Sagnac Idiocy
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From: Androcles on 7 Oct 2007 14:14 "Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:1191771446.895888.41700(a)50g2000hsm.googlegroups.com... : On Oct 6, 12:25 pm, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: : : > http://www.georgedishman.f2s.com/Sagnac/sagnac1913cras157b.pdf : : Please check my Babel Fish-assisted translation of the : interferometer description. I allowed myself a little looseness : in spots. : : The interferometer, already described briefly, is schematically : illustrated in the figure: a horizontal rotary table (50 cm in : diameter) carries, firmly screwed on it (the adjustment screws : being secured by lock screws), all the optical parts as well as : the source of light O, a small flashlight with a horizontal metal : filament. A microscope objective Co projects the image of this : filament through a Nicol prism N onto the horizontal slit F in : the focal plane of the collimating objective C; m is a reference : mirror. The vertically (per Fresnel's convention) polarized : parallel beam is divided by an air gap [frustrated total internal : reflection] beam splitter J, as in the usual interferometer of my : research (Comptes rendus, v. 150, p. 1676 (1910)), which I : applied to the optical study of the movements of the Earth : (Congress of Brussels, Sept. 1910, v. I, p. 207; Comptes rendus, : v. 152, p. 310 (1911); Le Radium, 1911, p. 1): the beam T : transmitted through the air gap J reflects successively on four : mirrors M and traverses the closed loop J-a1-a2-a3-a4-J of area S. : The beam R which the same air gap reflects traverses the same : circuit in the opposite direction. Returning to J, the beam T, : again transmitted, and the beam R, again reflected, are : superimposed in the same direction along T2 and R2, and form : interference fringes at the principal focus of the lens L on the : fine-grained photographic plate pp'. : : It should be obvious to everybody except Androcles that : "all the optical parts" which are "firmly screwed" onto the : rotary table includes the camera and lens assembly. : : Thanks, : Jerry You should simulate it then.
From: Dr. Henri Wilson on 7 Oct 2007 17:47 On Sun, 7 Oct 2007 12:01:48 +0100, "George Dishman" <george(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 >>>by Henry's "absolute wavelength" of course. >> >> It is the circumference + vt. > >Look at the animation when it has stopped and >think about what you are saying very carefully. 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. The difference between the two is what determines the fringe displacement. ...but this is probably far too hard for an engineer or a trainee nurse...... .. >>>> 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. >>>Actually the beam splitter >>>is the key point illustrated by Jerry's radial >>>black line (but the paths are common before >>>splitting and after recombination) so it is the >>>phase of the waves when they meet back at the >>>radial line that 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. ....and path length is speed dependent. I have improved the presentation of: www.users.bigpond.com/hewn/ringgyro.htm Have another look. >>>> They are using them correctly, you are the one that is wrong. >>> >>>Thanks for doing Henry's check for him, he seems >>>to be having some difficulty with this one. >> >> I am having NO difficulty. > >You have got it wrong, Jerry's simulation is >correct according to ballistic theory. Jerry omitted the most important feature. Jerry's model is not that of Sagnac... Jerry is worse than Androcles.... ....and you aren't much better.... >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 7 Oct 2007 18:04 On Sun, 7 Oct 2007 09:29:50 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:nt2gg3t7o2ev7nmamh2k0smgg97k0lm5n6(a)4ax.com... >> On Sat, 6 Oct 2007 13:37:40 +0100, "George Dishman" >>> >>>Clueless, it is drawn in the inertial frame, that's >>>why the short black radial line representing the >>>beam splitter moves anti-clockwise as the waves are >>>emitted. >> >> ..but you haven't changed the number of waves to match the different path >> lengths.... > >The waveforms are drawn by producing a sine wave series >of values and propagating it forward at the correct >speed. The resulting wavelength is therefore whatever it >should be, not an assumption. You are completely ignoring the fact that the source moves during the tavel The sagnac effect is a measure of the small difference the source moves during that time. >> In the rotating frame, wavelength is sped dependent. > >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 fact that they both take the same time to reach the detector in BaTH is irrelevant. >> 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 >>>> You are not accommodating the different number of wavelengths kin your >>>> paths. >>> >>>The blue and red lines are in phase at the source >>>as you can see most easily during the early part >>>of the animation. >>> >>>The waves move at c+v one way and c-v the other as >>>ballistic theory requires. >>> >>>There are 9.5 waves in each direction as you can >>>count when the animation completes. >> >> 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 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 >>>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. >> The phase depends entirely on the number of wavelengths around each path. >> Since >> the paths are not equal, the number is also different. >> I can't make this any more obvious.... > >Oh, what you are saying is perfectly clear, it is just >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. >> Do you or do you not agree that the path lengths are different >> George...and >> that difference is speed dependent? > >In the inertial frame, the distance from where the wave >is emitted to where it hits the detector is speed dependent >and so is the speed at which the wave travels so the time >taken is the same as you wrote on your web page, but you >didn't want to talk about the travel time. the travel times are equal..see: http://www.users.bigpond.com/hewn/ringgyro.htm >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 >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. Geore, I ambecoming quite embarrassed at your inability to see your error. It is so simple... >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 8 Oct 2007 03:51 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 > >>>by Henry's "absolute wavelength" of course. > > >> It is the circumference + vt. > > >Look at the animation when it has stopped and > >think about what you are saying very carefully. > > 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. > 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. 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. That is why there is no fringe displacement, the detector can only be aware of what is falling on it at any instant. > ..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. > >>>> 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. > >>>Actually the beam splitter > >>>is the key point illustrated by Jerry's radial > >>>black line (but the paths are common before > >>>splitting and after recombination) so it is the > >>>phase of the waves when they meet back at the > >>>radial line that 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 > ...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. > >>>> They are using them correctly, you are the one that is wrong. > > >>>Thanks for doing Henry's check for him, he seems > >>>to be having some difficulty with this one. > > >> I am having NO difficulty. > > >You have got it wrong, Jerry's simulation is > >correct according to ballistic theory. > > Jerry omitted the most important feature. Jerry's model is not that of > Sagnac... Jerry omitted nothing, se her new version where the yellow line makes it explicit. > Jerry is worse than Androcles.... He has got the maths and the PoR correct. He said this in considering the possibility of having the detector on the table: [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. George
From: George Dishman on 8 Oct 2007 13:53
"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: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:nt2gg3t7o2ev7nmamh2k0smgg97k0lm5n6(a)4ax.com... >>> On Sat, 6 Oct 2007 13:37:40 +0100, "George Dishman" >>>>Clueless, it is drawn in the inertial frame, that's >>>>why the short black radial line representing the >>>>beam splitter moves anti-clockwise as the waves are >>>>emitted. >>> >>> ..but you haven't changed the number of waves to match the different >>> path >>> lengths.... >> >>The waveforms are drawn by producing a sine wave series >>of values and propagating it forward at the correct >>speed. The resulting wavelength is therefore whatever it >>should be, not an assumption. > > 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. > The sagnac effect is a measure of the small difference the source moves > during > that time. > >>> In the rotating frame, wavelength is sped dependent. >> >>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. > 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. >>> 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. >>>>> You are not accommodating the different number of wavelengths kin your >>>>> paths. >>>> >>>>The blue and red lines are in phase at the source >>>>as you can see most easily during the early part >>>>of the animation. >>>> >>>>The waves move at c+v one way and c-v the other as >>>>ballistic theory requires. >>>> >>>>There are 9.5 waves in each direction as you can >>>>count when the animation completes. >>> >>> 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. > 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. >>>>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. >>> The phase depends entirely on the number of wavelengths around each >>> path. >>> Since >>> the paths are not equal, the number is also different. >>> I can't make this any more obvious.... >> >>Oh, what you are saying is perfectly clear, it is just >>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. >>> Do you or do you not agree that the path lengths are different >>> George...and >>> that difference is speed dependent? >> >>In the inertial frame, the distance from where the wave >>is emitted to where it hits the detector is speed dependent >>and so is the speed at which the wave travels so the time >>taken is the same as you wrote on your web page, but you >>didn't want to talk about the travel time. > > the travel times are equal..see: > http://www.users.bigpond.com/hewn/ringgyro.htm Yes, I just said that. We agreed it in November 2005. >>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. > 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. George |