How can the MMx math be corrected?
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From: Paul Stowe on 1 May 2010 15:12 On May 1, 11:56 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > harald wrote: > > Paul, this started apparently with a misunderstanding by Tom Roberts > > of what DDRR meant; and he insisted in keeping that misunderstanding > > (as MMX is repeated at different times of the year, at least at some > > times the interferometer has a speed v relative to the chosen frame > > and the arms are then *not* identical in the sense that one is shorter > > than the other). > > I don't know what you mean by "DDRR", nor what "misunderstanding" you mean, but > all you need to do is LOOK at the MMX apparatus -- the two arms are indeed > identical in the sense that they have the same length. As I have stated before, > and is implicit in what we mean by "identical", the comparison is performed in > the rest frame of the arms. > > You mention a "chosen frame". The simplest and most obvious inertial frame to > choose is the rest frame of the apparatus. Do this for each and every > measurement (meaning a single rotation of the interferometer around its axis). > SR immediately implies no fringe shift as the interferometer is rotated, and > that's what is observed. > > > Lorentz and Einstein based their derivations on the assumption of an > > isotropic, constant speed of light in whatever "rest" system; that > > stemmed logically from Maxwell's theory and Einstein turned it into a > > postulate. > > Yes, but Einstein used the phrase "rest system" merely as a means of > distinguishing this frame from others, and ANY inertial frame can be used as his > "rest system". This differs from Maxwell and Lorentz, for whom "rest system" > implicitly means the ether frame of their respective theories. The difference is > essential, and profound. > > Tom Roberts Essential yes, as it is to Einstein's work but, in what way practical way is it profound to Lorentz's version?
From: eric gisse on 1 May 2010 15:27 Paul Stowe wrote: > On May 1, 7:45 am, PD <thedraperfam...(a)gmail.com> wrote: >> On Apr 30, 9:31 pm, PaulStowe<theaether...(a)gmail.com> wrote: >> >> >> >> >> >> > On Apr 29, 11:51 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> >> > > Da Do Ron Ron wrote: >> >> > > > [T. Roberts wrote:] >> > > >> SR predicts a null result, with identical legs. >> >> > > > That is physically impossible. >> >> > > Nonsense! This is OBSERVED -- Michelson interferometers with >> > > identical legs give null results (i.e. a fringe shift of zero within >> > > resolutions). >> >> > > > It is also theoretically impossible. >> >> > > More nonsense. It is essentially trivial for SR to predict a null >> > > result for the MMX. Look in any SR textbook. >> >> > > Summary: in SR the speed of light in any inertial frame is >> > > isotropically c, so for a Michelson interferometer with its >> > > center of rotation at rest in any inertial frame, the position >> > > of the fringes is independent of orientation, hence zero fringe >> > > shift as the instrument is rotated. It's easy to show that the >> > > non-inertial effects due to a laboratory on earth are vastly >> > > smaller than the resolution of the best such measurement to date. >> >> > > Tom Roberts >> >> > If light is always isotropic then what is the basis for the roots (1 - >> > v/c) & (1 + (v/c) in (1 - [v/c]^2)? >> >> > In Lorentz's version it's logically explained, and how it works but >> > light isn't isotropic c. BUT!... you can pretend it is by using >> > Einstein's clock synchronization definition and the inherent symmetry >> > of the contraction. >> >> > However, if you 'assume' this the form of the Lorentz transform is >> > just PFA a fudge that just happens to have an asymmetry for its >> > roots. >> >> Oh good heavens, Paul, are you serious? >> If you see a term in an expression that can be written (1-v/c) or (1+v/ >> c), this means to you that anisotropy of the speed of light must be >> implied by the theory? Even if the derivation DIRECTLY STEMS from an >> assumption of the isotropy of the speed of light? You don't see a >> problem with that kind of analysis??? > > So, there's a conflict between the 'assumption' and mathematical > form. I'll take the math over 'assumption' every time. Especially > given that in Poincare/Lorentz version of the derivation it has > logical, derivable, basis. As, you should be aware, that basis isn't > isotropy of light speed. > > Paul Stowe This is precious.
From: Paul Stowe on 1 May 2010 15:57 On May 1, 9:26 am, harald <h...(a)swissonline.ch> wrote: > On May 1, 4:59 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > > More nonsense. It is essentially trivial for SR to predict a null result for the > > > > > MMX. Look in any SR textbook. > > > > > > Summary: in SR the speed of light in any inertial frame is > > > > > isotropically c, so for a Michelson interferometer with its > > > > > center of rotation at rest in any inertial frame, the position > > > > > of the fringes is independent of orientation, hence zero fringe > > > > > shift as the instrument is rotated. It's easy to show that the > > > > > non-inertial effects due to a laboratory on earth are vastly > > > > > smaller than the resolution of the best such measurement to date. > > > > > > Tom Roberts > > > > > If light is always isotropic then what is the basis for the roots (1 - > > > > v/c) & (1 + (v/c) in (1 - [v/c]^2)? > > > > > In Lorentz's version it's logically explained, and how it works but > > > > light isn't isotropic c. BUT!... you can pretend it is by using > > > > Einstein's clock synchronization definition and the inherent symmetry > > > > of the contraction. > > > > > However, if you 'assume' this the form of the Lorentz transform is > > > > just PFA a fudge that just happens to have an asymmetry for its > > > > roots. > > > > Oh good heavens, Paul, are you serious? > > > If you see a term in an expression that can be written (1-v/c) or (1+v/ > > > c), this means to you that anisotropy of the speed of light must be > > > implied by the theory? Even if the derivation DIRECTLY STEMS from an > > > assumption of the isotropy of the speed of light? You don't see a > > > problem with that kind of analysis??? > > > So, there's a conflict between the 'assumption' and mathematical > > form. I'll take the math over 'assumption' every time. Especially > > given that in Poincare/Lorentz version of the derivation it has > > logical, derivable, basis. As, you should be aware, that basis isn't > > isotropy of light speed. > > > Paul Stowe > > Paul, this started apparently with a misunderstanding by Tom Roberts > of what DDRR meant; and he insisted in keeping that misunderstanding > (as MMX is repeated at different times of the year, at least at some > times the interferometer has a speed v relative to the chosen frame > and the arms are then *not* identical in the sense that one is shorter > than the other). Not sure what you mean here since material systems consist of a stable equilibrium lattice of interlocked EM fields those fields will always be asymmeterical because nothing we can access is actually 'at rest'. The problem isn't that because this causes every fiber of every substance to 'be shorter' or not identical in the direction of motion. > Next you added to the confusion by simply talking about "isotropy" of > the speed light, while you probably (and as I now see, certainly) mean > that the one-way closing speed of light is anisotropic.) No, I mean that, c'^2 = c^2 - v^2 and given v is along the x axis, the magnitude of v at any angle t (v') relative to t = 0 in the direction of motion is v' = v Cos t Thus the above equation for any arbitrary plane is, c' = c[Sqrt(1 - [v'/c]^2)] What happens as a result of this physical fact is, that to remain internally consistent, all fields take a form consistent to this, a.k.a., the Lorentz contraction. Since c is, by definition, dx/dt light speed is alway dx'/dt' or, dx[Sqrt(1 - [v'/c]^2)] -------------------- dt[Sqrt(1 - [v'/c]^2)] which, in turn, make c 'appear' invariant since clearly the distortion factors cancel each other out. HOWEVER! the mathematical roots of what is actually happening is rather apparent & obvious if you examine the process. > Usually people > talk in this context about the operational two-way speed of light, > which is supposed to remain isotropic. > And that is again different from what is meant in a general > discussion: > Lorentz and Einstein based their derivations on the assumption of an > isotropic, constant speed of light in whatever "rest" system; that > stemmed logically from Maxwell's theory and Einstein turned it into a > postulate. While Einstein apparently came to realize this many still do not. So, while OWLS is actually anisotropic and experiments like Roland DeWitte's clearly showed this, one can come up with a defined system of measure that can define this away. Lorentz showed how in 1904, nearly a year earlier than Einstein. What's the really sad part of this? That rational people will accept logical inconsistencies, like actually believing that c can actually be isotropic in differently moving systems at the very same time and also be a gobally invariant value which is independent of emission speed. Hell, at least the ballistic advocates understand that both can't be true simulatneously. Regards, Paul Stowe
From: Tom Roberts on 1 May 2010 18:21 Paul Stowe wrote: > On May 1, 11:56 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> Yes, but Einstein used the phrase "rest system" merely as a means of >> distinguishing this frame from others, and ANY inertial frame can be used as his >> "rest system". This differs from Maxwell and Lorentz, for whom "rest system" >> implicitly means the ether frame of their respective theories. The difference is >> essential, and profound. > > Essential yes, as it is to Einstein's work but, in what way practical > way is it profound to Lorentz's version? Look right up there, my last sentence quoted: I said "The DIFFERENCE is essential, and profound [emphasis added]." You need to READ MORE CAREFULLY, because as I said, it is the DIFFERENCE that is profound -- thus it is not "profound to Lorentz's version", but rather what is profound is the COMPARISON between Einstein and Maxwell & Lorentz, and how they use similar phrases in VERY DIFFERENT WAYS. Not only did you miss my meaning completely, you also show that you failed to understand this difference in the original authors' words and approach. Ask yourself whether it really makes sense for you to try to discuss complicated and subtle theories, when you repeatedly make such elementary mistakes in reading. AFAICT most if not all of your questions and confusions around here are due to such simple mistakes in reading (both of original books and papers, and of posts around here). It seems to me that much of what you think you know about relativity is just plain wrong. Unlike many around here, you seem to make an honest effort, but that is not sufficient -- you must learn how to make an ACCURATE effort. Tom Roberts
From: harald on 1 May 2010 19:02
On May 1, 8:56 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > harald wrote: > > Paul, this started apparently with a misunderstanding by Tom Roberts > > of what DDRR meant; and he insisted in keeping that misunderstanding > > (as MMX is repeated at different times of the year, at least at some > > times the interferometer has a speed v relative to the chosen frame > > and the arms are then *not* identical in the sense that one is shorter > > than the other). > > I don't know what you mean by "DDRR", The OP with his funny name. > nor what "misunderstanding" you mean, It should be clear since I spelled it out.. In fact it should have been clear *before* I spelled it out, but I let it to the OP to try to explain to you what he exactly meant... > but > all you need to do is LOOK at the MMX apparatus -- the two arms are indeed > identical in the sense that they have the same length. As I have stated before, > and is implicit in what we mean by "identical", the comparison is performed in > the rest frame of the arms. As I stated before, the essence of the MMX was that the test must be done at different times of the year, in order to guarantee at least some of the time a significant velocity v, according to stationary ether theory. Looking back at it through SRT that translates to at least some of the time a significant velocity v relative to the frame of choice (usually the solar frame, but any inertial frame will do). > You mention a "chosen frame". The simplest and most obvious inertial frame to > choose is the rest frame of the apparatus. Do this for each and every > measurement (meaning a single rotation of the interferometer around its axis). > SR immediately implies no fringe shift as the interferometer is rotated, and > that's what is observed. Again, it's an essential aspect of MMX that it works in different seasons; if you were right then the Earth would be always considered to be "in rest" so that stellar aberration would be zero! > > Lorentz and Einstein based their derivations on the assumption of an > > isotropic, constant speed of light in whatever "rest" system; that > > stemmed logically from Maxwell's theory and Einstein turned it into a > > postulate. > > Yes, but Einstein used the phrase "rest system" merely as a means of > distinguishing this frame from others, and ANY inertial frame can be used as > his "rest system". Yes - as I pointed out. Harald > This differs from Maxwell and Lorentz, for whom "rest system" > implicitly means the ether frame of their respective theories. The difference > is essential, and profound. > > Tom Roberts |