Speed of Light: A universal Constant?
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From: kenseto on 1 Apr 2005 08:18 "bz" <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message news:Xns962AA7555C0E2WQAHBGMXSZHVspammote(a)130.39.198.139... > "kenseto" <kenseto(a)erinet.com> wrote in > news:33_2e.766$Fh4.242(a)fe2.columbus.rr.com: > > >> >> > >> >> > >> >> I don't define one way light speed by clocks. > >> >> The speed of light is how long it takes light to travel between two > >> >> point. ds/dt (where s is position and t is time) > >> >> > >> >> I measure it by measureing how long it takes light to travel between > >> >> two points. delta s/delta t. > >> > > >> > So how do you do that without a clock at each of the two points? > >> > BTW, delta s/delta t is velocity. Not how long it takes light to > >> > travel between two points. > >> > > >> > >> Light passes point 1 and some of it hits a photon detector. > >> A signal is sent to my oscilliscope through the cable from the > >> detector. It starts a clock in my scope and starts the trace on the > >> scope . > >> > >> When the light passes point 2, it hits a photon detector. > >> A signal is sent to my oscilliscopy through the cable from THAT > >> detector. This puts a signal on the screen of my scope. > >> The time between start of trace and signal is the delta t > >> I know delta s because I measured it. > > > > ROTFLOL...don't you know how stupid this sound?? The cable that carries > > the signal that started the clock is an outgoing leg and the cable that > > return the signal is the return leg. > > Have you ever used an oscilliscope? > > There is no cable 'that return the signal'. There is one cable between > detector 1 and the scope. There is one cable between detector 2 and the > scope. There is no return leg. What you are doing is replacing the speed of light with the speed of signal through the cable from two different directions. This will guarantee that your measured speed is less than c because the speed of light is faster than the speed of the signal through the cable. BTW your procedure is exactly a TWLS measurement of the speed of the signal through the cables. > > > So what you have is a TWLS measurement for > > delta t. > > You don't even understand that the cable that carries the signal is not > > instantaneous. > > > In a previous article in this thread I already addressed this. > I know how long the cables are and exactly how long it takes a pulse to get > from one end to the other. So you are assuming that you know the speed of the signal through the cable and that's why you know how long it takes a light pulse to get from one end to the other. You really don't know how stupid this sound?? ....using the speed of the signal through the cable to measure the speed of light which is known to be faster than the speed of signal through the cables??? Ken Seto
From: kenseto on 1 Apr 2005 08:29 "Tom Roberts" <tjroberts(a)lucent.com> wrote in message news:Z413e.16705$DW.8633(a)newssvr17.news.prodigy.com... > kenseto wrote: > > What you described is based on the bogus assumption that the leading edge of > > the light beam will hit where it is aimed. This bogus assumption is based on > > the assumption that there is no absolute motion of the target. > > What changes so that the leading edge misses but later portions of the > light pulse hit the detector? Absolute motion of the detector in the vertical direction during the flight time of light (in the horizontal direction) from the source to the detector. The speed of light is much faster than the absolute motion of the detector and that's why the later portion of the light pulse is able to hit the detector. This interpretation is supported by the observed gravitational red shift. Also this interpretation is what causes the observed time dilation. Ken Seto
From: George Dishman on 1 Apr 2005 08:39 "Henri Wilson" <H@..> wrote in message news:toup41thdkoj0d61n5tso57amfdvgr392h(a)4ax.com... > On Wed, 30 Mar 2005 22:13:28 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"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: >>> > > >>>>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. > > Yes and No. > > If that were true and the beams were perfectly parallel, you would see the > same > shade over the entire image. You wouldn't get rings at all. Yes, that _is_ what you get. > These are caused by the different path lengths at slightly different > viewing > angles. > > One form of the MMX deliberately slants the top mirror slightly so that it > forms a kind of optical wedge. That results in straight fringes rather > than > circular ones. That's right, and you calculate the fringe separation by finding the distance between points of constructive interference, i.e. where the phase difference is zero. If one beam falls perpendicularly on the screen it cerates a reference of the same phase everywhere. If the second beam falls at some angle to that, the relative phase depends on the location, hence you get fringes. >>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 gather these are produced by slightly changing the angle between two > coherent > laser beams. No, the circular patterns in the left column are all produced with beams that are exactly parallel. Those in the other columns show the effect of increasing lateral displacement of the sources. To see how the circular rings arise, look at figure 6 on the page I indicated earlier: http://www.sciences.univ-nantes.fr/physique/enseignement/english/s1s2.html Call x1 the distance from S1 to O and x2 the distance from S2 to O and y the distance from O to the point M on the screen. Let d1 be the path length from S1 to M and let d2 be the path length from S2 to M. By Pythagoras: d1^2 = x1^2 + y^2 d2^2 = x2^2 + y^2 It should be clear that as y changes, d1 will vary more rapidly than d2 hence there will be a path length difference and you get bright fringes when that is a multiple of the wavelength. Now I'm sure you understand that part but note carefully that if the line O-S1-S2 is perpendicular to the screen, the fringe will have the form of a circle of radius y about O so it passes through points +y and -y. If you offset S2 vertically upwards in an arc centered at O (so that the distance from S2 to O does not change), it should be clear that the distance to the +y point will be reduced while that to the -y point will increase. The fringe locations are no longer symmetrical about O and you no longer get circular fringes. >>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. > > I think there is a lot more to it than that. > The angle at which light enters the viewer is very important. It is important but only indirectly because it affects the path length from the source to the point of intererence, it has no direct effect. >>>>> 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. > > I can and it shows clearly that the two beams deflect in opposite > directions. > That should be obvious. Can you put it on the web so I can see if I can understand your view. It is obvious from mine that they turn the same way keeping the angle between them constant so there must be a difference in the diagrams. > Maybe we are not discussing the same problem. Perhaps, it seems that way. Another item supporting what I am saying is that, if they moved in opposite directions, it would create a lateral displacement so motion of the table would cause the rings to break up and look like the diagrams on the right of the Wiki page. http://en.wikipedia.org/wiki/Interference This doesn't happen. >>>>> 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. > > That's identical to moving one mirror slightly. It effectively changes the > path > length of one arm. Right. >>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. > > That's an aether concept. No, it is a "time = distance / speed" concept regardless of the mechanism. Changing either speed or distance affects the time taken. I'm drawing something to illustrate why this is so in SR but I'm using a new program so it may take me a few days. >>>>> 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. > > It does n\matter. ..and it also matters that they arrive back at the > detector > at slightly different angles. Only the time taken affects the phase for constant speed of rotation. (Angular acceleration is more complex but needn't concern us now.) If you can show that the angle affects the path length then you have a case. >>>>> >>>>> 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. > > Yes. Because of the small angular differences. No, because of the varying path difference. The angle does not change the value produced by superposition of the two fields, only the phase. >>> 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 main point is that there must be different path lengths to produce > circles > and these result from different angular effects, not just phase > differences. The point is that it is the difference in path lengths that produces the fringes. The angle only plays a part because of its influence on the path length. >>>>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. > > True. Yes I have actually looked in an MM interferometer. I once made one. Excellent :-) > Howeevr I still maintain that iof the two beams consisted of perfectly > parallel > light, no circles would appear even when the path length was varied. This diagram explains why they do: http://www.sciences.univ-nantes.fr/physique/enseignement/english/s1s2.html If you made one, then surely you must have played at finding the mirror location where the pattern flipped over through the surface of uniform brightness, no? That is where the two apparent sources are co-located, S1 = S2 and there can be no fringes. >>>>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. > > Each circular fringe subtends a different angle from the light source. Right, each is where the path length difference is a multiple of the wavelength. > I think the eyepiece focusses on the 45 mirror, which is where the fringes > are > formed. No, to get the sharpest image you focus about halfway between the two apparent sources. Since the distance between them is usually only a few wavelengths and the path is of the order of a metre in bench equipment, the focus is on the distant source. If you use a diffuser, I think you focus at infinity but I'm less sure of that. >>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. > > I will not agree. I can see that fringes move for the reasons I have > given. Your reason relies on a direct effect of angle where none exists, only phase affects the result of adding the waves by the rules of superposition. >>Here's another from a manufacturer: >> >>http://www.xbow.com/Support/Support_pdf_files/RateSensorAppNote.pdf > I'll read that later. It's fairly similar. We really must trim this a bit next time but there was a lot of needed context. Feel free to chop anything you agree with. George
From: kenseto on 1 Apr 2005 08:30 "Randy Poe" <poespam-trap(a)yahoo.com> wrote in message news:1112280111.279839.303150(a)f14g2000cwb.googlegroups.com... > > kenseto wrote: > > What you described is based on the bogus assumption that the leading > edge of > > the light beam will hit where it is aimed. This bogus assumption is > based on > > the assumption that there is no absolute motion of the target. > Throwing a > > base ball is different. The base ball will possess the absolute > motion of > > the thrower before he threw it. A pulse of light will not possess the > > absolute motion of the source before it is emitted. > > Why, in your universe, is light exempt from conservation > of momentum? It does not. The a detected photon will have the same momentum. Ken Seto
From: bz on 1 Apr 2005 08:53
"kenseto" <kenseto(a)erinet.com> wrote in news:3Tb3e.1147$tI6.789(a)fe2.columbus.rr.com: >> >> There is no cable 'that return the signal'. There is one cable between >> detector 1 and the scope. There is one cable between detector 2 and the >> scope. There is no return leg. > > What you are doing is replacing the speed of light with the speed of > signal through the cable from two different directions. This will > guarantee that your measured speed is less than c because the speed of > light is faster than the speed of the signal through the cable. BTW your > procedure is exactly a TWLS measurement of the speed of the signal > through the cables. I am measuring the difference in time between two signals. One coming down each cable. As long as the time in the cable is the same for both signals, the difference is measured correctly. >> >> > So what you have is a TWLS measurement for >> > delta t. >> > You don't even understand that the cable that carries the signal is >> > not instantaneous. >> >> >> In a previous article in this thread I already addressed this. >> I know how long the cables are and exactly how long it takes a pulse to > get >> from one end to the other. > > So you are assuming that you know the speed of the signal through the > cable NO! I KNOW the speed because I can measure it. TDR, time domain reflectometers will tell you how long a cable is in time units and length units(once you account for the velocity factor of the cable). But, as I have said, I don't need to know the length of the cable or the propagation speed, I only need to know that it is constant. > and that's why you know how long it takes a light pulse to get > from one end to the other. You really don't know how stupid this sound?? I know exactly how stupid it sounds. Have you ever used an oscilliscope? > ....using the speed of the signal through the cable to measure the speed > of light which is known to be faster than the speed of signal through > the cables??? The signal through the cable could move at 1% of the speed of light. It doesn't matter. Both cables are the same length. That can be tested by swapping the cables. t2+cable2delay - t1+cable1delay = t2-t1 provided that cable2delay=cable1delay I am interested in t2-t1 as I vary the rotation rate (speed) of the source. I am measuring owls from a moving source. The source moves toward the detectors for a very short time during each revolution of the shaft that the light source is mounted on. You still have failed to point out any real flaws in the experiment. If there was a gross error, a bunch of people would have jumped in and pointed it out. So far the ONLY rational objection was that the source speed might be too slow to clearly show relativistic effects. Since I am not trying to show such effects, just to show that the doppler effect is NOT due to changes in the speed of light, I don't think there is a problem. If we want relativistic effects, we might substitute a rotating beam of eletrons for the LED on a rotating device. In any case, we can use the 'time of flight' measurement of the beam of photons to verify that the speed of light has not been changed by the velocity of the source of the light. [aside: did you know that time-of-flight is used to measure the mass of atoms?] -- 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 |