GETTING RID OF EINSTEIN RELATIVITY
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From: Dr. Henri Wilson on 18 Oct 2007 20:23 On Fri, 19 Oct 2007 00:38:59 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... >> On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" >> Thiis might help. >> www.users.bigpond.com/hewn/toothwheel.exe >> (It is not complete and doesn't seem to want to run properly on windows >> Vista...but not many things do). >> >> The toothed wheel represents a snapshot of BOTH rays when they are in >> phase at >> the emission point. > >Useless, the whole point is that they move at >different speeds. Show that and see what happens. > >> The teeth represent absolute wavelengths. > >That part is correct but just make each part move >at the right speed and it falls out automatically. The red and blue lines move at c+v and c-v. They represet the light rays in the non rotating frame. They travel for the same time and meet at the same instant at the detector. They travel different distances and do not meet in the same phase. >> It is not shown >> moving because the positions of the teeth at that instant is independent >> of >> light speed. > >That's your error, the waves have to get from source >to detector and that takes time. Both rays take the same time to get there. That is clearly shown in the animation. >> The red and blue lines show the rays moving at c+v and c-v. They travel >> for the >> same time and meet at the detector together. > >You need to show them as moving waves. They are photons not sqiggly lines. >> If you vary the ring rotation speed via the combo box, you will see that >> the >> phases of the rays when they reunite are not the same. >> >> If you want to argue against this approach, I remind you that it gives the >> experimentally verified answer. > >Irrelevant, it is wrong. As I said before, it gives >you the answer you WANT. It isn't the CORRECT answer. George you are now speaking out of sheer desperation . My result is exactly the same as that of SR. Are you admitting YOU and Einstein have been wrong all along? >I would also remind you that SR gives the right answer >if you want to use that argument. Now you are contradicing yourself. >>> >>>You forgot your "K" factor which says the photons _don't_ >>>get compressed. >> >> I didnt forget. > >Yes, you just conveniently ignored it. It's too small to worry about over the distances involved here. It occurs only during an acceleration anyway, when the number of wavelengths in the paths change and the fringes MOVE. >>>> 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 said the current DISPLACEMENT is integrated George. Can't you read? > >Yes, and the displacement is proportional to acceleration >according to ballistic theory, not speed. That's according to your misguided view of BaTh. This tells you what BaTh predicts: http://www.users.bigpond.com/hewn/ringgyro.htm >> The time >> periods for integration are also very small to account for any >> acceleration. >> >> Displacement is proportional to ring speed. > >Wrong. so 'wrong' it produces the SR answer. So that makes SR wrong too, eh George? > >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 18 Oct 2007 20:24 On Thu, 18 Oct 2007 23:48:23 GMT, "Androcles" <Engineer(a)hogwarts.physics> wrote: > >"George Dishman" <george(a)briar.demon.co.uk> wrote in message >news:1oCdnQCQyo-RcIranZ2dnUVZ8q2dnZ2d(a)pipex.net... >: >: "Clueless Henri Wilson" <HW@....> wrote in message >: news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... >: > On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" >: > <george(a)briar.demon.co.uk> wrote: >: >>"Clueless Henri Wilson" <HW@....> wrote in message >: >>news:rfo7h31d5h0pk32rjd8tnofil9pe357v4m(a)4ax.com... >: > >: >>>>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. >: > >: > Thiis might help. >: > www.users.bigpond.com/hewn/toothwheel.exe >: > (It is not complete and doesn't seem to want to run properly on windows >: > Vista...but not many things do). >: > >: > The toothed wheel represents a snapshot of BOTH rays when they are in >: > phase at >: > the emission point. >: >: Useless, > >Knee jerking fuckhead, cant stand being beaten. George is currently in a state of severe shock. By nice to him... > > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Androcles on 18 Oct 2007 21:10 "Dr. Henri Wilson" <HW@....> wrote in message news:9ktfh35dkoaltrga720c8ccd480b00v2s7(a)4ax.com... : On Fri, 19 Oct 2007 00:38:59 +0100, "George Dishman" <george(a)briar.demon.co.uk> : wrote: : : > : >"Clueless Henri Wilson" <HW@....> wrote in message : >news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... : >> On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" : : >> Thiis might help. : >> www.users.bigpond.com/hewn/toothwheel.exe : >> (It is not complete and doesn't seem to want to run properly on windows : >> Vista...but not many things do). : >> : >> The toothed wheel represents a snapshot of BOTH rays when they are in : >> phase at : >> the emission point. : > : >Useless, the whole point is that they move at : >different speeds. Show that and see what happens. : > : >> The teeth represent absolute wavelengths. : > : >That part is correct but just make each part move : >at the right speed and it falls out automatically. : : The red and blue lines move at c+v and c-v. Lines do not move. The tip of the line moves. : They represet the light rays in the non rotating frame. Yes. The tip of the line represents ONE photon. : They travel for the same time and meet at the same instant at the detector. Sure, but how far do they travel? Is it a) piR, half way around? b) 2piR, all the way around? c) 3piR? d) (2pi+alpha) R? e) Other? f) all of the above? g) none of the above? : They travel different distances and do not meet in the same phase. Phase is too difficult for you or Dishpan. Stick to travelling different distances. : : >> It is not shown : >> moving because the positions of the teeth at that instant is independent : >> of : >> light speed. : > : >That's your error, the waves have to get from source : >to detector and that takes time. : : Both rays take the same time to get there. That is clearly shown in the : animation. Yes, that is correct. Dishpan is a fuckhead to mention time. That's his error. : : >> The red and blue lines show the rays moving at c+v and c-v. They travel : >> for the : >> same time and meet at the detector together. : > : >You need to show them as moving waves. : : They are photons not sqiggly lines. Correct. : : >> If you vary the ring rotation speed via the combo box, you will see that : >> the : >> phases of the rays when they reunite are not the same. : >> : >> If you want to argue against this approach, I remind you that it gives the : >> experimentally verified answer. : > : >Irrelevant, it is wrong. As I said before, it gives : >you the answer you WANT. It isn't the CORRECT answer. : : George you are now speaking out of sheer desperation . Of course he is. He's a dumbfuck like Jeery. : My result is exactly the same as that of SR. Ah! So you ARE a closet Einstein Dingleberry! : Are you admitting YOU and Einstein have been wrong all along? Of course not, your result (whatever it is) is exactly the same as that of SR. : : >I would also remind you that SR gives the right answer : >if you want to use that argument. : : Now you are contradicing yourself. And you are going to as well, Einstein Dingleberry. : : >>> : >>>You forgot your "K" factor which says the photons _don't_ : >>>get compressed. : >> : >> I didnt forget. : > : >Yes, you just conveniently ignored it. : : It's too small to worry about over the distances involved here. Fringe shifts are very small, Georges Sagnac used a lot of magnification. Just ask the fuckhead Dishpan. : It occurs only during an acceleration anyway, when the number of wavelengths in : the paths change and the fringes MOVE. No, the shift is radians per second dependent by the coriolis effect. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/RLG.gif : : >>>> 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 said the current DISPLACEMENT is integrated George. Can't you read? : > : >Yes, and the displacement is proportional to acceleration : >according to ballistic theory, not speed. : : That's according to your misguided view of BaTh. BaTh has h-aether, unifuckation, Wombat Wilson's Wedge-on Worbits, tick fairies, 30 postulates. What other view is there? : This tells you what BaTh predicts: : http://www.users.bigpond.com/hewn/ringgyro.htm Oh yes, counting the distance R to R' twice as the tick fairy said, then taking the extra tick from the red and giving it to the blue. That's BaTh... : : >> The time : >> periods for integration are also very small to account for any : >> acceleration. : >> : >> Displacement is proportional to ring speed. : > : >Wrong. : : so 'wrong' it produces the SR answer. So that makes SR wrong too, eh George? Yeah, but what's the question?
From: George Dishman on 19 Oct 2007 09:45 "Clueless Henri Wilson" <HW@....> wrote in message news:9ktfh35dkoaltrga720c8ccd480b00v2s7(a)4ax.com... > On Fri, 19 Oct 2007 00:38:59 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:08ffh39c9j2ac1pligl8k1vi8fnmd3ba45(a)4ax.com... >>> On Thu, 18 Oct 2007 11:46:08 +0100, "George Dishman" >>> >>> The toothed wheel represents a snapshot of BOTH rays when they are in >>> phase at the emission point. >> >>Useless, the whole point is that they move at >>different speeds. Show that and see what happens. >> >>> The teeth represent absolute wavelengths. >> >>That part is correct but just make each part move >>at the right speed and it falls out automatically. > > The red and blue lines move at c+v and c-v. The waves move, the job of ballistic theory is to say how fast. Your program has no propagation whatsoever. George
From: George Dishman on 19 Oct 2007 10:12
"sean" <jaymoseley(a)hotmail.com> wrote in message news:1192460319.117732.178540(a)q5g2000prf.googlegroups.com... .... > Its difficult to model the source when it doesnt exist. But your > right,.. > Ive been inconsistent when saying it doesnt effect the light when its > off. Ill retract that statement and put it this way.. > " In the source frame the light travels away from where the source > was located at c in that frame, even when the source no longer > emits light..." So what is your answer to my question then Sean, does the light get dragged round by the torch if you turn it to point in a different direction after the light has been emitted and the torch has been switched off: "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:YIqdnQWzWOTlAGfbnZ2dneKdnZydnZ2d(a)pipex.net... > > ... Suppose you decide you don't like me and throw a > ball to hit me. To do that you face me. I am 50m away > and you throw it at 10m/s. "S" is Sean, "G" is George > and "B" is the ball: After 1s it looks like this: > > S B . . . G > > The dots are just to mark off units of 10m. After 2s: > > S . B . . G > > After 3s you change your mind and decide I'm OK after > all so you turn 90 degrees to your left. You are saying > the picture is now like this: > > . > . > B > . > . > S . . . . G > > and after 4s like this > > . > B > . > . > . > S . . . . G > > and after 5s, when the ball should have hit me, it is > like this: > > B > . > . > . > . > S . . . . G > > Do you see how silly that sounds? > > Now instead of a ball, imagine a flash of light leaving > a laser. The flash lasts just 0.1ns so is just 30mm long. > You fire the laser at me and after about 33ns it has > moved 10m: > > S - . . . G > > After 67ns you are still facing me and it has moved this > far: > > S . - . . G > > After 100ns you decide you wish you hadn't fired it at > me so you turn through 90 degrees in a fraction of a > nanosecond. Does the laser flash swing round like the > ball above so it is now moving up the screen and I am > safe like this: > > . > . > | > . > . > S . . . . G > > Then this: > > . > | > . > . > . > S . . . . G > > and so on, or is it going to carry on towards me and put > a hole through me like this: > > S . . - . G > > My opinion is that it keeps going regardless of what > you do and I should duck or else. George |