A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: Henry Wilson DSc on 14 Feb 2010 18:07 On Mon, 15 Feb 2010 09:44:32 +1100, "Inertial" <relatively(a)rest.com> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >news:crrgn51ol0crcbpjdeff497l466fvvbsor(a)4ax.com... >[snip] >> I admitted my mistakes and corrected them in the next post. > >If so, that is a pleasant change for you .. Where is the next post? Just below the first one. posted on 14th here's a copy. *************** On Sat, 13 Feb 2010 09:35:45 +1100, "Inertial" <relatively(a)rest.com> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >news:0uhbn5tpit1ss0m8falvch7hg2tieuj1a0(a)4ax.com... >> On Thu, 11 Feb 2010 23:33:09 GMT, ..@..(Henry Wilson DSc) wrote: >> >>>Paul and Jerry are standing together on the equator of an Earthlike >>>planet, >>>next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. >>>Its >>>circumference is 40 million metres. >>> >>>//===================C======================// optical fibre around >>>equator >>>_____________v<-_____p___q_______________________surface (actually curved) > >What is q and v? > >>>(the surface is drawn flat for convenience. The fibre goes right around >>>the >>>planet) >>> >>>Paul's clock emits light pulses every 3.33 nanoseconds through both ends >>>of the >>>fibre. The pulses are labelled according to their emission times and move >>>at c >>>wrt the fibre and the clock. The linear distance between each pulse is >>>therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass >>>through >>>the fibre in either direction. >>> >>>Since the planet is not rotating, the pulses returning to C were emitted >>>and >>>detected by Jerry at point p. > >Doesn't matter if they are detected at p or at C I'll withdraw my previous reply because it was wrong. So was my initial message because I stuffed up what I wanted to say. p is the point where the pulses emitted at q are received. >>> observes that the pulses arriving simultaneously from both fibres carry >>>identical labels. >>> >>>Using powerful rockets, the planet is sent into rotation with a period of >>>1 >>>day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating >>>frame >>>is 465 m/s. > >OK .. assuming you got the math right > >>>In this new situation, the pulses that arrive at C when it is at point p, >>>were >>>not emitted at that point but rather from an imaginary point q, > >Not an imaginary point at all. It is as real and physical point as p is. >All you are saying is that C has moved (and continues to move) It is imaginary in the R frame...REAL in the nonR frame. >>> which is at >>>rest in the nonR frame and moving away from C at speed -v in C's R frame. > >Angular speed I think you would mean, it is moving around the ring according >to C's rotating frame. yes >>>Each pulse takes still takes (0.133) seconds to pass through the fibre in >>>either direction. > >So you're using an emission theory analysus, which does say that the time >taken for the two paths is the same. Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the other at c-v, so they both arrive at p together. >>>Therefore the distance p-q = 465.(4/30) metres.....or 62 metres. > >I'll take your word for it. > >>>Since the distance between pulses is still 1 metre in the C's frame, > >That's what emission theory would say > >>> there are >>>now 40000062 pulses moving through the fibre in one direction but only >>>39999938 >>>moving in the other. > >WRONG. Emmision theory does not say that. It says that the distance in C's >frame between pulses is one meter. The distance the pulses travel in C's >frame is the same. So the number of pulses in the fibre at any given time >is the same. But I'm talking about the total number of pulses in transit in each path at any instant. there are 124 more in one path than the other. >You can also tell it is wrong because the same number of pulses has been >emitted in each direction and you just said earlier that they take the same >time to travel around the ring. So there must be the same number of them in >there. Or does your explanation have pulse fairies that eat up the pulses? Now you are being silly. I made a mistake in my original post. This is regrettable but doesn't present a problem for Bath. As you rightly said, the pulses that leave together DO always arrive together at the detector. >>>"That's odd", says Tom, "now, for some reason, the pulses coming from the >>>left >>>fiber left Paul's clock 0.266 seconds ahead of those arriving from the >>>right. >>>Which one should I use to synch my clock?" This is wrong....and is not what I wanted to point out. >Emmision theory doesn't say that will happen at all. It says what you said >earlier .. the time take for them to travel is the same. It is.....and the label on each arriving pulse is the same. >What you are now suddenly describing is what SR or an aether theory would >predict. You've changed theories in mid analysis !!! Yes, basically I did. I apologise for any inconvenience. >>>"This is great", says Jerry after some deliberation, "now we can build a >>>ring >>>gyro that will detect absolute rotation, based entirely on Henry Wilson's >>>BaTh". > >No .. because your analysis above used SR at the end. You switched theories >mid way thru. Yes I did. >You started saying the time for the pulses to travel in both directions >(with the rotating tube) was the same. That is what emission theory says. >And you said the distance between the pulses (in the rotating frame) is the >same. That is what emission theory says. It is. >The consequence of that is that >there MUST be the same number of pulses in each direction in the tube at any >given time. No that is YOUR mistake. The path lengths are different so there are more pulses in one path than the other. >And they MUST arrive at the detector with the same label. They do. I was wrong, before. >And >so emission theory says that a sagnac device will NOT detect rotation. > >You've just admitted that your theory is refuted. Now I'll tell you why that is wrong. >The *real* Sagnac devices have proven that THAT theory is NOT correct, and >it is NOT the same time required for the pulses (which is what SR predicts). > >BAHAHAHHA The true fact is that because the number of pulses is different in each path, THE RATE at which they arrive at the detector is DIFFERENT for each path. Between the instant of emission from point q and the detection at point p, 40000062 pulses arrive from one direction but only 39999938 from the other. Those numbers will change only during a rotational speed change. It is that difference in number that is analogous to the classical 'phase difference' of the two light rays in a ring gyro. Two different arrival rates produces a beat frequency at the detector. Exactly how that gives rise to interference fringes I'm not sure at this stage but the fact that BaTh predicts such a difference and produces the correct equation for fringe displacement is sufficient to demonstrate that Sagnac definitely does NOT refute BaTh. Henry Wilson... ........provider of free physics lessons Henry Wilson... ........provider of free physics lessons
From: Inertial on 14 Feb 2010 18:21 "Henry Wilson DSc" <..@..> wrote in message news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... > On Mon, 15 Feb 2010 09:44:32 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message >>news:crrgn51ol0crcbpjdeff497l466fvvbsor(a)4ax.com... >>[snip] >>> I admitted my mistakes and corrected them in the next post. >> >>If so, that is a pleasant change for you .. Where is the next post? > > Just below the first one. posted on 14th > > here's a copy. > *************** > On Sat, 13 Feb 2010 09:35:45 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message >>news:0uhbn5tpit1ss0m8falvch7hg2tieuj1a0(a)4ax.com... >>> On Thu, 11 Feb 2010 23:33:09 GMT, ..@..(Henry Wilson DSc) wrote: >>> >>>>Paul and Jerry are standing together on the equator of an Earthlike >>>>planet, >>>>next to an optical fibre that encircles the planet, WHICH IS NOT >>>>ROTATING. >>>>Its >>>>circumference is 40 million metres. >>>> >>>>//===================C======================// optical fibre around >>>>equator >>>>_____________v<-_____p___q_______________________surface (actually >>>>curved) >> >>What is q and v? >> >>>>(the surface is drawn flat for convenience. The fibre goes right around >>>>the >>>>planet) >>>> >>>>Paul's clock emits light pulses every 3.33 nanoseconds through both ends >>>>of the >>>>fibre. The pulses are labelled according to their emission times and >>>>move >>>>at c >>>>wrt the fibre and the clock. The linear distance between each pulse is >>>>therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass >>>>through >>>>the fibre in either direction. >>>> >>>>Since the planet is not rotating, the pulses returning to C were emitted >>>>and >>>>detected by Jerry at point p. >> >>Doesn't matter if they are detected at p or at C > > I'll withdraw my previous reply because it was wrong. So was my initial > message > because I stuffed up what I wanted to say. That's fine. Everyone makes mistakes at some time or another. I am always willing and happy to take that into account when it is admitted in that way.
From: Inertial on 14 Feb 2010 18:28 "Henry Wilson DSc" <..@..> wrote in message news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... > Two different arrival rates produces a beat frequency at the detector. There is no beat frequency in a Sagnac device. The two beams have the same frequency at the detector. This is the case in both emission theory and SR. SR, however, predicts differing arrival TIMES for pulses emitted at the same time .. hence the shift. Emission theory says they arrive at the same time (as you agree) .. hence there is no shift predicted (this refuting the theory). That your theory is even worse and somehow predicts differing arrival rates and a beat frequency that does not occur is enough proof to refute it.
From: Henry Wilson DSc on 14 Feb 2010 19:32 On Mon, 15 Feb 2010 10:28:50 +1100, "Inertial" <relatively(a)rest.com> wrote: >"Henry Wilson DSc" <..@..> wrote in message >news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... >> Two different arrival rates produces a beat frequency at the detector. > >There is no beat frequency in a Sagnac device. The two beams have the same >frequency at the detector. This is the case in both emission theory and SR. >SR, however, predicts differing arrival TIMES for pulses emitted at the same >time .. hence the shift. Emission theory says they arrive at the same time >(as you agree) .. hence there is no shift predicted (this refuting the >theory). > >That your theory is even worse and somehow predicts differing arrival rates >and a beat frequency that does not occur is enough proof to refute it. You still don't get it. The two beams DO NOT have the same frequency at the detector. In the clock example I gave, 40000062 pulses arrive at the detector from one direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any particular split pulse. Those numbers change during a speed CHANGE. Convert that to 'wavecrests', in the case of light, and you get a beat frequency that is somehow related to fringe displacement. Don't ask me how at the moment but the reason will reveal quite a lot about the nature of a photon. The fact that the frequencies in each direction are different annihilates the Roberts, Jerry, Andersen 'rotating frame' argument as well as your own. Henry Wilson... ........provider of free physics lessons
From: Inertial on 14 Feb 2010 19:50
"Henry Wilson DSc" <..@..> wrote in message news:a05hn55udgqmbu5rs41ap1shceh27dki2d(a)4ax.com... > On Mon, 15 Feb 2010 10:28:50 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >>"Henry Wilson DSc" <..@..> wrote in message >>news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... >>> Two different arrival rates produces a beat frequency at the detector. >> >>There is no beat frequency in a Sagnac device. The two beams have the >>same >>frequency at the detector. This is the case in both emission theory and >>SR. >>SR, however, predicts differing arrival TIMES for pulses emitted at the >>same >>time .. hence the shift. Emission theory says they arrive at the same >>time >>(as you agree) .. hence there is no shift predicted (this refuting the >>theory). >> >>That your theory is even worse and somehow predicts differing arrival >>rates >>and a beat frequency that does not occur is enough proof to refute it. > > You still don't get it. I get it just fine > The two beams DO NOT have the same frequency at the detector. Yes, they do in reality. That your theory predicts otherwise shows it to be wrong. > In the clock example I gave, 40000062 pulses arrive at the detector from > one > direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of > any > particular split pulse. Those numbers change during a speed CHANGE. And that is nonsense .. as the time taken for each pulse to make the trip is the same (in both directions), and the time between pulses being emitted is the same (in both directions), then in a given time, the same number of pulses arrive (in both directions). > Convert that to 'wavecrests', in the case of light, and you get a beat > frequency that is somehow related to fringe displacement. Don't ask me how > at > the moment but the reason will reveal quite a lot about the nature of a > photon. Because it is nonnsese. There is no beat frequency. > The fact Lie > that the frequencies in each direction are different They are not. > annihilates the > Roberts, Jerry, Andersen 'rotating frame' argument as well as your own. Nope .. you've just proven your own argument wrong. Well done. |