Sagnac Idiocy
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From: George Dishman on 1 Oct 2007 17:56 "Clueless Henri Wilson" <HW@....> wrote in message news:g0p2g31piokdmiknndufe8548miqpm6a2l(a)4ax.com... > On Mon, 1 Oct 2007 20:21:40 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:vad0g39vvhr4e3evat3birni87hr020et5(a)4ax.com... > .... >>> George, let's get our definitions straight so we can get on with what >>> matters . >>> >>> Fringe 'displacement' is the current distance of a fringe from its >>> defined >>> position at zero rotation speed. >> >>Yes, that's the key one. >> >>> Fringe 'movement' or 'shifting' is a dynamic term describing the process >>> of >>> changing displacements. It has units of velocity. >> >>No, it has units of fringes per second and since >>fringes are dimensionless, that is actually Hz. > > No George, 'fringes' are actually distances.....small ones. No, the number is a count of fringes. You can count three fringes passing a point and it tells you nothing about a distance. You need to multiply by the fringe spacing which _is_ a distance to get the distance the pattern moved. > The pattern > actually moves across the screen even though fringe movement is detected > and > counted as just the way a particular spot on the screen goes alternatively > dark > and light.. > Displacement has units of length. Fringe movement is rate of change of > displacement dx/dt. If it has value zero, there is no change in > displacement..ie., no movement..signifying no angular aceleration. > >>> Having said that, I repeat, what you said is wrong. Travel time is not >>> the >>> criterion. Path length difference is what matters.... >> >>Actually, what matters is the relative phase of the >>two signals. That can be calculated easily as the >>time difference divided by the source period (inverse >>of frequency) or with some care from the path length >>difference and the two different distances moved by >>the waves along their respective paths during the >>period. > > Phase is irrelevant....except insofar as it creates the actual fringes. Exactly, it creates the fringes, so one fringe displacement is a phase change of 2 pi radians (note, also dimensionless). >>> and that varies with >>> angular velocity. >> >>No. I think you had a problem following my explanation >>of your error above so I have marked up a copy of your >>diagram. Only the bit near the source matters, you >>should be able to extrapolate the rest of the path >>yourself. I've not trimmed as this also illustrates the >>text of my previous post. >> >> http://www.georgedishman.f2s.com/Henri/Sagnac_wavelength.png > > No, you have made the common mistake of thinking there is a doppler shift > at > the source. No, there is no Doppler change of the frequency, the increased length is solely due to the increase speed of the wave away from the source. The wavelength remains lambda. The line segment shown as (1 + v*sqrt(2)/c)*lambda is _not_ the wavelength, it is the distance moved by the wave in the period of the source oscillation. You are making the mistake of thinking those are the same which is true in conventional physics but not in ballistic theory. > In BaTh, that doesn't happen. If the source is inertial..and in > this case it's near enough..the absolute wavelength of the emitted light > is the > same no matter what the source is doing.... or whatever frame is used to > measure it.. Yes, that is shown by the line segment from C to B and marked as lambda. That is the wavelength and it has the same length whether the table is turning or not (for constant velocity). >>Imagine one positive zero-crossing (or wave crest or >>whatever marker you like) leaves the source S when it is >>at point A, the start of your blue path. After one source >>cycle time, that marker has moved to point B on the blue >>path, a distance of: >> >> (1 + v*sqrt(2)/c)*lambda >> >>as shown. The same works for the green path but with a >>minus sign of course. > > It's wrrong. Nope, it is correct. Think about it carefully before posting a knee jerk next time. >>The source in this time has moved to point C (roughly >>the diagram gives about three waves per leg arbitrarily). >>At that time, the next cycle starts. Your "absolute >>wavelength" is thus from point C to point B, and you are >>right in that it has the same value whether the table is >>rotating or not, but the first cycle has moved from point >>A to point B which is farther. If you want to work out >>the number of waves in the system, you can use the length >>of the blue line and the distance from A to C. >> >>Alternatively, if you imagine repeating the step shown >>12 times, the first wave arrives at S' just as the >>twelfth is being emitted by the source which is also >>at S'. The wavelength you use is your invariant lambda >>but the path length along which it is measured is now >>from S' round the loop and back to S' which is the same >>as when the equipment was at rest. (Essentially you are >>working in the co-rotating frame since Galilean Relativity >>applies). >> >>All the methods give the same conclusion, the number of >>waves is unchanged by the velocity and the time difference >>is zero. > > George, I am correct. Nope, you screwed up again, you forgot the source was moving. If you can't follow my description, try animating it with a sequence of say ten red pixels then ten green pixels for positive/negative half cycles and track each to the detector at the appropriate speed. You'll find the same colour arrives from each beam at the same time no matter what the table speed and if you freeze the screen you can count the number of waves. George
From: Dr. Henri Wilson on 1 Oct 2007 18:50 On Mon, 1 Oct 2007 22:56:33 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:g0p2g31piokdmiknndufe8548miqpm6a2l(a)4ax.com... >> On Mon, 1 Oct 2007 20:21:40 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> wrote: >>>"Clueless Henri Wilson" <HW@....> wrote in message >>>news:vad0g39vvhr4e3evat3birni87hr020et5(a)4ax.com... >> >... >>>> George, let's get our definitions straight so we can get on with what >>>> matters . >>>> >>>> Fringe 'displacement' is the current distance of a fringe from its >>>> defined >>>> position at zero rotation speed. >>> >>>Yes, that's the key one. >>> >>>> Fringe 'movement' or 'shifting' is a dynamic term describing the process >>>> of >>>> changing displacements. It has units of velocity. >>> >>>No, it has units of fringes per second and since >>>fringes are dimensionless, that is actually Hz. >> >> No George, 'fringes' are actually distances.....small ones. > >No, the number is a count of fringes. You can count >three fringes passing a point and it tells you nothing >about a distance. You need to multiply by the fringe >spacing which _is_ a distance to get the distance >the pattern moved. You've obviously never seen an interferometer or its fringe movement. >> The pattern >> actually moves across the screen even though fringe movement is detected >> and >> counted as just the way a particular spot on the screen goes alternatively >> dark >> and light.. >> Displacement has units of length. Fringe movement is rate of change of >> displacement dx/dt. If it has value zero, there is no change in >> displacement..ie., no movement..signifying no angular aceleration. >> >>>> Having said that, I repeat, what you said is wrong. Travel time is not >>>> the >>>> criterion. Path length difference is what matters.... >>> >>>Actually, what matters is the relative phase of the >>>two signals. That can be calculated easily as the >>>time difference divided by the source period (inverse >>>of frequency) or with some care from the path length >>>difference and the two different distances moved by >>>the waves along their respective paths during the >>>period. >> >> Phase is irrelevant....except insofar as it creates the actual fringes. > >Exactly, it creates the fringes, so one fringe >displacement is a phase change of 2 pi radians >(note, also dimensionless). So what? >>>> and that varies with >>>> angular velocity. >>> >>>No. I think you had a problem following my explanation >>>of your error above so I have marked up a copy of your >>>diagram. Only the bit near the source matters, you >>>should be able to extrapolate the rest of the path >>>yourself. I've not trimmed as this also illustrates the >>>text of my previous post. >>> >>> http://www.georgedishman.f2s.com/Henri/Sagnac_wavelength.png >> >> No, you have made the common mistake of thinking there is a doppler shift >> at >> the source. > >No, there is no Doppler change of the frequency, >the increased length is solely due to the increase >speed of the wave away from the source. The wavelength >remains lambda. The line segment shown as > > (1 + v*sqrt(2)/c)*lambda > >is _not_ the wavelength, it is the distance moved by >the wave in the period of the source oscillation. You >are making the mistake of thinking those are the same >which is true in conventional physics but not in >ballistic theory. The ray moves at (very nearly) c+v/root2 in the inertial frame. The absolute wavelength of all the light that source emits in any direction is the same lambda. Length is the same in all frames in BaTh. >> In BaTh, that doesn't happen. If the source is inertial..and in >> this case it's near enough..the absolute wavelength of the emitted light >> is the >> same no matter what the source is doing.... or whatever frame is used to >> measure it.. > >Yes, that is shown by the line segment from C to B >and marked as lambda. That is the wavelength and it >has the same length whether the table is turning or >not (for constant velocity). > >>>Imagine one positive zero-crossing (or wave crest or >>>whatever marker you like) leaves the source S when it is >>>at point A, the start of your blue path. After one source >>>cycle time, that marker has moved to point B on the blue >>>path, a distance of: >>> >>> (1 + v*sqrt(2)/c)*lambda >>> >>>as shown. The same works for the green path but with a >>>minus sign of course. >> >> It's wrrong. > >Nope, it is correct. Think about it carefully before >posting a knee jerk next time. > >>>The source in this time has moved to point C (roughly >>>the diagram gives about three waves per leg arbitrarily). >>>At that time, the next cycle starts. Your "absolute >>>wavelength" is thus from point C to point B, and you are >>>right in that it has the same value whether the table is >>>rotating or not, but the first cycle has moved from point >>>A to point B which is farther. If you want to work out >>>the number of waves in the system, you can use the length >>>of the blue line and the distance from A to C. >>> >>>Alternatively, if you imagine repeating the step shown >>>12 times, the first wave arrives at S' just as the >>>twelfth is being emitted by the source which is also >>>at S'. The wavelength you use is your invariant lambda >>>but the path length along which it is measured is now >>>from S' round the loop and back to S' which is the same >>>as when the equipment was at rest. (Essentially you are >>>working in the co-rotating frame since Galilean Relativity >>>applies). >>> >>>All the methods give the same conclusion, the number of >>>waves is unchanged by the velocity and the time difference >>>is zero. >> >> George, I am correct. > >Nope, you screwed up again, you forgot the source >was moving. It isn'tmoving in its own frame. It emits light with wavelength lambda in its own and every other frame. You can regard a 'wavelength' as being the same as a rigid rod. >If you can't follow my description, >try animating it with a sequence of say ten red >pixels then ten green pixels for positive/negative >half cycles and track each to the detector at the >appropriate speed. You'll find the same colour >arrives from each beam at the same time no matter >what the table speed and if you freeze the screen >you can count the number of waves. George, I think the ring gyro diagram clears this up. > >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 2 Oct 2007 03:08 On 1 Oct, 23:50, HW@....(Clueless Henri Wilson) wrote: > On Mon, 1 Oct 2007 22:56:33 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message news:g0p2g31piokdmiknndufe8548miqpm6a2l(a)4ax.com... > >> On Mon, 1 Oct 2007 20:21:40 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: > >>>"Clueless Henri Wilson" <HW@....> wrote in message news:vad0g39vvhr4e3evat3birni87hr020et5(a)4ax.com... .... > >>>> Fringe 'movement' or 'shifting' is a dynamic term describing the process > >>>> of changing displacements. It has units of velocity. > > >>>No, it has units of fringes per second and since > >>>fringes are dimensionless, that is actually Hz. > > >> No George, 'fringes' are actually distances.....small ones. > > >No, the number is a count of fringes. You can count > >three fringes passing a point and it tells you nothing > >about a distance. You need to multiply by the fringe > >spacing which _is_ a distance to get the distance > >the pattern moved. > > You've obviously never seen an interferometer or its fringe movement. You've obviously realised I am correct, a count of the number of fringes is just that, a number, not a length. > >>>Actually, what matters is the relative phase of the > >>>two signals. That can be calculated easily as the > >>>time difference divided by the source period (inverse > >>>of frequency) or with some care from the path length > >>>difference and the two different distances moved by > >>>the waves along their respective paths during the > >>>period. > > >> Phase is irrelevant....except insofar as it creates the actual fringes. > > >Exactly, it creates the fringes, so one fringe > >displacement is a phase change of 2 pi radians > >(note, also dimensionless). > > So what? What determines the intensity, hence the fringes, is phase difference. To work out the phase difference, you can use 2 pi times the time difference between the paths divided by the source period, or you can use the difference in the number of cycles along the paths. The former is zero as you agree on your web page, and the latter must be the same since these are just two different ways to work out the same value, there is only one phase difference. > >>>> and that varies with > >>>> angular velocity. > > >>>No. I think you had a problem following my explanation > >>>of your error above so I have marked up a copy of your > >>>diagram. Only the bit near the source matters, you > >>>should be able to extrapolate the rest of the path > >>>yourself. I've not trimmed as this also illustrates the > >>>text of my previous post. > > >>>http://www.georgedishman.f2s.com/Henri/Sagnac_wavelength.png > > >> No, you have made the common mistake of thinking there is a doppler shift > >> at the source. > > >No, there is no Doppler change of the frequency, > >the increased length is solely due to the increase > >speed of the wave away from the source. The wavelength > >remains lambda. The line segment shown as > > > (1 + v*sqrt(2)/c)*lambda Typo, that should be (1 + v/sqrt(2)/c)*lambda I'll need to fix the graphic too, it has the same slip. > >is _not_ the wavelength, it is the distance moved by > >the wave in the period of the source oscillation. You > >are making the mistake of thinking those are the same > >which is true in conventional physics but not in > >ballistic theory. > > The ray moves at (very nearly) c+v/root2 in the inertial frame. Yes, that's what the above formula gives (I put the sqrt on the top by mistake, careless of me). > The absolute > wavelength of all the light that source emits in any direction is the same > lambda. Length is the same in all frames in BaTh. Yes, that's shown by the lambda distance from point C to point B in my diagram, but what you missed is that lambda doesn't lie along the blue path or start from the same place. You don't get the number of cycles by dividing the blue path length by lambda, you need to use the above formula. > >> In BaTh, that doesn't happen. If the source is inertial..and in > >> this case it's near enough..the absolute wavelength of the emitted light > >> is the > >> same no matter what the source is doing.... or whatever frame is used to > >> measure it.. > > >Yes, that is shown by the line segment from C to B > >and marked as lambda. That is the wavelength and it > >has the same length whether the table is turning or > >not (for constant velocity). > > >>>Imagine one positive zero-crossing (or wave crest or > >>>whatever marker you like) leaves the source S when it is > >>>at point A, the start of your blue path. After one source > >>>cycle time, that marker has moved to point B on the blue > >>>path, a distance of: > > >>> (1 + v*sqrt(2)/c)*lambda > > >>>as shown. The same works for the green path but with a > >>>minus sign of course. > > >> It's wrrong. > > >Nope, it is correct. Think about it carefully before > >posting a knee jerk next time. > > >>>The source in this time has moved to point C (roughly > >>>the diagram gives about three waves per leg arbitrarily). > >>>At that time, the next cycle starts. Your "absolute > >>>wavelength" is thus from point C to point B, and you are > >>>right in that it has the same value whether the table is > >>>rotating or not, but the first cycle has moved from point > >>>A to point B which is farther. If you want to work out > >>>the number of waves in the system, you can use the length > >>>of the blue line and the distance from A to C. > > >>>Alternatively, if you imagine repeating the step shown > >>>12 times, the first wave arrives at S' just as the > >>>twelfth is being emitted by the source which is also > >>>at S'. The wavelength you use is your invariant lambda > >>>but the path length along which it is measured is now > >>>from S' round the loop and back to S' which is the same > >>>as when the equipment was at rest. (Essentially you are > >>>working in the co-rotating frame since Galilean Relativity > >>>applies). > > >>>All the methods give the same conclusion, the number of > >>>waves is unchanged by the velocity and the time difference > >>>is zero. > > >> George, I am correct. > > >Nope, you screwed up again, you forgot the source > >was moving. > > It isn'tmoving in its own frame. It emits light with wavelength lambda in its > own and every other frame. You can regard a 'wavelength' as being the same as a > rigid rod. Right, and in its own frame the path length it travels is the same as it is when the table is not rotating. In the inertial frame, the overall path length is extended as you have correctly calculated but you then need to divide by the larger distance above to get the number of cycles along that extended path, not by the wavelength. > >If you can't follow my description, > >try animating it with a sequence of say ten red > >pixels then ten green pixels for positive/negative > >half cycles and track each to the detector at the > >appropriate speed. You'll find the same colour > >arrives from each beam at the same time no matter > >what the table speed and if you freeze the screen > >you can count the number of waves. > > George, I think the ring gyro diagram clears this up. No, a ring gyro works in a completely different way and gives an output proportional to the angle moved, not the angular velocity. I had a quick look at your sketch but you don't show the standing wave pattern which gets measured so I doubt it is of any use at all. I'll have another look if I get time later. In the meantime, stick to the current subject. George
From: George Dishman on 2 Oct 2007 03:22 On 1 Oct, 23:50, HW@....(Clueless Henri Wilson) wrote: > On Mon, 1 Oct 2007 22:56:33 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: .... > George, I think the ring gyro diagram clears this up. http://www.users.bigpond.com/hewn/ringgyro1.jpg I've had another brief look, your title is inappropriate and I was trying to see how it related to a ring gyro. The answer is it doesn't, it again shows the Sagnac experiment. Anyway, your "Einstein" column is valid and the "Ballistic" column is ok down to the time difference row where you show zero difference. Below that you have a problem because the phase is determined by the path length divided by the distance moved per cycle, not the wavelength, and that distance isn't common to the two paths. That is just the same error I pointed out before. George
From: Jeckyl on 2 Oct 2007 03:59
"George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1191309765.557105.320160(a)r29g2000hsg.googlegroups.com... > On 1 Oct, 23:50, HW@....(Clueless Henri Wilson) wrote: >> On Mon, 1 Oct 2007 22:56:33 +0100, "George Dishman" >> <geo...(a)briar.demon.co.uk> wrote: > ... >> George, I think the ring gyro diagram clears this up. > > http://www.users.bigpond.com/hewn/ringgyro1.jpg > > I've had another brief look, your title is inappropriate > and I was trying to see how it related to a ring gyro. > The answer is it doesn't, it again shows the Sagnac > experiment. > > Anyway, your "Einstein" column is valid and the > "Ballistic" column is ok down to the time difference > row where you show zero difference. > > Below that you have a problem because the phase is > determined by the path length divided by the distance > moved per cycle, not the wavelength, and that distance > isn't common to the two paths. That is just the same > error I pointed out before. It must be the same distance in the moving frame, as they arrive back at the same position at the same time in ballistic theory .. ballistic theory is clearly refuted by Sagnac. |