Speed of Light: A universal Constant?
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From: bz on 31 Mar 2005 17:26 "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. > 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. But, it doesn't matter how long the cables are as long as both cables are the same length. Lets say cable 1 adds 1 microsecond to the start signal cable 2 adds 1 microsecond to the stop signal The time between start and stop is still the same. This is because (b+d)-(a+d)=(b-a) I don't _care_ what the value of d is as long as the delay is the same in both cables. I am interested in b-a. The time that the light reaches the second detector minus the time it reaches the first detector. And, as I said before, if you say the cables (or the detectors) have different dealys, we can dispense with that objection by swapping detectors and cables around. We will see if there are any discrepancies due to the equipment and we can average the speeds two different values. Of course, the tests would be run several times each way and at each fan speed. >> One clock. Why in the world would I need two clocks since my detectors >> are in the same room? >> >> When I want to see if the velocity of the light source makes a >> difference in the speed of light I spin the fan that the light source >> is mounted on. >> >> Much simpler than your 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 31 Mar 2005 18:34 bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote in message news:<Xns9629C9EDAD888WQAHBGMXSZHVspammote(a)130.39.198.139>... > jgreen(a)seol.net.au (Jim Greenfield) wrote in > news:e7b5cc5d.0503301623.679f97c0(a)posting.google.com: > > > 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. > > > > > > The thoughtful scientist looks for ways to test things and figure out if > the profile of the fence altered or not. See! You (and your "thoughtful scientist") have ALREADY discounted the OTHER possibility. Why should not the fence be going faster (or the roller blader)? Where is the 'science' in that? > > See, in your example, the guy has been roller blading along this same fence > for years. He knows that the spacing is constant (speed of light). There has NEVER been an experiment done to measure / compare the velocity of light from a moving source: NEVER!! The ONLY tests done are from sources stationary (co-moving) with the receiver, or highly questionable data from Sne's and accellerators. Long before roller blades were invented, potential riders held the entirely wrong and misplaced belief that the change of frequency was ALWAYS due to the profile, and NEVER the speed. It has been taken as a 'given'- an ASSUMPTION: IT IS WRONG! > > So he is reasonably sure that if the frequency has changed then SOMETHING > has changed his speed (frequency of light). He checks for a tail wind or > something else that might have changed his speed. Looking, looking; but never for the REAL reason. > > If he eliminates all reasonable explanations, THEN he looks closely to see > if someone has removed the old fence and replaced it with one that looks > the same but has 20 flutes per meter, and painted it to look just like the > old fence and even put scratch marks on to match the ones he makes each > time he goes by. > > Extraordinary claims require extraordinary evidence. Show me the evidence. I have an hypothesis: If I compare the light from a source stationary ref me, with that of one which is moving away, and both use the same crystal to produce said light, I will observe a redshift in that from the moving source, as c'=c+v and c= fu Voila'! It is so! In order to show c=c+v then it necessary to employ magic to the light, which must alter its wavelength correspondingly to its frequency. And this imbues a photon with a very high intelligence, because how otherwise does the wavelength KNOW to alter from that which the crystal produces, just because there is relative motion with the receiver? Or is it the crystal with the smarts? It suddenly alters its emitted wavelength? I have the observation explainably pretty simply (and tested) You need to employ magic! So where lies the onus of proof? > > Not just 'well this theory explains the red shift of distant galaxies > almost as well as current explanations and it matches my outlook on the > world(to heck with the fact that there is plenty of data collected at short > range that say my theory is wrong). And as above, this last claim is rubbish. ASSUMPTION has too long been accepted as fact- go find the real causes of Doppler for light. ....and think on some ways to devise races between slugs of emr which are not "cost prohibitive" Jim G c'=c+v
From: bz on 31 Mar 2005 19:12 jgreen(a)seol.net.au (Jim Greenfield) wrote in news:e7b5cc5d.0503311534.1c6bc48c(a)posting.google.com: > See! You (and your "thoughtful scientist") have ALREADY discounted the > OTHER possibility. Why should not the fence be going faster (or the > roller blader)? > Where is the 'science' in that? The skater can see that the fence is not moving. >> See, in your example, the guy has been roller blading along this same >> fence for years. He knows that the spacing is constant (speed of >> light). > > There has NEVER been an experiment done to measure / compare the > velocity of light from a moving source: NEVER!! How would you like a test that measures the velocity of light from a moving source? Have you seen the test I suggested? 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. The cables from each detector to the scope are equal in length. 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. -- 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: Tom Roberts on 31 Mar 2005 20:02 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? Tom Roberts tjroberts(a)lucent.com
From: Henri Wilson on 1 Apr 2005 03:09
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. 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. > >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. > >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. > >>>> 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. Maybe we are not discussing the same problem. > >>>> 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. > >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. > >>>> 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. >>>> >>>> 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. > >> 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 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. 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. > >>>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. I think the eyepiece focusses on the 45 mirror, which is where the fringes are formed. > >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. > >Here's another from a manufacturer: > >http://www.xbow.com/Support/Support_pdf_files/RateSensorAppNote.pdf I'll read that later. > >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. |