Sagnac Idiocy
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From: Dr. Henri Wilson on 3 Oct 2007 17:51 On Wed, 03 Oct 2007 00:14:57 -0700, George Dishman <george(a)briar.demon.co.uk> wrote: >On 2 Oct, 22:41, HW@....(Clueless Henri Wilson) wrote: >> On Tue, 02 Oct 2007 00:22:45 -0700, George Dishman <geo...(a)briar.demon.co.uk> wrote: >> >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. >> >> http://www.mathpages.com/rr/s2-07/2-07.htm >> >> Optic fibre gyros are just this with lots of turns. >> They use the sagnac effect and the output is exactly as predicted on purely >> ballstic grounds. > >Right, those are the iFOG devices we discussed >months ago. A ring laser gyro on the other hand >is quite different. From the same page: > > "A ring interferometer typically consists > of many windings of fiber optic lines, > conducting light (of a fixed frequency) > in opposite directions around a loop, and > then recombining them to measure the phase > difference, just as in the original Sagnac > apparatus, but with greater efficiency and > sensitivity. A ring laser, on the other > hand, consists of a laser cavity in the > shape of a ring, which allows light to > circulate in both directions, producing > two standing waves with the same number > of nodes in each direction." > >I thought your graphic was supposed to be >related to the latter. > >> >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. >> >> You are raving pure relativistic nonsense out of sheer desperation. >> The wavelenght is absolute and exactly the same in both rays. > >Your naive assumption that you divide by the >wavelength is your error. It applies to your >analysis of both graphics. George, a saw blade has the same number of teeth per inch no matter how fast it moves past you. >George Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 3 Oct 2007 18:37 "Clueless Henri Wilson" <HW@....> wrote in message news:nl38g3hcb61seme686kltjhtm6ui8v70t3(a)4ax.com... > On Wed, 03 Oct 2007 00:14:57 -0700, George Dishman > <george(a)briar.demon.co.uk> wrote: .... >>Your naive assumption that you divide by the >>wavelength is your error. It applies to your >>analysis of both graphics. > > George, a saw blade has the same number of teeth per inch no matter how > fast it > moves past you. Indeed, but if you want to know how long it will will take to reach New York, you don't divide the distance to New York by the distance between of the teeth, you divide by the distance it travels per second. The phase difference is controlled by the time of arrival. George
From: Dr. Henri Wilson on 3 Oct 2007 19:11 On Wed, 3 Oct 2007 23:37:55 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:nl38g3hcb61seme686kltjhtm6ui8v70t3(a)4ax.com... >> On Wed, 03 Oct 2007 00:14:57 -0700, George Dishman >> <george(a)briar.demon.co.uk> wrote: >... >>>Your naive assumption that you divide by the >>>wavelength is your error. It applies to your >>>analysis of both graphics. >> >> George, a saw blade has the same number of teeth per inch no matter how >> fast it >> moves past you. > >Indeed, but if you want to know how long it will >will take to reach New York, you don't divide the >distance to New York by the distance between of >the teeth, you divide by the distance it travels >per second. The phase difference is controlled by >the time of arrival. but I don't want to know that. I want to know how many teeth are involved...and that number is independent of speed. .....agree george? >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 4 Oct 2007 03:01 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: > >"Clueless Henri Wilson" <HW@....> wrote in message > >news:h638g3hgnraifpinqsbl5hbgi602u414sd(a)4ax.com... > >> On Wed, 03 Oct 2007 00:04:00 -0700, George Dishman > >> <geo...(a)briar.demon.co.uk> wrote: > >>>On 2 Oct, 22:35, HW@....(Clueless Henri Wilson) wrote: > >>>> On Tue, 02 Oct 2007 00:08:20 -0700, George Dishman > >>>> <geo...(a)briar.demon.co.uk> wrote: > > >>>> >> 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. > > >>>> George, George, I can see I will have to educate you on the very basics. > >>>> For constant rotation, the travel times are indeed the same and the > >>>> fringe > >>>> pattern remains static. During a speed change, the travel times are NOT > >>>> the > >>>> same, pahse changes occcur AND THE FRINGE PATTERN MOVES SIDEWAYS. > > >>>> THIS IS WHAT HAPPENS. > > >>>Right, and if you divide the distance moved by > >>>the distance between the fringes and multiply > >>>by 2 pi, you get the phase difference between > >>>the waves for that constant rotational velocity. > > >>>That phase difference is equal to the time > >>>difference divided by the period of the source, > >>>or you can use path length difference but you > >>>_don't_ use the wavelength to do that calculation, > >>>that is where you make your mistake. > > >> The phase difference has nothing to do with travel times. > >> It is purely a function of path LENGTH DIFFERENCES. > > >It is what you divide that difference by that > >you are getting wrong, it shouldn't be the > >wavelength. > > I want to know the number of absolute wavelengths (l) in the path length (L). What you really want to know is the phase difference at the detector. > The answer is L/l.....primary school stuff.... but obviously too hard for you > George.... The answer is L/l in the rotating frame. > >> An absolute length L contains L/lambda absolute wavelengths no matter how > >> fast > >> the light is moving through it. > > >> That should be obvious to anyone. > > >The number of waves is the time of flight divided > >by the period, that should also be obvious, however > >those sms give different answers so while both may > >be obvious, one is wrong. You have divided by the > >wrong number. > > You are merely calculating the number of waves emitted by the source during the > time taken for light to complete one loop. ..that's irrelevant That is the number that can be drawn along the blue line. Since you calculated the length of the blue line to get your path length, you need the corresponding number. > The number lying AROUND the loop depends solely on the length of the loop and > the absolute wavelength. 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 the physical effect of fringes, nothing else. George
From: George Dishman on 4 Oct 2007 03:04
On 4 Oct, 00:11, HW@....(Dr. Henri Wilson) wrote: > On Wed, 3 Oct 2007 23:37:55 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> > wrote: > > > > > > > > >"Clueless Henri Wilson" <HW@....> wrote in message > >news:nl38g3hcb61seme686kltjhtm6ui8v70t3(a)4ax.com... > >> On Wed, 03 Oct 2007 00:14:57 -0700, George Dishman > >> <geo...(a)briar.demon.co.uk> wrote: > >... > >>>Your naive assumption that you divide by the > >>>wavelength is your error. It applies to your > >>>analysis of both graphics. > > >> George, a saw blade has the same number of teeth per inch no matter how > >> fast it > >> moves past you. > > >Indeed, but if you want to know how long it will > >will take to reach New York, you don't divide the > >distance to New York by the distance between of > >the teeth, you divide by the distance it travels > >per second. The phase difference is controlled by > >the time of arrival. > > but I don't want to know that. I want to know how many teeth are involved... No you don't. What you really need to know is the phase difference at any given point on the screen. For that you need to know the delay between the emission of some part of the wave and its arrival at the detector. If the delay along the two paths is equal you ge constructive interference while if it is half a period you get destructive. That is what creates the fringes. > and > that number is independent of speed. The phase difference is zero, independent of speed. George |