GETTING RID OF EINSTEIN RELATIVITY
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From: Androcles on 16 Oct 2007 20:47 "Dr. Henri Wilson" <HW@....> wrote in message news:s6fah3hr1l27bkm48mtodblbhl79afkk5u(a)4ax.com... : On Mon, 15 Oct 2007 22:41:48 GMT, "Androcles" <Engineer(a)hogwarts.physics> : wrote: : : > : >"Dr. Henri Wilson" <HW@....> wrote in message : >news:u8p7h3l41okcp8vsnitaikbp10e04fipan(a)4ax.com... : : >: >>>You get the same wavelength by either method. : >: > : >: >Get it yet? : >: > : >: >"You get the same wavelength by either method." : >: > : >: >I am saying your claim is _correct_ so stop and : >: >_think_ before disagreeing! : >: : >: I'm basicaly disagreeing with him. : > : > : >That's good, Wilson, I want it on record that you disagreed with me. : > : >"There are about 13 waves in the blue and 6 in the red." --Wilson, 12 Oct : >2007. : >"the number of waves between the source and detector is : >the same in both paths and constant." -- Wilson, 15 Oct 2007. : : Both above statements are correct. : It is no fault of mine if you cannot get into your head that the 'stationary : emission point' is NOT the same thing as the 'moving source'. It is no fault of mine if you cannot count. http://www.androcles01.pwp.blueyonder.co.uk/tickfairy.gif How many teeth on each wheel have I drawn? Is it A) 20 B) 21 C) 22 D) 23 E) 24 F) 25 G) 26 H) 27 I) 28 J) 29 K) 30 L) Other ......................... (specify) : >"That's the kind of argument I'd expect from a desperate : >person....completely out of ideas... ahahahaha!" -- Wilson. : > : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA : >AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA! : : get off the whisky.... "That's the kind of argument I'd expect from a desperate person....completely out of ideas... ahahahaha!" -- Wilson.
From: George Dishman on 17 Oct 2007 03:37 On 16 Oct, 23:45, HW@....(Clueless Henri Wilson) wrote: > On Tue, 16 Oct 2007 00:50:38 -0700, George Dishman <geo...(a)briar.demon.co.uk> wrote: > >On 15 Oct, 23:24, HW@....(Clueless Henri Wilson) wrote: > >> On Sun, 14 Oct 2007 14:52:37 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: > >> >"Clueless Henri Wilson" <HW@....> wrote in message > >... > >> >Henry, have you got the applet running yet? If not, > >> >do so before you go any farther. I'll wait until you > >> >can see it before replying to any of your other posts. > > >> I have the applet running.... > > >Excellent. > > >> and it's quite useless. > >> I'm creating a better one. > > >> >If (when) you can, watch what happens. The yellow line > >> >marks the source when the signal is first emitted and > >> >you can watch the discontinuity at the start of the > >> >waves moving round. While that is happening you can > >> >see the source emitting waves in sync. However, there > >> >is but no equivalent detection mark. > > >> >The yellow dot marks the coordinates in the inertial > >> >frame where the magenta dots are emitted and in fact > >> >Jerry could have removed the yellow line when she added > >> >the dots. The green dot marks the location where the > >> >magenta dots are detected. The idea is that you can > >> >see that the crests marked by the magenta dots move > >> >away from the yellow dot at the right speeds for > >> >ballistic theory. If the pixels-per-second for the > >> >source line from the yellow dot is v then the magenta > >> >dots need to move at c+v and c-v pixels-per-second. > > >> I suspected that was the case. > >> Now count the wavelengths in each path. They are different and proportional to > >> speed. > >> This does not mean the number of wavelengths is NOT the same between source and > >> detector. It always remains the same (at any constant speed) > > >So Jerry's simulation is not useless, you finally > >learned what it shows. You can also now watch the > >phase of the two signals as they pass the yellow > >dot. Can you see that their phase changes with > >time? Can you work out why? If so, you should > >understand why trying to use the path length > >difference in the inertial frame doesn't easily > >tell you the phase at the detector. That's why > >the math on your web page is wrong. > > George, your mistake is in assuming that the rays are always in phase when they > meet. they are not. The signal emitted is of the form Ve = A * sin(w*t + phi) for both rays. Both take the same time to reach the detector, say T. The signal received is therefore Vr = A * sin(w*t + phi - T) for both rays, they are in phase. > >> Fringe movement and displacement change occurs only during an acceleration, > >> when 'wavelength' in each path is NOT the same. > > >No, fringe displacement is proportional to the > >speed-induced phase difference at the detector. > > That's what I said. You said ".. displacement change occurs .." when you should have said ".. displacement occurs ..". > The is a phase difference at the detector.....determined by the difference in > number of wavelengths in each path. No, look at Jerry's simulaton again. > >Jerry's code is correct. > > It might be...but it doesn't model a ring gyro. True, it models Sagnac. > >> >Try to think that through Henry, I'm sure it will > >> >give you some difficulty. > > >> I have already George. Much of what you say is basically correct except that > >> you are completely overlooking the vital question of what happens during an > >> acceleration. > >> The answer is, "everything". > > >The fringes will vary in a complex way but there > >is no memory in the system. Sagnac didn't switch > >on the light bulb until the speed was stable so > >what would have happened to fringes had he switched > >on the bulb is moot, only the phase difference > >while the bulb was on matters. > > George, the rays are not in phase when they reunite. > You seem to think this should be the case just because the travel times are the > same. That does not follow. It does Henry, see the maths above. > Think again George. Look again at jerry's animation, look specifically at the leading edge when the waves start and watch it arrive at the detector. Notice that the other wave arrives at the same time but more importantly the radius of the two waves is the same and they stay the same thereafter. Pause the display and look back round the circumference until you reach the source and note they are being emitted in phase 9.5 cycles earlier. Ballistic theory requires that they be in phase, they hit the detector 9.5 cycles after they were emitted. From the yellow dot, they _move_ a distance greater than 9.5 wavelengths but they do that in the time of 9.5 cycles because they move at c+v while the wavelength is independent of the speed. George
From: Androcles on 17 Oct 2007 05:06 "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1192606637.531895.161470(a)q5g2000prf.googlegroups.com... : On 16 Oct, 23:45, HW@....(Clueless Henri Wilson) wrote: : > On Tue, 16 Oct 2007 00:50:38 -0700, George Dishman <geo...(a)briar.demon.co.uk> wrote: : > >On 15 Oct, 23:24, HW@....(Clueless Henri Wilson) wrote: : > >> On Sun, 14 Oct 2007 14:52:37 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: : > >> >"Clueless Henri Wilson" <HW@....> wrote in message : > >... : > >> >Henry, have you got the applet running yet? If not, : > >> >do so before you go any farther. I'll wait until you : > >> >can see it before replying to any of your other posts. : > : > >> I have the applet running.... : > : > >Excellent. : > : > >> and it's quite useless. : > >> I'm creating a better one. : > : > >> >If (when) you can, watch what happens. The yellow line : > >> >marks the source when the signal is first emitted and : > >> >you can watch the discontinuity at the start of the : > >> >waves moving round. While that is happening you can : > >> >see the source emitting waves in sync. However, there : > >> >is but no equivalent detection mark. : > : > >> >The yellow dot marks the coordinates in the inertial : > >> >frame where the magenta dots are emitted and in fact : > >> >Jerry could have removed the yellow line when she added : > >> >the dots. The green dot marks the location where the : > >> >magenta dots are detected. The idea is that you can : > >> >see that the crests marked by the magenta dots move : > >> >away from the yellow dot at the right speeds for : > >> >ballistic theory. If the pixels-per-second for the : > >> >source line from the yellow dot is v then the magenta : > >> >dots need to move at c+v and c-v pixels-per-second. : > : > >> I suspected that was the case. : > >> Now count the wavelengths in each path. They are different and proportional to : > >> speed. : > >> This does not mean the number of wavelengths is NOT the same between source and : > >> detector. It always remains the same (at any constant speed) : > : > >So Jerry's simulation is not useless, you finally : > >learned what it shows. You can also now watch the : > >phase of the two signals as they pass the yellow : > >dot. Can you see that their phase changes with : > >time? Can you work out why? If so, you should : > >understand why trying to use the path length : > >difference in the inertial frame doesn't easily : > >tell you the phase at the detector. That's why : > >the math on your web page is wrong. : > : > George, your mistake is in assuming that the rays are always in phase when they : > meet. they are not. : : The signal emitted is of the form : : Ve = A * sin(w*t + phi) : : for both rays. Bullshit, Dishpan! The signals emitted are of the form V1(t) = A * sin( (w+alpha)*t + phi) V2(t) = A * sin( (w-alpha)*t + phi) for each ray.
From: George Dishman on 17 Oct 2007 11:11 On 17 Oct, 10:06, "Androcles" <Engin...(a)hogwarts.physics> wrote: > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message > news:1192606637.531895.161470(a)q5g2000prf.googlegroups.com... > > : > : The signal emitted is of the form > : > : Ve = A * sin(w*t + phi) > : > : for both rays. > > Bullshit, Dishpan! > > The signals emitted are of the form > V1(t) = A * sin( (w+alpha)*t + phi) > V2(t) = A * sin( (w-alpha)*t + phi) > for each ray. Assuming you are using "alpha" to be the angle Henry described by that symbol, you are wrong. The signals are emitted from the source, not some distance away. Ve is the signal at the source at time t. George
From: Androcles on 17 Oct 2007 13:19
"George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1192633862.967267.136760(a)i13g2000prf.googlegroups.com... : On 17 Oct, 10:06, "Androcles" <Engin...(a)hogwarts.physics> wrote: : > "George Dishman" <geo...(a)briar.demon.co.uk> wrote in message : > news:1192606637.531895.161470(a)q5g2000prf.googlegroups.com... : > : > : : > : The signal emitted is of the form : > : : > : Ve = A * sin(w*t + phi) : > : : > : for both rays. : > : > Bullshit, Dishpan! : > : > The signals emitted are of the form : > V1(t) = A * sin( (w+alpha)*t + phi) : > V2(t) = A * sin( (w-alpha)*t + phi) : > for each ray. : : Assuming you are using "alpha" to be the angle Henry : described by that symbol, you are wrong I am using alpha as the angle alpha in http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/SagnacIdiocy.htm and as the angle phi in http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf and you are an ignorant fuckhead. See http://home.c2i.net/pb_andersen/pdf/sagnac_ring.pdf to discover why Tusseladd's Sagnac analysis is wrong. According to Tusseladd, t_f = (2 * pi * R)/(c+v) but he's not allowed to use c+v (Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img76.gif) and he left out his angle phi. Now THAT is a crank who cannot manage schoolboy geometry. It should be t = (2 * pi + phi)* R/(c+v) and of course t = (2 * pi - phi)* R/(c-v) If he recalls, x' = x-vt, so the equivalent here is x = 2piR = circumference vt = phiRt (pronounced "fart", remember this is in Norwegian) or circumference' = circumference - phiRt and to mimic Einstein's coordinate system he should have 1/2 [ tau(0,0,0,t) + tau (0,0,0, t+ circumference'/(c+v) + circumference'/(c-v) ] = tau (circumference',0,0, t+circumference'/(c-v)) to match http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif from which he can derive his cuckoo malformations. Androcles, hoping to be amused by Tusseladd for another 8 years. |