How can the MMx math be corrected?
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From: Tom Roberts on 3 May 2010 13:22 Paul Stowe wrote: > On May 3, 9:11 am, Tom Roberts <tjrob...(a)sbcglobal.net> wrote: >> As you seem to forget it, I'll repeat: the SR explanation is >> that the local structure of spacetime and the behavior of >> electromagnetic fields combine to make the vacuum speed of >> light be isotropically c in any locally inertial frame. Since >> the interferometer is at rest in such a frame to sufficient >> accuracy, as long as the arms remain unchanged while rotating >> (i.e. DO NOT "contract") the MMX will yield a null result. > > WOW, OK, so according to you, there is no Lorentz contraction. I repeat: YOU NEED TO LEARN HOW TO READ MORE ACCURATELY! Nowhere did I say or imply that there is no "Lorentz contraction" (aka "length contraction"). But you seem not to know what "Lorentz contraction" is. In SR, it is merely the consequence of geometric projection of an object's invariant proper length onto coordinates moving relative to the object. In particular, THE OBJECT ITSELF never "contracts" (which was your earlier claim), it is just that the MEASUREMENT of its length using such moving coordinates yields a value smaller than the object's proper length. > it seems to fly in the face of the experiments > that suggest that such a cotraction exists and is quite real... I am VERY INTERESTED in any experiment that has actually measured "Lorentz contraction" -- please give references. AFAIK there have been none (and I have looked). There are many experiments showing various invariances, and several experiments showing the behavior of particle distributions looked at from different frames; there are many experiments showing the effects of geometric projection on clocks, but none for rulers (length) so far. And your "real" is ambiguous -- in SR this is "real" in that it affects measurements and has physical consequences, but is is also "not real" in that the object itself is unaffected by it. I repeat: YOU NEED TO LEARN HOW TO WRITE MORE ACCURATELY. AND MOST ESPECIALLY: YOU NEED TO LEARN HOW TO THINK MORE ACCURATELY. It is clear to me that you have a number of implicit assumptions and beliefs about the world that you are unable to question. Most of them are just plain wrong. You REALLY need to ask yourself whether it makes sense for you to attempt to discuss physics with so much invalid baggage. Or whether you can discard that junk and actually LEARN some modern physics. Tom Roberts
From: Paul Stowe on 3 May 2010 13:33 On May 3, 9:37 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 3, 11:18 am, 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. > 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... > 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 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. 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 obsered to be 'at rest' relative to it? Paul Stowe > See some of the astronomical papers for the mass of the photon:http://pdg.lbl.gov/2009/listings/rpp2009-list-photon.pdf > > > > > So, take the CMB dipole = > > 0 frame and, wrt to it, is 'relative' light speed 'isotropic' wrt to > > systems moving wrt to it? > > > > > Otherwise, my original question remains, Paul Stowe
From: Da Do Ron Ron on 3 May 2010 14:56 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~
From: PD on 3 May 2010 15:26 On May 3, 1:56 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?) Since this is discussed in so many places on the freeweb, why on earth are you asking for it here? > > Please, no more word salad. > > (snip) > > > > > Tom Roberts > > ~RA~
From: PD on 3 May 2010 15:32
On May 3, 12:33 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > On May 3, 9:37 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 3, 11:18 am, 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? > > > 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. > > > 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. > > > > > > 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. > 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 obsered to be 'at rest' relative to it? I don't know why that would be important one way or the other. > > Paul Stowe > > > See some of the astronomical papers for the mass of the photon:http://pdg.lbl.gov/2009/listings/rpp2009-list-photon.pdf > > > > So, take the CMB dipole = > > > 0 frame and, wrt to it, is 'relative' light speed 'isotropic' wrt to > > > systems moving wrt to it? > > > > > > Otherwise, my original question remains, > > Paul Stowe |