Speed of Light: A universal Constant?
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From: PD on 5 Apr 2005 15:57 Jim Greenfield wrote: > The Ghost In The Machine <ewill(a)sirius.athghost7038suus.net> wrote in message news:<l9a9i2-f6i.ln1(a)sirius.athghost7038suus.net>... > > In sci.physics, H@..(Henri Wilson) > > <H@> > > wrote > > on Mon, 04 Apr 2005 21:31:29 GMT > > <u6c351du1rm845dlvhj309smegtid0gnm9(a)4ax.com>: > > > On Mon, 4 Apr 2005 12:09:32 +0000 (UTC), bz <bz+sp(a)ch100-5.chem.lsu.edu> wrote: > > > > > >>H@..(Henri Wilson) wrote in news:gh4251dpkork18r2kknvn2gu6lt979b8m3@ > > >>4ax.com: > > >> > > >>> Ghost, is not velocity always specified relative to something? > > >>> > > >>> Is not the speed of light always 'c' wrt its source? > > >>> > > >>> > > >> > > >>The velocity of light is always c with respect to the observer. > > > > > > Proof please! > > > > No proof available. At best, there are several experiments > > that show evidence for this statement, a number of indirect > > experiments that show evidence for related concepts, and > > a number of observations of astrophysical phenomena that > > show evidence for other related concepts given certain > > assumptions. > > Primary assumption that has mired physics / astronomy for decades > being that > Doppler is falsely attributed to a magical wavelength alteration, thus > skewing > many measurements as to distance, velocity and composition (spectra) And this shows you know nothing about how light's wavelength is measured. One approach that's been around for years is the diffraction grating. With a diffraction grating, light of a particular wavelength is scattered and shows constructive interference at an angle that is a function of the ratio of the light's wavelength and the spacing of the etching in the grating. Nothing else -- no c's, no frequencies, no other buried physics -- just the ratio of the light's wavelength to the spacing of the etching the grating, a ratio of two distances. If what you say were true, that the wavelength stayed the same but the speed and frequency changed, then a blue line shifted to green by the Doppler effect would emerge from the grating at exactly the same angle as the unshifted blue line. Why? Because, if what you say were true, the wavelength would be the same and the spacing of the etching would be the same, so the ratio of those two distances would be the same. This is demonstrably NOT the case. In spectrometers, we have verification that the angle for a blue line shifted to green (and seen to be green by taking a color film plate) falls exactly where an unshifted *green* line should fall, not where the unshifted blue line should fall. Thus, we have measurement of both frequency and wavelength, showing that both are shifted. The product of the wavelength and frequency, even for the shifted lines, is (miraculously) c. There are no holes, no hidden assumptions, Jim. What you propose is flat-out ruled out experimentally. It does not hold water. PD [snip]
From: David Cross on 5 Apr 2005 15:51 On Mon, 04 Apr 2005 21:31:29 GMT, H@..(Henri Wilson) wrote: >>The velocity of light is always c with respect to the observer. > >Proof please! Several different measurements done different ways haven't convinced you of this? There is doubt of empirical observation due, perhaps, to inherent error in the measurements, and then there is just plain die-hard refusal to face facts. >>The wavelength, on the other hand, will not be a constant if the observer >>is in motion with respect to the source. > >what causes the phenomenon we call 'wavelength' in single photons? You can get it from the de Broglie relationship and the fact that a photon's momentum and energy are interrelated. >>As far as I know, over 100 years of observations confirm this. >>Do you have any data that invalidates this? > >Doppler shift is caused by varying relative light speed > >Do you know of anyone who has observed a doppler shift in a gamma particle? Doppler shifting is how the Mossbauer Effect works. --- David Cross dcross1 AT shaw DOT ca
From: George Dishman on 5 Apr 2005 18:51 Henri, there's a lot we have agreed. I've trimmed most out to get the size down. I'm going to try to bottom out this angle business, I think we are almost saying the same thing and it just needs a little clarification. "Henri Wilson" <H@..> wrote in message news:te4u41t0rg5so58japss9scbcsn9njcjcg(a)4ax.com... > 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: <much snipped> >>>> 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? Each is a source of spherical wavefronts, just the distance between them changes from diagram to diagram. Of course the angle changes as a function of the point on the screen within each diagram but there is no change of angle of the laser beams. Each has to be spread wide enough to illuminate most of the screen anyway so the term "angle between the beams" isn't very well defined. Let me try another way to express this, perhaps you can agree this approach instead. I'll be a bit vague about the details but hopefully you'll allow me a little leeway. Take a point source S and a screen and create fringes by reflecting the source off a mirror M onto the screen as well as the direct beam. I've shown the beams as lines though they would need to illuminate most of the screen as usual: X S /| / | M / | R__|/ | |\ | \ | \A| \| ---+-- D screen There is a detector at point D on the screen and the reflected beam hits the mirror at point R and the angle between the beams is A at that point. So far, so good. Now suppose we move and tilt the mirror such that the point where the light reflects moves along an ellipse with S and D as its foci. The path length from S via R to D will remain the same but the angle A will change. What I am saying is that the brightness at the detector will be unchanged even though the angle is changing. What will be affected is the fringe spacing either side of the detector but that is because the path length anywhere else cannot be held constant as well as that to point D. Now consider the converse, suppose we move the mirror along the line from D to X (top left) again tilting the mirror as required to keep the point of reflection exactly on the X-D line. This time there will be a change of brightness at D due to the change of path length even though the angle A is being held constant. <more snipped> >>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'm sure we are close to a mutual understanding. >>>>>>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. OK, I have been trying out some new software that makes it easy to write Java simulations and I'd like to try it out if you are willing. There is a simulation of the angles here: http://www.briar.demon.co.uk/Henri/SagnacAngles.html Can you let me know if you can see it or whether you get some horrible warning please. It is a Java applet so in theory cannot do any harm as it is stuck in a 'sandbox' with no access to your system, but some ISPs like AOL are a bit overzealous at times so you might get some sort of warnings. I don't, but then I wrote it :-) If it works, switch on one beam at a time using the check boxes and move the location of the detector using the slider. Wiggle it about and see if you agree with the beam paths. Then switch both on and look at the angle between the beams at the detector. Let me know what you think. <much snipped> > but you are still basing that on the assumption that both beams are > defleted in > the same direction. They are not. Let's see if the simulation can sort this out. <more snips> >>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. I am part way through writing another sim that can form a basis for discussion on the rest of your post but I'd like to know if I'm wasting my time and you cannot see these applets and sorting out the angles first will be useful too. I'll reply to the rest next time. George p.s. I should add a circle to the diagram showing the edge of the turntable but I haven't found out how to do that yet :-(
From: bz on 5 Apr 2005 20:30 "PD" <pdraper(a)yahoo.com> wrote in news:1112731023.977565.318940 @z14g2000cwz.googlegroups.com: > If what you say were true, that the wavelength stayed the same but the > speed and frequency changed, then a blue line shifted to green by the > Doppler effect would emerge from the grating at exactly the same angle > as the unshifted blue line. Why? Because, if what you say were true, > the wavelength would be the same and the spacing of the etching would > be the same, so the ratio of those two distances would be the same. > > GOOD argument! Now, tell it to kenseto, he ALSO thinks the doppler shift is due to speed changes and the wavelenght stays the same. -- 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 6 Apr 2005 01:00
On Tue, 5 Apr 2005 23:51:10 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: >Henri, there's a lot we have agreed. I've >trimmed most out to get the size down. > >I'm going to try to bottom out this angle >business, I think we are almost saying the >same thing and it just needs a little >clarification. OK, it isn't all that easy to understand this. >>> >>>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? > >Each is a source of spherical wavefronts, just >the distance between them changes from diagram >to diagram. Of course the angle changes as a >function of the point on the screen within each >diagram but there is no change of angle of the >laser beams. Each has to be spread wide enough >to illuminate most of the screen anyway so the >term "angle between the beams" isn't very well >defined. That's correct. There is a certain amount of dispersion present in the beams. In my sagnac model, there is both sideways displacement AND angular change. > >Let me try another way to express this, perhaps >you can agree this approach instead. I'll be a >bit vague about the details but hopefully you'll >allow me a little leeway. Take a point source S >and a screen and create fringes by reflecting >the source off a mirror M onto the screen as >well as the direct beam. I've shown the beams >as lines though they would need to illuminate >most of the screen as usual: > > > X S > /| > / | > M / | > R__|/ | > |\ | > \ | > \A| > \| > ---+-- > D > screen > >There is a detector at point D on the screen and >the reflected beam hits the mirror at point R >and the angle between the beams is A at that >point. > >So far, so good. > >Now suppose we move and tilt the mirror such that >the point where the light reflects moves along an >ellipse with S and D as its foci. The path length >from S via R to D will remain the same but the >angle A will change. What I am saying is that the >brightness at the detector will be unchanged even >though the angle is changing. What will be >affected is the fringe spacing either side of the >detector but that is because the path length >anywhere else cannot be held constant as well as >that to point D. > >Now consider the converse, suppose we move the >mirror along the line from D to X (top left) >again tilting the mirror as required to keep >the point of reflection exactly on the X-D line. >This time there will be a change of brightness >at D due to the change of path length even though >the angle A is being held constant. I undestand the experiment but I cannot see how it relates to the problem. > ><more snipped> > >>>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'm sure we are close to a mutual understanding. Not all that close yet. > > >>>>>>>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. > >OK, I have been trying out some new software >that makes it easy to write Java simulations >and I'd like to try it out if you are willing. Java is terrible. What is the program? >There is a simulation of the angles here: > >http://www.briar.demon.co.uk/Henri/SagnacAngles.html > >Can you let me know if you can see it or >whether you get some horrible warning please. >It is a Java applet so in theory cannot do >any harm as it is stuck in a 'sandbox' with >no access to your system, but some ISPs like >AOL are a bit overzealous at times so you >might get some sort of warnings. I don't, but >then I wrote it :-) No, only the text appears. > >If it works, switch on one beam at a time >using the check boxes and move the location >of the detector using the slider. Wiggle it >about and see if you agree with the beam >paths. Then switch both on and look at the >angle between the beams at the detector. Let >me know what you think. Have another attempt. I gave up on Java. There has to be an easier way. Visual basic is SO simple. > ><much snipped> > >> but you are still basing that on the assumption that both beams are >> defleted in >> the same direction. They are not. > >Let's see if the simulation can sort this out. > ><more snips> > >>>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. > >I am part way through writing another sim that >can form a basis for discussion on the rest of >your post but I'd like to know if I'm wasting >my time and you cannot see these applets and >sorting out the angles first will be useful >too. I'll reply to the rest next time. > >George >p.s. I should add a circle to the diagram >showing the edge of the turntable but I haven't >found out how to do that yet :-( > Keep at it George. You'll get there eventually. 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. |