Ballistic Theory, Progress report...Suitable for 5yo Kids
in [Relativity]
|
Prev: OWLS is not equal to c
Next: Mathematical Inconsistencies in Einstein's Derivation of the Lorentz Transformation
From: Timo Nieminen on 22 Sep 2005 20:25 On Thu, 22 Sep 2005, it was written: > Timo Nieminen wrote: >> >> Dirt cheap compared to any space-based experiment. Emission theories mean >> that the Maxwell equations are wrong. > > No they don't. > Maxwell's equations apply to an aether. At best, they describe the speed of > light EMITTED by the observer who measures the two constants. > They say nothing about the speed of light from moving sources. > They say nothing about light speed in 'no aether'. So you are saying that the Maxwell equations don't describe electromagnetic phenomena correctly? OK, all you need to do is figure out exactly how your theory of electromagnetism differs from classical EM theory, find where the difference is experimentally accessible, and do the experiment (moving source, maybe - what does your theory of EM say about synchrotron radiation, Brehmsstrahlung, Cerenkov radiation etc?). If you want your theory to be tested, go for it. [cut] >> Basically, will extinction by the interplanetary medium mean that such an >> experiment would be useless? If you can't give a definite answer, then can >> you convince a donor to fund an experiment that might well be useless? > > There is probably enough 'extinction free' space between earth and moon to make > this work. However an unlikely null result would not rule out the BaT. So, you say that a null result would be useless. Have fun trying to get somebody to provide the funding! Can't you provide a more quantitative answer than "probably"? >>> I have realised that extinction (or 'speed unification') would also cause >>> corresponding errors in radial speed measurements and estimations. >>> These is normally based on observed doppler shifts. If speed is unified, then >>> so too would be doppler shift. >> >> Why? Emission theories, special relativity, and Galilean ether theories >> all give the same 1st order Doppler shift, assuming that light is an >> electromagnetic wave with frequency given by the rate of oscillation of >> fields. If your theory has a different mechanism for Doppler shift, then, >> great! That's even more deviations from currently accepted classical >> electrodynamics, so it should be even easier to do a cheap tabletop >> experiment. > > At practical speeds, the Doppler equation is virtually the same for all three > theories. So now you are saying that the Doppler shift _won't_ be unified? -- Timo
From: Eric Gisse on 22 Sep 2005 21:42 Timo Nieminen wrote: > On Thu, 22 Sep 2005, it was written: > > > Timo Nieminen wrote: > >> > >> Dirt cheap compared to any space-based experiment. Emission theories mean > >> that the Maxwell equations are wrong. > > > > No they don't. > > Maxwell's equations apply to an aether. At best, they describe the speed of > > light EMITTED by the observer who measures the two constants. > > They say nothing about the speed of light from moving sources. > > They say nothing about light speed in 'no aether'. > > So you are saying that the Maxwell equations don't describe > electromagnetic phenomena correctly? OK, all you need to do is figure out > exactly how your theory of electromagnetism differs from classical EM > theory, find where the difference is experimentally accessible, and do the > experiment (moving source, maybe - what does your theory of EM say about > synchrotron radiation, Brehmsstrahlung, Cerenkov radiation etc?). Henri does not do experiments, I have asked. He says he is an "idea man". Henri does not derive anything, I have asked. His response is that he is "too busy". Obviously not busy enough to carry on multiple conversations, however. > > If you want your theory to be tested, go for it. Reality is a scary place, Henri likes to isolate himself from it. [snippy] Don't take him seriously, he can't back back up anything he says.
From: jgreen on 23 Sep 2005 07:00 George Dishman wrote: > <jgreen(a)seol.net.au> wrote in message > news:1127004753.829320.91010(a)z14g2000cwz.googlegroups.com... > > > > George Dishman wrote: > So far you haven't shown anything wrong with it. > If you want to do that, there have been two recent > papers which you could have cited that you haven't > even noticed, you are just going back to repeating > an error you made months ago when you misread a > post in the group. You seem to have forgotten you > got it wrong and only remember your glee at an > imagined problem. That would be the one where stars 8Gya old have been seen at 12.7 TERMINAL CANCER!!!!!!!!! Jim
From: George Dishman on 23 Sep 2005 07:12 <jgreen(a)seol.net.au> wrote in message news:1127473221.176343.176140(a)f14g2000cwb.googlegroups.com... > > George Dishman wrote: >> <jgreen(a)seol.net.au> wrote in message >> news:1127004753.829320.91010(a)z14g2000cwz.googlegroups.com... >> > >> > George Dishman wrote: >> So far you haven't shown anything wrong with it. >> If you want to do that, there have been two recent >> papers which you could have cited that you haven't >> even noticed, you are just going back to repeating >> an error you made months ago when you misread a >> post in the group. You seem to have forgotten you >> got it wrong and only remember your glee at an >> imagined problem. > > That would be the one where stars 8Gya old have been seen at 12.7 No such paper exists. > TERMINAL CANCER!!!!!!!!! Nope, it's the one that actually said the galaxies were less than 2Ga old and 8Ga _away_ which you misread as saying they were 8Ga _old_. Still that suits your bias so I guess accuracy doesn't matter to you as long as you can bolster your fantasy. George
From: Paul B. Andersen on 23 Sep 2005 07:47
Henri Wilson wrote: > On Tue, 20 Sep 2005 14:24:41 +0200, "Paul B. Andersen" > <paul.b.andersen(a)deletethishia.no> wrote: > > >>Henri Wilson wrote: >> >>>On Sun, 18 Sep 2005 21:50:03 +0200, "Paul B. Andersen" >>><paul.b.andersen(a)deletethishia.no> wrote: >>> > > >>>The fringe shifts are caused by he different angles of approach by both beams >>>at the eyepiece. >>>Light has its own built-in gyro in the form of an 'axis'. >> >>Don't obfuscate the matter with nonsense, please. >> >> >>>>Sagnac falsifies the ballistic theory. >>> >>> >>>To the unwary it might. >>> >>> >>> >>>>>>>It does not rfute the BaT because the light emitted by the source is moving >>>>>>>normal to hte next mirror IN THAT MIRROR'S FRAME. It is NOT moving at c+v wrt >>>>>>>that mirror at all. >>>>>> >>>>>>Well said. >>>>>>That's why Sagnac falsifies BaT. >>>>> >>>>> >>>>>It has nothing to do with the BaT. >>>>>Each member is moving at right angles to the next member in the frame of that >>>>>next member. >>>> >>>>What has nothing to do with the ballistic theory? >>>>Because - as you correctly state - the light according >>>>to the ballistic theory is NOT moving at c+v wrt that mirror >>>>at all, but is moving with the speed c in the mirror frame, >>>>will the light according to the ballistic theory use the same >>>>time in either direction regardless of the rotation of the mirror >>>>frame. So the ballistic theory predicts no Sagnac effect. >>> >>> >>>The ballistic theory does not encompass the sagnac effect. >> >>What a strange way to state that >>the ballistic theory predicts no Sagnac effect. :-) >> >> >>>>Have the wrong predictions of the ballistic theory >>>>nothing with the ballistic theory to do? :-) >>>> >>>>But I think I got your point. >>>>It is that the motion of mirrors will affect the light >>>>path drawn in the mirror frame regardless of which >>>>theory you use to explain it, and that it is this that >>>>is responsible for the Sagnac effect. >>>>It is true that such an effect exists, but this effect >>>>is NOT the Sagnac effect. >>>> >>>>So what is this effect? >>>>Let's draw the light path between the mirrors >>>>when the ring is not rotating. The light path is >>>>then the same for both beams. >>>> >>>> / \ >>>>/-----------\ >>>>/| |\ >>>>| | >>>>| | >>>>| | >>>>\| |/ >>>>\-----------/ >>>> \ / >>>> >>>>So what happens when the ring is rotating? >>>>(Remember that we are talking about the shape of the >>>>light paths as viewed in the mirror frame.) >>>>The light paths will not be the same in both directions. >>>>The light path of the beam going in the same direction >>>>as the rotation will be slightly curved inwards >>>>(concave) while the light path of the light beam going >>>>in the opposite direction will be slightly curved outwards >>>>(convex). So the lengths of the light paths will be very slightly >>>>longer than when it is not rotating, and the difference of >>>>the lengths of the two light paths will be extremely little different. >>>>But this effect is extremely small, it is no first order effect, >>>>like the Sagnac effect is. >>> >>> >>>The angle between the two beams when they reunite is considerable. >>>It is that angle which causes the fringe shift. >> >>Nonsense. >>Even utter nonsense. >> >>You seem to be completely ignorant of how an >>interference pattern is formed, and why fringes shifts. > > > I'm not. Yes, you are. Which you prove yet again in this posting. >>Please take the time to read the following properly, >>I am using time to write it. >> >>To get an interference pattern with fringes, >>the beams must be diverging and overlapping. >> >>Consider this simple figure: >> >> 1 2 >> * * Two correlated (in phase) point sources >> emitting monochromatic, coherent light. >> (Laser and a beam splitter) >> >> >> >>--|--|--|---- screen >> A B C >> >>The point B is equidistant to source 1 and 2. >>We get a bright fringe through B. The fringe >>will be a straight line. >>The distance from the point A to point 2 is >>half a wavelength longer than the distance >>to point 1. We get a dark fringe through A. >>This fringe will be a bit curved. >>Likewise for point C, a dark fringe. > > > Note that the angle to the observer's eyepiece is changing as you go from > A->B->C. I typical Wilsonian nonsensical comment, and a strong indication that you have no idea of what you are talking about. You don't "go from A->B->C". You don't move the screen, and you don't move the eyepiece, so what the hell is your comment supposed to mean? I would advice you to forget the eyepiece. Think that the interference pattern is formed on a screen. Using an eyepiece doesn't change the principle. All it does is to project the light into your eye so that the interference pattern is formed on your retina. > >>Note that the reason why there are fringes >>at all is that the beams are diverging, so >>the distance from the source to the screen >>is different on different parts of the screen. >> >>The angle of the beams when they unite >>has obviously nothing whatsoever with >>the matter to do. The only thing that matters >>is the difference in the light path lengths >>to the two sources measured in wavelengths. > > > In tyhe sagnac, the two path lengths are different, whether or not each > component's sped relative to the next IS considered. You are babbling. I will take the typos as an indication that you are writing without thinking. >>The only way to make the fringes move, >>is to change the distance to one of the sources. >>If we move source 1 a bit upwards, the fringes >>will move to the left, and vice versa. > > > If the two beams reunite at different angles, it is implied that they have > traveled diferent lengths. Nonsense. Read below. >>So when fringes move, the difference between >>the two path lengths measured in wavelengths >>changes. >> >>Now let this "interferometer" rotate. >>Observed in the interferometer frame, >>the light paths will be slightly curved, >>so the angle with which the light hits >>the screen is slightly altered. >>But the fringes will not move, because >>the slightly curved light paths from >>point 1 to B and point 2 to B will still >>be equally long. The angle at which the beams >>hit the screen is utterly irrelevant. > > > No. Yes. Read below. >>And please don't say something like >>"the beam will no longer hit point B, >>because it is deflected." This IS what you are saying. So read carefully: >>That is irrelevant. The beams are diverging >>and overlapping, and what happens in point B >>depend only on the lengths of the paths of >>the light that hit point B, obviously. THINK about it. Did you get it? >>The same applies for the four mirror set up. >>It is stupid to say something like "the two >>contra going beams will no longer combine >>at the same point on the mirror." Henri, read carefully: >>Some light will always hit at "the midpoint" >>of the combining mirror, and what happens in >>that point is only determined by the phase >>difference of the two light paths that >>actually hit that point. If the fringes shifts, >>it means that the phase difference changes, >>which only can mean that the length difference >>of the light paths have changed. > > > Paul, George and I discused this at length. I have an animation of the way the > beams move. I am content to believe that I won the argument. You have obviously not even understood how an interefrence pattern is formed. I don't think you read my posting properly. Please read it agin, carefully. >>The ballistic theory predicts no length difference >>of the light paths (measured in wavelengths) >>and thus no fringe shifts when the Sagnac ring rotates. >> >>But the fringes do shift when the Sagnac ring rotates. >> >>Sagnac falsifies the ballistic theory. >> >>No other conclusion is possible. > > > Rubbish. It is the fact whether you understand it or not. "The ballistic theory predicts no length difference of the light paths (measured in wavelengths)." So far you agree. "..and thus no fringe shifts when the Sagnac ring rotates." Here is where you differ. But you cannot explain this away by claiming that interferometers work in an entirely different way than they actually do. Interferometers are used in a lot of different applications, and they have been used for centuries. It is very well know how they work. Disputing this is idiocy. You are not even able to state what your alternative explanation for how interferometers work, is. You are only babbling some incoherent nonsense about changing angles. You are indeed living in Wonderland. To us in the real world, interferometers are no mystery. We know how they work. Paul |