A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
in [Relativity]
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From: Inertial on 16 Feb 2010 06:33 "Henry Wilson DSc" <..@..> wrote in message news:m8rkn5hhe61jtq4rcne9an6v7i3sm3pnc4(a)4ax.com... > On Tue, 16 Feb 2010 17:31:51 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message >>news:ne2kn59rbmokiv1revbefkn1m3npnbq3k7(a)4ax.com... >>> On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry >>> <Cephalobus_alienus(a)comcast.net> wrote: >>> >>>>On Feb 15, 4:20 pm, ..@..(Henry Wilson DSc) wrote: >>>>> On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry >>>>> >>> >>>> >>>>Your "rayphases" model has been refuted many times over, by me >>>>and by many others employing different arguments and methods of >>>>analysis. It is self-contradictory and predicts effects that have >>>>never been seen. In this thread alone, I've seen you squirm and >>>>deny assertions that you made mere hours previously. Your model >>>>is HOPELESS. >>>>http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm >>> >>> Well, this experiment requires no light rays or interferometers. >> >>So a light pulse isn't just a short light ray? >> >>> It is simple. Just wrap two optical fibres around the equator and send >>> time >>> labelled light pulses every 3.33 nanoseconds down each one in opposite >>> directions. The pulses will be 1 metre apart in both fibres. >> >>OK >> >>> According to BaTh, the Earth rotates by 62 metres during the time it >>> takes >>> for >>> a pulse to travel around the ring. So one light path will be 4000062 >>> metres >>> long and the other 3999938. >> >>But it will travel the longer path quicker, so they arrive at the same >>time >>with the same arrival rate >> >>> Therefore, between the time pulse N is emitted and received, 4000062 >>> pulses >>> arrive from one fibre and only 3999938 from the other. >> >>WRONG. >> >>If the pulses leave in pairs (as you imply), and take the same time to >>travel all the way around their path and arrive back again (as emission >>theory says), then they MUST arrive back in the same pairs they were >>emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path >>in any given time interval. >> >>You cannot determine the rotation of the device (according to emission >>theory) by looking at the arrival times and rates of the pulses. And the >>fact that your CAN in reality (with a real sagnac device) is resounding >>and >>refutation of emission theory. >> >>Why do you keep lying, Henry? >> >>> All you need is a >>> counter. >> >>All you need is SR .. which DOES let you determine the absolute rotation. > > You will remain plonked until your brain decides to function When will yours .. it hasn't shown any sign of rational though in the last couple of days (in particular). I've refuted your theory .. get a new hobby.
From: Jerry on 16 Feb 2010 07:48 On Feb 16, 6:43 am, Jerry <Cephalobus_alie...(a)comcast.net> wrote: Left out a parameter: > Work it out. If you spin a four-mirror Sagnac ring with mirrors spaced 1 meter apart > at 1 m/s, then > Coriolis curvature adds approximately 1e-18 meters to the path > length. That's a tiny, tiny, tiny, TINY fraction of the typical > distance between atoms in a solid. Jerry
From: Androcles on 16 Feb 2010 08:43 "Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:74fc8891-d6ce-4b60-a908-c37c81ed0332(a)b7g2000yqd.googlegroups.com... On Feb 16, 6:43 am, Jerry <Cephalobus_alie...(a)comcast.net> wrote: Left out a parameter: Yes, you did. Coriolis is ably and undeniably demonstrated here, bigot. http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov and here, arsehole: http://www.youtube.com/watch?v=Rt-uJGEnIcA You are a third order ignoramus and first order troll.
From: Jerry on 16 Feb 2010 15:22 On Feb 16, 7:36 am, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message >> The only reason Coriolis curvature SEEMS to add significantly to >> path length in your animation is that you spin the wheel at some >> huge fraction of the speed of light. >> >> Coriolis curvature is a second-order effect. > ======================================= > Liar. > <Irrelevant rant snipped> Consider the path of light between two mirrors in a high speed Sagnac apparatus spinning at 72,000 rpm. This extremely high rate of rotation results in the beam path following an arc with an apparent radius of 250000 meters as observed from the rotating frame. If we assume that the two mirrors are 1 meter apart, the distance from the center of the chord representing the straight line distance between the two mirrors to the center of the arc representing the apparent path of the light between the two mirrors would be 1 micron. Q: What is the total distance traversed by the beam of light between the two mirrors as measured in the rotating frame? A: The half-angle between the two endpoints of the arc has a sine of 0.5/250000. Setting this equal to the first two terms of the Taylor series expansion for sine x - x^3/6 = 0.000002 we find that the half-angle between the two endpoints of the arc is (0.000002 + 1.33e-18) radians. Multiplying this by twice the radius, we find that the apparent total distance traversed by the beam of light between the two mirrors as measured in the rotating frame is (1 + 6.7e-13) meters, which is immeasurably different from 1 meter even though we are spinning the turntable at such high speed that any real device would probably fly apart. Jerry
From: Jerry on 16 Feb 2010 16:23
On Feb 16, 3:10 pm, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message > > news:967fd8ad-c352-4bb8-bb8c-2d7fbb7b1e39(a)a1g2000vbl.googlegroups.com... > > > On Feb 16, 7:36 am, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > >> "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message > > >>> The only reason Coriolis curvature SEEMS to add significantly to > >>> path length in your animation is that you spin the wheel at some > >>> huge fraction of the speed of light. > > >>> Coriolis curvature is a second-order effect. > >> ======================================= > >> Liar. > >> <Irrelevant rant snipped> > > > Consider the path of light between two mirrors in a high speed > > Sagnac apparatus spinning at 72,000 rpm. > > No, I will not. Coward. > The only reason you get any fringe shift is that the > frequency of the light is in the order of 10^14 Hz. > Coriolis is a first-order effect, you lying incompetent tord, The horizontal displacement of the beam would be first order. The increase in path length would be second order. What is relevant here is the increase in path length. Consider the path of light between two mirrors in a high speed Sagnac apparatus spinning at 72,000 rpm. This extremely high rate of rotation results in the beam path following an arc with an apparent radius of 250000 meters as observed from the rotating frame. If we assume that the two mirrors are 1 meter apart, the distance from the center of the chord representing the straight line distance between the two mirrors to the center of the arc representing the apparent path of the light between the two mirrors would be 1 micron. Q: What is the total distance traversed by the beam of light between the two mirrors as measured in the rotating frame? A: The half-angle between the two endpoints of the arc has a sine of 0.5/250000. Setting this equal to the first two terms of the Taylor series expansion for sine x - x^3/6 = 0.000002 we find that the half-angle between the two endpoints of the arc is (0.000002 + 1.33e-18) radians. Multiplying this by twice the radius, we find that the apparent total distance traversed by the beam of light between the two mirrors as measured in the rotating frame is (1 + 6.7e-13) meters, which is immeasurably different from 1 meter even though we are spinning the turntable at such high speed that any real device would probably fly apart. Jerry |