Sagnac Idiocy
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From: Jerry on 30 Sep 2007 21:19 On Sep 30, 6:57 pm, HW@....(Dr. Henri Wilson) wrote: > Fringe 'displacement' is the current distance of a fringe from > its defined position at zero rotation speed. > Fringe 'movement' or 'shifting' is a dynamic term describing > the process of changing displacements. It has units of velocity. You have been WOEFULLY INCONSISTENT in your use of these terms in the course of our discussions. Question 1: ====================== I wrote: * When the turntable reaches steady state constant rotation, * all the wavecrests will be traveling at c +/- v, and the * fringe shifts will be zero again! You replied: * That's what one would expect, you dope. That's what happens. * Fringes don't move during a period of CONSTANT angular * velocity. http://groups.google.com/group/sci.physics.relativity/msg/8f2bb996a5941d6d In the above, I meant "fringe shift" to be synonymous with "fringe displacement". In the above, you answered my statement concerning "fringe shift" with a statement about "fringe movement". GIVEN THE CONTEXT OF YOUR ANSWER, it was natural for me to assume that you meant by "fringe movement" the same as what I meant by "fringe shift", is that correct? ====================== Question 2: ====================== In the same post, you stated: "Both path lengths change during - and only during - an acceleration and that's when the number of wavelengths in each beam changes." The above is plainly false. There is a TIME LAG between acceleration and any observed changes in fringe displacement due to the finite speed of light. Consider the following thought experiment: Two counterrotating rays meet in phase after traveling the same distance around a stationary Sagnac turntable. Let us use your diagram. http://www.users.bigpond.com/hewn/sagnac.jpg At time T0, instantaneously accelerate the turntable to speed v. The green path light impinging on S will be temporarily blue-shifted. The blue path light impinging on S will be temporarily red-shifted. During the initial phase of this dynamic transitional period, the phase of the green path light continuously advances relative to the phase of the blue path light, so there will be continuous fringe movement at the screen S. At time = T0 + P/c, where P is the path length around the Sagnac ring, the "c-v" green path light emitted from S and the "c+v" blue path light emitted from S will simultaneously catch up to the screen. I use parentheses to indicate that yes, I am perfectly aware of the sqrt(2) term in v. The "c-v" light traverses a shorter distance, while the "c+v" light traverses a longer distance, so that their arrival at S will be simultaneous. The green path light impinging on the screen S will no longer be blue-shifted, and the blue path light will no longer be red-shifted. The green path light and the blue path light will once again be precisely in phase, AND THE FRINGE DISPLACEMENT (FRINGE SHIFT) WILL DROP TO ZERO. In the following graph, the horizontal axis represents time, and the vertical axis represents the fringe displacement observed at screen S. _/| _/ | _/ | ________/ |_______ T0 T0+P/c Since wavelength is invariant in ballistic theory, and the number of waves in each path equals length/L, the fringe displacement after steady state is reached MUST EQUAL ZERO. Except for a brief transitional period, the green path light remains in phase with the blue path light. BaTh remains disproven, is that not correct? ====================== Question 3: ====================== What do you mean when you state that Michelson-Gale "returned a null result...pure ballistic stuff also.." http://adsabs.harvard.edu/abs/1925ApJ....61..140M ====================== Jerry Henri Wilson's Lies (1)Fakes Diploma (2)Uses Deceptive Language (3)Fakes Program (4)Intentionally Misquotes (5)Snips (6)Accuses Others of Lying 1 http://mysite.verizon.net/cephalobus_alienus/henri/diploma.htm 2 http://mysite.verizon.net/cephalobus_alienus/henri/deception.htm 3 http://mysite.verizon.net/cephalobus_alienus/henri/rt_aurigae.htm 4 http://mysite.verizon.net/cephalobus_alienus/henri/history.htm 5 http://mysite.verizon.net/cephalobus_alienus/henri/snips.htm 6 http://mysite.verizon.net/cephalobus_alienus/henri/accuses.htm
From: Dr. Henri Wilson on 30 Sep 2007 21:51 On Sun, 30 Sep 2007 18:19:16 -0700, Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Sep 30, 6:57 pm, HW@....(Dr. Henri Wilson) wrote: > >> Fringe 'displacement' is the current distance of a fringe from >> its defined position at zero rotation speed. >> Fringe 'movement' or 'shifting' is a dynamic term describing >> the process of changing displacements. It has units of velocity. > >You have been WOEFULLY INCONSISTENT in your use of these terms >in the course of our discussions. > >Question 1: >====================== >I wrote: >* When the turntable reaches steady state constant rotation, >* all the wavecrests will be traveling at c +/- v, and the >* fringe shifts will be zero again! > >You replied: >* That's what one would expect, you dope. That's what happens. >* Fringes don't move during a period of CONSTANT angular >* velocity. >http://groups.google.com/group/sci.physics.relativity/msg/8f2bb996a5941d6d > >In the above, I meant "fringe shift" to be synonymous >with "fringe displacement". All right, I'm sorry if there has been some confusion. My illustration does not contain this ambiguity and I assumed that people would understand that the expression 'fringes moved' was synonymous with fringe displacement. Let's use these definitions. 1) �Fringe displacement� refers to the distance between the current fringe position and the one that existed at the defined zero rotation position. Units are typically �fringes�. 2) �Fringe movement� is a dynamic term associated with a rate of change of displacement. Units are typically �fringes per second�. A fringe is �moving� when it is changing position. 3) �Fringes moved� is a measure of the difference between the displacement of the fringe pattern before and after a fringe movement. >In the above, you answered my statement concerning "fringe >shift" with a statement about "fringe movement". GIVEN THE >CONTEXT OF YOUR ANSWER, it was natural for me to assume >that you meant by "fringe movement" the same as what I >meant by "fringe shift", is that correct? >====================== > > >Question 2: >====================== >In the same post, you stated: >"Both path lengths change during - and only during - an >acceleration and that's when the number of wavelengths in >each beam changes." > >The above is plainly false. There is a TIME LAG between >acceleration and any observed changes in fringe displacement >due to the finite speed of light. Correct...and that lag is the very process which enables wavecrests are added and subtracted. >Consider the following thought experiment: >Two counterrotating rays meet in phase after traveling the >same distance around a stationary Sagnac turntable. > >Let us use your diagram. >http://www.users.bigpond.com/hewn/sagnac.jpg > >At time T0, instantaneously accelerate the turntable to >speed v. > >The green path light impinging on S will be temporarily >blue-shifted. > >The blue path light impinging on S will be temporarily >red-shifted. Don't bring doppler shift into this. Wavelength is absolute and the number in each path is all you have to worry about. >During the initial phase of this dynamic transitional >period, the phase of the green path light continuously >advances relative to the phase of the blue path light, so >there will be continuous fringe movement at the screen S. forget it...you're preaching relativity again. There is no change in wavelength in this situation according to BaTh. >At time = T0 + P/c, where P is the path length around the >Sagnac ring, the "c-v" green path light emitted from S and >the "c+v" blue path light emitted from S will simultaneously >catch up to the screen. I use parentheses to indicate that >yes, I am perfectly aware of the sqrt(2) term in v. > >The "c-v" light traverses a shorter distance, while the >"c+v" light traverses a longer distance, so that their >arrival at S will be simultaneous. > >The green path light impinging on the screen S will no longer >be blue-shifted, and the blue path light will no longer be >red-shifted. > >The green path light and the blue path light will once >again be precisely in phase, AND THE FRINGE DISPLACEMENT >(FRINGE SHIFT) WILL DROP TO ZERO. you have assumed there is a wavelength change due to doppler. There isn't. >In the following graph, the horizontal axis represents >time, and the vertical axis represents the fringe >displacement observed at screen S. > > _/| > _/ | > _/ | >________/ |_______ > T0 T0+P/c > >Since wavelength is invariant in ballistic theory, and the >number of waves in each path equals length/L, the fringe >displacement after steady state is reached MUST EQUAL ZERO. In steady rotation, the path lengths are different (except at zero speed.) >Except for a brief transitional period, the green path >light remains in phase with the blue path light. Forget phase, forget doppler...just worry about path length differences and divide it by the ABSOLUTE wavelength, which is the same in both rays.. >BaTh remains disproven, is that not correct? No it is not correct. There is no doppler shift AT THE SOURCE END in BaTh. ..a point that is regularly missed by its opponents. >====================== > > >Question 3: >====================== >What do you mean when you state that Michelson-Gale >"returned a null result...pure ballistic stuff also.." See my other answers. It was supposed to show a cyclic change in fringe displacement...it showed a small and almost constant displacement consistent with sagnac. Variations in the displacement were not statistically significant. happy now? I wont insult you any more Crank. >http://adsabs.harvard.edu/abs/1925ApJ....61..140M Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Jerry on 30 Sep 2007 22:24 On Sep 30, 8:51 pm, HW@....(Dr. Henri Wilson) wrote: > On Sun, 30 Sep 2007 18:19:16 -0700, Jerry <Cephalobus_alie...(a)comcast.net> > wrote: > >Question 2: > >====================== > >In the same post, you stated: > >"Both path lengths change during - and only during - an > >acceleration and that's when the number of wavelengths in > >each beam changes." > > >The above is plainly false. There is a TIME LAG between > >acceleration and any observed changes in fringe displacement > >due to the finite speed of light. > > Correct...and that lag is the very process which enables > wavecrests are added and subtracted. > > >Consider the following thought experiment: > >Two counterrotating rays meet in phase after traveling the > >same distance around a stationary Sagnac turntable. > > >Let us use your diagram. > >http://www.users.bigpond.com/hewn/sagnac.jpg > > >At time T0, instantaneously accelerate the turntable to > >speed v. > > >The green path light impinging on S will be temporarily > >blue-shifted. > > >The blue path light impinging on S will be temporarily > >red-shifted. > > Don't bring doppler shift into this. Understanding doppler shift is ESSENTIAL. > Wavelength is absolute and the number in > each path is all you have to worry about. Frequency is NOT invariant. Interference between the green path light and the blue path light will result in dynamically shifting interference fringes...until the "c+v" and "c-v" light catches up... > > >During the initial phase of this dynamic transitional > >period, the phase of the green path light continuously > >advances relative to the phase of the blue path light, so > >there will be continuous fringe movement at the screen S. > > forget it...you're preaching relativity again. There is no > change in wavelength > in this situation according to BaTh. There IS a change in FREQUENCY. > > >At time = T0 + P/c, where P is the path length around the > >Sagnac ring, the "c-v" green path light emitted from S and > >the "c+v" blue path light emitted from S will simultaneously > >catch up to the screen. I use parentheses to indicate that > >yes, I am perfectly aware of the sqrt(2) term in v. > > >The "c-v" light traverses a shorter distance, while the > >"c+v" light traverses a longer distance, so that their > >arrival at S will be simultaneous. > > >The green path light impinging on the screen S will no longer > >be blue-shifted, and the blue path light will no longer be > >red-shifted. > > >The green path light and the blue path light will once > >again be precisely in phase, AND THE FRINGE DISPLACEMENT > >(FRINGE SHIFT) WILL DROP TO ZERO. > > you have assumed there is a wavelength change due to doppler. > There isn't. During the transitional period, there is a FREQUENCY difference between the blue path light and the green path light. > >In the following graph, the horizontal axis represents > >time, and the vertical axis represents the fringe > >displacement observed at screen S. > > > _/| > > _/ | > > _/ | > >________/ |_______ > > T0 T0+P/c > > >Since wavelength is invariant in ballistic theory, and the > >number of waves in each path equals length/L, the fringe > >displacement after steady state is reached MUST EQUAL ZERO. > > In steady rotation, the path lengths are different (except > at zero speed.) ....and the number of waves going in each direction is given by length/L. Since length_cw + length_ccw is constant, this implies that in the steady state situation, the relative phases of the clockwise and counterclockwise beams will be constant. > >Except for a brief transitional period, the green path > >light remains in phase with the blue path light. > > Forget phase, forget doppler...just worry about path length > differences and divide it by the ABSOLUTE wavelength, which > is the same in both rays.. No. You MUST consider phase. Since the length_cw + length_ccw equals a constant, and since BaTh asserts wavelength to be invariant, any advance in phase over the shorter path is precisely matched by a retardation in phase of the counter- rotating beam over the longer path. BaTh predicts that the phase relationship remains constant. Therefore, BaTh predicts no fringe displacement as a result of steady-state rotation. BaTh is falsified. > >BaTh remains disproven, is that not correct? > > No it is not correct. > There is no doppler shift AT THE SOURCE END in BaTh. ..a point > that is regularly missed by its opponents. > > >====================== > > >Question 3: > >====================== > >What do you mean when you state that Michelson-Gale > >"returned a null result...pure ballistic stuff also.." > > See my other answers. It was supposed to show a cyclic change > in fringe displacement... Wrong, Henri. > it showed a small and almost constant displacement consistent > with sagnac. Variations in the displacement were not > statistically significant. Jerry Henri Wilson's Lies (1)Fakes Diploma (2)Uses Deceptive Language (3)Fakes Program (4)Intentionally Misquotes (5)Snips (6)Accuses Others of Lying 1 http://mysite.verizon.net/cephalobus_alienus/henri/diploma.htm 2 http://mysite.verizon.net/cephalobus_alienus/henri/deception.htm 3 http://mysite.verizon.net/cephalobus_alienus/henri/rt_aurigae.htm 4 http://mysite.verizon.net/cephalobus_alienus/henri/history.htm 5 http://mysite.verizon.net/cephalobus_alienus/henri/snips.htm 6 http://mysite.verizon.net/cephalobus_alienus/henri/accuses.htm
From: Dr. Henri Wilson on 1 Oct 2007 00:05 On Sun, 30 Sep 2007 19:24:57 -0700, Jerry <Cephalobus_alienus(a)comcast.net> wrote: >On Sep 30, 8:51 pm, HW@....(Dr. Henri Wilson) wrote: >> On Sun, 30 Sep 2007 18:19:16 -0700, Jerry <Cephalobus_alie...(a)comcast.net> >> wrote: > >> >The green path light impinging on the screen S will no longer >> >be blue-shifted, and the blue path light will no longer be >> >red-shifted. >> >> >The green path light and the blue path light will once >> >again be precisely in phase, AND THE FRINGE DISPLACEMENT >> >(FRINGE SHIFT) WILL DROP TO ZERO. >> >> you have assumed there is a wavelength change due to doppler. >> There isn't. > >During the transitional period, there is a FREQUENCY difference >between the blue path light and the green path light. > >> >In the following graph, the horizontal axis represents >> >time, and the vertical axis represents the fringe >> >displacement observed at screen S. >> >> > _/| >> > _/ | >> > _/ | >> >________/ |_______ >> > T0 T0+P/c >> >> >Since wavelength is invariant in ballistic theory, and the >> >number of waves in each path equals length/L, the fringe >> >displacement after steady state is reached MUST EQUAL ZERO. >> >> In steady rotation, the path lengths are different (except >> at zero speed.) > >...and the number of waves going in each direction is given by >length/L. > >Since length_cw + length_ccw is constant, this implies that >in the steady state situation, the relative phases of the >clockwise and counterclockwise beams will be constant. that's right. >> >Except for a brief transitional period, the green path >> >light remains in phase with the blue path light. >> >> Forget phase, forget doppler...just worry about path length >> differences and divide it by the ABSOLUTE wavelength, which >> is the same in both rays.. > >No. You MUST consider phase. Since the length_cw + length_ccw >equals a constant, and since BaTh asserts wavelength to be >invariant, any advance in phase over the shorter path is >precisely matched by a retardation in phase of the counter- >rotating beam over the longer path. BaTh predicts that the phase >relationship remains constant. There's your mistake....It's a doubling effect...not a cancellation. As one increases in length the other decreases. >Therefore, BaTh predicts no fringe displacement as a result >of steady-state rotation. > >BaTh is falsified. v is the peripheral speed. path1= 4(c+(v/root2).R.root2/c) path2= 4(c-(v/root2).R.root2/c) Path length difference = 8vR/c = 4Aw/c correct eh? ..and terribly simple.... Incidentally, the SR explanation is actually quite amusing. It states that light moves at c both the source and rest frame then concocts a way of turning that into a travel time difference which in turn is translated into wavelengths. It is nothing but a disgused version of MY explanation. Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Jerry on 1 Oct 2007 00:22
On Sep 30, 11:05 pm, HW@....(Dr. Henri Wilson) wrote: > On Sun, 30 Sep 2007 19:24:57 -0700, Jerry <Cephalobus_alie...(a)comcast.net> > wrote: > > >On Sep 30, 8:51 pm, HW@....(Dr. Henri Wilson) wrote: > >> On Sun, 30 Sep 2007 18:19:16 -0700, Jerry <Cephalobus_alie...(a)comcast.net> > >> wrote: > > >> >The green path light impinging on the screen S will no longer > >> >be blue-shifted, and the blue path light will no longer be > >> >red-shifted. > > >> >The green path light and the blue path light will once > >> >again be precisely in phase, AND THE FRINGE DISPLACEMENT > >> >(FRINGE SHIFT) WILL DROP TO ZERO. > > >> you have assumed there is a wavelength change due to doppler. > >> There isn't. > > >During the transitional period, there is a FREQUENCY difference > >between the blue path light and the green path light. > > >> >In the following graph, the horizontal axis represents > >> >time, and the vertical axis represents the fringe > >> >displacement observed at screen S. > > >> > _/| > >> > _/ | > >> > _/ | > >> >________/ |_______ > >> > T0 T0+P/c > > >> >Since wavelength is invariant in ballistic theory, and the > >> >number of waves in each path equals length/L, the fringe > >> >displacement after steady state is reached MUST EQUAL ZERO. > > >> In steady rotation, the path lengths are different (except > >> at zero speed.) > > >...and the number of waves going in each direction is given by > >length/L. > > >Since length_cw + length_ccw is constant, this implies that > >in the steady state situation, the relative phases of the > >clockwise and counterclockwise beams will be constant. > > that's right. > > >> >Except for a brief transitional period, the green path > >> >light remains in phase with the blue path light. > > >> Forget phase, forget doppler...just worry about path length > >> differences and divide it by the ABSOLUTE wavelength, which > >> is the same in both rays.. > > >No. You MUST consider phase. Since the length_cw + length_ccw > >equals a constant, and since BaTh asserts wavelength to be > >invariant, any advance in phase over the shorter path is > >precisely matched by a retardation in phase of the counter- > >rotating beam over the longer path. BaTh predicts that the phase > >relationship remains constant. > > There's your mistake....It's a doubling effect...not a > cancellation. > As one increases in length the other decreases. The beams are COUNTERROTATING, Henri. The effect is a cancellation. BaTh is falsified by your own diagram. Jerry Henri Wilson's Lies (1)Fakes Diploma (2)Uses Deceptive Language (3)Fakes Program (4)Intentionally Misquotes (5)Snips (6)Accuses Others of Lying 1 http://mysite.verizon.net/cephalobus_alienus/henri/diploma.htm 2 http://mysite.verizon.net/cephalobus_alienus/henri/deception.htm 3 http://mysite.verizon.net/cephalobus_alienus/henri/rt_aurigae.htm 4 http://mysite.verizon.net/cephalobus_alienus/henri/history.htm 5 http://mysite.verizon.net/cephalobus_alienus/henri/snips.htm 6 http://mysite.verizon.net/cephalobus_alienus/henri/accuses.htm |