Wang / Sagnac Devices
in [Relativity]
|
Prev: Relativity ring problem - what shape is this?
Next: BUY CHEAP TEXTBOOKS | College Textbooks | Used Textbooks |
From: Androcles on 25 Oct 2009 03:21 "Sam" <shayiam(a)yahoo.com> wrote in message news:00b22a79-bb5f-489e-bcc9-0d49b4982a85(a)e18g2000vbe.googlegroups.com... On Oct 24, 7:55 pm, Sam <shay...(a)yahoo.com> wrote: > ...the particles continue to move at c relative to > the "original source" at the time they were emitted. > However, this hypothesis leads again to a prediction > of zero Sagnac effect... Correction: ================================= You are still OBVIOUSLY an idiot.
From: Androcles on 25 Oct 2009 03:29 "Henry Wilson DSc ." <HW@..> wrote in message news:q2p7e5plp2m6tmsqfohv6979534vk0lshj(a)4ax.com... > On Sun, 25 Oct 2009 00:19:21 +0100, "Androcles" > <Headmaster(a)Hogwarts.physics_p> > wrote: > >> >>"Henry Wilson DSc ." <HW@..> wrote in message >>news:o307e5hce7ic1g0mopcv46so92a7ipmjnr(a)4ax.com... >>> On Sat, 24 Oct 2009 12:43:01 +0100, tominlaguna(a)yahoo.com wrote: >> >>>>And, surprise surprise the 2v value appears. It is the same value >>>>that appears in the Doppler radar formula. So could it be that every >>>>traffic cop with a radar gun is validating the Ballistic theory with >>>>each speeding ticket issued? >>>> >>>>B. Equation 74 looks vaguely familiar... Oh yes, it's the equation >>>>for Snell's Law! Tom Roberts claimed that the Ballistic theory >>>>violated Snell's Law and therefore was not consistent with the Sagnac >>>>Effect. That appears not to be the case, does it? >>>> >>>>I think the bigger issue for Tom Roberts is to find a SRT-consistent >>>>formulation for the angle of reflection from a moving mirror. I cited >>>>two different equations for the angle of reflection by mainstream >>>>physics authors. I know if I do some digging, I will come up with >>>>other variations. This issue could be interesting to follow. >>> >>> This analysis is OK >> >> >>No it isn't. The moving transponder can only emit at c and the photon >>can only hit the cop at c+v, not c+2v as La Goona Tommy is claiming. >>You are not paying attention, you've got one eye on the Shiraz and >>gout on the brain. > > Why don't you read what he said. I did read what he said, I've been following the thread. He said "Almost correct. For example, in the situation where a mirror is moving normally toward a source at velocity v, the mirror will experience the light as arriving at c + v. Upon reflection, the light will be traveling at c + 2v with respect to the source; and, as you state, at c + v with respect to the mirror." I said he was wrong, he can't add v twice. He then gave some bullshit arguments to back up his crank claim. WHY DON'T YOU READ WHAT HE SAID, YOU STOOOPID OLD SHEEP SHAGGER?
From: Jonah Thomas on 25 Oct 2009 03:35 Darwin123 <drosen0000(a)yahoo.com> wrote: > Jonah Thomas <jethom...(a)gmail.com> wrote: > > Darwin123 <drosen0...(a)yahoo.com> wrote: > > > Jonah Thomas <jethom...(a)gmail.com> wrote: > > > > Darwin123 <drosen0...(a)yahoo.com> wrote: > > > So without arguing about it, I don't understand why you can't > > consider the detector from an inertial point of view because it's > > rotating. > You can't because of Doppler effect. Very nice! Most Sagnac explanations ignore doppler effect, but it might be central. Thank you! > From the standpoint of the > inertial detector, the light that reflects off a mirror undergoes > Doppler effect. Consider the emitter is traveling at a constant speed, > emitting at a certain frequency as seen by the detector. Note: I am > not talking about the rest frame frequency of the emitter. The light > reflecting off of the mirror will be at a different frequency than the > original emitter. > You may not be convinced that light reflecting off a moving mirror > has to undergo Doppler shift. If you don't believe that mirrors > produce Doppler shift, the Sagnac effect could appear like magic. To > convince you, I may need to refer to microscopic details of the mirror > surface. I'm certainly not ready to dismiss it. I don't understand well enough to see just how it would go, but this is the most exciting Sagnac idea I've seen in weeks. > > When > > the question is whether two parts of a split light beam arrive at > > the detector at the same time or not, if it's the same time in one > > frame it ought to be the same time in all other frames, shouldn't > > it? > However, the two beams may not have the same frequency as seen in > a true inertial frame. Good! Most of the debaters here have insisted that it must have precisely the same frequency. > Even if the beams arrive at the same "time" at > the same "place", they can be a different frequencies. The word "beam" > convinced you that there is no frequency in the wave. The nodes in one > beam are closer together than the nodes in the other beam. So the > difference in frequencies is a beat frequency. Is that what's observed? > >If it was a > > question when they arrived at two different places then observers > > might disagree which was first. > We are not measuring difference in arrival time of "photons." We > are measuring the beats between two beams of different frequency. At > least they are different frequencies as seen in an inertial frame. I don't yet see that this has to be true, but I can sort of imagine it being true. Thank you. > > But how can they disagree about whether or not > > they arrive at the same place and the same time? > What is arriving at the same place and time? I think you mean two > wave fronts arriving at the same place and time. It is quite possible > for a wave front to arrive at a given place without the other wave > front arriving. OK. The usual analysis looked at the time of arrival of two wavefronts at the detector, and concluded that they arrive at different times. If that is not an appropriate way to analyse it, then surely some better way can be found. > From the POV of the inertial observer the two beams can't have the > same frequency. One beam is having its frequency raised by Doppler > shift, and the other is having its frequency lowered by Doppler shift. Is that true? I've lost track of which theory you're talking about. If the light actually travels at c for inertial observers, then it will arrive at the detector at two different times according to them because it must travel different distances. Why would an inertial observer see a doppler effect? Here's a thought experiment. Take a giant fiber-optic coil and spin it as fast as you can. Put a laser on one end in the direction that gets a blue-shift doppler. Send the light around the coil as any times as you can. Could you get UV light that way? X-rays? Well no, not in practice, because the fiber optic cable is tuned for some particular frequencies and when you get higher frequencies they leak and absorb. But as a thought.... > > > My argument was that the cavity itself is accelerating in the > > > sense of special relativity. The atoms and electrons of the > > > reflective cavity are accelerating. So even if the source and > > > detector are not accelerating (the usual case), the Sagnac effect > > > will still occur (the always case). > > > > I don't think so. (Though I could be wrong, and if the experiments > > have been done I'll be very interested.) I think the Sagnac effect > > is due to the source and detector moving and not the medium. If it's > > the medium that moves then you might get some sort of Fizeau effect. > Are you calling the reflective material a medium? If so, I > agree. If the light wave is traveling as an effenescent wave on a > "reflective" surface, then you could technically call it a Fizeau > effect. > > > > Ives's thought experiment could be done. You make a cylindrical > > mirror with carefully-spaced holes. You arrange that light shining > > into the cylinder through one hole will bounce repeatedly across the > > inner surface of the mirror until it leaves by another hole. > > Position your stationary source and detector just so, and then spin > > the mirror. The spinning mirror will alternately block the emitter > > (and detector) and let the light through. > Suppose you do that. > 1) Take into account Doppler effect. > 2) Take into account the diffraction caused by the curvature of the > metal. > >You should see no Sagnac effect when the light > > gets through. > Basically, the same experiment that you described has been done > with fiber optics. The Sagnac effect was observed. Was it the same experiment? Did they have a way to start with a stationary light source and a stationary detector, and somehow insert light into the cable at random locations, and extract light for the detector from random locations? If the light only comes in and out at the ends of the cable and particularly if the source and detector move with the cable, then it doesn't look to me like the Ives thought experiment. > > If you do get an effect I think it will deserve some other > > name, > The soggy effect? > > and there will be a reason that it is not observed during the > > usual Sagnac experiment. > You could call it anything you want. That won't make it go away. > Herbert E. Ives was wrong. He predicted the result of a thought experiment? Yes, he was quite plausibly wrong. > > The reason I predict that will have no effect is that the > > traditional Sagnac explanation (the one that says the detector moves > > so that the light paths are in proportion c+v:c-v) gives the right > > answer. >It does not depend in any way on the movement of the > > mirrors along the path, but only on the movement of the detector. If > > you get the same effect by moving the mirrors and leaving the > > detector alone, then this explanation gives the correct result > > entirely by accident. > I don't follow. The traditional explanation http://www.mathpages.com/rr/s2-07/2-07.htm is very simple. Sometimes there is some play with mirrors, but the effect is explained entirely without them, starting simply with the idea that the light arrives at the detector at the ratio c+v:c-v. Since the explanation does not depend on the mirrors moving, it predicts the Sagnac effect even when the mirrors don't move at all but only the source and detector move. Your explanation does depend on mirrors creating a doppler effect. So your explanation should give a different result when the mirrors move relative to the source and detector, versus when they all move together. If you are right then this other explanation is wrong. My natural hunch is to suppose that the explanation that most people believe is likely to be true, while the untested new contender is probably not. But you could easily be right and most of the others wrong.
From: Androcles on 25 Oct 2009 03:31 "Henry Wilson DSc ." <HW@..> wrote in message news:7qp7e5p8cjaej5r5gleg6fpbmg1hfmmh3i(a)4ax.com... > On Sun, 25 Oct 2009 00:00:49 +0100, "Androcles" > <Headmaster(a)Hogwarts.physics_p> > wrote: > >> >>"Henry Wilson DSc ." <HW@..> wrote in message >>news:0ov6e55o8bk6ppnqt55v76k8iluip5t774(a)4ax.com... >>> On Sat, 24 Oct 2009 10:58:53 +0100, "Androcles" >>> <Headmaster(a)Hogwarts.physics_p> >>> wrote: > >>>>>>> >>>>>>> Einstein...World's greatest SciFi writer.. >>>>>>and Wilson wants to be better than Einstein, but headless crocodiles >>>>>>flying kites in thunderstorms doesn't cut the mustard. >>>>> >>>>> Why don't you try it yourself...then tell us all about what happens. >>>> >>>>You brought it up, it's all yours. I'm content connecting batteries >>>>across transformers and telling you what happens. >>> >>> I'd rather disconnect a battery from a spark coil. >> >>Yeah, I'd expect a used VW camper van dealer to graduate to junkyard >>dog breaking up old cars for their parts, that is just the right >>occupation >>for you. Glad it makes you happy. > > How many empty barrels today? Why don't you read what he said?
From: Sam on 25 Oct 2009 16:57
On Oct 24, 10:57 pm, Jonah Thomas <jethom...(a)gmail.com> wrote: > If a ballistic theory predicts that the speed of light > depends on the speed of its source, then in every early > Sagnac experiment the light went the same speed in two > different directions, according to an inertial observer. Right, relative to the instantaneous rest frame of the emission point. So, relative to the Galilean rest frame of the hub, the light is going faster in the direction of rotation than in the opposite direction. > Jonah Thomas <jethom...(a)gmail.com> wrote: > The only way to get a change in lightspeed in that experiment > comes if somehow mirrors bounce light at different speeds, > or a change in direction changes the speed, etc. There's no change in speed (to the first order). Each pulse has the speed c relative to the rest frame of it's emitter at the instant of emission, and this is true for each of the successive emissions from the mirrors along the path. As a result, relative to the hub frame, the forward pulse is always moving faster than the rearward pulse, because the mirrors are moving in the forward direction, and the pulses are moving at c relative to the mirrors. In addition, relative to the hub frame, the forward pulse has a greater distance to travel, exactly proportional to its greater speed, so the two pulses arrive at the detector simultaneously. >Jonah Thomas <jethom...(a)gmail.com> wrote: > Simple straighforward ballistic theories would say that > the light travels at the same speed relative to an inertial > observer... But not just any inertial observer. As I mentioned before, when people talk about simple ballistic theory they generally mean a theory based on Galilean relativity, and one that says light travels at the same speed, c, relative to the inertial rest frame coordinates of the emitter at the time of emission. This implies that light does not have the speed c in terms of other inertial coordinate systems in Galilean spacetime. In particular, for a Sagnac device with the mirrors moving at v relative to the hub, light pulses move at c+v and c-v relative to the hub, and so arrive simultaneously at the detector. |