GETTING RID OF EINSTEIN RELATIVITY
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From: George Dishman on 12 Oct 2007 10:26 "Androcles" <Engineer(a)hogwarts.physics> wrote in message news:XV%Oi.307600$xp6.305213(a)fe3.news.blueyonder.co.uk... > "Clueless Henri Wilson" <HW@....> wrote in message > news:e7gmg3hgn3dqejmkarjnb1l83m5hp2q5rh(a)4ax.com... > : On Tue, 09 Oct 2007 00:13:59 -0700, George Dishman > <george(a)briar.demon.co.uk> wrote: > : >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: > : > : >> >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. > : > : The number of waves between the static emission point and the end point > is > : certainly not the same in both rays. > > Yes it IS. You are showing your confusion. > THE WAVELENGTH is not the same in boh rays http://www.georgedishman.f2s.com/Henri/SameWavelength.png Yes it is, Henry is correct on that point. > lambda_blue =(2pi+alpha)R/9.5 > lambda_red = (2pi - alpha)R/9.5 > Changing the NUMBER of waves is Wilson tick fairy work. George
From: George Dishman on 12 Oct 2007 10:40 "Dr. Henri Wilson" <HW@....> wrote in message news:e7gmg3hgn3dqejmkarjnb1l83m5hp2q5rh(a)4ax.com... > On Tue, 09 Oct 2007 00:13:59 -0700, George Dishman > <george(a)briar.demon.co.uk> > wrote: > >>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: > >>> >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. > > The number of waves between the static emission point and the end point is > certainly not the same in both rays. The above number is from source to detector at any instant. >>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. > > :) No George, that's not how it works out. > The distance between the emission point and the detector increses with > ring > speed. Jerry hasn't shown that. The numberof absolute wavelengths in that > distance is D/lambda = 2piR+vt /lambda. Yes, but that doesn't tell you the end result because the waves move past that point while the light is in flight. > It matters not how fast the rays move. Yes it does, you need that to work out the phase of the wave passing your point. > Jerry has not included ring speed at all. Yes she has. >>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. > > Nice to see you have Adrocles on side... :) Yes, and unusual. > ,....that's very reassuring.. for me... Not really, he confirmed Jerry _did_ include the ring speed so he proved you wrong. Still, you have now admitted she was right all along since you posted this. >>> >> 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. > > that's only because Jerry programmed them to do just that. Nope, she programmed the speeds correctly and that is the result, look at the code and see for yourself. >>> 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 >>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. > > George, your original 'proof' that Sagnac refuted BaTh was based on the > simple > rotating frame idea. I have now found the error in that argument. The > emission > point moves (backward) in the rotating frame... The reference point where the emission is in-phase is the current location of the source. > You were wrong on that...you are wrong again about this. I was right, you are wrong again. >>> >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. > > George, the number of waves in each path is different. Not from source to detector, and the source is where the waves start out in phase. > Jerry's animation shows that. The emission point is at 3 O'Clock. There > are 3.8 > waves in one ray and 6 in the other...of course you have to add the 9.5 to > those numbers. http://www.georgedishman.f2s.com/Henri/NumberOfWaves.png Follow the red wave back from the detector anti-clockwise to the yellow dot and you get about 7.8 waves. Follow the blue wave clockwise (the last bit has been overlayed by the red) and you get about 11.2 waves. However, notice that the waves at this time are NOT in phase at the yellow dot because they move at differentn speeds past it. The waves ARE in phase at the detector because they are both 9.5 waves away from the source which is where they start out in phase. >>> >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. > > Still you don't get it... Still you cannot even look at a simple picture and see what is obvious, the waves hit the detector in phase which means NO fringe displacement. George
From: Androcles on 12 Oct 2007 13:22 "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:yaOdnXJtlK-nHpLanZ2dnUVZ8v-dnZ2d(a)pipex.net... : : "Androcles" <Engineer(a)hogwarts.physics> wrote in message : news:XV%Oi.307600$xp6.305213(a)fe3.news.blueyonder.co.uk... : > "Clueless Henri Wilson" <HW@....> wrote in message : > news:e7gmg3hgn3dqejmkarjnb1l83m5hp2q5rh(a)4ax.com... : > : On Tue, 09 Oct 2007 00:13:59 -0700, George Dishman : > <george(a)briar.demon.co.uk> wrote: : > : >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: : > : : > : >> >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. : > : : > : The number of waves between the static emission point and the end point : > is : > : certainly not the same in both rays. : > : > Yes it IS. You are showing your confusion. : > THE WAVELENGTH is not the same in boh rays : : http://www.georgedishman.f2s.com/Henri/SameWavelength.png : : Yes it is, Henry is correct on that point. NO IT ISN'T, WILSON IS WRONG. Another believer in tick fairies. http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif 24 ticks blue, 24 ticks red, same frequency, different distance, different speed, different Dopper shift. That's two that idiots that can't count. I wonder how many more there are... : : > lambda_blue =(2pi+alpha)R/9.5 : > lambda_red = (2pi - alpha)R/9.5 : > Changing the NUMBER of waves is Wilson tick fairy work. : : George : :
From: George Dishman on 12 Oct 2007 17:21 "Androcles" <Engineer(a)hogwarts.physics> wrote in message news:u3OPi.322498$xp6.195098(a)fe3.news.blueyonder.co.uk... > "George Dishman" <george(a)briar.demon.co.uk> wrote in message > news:yaOdnXJtlK-nHpLanZ2dnUVZ8v-dnZ2d(a)pipex.net... > : "Androcles" <Engineer(a)hogwarts.physics> wrote in message > news:XV%Oi.307600$xp6.305213(a)fe3.news.blueyonder.co.uk... .... > : > Yes it IS. You are showing your confusion. > : > THE WAVELENGTH is not the same in boh rays > : > : http://www.georgedishman.f2s.com/Henri/SameWavelength.png > : > : Yes it is, Henry is correct on that point. > > NO IT ISN'T, WILSON IS WRONG. He is correct, I already sketched it in another reply so see the copy below. In time 1/f, the first wave moves a distance (c+v)/f while the source moves a distance v/f. The second is therefore emitted c/f behind the first and they then move with the same speed so the distance between the waves remains that value as it propagates. That is the wavelength. > Another believer in tick fairies. > http://www.androcles01.pwp.blueyonder.co.uk/Dualwave.gif > 24 ticks blue, 24 ticks red, same frequency, > different distance, different speed, different Dopper shift. You cited the wrong diagram, this one is correct: http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif The teeth on the wheels are equally spaced. > That's two that idiots that can't count. I wonder how many more > there are... No, you're own your own this time, even Henry got it right. > : > lambda_blue =(2pi+alpha)R/9.5 > : > lambda_red = (2pi - alpha)R/9.5 > : > Changing the NUMBER of waves is Wilson tick fairy work. The number 9.5 is from the source to the detector so it is (2*pi*R)/9.5 for both waves. http://www.georgedishman.f2s.com/Henri/SameWavelength.png Or you can count from the yellow dot to the green dot: http://www.georgedishman.f2s.com/Henri/NumberOfWaves.png lambda_blue =(2pi + alpha)R / 11.2 lambda_red = (2pi - alpha)R / 7.8 You get the same wavelength by either method. "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:V4-dnelk2qlSlpnanZ2dneKdnZydnZ2d(a)pipex.net... > > ... Consider a stationary source: > > _ > / \ > S+---+---+---+---+- > | \_/ > | > |<--8-->| > _ | _ > / \ |/ \ > S+---+---+---+---+- > \_/ \_/ > |<--8-->| > > The wavelength above is 8 characters and in one > period the wave will move 8 characters too. Now > compare that with the moving source. During one > period, the source moves 5 characters from S to X: > _ _ > / \ / \ > S+---+---+---+---+- > | \_/ \_/ > | > |<----13---->| > | | > |<5>| | > | _ | _ > | / \ |/ \ > ----X+---+---+---+---+- > \_/ \_/ > |<--8-->| > > The wavelength remains 8 characters but the > distance moved by the wave is now 13 characters.
From: Dr. Henri Wilson on 13 Oct 2007 17:53
On Fri, 12 Oct 2007 22:21:51 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Androcles" <Engineer(a)hogwarts.physics> wrote in message >news:u3OPi.322498$xp6.195098(a)fe3.news.blueyonder.co.uk... > > http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif > >The teeth on the wheels are equally spaced. > >> That's two that idiots that can't count. I wonder how many more >> there are... > >No, you're own your own this time, even Henry got it right. > >> : > lambda_blue =(2pi+alpha)R/9.5 >> : > lambda_red = (2pi - alpha)R/9.5 >> : > Changing the NUMBER of waves is Wilson tick fairy work. > >The number 9.5 is from the source to the detector so >it is (2*pi*R)/9.5 for both waves. The number from the emision point is 13 blue, 6 red. > http://www.georgedishman.f2s.com/Henri/SameWavelength.png > >Or you can count from the yellow dot to the >green dot: > > http://www.georgedishman.f2s.com/Henri/NumberOfWaves.png > > lambda_blue =(2pi + alpha)R / 11.2 > lambda_red = (2pi - alpha)R / 7.8 > >You get the same wavelength by either method. > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |