GETTING RID OF EINSTEIN RELATIVITY
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From: George Dishman on 10 Oct 2007 10:52 On 10 Oct, 09:13, "Androcles" <Engin...(a)hogwarts.physics> wrote: > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message > > news:1191914039.526807.231170(a)22g2000hsm.googlegroups.com... > : On 9 Oct, 00:59, HW@....(Dr. Henri Wilson) wrote: > : > On Mon, 8 Oct 2007 23:44:18 +0100, "George Dishman"<geo...(a)briar.demon.co.uk> wrote: > > : > >"Clueless Henri Wilson" <HW@....> wrote in messagenews:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... > : > >> On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" > : > > : > >>>The numbers are equal. > : > > : > >> George, I know this is difficult. > : > > : > >Clearly, since you can't even look at the simulation > : > >and see that it already shows exactly what you are > : > >telling me. > : > > : > >> Go back to the spinning wheel idea. > : > >> Consider a wheel with 1000000 teeth around the edge. > : > >> It will represent the light as it moves around the ring. Its 'edge > speed' > : > >> will > : > >> be c+v, where v is still to be defined. > : > >> You will agree that no matter how fast it spins, the number of teeth > : > >> remains > : > >> 1000000. > : > > : > >Just as the number of waves round Jerry's animation > : > >is 9.5 no matter how fast it spins, no difference > : > >there. > : > > : > That's correct. The number is independent of LIGHT speed BUT NOT RING > SPEED. > : > : Look at the simulation once it has stopped. Count > : the number of waves, it is 9.5 in both directions. > : Ask yourself whether it mattered how fast the > : table moved getting to where it is. If the table > : moved faster then the waves move faster (c+v) and > : that number stays the same. > : > : > Jerry has not included ring speed at all. > : > : Yes she has, count how many pixels the magenta > : dots move on the wave between the first two of > : my screenshots and you will find it is c+v for > : the blue wave, c-v for the red and the black > : line moves at v. Androcles did the sums for you > : already and confirme they were right to within > : pixel rounding. > > It's like pulling teeth with him. He doesn't believe the arithmetic > if it's counter to his theory, and he doesn't believe the arithmetic > if it agrees with his theory. It is worse than that, he will refuse to count the pixels because he knows the number refutes his claim that the speeds are wrong. > : > >> Now draw two lines in the rest frame on either side of the wheel. The > : > >> number of > : > >> teeth between the two lines is 500000, no matter how fast the wheel > spins. > : > >> One line represents the source at the instant of emission of a > particular > : > >> photon or wavecrest. The source is moving but the line isn't. The > other > : > >> line > : > >> will represent the position of the detector when the wavecrest > arrives. > : > > : > >> The first line is a chosen starting point and is static. The position > of > : > >> the > : > >> second line varies with rotation speed of the ring (v). If the ring > moves > : > >> at > : > >> 0.000001c, the second line will be 500000.5 teeth width from the > start > : > >> point, > : > >> IRREPECTIVE OF WHEEL SPEED. ...So if the wheel speed is increased to > c+v, > : > >> the > : > >> number of teeth in the path remains at 500000.5. > : > > : > >> Similarly, the path in the reverse direction has 499999.5 teeth. > : > > : > >> I hope you can digest that properly. > : > > : > >Sure, but those numbers do not determine the > : > >phase difference. > : > > : > Of course they do. If one ray has 500000.1 and the other 499999.9 then > the > : > phasing is obvously different. > : > : No, again look at the simulation as you read this. > : The number of waves to the detector from where the > : light was emitted differs for the two paths but > : they are in-phase at the detector. > > Typical relativist trick, that is. Throw in bigger numbers to confuse > the matter, then say it's obvious. The big numbers were Henry's toy example. The fact that the waves are in phase at the detector is easy to see as they move along the black radial line together. > : > In light of what I have now revealed about the 'emission point moving > : > backwards' in the rotating frame and since we know the phasing must be > : > identical at the source, Jerry's program is not representative of the > facts. > : > : What you have to remember is that the waves move > > Ah... What you have to remember is the waves do NOT move, they > are standing waves. Not in the Sagnac version, standing waves occur in the laser ring gyro but for Sagnac's construction you have two separate paths with a simple wave moving along each. The two are recombined at the splitter. > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/eightframe.gif > The entire wheel is turning as shown by the broken spoke, but > the spokes are not turning. Same thing here: > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif > There are only 6 frames, beat frequency shows up when the teeth are aligned. > The blue wheel moves at v=4 , > the yellow at v+c = (4+1) = 5,and > the red at v-c = (4-1) = 3. > > OR... > The blue wheel moves at v=4 , > the yellow at v+c = (4+61) modulo 60 = 5, and > the red at v-c = (4-61) modulo 60 = 3 or -57. > There are 60 teeth. That's not bad for the ring gyro configuration but I'll leave that for another time, Henry has enough trouble with George's version without looking at alternatives. > : and the phasing as you say is identical at the > : source but only when the light is emitted. If > : you look at the location where the light is > : emitted, after the radial line passes, the > : waves become out of phase because they move past > : at different speeds. By the time the light hits > : the detector it is no longer in-phase at the > : emission location but is still in phase at the > : _current_ location of the source. > : > : > >Look at Jerry's simulation > : > >which illustrates precisely what you just said > : > >and note that the signals arrive in phase. That > : > >is what you are trying to calculate and the > : > >method you are attempting doesn't give you the > : > >answer. > : > > : > The signals that Jery has drawn arriving in phase did not start out in > : > phase...AS REQUIRED. > : > : Look at the waves coming from the radial line, you > : can see that they _do_ start in phase all the time. > : There is no flaw in the simulation. > > There is, the waves are not shown in the stationary frame, > which is why they APPEAR TO MOVE. > The wave does NOT move in the stationary frame. Yes they do, they propagate along the path because the light goes from source to detector only, there is no reflected wave. Of course in the original construction they moved along straight lines between the mirros rather than round the circumference and Jerry's diagram is closer to the modern fibre implementation, but you can view it as showing a build with the light moving at grazing incidence round the inside of a cylindrical mirror. > This wave does NOT move unless the mass moves in x. > http://www.kettering.edu/~drussell/Demos/SHO/damp.html > The idiot Wilson just mumbles "irrelevant" because he's stupid. I don't see the relevance either, in Sagnac's construction you have two simple point-to-point one way paths. > : > >If you take a snapshot at any instant, there will > : > >be 50000 waves between the CURRENT locations of > : > >the source and the detector, the same number on > : > >both paths therefore the signals must arrive in > : > >phase because they were emitted in phase. > : > > : > >The location where the light was emitted doesn't > : > >help you calculate the arrival phase, your approach > : > >to the maths is flawed. > : > > : > the math is simple. > : > Path lengths are different. Number of wavelengths = L/lambda = > : > (2piR+/-vt)/lambda > : > : Still the same error, see above for details. > > Correct, so there are two values for lambda, > lambda_blue = L1/N = (2piR+vt)/9.5 > lambda_red = L2/N = (2piR-vt)/9.5 > which is NOT shown in Jeery's model and why > Wilson can't see it (nor can you or Jeery). > You shouldn't mix vt with alpha (length with angle), either. > > lambda_blue = L1/N = (2pi+alpha)R/9.5 > lambda_red = L2/N = (2pi - alpha)R/9.5 Nah. In the inertial frame a peak on the blue wave moves a distance of (c+v)/f in the period of the source. The source itself move v/f, both measured along the circumference. A second peak is therefore emitted c/f behind the first and therefter travels at the same speed so the waves are that constant distance apart, that is the wavelength. For Jerry's animation lambda = circumference / 9.5 in both directions. Hit the pause button when the two waves coincide and you can see they are the same as they should be. George
From: Androcles on 10 Oct 2007 14:39 "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1192027957.218767.273550(a)57g2000hsv.googlegroups.com... : On 10 Oct, 09:13, "Androcles" <Engin...(a)hogwarts.physics> wrote: : > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message : > : > news:1191914039.526807.231170(a)22g2000hsm.googlegroups.com... : > : On 9 Oct, 00:59, HW@....(Dr. Henri Wilson) wrote: : > : > On Mon, 8 Oct 2007 23:44:18 +0100, "George Dishman"<geo...(a)briar.demon.co.uk> wrote: : > : > : > >"Clueless Henri Wilson" <HW@....> wrote in messagenews:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... : > : > >> On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" : > : > : > : > >>>The numbers are equal. : > : > : > : > >> George, I know this is difficult. : > : > : > : > >Clearly, since you can't even look at the simulation : > : > >and see that it already shows exactly what you are : > : > >telling me. : > : > : > : > >> Go back to the spinning wheel idea. : > : > >> Consider a wheel with 1000000 teeth around the edge. : > : > >> It will represent the light as it moves around the ring. Its 'edge : > speed' : > : > >> will : > : > >> be c+v, where v is still to be defined. : > : > >> You will agree that no matter how fast it spins, the number of teeth : > : > >> remains : > : > >> 1000000. : > : > : > : > >Just as the number of waves round Jerry's animation : > : > >is 9.5 no matter how fast it spins, no difference : > : > >there. : > : > : > : > That's correct. The number is independent of LIGHT speed BUT NOT RING : > SPEED. : > : : > : Look at the simulation once it has stopped. Count : > : the number of waves, it is 9.5 in both directions. : > : Ask yourself whether it mattered how fast the : > : table moved getting to where it is. If the table : > : moved faster then the waves move faster (c+v) and : > : that number stays the same. : > : : > : > Jerry has not included ring speed at all. : > : : > : Yes she has, count how many pixels the magenta : > : dots move on the wave between the first two of : > : my screenshots and you will find it is c+v for : > : the blue wave, c-v for the red and the black : > : line moves at v. Androcles did the sums for you : > : already and confirme they were right to within : > : pixel rounding. : > : > It's like pulling teeth with him. He doesn't believe the arithmetic : > if it's counter to his theory, and he doesn't believe the arithmetic : > if it agrees with his theory. : : It is worse than that, he will refuse to count the : pixels because he knows the number refutes his claim : that the speeds are wrong. Yes, I know. He claims there are 13 and 6 waves, 7 made by the tick fairy, but won't change the wavelength ... but then, neither will Jeery. http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif : : > : > >> Now draw two lines in the rest frame on either side of the wheel. The : > : > >> number of : > : > >> teeth between the two lines is 500000, no matter how fast the wheel : > spins. : > : > >> One line represents the source at the instant of emission of a : > particular : > : > >> photon or wavecrest. The source is moving but the line isn't. The : > other : > : > >> line : > : > >> will represent the position of the detector when the wavecrest : > arrives. : > : > : > : > >> The first line is a chosen starting point and is static. The position : > of : > : > >> the : > : > >> second line varies with rotation speed of the ring (v). If the ring : > moves : > : > >> at : > : > >> 0.000001c, the second line will be 500000.5 teeth width from the : > start : > : > >> point, : > : > >> IRREPECTIVE OF WHEEL SPEED. ...So if the wheel speed is increased to : > c+v, : > : > >> the : > : > >> number of teeth in the path remains at 500000.5. : > : > : > : > >> Similarly, the path in the reverse direction has 499999.5 teeth. : > : > : > : > >> I hope you can digest that properly. : > : > : > : > >Sure, but those numbers do not determine the : > : > >phase difference. : > : > : > : > Of course they do. If one ray has 500000.1 and the other 499999.9 then : > the : > : > phasing is obvously different. : > : : > : No, again look at the simulation as you read this. : > : The number of waves to the detector from where the : > : light was emitted differs for the two paths but : > : they are in-phase at the detector. : > : > Typical relativist trick, that is. Throw in bigger numbers to confuse : > the matter, then say it's obvious. : : The big numbers were Henry's toy example. The fact : that the waves are in phase at the detector is easy : to see as they move along the black radial line : together. Here's N = 24: http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif : : > : > In light of what I have now revealed about the 'emission point moving : > : > backwards' in the rotating frame and since we know the phasing must be : > : > identical at the source, Jerry's program is not representative of the : > facts. : > : : > : What you have to remember is that the waves move : > : > Ah... What you have to remember is the waves do NOT move, they : > are standing waves. : : Not in the Sagnac version Yes in the Sagnac version. http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif , standing waves occur in : the laser ring gyro Not really, there is no medium to wave. Jeery's water waves are nonsense. All you have is an inertial trace where a photon has been. (Inertial is the modern term for stationary). but for Sagnac's construction : you have two separate paths with a simple wave : moving along each. The two are recombined at the : splitter. What you have is this: http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG at any point around the path. We engineers call that a beat frequency but you and Wilson can call it a long enveloping wavelength if you like. It is non-inertial, it non-inertials as the turntable non-inertials. : : > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/eightframe.gif : > The entire wheel is turning as shown by the broken spoke, but : > the spokes are not turning. Same thing here: : > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif : > There are only 6 frames, beat frequency shows up when the teeth are aligned. : > The blue wheel moves at v=4 , : > the yellow at v+c = (4+1) = 5,and : > the red at v-c = (4-1) = 3. : > : > OR... : > The blue wheel moves at v=4 , : > the yellow at v+c = (4+61) modulo 60 = 5, and : > the red at v-c = (4-61) modulo 60 = 3 or -57. : > There are 60 teeth. : : That's not bad for the ring gyro configuration : but I'll leave that for another time, Henry : has enough trouble with George's version without : looking at alternatives. Until Jeery understands the wave is an inertial trace of where the photon WAS, she's never going to understand physics. : : > : and the phasing as you say is identical at the : > : source but only when the light is emitted. If : > : you look at the location where the light is : > : emitted, after the radial line passes, the : > : waves become out of phase because they move past : > : at different speeds. By the time the light hits : > : the detector it is no longer in-phase at the : > : emission location but is still in phase at the : > : _current_ location of the source. : > : : > : > >Look at Jerry's simulation : > : > >which illustrates precisely what you just said : > : > >and note that the signals arrive in phase. That : > : > >is what you are trying to calculate and the : > : > >method you are attempting doesn't give you the : > : > >answer. : > : > : > : > The signals that Jery has drawn arriving in phase did not start out in : > : > phase...AS REQUIRED. : > : : > : Look at the waves coming from the radial line, you : > : can see that they _do_ start in phase all the time. : > : There is no flaw in the simulation. : > : > There is, the waves are not shown in the stationary frame, : > which is why they APPEAR TO MOVE. : > The wave does NOT move in the stationary frame. : : Yes they do, No they don't. http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif : they propagate along the path because : the light goes from source to detector only, The photon goes from source to detector via the blue path, the anti-photon goes from source to detector via the red path, for every photon there is an equal and opposite re-photon. (Androcles' third law). http://www.androcles01.pwp.blueyonder.co.uk/rephoton.gif : there : is no reflected wave. Of course in the original : construction they moved along straight lines between : the mirros rather than round the circumference and : Jerry's diagram is closer to the modern fibre : implementation, That's what I said, Jeery has a FOG, not a Sagnac. It is much easier to implement a FOG, Grandpa can take a back seat now that his work is done. The Sagnac effect is the Coriolis effect. : but you can view it as showing a : build with the light moving at grazing incidence : round the inside of a cylindrical mirror. It is easier to view the Coriolis effect from Grandpa's position, the kids on the roundabout might think Newton's laws were not obeyed and day dream of time dilation idiocy, claming the velocity is c in all frame of reference which they say are stationary. : : > This wave does NOT move unless the mass moves in x. : > http://www.kettering.edu/~drussell/Demos/SHO/damp.html : > The idiot Wilson just mumbles "irrelevant" because he's stupid. : : I don't see the relevance either, in Sagnac's : construction you have two simple point-to-point : one way paths. The wave is a trace along the TIME axis. The position of the photon at any TIME is inertial (the modern term for stationary) : : > : > >If you take a snapshot at any instant, there will : > : > >be 50000 waves between the CURRENT locations of : > : > >the source and the detector, the same number on : > : > >both paths therefore the signals must arrive in : > : > >phase because they were emitted in phase. : > : > : > : > >The location where the light was emitted doesn't : > : > >help you calculate the arrival phase, your approach : > : > >to the maths is flawed. : > : > : > : > the math is simple. : > : > Path lengths are different. Number of wavelengths = L/lambda = : > : > (2piR+/-vt)/lambda : > : : > : Still the same error, see above for details. : > : > Correct, so there are two values for lambda, : > lambda_blue = L1/N = (2piR+vt)/9.5 : > lambda_red = L2/N = (2piR-vt)/9.5 : > which is NOT shown in Jeery's model and why : > Wilson can't see it (nor can you or Jeery). : > You shouldn't mix vt with alpha (length with angle), either. : > : > lambda_blue = L1/N = (2pi+alpha)R/9.5 : > lambda_red = L2/N = (2pi - alpha)R/9.5 : : Nah. In the inertial frame a peak on the blue : wave moves a distance of (c+v)/f in the period : of the source. That peak becomes a trough by the time it has got halfway there. A new peak appears brought into existence by the peak fairy. As with Wilson, you don't believe the arithmetic if it's counter to your theory, and you don't believe the arithmetic if it agrees with your theory. It is worse than that, you will refuse to measure the wavelength because you know the value refutes your claim that the speed is c in all stationary frames of reference (stationary is the ancient term for inertial). The source itself move (non-inertials, use the moder term) v/f, Yeah, but the photon non-inertials faster. You need a trace to show how the source non-inertials from where it was to where it is. : both : measured along the circumference. A second peak : is therefore emitted c/f behind the first We've no idea when the next photon is emitted. It could be a nanosecond later or a year later. You are an aetherialist, I'm sure of it. Have you been talking to Glowbar? and : therefter travels at the same speed so the : waves are that constant distance apart, that : is the wavelength. For Jerry's animation : : lambda = circumference / 9.5 Can't be, the red ray non-inertials 144 degrees and the blue ray non-inertials 216 degrees. lamba 1 <> lambda2. Perhaps Jeery non-inertialled the reference point. : : in both directions. Hit the pause button when : the two waves coincide and you can see they : are the same as they should be. I did, the red ray non-inertials 144 degrees and the blue ray non-inertials 216 degrees. lambda 1 <> lambda2. Perhaps Jeery non-inertialled the reference point. It is worse than that, you will refuse to measure the wavelength because you know the value refutes your claim that the speed is c in all stationary frames of reference (stationary is the ancient term for inertial).
From: Androcles on 10 Oct 2007 18:49 "Dr. Henri Wilson" <HW@....> wrote in message news:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... : On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" <george(a)briar.demon.co.uk> : wrote: : : > : >"Clueless Henri Wilson" <HW@....> wrote in message : >news:ftrig3h6nri7ra2t6d2hm7sehuvc7iaqnd(a)4ax.com... : >> On Sun, 7 Oct 2007 12:31:28 +0100, "George Dishman" : >> <george(a)briar.demon.co.uk> wrote: : >>>"Clueless Henri Wilson" <HW@....> wrote in message : >>>news:hn0gg3le7a72fm1mqmsmn16fi6lgh4fd2r(a)4ax.com... : >>>> On Sat, 6 Oct 2007 12:34:59 +0100, "George Dishman" : >> : >>>>>> www.users.bigpond.com/hewn/ringgyro.htm : >>>>>> : >>>>>> Phase is taken care of. : >>>>> : >>>>>The phase isn't even shown. : >>>> : >>>> It is obvious what happens to phase. : >>>> At constant rotation speed, the phase relationship between the two is : >>>> constant. : >>> : >>>Yes, but also it is zero. You cannot see that from the : >>>diagram but you can in Jerry's simulaton because the : >>>two waves arriving at the detector hit the radial line : >>>at exactly the same point at all times. : >> : >> :) : >> It isn't zero George, It depends solely on the relative numbers of : >> wavelengths : >> in the two paths. : > : >The numbers are equal. : : George, I know this is difficult. : : Go back to the spinning wheel idea. : Consider a wheel with 1000000 teeth around the edge. : It will represent the light as it moves around the ring. Its 'edge speed' will : be c+v, where v is still to be defined. : You will agree that no matter how fast it spins, the number of teeth remains : 1000000. : : Now draw two lines in the rest frame on either side of the wheel. The number of : teeth between the two lines is 500000, no matter how fast the wheel spins. : One line represents the source at the instant of emission of a particular : photon or wavecrest. The source is moving but the line isn't. The other line : will represent the position of the detector when the wavecrest arrives. : : The first line is a chosen starting point and is static. The position of the : second line varies with rotation speed of the ring (v). If the ring moves at : 0.000001c, the second line will be 500000.5 teeth width from the start point, : IRREPECTIVE OF WHEEL SPEED. ...So if the wheel speed is increased to c+v, the : number of teeth in the path remains at 500000.5. Half-a-tick fairy. Probably the tooth fairy when she's off duty. : : Similarly, the path in the reverse direction has 499999.5 teeth. The other half-a-tick fairy took the half tick away. What happens when v is increased to 0.1c? : I hope you can digest that properly. I'm sure he can. Wilson, I know this is difficult, but learn to count. : : >>>>>> I told you, the error lies in the fact that wavelength is not constant : >>>>>> in : >>>>>> the rotating frame. : >>>>> : >>>>>You told me it was absolute, and it is. : >>>> : >>>> ..in inertial frames. : >>> : >>>All frames Henry, and you didn't make that : >>>distinction previously. : >> : >> It isn't absolute OR constant in a rotating frame. : >> Straight lines become curved for one thing.. : > : >Yes Henry, so you measure along the curved path : >since your speed of c+v is likewise along that : >path. The distance moved by the wave is (c+v)/f : >measured along whatever path it takes (including : >reflecting from mirrors) while the source moves : >v/f so they are separated by c/f when the next : >wave is emitted, hence that is the wavelength. : > : >> ..best not to use rotating frames George. : > : >Jerry's animation is in the inertial frame, your : >web page has a division that is half in each! : > : >>>>>> You are missing the fact that the path lengths are different as shown : >>>>>> in : >>>>>> www.users.bigpond.com/hewn/ringgyro.htm : >>>>> : >>>>>Nope, I am well aware of that, I told you your : >>>>>error lies elsewhere. : >>>> : >>>> George, I don't have an error. The number of wavelengths around the : >>>> paths : >>>> is : >>>> (2piR+/-vt)/lambda. I've done the rest of the maths for you. : >>> : >>>It is 2piR/lambda regardless of which frame : >>>you calculate in, that is your error. : >> : >> Just have another look at: http://www.users.bigpond.com/hewn/ringgyro.htm : >> : >> where's the error there, George? : > : >You divide the path length by lambda when it should : >be (c+v)/f or (c-v)/f for the opposing beam. The : >result is 2piR/lambda in the inertial frame. : > : >George : > : : : : Henri Wilson. ASTC,BSc,DSc(T) : : www.users.bigpond.com/hewn/index.htm
From: George Dishman on 11 Oct 2007 06:00 On 10 Oct, 19:39, "Androcles" <Engin...(a)hogwarts.physics> wrote: > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message > > news:1192027957.218767.273550(a)57g2000hsv.googlegroups.com... > : On 10 Oct, 09:13, "Androcles" <Engin...(a)hogwarts.physics> wrote: > : > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message > : > > : >news:1191914039.526807.231170(a)22g2000hsm.googlegroups.com... > : > : On 9 Oct, 00:59, HW@....(Dr. Henri Wilson) wrote: > : > : > On Mon, 8 Oct 2007 23:44:18 +0100, "GeorgeDishman"<geo...(a)briar.demon.co.uk> wrote: > > : > > : > : > >"Clueless Henri Wilson" <HW@....> wrote in > messagenews:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... > : > : > >> On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" > : > : > > : > : > >>>The numbers are equal. > : > : > > : > : > >> George, I know this is difficult. > : > : > > : > : > >Clearly, since you can't even look at the simulation > : > : > >and see that it already shows exactly what you are > : > : > >telling me. > : > : > > : > : > >> Go back to the spinning wheel idea. > : > : > >> Consider a wheel with 1000000 teeth around the edge. > : > : > >> It will represent the light as it moves around the ring. Its > 'edge > : > speed' > : > : > >> will > : > : > >> be c+v, where v is still to be defined. > : > : > >> You will agree that no matter how fast it spins, the number of > teeth > : > : > >> remains > : > : > >> 1000000. > : > : > > : > : > >Just as the number of waves round Jerry's animation > : > : > >is 9.5 no matter how fast it spins, no difference > : > : > >there. > : > : > > : > : > That's correct. The number is independent of LIGHT speed BUT NOT > RING > : > SPEED. > : > : > : > : Look at the simulation once it has stopped. Count > : > : the number of waves, it is 9.5 in both directions. > : > : Ask yourself whether it mattered how fast the > : > : table moved getting to where it is. If the table > : > : moved faster then the waves move faster (c+v) and > : > : that number stays the same. > : > : > : > : > Jerry has not included ring speed at all. > : > : > : > : Yes she has, count how many pixels the magenta > : > : dots move on the wave between the first two of > : > : my screenshots and you will find it is c+v for > : > : the blue wave, c-v for the red and the black > : > : line moves at v. Androcles did the sums for you > : > : already and confirme they were right to within > : > : pixel rounding. > : > > : > It's like pulling teeth with him. He doesn't believe the arithmetic > : > if it's counter to his theory, and he doesn't believe the arithmetic > : > if it agrees with his theory. > : > : It is worse than that, he will refuse to count the > : pixels because he knows the number refutes his claim > : that the speeds are wrong. > > Yes, I know. He claims there are 13 and 6 waves, 7 made by the tick fairy, > but won't change the wavelength ... but then, neither will Jeery. > http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif > > : > : > : > >> Now draw two lines in the rest frame on either side of the wheel. > The > : > : > >> number of > : > : > >> teeth between the two lines is 500000, no matter how fast the > wheel > : > spins. > : > : > >> One line represents the source at the instant of emission of a > : > particular > : > : > >> photon or wavecrest. The source is moving but the line isn't. The > : > other > : > : > >> line > : > : > >> will represent the position of the detector when the wavecrest > : > arrives. > : > : > > : > : > >> The first line is a chosen starting point and is static. The > position > : > of > : > : > >> the > : > : > >> second line varies with rotation speed of the ring (v). If the > ring > : > moves > : > : > >> at > : > : > >> 0.000001c, the second line will be 500000.5 teeth width from the > : > start > : > : > >> point, > : > : > >> IRREPECTIVE OF WHEEL SPEED. ...So if the wheel speed is > increased to > : > c+v, > : > : > >> the > : > : > >> number of teeth in the path remains at 500000.5. > : > : > > : > : > >> Similarly, the path in the reverse direction has 499999.5 teeth. > : > : > > : > : > >> I hope you can digest that properly. > : > : > > : > : > >Sure, but those numbers do not determine the > : > : > >phase difference. > : > : > > : > : > Of course they do. If one ray has 500000.1 and the other 499999.9 > then > : > the > : > : > phasing is obvously different. > : > : > : > : No, again look at the simulation as you read this. > : > : The number of waves to the detector from where the > : > : light was emitted differs for the two paths but > : > : they are in-phase at the detector. > : > > : > Typical relativist trick, that is. Throw in bigger numbers to confuse > : > the matter, then say it's obvious. > : > : The big numbers were Henry's toy example. The fact > : that the waves are in phase at the detector is easy > : to see as they move along the black radial line > : together. > Here's N = 24: > http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif Sorry, you show the waves as static, they actually move, the blue at c+v and the red at c, and they should have the same wavelength. If you sum the red and blue you could get a 'standing wave' effect but individually they move. > : > : > In light of what I have now revealed about the 'emission point > moving > : > : > backwards' in the rotating frame and since we know the phasing must > be > : > : > identical at the source, Jerry's program is not representative of > the > : > facts. > : > : > : > : What you have to remember is that the waves move > : > > : > Ah... What you have to remember is the waves do NOT move, they > : > are standing waves. > : > : Not in the Sagnac version > > Yes in the Sagnac version. > http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif No, see above. > , standing waves occur in > : the laser ring gyro > > Not really, there is no medium to wave. I agree, light is really particulate but Henry's understanding only copes with fields. > Jeery's water waves are > nonsense. Jerry's animation is a polar coordinate plot of the classical electric field strength round the path at any given instant, it doesn't show any history other than the yellow and green marker points added to help Henry. > All you have is an inertial trace where a photon has been. That isn't shown, but you could imagine such a trace added to the magenta dots, it is not unreasonable to view them as a photon. > (Inertial is the modern term for stationary). > > but for Sagnac's construction > : you have two separate paths with a simple wave > : moving along each. The two are recombined at the > : splitter. > > What you have is this: > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG > at any point around the path. At any point fixed in the inertial coordinate system, yes. > We engineers call that a beat frequency but you and Wilson can > call it a long enveloping wavelength if you like. It is non-inertial, > it non-inertials as the turntable non-inertials. Yes. > : > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/eightframe.gif > : > The entire wheel is turning as shown by the broken spoke, but > : > the spokes are not turning. Same thing here: > : > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif > : > There are only 6 frames, beat frequency shows up when the teeth are > aligned. > : > The blue wheel moves at v=4 , > : > the yellow at v+c = (4+1) = 5,and > : > the red at v-c = (4-1) = 3. > : > > : > OR... > : > The blue wheel moves at v=4 , > : > the yellow at v+c = (4+61) modulo 60 = 5, and > : > the red at v-c = (4-61) modulo 60 = 3 or -57. > : > There are 60 teeth. > : > : That's not bad for the ring gyro configuration > : but I'll leave that for another time, Henry > : has enough trouble with George's version without > : looking at alternatives. > > Until Jeery understands the wave is an inertial trace of where > the photon WAS, she's never going to understand physics. It isn't, it is the aggregate effect of vast numbers of photons expressed as a field strength. The magenta dots are the closest thing to an individual photon. > : > : and the phasing as you say is identical at the > : > : source but only when the light is emitted. If > : > : you look at the location where the light is > : > : emitted, after the radial line passes, the > : > : waves become out of phase because they move past > : > : at different speeds. By the time the light hits > : > : the detector it is no longer in-phase at the > : > : emission location but is still in phase at the > : > : _current_ location of the source. > : > : > : > : > >Look at Jerry's simulation > : > : > >which illustrates precisely what you just said > : > : > >and note that the signals arrive in phase. That > : > : > >is what you are trying to calculate and the > : > : > >method you are attempting doesn't give you the > : > : > >answer. > : > : > > : > : > The signals that Jery has drawn arriving in phase did not start out > in > : > : > phase...AS REQUIRED. > : > : > : > : Look at the waves coming from the radial line, you > : > : can see that they _do_ start in phase all the time. > : > : There is no flaw in the simulation. > : > > : > There is, the waves are not shown in the stationary frame, > : > which is why they APPEAR TO MOVE. > : > The wave does NOT move in the stationary frame. > : > : Yes they do, > > No they don't. > http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif That is not correct, the only 'static pattern' you can form is a standing wave in the rotating frame which you get as the sum of the red and blue waves. > : they propagate along the path because > : the light goes from source to detector only, > > The photon goes from source to detector via the blue path, > the anti-photon goes from source to detector via the red path, > for every photon there is an equal and opposite re-photon. Almost, the photon goes from source to detector via both the red and blue paths, the anti-photon goes from detector to source via both paths but those are two equivalent descriptions of the same particle. > (Androcles' third law). > http://www.androcles01.pwp.blueyonder.co.uk/rephoton.gif > > : there > : is no reflected wave. Of course in the original > : construction they moved along straight lines between > : the mirros rather than round the circumference and > : Jerry's diagram is closer to the modern fibre > : implementation, > > That's what I said, Jeery has a FOG, not a Sagnac. The physics is the same, it is the ring laser gyro that differs. Sagnac's table and an iFOG both give an output proportional to angular speed while the ring gyro gives an output proportional to the angle turned from some reference. > It is much easier to implement a FOG, Grandpa can take > a back seat now that his work is done. The Sagnac effect is > the Coriolis effect. > > : but you can view it as showing a > : build with the light moving at grazing incidence > : round the inside of a cylindrical mirror. > > It is easier to view the Coriolis effect from Grandpa's position, > the kids on the roundabout might think Newton's laws were not > obeyed and day dream of time dilation idiocy, claming the velocity > is c in all frame of reference which they say are stationary. > > : > : > This wave does NOT move unless the mass moves in x. > : > http://www.kettering.edu/~drussell/Demos/SHO/damp.html > : > The idiot Wilson just mumbles "irrelevant" because he's stupid. > : > : I don't see the relevance either, in Sagnac's > : construction you have two simple point-to-point > : one way paths. > > The wave is a trace along the TIME axis. No, you misunderstand Jerry's plot, it shows the fields versus distance along the circumference at a single instant, it isn't a historical trace. > The position of the > photon at any TIME is inertial (the modern term for stationary) Think of the photon as the magenta dot, or imagine a radial line from the circumference to the magenta dot as a voltage sample that moves round the path at ballistic speed. Incidentally, can you see that replacing the sine wave source with a random noise generator would still have the signals arriving at the detector "in phase", in other words the same sample value arrives from both directions at the same time. > : > : > >If you take a snapshot at any instant, there will > : > : > >be 50000 waves between the CURRENT locations of > : > : > >the source and the detector, the same number on > : > : > >both paths therefore the signals must arrive in > : > : > >phase because they were emitted in phase. > : > : > > : > : > >The location where the light was emitted doesn't > : > : > >help you calculate the arrival phase, your approach > : > : > >to the maths is flawed. > : > : > > : > : > the math is simple. > : > : > Path lengths are different. Number of wavelengths = L/lambda = > : > : > (2piR+/-vt)/lambda > : > : > : > : Still the same error, see above for details. > : > > : > Correct, so there are two values for lambda, > : > lambda_blue = L1/N = (2piR+vt)/9.5 > : > lambda_red = L2/N = (2piR-vt)/9.5 > : > which is NOT shown in Jeery's model and why > : > Wilson can't see it (nor can you or Jeery). > : > You shouldn't mix vt with alpha (length with angle), either. > : > > : > lambda_blue = L1/N = (2pi+alpha)R/9.5 > : > lambda_red = L2/N = (2pi - alpha)R/9.5 > : > : Nah. In the inertial frame a peak on the blue > : wave moves a distance of (c+v)/f in the period > : of the source. > > That peak becomes a trough by the time it has got halfway there. No, the peak remains a peak and the trough is following half a wave behind. > A new peak appears brought into existence by the peak fairy. New peaks and troughs are being emitted from the source all the time as Jerry shows. > As with Wilson, you don't believe the arithmetic if it's counter to > your theory, and you don't believe the arithmetic if it agrees with > your theory. It is worse than that, you will refuse to measure the > wavelength because you know the value refutes your claim that > the speed is c in all stationary frames of reference (stationary > is the ancient term for inertial). > > The source itself move (non-inertials, use the moder term) v/f, > > Yeah, but the photon non-inertials faster. You need a trace to show > how the source non-inertials from where it was to where it is. Photons, peaks and troughs move at (c+v)/f according to ballistic theory so peaks and troughs are [(c+v)/f] - [c/f] or c/f apart, that is why Henry says the wavelength is "absolute", it has the same value regardless of the speed of the source. > : both > : measured along the circumference. A second peak > : is therefore emitted c/f behind the first > > We've no idea when the next photon is emitted. There are vast numbers emitted in every period, I am talking about the macroscopic aggregate of their effect which is classically described as the electric and magnetic fields. It is peaks of the voltage detected with an array of antennas round the table that move at (c+v) and a distance of c/f apart. > It could be > a nanosecond later or a year later. You are an aetherialist, I'm sure > of it. No, I go for QED. > Have you been talking to Glowbar? > > and > : therefter travels at the same speed so the > : waves are that constant distance apart, that > : is the wavelength. For Jerry's animation > : > : lambda = circumference / 9.5 > > Can't be, the red ray non-inertials 144 degrees and > the blue ray non-inertials 216 degrees. Yes, at different speeds but with the same wave spacing, Jerry's graphics are correct. > lamba 1 <> lambda2. > Perhaps Jeery non-inertialled the reference point. Of course, it is the source. > : in both directions. Hit the pause button when > : the two waves coincide and you can see they > : are the same as they should be. > > I did, the red ray non-inertials 144 degrees and > the blue ray non-inertials 216 degrees. Yes. > lambda 1 <> lambda2. No, they match exactly. > Perhaps Jeery non-inertialled the reference point. Yes, the source moves. > It is worse than that, you will refuse to measure the wavelength I'll put a screengrab on my website tonight if you doubt it, but if you really did use the pause, I shouldn't need to. > because you know the value refutes your claim that the speed is > c in all stationary frames of reference (stationary is the ancient > term for inertial). The simulation shows ballistci theory, not SR. George
From: George Dishman on 12 Oct 2007 10:24
"George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1192027957.218767.273550(a)57g2000hsv.googlegroups.com... > On 10 Oct, 09:13, "Androcles" <Engin...(a)hogwarts.physics> wrote: .... >> Correct, so there are two values for lambda, >> lambda_blue = L1/N = (2piR+vt)/9.5 >> lambda_red = L2/N = (2piR-vt)/9.5 >> which is NOT shown in Jeery's model and why >> Wilson can't see it (nor can you or Jeery). >> You shouldn't mix vt with alpha (length with angle), either. >> >> lambda_blue = L1/N = (2pi+alpha)R/9.5 >> lambda_red = L2/N = (2pi - alpha)R/9.5 > > Nah. In the inertial frame a peak on the blue > wave moves a distance of (c+v)/f in the period > of the source. The source itself move v/f, both > measured along the circumference. A second peak > is therefore emitted c/f behind the first and > therefter travels at the same speed so the > waves are that constant distance apart, that > is the wavelength. For Jerry's animation > > lambda = circumference / 9.5 > > in both directions. Hit the pause button when > the two waves coincide and you can see they > are the same as they should be. Here is a screenshot of the applet proving the point: http://www.georgedishman.f2s.com/Henri/SameWavelength.png Jerry's code is correct. George |