Sagnac Idiocy
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From: Dr. Henri Wilson on 6 Oct 2007 18:34 On Sat, 6 Oct 2007 13:37:40 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:2padg3teh43srqo1s7u7m1quh0943nfpvl(a)4ax.com... >> On Fri, 5 Oct 2007 16:22:40 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> wrote: >>>"Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message >>>news:1191593421.613830.159110(a)50g2000hsm.googlegroups.com... >> >>>> http://groups.google.com/group/sci.physics.relativity/msg/1c03f332f90d65b5 >>>> >>>> which links to >>>> >>>> http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm >>> >>>Perfect! that's _exactly_ what I had in mind. >> >> It seems two minds are better than one....EXCEPT when they belong to >> relativists. >> You have both fallen into the same trap of trying to use the rotating >> frame. > >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.... In the rotating frame, wavelength is sped dependent. Lesson: don't dabble with rotating frames. You will probably make errors and a fool of oneself.... >> 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. >Your conclusion that the number of wavelengths would >be different is wrong (because you divided the path >lengths by the wrong number). Wavelength IS the saeme 'number' in all inertial frames. YOU are using the same number in the rotating frame...which is wrong. >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. 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.... Do you or do you not agree that the path lengths are different George...and that difference is speed dependent? >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 6 Oct 2007 18:37 On Sat, 6 Oct 2007 17:48:01 +0100, John Kennaugh <JKNG(a)kennaugh2435hex.freeserve.co.uk> wrote: >George Dishman wrote: >>On 4 Oct, 09:29, HW@....(Clueless Henri Wilson) wrote: >>> On Thu, 04 Oct 2007 00:01:31 -0700, George Dishman >>><geo...(a)briar.demon.co.uk> wrote: >>> >On 4 Oct, 00:27, HW@....(Clueless Henri Wilson) wrote: >>> >> On Wed, 3 Oct 2007 23:34:24 +0100, "George Dishman" >>> >><geo...(a)briar.demon.co.uk> wrote: >>... >>> >The path length in the rotating frame is not >>> >the same as that in the inertial frame yet the >>> >wavelength is the same in both. Can there be a >>> >different number of waves round the path >>> >depending on which frame you choose? Think about >>> >it, you are wrong. >>> >>> >If you cannot see why, go back to basics and work >>> >out the phase difference at the detector, that's >>> >what causes interference. >>> >>> George, i'm not interested in your childish drivel. >>> I know you have a vested interest in keeping Einstein alive. >>> I have a vested interest in science. >> >>You don't even recognise science when you have >>it thrust under your nose, your approach never >>gets beyond seventeenth century philosophy. >> >>The science is simple in this case, a bright fringe >>occurs where the waves on the two paths arrive "in >>phase". That means a peak from one path arrives at >>the same time as a peak on the other. That happens >>if the time it takes that peak to reach the detector >>from the beam splitter is the same for both paths >>which is always the case for constant angular velocity >>in ballistic theory. > >I have not been following this thread but disagree with your above >statement. It is simplistic as was the arguments put forward by Sagnac >who was convinced that his experiment invalidated relativity. > >Ballistic Theory > > > A->v > > >Inertial source A is moving in your FoR. Light emitted by A spreads out >in a circle and in ballistic theory the centre of that circle remains >with A. (in SR it remains at the fixed point in your FoR where A was >when the light was emitted). > >OK now let us assume A is not inertial. Let us assume that just after >the light is emitted something collides with A and knocks it off its >track. That cannot possibly effect the light which has already been >emitted and the centre of the circle of light will continue to be where >A *would* have been *if* A had not been accelerated. In ballistic theory >where the source is not inertial the speed of light is c in the >*inertial* frame in which the light was emitted. > > A->v > C | B > | > | > | > | > | > | > | > | > | > O > >Suppose A is travelling in a circle centre O and it emits light at the >point shown. > > C--------A' > | A > | > | B > | > | > | > | > | > | > | > O > >Suppose at time t later that light reaches B. While A is at the position >shown the centre of the circle of light reaching B is at A' not A, which >is the point A *would* have reached *if* it had been inertial. The light >reaching B has effectively come from A' and not from A. >If A is equidistant from B and C and the table is stationary A and A' >are the same point and light will reach B and C simultaneously. If the >table moves A' is not the same point as A and is not equidistant from B >and C so the light does not reach then simultaneously. There is your >mechanism for fringe shift with ballistic theory. > >My admittedly limited experience is that ballistic theory and relativity >always give the same answer despite often widely different descriptions >of what is going on. My assumption is that Ballistic theory and SR will >give identical predictions. The assumption that if relativity gives the >right answer ballistic theory won't is totally false. I of course am not >the first to notice this. It's all very simple John. see: http://www.users.bigpond.com/hewn/ringgyro.htm Remember, wavelength is absolute and invariant in all inertial frames in BaTh....just like a rigid rod. >Reviewer: Walter G. Hecker >Book 'The Wave and Ballistic Theories of Light - A Critical Review', >Author R A Waldron Published Frederick Muller Ltd., London 1977 > >"Waldron developed a ballistic theory of light on his own before >learning that Walter Ritz had already done it at about the time that A. >Einstein developed his 'Special Relativity'. Waldron shows with >mathematical accuracy and in excruciating detail, one by one, that all >the so called proofs of Einsteinian relativity (approx. 20) aren't proof >at all but that the experimental and observational results can be just >as well explained with the more pedestrian Ritzian relativity. >Unfortunately the books writing style is extremely dry and the back >referencing to figures and what was said earlier is hard to follow >(References are to sections, but the pages have no section headers). But >he is so detailed and conscientious in his proofs, that its a book worth >having for any 'dissenter' regarding Einsteinian Relativity, special or >general." > >I do not claim to have the necessary knowledge nor mathematical >expertise to properly evaluate ballistic theory and I seriously doubt >that Henri is up to it either :o). Very few with that capability have >ever looked at it objectively. One such was Fox. The following quote is >of interest: > > "Fox claims to have invalidated the majority, if not all, of the speed- > of-light experiments (including binary star observations) that have >been conducted to help us choose between Ritz and Einstein.... Fox gave >a decision in favour of Einstein, but did so in a manner that seems to >suggest that the final verdict is not in. In private correspondence Fox >says: >'...it is of interest for the general philosophy of science that Ritz's >theory, so different in structure from that of Maxwell, Lorentz and >Einstein, could come so close to describing correctly the vast quantity >of phenomena described today by relativistic electromagnetic theory.'" > >If you study the provenance of relativity without the sanitizing modern >spin then quite simply Einstein ignored the fact that light was >particulate and assumed that Maxwell's waves in aether theory was in no >ways compromised by the discovery that light is not in fact waves. If >you assume the validity of Maxwell's ET, as he did, then the MMX is a >valid experiment measuring the speed of the observer relative to the >aether and the result that it is always zero is a valid result i.e. the >observer is always stationary w.r.t. the aether. The second postulate >simply describes what an observer stationary w.r.t the aether would >experience. What I did not realise for some time was that the reason >Einstein justified in detail his first postulate in his 1905 paper but >simply announces his second without justification was that at the time >the second postulate simply put into words what everyone at the time >accepted as being the case. To Einstein it was his first postulate which >was controversial. > I find it hard to believe that such thinking could possibly >result in correct theory. There was an alternative which is a far better >match with the particulate nature of light and far more in keeping with >the 'no aether' doctrine of modern physics. One which gives a simpler >explanation of the MMX, retains Euclidean geometry and the 2 axioms of >physics which Einstein had to jettison in order to retain his waves in >aether belief. If in 1905 one accepted the particulate nature of light >as being more fundamental that its wavelike nature and if one decided >that space was indeed empty then there is no conceivable reason why one >would assume source independence and jettison two axioms of physics in >order to keep the idea alive. > > With that as the absurd basis of relativity, the possibility, no >matter how small that Ballistic theory might be made to work, cannot be >ignored. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 7 Oct 2007 04:29 "Dr. Henri Wilson" <HW@....> wrote in message news:nt2gg3t7o2ev7nmamh2k0smgg97k0lm5n6(a)4ax.com... > On Sat, 6 Oct 2007 13:37:40 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:2padg3teh43srqo1s7u7m1quh0943nfpvl(a)4ax.com... >>> On Fri, 5 Oct 2007 16:22:40 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> wrote: >>>>"Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message >>>>news:1191593421.613830.159110(a)50g2000hsm.googlegroups.com... >>> >>>>> http://groups.google.com/group/sci.physics.relativity/msg/1c03f332f90d65b5 >>>>> >>>>> which links to >>>>> >>>>> http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm >>>> >>>>Perfect! that's _exactly_ what I had in mind. >>> >>> It seems two minds are better than one....EXCEPT when they belong to >>> relativists. >>> You have both fallen into the same trap of trying to use the rotating >>> frame. >> >>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. > 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. > 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. >>> 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. >>Your conclusion that the number of wavelengths would >>be different is wrong (because you divided the path >>lengths by the wrong number). > > Wavelength IS the saeme 'number' in all inertial frames. YOU are using the > same > number in the rotating frame...which is wrong. The simulation is entirely done in the inertial frame. >>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. > 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. > 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 distance between the source and the detector is the length of the circumference which doesn't change with speed, do you not agree Henry? 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
From: George Dishman on 7 Oct 2007 07:01 "Clueless Henri Wilson" <HW@....> wrote in message news:ku1gg39nq2eg50rdptt8pj2i6h4v1hfuc4(a)4ax.com... > On Sat, 06 Oct 2007 06:07:23 -0700, George Dishman > <george(a)briar.demon.co.uk> wrote: >>On 5 Oct, 23:28, "Androcles" <Engin...(a)hogwarts.physics> wrote: >>> "Clueless Henri Wilson" <HW@....> wrote in >>> messagenews:r5bdg3pbhpkb7ci3fe7opas6rhgg5qk58e(a)4ax.com... >>> : On Fri, 05 Oct 2007 15:47:25 GMT, "Androcles" >>> <Engin...(a)hogwarts.physics>: wrote: >>> : >"George Dishman" <geo...(a)briar.demon.co.uk> wrote in message >>> news:ermdnYwTBeXVyJvaRVnytQA(a)pipex.net... .... >>> : >: All he needs to do is confirm for himself that the >>> : >: speed of the waves is correct >>> : > >>> : >It is: >>> : > In Jeery's model the speeds are: >>> : >144 degrees in 27 seconds for the rotating frame (5.3 degrees/sec) >>> : >720 + 144 in 27 seconds for the blue ray (32 degrees/sec) >>> : >720 - 144 in 27 seconds for the red ray ( 21.3 degrees/sec) >>> : >21.3 + 2 * 5.3 = 31.9, close enough to 32 with my rounding error. >>> : > >>> : >Excellent model, two speeds of light in the frame of the screen, >>> : >one in the rotating frame, 720 degrees in 27 seconds or 26.7 >>> degrees/sec >>> : >>> : They are trying to use rotating frames..... and using them wrongly... >>> >>> The time is the same in both frames. The NUMBER of wavelengths >>> is the same in both frames. >> >>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. .... >>> 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. >>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. >>> 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. George
From: Jerry on 7 Oct 2007 11:37
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 |