Speed of Light: A universal Constant?
in [Relativity]
|
Prev: Joan-Claude van Dirk Helps to Trivialize Special Relativity
Next: GOD=G_uv Measure your IQ in 30 seconds
From: bz on 2 Apr 2005 07:45 H@..(Henri Wilson) wrote in news:m7js41592c9au3a6sm78losoh02h5knpho(a)4ax.com: > Bz assumes the signals travel at the same speed through the (identical) > cables irrespective of their direction of travel. > I assume no such thing. First of all, the signals are traveling in ONLY ONE DIRECTION, from the detector to the oscilliscope. I have signals from two different detectors I have those signals travel down different cables to my scope. I measure the difference in arrival time of the two signals I then swap the cables around so that signal 1 travels down cable 2 and signal 2 travels down cable 1, I will be able to accurately determine the difference in time between signal 1 and signal 2 by averaging the times. Have you ever used an oscilliscope? -- 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: Henri Wilson on 2 Apr 2005 17:11 On Fri, 1 Apr 2005 14:39:33 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"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. But not angle? > >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 That shows what I have been trying to tell you. It is the angular difference that causes the fringes. If light was perfectly parallel and if Huygens principle didn't operate, none would be producd. > >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. Fair enough. > >>>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. Except that there is also a change in angle, as well as the displacement. Maybe that somehow cancels the effect. > >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. What I cannot understand is why the fringes should move at all when the apparatus is rotating at constant angular velocity. But that is what happens...and the number of fringes moved is a measure of angular displacement. I cannot see that any theory other than some kind of 'local aether' one can account for this. SR certainly doesn't and I cannot yet see how it fits in with the ballistic theory. >>> >>>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. Effectively, it does. If you consider any one point in the image plane, the sideways displacement and change in angle of one beam will mean that the light ariving at that point will have left the source at a different time. ie, it has a different path length. >>>> 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. but you are still basing that on the assumption that both beams are defleted in the same direction. They are not. > >>>> 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. Yes. Of course. That's how all interference works. > >>>>>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. Yes. However in the diagram above, the two beams are NOT parallel to the x axis. >>>> 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. Oh yes, that is correct, the focus is actually on the two reflecting mirrors, which is half the path length. > >>>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. But the phase also changes. > >>>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. I still want to know why Sagnac fringes move during constant angular rotation speed. > >George > 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: Henri Wilson on 2 Apr 2005 18:21 On Sat, 2 Apr 2005 12:45:15 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >H@..(Henri Wilson) wrote in >news:m7js41592c9au3a6sm78losoh02h5knpho(a)4ax.com: > >> Bz assumes the signals travel at the same speed through the (identical) >> cables irrespective of their direction of travel. >> > >I assume no such thing. > >First of all, the signals are traveling in ONLY ONE DIRECTION, from the >detector to the oscilliscope. > >I have signals from two different detectors >I have those signals travel down different cables to my scope. >I measure the difference in arrival time of the two signals > >I then swap the cables around so that signal 1 travels down cable 2 and >signal 2 travels down cable 1, I will be able to accurately determine the >difference in time between signal 1 and signal 2 by averaging the times. > >Have you ever used an oscilliscope? I was using CROs before you were born. You are a fool. You understand nothing about this subject. HOW DO YOU KNOW THAT THE SPEED OF YOUR SIGNALS IS NOT DEPENDENT ON DIRECTION OF THE CABLES? TRY USING AN OPTIC FIBRE. 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: RichD on 2 Apr 2005 18:33 robert j. kolker wrote: > > So what?? Everybody knows what the term universal time means. > > Really? Everytime you use it, I think you ought to say proper > time which is the term used in the literature. There is no > evidence that a Master Clock exists and all other > clocks are slaved to it. Yes, but now we have the theory of a "multiverse", a la Andrei Linde, which contends there are new 'baby' universes being continually spawned, through inflation. Now what does "continually" mean in this context? Like, every 3 minutes, or 2 million years, or what? Whatever it means, it posits a sequence, which implies a time reference outside our universe, which implies some kind of master clock. It's especially bizarre, in view of the claim that time was created by the Big Bang. Now we have Big Bangs occurring one after another, but what's "before" and "after"? -- Rich
From: bz on 2 Apr 2005 19:22
H@..(Henri Wilson) wrote in news:f5au41p1m4h5pjacaresa5e6082hcuro8q(a)4ax.com: > On Sat, 2 Apr 2005 12:45:15 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> > wrote: > >>H@..(Henri Wilson) wrote in >>news:m7js41592c9au3a6sm78losoh02h5knpho(a)4ax.com: >> >>> Bz assumes the signals travel at the same speed through the >>> (identical) cables irrespective of their direction of travel. >>> >> >>I assume no such thing. >> >>First of all, the signals are traveling in ONLY ONE DIRECTION, from the >>detector to the oscilliscope. >> >>I have signals from two different detectors >>I have those signals travel down different cables to my scope. >>I measure the difference in arrival time of the two signals >> >>I then swap the cables around so that signal 1 travels down cable 2 and >>signal 2 travels down cable 1, I will be able to accurately determine >>the difference in time between signal 1 and signal 2 by averaging the >>times. >> >>Have you ever used an oscilliscope? > > I was using CROs before you were born. Sorry Henry, I confused you with kenseto. I have asked him several times the same question and he never answered it. I was born in 1945. Are you sure you were using a CRO before I was born. I have been playing with electronics since I was 8. I built a scope when I was 16. Got my ham license the same year. Tried to build a Ne laser in the 60's but couldn't get it to lase. Worked with YAG and CO2 lasers in the 70's while I was designing and building resistors and capacitors for Sprague. > > You are a fool. You understand nothing about this subject. I understand that name calling is the refuge of those who have no good arguments to make. > > HOW DO YOU KNOW THAT THE SPEED OF YOUR SIGNALS IS NOT DEPENDENT ON > DIRECTION OF THE CABLES? > No need to shout. As for the direction of the cables, it doesnt matter. > TRY USING AN OPTIC FIBRE. So that someone can say 'how do you know that the speed of your signals is not dependent on the direction of the fiber optics?' Here is my suggest set up: moving source of photons second detector >>-----------------------|-----------------------------| << first detector above is a diagram of the test set up. On the left is a disk that can be spun. It has an LED at one point on the disk. 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. There is a coax cable from first detector to channel 1 of the scope There is a coax cable from second detector to channel 2 of the scope These cables are identical in length and construction. TDR (time domain reflectometer) shows them identical in length. First we trigger on channel 1 and measure time delay to signal on channel 2. Then we swap coax cables and trigger on channel 2 and measure time delay to signal on channel 1. In each case we are measuring the time that it takes light to go from first to second detector. Swapping cables shouldnt make any difference because the cables are the same length. But if it does, we average the times and correct for any extra delay in one cable. Then we spin the disk with the LED and measure the one way speed of light from a moving source. We can do this with many different speeds, if we want. I am certain that I can show that the doppler shift is NOT due to changes in the speed of the photons with this set up. kenseto keeps saying that I can't measure the owls with the setup and talking about signals going both ways on the coax cables, and talking about two clocks and the need to syncronize them. I don't have two clocks. My cables are the same length. Tell me why I can't measure the one way speed of light with my set up. -- 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 |