How can the MMx math be corrected?
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From: John Polasek on 30 Apr 2010 14:36 snip >>>>>>>>>> >>>>>>>>>> As I recall the purpose of the MM experiment was to measure the >>>>>>>>>> velocity with respect to ether. The finding was a null, for which >>>>>>>>>> the >>>>>>>>>> logical conclusion would be that there is no measurable ether. QED >>>>>>>>>> >>>>>>>>>> It seems to me there is no place for talk about length contraction >>>>>>>>>> or >>>>>>>>>> time dilation because these adjustments only need to be brought to >>>>>>>>>> bear if you believe that there is an ether and that v is a >>>>>>>>>> measurable >>>>>>>>>> quantity and that the expressions c+v and c-v make sense. I think >>>>>>>>>> it's generally agreed that they don't make sense. >>>>>>>>>> John Polasek >>>>>>>>> >>>>>>>>>No that's wrong too. Over the period of one year (which is the time >>>>>>>>>frame under discussion), relative to any inertial coordinate system >>>>>>>>>there is talk about length contraction as well as c+v and c-v. Time >>>>>>>>>dilation doesn't matter for MMX; however, for that we have KTX as I >>>>>>>>>already pointed out. >>>>>>>>> >>>>>>>>>And although in his 1905 paper Einstein didn't mention MMX directly, >>>>>>>>>Lorentz did so in 1904 and Einstein did the same in his 1907 >>>>>>>>>overview. >>>>>>>>>Moreover, Einstein derives in 1905 the LT by discussing a similar >>>>>>>>>setup as MMX, complete with c+v and c-v; he calls in that paper >>>>>>>>>length >>>>>>>>>contraction "physical". And it is generally agreed that all that >>>>>>>>>does >>>>>>>>>make sense. >>>>>>>>> >>>>>>>>>Cheers, >>>>>>>>>Harald >>>>>>>> You don't really think there is an ether with respect to which v >>>>>>>> could >>>>>>>> be measured but whose presence has been masked by the Lorenz >>>>>>>> transform? >>>>>>>> Why would the Lorenz transform apply here? There's no relative >>>>>>>> motion >>>>>>>> of a second moving frame or anything: the observer is sitting right >>>>>>>> next to the mirrors and the only active phenomenon is the purported >>>>>>>> ether wind that would affect both mirrors. >>>>>>>> The overall conclusion, which is quite generally held, is that no >>>>>>>> fringeshift means no ether wind. It's part of relativity >>>>>>>> Einstein didn't get everything right; the contraction is only an >>>>>>>> apparent one. >>>>>>>> John Polasek >>>>>>> >>>>>>>What "contraction"? >>>>>>>xi = (x-vt)/sqrt((c-v)*(c+v) /c^2) is an expansion. >>>>>>>Obviously you don't understand algebra. >>>>>>> >>>>>>>MMX is done today on a far greater scale than you realise, >>>>>>>and has been for 40 years. >>>>>>> http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment >>>>>>> >>>>>>>The time for laser light to reach the new moon and return is one >>>>>>>leg, and the time for it to reach the moon and return at last quarter >>>>>>>one week later is the other leg of MMX. The Earth-Moon system >>>>>>>moves around the Sun so if Einstein's ridiculous postulate was true >>>>>>>then the light wound travel to the new moon and back at c and to >>>>>>>the quarter moon at c+v, coming back at c-v, where v is the >>>>>>> speed of the Earth-Moon system going around the Sun. This >>>>>>>has the relativistic value of 0.0001c, so someone would notice >>>>>>>it by now. Instead, we measure the distance to the Moon because >>>>>>>the speed is c both ways always. >>>>>>> >>>>>> What are you babbling about? >>>>> >>>>>I'm babbling about asking you a question, what "contraction"? >>>>> >>>>>xi = (x-vt)/sqrt((c-v)*(c+v) /c^2) is an expansion. >>>>> >>>>>Can't you read, cretin? >>>>> >>>>>> Others are speaking of using the Lorentz >>>>>> transform (contraction)-I am not-and they must be doing it to adjust >>>>>> the leg lengths so as to get a null, apparently in the face of an >>>>>> ether wind. >>>>>> There is no ether wind, there is no c+v, it forms the basis of >>>>>> relativity. >>>>> >>>>>What are you babbling about? >>>>>As I said, you are totally incapable of grasping algebra, you babbling >>>>>cretin. >>>>> >>>> I admit that I came in late to the discussion and I have not read all >>>> of the notes and particularly never looked at that verbose equation of >>>> which you appear to be overly fond. Let me simplify it for you: >>>> xi = (x-vt)*gamma >>> >>> What are you babbling about ? >>> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif >>>where >>> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif >>> >>>xi = beta * (x-vt), learn to read the Greek alphabet. >>> >>>beta = 1/ sqrt[{(c-v) / c} * {(c+v) / c}], >>> = 1/ sqrt[(c-v) * (c+v) / c^2], >>> = 1/ sqrt[(c^2 -v^2) / c^2], >>> = 1/ sqrt[c^2/c^2 - v^2/c^2], >>> = 1/ sqrt[1 - v^2/c^2], >> Those conversant in relativity will immediately recognize that, to the >> contrary, beta = v/c, from time immemorial. > >As I previously babbled, you can't read. >Come back when you've passed first grade and learned to click > on a link with a mouse, you stooopid babbling cretin. > > http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif > {click here-------------^^^^^ } > Just Google "gamma + relativity" and at least the first 5 topics (I did not go any further) will echo my definitions of gamma and Beta. I remember this as being true since I first encountered relativity back in the1940's. That knowledge should be part of your kit, as a designated relativity basher. I am surprised by your resilience, in getting back up from the canvas like Rocky Balboa, after repeated rebuttals, but again point out that it appears that your petulance exceeds your competence by a comfortable margin. Oh, I'm Czech, not Polish (not that there's anything wrong with being Polish!). You're pretty insidious-I see how you redirected the header in this message. John Polasek
From: waldofj on 30 Apr 2010 15:46 > Those conversant in relativity will immediately recognize that, to the > contrary, beta = v/c, from time immemorial. > gamma is the object you identified above as beta = ---it's not beta, > it's gamma. It's just a nitpick but the use of those terms has changed over the years. In E's 1905 paper there is no gamma and beta = 1/sqrt(1 - v^2/c^2) In a paper I saw in the 70's (can't remember the details) the author but beta = sqrt(1 - v^2/c^2) so gamma = 1/beta nowadays beta = v/c and so gamma = 1/(1 - beta^2) Bloody inconvenient of "them" to keep changing the meanings of those terms. I keep wondering what's next!
From: John Polasek on 30 Apr 2010 16:16 snip >>>>> I admit that I came in late to the discussion and I have not read all >>>>> of the notes and particularly never looked at that verbose equation of >>>>> which you appear to be overly fond. Let me simplify it for you: >>>>> xi = (x-vt)*gamma >>>> >>>> What are you babbling about ? >>>> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img53.gif >>>>where >>>> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif >>>> >>>>xi = beta * (x-vt), learn to read the Greek alphabet. >>>> >>>>beta = 1/ sqrt[{(c-v) / c} * {(c+v) / c}], >>>> = 1/ sqrt[(c-v) * (c+v) / c^2], >>>> = 1/ sqrt[(c^2 -v^2) / c^2], >>>> = 1/ sqrt[c^2/c^2 - v^2/c^2], >>>> = 1/ sqrt[1 - v^2/c^2], >>> Those conversant in relativity will immediately recognize that, to the >>> contrary, beta = v/c, from time immemorial. >> >>As I previously babbled, you can't read. >>Come back when you've passed first grade and learned to click >> on a link with a mouse, you stooopid babbling cretin. >> >> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img54.gif >> {click here-------------^^^^^ } >> >Just Google "gamma + relativity" and at least the first 5 topics (I >did not go any further) will echo my definitions of gamma and Beta. I >remember this as being true since I first encountered relativity back >in the1940's. That knowledge should be part of your kit, as a >designated relativity basher. >I am surprised by your resilience, in getting back up from the canvas >like Rocky Balboa, after repeated rebuttals, but again point out that >it appears that your petulance exceeds your competence by a >comfortable margin. >Oh, I'm Czech, not Polish (not that there's anything wrong with being >Polish!). >You're pretty insidious-I see how you redirected the header in this >message. >John Polasek Your na�vet� regarding relativity is both refreshing and appalling. You're sticking with Fourmilabs definition of data which you have stumbled upon, and which is incorrect according to numerous citations available on Google of which I again invite you to avail yourself. Try Wiki's special relativity. Oh heck, Google gamma + special relativity and pick something you like-maybe you can even find one that's wrong to match your version. I'll bet you were surprised to find that they had a whole Greek alphabet right in plain English weren't you?
From: Paul Stowe on 30 Apr 2010 22:31 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.
From: PD on 1 May 2010 10:45
On Apr 30, 9:31 pm, Paul Stowe <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??? |