A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
in [Relativity]
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From: Henry Wilson DSc on 16 Feb 2010 16:57 On Tue, 16 Feb 2010 22:39:14 +1100, "Inertial" <relatively(a)rest.com> wrote: >"Henry Wilson DSc" <..@..> wrote in message >news:fpskn5h4k9g5nl40ihimn5f3ehhbsngbu5(a)4ax.com... >> On Tue, 16 Feb 2010 10:22:09 +1100, "Inertial" <relatively(a)rest.com> >> >> I see I'll have to unplonk you for your own sake. You need help. >> >> 1st question: Do you understand how and why the path lengths are different >> in >> the nonR frame? > >Of course .. I understand all this far better than you. So you should understand that since the distance between pulses is always 1 metre, there are 124 more pulses in one frame than in the other. >Do YOU understand that the number of particles (at any given time) in the >tube/fibre going in one direction is the same number as going in the other? Definitely not. > >Do YOU understand that every pulse has its own path? I certainly do...and I'm glad your brain cells are now showing signs of life. >Do YOU understand that the pulses with the longer paths travel faster than >the ones with the slower paths so that the time to travel the whole path is >the same in either direction? I certainly do. >Do YOU understand that a pair of pulses emitted at the same time in opposite >directions will therefore take the same time to travel along their >respective paths and will arrive at the same time? I certainly do. >Do YOU understand the pairs of pulses arriving at the same time means the >arrival rates are the same for both paths? NO. Where did the 62 pulses go? Did your fairies gobble them up? >>Do YOU understand that there is only one pulse on each path, because every >pulse has its own path? That's not a good way to look at it. the paths are the same but the start and end points advance 1 metre per pulse in the nonR frame. In that wiki diagram I referred to, the whole pattern is rotating with the Earth. What is interesting and where Paul, Tom and Jerry have gone wrong is that the distance vt and the 62 pulses don't magically disappear in the frame of the source. D=============62===============|============================== =========================================S __________________________________<-0.465km/s______________________________Ground The nonR observer sees the above picture and sees it ROTATING with the earth. An observer at the pole axis and rotating with the earth still sees the above picture but now it is NOT rotating with the Earth. Get it now? Henry Wilson... ........provider of free physics lessons
From: Androcles on 16 Feb 2010 17:18 "Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:0f8f836f-78c6-43c0-95ce-6e2257be4467(a)b18g2000vba.googlegroups.com... On Feb 16, 4:05 pm, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: You can't win, coward, I challenged you first. I'm not interested in your strawman. I'm bored now, I'll repeat the challenge tomorrow.
From: Jerry on 16 Feb 2010 17:27 On Feb 16, 4:18 pm, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message > > news:0f8f836f-78c6-43c0-95ce-6e2257be4467(a)b18g2000vba.googlegroups.com... > On Feb 16, 4:05 pm, "Androcles" <Headmas...(a)Hogwarts.physics_u> wrote: > > You can't win, coward, I challenged you first. So what? Your challenge is meaningless. You do not understand the difference between a simulation and an animation. A realistic simulation attempts to reproduce the actual conditions of an experiment, not some impossible setup where you are rotating a Sagnac turntable at a large fraction of the speed of light. > I'm not interested in > your strawman. I'm bored now, I'll repeat the challenge tomorrow. In other words, you've lost and you know that you've lost. 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: Paul B. Andersen on 17 Feb 2010 07:10 On 15.02.2010 23:24, Henry Wilson DSc wrote: > On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen" > <paul.b.andersen(a)somewhere.no> wrote: > >> On 12.02.2010 00:33, Henry Wilson DSc wrote: >>> Paul and Jerry are standing together on the equator of an Earthlike planet, >>> next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its >>> circumference is 40 million metres. >>> >>> //===================C======================// optical fibre around equator >>> _____________v<-_____p___q_______________________surface (actually curved) >>> > > I recognized my mistake and immediately corrected it. > > The pulses that leave together arrive together. > > If you want to comment, use my later version in which the pulse arrival rates > are different from both directions. The theory should be obvious to anyone with > a higher IQ than inertial. "The pulses that leave together arrive together", but the pulses are arriving at a different rate than those they arriving together with. It sure takes a high IQ to understand that! :-) BTW, in my previous posting I wrote: | Yet again you have made a giant fool of yourself. | I wonder how long it will take this time before you | realize that you have made a giant, unbelievable stupid blunder. | | I bet it will take a very long time, and if it eventually should | dawn to you, you will never admit it. What will you do then? | Start a new thread with a modified 'experiment'? :-) Now we know what you did! :-) You admitted your blunder, but couldn't admit that your conclusion based on that blunder was wrong, so you had to make an even bigger blunder to defend your wrong conclusion. Good to see that you still are able to invent new stupidities and not only are recycling old ones. This was definitely one of your better. Keep it up, the sky is the limit! -- Paul, still able to be amused by Ralph Rabbidge's stupidites http://home.c2i.net/pb_andersen/
From: Androcles on 17 Feb 2010 10:42
"Jerry" <Cephalobus_alienus(a)comcast.net> wrote in message news:c8017401-fcc6-4d73-871c-6d00c836cad8(a)g28g2000yqh.googlegroups.com... BTW, checking over my calculations, I discovered a math error. CHALLENGE: FIND THE ERROR IN THE FOLLOWING: 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 =========================================== So you want to know the error in YOUR calculation to show why Sagnac PROBABLY doesn't work, when all we are looking for is a change in phase of two signals each with a wavelength measured in Angstroms, because a fraction of an Angstrom plus a kilometre is immeasurably different from a kilometre plus an Angstrom. You've never heard of Vernier, either. http://en.wikipedia.org/wiki/Vernier_scale You are immeasurably different from an ignorant lunatic. |