>> <Headmaster(a)Hogwarts.physics_p> wrote:
>>[snip]
>>>>>>Both contain the same blunder, namely, there are two angles alpha
>>>>>>and -alpha not one, as shown here,
>>>>>> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/SagnacIdiocy.htm
>>
>>There are lots of angles, alpha and -alpha, made by the beams relative to
>>the start/end point in the rotating frame .. which is what Androcles
>>animation in that page shows. Those angles both end up as zero when the
>>beams meet at the detector.
>>
>>Of course, this is not the alpha described in the article. Androcles is
>>confused. The article describes alpha as the amount the start/end has
>>rotated from its initial position, in the non-rotating frame. Androcles
>>doesn't understand that, as is obvious from his irrelevant animation and
>>nonsensical questions in red on that page.
>
> I agree with most all that you have said regarding the analysis of the
> Mathpages diagram. I don't understand the Androcles animation.

Not surprising .. he doesn't understand Sagnac :)

>>In emission theory, there is only one alpha value (ioe one angle through
>>whic h the start/end point has rotated in the non-rotating frame) when the
>>beams arrive back at the start/end location. Because the rays meet
>>simultaneously, there is no phase shift.
>
> I do disagree with your comments about the emission theory. You
> appear to be claiming that photons are travelling at c +/- omega*R.

They travel in emission theory at c wrt the source .. which makes it c+/-v
in the non-rotating frame (because the source is travelling at v at the time
of emission)

> If
> they were, you would be correct there would be no phase difference.
> But, they only travel at c in each direction.

In which frame of reference?

> There is no mechanism
> for them to pick up or lose an omega*R component.

They get the velocity of the source (in emission theory)

>>In SR, the beams do not arrive simultaneously, so there are two alpha
>>values
>>.. one alpha value when the first arrives, and a slightly larger alpha
>>value
>>when the second other arrives.
>>
>>
>>>>> Sorry, I don't see your reasoning.
>>
>>Noone does
>
> Disagree. I think he is on-track most of the time.

Nope .. he lies and tries to deceive.

>>>>> The end point has moved for both
>>>>> beams to the 1 o'clock position
>>
>>Androcles is rotating the clock.
>>
>>>>Both beams have started from and returned to the 12 o'clock position,
>>
>>When your clock rotates, yes.
>>
>>>>and the start position is now at the 11 o'clock position.
>>
>>More precisely .. where the start position WAS in the non-rotating frame
>>is
>>now at the 11 o'clock position in the rotating frame.
>>
>>>> The position
>>>>of the start is history and not relevant to the simultaneous meeting of
>>>>beams.
>>
>>Yeup .. and it is that simultaneous meeting which refutes emission theory.
>>SR does not have a simultaneous meeting, because the light travels two
>>different length paths at the same speed, and so they arrive at the
>>detector
>>at different times, giving a phase shift. So far Androcles has failed to
>>grasp that.
>
> His animation appears to have +/- omega*R in each ray. That is not
> how I view the emission theory.

Then your view is not correct Androcles animation shows that emission
theory says .. that the light leaves the source with a separation speed of
c. AS the source has a speed v in the non-rotating frame, that means the
light travel and c+v and c-v in that frame

>>>> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/ring.gif
>>
>>And that animation shows how the light in an emission theory arrives at
>>the
>>detectors at the same time and in phase, hence refuting emission theory.
>>There's no relevant meaning for those two alpha values, nor for the
>>animation to pause where it does .. other than Androcles own confusion
>>about
>>what is happening.
>>
>>>>The end point hasn't moved at all, it is still at the 12 o'clock
>>>>position.
>>
>>The end point in the rotating frame is ALWAYS at the 12 o'clock position
>>in
>>the rotating frame .. and the start point in the rotating frame is ALSO
>>ALWAYS at the 12 o'clock position in the rotating frame. The start and
>>end
>>points in the rotating frame remain at the same point in the rotating
>>frame
>>all the time.
>>
>>However, the start and end points in the non-rotating frame (two different
>>fixed point in that frame) DO move in the rotating frame. The start point
>>in the non-rotating frame moves from the 12 o'clock to 11 o'clock position
>>in the rotating frame, and the end point in the non-rotating frame moves
>>from the 2 o'clock to the 12 o'clock in the rotating frame.
>>
>>It all depends on where you mark your start and end points .. on the
>>rotating frame, or on the non-rotating lab frame .. and in which frame you
>>measure their positions over time.
>>
>>>>What part of that reasoning do you not understand? It is clear enough
>>>>in the gif I drew!
>>
>>Its clear that Androcles is confused.
>>
>>> Sorry, I still don't get it... The start and end points remain
>>> together when the device rotates.
>>
>>Yes. In the rotating device frame. In the lab frame they both rotate
>>together
>
> Agree.
>
>>> Rays should be shown leaving the
>>> start place and returning to the now displaced, start place.
>>
>>Yes .. at the same time according to emission theory .. so no phase shift.
>
> Disagree. That only applies to the re-emission theory.

How does your pet emission theory differ?


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