How can the MMx math be corrected?
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From: Paul Stowe on 3 May 2010 17:07 On May 3, 12:32 pm, PD <thedraperfam...(a)gmail.com> wrote: > On May 3, 12:33 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > > > Sigh, the quote provided WAS FROM Einstein's 1905 paper. > > > > > > Yes, I know. This is an exquisite demonstration that your eyes can > > > > > fixate on a line or two from a paper, having completely forgotten what > > > > > was said earlier or later in the paper. > > > > > I haven't forgotten, but words don't trump mathematics. Introduce any > > > > non-zero velocity (a vector) AND! Einstein's second postulate, and the > > > > result is (or should be) obvious. It is physically impossible for the > > > > 'relative' speed of light to be isotropic in moving systems. Note, I > > > > did NOT! say measured speed. > > > > I don't know why you'd say this. > > > You may have some buried assumptions that are leading you to conclude > > > that speed of light cannot be isotropic. > > > Here are some candidate guess as to what those assumptions are: > > > > 1. That if there is an object in frame A that is moving with velocity > > > v, and there is another frame B moving at u relative to A (in the same > > > line as the object's motion in A), then the object must be moving in > > > frame B at (v-u) or (v+u). Those are the only physical possibilities. > > > Yes, and given the second postulate, namely that light alway > > propagates isotropically at c irrespective of the motion of the > > source, AND, nothing, and I mean nothing, is 'at rest' in our universe > > the the relative speed of light wrt to all sources MUST BE c + (v Cos > > t) where t is the angle relative to the motion vector. Now I am NOT!, > > nor was I ever, talking about measurement of round trip speeds. That > > will 'measure' invariant and can, by judicious definitions, be > > declared measurably so. > > First, given my simplifying stipulation of colinearity, let's dispose > of the cos t term, which would be 1 in that case. > Now I would ask WHY you believe that the only possible value is v+u > (or v-u) as opposed to any other possibility. > In particular, do you have experimental evidence that velocity > transforms as v+u? Now we come full circle BACK to my original comment, namely the roots of the Lorentz Transform itself. In the direction of the motion of the source, c - v and the opposing direction, c + v. Now Einstein only dealt with delta v, but, the mathematical form comes from the underlying physical process (function). Which Poincare clearly showed. > > > 2. That if something is moving in a frame A with a velocity v, then in > > > any other frame moving relative to A, the same something must be > > > moving at a velocity different than v, regardless of the something or > > > the value of v. > > > That too... > > And again, I would ask how you know this is true regardless of the > "something" and regardless of the value of v. The assumption of existence an independent physical universe which is not dependent upon the existence of observation. This article reflects this prespective... http://arxiv.org/abs/physics/0603267 > > > Any conclusion that is at variance with one of those assumptions is > > > therefore taken as impossible. > > > > Have I got that right? > > > With the clarification above, yes. > > > > > > > And > > > > > > Einstein's postulate that light speed is invariant and not affected by > > > > > > the speed of the emitter/receiver pretty much negates isotropy. > > > > > > Uh, no. If light speed is invariant with respect to the speed of the > > > > > receiver, then the receiver will ALWAYS measure light speed to be > > > > > isotropic. Perhaps there is a confusion in terms here. > > > > > Forget measurements (and the limitations on same) for the moment. For > > > > given that we 'define' speed as distance traveled per unit time then, > > > > if both distance traveled and time it takes to do so (which it would > > > > for independent propagation) increases at the very same rate with > > > > increasing motion, > > > > You mean, as in the (v+u) assumption? > > > Yes, I say all non-zero v... > > Then again, I would ask where that assumption stems from in your mind, > being especially careful to cite measurements please. Measurement? I thought I made it clear that this was about basis behavior. In fact I clearly stated that one can establish by convention that for round trip processing one can set clocks to establish apparent isotropy. I've said this all along... This whole discussion is about necessary consequences of the second postulate. > > > > then the computational result is invariant. > > > > However, this does NOT! address the issue of isotropy of propagation > > > > speed wrt a moving system. > > > > > As far as I can tell, THE ONLY process that can result in 'actual' > > > > isotropic c in moving systems is the ballistic theory of light. > > > > However, by definition, then Einstein's second posulate would have to > > > > be false since light would be emitted at c + (v Cos t). Even then, it > > > > certainly isn't for any observer moving wrt to the emitting > > > > element(s). > > > > > > Again, I go back to my earlier question about how, if you can derive > > > > > the very same LTs from two opposite claims about light speed isotropy, > > > > > you then are able to discern that one of them is right and the other > > > > > is wrong. > > > > > I'm pointing logical inconsistencies in the basis. The fact is, even > > > > SRT must deal with the anisotropies introduced by motion... For > > > > example, we know the Earth, the solar system, our galaxy is not at > > > > rest by the 'measured' doppler in the CMBR. > > > > It is not at rest with respect to the thermal horizon that EMITTED the > > > CMBR. This is no different than saying that we are not at rest with > > > respect to some galaxy. This does not at all imply that the light from > > > this horizon or from the galaxy is received by us at anything other > > > than c. And in fact, measurement indicates that it is indeed traveling > > > at c. > > > It's the opposite, since the CMB IS LIGHT, and uniform throughout our > > universe, the apparent frame in which it is measured isotropic (no > > dominant doppler shift) is logically the rest frame FOR! light. > > No sir. Doppler shift is a comparison of wavelength received at the > receiver and wavelength emitted from the source. As I said, the > measured Doppler shift is with respect to emission from the thermal > horizon that is the source of the CMBR. Thermal horizon? So, I take that you do not agree that the CMB is uniform throughout the universe? > > It is > > because of these facts that more and more GR analysts are considering > > it the preferred frame of reference. I consider it to be the > > universe's rest frame. It provides a common backdrop from which all > > motion can be uniformly quantified. Those GR analyst know this. Now > > tell me of ANY OBJECT that is observed to be 'at rest' relative to it? > > I don't know why that would be important one way or the other. Well if you cannot then, given the second postulate, and, the very nature of the CMB, you are one of the very few thar can't. Paul Stowe
From: PD on 3 May 2010 17:24 On May 3, 4:07 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > On May 3, 12:32 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 3, 12:33 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > > > > Sigh, the quote provided WAS FROM Einstein's 1905 paper. > > > > > > > Yes, I know. This is an exquisite demonstration that your eyes can > > > > > > fixate on a line or two from a paper, having completely forgotten what > > > > > > was said earlier or later in the paper. > > > > > > I haven't forgotten, but words don't trump mathematics. Introduce any > > > > > non-zero velocity (a vector) AND! Einstein's second postulate, and the > > > > > result is (or should be) obvious. It is physically impossible for the > > > > > 'relative' speed of light to be isotropic in moving systems. Note, I > > > > > did NOT! say measured speed. > > > > > I don't know why you'd say this. > > > > You may have some buried assumptions that are leading you to conclude > > > > that speed of light cannot be isotropic. > > > > Here are some candidate guess as to what those assumptions are: > > > > > 1. That if there is an object in frame A that is moving with velocity > > > > v, and there is another frame B moving at u relative to A (in the same > > > > line as the object's motion in A), then the object must be moving in > > > > frame B at (v-u) or (v+u). Those are the only physical possibilities. > > > > Yes, and given the second postulate, namely that light alway > > > propagates isotropically at c irrespective of the motion of the > > > source, AND, nothing, and I mean nothing, is 'at rest' in our universe > > > the the relative speed of light wrt to all sources MUST BE c + (v Cos > > > t) where t is the angle relative to the motion vector. Now I am NOT!, > > > nor was I ever, talking about measurement of round trip speeds. That > > > will 'measure' invariant and can, by judicious definitions, be > > > declared measurably so. > > > First, given my simplifying stipulation of colinearity, let's dispose > > of the cos t term, which would be 1 in that case. > > Now I would ask WHY you believe that the only possible value is v+u > > (or v-u) as opposed to any other possibility. > > In particular, do you have experimental evidence that velocity > > transforms as v+u? > > Now we come full circle BACK to my original comment, namely the roots > of the Lorentz Transform itself. In the direction of the motion of > the source, c - v and the opposing direction, c + v. Now Einstein > only dealt with delta v, but, the mathematical form comes from the > underlying physical process (function). Which Poincare clearly > showed. I'm sorry, but you'll notice that the very same LT is derivable, as I told you earlier, from assumptions that are completely contrary to that. To wit: It is derivable from the assumption that the speed of light is isotropic and frame-independent, and it is fully consistent with a velocity transformation rule that is not v+u. Since the very same LT comes from two completely inconsistent sets of base assumptions, one cannot use the LT to distinguish one set over the other. In conjugate fashion, one cannot start at the LT and systematically *derive* (by reverse algebra) the (v+u) transform or light-speed anisotropy from them, at least not without some additional assumptions. Therefore, since the selection cannot be made on this basis, there must be SOME OTHER rationale for assuming the (v+u) transform. What would that be? > > > > > 2. That if something is moving in a frame A with a velocity v, then in > > > > any other frame moving relative to A, the same something must be > > > > moving at a velocity different than v, regardless of the something or > > > > the value of v. > > > > That too... > > > And again, I would ask how you know this is true regardless of the > > "something" and regardless of the value of v. > > The assumption of existence an independent physical universe which is > not dependent upon the existence of observation. This article > reflects this prespective... I'm sorry, but the statement is "if something is moving in frame A with velocity v, then in any other frame moving relative to A, the same something must be moving at a velocity DIFFERENT THAN v, regardless of the something or the value of v." Kindly explain how independent physical existence of the universe, regardless of observation, would support this statement above. > > http://arxiv.org/abs/physics/0603267 > > > > > > > Any conclusion that is at variance with one of those assumptions is > > > > therefore taken as impossible. > > > > > Have I got that right? > > > > With the clarification above, yes. > > > > > > > > And > > > > > > > Einstein's postulate that light speed is invariant and not affected by > > > > > > > the speed of the emitter/receiver pretty much negates isotropy. > > > > > > > Uh, no. If light speed is invariant with respect to the speed of the > > > > > > receiver, then the receiver will ALWAYS measure light speed to be > > > > > > isotropic. Perhaps there is a confusion in terms here. > > > > > > Forget measurements (and the limitations on same) for the moment. For > > > > > given that we 'define' speed as distance traveled per unit time then, > > > > > if both distance traveled and time it takes to do so (which it would > > > > > for independent propagation) increases at the very same rate with > > > > > increasing motion, > > > > > You mean, as in the (v+u) assumption? > > > > Yes, I say all non-zero v... > > > Then again, I would ask where that assumption stems from in your mind, > > being especially careful to cite measurements please. > > Measurement? I thought I made it clear that this was about basis > behavior. I'm sorry, but I'm not interested in opining on the existence of physical entities that resist all attempts at detection by measurement. That is the realm of leprechauns and angels. Nor am I talking about a round trip measurement. I asked you for ANY measurement that supports the (v+u) transformation model of velocities. You clearly believe this model is correct, and I'm asking on what basis you believe it is correct, other than blind preference or an appeal to common sense. This is why I'm asking for experimental evidence of ANY KIND that this manner of transforming velocities is reflected in nature. > In fact I clearly stated that one can establish by > convention that for round trip processing one can set clocks to > establish apparent isotropy. I've said this all along... This whole > discussion is about necessary consequences of the second postulate. > > > > > > > > then the computational result is invariant. > > > > > However, this does NOT! address the issue of isotropy of propagation > > > > > speed wrt a moving system. > > > > > > As far as I can tell, THE ONLY process that can result in 'actual' > > > > > isotropic c in moving systems is the ballistic theory of light. > > > > > However, by definition, then Einstein's second posulate would have to > > > > > be false since light would be emitted at c + (v Cos t). Even then, it > > > > > certainly isn't for any observer moving wrt to the emitting > > > > > element(s). > > > > > > > Again, I go back to my earlier question about how, if you can derive > > > > > > the very same LTs from two opposite claims about light speed isotropy, > > > > > > you then are able to discern that one of them is right and the other > > > > > > is wrong. > > > > > > I'm pointing logical inconsistencies in the basis. The fact is, even > > > > > SRT must deal with the anisotropies introduced by motion... For > > > > > example, we know the Earth, the solar system, our galaxy is not at > > > > > rest by the 'measured' doppler in the CMBR. > > > > > It is not at rest with respect to the thermal horizon that EMITTED the > > > > CMBR. This is no different than saying that we are not at rest with > > > > respect to some galaxy. This does not at all imply that the light from > > > > this horizon or from the galaxy is received by us at anything other > > > > than c. And in fact, measurement indicates that it is indeed traveling > > > > at c. > > > > It's the opposite, since the CMB IS LIGHT, and uniform throughout our > > > universe, the apparent frame in which it is measured isotropic (no > > > dominant doppler shift) is logically the rest frame FOR! light. > > > No sir. Doppler shift is a comparison of wavelength received at the > > receiver and wavelength emitted from the source. As I said, the > > measured Doppler shift is with respect to emission from the thermal > > horizon that is the source of the CMBR. > > Thermal horizon? So, I take that you do not agree that the CMB is > uniform throughout the universe? Perhaps you don't know what the thermal horizon is. Are you thinking it is some 2D surface sitting somewhere in space? Google is your friend. > > > > It is > > > because of these facts that more and more GR analysts are considering > > > it the preferred frame of reference. I consider it to be the > > > universe's rest frame. It provides a common backdrop from which all > > > motion can be uniformly quantified. Those GR analyst know this. Now > > > tell me of ANY OBJECT that is observed to be 'at rest' relative to it? > > > I don't know why that would be important one way or the other. > > Well if you cannot then, given the second postulate, and, the very > nature of the CMB, you are one of the very few thar can't. ????? > > Paul Stowe
From: Tom Roberts on 4 May 2010 01:33 Da Do Ron Ron wrote: > On Apr 29, 2:51 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> Michelson interferometers with identical legs give >> null results (i.e. a fringe shift of zero within resolutions). > > 1. Have you any proof that the legs are identical? "Proof" does not apply -- this is physics, not math. But it's easy to verify that the arms have identical lengths -- here are three different ways: A) measure their lengths (in their rest frame, of course) and verify they are identical. B) remove one arm and hold it next to the other, verifying their endpoints match up. C) Use white light and verify that the fringes at the center are black and white (i.e. not washed out and colored). (A) and (B) are, after all, what we mean by "identical length" (your parenthetical statement below notwithstanding). Michelson and Morley used (C), as did most repetitions except those using lasers. In fact it is not really necessary that the arms have identical lengths, unless that is required by the measurement technique (e.g. using white light). All that is required is that the interferometer rotates rigidly (i.e. each arm has a constant length). NOTE: Whenever I mention the length of an object, I mean its proper length (i.e. measured in the rest frame of the object). You guys get yourselves all confused when you attempt to think about objects "contracting" due to some unmeasurable motion relative to some unknown and unobservable aether frame -- such as in your next statement: > (Mere measurement with a ruler does not constitute proof > because the ruler would also contract.) You are confused. Such measurements are what we mean by "identical length". You are attempting to apply some mystical new meaning to "length" -- don't do that. "Length of an object" means what it always did: the value obtained by holding a ruler up to the object and measuring it (which implicitly means the ruler is at rest relative to the object being measured). After all, kids in kindergarten are taught not to move the ruler when making a measurement; do you need to go back and learn that lesson? > 2. Can you show it on paper? (Can you show the > MMx null result on paper using identical legs?) It is trivial, and no math is needed; I'll use the inertial frame in which the center of the interferometer is at rest [#]. In SR the speed of light is isotropically c in any inertial frame. Since the arms have constant lengths, the round-trip time for the light to traverse each arm is independent of the orientation of the instrument. So the location of the visible fringes does not vary with orientation -- a null result. [#] I point out that the revolution, rotation, and gravity of the earth and other celestial objects have effects far smaller than the measurement resolution, and are neglected. Tom Roberts
From: harald on 4 May 2010 03:28 On May 3, 5:19 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > On May 3, 7:41 am, harald <h...(a)swissonline.ch> wrote: > > > > > On May 2, 7:00 am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > > > > harald wrote: > > > > On May 1, 8:56 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > > > >> 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).. > > > > In SR you can select ANY inertial frame. There is no need to select one frame > > > and use it throughout the year. > > > Indeed it's not; and when you change inertial frame, you also declare > > your earlier measurement as having been done in a "moving" frame. > > > > As I said, it is simplest to select the rest > > > frame of the instrument for any given measurement; that is necessarily a > > > different frame for each measurement. But still, the prediction of a null result > > > FOR EACH MEASUREMENT is clear and obvious. If each measurement is null, > > > that applies for measurements done at different times of the year. > > > Sure - which is either misleading or completely besides the point of > > the OP (and this thread), just as it is either misleading or > > completely besides the point in a discussion on stellar aberration. > > > > > 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! > > > > That is not essential to the MMX at all, because if the motion of the instrument > > > relative to the ether cannot be measured at any time of the year, time of year > > > is irrelevant. Yes it is essential for stellar aberration -- it is the > > > non-inertial nature of earth's orbit that gives rise to stellar aberration > (or rather to its measurability). > > > It's reassuring that you understand that the variation of velocity of > > the Earth is relevant for stellar aberration. But that makes it even > > more amazing that you cannot (or refuse to) understand that the > > variation in velocity is equally relevant for Lorentz contraction in > > SRT (which the OP called "SRT math"). No motion = nothing to discuss > > or consider; the OP (DDRR) obviously meant that it is "physically > > impossible" to get a null result with a *moving* interferometer > > without a modified length of at least one of the arms, assuming that > > the speed of light is completely unaffected by the speed of the Earth. > > But never mind, he/she seems to have abandoned this discussion by > > now! > > > Harald > > Not one of its arms Harald, both of its arms, since the apparatus is > rotated. Right - at least we both seem to understand now what the OP was talking about. ;-) > Further, it makes no difference that Earth's speed varies > with the seasons, the magnitude of the physical field distortion > (Contraction in the direction of motion) is solely a result of the > instantaneous velocity affecting the field at any given moment. Sure - but here you miss two subtle points which I tried to make clear, but here once more: 1. M-M tested at different seasons to guarantee that at least at one time the "absolute speed" would be significantly different from zero. 2. when discussing SRT, we may choose a frame in which the Earth is momentarily in rest; the fact that MMX is performed with the same result in different seasons is therefore pertinent for the topic of this thread. > When > the fields constituting the material arms of an interferometer are > rotated they maintain their physical orientation wrt to the motion, > thus the overall length changes as the angle to the motion does. That > is, specifically, what I was referring to with the equation Sqrt(1 - > [v'/c]^2) and v' = v Cos t. Let angle t be zero in the direction of > motion and you'll see that v' disappears at 90 & 270 degrees and > becomes -v at 180. This is a necessary physical aspect of the system > which maintains continuity of the field wrt to all of its elements > given the actual invariance and source independence of c. > > Yes, it is physically impossible to get a null result without this > effect and I don't think Tom disputes this. So, what, exactly, is > your point? My point was what I stated that it was: Tom misunderstood what DDRR was talking about and you next jumped in with more confusion, as I explained; that is not helpful for him. But as he is clearly gone now, we're talking to nobody. End of effort! Cheers, Harald
From: harald on 4 May 2010 03:40
On May 3, 8:56 pm, Da Do Ron Ron <ron_ai...(a)hotmail.com> wrote: > On Apr 29, 2:51 pm, 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).. > > 1. Have you any proof that the legs are identical? > (Mere measurement with a ruler does not constitute proof > because the ruler would also contract.) > > 2. Can you show it on paper? (Can you show the > MMx null result on paper using identical legs?) > > Please, no more word salad. > > (snip) > > > Tom Roberts > > ~RA~ DDRR I thought that you had left! I'm still waiting for your comments on my debunking of your argument. And as you should know by now from the little discussion on the side, Tom does *not want* to understand what you are talking about (as he now even puts Einstein in the kindergarten, he clearly does understand what you are talking about). Perhaps you should stress that you are (of course!) *not* discussing the viewpoint of an inertial frame in which the interferometer is in rest. Regards, Harald |