How can the MMx math be corrected?
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From: Tom Roberts on 10 May 2010 19:55 Da Do Ron Ron wrote: > [from a Roberts prior post] >> It boggles the mind to think you don't accept that two legs have >> identical lengths if when you hold them up next to each other their >> ends line up [or when a ruler measures them as equal]. > > Let's try to attain some common ground by using the math: OK. But please remember I am using the context of SR, not some unspecified and nebulous "aether" theory. And I am using SR as known in 2010, not as discussed in 1905-20 (the meanings of some important words have changed since then). > The horizontal leg time is t = (2L/c)(1/(1-v^2/c^2)), whereas the > vertical leg time is t' = (2L'/c)(1/sqrt(1-v^2/c^2)), a time that > is shorter than t. (L = horiz. leg, L' = vert. leg) > > You seem to be saying that L = L'. Not with those meanings. Where do you get a "vertical" leg???? The MMx CLEARLY has none. Let me change to a more standard modern notation: use primes for different frames, and use 1 and 2 for the two arms. In the rest frame of the interferometer, L1 = L2 by construction, so: t1 = 2*L1/c, t2 = 2*L2/c, and thus t1 = t2. Here L1 is the length of arm 1 in this frame, c is the speed of light (in any inertial frame including this one), and t1 is the round-trip travel time for light in arm 1; 1->2 for arm 2. Note there is no "v", because the interferometer is at rest in this frame. Note I did not specify the orientation of the arms, because these equations are valid for any orientation (including vertical, as long as it is less than a few dozen meters long) -- it is the orientation independence that really shows this is a null result (i.e. no fringe movement as the orientation changes). Note also the absence of any primes -- I am using one AND ONLY ONE frame for the analysis of a given measurement. One must use different frames for different measurements because the equations I used require one to use the rest frame of the interferometer. No matter, because the prediction is for a null result, which is INDEPENDENT of frame in two different ways: one can analyze any single measurement in any frame and obtain the same null result [#]; one can place the interferometer at rest in any frame and apply this analysis to obtain the null result. [#] Using any other frame the analysis is much more complicated. But since the fringe position is a measurement, it is guaranteed that the result is independent of the analysis frame. > However, if L = L', then the travel times are different, which is > a positive result, not a null result. > And, as we all know, the MMx had a null result. You are attempting to apply my comments from the context of SR in the rest frame of the interferometer, to your approach which is not SR and does not seem to be at rest. So it's no wonder you are confused. > [gleaned from Roberts' current comments] >> Irrelevant -- there is no "outside viewpoint" here. > > Einstein disagrees. No, he didn't. He makes no mention of "outside viewpoint" in your quote. Moreover, no matter which frame is used, one must look "inside" the interferometer to measure its legs. > Thus for a co-ordinate system moving with the earth the mirror system > of Michelson and Morley is not shortened, but it is shortened for a > co-ordinate system which is at rest relatively to the sun. > http://www.bartleby.com/173/16.html Well DUH!!!! In the last phrase Einstein was using a different frame than I was. And he was also using older meanings of words while speaking in a NON-TECHNICAL book (i.e. one intended for general audiences, so one cannot expect him to list every nuance and caveat). We have LEARNED since then that one MUST be more precise in terminology that this. Part of your confusion is the misleading intermixing of similar-looking but rather different phrases in this thread, along with your inaccurate reading that does not differentiate among them: "length" (by itself) is rather too ambiguous to know what is meant. For instance, I have no idea how to measure "length", or even what it means, but I do know how to measure the "length of an object", and its meaning. "length of an arm" can only mean the INTRINSIC length of the arm; its proper length. In particular, no frame is mentioned, so this cannot possibly depend on frame (in keeping with historical meanings). Yes, all too many authors are quite loose about this -- that's part of what I meant above by the need to be more precise in terminology. "length of an arm measured in frame S" means a measurement in frame S, and if S is moving relative to the arm then the value can be different from the proper length of the arm. Note also this is a PUN on "length" as this is not the usual meaning of that word (the previous paragraph gives the usual meaning of "length of an arm"). When Einstein said "shortened" in that quote, he was clearly discussing "length of the arm in frame S", where he mentioned two different frames for S. But he is speaking too loosely for anyone to make solid conclusions from his words. Many people around here insist on speaking excessively loosely, and then wonder why they are so confused. Indeed, many are so confused that they don't even realize they are confused. Such as yourself. Tom Roberts
From: Paul Stowe on 10 May 2010 20:33 On May 10, 8:09 am, Tom Roberts <tjrob...(a)sbcglobal.net> wrote: > Paul Stowe wrote: > > On May 8, 4:17 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > >> Paul Stowe wrote: > > >>> Detection of absolute motion is easy, > >> This depends STRONGLY on what one means by "absolute motion". > >> But if one means this, then it is not easy at all, and has never been done: > >> * motion relative to a frame that participates in the dynamics > >> (e.g. an "aether frame") > > > What the heck does 'frame that participates in the dynamics' mean? > > It means just what it says: a unique frame that in some way or other appears in > the dynamical equations (here presumably of light propagation). ANY aether with > a rest frame would be in this class. You make no sense... > Maxwell's original theory has such a frame; the modern synthesis > called classical electrodynamics does not. > > > Light itself 'participates in the dynamics' of itself... > > No. You are mixing up model and world. Dynamics is a MODEL. Funny, here I thought physics was about 'the world, universe, & everything'... If we get it right there is no difference... > > By any > > reasonable definition of aether a uniformly distributed light source > > would 'illuminated' the aether rest frame.. > > How? Why? What do you mean? A searchlight "illuminated" the clouds -- why is it > we can see the clouds but not the aether? Indeed, why is it we cannot detect the > aether by any means whatsoever (or at least have not yet done so)? Your ignorance is not my problem. We detect it all the time, in every measurement we make. But you should know that light, like sound in other mediums, is the native waveforms of the aether medium. > The ether of LET cannot be "illuminated" or observed in any way. Again, light IS! the illuminator of LET. The CMB isn't 'at the horizon' it is uniformly distributed throughout. Funny ho9w the frequency of this 'just happens' to be the mass of an lowly electron. > > If one exists, then > > that's all you need to determine absolute motion, period! > > That's a mighty big "if". And one not in evidence -- there has not been a single > observation of any aether or any aether rest frame (counting only those that a > scientist would accept as being both reliable and significant). Just because 'you' are in denial doesn't mean others cannot see that established aether theory had accounted properly for the behavior of many things, like Maxwell's model for the actual characteristics of electric and magnetic effects, light, the density & modulus of the medium, etc. In fact, his model was the first quantum superfluid model. > >> One of your problems is that you intermix these different meanings; such PUNs > >> destroy your arguments. > > > No Tom, I am consistent in the use of the meaning of words and > > phrases... I do not redefine them to suit my fancy... > > Yes, you do, but are completely unaware of doing so. You repeatedly intermix > motion wrt the CMBR with motion wrt the "aether". This is your problem, not mine. Waves in a medium like phonons in superfluids are part & parcel of the medium. Compressible mediums have a natural background acoustical signatiure that is never zero... Since aether is, by definition, a medium why try to deny that light (its sound) is a natural characteristic that would, for a uniformly distribution, only be without a dipole when one were 'at rest' in the medium??? The problem isn't mine, it never was. > >>> All one has to do is open their eyes and look around :)... > >> First one must learn to think and speak more accurately, so one knows what one > >> is looking for. > > > Indeed you do... > > Again, your problem, not mine. Your denial of basic aether characteristics is irrational, thus it really is, your problem not mine Tom. Paul Stowe
From: Inertial on 10 May 2010 22:04 "Paul Stowe" <theaetherist(a)gmail.com> wrote in message news:cbde034a-ab2f-483c-bfe8-7ad8d44f435a(a)z13g2000prh.googlegroups.com... > Your denial of basic aether characteristics is irrational, thus it > really is, your problem not mine Tom. What are the basic aether characteristics How do we observe and measure them? What are the properties of the aether .. mass? charge? density? viscosity?
From: Paul Stowe on 10 May 2010 23:01 On May 10, 7:04 pm, "Inertial" <relativ...(a)rest.com> wrote: > "PaulStowe" <theaether...(a)gmail.com> wrote in message > > news:cbde034a-ab2f-483c-bfe8-7ad8d44f435a(a)z13g2000prh.googlegroups.com... > > > Your denial of basic aether characteristics is irrational, thus it > > really is, your problem not mine Tom. > > What are the basic aether characteristics > > How do we observe and measure them? > > What are the properties of the aether .. mass? charge? density? > viscosity? In the classical model of aether which includes the efforts of Kelvin, Helmholtz, Maxwell, Lorentz, Ohms, Gauss, Green, and many, many others the basic properties of the aether medium are, in SI system of measure, 1. a vortex momentum quantum (P) 5.15173E-27 kg-m/sec 2. a vortex distance quantum (L) 6.43092E-08 m 3. a wave propagation velocity (c) 2.99792E+08 m/sec From these primitives all constant of nature can be derived including density, modulus, charge, action, ... etc. Paul Stowe
From: Inertial on 10 May 2010 23:55
"Paul Stowe" <theaetherist(a)gmail.com> wrote in message news:b73c1b0b-5f5d-4b66-b8b6-0bf2a4fb3abe(a)t14g2000prm.googlegroups.com... > On May 10, 7:04 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "PaulStowe" <theaether...(a)gmail.com> wrote in message >> >> news:cbde034a-ab2f-483c-bfe8-7ad8d44f435a(a)z13g2000prh.googlegroups.com... >> >> > Your denial of basic aether characteristics is irrational, thus it >> > really is, your problem not mine Tom. >> >> What are the basic aether characteristics >> >> How do we observe and measure them? >> >> What are the properties of the aether .. mass? charge? density? >> viscosity? > > In the classical model of aether which includes the efforts of Kelvin, > Helmholtz, Maxwell, Lorentz, Ohms, Gauss, Green, and many, many others Name dropping doesn't help > the basic properties of the aether medium are, in SI system of > measure, > > 1. a vortex momentum quantum (P) 5.15173E-27 kg-m/sec > 2. a vortex distance quantum (L) 6.43092E-08 m > 3. a wave propagation velocity (c) 2.99792E+08 m/sec So basically, because (for example) that we know that light travels at c, then you ascribe the propagation velocity of aether to be that, an then claim that that is evidence that the aether exists? > From these primitives all constant of nature can be derived including > density, modulus, charge, action, ... etc. So .. what are they? How do we know the aether exists? That light propagates is not a reason, as theories preit light propagation without an aether. How can we detect it? How is it any different to the empty spacetime of SR/GR? If its not any different, then how can we say it exists? |