A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: Androcles on 14 Feb 2010 02:31 "Henry Wilson DSc" <..@..> wrote in message news:fr5fn51k8n6kdtvq0gpc2q9fvh0km0luqf(a)4ax.com... > On Sun, 14 Feb 2010 01:20:44 -0000, "Androcles" > <Headmaster(a)Hogwarts.physics_u> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message >>news:3vden5tuq5ctn9ipglsp844a5thb45cb9f(a)4ax.com... >>> On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles" > >>>> >>>>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 >>> >>> In the case of light it will be the difference that creates the pattern. >> >>Yeah, >> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG > > Yes. That shows the difference. > >>> ....and fringe DISPLACEMENT will be proportional to beat frequency. >> >>Bullshit. You've never worked with an oscilloscope. >>How come you never produce a web page to back up your stupid claims? > > ...you're starting to sound more like little eric every day..... > ...you're starting to sound more like little Einstein every day..... -- Androcles ........provider of expensive physics lessons Awilson can't afford.
From: eric gisse on 14 Feb 2010 04:13 ...@..(Henry Wilson DSc) wrote: > On Sat, 13 Feb 2010 23:19:17 -0800, eric gisse <jowr.pi.nospam(a)gmail.com> > wrote: > >>..@..(Henry Wilson DSc) wrote: >>[...] >> >>> >>>>> ....and fringe DISPLACEMENT will be proportional to beat frequency. >>>> >>>>Bullshit. You've never worked with an oscilloscope. >>>>How come you never produce a web page to back up your stupid claims? >>> >>> ...you're starting to sound more like little eric every day..... >> >>Which makes me wonder why you post here at all. Nobody here likes you, you >>don't like anyone here, and you have your own webspace. Why not just post >>to it and be smug in the knowledge that you are right and the world is >>wrong? > > when are you going to say something intelligent? When are you going to apologize for posting forged degrees? > >> >>> >>> Henry Wilson... >>> >>> .......provider of free physics lessons > > > Henry Wilson... > > .......provider of free physics lessons
From: Inertial on 14 Feb 2010 06:25 "Henry Wilson DSc" <..@..> wrote in message news:7tccn59hs2oamj9iv4lj36q3lup5o1iran(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 > > C is at point p. Yeup >>>>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. You just muddled up your example by changing Jerry's name :) Guess you're not able to admit something like that though. >>>> 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? Why do you think points need to be marked in order not to be imaginary? >>>> 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. Fine .. I'm happy with that. Though a constant speed, it is not a constant velocity. >>>>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. Good >>>>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. Good >>>> 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 In which frame? >>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. I meant in each direction. >>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. Yes it does. > I have explained perfectly well what happens. And it is self-contradictory > It is only during a CHANGE of speed that pulses move in and out of the two > paths. Irrelevant >>Or does your explanation have pulse fairies that eat up the pulses? > > No the explanation is that you are plain dopey. No .. it is that you can't put forward a logically consistent argument >>>>"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. So you cannot measure the absolute rotation according to emission theory. Real sagnac devices operate as SR predicts, and you CAN measure it >>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. I understand just fine. Which is why i know you are wrong .. yet again >>>>"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. Not according to emission theory. if the pulses take the same time to get back to point C, and they travel at the same speed in C's frame, then they must have the same time stamp on them and must have travelled the same distance (in that frame) >>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. No .. yhou've just been getting it wrong for years .. and continue to post your nonsese >>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 At least I have one that works
From: Henry Wilson DSc on 14 Feb 2010 16:49 On Sun, 14 Feb 2010 22:25:20 +1100, "Inertial" <relatively(a)rest.com> wrote: > > >"Henry Wilson DSc" <..@..> wrote in message >news:7tccn59hs2oamj9iv4lj36q3lup5o1iran(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 >> >> C is at point p. > >Yeup > >>>>>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. > >You just muddled up your example by changing Jerry's name :) Guess you're >not able to admit something like that though. Tom has always believed that is true. >>>>> 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? > >Why do you think points need to be marked in order not to be imaginary? > >>>>> 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. > >Fine .. I'm happy with that. Though a constant speed, it is not a constant >velocity. > >>>>>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. > >Good > >>>>>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. > >Good > >>>>> 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 > >In which frame? > >>>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. > >I meant in each direction. > >>>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. > >Yes it does. > >> I have explained perfectly well what happens. > >And it is self-contradictory > >> It is only during a CHANGE of speed that pulses move in and out of the two >> paths. > >Irrelevant > >>>Or does your explanation have pulse fairies that eat up the pulses? >> >> No the explanation is that you are plain dopey. > >No .. it is that you can't put forward a logically consistent argument > >>>>>"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. > >So you cannot measure the absolute rotation according to emission theory. >Real sagnac devices operate as SR predicts, and you CAN measure it > >>>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. > >I understand just fine. Which is why i know you are wrong .. yet again > >>>>>"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. > >Not according to emission theory. if the pulses take the same time to get >back to point C, and they travel at the same speed in C's frame, then they >must have the same time stamp on them and must have travelled the same >distance (in that frame) > >>>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. > >No .. yhou've just been getting it wrong for years .. and continue to post >your nonsese > >>>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 > >At least I have one that works I admitted my mistakes and corrected them in the next post. the labels are the same at all speeds. Henry Wilson... ........provider of free physics lessons
From: Inertial on 14 Feb 2010 17:44
"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? |