A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: Henry Wilson DSc on 11 Feb 2010 18:33 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
From: Henry Wilson DSc on 12 Feb 2010 16:27 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) > >(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 Not ONE ANSWER!!!!!!!! Henry Wilson... ........provider of free physics lessons
From: Inertial on 12 Feb 2010 17:35 "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 >>At any instant, 40 million pulses are in transit >>around the planet in each direction. >> >>Tom Tom? I thought it was Paul and Jerry? >> 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) >> 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. >>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. >>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. 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? >>"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?" 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. What you are now suddenly describing is what SR or an aether theory would predict. You've changed theories in mid analysis !!! >>"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. 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. The consequence of that is that there MUST be the same number of pulses in each direction in the tube at any given time. And they MUST arrive at the detector with the same label. And so emission theory says that a sagnac device will NOT detect rotation. You've just admitted that your theory is refuted. 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
From: BURT on 12 Feb 2010 17:47 On Feb 12, 2:35 pm, "Inertial" <relativ...(a)rest.com> wrote: > "Henry Wilson DSc" <..@..> wrote in messagenews: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 > > >>At any instant, 40 million pulses are in transit > >>around the planet in each direction. > > >>Tom > > Tom? I thought it was Paul and Jerry? > > >> 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) > > >> 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. > > >>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. > > >>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. > > 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? > > >>"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?" > > 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. > > What you are now suddenly describing is what SR or an aether theory would > predict. You've changed theories in mid analysis !!! > > >>"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. > > 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. The consequence of that is that > there MUST be the same number of pulses in each direction in the tube at any > given time. And they MUST arrive at the detector with the same label. And > so emission theory says that a sagnac device will NOT detect rotation. > > You've just admitted that your theory is refuted. > > 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- Hide quoted text - > > - Show quoted text - Einstein could never reconcile the photon with the light wave. I ask which wave is the photon in? The magnetic or electric? Light is electric energy oscillating with magnetism in a Dual Unfied force. Mitch Raemsch
From: Henry Wilson DSc on 13 Feb 2010 00:18
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 C is at point p. >>>At any instant, 40 million pulses are in transit >>>around the planet in each direction. >>> >>>Tom > >Tom? I thought it was Paul and Jerry? Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong. >>> 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) Hey dopey, can you see a point marked where you were 10 seconds ago? >>> 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. v is its peripheral speed wrt the nonR frame. >>>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. It certainly is. >>>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 that's what is happening. >>> 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. No dopey, the travel times are the same. The path lengths are different by +/- 62m >So the number of pulses in the fibre at any given time >is the same. that's correct. that is why the fringes on a sagnac interferometer do not move during constant rotation. >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. No it doesn't mean that at all. I have explained perfectly well what happens. It is only during a CHANGE of speed that pulses move in and out of the two paths. >Or does your explanation have pulse fairies that eat up the pulses? No the explanation is that you are plain dopey. >>>"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?" > >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. Correct. >What you are now suddenly describing is what SR or an aether theory would >predict. You've changed theories in mid analysis !!! No dopey. You are incapable of understanding what i said. >>>"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. > >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. The consequence of that is that >there MUST be the same number of pulses in each direction in the tube at any >given time. Only when it is not rotating. >And they MUST arrive at the detector with the same label. And >so emission theory says that a sagnac device will NOT detect rotation. No dopey , it says the sagnac effect is too hard for you to understand. >You've just admitted that your theory is refuted. > >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 > come back when you have learnt to use your brain Henry Wilson... ........provider of free physics lessons |