Sagnac Idiocy
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From: Dr. Henri Wilson on 8 Oct 2007 18:21 On Mon, 08 Oct 2007 00:51:41 -0700, George Dishman <george(a)briar.demon.co.uk> wrote: >On 7 Oct, 22:47, HW@....(Clueless Henri Wilson) wrote: >> On Sun, 7 Oct 2007 12:01:48 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: >> >"Clueless Henri Wilson" <HW@....> wrote in message news:ku1gg39nq2eg50rdptt8pj2i6h4v1hfuc4(a)4ax.com... >> >>>Right, it is just the circumference divided >> I have thought about it...you obviously haven't. >> When the animation has stopped, only the detector position is shown. The >> position of the source when the arriving light WAS EMITTED must also be shown. > > http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm > >Jerry has now added a yellow line mrking the >location of the source at the time of emission. It still doesn't show in windows Vista. >> The difference between the two is what determines the fringe displacement. > >No, what determines the fringes is the relative >phase of the two waves when they reach the >detector. that is correct....and that phase is dtermined solely by the path length difference, since wavelength is absolute. >If you wait until the waves have made >the first complete circuit then watch how they >hit the radial line, you will see they are exactly >in phase as they reach it. they are in phase because Jerry has not programmed a ring gyro. >That is why there is >no fringe displacement, the detector can only be >aware of what is falling on it at any instant. George, animate my 'toothed wheel' model...that is the correct one. >> ..but this is probably far too hard for an engineer or a trainee nurse...... > >It was always correct, you just need it spelled >out for you. Think about the wheel model George. It clarifies the picture. The number of teeth betwen the two lines is dependent solely on path length. >> >>>> The DISTANCES are NOT the same >> >>>> in both frames. >> >>>> http://www.androcles01.pwp.blueyonder.co.uk/JerrySagnac.GIF >> >> >>>Also correct, but you forgot the detector is >> >>>also on the table. >> >> >> that doesn't affec the path length.. >> >> >The path length alone doesn't tell you the >> >resulting phase at the detector. >> >> Sorry George, it does. > >Nope, you need to convert it by a division and >you used the wrong denominator. Light speed makes no difference. The number of teeth between the two lines is independent of light speed. >> >> The phase is determined solely by the difference in path lengths. >> >> >Wrong, it also depends on the speed with which the >> >waves traverse that path. >> >> No George. ..number of wavelengths in a path = path length /lambda. > >No, you are making the mistake of mixing frames. >In the inertial frame: > > number = path length / distance per cycle > >In the rotating frame: > > number = circumference / wavelength George, you have forgotten that in the rotating fame, the source position AT EMISSION appears to move backwards....THERE IS YOU ERROR. number = (circumference + vt)/wavelength >> ...and path length is speed dependent. >> I have improved the presentation of:www.users.bigpond.com/hewn/ringgyro.htm >> >> Have another look. > >You are still making the same mistake, you >are dividing by the wavelength when you are >supposed to be working in the inertial frame. >The wavelength is the same in both frames but >the waves _move_ so you can't just blindly use >that number. I think you will have recognised your mistakes by now... >[I wrote:] >> : Spin up the table to constant speed, apply >> : power to the slip rings to light up the source, >> : switch it off when the exposure is complete >> : and slow the table to a stop. Take the plate >> : off the table and develop it. What could be >> : easier. >[Androcles wrote:] >> You won't see any fringe shift with the camera >> on the table, except during acceleration. You'll see >> a fringe, but it won't be shifted. What was his exposure time? >> Did Georges Sagnac use slip rings to operate the camera? > >He understands the consequences, and he denies >that the experiment was built the way it was >because of those consequences. You coorectly >understand the construction but can't do the >maths. If you could work as a team, you could >understand as much as Jerry. You should know your mistake by now. > >George Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 8 Oct 2007 18:44 On Mon, 8 Oct 2007 18:53:58 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:g0lig39qidff8g2rmqru6nkg7matkqorr1(a)4ax.com... >> On Sun, 7 Oct 2007 09:29:50 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> wrote: >> >> You are completely ignoring the fact that the source moves during the >> tavel > >The animation always included it but Jerry has added >a yellow marker line for the dul witted. The location >of the source at the moment of emission of the start >of the wave is shown by that and its subsequent >position at any time is shown by the black line. Jerry's animation is a joke... >>>In the _inertial_ frame, in one source period T=1/f, a >>>wave moves a distance of (c+v)T round the circumference >>>while the source moves a distance of vT. The next wave >>>is therefore emitted a distance of cT behind the previous >>>one, that is the wavelength. That applies whether v is >>>positive or negative so the wavelength is cT = c/f for >>>both beams. As you have said many times, for constant >>>speed, the wavelength is not affected by the speed. >> >> George, one ray moves 2piR+vt the other moves 2piR-vt. >> In Jerry's animation, they both shown to move the same distance. > >The distance they move is from the yellow line to the >black line at the time the start of the wave reaches it. which is as I said above. >> The fact that they both take the same time to reach the detector in BaTH >> is >> irrelevant. > >It determines the relative phase hence the fringes. the travel times AND WAVELENGTHS are the same. the paths lengths are different. >>>> Lesson: don't dabble with rotating frames. You will probably make errors >>>> and a fool of oneself.... >>> >>>The source is moving and the simulation is drawn in the >>>inertial frame. >>> >>>Lesson: calculate things, don't guess or you will probably >>>make errors and a fool of yourself. >> >> All the calculations you need are here: >> http://www.users.bigpond.com/hewn/ringgyro.htm > >It still has the same error I pointed out before. It doesn't have an error. Your mistake is that you forget that the emission point moves backwards in the rotating frame. >>>> There is your error. >>> >>>No, that is what ballistic theory predicts. If the >>>length of the circumference is L, the number of >>>waves is N = Lf/c in both directions for constant >>>angular velocity. >> >> The number is (L+vt)/lambda > >No, it is (L+vt)/(distance moved per cycle) if >you are working in the inertial frame. You only >get lambda as the denominator if you work in the >rotating frame but your numerator is from the >inertial frame. You can't mix frames Henry. Your mistake is that you forget that the emission point moves backwards in the rotating frame. >> You still don't get it. >> The path length difference is virtually the same in BaTh as in the SR >> treatment. see: http://www.users.bigpond.com/hewn/ringgyro.htm > >Sure but the speeds are different hence you get >a different result. Light speed makes no diffrence to the number of 'teeth' between the two lines. >>>>>The bottom line is that when the waves get back to >>>>>the splitter, they arrive in phase, the same as if >>>>>the table was static, and hence there is NO fringe >>>>>DISPLACEMENT. >>>> >>>> Rubbish. >>> >>>Sorry Henry, whether you use a simulation or equations, >>>the number of waves is N = Lf/c and the travel time is >>>L/c for _both_ beams. >> >> You are completely ignoring the main factor involved. ..the difference >> between >> the source and detector positions for a particular photon..or wavecrest. > >No, that is taken into account and appears as a ratio >in both numerator and denominator terms so cancels. Your mistake is that you forget that the emission point moves backwards in the rotating frame. >>>wrong. >> >> Sorry George, I know this hurts because it makes you, Andersen Jerry and >> Co >> look quite stupid..which of course is true.... >> YOU STILL can't get it into your head that the source and detector for a >> particular 'wavecrest' should not be drawn at the same point. > >Look at Jerry's new version and see if you can follow >it this time, Either that or get Androcles to explain >it to you, he understands what would happen but he >thinks it is a hypothetical question since the detector >is not on the table in his opinion. Your mistake is that you forget that the emission point moves backwards in the rotating frame. > >>>The distance between the source and the detector is the >>>length of the circumference which doesn't change with >>>speed, do you not agree Henry? >> >> George, the two rays travel distances equal to the circumference +/-vt > >I didn't ask how far they travelled, I asked how >far apart they were at any instant. You seem to be >confusing these two different numbers. > >>>The wavelength does not depend on speed either (for constant >>>speed) as explained above so the number of waves that fit >>>between source and detector is also independent of speed and >>>the same in both directions. Do you agree? >> >> George, you are still talking about what happens if you make 1000 marks on >> a >> wheel. There will still be 1000 marks on it no matter how fast it spins. > >Exactly. Think about it, one wave is emitted and moves >at c relative to the source, so in time t=1/f is is >ct = c/f ahead of the source when the next wave is >emitted. That means they are c/f apart and that pattern >moves round at a speed of c+v. George, we can use the linear analogy. ...........................................................................................................->c A_____________________B The line of dots represents the wavecrests of a light ray moving at c. A and B are the end points of a rigid rod. There are N dots between points A nd B, NO MATTER HOW fast the rod moves past them...or no matter how fast the light ray moves past the rod.. Next, let the position of B move by vt, where v is the speed of the rod and t is the time taken for a wavecrest to travel from A to B. Note, A, representing the source, does not move. AB' = AB.c/(c-v) The number of dots between A and B is increased to Nc/(c-v)...irrespective of light speed. In the gyro, that speed becomes c+v. ......the rest should be obvious.... >> Geore, I ambecoming quite embarrassed at your inability to see your error. >> It >> is so simple... > >It is extremely simple and the error is yours. The >waves are a distance of c/f apart and move round >like the teeth on a cog at a speed of c+v, what >could be simpler. That is exactly what Jerry's >simulation shows. Too simple for YOU two apparently.... >>George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 8 Oct 2007 18:52 "Clueless Henri Wilson" <HW@....> wrote in message news:voalg3dlnsb0odv14rh9juo0ifhm82qu3d(a)4ax.com... > On Mon, 08 Oct 2007 00:51:41 -0700, George Dishman > <george(a)briar.demon.co.uk> wrote: >>On 7 Oct, 22:47, HW@....(Clueless Henri Wilson) wrote: >>> On Sun, 7 Oct 2007 12:01:48 +0100, "George Dishman" >>> <geo...(a)briar.demon.co.uk> wrote: >>> >"Clueless Henri Wilson" <HW@....> wrote in message >>> >news:ku1gg39nq2eg50rdptt8pj2i6h4v1hfuc4(a)4ax.com... >>> >>>Right, it is just the circumference divided > >>> I have thought about it...you obviously haven't. >>> When the animation has stopped, only the detector position is shown. The >>> position of the source when the arriving light WAS EMITTED must also be >>> shown. >> >> http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm >> >>Jerry has now added a yellow line mrking the >>location of the source at the time of emission. > > It still doesn't show in windows Vista. Make sure you have the Sun VM installed and do a hard refresh. >>> The difference between the two is what determines the fringe >>> displacement. >> >>No, what determines the fringes is the relative >>phase of the two waves when they reach the >>detector. > > that is correct....and that phase is dtermined solely by the path length > difference, since wavelength is absolute. No, phase is determined by time. >>If you wait until the waves have made >>the first complete circuit then watch how they >>hit the radial line, you will see they are exactly >>in phase as they reach it. > > they are in phase because Jerry has not programmed a ring gyro. She has programmed what you are describing. >>That is why there is >>no fringe displacement, the detector can only be >>aware of what is falling on it at any instant. > > George, animate my 'toothed wheel' model...that is the correct one. Jerry's animation has 9.5 teeth, other than that, it is what you want, you cannot identify any error. >>> ..but this is probably far too hard for an engineer or a trainee >>> nurse...... >> >>It was always correct, you just need it spelled >>out for you. > > Think about the wheel model George. It clarifies the picture. > The number of teeth betwen the two lines is dependent solely on path > length. That is unrelated to the phase though unless you take account of the different speeds. >>> >>>> The DISTANCES are NOT the same >>> >>>> in both frames. >>> >>>> http://www.androcles01.pwp.blueyonder.co.uk/JerrySagnac.GIF >>> >>> >>>Also correct, but you forgot the detector is >>> >>>also on the table. >>> >>> >> that doesn't affec the path length.. >>> >>> >The path length alone doesn't tell you the >>> >resulting phase at the detector. >>> >>> Sorry George, it does. >> >>Nope, you need to convert it by a division and >>you used the wrong denominator. > > Light speed makes no difference. The number of teeth between the two lines > is > independent of light speed. The _difference_ in the speeds matters. >>> >> The phase is determined solely by the difference in path lengths. >>> >>> >Wrong, it also depends on the speed with which the >>> >waves traverse that path. >>> >>> No George. ..number of wavelengths in a path = path length /lambda. >> >>No, you are making the mistake of mixing frames. >>In the inertial frame: >> >> number = path length / distance per cycle >> >>In the rotating frame: >> >> number = circumference / wavelength > > George, you have forgotten that in the rotating fame, the source position > AT > EMISSION appears to move backwards....THERE IS YOU ERROR. But the light is already in transit so doesn't care what happens to the source after emission, that is _your_ error. > number = (circumference + vt)/wavelength No, in the rotating frame it is number = circumference/wavelength >>> ...and path length is speed dependent. >>> I have improved the presentation >>> of:www.users.bigpond.com/hewn/ringgyro.htm >>> >>> Have another look. >> >>You are still making the same mistake, you >>are dividing by the wavelength when you are >>supposed to be working in the inertial frame. >>The wavelength is the same in both frames but >>the waves _move_ so you can't just blindly use >>that number. > > I think you will have recognised your mistakes by now... I have again identified your mistake, you are still making the same one and just repeating it, not thinking about what is being said to you. >>[I wrote:] >>> : Spin up the table to constant speed, apply >>> : power to the slip rings to light up the source, >>> : switch it off when the exposure is complete >>> : and slow the table to a stop. Take the plate >>> : off the table and develop it. What could be >>> : easier. >>[Androcles wrote:] >>> You won't see any fringe shift with the camera >>> on the table, except during acceleration. You'll see >>> a fringe, but it won't be shifted. What was his exposure time? >>> Did Georges Sagnac use slip rings to operate the camera? >> >>He understands the consequences, and he denies >>that the experiment was built the way it was >>because of those consequences. You coorectly >>understand the construction but can't do the >>maths. If you could work as a team, you could >>understand as much as Jerry. > > You should know your mistake by now. The mistake is unchanged, your maths is wrong and Jerry's simulation is accurate. George
From: George Dishman on 8 Oct 2007 18:59 "Clueless Henri Wilson" <HW@....> wrote in message news:vbblg3ht7ve0shkjc0cc2mpvhaqvarkg5t(a)4ax.com... > On Mon, 8 Oct 2007 18:53:58 +0100, "George Dishman" > <george(a)briar.demon.co.uk>> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:g0lig39qidff8g2rmqru6nkg7matkqorr1(a)4ax.com... >>> On Sun, 7 Oct 2007 09:29:50 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> wrote: > >>> >>> You are completely ignoring the fact that the source moves during the >>> tavel >> >>The animation always included it but Jerry has added >>a yellow marker line for the dul witted. The location >>of the source at the moment of emission of the start >>of the wave is shown by that and its subsequent >>position at any time is shown by the black line. > > Jerry's animation is a joke... Then why can't you find a flaw in it? >>The distance they move is from the yellow line to the >>black line at the time the start of the wave reaches it. > > which is as I said above. Exactly, the animation already shows what you complained about. >>> The fact that they both take the same time to reach the detector in BaTH >>> is >>> irrelevant. >> >>It determines the relative phase hence the fringes. > > the travel times AND WAVELENGTHS are the same. the paths lengths are > different. And travel time determines phase. >>> http://www.users.bigpond.com/hewn/ringgyro.htm >> >>It still has the same error I pointed out before. > > It doesn't have an error. > Your mistake is that you forget that the emission point moves backwards in > the > rotating frame. Your mistake is that Jerry's animation is in the inertial frame. > Your mistake is that you forget that the emission point moves backwards in > the > rotating frame. Your mistake is that Jerry's animation is in the inertial frame. > Your mistake is that you forget that the emission point moves backwards in > the > rotating frame. Your mistake is that Jerry's animation is in the inertial frame. > Your mistake is that you forget that the emission point moves backwards in > the > rotating frame. Your mistake is that Jerry's animation is in the inertial frame. >>> wheel. There will still be 1000 marks on it no matter how fast it spins. >> >>Exactly. Think about it, one wave is emitted and moves >>at c relative to the source, so in time t=1/f is is >>ct = c/f ahead of the source when the next wave is >>emitted. That means they are c/f apart and that pattern >>moves round at a speed of c+v. > > George, we can use the linear analogy. Why not just look at the maths above. > ..........................................................................................................->c > A_____________________B > > The line of dots represents the wavecrests of a light ray moving at c. > A and B are the end points of a rigid rod. > > There are N dots between points A nd B, NO MATTER HOW fast the rod moves > past > them...or no matter how fast the light ray moves past the rod.. > > Next, let the position of B move by vt, where v is the speed of the rod > and t > is the time taken for a wavecrest to travel from A to B. Note, A, > representing > the source, does not move. Wrong, the source _does_ move, there is your error. You are as bad as Androcles only you want the source to be off the table. Just look at Jerry's simulation and try to find an error, there isn't one. George
From: Dr. Henri Wilson on 8 Oct 2007 20:08
On Mon, 8 Oct 2007 23:52:16 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:voalg3dlnsb0odv14rh9juo0ifhm82qu3d(a)4ax.com... >> On Mon, 08 Oct 2007 00:51:41 -0700, George Dishman >> <george(a)briar.demon.co.uk> wrote: >>>On 7 Oct, 22:47, HW@....(Clueless Henri Wilson) wrote: >>>> On Sun, 7 Oct 2007 12:01:48 +0100, "George Dishman" >>>> <geo...(a)briar.demon.co.uk> wrote: >>>> >"Clueless Henri Wilson" <HW@....> wrote in message >>>> >news:ku1gg39nq2eg50rdptt8pj2i6h4v1hfuc4(a)4ax.com... >>>> >>>Right, it is just the circumference divided >> >>>> I have thought about it...you obviously haven't. >>>> When the animation has stopped, only the detector position is shown. The >>>> position of the source when the arriving light WAS EMITTED must also be >>>> shown. >>> >>> http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm >>> >>>Jerry has now added a yellow line mrking the >>>location of the source at the time of emission. Look at the phasing at the YELLOW line when the thing stops. Is it the same George? ..AS REQUIRED. How does it vary with ring rotation speed? You don't know because Jerry hasn't included ring rotation speed at all. Jerry hasn't a clue.... >> It still doesn't show in windows Vista. > >Make sure you have the Sun VM installed and do >a hard refresh. I'll try my other computer...but I know what has been done... >>>> The difference between the two is what determines the fringe >>>> displacement. >>> >>>No, what determines the fringes is the relative >>>phase of the two waves when they reach the >>>detector. >> >> that is correct....and that phase is dtermined solely by the path length >> difference, since wavelength is absolute. > >No, phase is determined by time. When the animation stops, the phasing at the emission point is not the same...AS REQUIRED. >>>That is why there is >>>no fringe displacement, the detector can only be >>>aware of what is falling on it at any instant. >> >> George, animate my 'toothed wheel' model...that is the correct one. > >Jerry's animation has 9.5 teeth, other than that, >it is what you want, you cannot identify any error. Jerry has not included ring rotation speed at all. jerry is stupid.. >>> >>>It was always correct, you just need it spelled >>>out for you. >> >> Think about the wheel model George. It clarifies the picture. >> The number of teeth betwen the two lines is dependent solely on path >> length. > >That is unrelated to the phase though unless you >take account of the different speeds. see above >> George, you have forgotten that in the rotating fame, the source position >> AT >> EMISSION appears to move backwards....THERE IS YOU ERROR. > >But the light is already in transit so doesn't care >what happens to the source after emission, that is >_your_ error Not so george...read above again.... > >> number = (circumference + vt)/wavelength > >No, in the rotating frame it is > > number = circumference/wavelength You forgot that the emission point moves backwards in the rotating frame. >>>You are still making the same mistake, you >>>are dividing by the wavelength when you are >>>supposed to be working in the inertial frame. >>>The wavelength is the same in both frames but >>>the waves _move_ so you can't just blindly use >>>that number. >> >> I think you will have recognised your mistakes by now... > >I have again identified your mistake, you are >still making the same one and just repeating >it, not thinking about what is being said to >you. You forgot that the emission point moves backwards in the rotating frame. AND THE PHASING AT THE EMISSION POINT MUST BE THE SAME NOT DIFFRENT AS IN JERRY'S ANIMATION. >>>[I wrote:] >>>> : Spin up the table to constant speed, apply >>>> : power to the slip rings to light up the source, >>>> : switch it off when the exposure is complete >>>> : and slow the table to a stop. Take the plate >>>> : off the table and develop it. What could be >>>> : easier. >>>[Androcles wrote:] >>>> You won't see any fringe shift with the camera >>>> on the table, except during acceleration. You'll see >>>> a fringe, but it won't be shifted. What was his exposure time? >>>> Did Georges Sagnac use slip rings to operate the camera? >>> >>>He understands the consequences, and he denies >>>that the experiment was built the way it was >>>because of those consequences. You coorectly >>>understand the construction but can't do the >>>maths. If you could work as a team, you could >>>understand as much as Jerry. >> >> You should know your mistake by now. > >The mistake is unchanged, your maths is wrong >and Jerry's simulation is accurate. You forgot that the emission point moves backwards in the rotating frame. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |