A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: bert on 13 Feb 2010 13:48 On Feb 11, 6:33 pm, ..@..(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) > > (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. At any instant, 40 million pulses are in transit > around the planet in each direction. > > Tom 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. > > 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, which is at > rest in the nonR frame and moving away from C at speed -v in C's R frame. > > Each pulse takes still takes (0.133) seconds to pass through the fibre in > either direction. > Therefore the distance p-q = 465.(4/30) metres.....or 62 metres. > > Since the distance between pulses is still 1 metre in the C's frame, there are > now 40000062 pulses moving through the fibre in one direction but only 39999938 > moving in the other. > > "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 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". > > Henry Wilson... > > .......provider of free physics lessons My "Spin is in Theory" Tells it all. The universe is #1 energy,and rotational spin is in reality what all there is is all about TreBert
From: BURT on 13 Feb 2010 14:11 On Feb 13, 10:48 am, bert <herbertglazie...(a)msn.com> wrote: > On Feb 11, 6:33 pm, ..@..(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) > > > (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. At any instant, 40 million pulses are in transit > > around the planet in each direction. > > > Tom 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. > > > 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, which is at > > rest in the nonR frame and moving away from C at speed -v in C's R frame. > > > Each pulse takes still takes (0.133) seconds to pass through the fibre in > > either direction. > > Therefore the distance p-q = 465.(4/30) metres.....or 62 metres. > > > Since the distance between pulses is still 1 metre in the C's frame, there are > > now 40000062 pulses moving through the fibre in one direction but only 39999938 > > moving in the other. > > > "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 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". > > > Henry Wilson... > > > .......provider of free physics lessons > > My "Spin is in Theory" Tells it all. The universe is #1 energy,and > rotational spin is in reality what all there is is all about TreBert- Hide quoted text - > > - Show quoted text - Spin is misused for Rotation. A rotation speed flow that changes sizes is what spin is. Mitch Raemsch
From: bert on 13 Feb 2010 14:44 On Feb 13, 2:11 pm, BURT <macromi...(a)yahoo.com> wrote: > On Feb 13, 10:48 am, bert <herbertglazie...(a)msn.com> wrote: > > > > > > > On Feb 11, 6:33 pm, ..@..(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) > > > > (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. At any instant, 40 million pulses are in transit > > > around the planet in each direction. > > > > Tom 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. > > > > 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, which is at > > > rest in the nonR frame and moving away from C at speed -v in C's R frame. > > > > Each pulse takes still takes (0.133) seconds to pass through the fibre in > > > either direction. > > > Therefore the distance p-q = 465.(4/30) metres.....or 62 metres. > > > > Since the distance between pulses is still 1 metre in the C's frame, there are > > > now 40000062 pulses moving through the fibre in one direction but only 39999938 > > > moving in the other. > > > > "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 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". > > > > Henry Wilson... > > > > .......provider of free physics lessons > > > My "Spin is in Theory" Tells it all. The universe is #1 energy,and > > rotational spin is in reality what all there is is all about TreBert- Hide quoted text - > > > - Show quoted text - > > Spin is misused for Rotation. A rotation speed flow that changes sizes > is what spin is. > > Mitch Raemsch- Hide quoted text - > > - Show quoted text - Burt lectron Speed rotation is the heart of my c speed of c TreBert
From: Henry Wilson DSc on 13 Feb 2010 16:53 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
From: Androcles on 13 Feb 2010 17:37
"Henry Wilson DSc" <..@..> wrote in message news:jl4en5pdesnrl14qpop7u3gaovgrk0055b(a)4ax.com... > 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. His "Emmision" theory has nothing to do with emission theory. > >>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 It's simple enough. There are two beat frequencies, the sum and the difference. http://en.wikipedia.org/wiki/Circle_of_fifths So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't be very musical. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif > 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 |