GETTING RID OF EINSTEIN RELATIVITY
in [Relativity]
|
Prev: What is the Aether?
Next: Debunking Nimtz
From: Dr. Henri Wilson on 8 Oct 2007 19:59 On Mon, 8 Oct 2007 23:44:18 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... >> On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" >>> >>>The numbers are equal. >> >> George, I know this is difficult. > >Clearly, since you can't even look at the simulation >and see that it already shows exactly what you are >telling me. > >> Go back to the spinning wheel idea. >> Consider a wheel with 1000000 teeth around the edge. >> It will represent the light as it moves around the ring. Its 'edge speed' >> will >> be c+v, where v is still to be defined. >> You will agree that no matter how fast it spins, the number of teeth >> remains >> 1000000. > >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. Jerry has not included ring speed at all. >> Now draw two lines in the rest frame on either side of the wheel. The >> number of >> teeth between the two lines is 500000, no matter how fast the wheel spins. >> One line represents the source at the instant of emission of a particular >> photon or wavecrest. The source is moving but the line isn't. The other >> line >> will represent the position of the detector when the wavecrest arrives. >> >> 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. 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. >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. >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 >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 9 Oct 2007 03:13 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: > >"Clueless Henri Wilson" <HW@....> wrote in message news:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... > >> On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" > > >>>The numbers are equal. > > >> George, I know this is difficult. > > >Clearly, since you can't even look at the simulation > >and see that it already shows exactly what you are > >telling me. > > >> Go back to the spinning wheel idea. > >> Consider a wheel with 1000000 teeth around the edge. > >> It will represent the light as it moves around the ring. Its 'edge speed' > >> will > >> be c+v, where v is still to be defined. > >> You will agree that no matter how fast it spins, the number of teeth > >> remains > >> 1000000. > > >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. 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. > Jerry has not included ring speed at all. 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. > >> Now draw two lines in the rest frame on either side of the wheel. The > >> number of > >> teeth between the two lines is 500000, no matter how fast the wheel spins. > >> One line represents the source at the instant of emission of a particular > >> photon or wavecrest. The source is moving but the line isn't. The other > >> line > >> will represent the position of the detector when the wavecrest arrives. > > >> 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. > 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. > >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. > >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. George
From: Dr. Henri Wilson on 9 Oct 2007 05:08 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. >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. It matters not how fast the rays move. Jerry has not included ring speed at all. >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... :) ,....that's very reassuring.. for me... >> >> 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. >> 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... You were wrong on that...you are wrong again about this. >> >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. 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. >> >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... > >George Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Androcles on 10 Oct 2007 04:13 "George Dishman" <george(a)briar.demon.co.uk> wrote in message news:1191914039.526807.231170(a)22g2000hsm.googlegroups.com... : 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: : > >"Clueless Henri Wilson" <HW@....> wrote in message news:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... : > >> On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" : > : > >>>The numbers are equal. : > : > >> George, I know this is difficult. : > : > >Clearly, since you can't even look at the simulation : > >and see that it already shows exactly what you are : > >telling me. : > : > >> Go back to the spinning wheel idea. : > >> Consider a wheel with 1000000 teeth around the edge. : > >> It will represent the light as it moves around the ring. Its 'edge speed' : > >> will : > >> be c+v, where v is still to be defined. : > >> You will agree that no matter how fast it spins, the number of teeth : > >> remains : > >> 1000000. : > : > >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. : 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. : : > Jerry has not included ring speed at all. : : 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. It's like pulling teeth with him. He doesn't believe the arithmetic if it's counter to his theory, and he doesn't believe the arithmetic if it agrees with his theory. : : > >> Now draw two lines in the rest frame on either side of the wheel. The : > >> number of : > >> teeth between the two lines is 500000, no matter how fast the wheel spins. : > >> One line represents the source at the instant of emission of a particular : > >> photon or wavecrest. The source is moving but the line isn't. The other : > >> line : > >> will represent the position of the detector when the wavecrest arrives. : > : > >> 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. Typical relativist trick, that is. Throw in bigger numbers to confuse the matter, then say it's obvious. : : > 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 Ah... What you have to remember is the waves do NOT move, they are standing waves. http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/eightframe.gif The entire wheel is turning as shown by the broken spoke, but the spokes are not turning. Same thing here: http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif There are only 6 frames, beat frequency shows up when the teeth are aligned. The blue wheel moves at v=4 , the yellow at v+c = (4+1) = 5,and the red at v-c = (4-1) = 3. OR... The blue wheel moves at v=4 , the yellow at v+c = (4+61) modulo 60 = 5, and the red at v-c = (4-61) modulo 60 = 3 or -57. There are 60 teeth. : 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. : : > >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. There is, the waves are not shown in the stationary frame, which is why they APPEAR TO MOVE. The wave does NOT move in the stationary frame. This wave does NOT move unless the mass moves in x. http://www.kettering.edu/~drussell/Demos/SHO/damp.html The idiot Wilson just mumbles "irrelevant" because he's stupid. : : > >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. Correct, so there are two values for lambda, lambda_blue = L1/N = (2piR+vt)/9.5 lambda_red = L2/N = (2piR-vt)/9.5 which is NOT shown in Jeery's model and why Wilson can't see it (nor can you or Jeery). You shouldn't mix vt with alpha (length with angle), either. lambda_blue = L1/N = (2pi+alpha)R/9.5 lambda_red = L2/N = (2pi - alpha)R/9.5
From: Androcles on 10 Oct 2007 04:19
"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. Yes it IS. You are showing your confusion. THE WAVELENGTH is not the same in boh rays lambda_blue =(2pi+alpha)R/9.5 lambda_red = (2pi - alpha)R/9.5 Changing the NUMBER of waves is Wilson tick fairy work. |