How can the MMx math be corrected?
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From: Paul Stowe on 1 May 2010 12:27 On May 1, 9:09 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 1, 10:54 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > > > On May 1, 8:09 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 1, 9:59 am, PaulStowe<theaether...(a)gmail.com> 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. > > > > > PaulStowe > > > > So let's recap: > > > Given two derivations of a common mathematical form, you feel free to > > > interpret a term in the mathematical form to point to the premise of > > > one derivation over the other, even though they both generate the same > > > form from strict derivational deduction. Furthermore, you feel free to > > > choose the premise of one of those derivations as being favored, > > > because you like it better. > > > Concretely, given a derivation that assumes isotropy of the speed of > > > light and a derivation that assumes the anisotropy of the speed of > > > light, and given that both derivations produce the IDENTICAL > > > mathematical form, it seems obvious to you that the one that assumes > > > the anisotropy of the speed of light is the correct one, by inspection > > > of the mathematical form. > > > Hmmm.... > > > I guess I missed it, where does Einstein derive the gamma factor from > > assumed isotropy? What I see is, > > > "...it being borne in mind that light is always propagated along > > these axes, when viewed from the stationary system, with the > > velocity Sqrt(c^2 - v^2)..." > > > For any v > 0 that equation gives us a light velocity not equal to c > > in the stationary system. YES! c 'appears' to measure as invariant > > but, if we take that quote of Einstein verbatim it certainly IS NOT > > ISOTROPIC. It looks to me that he derived gamma from assumed > > anisotropy when moving. > > Uh, no. So if you don't understand how relativity was derived from an > assumption of isotropy, just say so. > It's actually pretty clear from the original paper in 1905. > > > > > Yes, both versions, Einstein's and Lorentz/Poincare's predict the same > > measurables but perhaps you can show how the LT gets derived from > > assumed isotropy. > > You really need it trotted out here? Would you like a reference? > It's done in the 1905 paper by Einstein. > It's done a slightly different way but from the same premises by > Taylor & Wheeler in Spacetime Physics, pgs 95-102. > It's done in yet a slightly different way but from the same premises > by Griffiths in his Intro to Electrodynamics, chapter 12. > Need more? Sigh, the quote provided WAS FROM Einstein's 1905 paper. And Einstein's postulate that light speed is invariant and not affected by the speed of the emitter/receiver pretty much negates isotropy. I don't argue with his other postulate either (nor did Lorentz or Poincare) one can indeed, set up their system of measure in such a way as to make it appear to be 'isotropic' for them. However, the very existence, and need for, the LT in transforming coordinates between moving systems along with the postulate about light speed invariance 'should be' indisputable proof that Poincare & Lorentz's take on this is the right one. Otherwise, my original question remains, Paul Stowe
From: Tom Roberts on 1 May 2010 13:14 Paul Stowe wrote: > On Apr 29, 11:51 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> Summary: in SR the speed of light in any inertial frame is >> isotropically c, [...] > > If light is always isotropic then what is the basis for the roots (1 - > v/c) & (1 + (v/c) in (1 - [v/c]^2)? That is simple and basic algebra; light has nothing whatsoever to do with the fact that the latter expression can be factored into the other two expressions you gave. If you wonder where the gamma factor in the Lorentz transform comes from, it is again basic math -- the set of transforms between inertial frames must form a group, and basic group theory restricts such transforms to three groups: the Euclid group, the Galilei group, and the Poincar� group. The algebraic formulas for each group (using Cartesian coordinates) are fully determined by group theory. So the form of the gamma factor comes from the requirement of logical consistency among the transforms. Which of the three groups applies in the world we inhabit is an EXPERIMENTAL issue. Experiments overwhelmingly show that the Poincar� group is the one that applies in our world. This derivation of the Lorentz transform came much later than 1904-5. SR inaugurated a new recognition of the power of such symmetries in fundamental theories of physics. We do not know why nature chooses to obey such symmetries, but the observed fact that she does is an ENORMOUS simplification in theoretical physics. > In Lorentz's version it's logically explained, [...] Boy do you have a warped sense of "logically explained"! Lorentz merely STIPULATED the form of his "change of variables", with no justification whatsoever, and then showed that it worked to give the same Maxwell's equations in the moving frame as in the ether frame. One must correct an error in his 1904 paper. It did not affect his presentation, because he considered only vacuum and his error was in charge density -- \rho = 0 hides the mistake. Poincar� got it right. Tom Roberts
From: Tom Roberts on 1 May 2010 13:34 Paul Stowe wrote: > So, there's a conflict between the 'assumption' and mathematical > form. There's no such "conflict". In the 1-d Lorentz transform you are discussing, the terms (1-v/c) and (1+v/c) appear only MULTIPLIED TOGETHER. That equals (1-(v/c)^2) which is isotropic. Duh! But you really need to look at the FULL formula in which v is a 3-vector. The formula for 1-d transforms you are using is INCOMPLETE, and "isotropy" has quite different meanings in 1-d and 3-d. In the correct formula, in which v is a 3-vector, (1-(v/c)^2) cannot be "factored" as you claim, so your entire claim is just plain wrong. > I'll take the math over 'assumption' every time. It's not about what "you'll take", but rather about the structure of the ACTUAL theory. You are attempting to discuss the wrong thing -- projecting the 3-d theory onto 1-d removes essential aspects of the ACTUAL theory, giving you a false sense of the structure. > 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. Hmmm. Poincar�'s "version" is incomplete, and not really SR (he logically derived the invariance group of Maxwell's equations); Lorentz's version is not "logical" at all -- he simply stated his "change of variables" without any justification whatsoever. But that's all old hat -- the modern foundation of SR is group theory (loosely: following Poincar�'s approach in a more general context). Your arguing about ancient papers is irrelevant to PHYSICS. The rest of us live in 2010, not 1900-7. Tom Roberts
From: Tom Roberts on 1 May 2010 14:56 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
From: Tom Roberts on 1 May 2010 15:11
Paul Stowe wrote: > Sigh, the quote provided WAS FROM Einstein's 1905 paper. And > Einstein's postulate that light speed is invariant and not affected by > the speed of the emitter/receiver pretty much negates isotropy. Obviously you do not really understand Einstein's 1905 paper. He EXPLICITLY SHOWED that his postulates are self-consistent, and thus do not "negate isotropy". THAT WAS THE MAIN POINT OF THE PAPER! BTW those are NOT Einstein's postulates. You REALLY need to learn how to read more accurately. > However, the very > existence, and need for, the LT in transforming coordinates between > moving systems along with the postulate about light speed invariance > 'should be' indisputable proof that Poincare & Lorentz's take on this > is the right one. Nope. Not to people who actually understand the papers and the underlying issues, and who have learned more than just what was known in 1907 or so (which appears to be the limit of your knowledge, all too much of which you got wrong). To modern physicists, the "right take" is a sort-of synthesis of all the above, using basic group theory. This approach does not use light at all, which is appropriate as SR is more general than just EM phenomena. > Otherwise, my original question remains, Nope. See my recent posts in this thread. Your question is based on your basic misunderstanding of the ACTUAL theory. Tom Roberts |