A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
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From: Henry Wilson DSc on 13 Feb 2010 18:43 On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles" <Headmaster(a)Hogwarts.physics_u> wrote: > >"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. .....do you know what 'typophobia' is? You should ask your doctor about it... >> >> 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 In the case of light it will be the difference that creates the pattern. .....and fringe DISPLACEMENT will be proportional to beat frequency. Henry Wilson... ........provider of free physics lessons
From: Androcles on 13 Feb 2010 20:20 "Henry Wilson DSc" <..@..> wrote in message news:3vden5tuq5ctn9ipglsp844a5thb45cb9f(a)4ax.com... > On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles" > <Headmaster(a)Hogwarts.physics_u> > wrote: > >> >>"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. > > ....do you know what 'typophobia' is? You should ask your doctor about > it... > > >>> >>> 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 > > 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 > ....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?
From: Henry Wilson DSc on 14 Feb 2010 01:27 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..... Henry Wilson... ........provider of free physics lessons
From: eric gisse on 14 Feb 2010 02:19 ...@..(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? > > Henry Wilson... > > .......provider of free physics lessons
From: Henry Wilson DSc on 14 Feb 2010 02:28
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? > >> >> Henry Wilson... >> >> .......provider of free physics lessons Henry Wilson... ........provider of free physics lessons |