GETTING RID OF EINSTEIN RELATIVITY
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From: Dr. Henri Wilson on 13 Oct 2007 18:35 On Fri, 12 Oct 2007 15:40:18 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"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: >>>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. Using that figure leads to your mistake. Do you think the waves emitted at the yellow line disappear somehow? There are about 13 waves in the blue and 6 in the red. >>>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. The blue one does. That doesn't matter. The detector has moved by vt also. >> 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. The demo is for only one speed of the ring. >>>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. It is a sure sign you are wrong. > >> ,....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. Jerry's demo shows only one ring speed. >>>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. Jerry has shown that for constant speed, the common phase at the detector differs from the common phase at the emission point. That is what is observed...fringe displacement but no fringe movement. >> >> 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. George, do you o do you not admit that in the rotating frame, the emission point moves (backwards)? Will you then admit that you have been deluded for years about the use of the rotating frame to refute Sagnac? >>>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. George, plot the movement of one wave. It moves 13 wavelengths in the blue and 6 in the red before it hits the detector. The travel times for both rays are the same. The length of the blue path is 13 wavelengths, that of the red, 6. The number of wavelengths in each path changes only during an acceleration. that is when fringe movement occurs. >> 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. This is absolute nonsense. What the hell the yellow and green dots are supposed to represent is anyone's guess. George, consider the distance moved by one 'wavecrest'. At constant rotation speed, it moves from the emission point (3 o'clock yellow line) to the detector. The blue path is 2piR+vt, the red is 2piR -vt is as I calculated. >>> >>>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. At constant speed, the waves leave the source in phase and arrive at the detector in phase..BUT AT A DIFFERENT PHASE ANGLE....CAN'T YOU SEE THAT? That means there IS fringe displacement. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 13 Oct 2007 18:40 On Fri, 12 Oct 2007 15:24:08 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"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: > >... >> >> 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 George, the emission point is the yellow line. ....but even if you use the yellow dot, you should be able to see that the (common) phase angle at the detector is different from the one at the emission point. >Jerry's code is correct. It might be, although the yellow dot is meaningless. It certanly shows that the number of wavelengths in each path is speed dependent, as observed. ....but it doesn't show how fringes move in and out of the paths during an acceleration. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Androcles on 13 Oct 2007 20:12 "Dr. Henri Wilson" <HW@....> wrote in message news:gff2h3t1b7usrumct7fq6ae36bdql9fbsl(a)4ax.com... : 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. You are counting the same 3.5 of them twice: http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/z4.gif Thirteen "wavelengths" (distance a photon has to move in one of its cycles) are needed to catch up to the moving source in one revolution when moving counterclockwise, eleven are needed in the clockwise direction. Yet at any given instant there are exactly twelve in the circumference. The tick fairy changed 1 of them in my model, 3.5 of them in Jeery's model. You pair of idiots cannot count and are oh-so-easy to fool. http://www.androcles01.pwp.blueyonder.co.uk/tickfairy.gif How many teeth on each wheel?
From: Androcles on 13 Oct 2007 20:12 "Dr. Henri Wilson" <HW@....> wrote in message news:50g2h39b2gg6pv4e1sm847cgeue23ubscq(a)4ax.com... : On Fri, 12 Oct 2007 15:40:18 +0100, "George Dishman" <george(a)briar.demon.co.uk> : wrote: : : > : >"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: : : >>>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. : : Using that figure leads to your mistake. : Do you think the waves emitted at the yellow line disappear somehow? : : There are about 13 waves in the blue and 6 in the red. You pair of morons cannot count. http://www.androcles01.pwp.blueyonder.co.uk/tickfairy.gif How many teeth in each wheel?
From: Dr. Henri Wilson on 14 Oct 2007 03:26
On Sat, 13 Oct 2007 17:27:08 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Androcles" <Engineer(a)hogwarts.physics> wrote in message > >When you realise Henry counts 11.2 over 424.4 degrees >and 7.8 waves over 205.6 degrees, you might understand >what he is trying to do. With a bit more help from me, >you might even understand why his approach is never >going to work. It DOES work....and it produces the right answer.. Sory George, sagnac is 100% supportive of BaTh. > >> : > : lambda_blue =(2pi + alpha)R / 11.2 >> : > : lambda_red = (2pi - alpha)R / 7.8 >> : > : >> : > : You get the same wavelength by either method. >> >> With 11.2 trains and 7.8 trains on the rollercoaster tracks >> that the Dishpan tick fairy put there. > > Wavelength/R = 424.4 / 11.2 = 37.9 degrees > Wavelength/R = 295.6 / 7.8 = 37.9 degrees > >> : > So the tick fairy sent 11.2 pulses from the source >> : > in one direction and 7.8 pulses from the source in >> : > the other direction. >> : >> : Wrong again, from the _SOURCE_, it sent 9.5 in each >> : direction. The source MOVES. >> >> So the tick fairy sent 11.2 pulses from the source >> in one direction and 7.8 pulses from the source in >> the other direction. > >Nope, the source emitted 11.2 waves over a longer >time than the 7.8, Henry adds in waves by counting >over more than the circumference ofor one direction >and doesn't count some that were emitted when working >the other direction, he is clueless. So clueless he has found all your previous mistakes and shown that Sagnac is fully supportive of BaTh. > >I don't need to learn Henry's mistakes, there are 9.5 in >both directions from source to detector regardless of >the table speed as long as it is constant. George, you hav ten fingers. Try counting the number of waves in each path....and don't use Jerry's stupid program, the detector and all movement should stop when the leading edge of the waves arrives there. >> Why the moving observer sees a shift that Jeery doesn't show: >> http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/RLG.gif > > >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |