Wang / Sagnac Devices
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From: Inertial on 27 Oct 2009 01:53 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20091027010207.305ff818.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > Darwin123 <drosen0000(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).
From: Jonah Thomas on 27 Oct 2009 03:15 "Inertial" <relatively(a)rest.com> wrote: > "Jonah Thomas" <jethomas5(a)gmail.com> wrote > > "Inertial" <relatively(a)rest.com> wrote: > >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote > >> > Darwin123 <drosen0000(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?
From: Inertial on 27 Oct 2009 08:06 "Jonah Thomas" <jethomas5(a)gmail.com> wrote in message news:20091027031535.6fab69f6.jethomas5(a)gmail.com... > "Inertial" <relatively(a)rest.com> wrote: >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> > "Inertial" <relatively(a)rest.com> wrote: >> >> "Jonah Thomas" <jethomas5(a)gmail.com> wrote >> >> > Darwin123 <drosen0000(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.
From: Inertial on 27 Oct 2009 08:37 "Henry Wilson DSc." <HW@..> wrote in message news:06lde5t4bdbmic7g3l36g75tkgqnt20gi0(a)4ax.com... > On Mon, 26 Oct 2009 22:38:41 -0500, Tom Roberts > <tjroberts137(a)sbcglobal.net> > wrote: > >>Jonah Thomas wrote: >>> I'll say it again: One beam of light leaves the emitter at speed c >>> relative to the emitter. It goes through a beam-splitter which does not >>> change its speed. It bounces off mirrors which do not change its speed. >>> So the light traveling in opposite directions travel at the same speed. >> >> [I assume you are still trying to discuss some sort of >> ballistic or emission theory.] >> >>The only frame in which this could possibly be true is the rotating >>frame. But in that frame you don't know how things like beam splitters >>and mirrors behave. You are ASSUMING that they behave in the rotating >>frame just like they do in an inertial frame, and that's just plain >>wrong -- they CANNOT possibly do that and still behave correctly in an >>inertial frame. > > ....this is amusing...it's the exact principle you use in your rotating > frame > 'disproof' of BaTh. You don't need a rotating frame analysis to refute ballistic theory >>As I have said before, you are being TOO IMPRECISE in your descriptions. >>You are not specifying in which frame you description holds, and thus >>end up with nonsense. You are not specifying your model, either, making >>it impossible for others to interpret your words or figure out what you >>are tying to say. Do not expect people to figure it out from context, >>because that is muddled; mention both frame and model explicitly, each >>and every time. >> >>You fool yourself by attempting to analyze in the rotating frame. It is >>straightforward to analyze the Sagnac interferometer in the inertial >>frame of its center. Consider any emission or ballistic theory that >>satisfies: >> A) light is emitted with speed c relative to the instantaneously- >> comoving inertial frame of its source >> B) Snell's law holds for a mirror in its instantaneously-comoving >> inertial frame, and the reflected light has the same speed as >> the incoming light in this frame >> C) a beam splitter emits light from both outputs with the same >> speed the incoming light had in the instantaneously-comoving >> inertial frame of the splitter >> D) Galilean relativity holds for light >> >>Do the analysis for a short light pulse, and you'll find that any such >>theory predicts the two halves of the split light pulse arrive at the >>detector simultaneously via the two counter-propagating paths. That >>implies no fringe shift, so any such theory is refuted. > > Well that would be true if light was a classical oscillator or traveling > wave. > But it isn't. Hence your confusion in your so-called analysis
From: Inertial on 27 Oct 2009 09:31
<tominlaguna(a)yahoo.com> wrote in message news:shpde598m1o1ksut08l49llkf9je92dvp0(a)4ax.com... > On Sun, 25 Oct 2009 16:02:15 -0000, "Androcles" > <Headmaster(a)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? |