Speed of Light: A universal Constant?
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From: George Dishman on 17 Apr 2005 07:50 "Henri Wilson" <H@..> wrote in message news:cnv261tp3cr5s0qk3l246r81307195hui8(a)4ax.com... > On Sat, 16 Apr 2005 10:14:11 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <H@..> wrote in message >>news:748161p3j4p7o776a6q3uq1ddlf7adi1l8(a)4ax.com... >>> On Fri, 15 Apr 2005 15:32:14 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>> >>>>Second part: >>>> >>>>"Henri Wilson" <H@..> wrote in message >>>>news:f5st519lab6a4ocvbi382jrkm5m9u17dtu(a)4ax.com... >>>>> On Thu, 14 Apr 2005 00:30:42 +0100, "George Dishman" >>>>> <george(a)briar.demon.co.uk> >>>>> wrote: >>>>... >>>>> >>>>> I'll try using a ball bouncing off a moving 45 wall. >>>> >>>>That's easy. Relative to the mirror the speed >>>>of the reflected beam equals the speed of the >>>>incident beam and Huygens then says the angles >>>>are equal either side of the normal to the >>>>mirror at the point of reflection. >>> >>> ...but consider a ball bouncing off a 'frictionless' wall that is moving >>> laterally as against one that imparts a velocity component to the ball. >> >>That's what I was describing. The conservation >>of energy tells you the speeds are equal in the >>mirror frame and Huygens then tells you the >>reflected direction. >> >>> Which ballistic model should be used for light? >> >>We were discussing the same one, sorry if >>that wasn't clear. If in doubt, there is >>nothing to stop you trying both :-) > > Well, in the case of a bouncing and perfectly elastic ball, its angle of > incidence would definitely equal the angle of reflection when bouncing off > a > 'frictionless' wall, no matter how fast the wall moved (in the wall > plane). > Would you agree? Yes, assuming the reflected speed was also the same as the incident speed but with the component normal to the wall reversed. If the speed changes then the angle can change like a frictionless ball bearing in a pinball machine. The speed component parallel to the wall remains the same but the normal component changes, hence an angle change. However, that would also affect reflection normal to the mirror giving a blue shift to the light. This is why I was suggesting Dopller shift might help. We know the fringes are static for constant speed of rotation so the frequency received back at the detector must be precisely identical to that emitted. > But in the case of a 100% interconnection at the instant of touch, the > equation > is very different. > Also, one might quite confidentally expect a grading of 'in between' > results > depending on the degree of 'friction'. Yes but that would also affect the angle for the static mirror would also change. Friction reduces the component parallel to the wall but not the (reversed) normal component if the ball is perfectly elastic, and if the ball isn't elastic you get loss of energy on a normal reflection, which would imply a red shift for normal reflection from a static mirror. > Where does that place light reflecting from a plane mirror? The above answers should constrain that, or Huygens can tell you once you know the speeds. > The reflection takes place over a very short distance but can we assume it > occurs in zero time? I don't think so. That's another matter entirely. I believe a classical analysis using Maxwell's equations will say there is no delay at all. > I hope I am making myself clear here. Yes, most of these are points I mentioned briefly some posts back but you are obviously thinking about them far more deeply now. Your thoughts bgenerally mirror my own. >>Really you need a situation where the light >>is reflected through 90 degrees from the >>rotating mirror but I just wanted to suggest >>that maybe you could find an experiment that >>had already been done. It was a useful >>technique but I don't know if any specific >>experiments meets your needs. > > A major problem with this type of experiment is the structural > stability of the mirror and wheel. It is a different class of experiment in many ways. I'd rather stick with the Sagnac discussion as there is enough complexity there without adding to it. > Well George, you are a welcome exception. Thanks. > Maybe some of the 'abusers' have > finally left this group and we can now have a few sensible > discussions like this one. > After all, that is what these newsgroups are for. The whole basis of science is that it should be possible to answer questions like this without subjective opinion entering into it so if I have to resort to abuse, I know I am admitting defeat. If you ever think I am replying that way, please do ask, it isn't intentional. regards George
From: George Dishman on 17 Apr 2005 08:14 "Henri Wilson" <H@..> wrote in message news:cnv261tp3cr5s0qk3l246r81307195hui8(a)4ax.com... > The reflection takes place over a very short distance but > can we assume it occurs in zero time? I don't think so. I gave this a little more thought. Try using a delay on reflection which is proportional to the angle of incidence. Since the angle for one beam increases while the other decreases, this should give you the right qualitative result. The problem then will be that when you derive the equation for the whole instrument, it will be independent of the size of the instrument when in reality we know it depends on the area enclosed. Take this line of thought a few more steps and you will find you are getting into Ralph Sansbury's territory :-o George
From: Henri Wilson on 17 Apr 2005 18:11 On Sun, 17 Apr 2005 01:01:35 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: >H@..(Henri Wilson) wrote in >news:sh0361pepm96fke2tvbim51f1ftv4ntja9(a)4ax.com: > >> OK, let's say half a fringe deflection corresponds to a rotational speed >> of 1 degree per hour. >> >> The instrument reading fluctuates between between 0.4 and 0.6 of a >> fringe shift over a period of a 30 minutes. >> >> How much has the thing rotated at the end of that period? >> >> If 100 fringes corresponded to say 1 degree per hour then it might be >> possible to produce accurate results but is that the kind of movement >> you get? > >http://www.fesg.tu-muenchen.de/us/Docs/Ring-SPb04.pdf Why don't you read these articles before you recommend them? >http://www.phys.canterbury.ac.nz/research/laser/ring_open.shtml What a useless article. 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 17 Apr 2005 18:17 On Sun, 17 Apr 2005 12:28:19 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:sh0361pepm96fke2tvbim51f1ftv4ntja9(a)4ax.com... >> On Sat, 16 Apr 2005 09:52:32 +0100, >> "George Dishman" <george(a)briar.demon.co.uk> wrote: >>>"Henri Wilson" <H@..> wrote in message >>>news:tn8161hfhml9om3ec4sdat9a4bsgt22aqt(a)4ax.com... >... >>>> Don't tell me the fringe movement is sensitive enough to involve >>>> an integration over time? >>> >>>That's right. That's why it they have only >>>become feasible with modern optical components >>>special fibre materials and digital integration. >>>Counters don't drift ;-) >> >> OK, let's say half a fringe deflection corresponds to a rotational speed >> of 1 >> degree per hour. >> >> The instrument reading fluctuates between between 0.4 and 0.6 of a fringe >> shift >> over a period of a 30 minutes. >> >> How much has the thing rotated at the end of that period? > >If the rate due to rotation is exactly 0.5 and >we see 0.4 to 0.6 then there is noise of +/-0.1. >You don't say how often it is being measured but >suppose this was 1800 readings so the +/-0.1 is >for 1 second long measurements. Assuming the >noise is Gaussian, the mean is 0.0 and the >readings uncorrelated so the total will be >1800 * 0.5 = 900 and the error will be >0.1 * sqrt(1800) = 4.2 or 0.47% or equivalent >to 0.5 +/- 0.0024 fringes. > >> If 100 fringes corresponded to say 1 degree per hour then it might be >> possible >> to produce accurate results but is that the kind of movement you get? > >I would need to know how many turns of fibre >were used but I'll research that later, or >maybe just guess for the purpose of >illustration. I have some other stuff to do >first but my gut feel is that it would be >smaller than you suggest. > >Actual techniques are much more complex, the >web page I gave a few posts back talked about >it. For example at zero rotation, you have the >same length in both paths since it is the same >fibre so you are in the regime where there are >no fringes but the uniform half brightness over >the whole screen. The sensor is a photodiode or >something like that which measures the >brightness at the centre spot hence my earlier >focus on point behaviour rather than fringes, >and the change of brightness either side of >zero speed is symmetrical hence an additional >modulation is superimposed. > >There is also a possible ambiguity if the simple >technique is used because a shift of 1/8 fringe >is indistinguishable from a shift of 3/8 fringe >(think of a sine wave). The maximum rate for the >DSP3000 is 375 degrees per second so I would >expect that is less than 1/4 fringe. The spec >is here: > > http://www.kvh.com/pdf/DSP3000_5.04.pdf It doesn't tell us anything about the actual design. that's what we want to know. > >Note the output formats are rate (primary), >incremental angle (change since some earlier >orientation) and integrated angle. Yes. > >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: George Dishman on 17 Apr 2005 18:50
"Henri Wilson" <H@..> wrote in message news:d0o561peg25rjssbo4h696r8vv3grnvbv1(a)4ax.com... > On Sun, 17 Apr 2005 12:28:19 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: >>"Henri Wilson" <H@..> wrote in message >>news:sh0361pepm96fke2tvbim51f1ftv4ntja9(a)4ax.com... .... >>> If 100 fringes corresponded to say 1 degree per hour then it might be >>> possible >>> to produce accurate results but is that the kind of movement you get? >> >>I would need to know how many turns of fibre >>were used but I'll research that later, or >>maybe just guess for the purpose of >>illustration. I have some other stuff to do >>first but my gut feel is that it would be >>smaller than you suggest. >> >>Actual techniques are much more complex, the >>web page I gave a few posts back talked about >>it. For example at zero rotation, you have the >>same length in both paths since it is the same >>fibre so you are in the regime where there are >>no fringes but the uniform half brightness over >>the whole screen. The sensor is a photodiode or >>something like that which measures the >>brightness at the centre spot hence my earlier >>focus on point behaviour rather than fringes, >>and the change of brightness either side of >>zero speed is symmetrical hence an additional >>modulation is superimposed. >> >>There is also a possible ambiguity if the simple >>technique is used because a shift of 1/8 fringe >>is indistinguishable from a shift of 3/8 fringe >>(think of a sine wave). The maximum rate for the >>DSP3000 is 375 degrees per second so I would >>expect that is less than 1/4 fringe. The spec >>is here: >> >> http://www.kvh.com/pdf/DSP3000_5.04.pdf > > It doesn't tell us anything about the actual design. > that's what we want to know. The details will be commercial so not published but there are other sources that give generic descriptions. I'll try to dig some out when I have time but given the specs and some basic facts, we can get some good estimates. George |