GETTING RID OF EINSTEIN RELATIVITY
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From: George Dishman on 18 Oct 2007 06:39 "Androcles" <Engineer(a)hogwarts.physics> wrote in message news:7v_Qi.174649$BW4.156988(a)fe1.news.blueyonder.co.uk... > > "George Dishman" <george(a)briar.demon.co.uk> wrote in message > news:1192520099.391084.63890(a)e34g2000pro.googlegroups.com... > : On 15 Oct, 19:45, "Androcles" <Engin...(a)hogwarts.physics> wrote: > : > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message > : > news:1192460718.255292.114600(a)v23g2000prn.googlegroups.com... > : > : <snip history> > : > : > : Not even close, the waves move at v+c and v-c > : > : while yours are static. > : > > : > The wave IS static. The photons moves at v+/-c. > : > That's the entire confusion about wave and particle duality. > : > : I see what you are saying now, this is the one > : that conveys your view; > > Good, you are learning. > > : > : http://www.androcles01.pwp.blueyonder.co.uk/AC/Photon.gif > : > : You haven't thought it through though. > > Oh yes I have. You are still in the process of thinking it through, > you haven't thought it yet, through. > > > > : Consider > : what is seen at one location versus time as the > : photon passes. Before it arrives there is no > : field and once it has passed you see a residual > : DC level. > > http://www.kettering.edu/~drussell/Demos/SHO/damp.html > > Consider what is seen at one location versus time as the > mass on a spring (say a car) passes. Before it arrives there is no > height above the road and once it has passed you see a residual > height level. Well that's what your thin line sine wave seems to be saying but technically it would be something close to a finite height delta function unless you imply some sort of memory effect. > I know you were crazy, Dishpan, all relativists are. I didn't realize > just how crazy. Whatever crazy idea you were trying to convey, it certainly doesn't describe the moving waves we see in real life. George
From: George Dishman on 18 Oct 2007 06:43 "Clueless Henri Wilson" <HW@....> wrote in message news:j8o7h3d33a3c1pvaue9a79810eqsshcsh8(a)4ax.com... > On Mon, 15 Oct 2007 00:35:51 -0700, George Dishman > <george(a)briar.demon.co.uk> wrote: .... >>Both source and detector are on the turntable. >>At constant angular speed, the path lengths ^^^^^^^^^^^^^^^^^^^^^^^^^ >>are constant so there is no Doppler shift, as >>many ticks arrive at the detector as left the >>source in the same time - no fairies needed. > > George, the path lengths are NOT constant. > The number of wavelengths between source and detector IS constant at > constant > speed. Your problem is you cannot read, see above. Stop wasting our time by posting knee-jerk contradictions. George
From: George Dishman on 18 Oct 2007 06:46 "Dr. Henri Wilson" <HW@....> wrote in message news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... > On Sun, 14 Oct 2007 15:30:44 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:kjg3h35ok4vf1n6ik43dl2ga79s1q0o327(a)4ax.com... >>> On Sat, 13 Oct 2007 17:27:08 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> wrote: >>>> >>>>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.. >> >>It gives you the answer you WANT. It isn't the CORRECT >>answer, you need to learn how an interferometer works. >>Start with the applet and instructions in my other post >>and I'll try to educate you when you immediately leap >>to the wrong conclusion. > > George, the path lengths are different..and change with rotation speed. And the waves are not in phase at the start point of that path, they are in phase at the end. > At constant speed, no matter what that speed is, the fringe pattern is the > same > and static......and the number of waves between the source and detector is > the > same in both paths and constant. Exactly. > The two path lengths change and fringes move during an acceleration. This > is > due to the fact that photons and their 'wavelengths' shrink or expand > during an > acceleration. You forgot your "K" factor which says the photons _don't_ get compressed. > The movement is monitored. In iFoGs there is a constant > integration over very short time intervals to turn current fringe > displacement > into total angle turned. Integrating something proportional to angular acceleration doesn't give you the angle turned. > I know you wont be able to understand any of this until I demonstrate it. > An > animation is on the way. I know you won't be able to understand any of this until you learn basic calculus. George
From: George Dishman on 18 Oct 2007 07:02 "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. > >>>>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. They both do but at different speeds so the nodes of common phase move past as well. You have to add the difference in phase caused by the different path lengths to the phase diffrerence at the start point to find the phase at the end point and you are ignoring that factor. That's why you get a value that conflicts with other ways of working out the same number. > That doesn't matter. The detector has moved by vt also. Yes but you have to account both both changes, you are ignoring one. >>> 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. Bollocks, press the button to change the speed. >>>>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. The spees are correct, count them yourself. >>> ,....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. Press the button to change it clueless. >>>>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. Fringe displacement does not depend on the phase at the emission point, the emulsion isn't there. >>> 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)? No. A point is a fixed set of coordinates. > Will you then admit that you have been deluded for years about the use of > the > rotating frame to refute Sagnac? Physically the waves are in phase at the source (technically the beam splitter) not the offset location you are thinking of. You can work out the experiment in either the rotating or inertial frames and you get the same answer, no fringe displacement for constant speed, constant displacement for constant acceleration and constant rate of fring movement for constant jerk. >>>>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. That's what Jerry's program does. > It moves 13 wavelengths in the blue and 6 in the red before it hits the > detector. 11.2 versus 7.8, you cannot count. > The travel times for both rays are the same. And that determines the phase, not the number of wavelengths. > 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. That is when fringe displacement 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. <sigh> Yet again .... the yellow dot shows where an arbitrary wave crest is emitted, the magenta dots are fixed to the crest to help you follow them round, the green dot marks where they hit the detector. You can see that they hit it simultaneously and since they both mark a crest it is obvious that the waves are arriving in sync. You can pause the applet twice and see how far the magenta dots move compared to how far the source moves to check the speeds - they are correct. > 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. Either magenta dot marks the wavecrest. The emission point is the yellow dot. The detection point is the green dot. Pause the display just as the magenta dots arrive at the green. You will see the waves are in phase at the detector but not at the yellow dot. You are right in that the number of waves differ but they are in phase at the detector end of the path, not at the yellow dot end. >>>>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? The relative phase at the source CANNOT affect the emulsion. > That means there IS fringe displacement. I can see you are clueless about how a photographic plate works. George
From: Androcles on 18 Oct 2007 08:00
"George Dishman" <george(a)briar.demon.co.uk> wrote in message news:7tydnbB4ntLZqoraRVnyhwA(a)pipex.net... : : "Androcles" <Engineer(a)hogwarts.physics> wrote in message : news:7v_Qi.174649$BW4.156988(a)fe1.news.blueyonder.co.uk... : > : > "George Dishman" <george(a)briar.demon.co.uk> wrote in message : > news:1192520099.391084.63890(a)e34g2000pro.googlegroups.com... : > : On 15 Oct, 19:45, "Androcles" <Engin...(a)hogwarts.physics> wrote: : > : > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message : > : > news:1192460718.255292.114600(a)v23g2000prn.googlegroups.com... : > : : > : <snip history> : > : : > : > : Not even close, the waves move at v+c and v-c : > : > : while yours are static. : > : > : > : > The wave IS static. The photons moves at v+/-c. : > : > That's the entire confusion about wave and particle duality. : > : : > : I see what you are saying now, this is the one : > : that conveys your view; : > : > Good, you are learning. : > : > : : > : http://www.androcles01.pwp.blueyonder.co.uk/AC/Photon.gif : > : : > : You haven't thought it through though. : > : > Oh yes I have. You are still in the process of thinking it through, : > you haven't thought it yet, through. : > : > : > : > : Consider : > : what is seen at one location versus time as the : > : photon passes. Before it arrives there is no : > : field and once it has passed you see a residual : > : DC level. : > : > http://www.kettering.edu/~drussell/Demos/SHO/damp.html : > : > Consider what is seen at one location versus time as the : > mass on a spring (say a car) passes. Before it arrives there is no : > height above the road and once it has passed you see a residual : > height level. : : Well that's what your thin line sine wave seems to : be saying but technically it would be something : close to a finite height delta function unless you : imply some sort of memory effect. Dishpan, if a car passed you yesterday it had a height level. That doesn't mean it has a residual one today, even if the graph you drew shows one for yesterday. No matter what it SEEMS like to you, the sine wave has TIME along its axis. Even the water wave you think is travelling toward the shore (why not away?) makes the little boats move UP AND DOWN, (giving them a height residual from yesterday). LOL! There is no memory effect, you have a brain defect. : > I know you were crazy, Dishpan, all relativists are. I didn't realize : > just how crazy. : : Whatever crazy idea you were trying to convey, it : certainly doesn't describe the moving waves we see : in real life. The little boats move UP AND DOWN, Dishpan, giving them a memory height residual. You really hallucinate that mass on a spring shown at http://www.kettering.edu/~drussell/Demos/SHO/damp.html has a sine wave along the position axis, Dishpan? When your fuddled brain works out the difference between x and t let us know and we'll discuss physics instead of your psychosis. |