Wang / Sagnac Devices
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From: Androcles on 27 Oct 2009 09:34 <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. >>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 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.
From: doug on 27 Oct 2009 10:41 Jonah Thomas wrote: > "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? 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. > > >>>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: Tom Roberts on 27 Oct 2009 10:02 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. [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: tominlaguna on 27 Oct 2009 11:15 On Wed, 28 Oct 2009 00:31:21 +1100, "Inertial" <relatively(a)rest.com> wrote: ><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? Because he won't sit still.
From: Androcles on 27 Oct 2009 11:26
<tominlaguna(a)yahoo.com> wrote in message news:3j3ee5ln86sapsruvl5f73f14o806gvsdo(a)4ax.com... > On Wed, 28 Oct 2009 00:31:21 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >><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? > > Because he won't sit still. Ants in my pants, I'm not inertial. |