Speed of Light: A universal Constant?
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From: George Dishman on 30 Mar 2005 16:13 "Henri Wilson" <H@..> wrote in message news:sr1h41he74rnqareanja2hfr8m0qu8fhnq(a)4ax.com... > On Mon, 28 Mar 2005 11:52:28 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <H@..> wrote in message >>news:hlie41lsobbs9jnn3l8ai3g08pi2e0se0o(a)4ax.com... >>> On Sat, 26 Mar 2005 00:10:05 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>> >>>> >>>>"Henri Wilson" <H@..> wrote in message >>> >>>> >>>>>>As we have discussed before, ballistic theory is >>>>>>disproved by the Sagnac Experiment. >>>>> >>>>> Bull. The reason for the fringe shift has nothing to do with light >>>>> speed. >>>>> It is due to the fact that the mirrors rotate slightly during the time >>>>> light >>>>> travels between them. The two opposite beams are deflected in opposite >>>>> directions. >>>> >>>>We went over that last time. You got the >>>>geometry wrong, the beams are deflected >>>>such that they maintain the angle between >>>>them at the interferometer. >>> >>> No they don't. >>> Just draw a simple diagram. >> >>Here's the one I drew a year ago: >> >>http://www.briar.demon.co.uk/Henri/lab.gif >> >>I have added a small indicator showing where >>the rotation maintains the 90 degree angle >>at the detector in the four-mirror case. >> >>Note also any change of angle would alter the >>gap between the fringes and not produce a >>lateral shift. > > Look, the reason fringes are observed at all is pretty complicated. > > If the apparatus is properly set up and not rotating, a series of circular > fringes will be observed through the eyepiece. That depends on the type of interferometer but this illustrates what you describe: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/michel.html This is a bit French in places but gives the clearest explanation I've found so far. http://www.sciences.univ-nantes.fr/physique/enseignement/english/separa.html In particular look at this diagram but do use the 'forward' link . http://www.sciences.univ-nantes.fr/physique/enseignement/english/s1s2.html > They appear as a result of viewing angle differences. They appear due to the phase difference between the two beams at any point which determines whether the beams add or cancel hence determining the brightness. The phase difference is a combination of the phase difference at the centre plus an amount that depends on the location on the screen. How the phase varies with location is in part determined by the angle between the beams but in the example you give of circular fringes the beams are exactly parallel and exactly perpendicular to the screen. The reason the brightness varies with radius is because the distance to the apparent source changes more slowly as the radius increases for the more distant source. They are of course the same source seen at different distances through the two paths. This applet is excellent: http://www.physics.uq.edu.au/people/mcintyre/applets/optics/michelsc.html > If the two interferometer beams are deflected sideways in opposite > directions > and end up with a small angular difference because of apparatus rotation, > then > then I believe the fringes will be seen to move. No, what will happen at the centre depends on how you move the mirror. If you keep the path length the same by moving it along an arc, the brightness will remain the same. However, if you move the mirror horizontally, the horizontal spacing of the fringes will change while the vertical is unchanged so the circular fringes will become elliptical. You also start to get a superimposed double-source pattern like Young's slits. The combination is complex as you say, see the photos on the right of this page: http://en.wikipedia.org/wiki/Interference I had hoped you understood the mechanism, it is outlined at the bottom of that page. The brightness depends on the cosine of the phase difference between the waves arriving at a point. The angle of the beam doesn't play any part in determining the brightnes at any point, but it does influence how the relative phase varies over the screen which is how it influences the fringes. >>> The angular deflection of the beam at each miror is double the angular >>> change >>> of the mirror during the travel time. >> >>Correct for each beam, but it is in the same >>direction for the two beams so cancels. > > NO!!!! > > It is in the opposite direction for each beam. The diagram I drew above shows this and is almost identical to one you drew yourself last year. The point is academic anyway but if you work it out you'll see I'm right, the only way they could go in opposite directions is if they hit the detector either side of the original location of the source, and of course there is only one detector. See if you can find your own drawing. >>> For three mirrors, the total angular difference is 16 x rotation during >>> total >>> light travel time. >>> >>> Like I said, whether c or c+v is used is of little consequence. >> >>On the contrary, since the detector is sensitive >>to the time difference and not the angle between >>the beams, speed is the critical factor. > > I agree the detector WOULD BE sensitive to time differences but it is > rotatiopn > that we are interested in. I say the fringe movement effect would be the > same > if light speed was 0.5c. Yes it would if _both_ beams were moving at 0.5c. I'll assume you have read the pages I cited above and played with the applet so you understand how the interference works. If you start with a bright spot in the centre and both beams are moving at the same speed, what will happen to the brightness if you slow one beam so that it takes half a cycle longer to travel the path? The answer is that all the bright rings become dark and vice versa. In the Michelson interferometer you can do that by putting a thin film of transparent material in one beam. The speed is reduced by the refractive index so it takes longer to cover the same distance. In the Sagnac experiment, the detector moves while the light is in flight and because the beams are counter-rotating, one path is lengthened while the other is reduced. >>> Don't use the rotating frame. There is too much room for >>> misinterpretation. >> >>True, we need to be careful, but here is the >>same drawing as above in the rotating frame. >> >>http://www.briar.demon.co.uk/Henri/paths.gif >> >>It produces the same conclusion. > > But you are basing this on the incorrect conclusion that the > deflections are in > the same direction and cancel. That is wrong. > The beams never meet again at any of the mirrors. Try drawing it yourself, but it really doesn't matter. >><snip> >>>>... show your >>>>calculations if you think you can explain >>>>how the first order output arises. >>> >>> Let the light travel time around the four mirror system be 10 ns. >>> >>> An apparatus rotation of (2pi/3600) radians/hour (0.1 deg/hr) should >>> result in >>> a deflection difference between the two beams of 12 x 3 (m) x >>> 10/3600/10^8 >>> metres/sec. >>> >>> That is 10^-9 m/sec. >>> >>> or 0.001 um. >>> >>> ~ 500 x wavelength? >>> >>> How wide is a fringe? >> >>You seem to be calculating the lateral shift of >>the beam but that is parallel to the wavefront >>so has no effect on the fringe pattern or the >>brightness at the centre. > > I don't agree.. for the reasons I gave above. > > If what you say is correct, there would be no fringes, just black or > white. Well to a degree you are right, the fringes are due to the difference in path length so if there is no path difference, no fringes. Try the applet and set the separation to minimum: http://www.physics.uq.edu.au/people/mcintyre/applets/optics/michelsc.html No fringes, just a uniform shade. In fact it isn't just black or white but a brightness given by the cosine of the phase difference which is uniform over the screen. Of course with zero separation you would get black but the applet only goes down to 8.94e-4mm (probably to avoid a division-by-zero error). Remember a properly set up Michelson Interferometer has the beams exactly parallel. You can tell that's true because the pattern is circular - it has to be symmetric under rotation which cannot be true if either source is off the axis. Another way to look at it - imagine we had one image of the source at infinity but the other nearby, As you move a photodetector across the screen, the distance to the source at infinity stays the same but to the closer image varies, hence the phase difference varies, hence you get alternating constructive and destructive interference - fringes. > It is the angle subtended by the light entering the eyepiece that matters. It does but only indirectly as one factor influencing the phase difference. >>The relative brightness at any point is a measure >>of the phase difference between the beams so you >>need to find the difference in propagation time >>as a fraction of the period (1/frequency of >>course). > > .and like I said, if that were true, you would get no fringes but a > uniform > light intensity across the whole viewing area. And that's what you get if you can set the path lengths exactly equal. I guess you have never actually done this. >>Referring to this: >>http://www.briar.demon.co.uk/Henri/lab.gif >> >>It should be obvious that the red path for the >>co-rotating beam is longer than the green path >>for the counter-rotating beam. If the speed is >>c in the lab frame, that length difference >>produces exactly the output observed while if >>it is c+v for the red beam and c-v for the green >>beam, the speed difference exactly cancels the >>path length difference so the propagation time >>for the two beams is identical meaning no output >>for the Ritzian model. >> >>Let me turn that around - if you measure the >>difference in propagation time and you know the >>speed of the table, you can calculate the speed >>of the beams. That answer is c in the lab frame >>regardless of the speed of the table even though >>the source, the detector and the mirrors (or the >>fibre in the case of an iFOG) are all moving. >>The experiment measures the speed of light from >>a moving source and the observed result is c, >>not c+v. > > But your diagram is wrong. > The beams never meet again at any mirror. My diagram is correct but it doesn't matter, interference depends only on phase difference and the beam angle is just one way of affecting the phase. In Sagnac-based devices, the path length is identical and the change of brightness is due to the time difference for the beams resulting from the movement of the detector during the flight time of the light. This is a simple explanation at the top but gets into a bit of a rant later: http://www.mathpages.com/rr/s2-07/2-07.htm The formula dt = 4Aw/c^2 for v << c is well confirmed and the reason is obvious, but if the speed were source dependent, dt would always be zero. Here's another from a manufacturer: http://www.xbow.com/Support/Support_pdf_files/RateSensorAppNote.pdf George
From: Henri Wilson on 30 Mar 2005 17:00 On Wed, 30 Mar 2005 08:03:29 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >H@..(Henri Wilson) wrote in >news:rquj41lbmant4unh16f4c7aje45363aiqq(a)4ax.com: > >>>I know the wave is launched from a moving source, an LED at the end of a >>>moving fan blade. >>> >>>I start a timer when the photons from the LED pass the first detector. >>>I stop timer and report time when the photons from the LED reach the >>>second detector. >>> >>>I know distance between detectors. >>>I know time >>>speed = distance/time >> >> How long does it take the signal from the second LED to stop the timer? > >1) this should be > how long does it take the signal from the FIRST LED to stop the timer? > >> Answer: D/(light speed). > >right. > >> You cannot be sure light speed is the same in both directions. That is >> what you are trying to determine. > >There is NOT two directions. There is only one direction. > moving source of photons second detector > >>-----------------------|-----------------------------| > << first detector > >above is a diagram of the test set up. On the left is a spinning fan with >an LED at the end of one fan blade. In the center is a half silvered mirror >that deflects half the passing photons to a detector, on the right is a >second detector that detects the rest of the photons. > >in between the LED and the first detector there are a couple of pin holes >to make sure that we only see photon that are emitted when the LED is >travelling straight toward the detectors. > >We are measuring time it takes the light to pass from the first detector to >the second detector. >We are using one clock to do this. >The light travels only one direction. >This is the one way speed of light. > >> >> You have described a typical TWLS experiment. > >there is NO two way light. The light only travels one direction. > >> >>> >>>I have just measured the time. >>> >>>What is so hard about that????? >>>Where is there a flaw in the experiment? >> >> You don't understand anything. > >I can count to two. >In this case there is only one way. > >Where is the flaw in the experiment? You are oblivious to the fact that your signals don't travel instantaneously. If you use two separated clocks, you have to be able to synch them to within the accuracy required by your experiment. The way you have designed it, you will be looking for time differences around 10^-20 seconds. ...best of luck. HW. www.users.bigpond.com/hewn/index.htm Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong.
From: bz on 30 Mar 2005 18:41 H@..(Henri Wilson) wrote in news:fn7m41lu2td5gr96berr2hqdtdn8c397v5(a)4ax.com: > On Wed, 30 Mar 2005 08:03:29 +0000 (UTC), bz > <bz+sp(a)ch100-5.chem.lsu.edu> wrote: > >>H@..(Henri Wilson) wrote in >>news:rquj41lbmant4unh16f4c7aje45363aiqq(a)4ax.com: >> >>>>I know the wave is launched from a moving source, an LED at the end of >>>>a moving fan blade. >>>> >>>>I start a timer when the photons from the LED pass the first detector. >>>>I stop timer and report time when the photons from the LED reach the >>>>second detector. >>>> >>>>I know distance between detectors. >>>>I know time >>>>speed = distance/time >>> >>> How long does it take the signal from the second LED to stop the >>> timer? >> >>1) this should be >> how long does it take the signal from the FIRST LED to stop the timer? >> >>> Answer: D/(light speed). >> >>right. >> >>> You cannot be sure light speed is the same in both directions. That is >>> what you are trying to determine. >> >>There is NOT two directions. There is only one direction. >> moving source of photons second detector >> >>-----------------------|-----------------------------| >> << first detector >> >>above is a diagram of the test set up. On the left is a spinning fan >>with an LED at the end of one fan blade. In the center is a half >>silvered mirror that deflects half the passing photons to a detector, on >>the right is a second detector that detects the rest of the photons. >> >>in between the LED and the first detector there are a couple of pin >>holes to make sure that we only see photon that are emitted when the LED >>is travelling straight toward the detectors. >> >>We are measuring time it takes the light to pass from the first detector >>to the second detector. >>We are using one clock to do this. >>The light travels only one direction. >>This is the one way speed of light. >> >>> >>> You have described a typical TWLS experiment. >> >>there is NO two way light. The light only travels one direction. >> >>> >>>> >>>>I have just measured the time. >>>> >>>>What is so hard about that????? >>>>Where is there a flaw in the experiment? >>> >>> You don't understand anything. >> >>I can count to two. >>In this case there is only one way. >> >>Where is the flaw in the experiment? > > You are oblivious to the fact that your signals don't travel > instantaneously. To the contrary. I know about propagation delays. I can easily measure the time delay in the cables. It is easy to make sure that the cable lengths are identical. But a TDR (Time domain reflectometer) can be used to check the cable and make sure the velocity factor is uniform along the length. (just in case you were going to raise that as an issue) > > If you use two separated clocks, you have to be able to synch them to > within the accuracy required by your experiment. There is only one clock. There is only one source. There are two detectors. Yes, detectors have time delays too, but those can be accounted for. Measure the transit time from detector 1 to detector 2. Then swap the detectors and the cables around. If the delays are unequal, it will show up. The light is STILL only going one way. We still have the same distance between the detectors. The source is still moving the same speed. > > The way you have designed it, you will be looking for time differences > around 10^-20 seconds. > ..best of luck. That depends on the distance between the detectors and the speed of rotation of the source, right? Where is the flaw in the experiment? -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: Jim Greenfield on 30 Mar 2005 19:23 H@..(Henri Wilson) wrote in message news:<rquj41lbmant4unh16f4c7aje45363aiqq(a)4ax.com>... > On Tue, 29 Mar 2005 01:16:34 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> > wrote: > > >H@..(Henri Wilson) wrote in > >news:bu5h4153amrok96f30g78n1i6t938kegt6(a)4ax.com: Henri, These DHR's need a breath of fresh air; tell them to take a walk..... down a lane with a corrugated iron fence! The profile of the fence, is the distance between flutes- say ten per meter. Now they run a stick along the fence as they walk at one meter per sec, and hear a frequency of 10. Now let's let the fence do the moving, which makes no difference (only relative motion). The well trained DHR, his eyes shut, suddenly notices the frequency change to 20/sec. An intelligent observer would say there are TWO possibilities why this occurred: the profile of the fence altered to 20 flutes per meter length, or the fence velocity past the stick INCREASED! (or a bit of both). Not so the DHR!! His mind awash with magical belief, he REJECTS ENTIRELY that the speed altered, and is adamant that the fluting changed! Pushed, he will point to the cut finger where the raised edge of the new iron caught him. Nothing like a little blood to "prove" something. Jim G c'=c+v
From: Sam Wormley on 30 Mar 2005 19:32
Jim Greenfield wrote: > c'=c+v Contracted by the empirical results of observation and experiment, Greenfield |