A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
in [Relativity]
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From: eric gisse on 14 Feb 2010 19:57 ...@..(Henry Wilson DSc) wrote: [...] > The fact that the frequencies in each direction are different annihilates > the Roberts, Jerry, Andersen 'rotating frame' argument as well as your > own. ....and your evidence that the frequencies are different is based on what? > > > > Henry Wilson... > > .......provider of free physics lessons
From: Henry Wilson DSc on 14 Feb 2010 21:21 On Mon, 15 Feb 2010 11:50:45 +1100, "Inertial" <relatively(a)rest.com> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >news:a05hn55udgqmbu5rs41ap1shceh27dki2d(a)4ax.com... >> On Mon, 15 Feb 2010 10:28:50 +1100, "Inertial" <relatively(a)rest.com> >> wrote: >> >>>"Henry Wilson DSc" <..@..> wrote in message >>>news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... >>>> Two different arrival rates produces a beat frequency at the detector. >>> >>>There is no beat frequency in a Sagnac device. The two beams have the >>>same >>>frequency at the detector. This is the case in both emission theory and >>>SR. >>>SR, however, predicts differing arrival TIMES for pulses emitted at the >>>same >>>time .. hence the shift. Emission theory says they arrive at the same >>>time >>>(as you agree) .. hence there is no shift predicted (this refuting the >>>theory). >>> >>>That your theory is even worse and somehow predicts differing arrival >>>rates >>>and a beat frequency that does not occur is enough proof to refute it. >> >> You still don't get it. > >I get it just fine > >> The two beams DO NOT have the same frequency at the detector. > >Yes, they do in reality. That your theory predicts otherwise shows it to be >wrong. > >> In the clock example I gave, 40000062 pulses arrive at the detector from >> one >> direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of >> any >> particular split pulse. Those numbers change during a speed CHANGE. > >And that is nonsense .. as the time taken for each pulse to make the trip is >the same (in both directions), and the time between pulses being emitted is >the same (in both directions), then in a given time, the same number of >pulses arrive (in both directions). Try thinking for once. I know this is difficult...but you should be able to understand. The path lengths are different. The distance between pulses is 1 metre in all frames. The number of pulses in each path differs by 124. When a pulse is split into two and sent down both fibres, it takes the same time to get to the detector in both directions. Both halves arrive together. So 124 more pulses have arrived from one direction than the other. IN OTHER WORDS, the pulse arrival rate is different from both paths. >> Convert that to 'wavecrests', in the case of light, and you get a beat >> frequency that is somehow related to fringe displacement. Don't ask me how >> at >> the moment but the reason will reveal quite a lot about the nature of a >> photon. > >Because it is nonnsese. There is no beat frequency. > >> The fact > >Lie > >> that the frequencies in each direction are different > >They are not. > >> annihilates the >> Roberts, Jerry, Andersen 'rotating frame' argument as well as your own. > >Nope .. you've just proven your own argument wrong. Well done. Use your brain... Henry Wilson... ........provider of free physics lessons
From: Inertial on 14 Feb 2010 21:32 "Henry Wilson DSc" <..@..> wrote in message news:0fbhn59irun7u51r5r9u2dnbtv4ehbi11d(a)4ax.com... > On Mon, 15 Feb 2010 11:50:45 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message >>news:a05hn55udgqmbu5rs41ap1shceh27dki2d(a)4ax.com... >>> On Mon, 15 Feb 2010 10:28:50 +1100, "Inertial" <relatively(a)rest.com> >>> wrote: >>> >>>>"Henry Wilson DSc" <..@..> wrote in message >>>>news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... >>>>> Two different arrival rates produces a beat frequency at the detector. >>>> >>>>There is no beat frequency in a Sagnac device. The two beams have the >>>>same >>>>frequency at the detector. This is the case in both emission theory and >>>>SR. >>>>SR, however, predicts differing arrival TIMES for pulses emitted at the >>>>same >>>>time .. hence the shift. Emission theory says they arrive at the same >>>>time >>>>(as you agree) .. hence there is no shift predicted (this refuting the >>>>theory). >>>> >>>>That your theory is even worse and somehow predicts differing arrival >>>>rates >>>>and a beat frequency that does not occur is enough proof to refute it. >>> >>> You still don't get it. >> >>I get it just fine >> >>> The two beams DO NOT have the same frequency at the detector. >> >>Yes, they do in reality. That your theory predicts otherwise shows it to >>be >>wrong. >> >>> In the clock example I gave, 40000062 pulses arrive at the detector from >>> one >>> direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME >>> of >>> any >>> particular split pulse. Those numbers change during a speed CHANGE. >> >>And that is nonsense .. as the time taken for each pulse to make the trip >>is >>the same (in both directions), and the time between pulses being emitted >>is >>the same (in both directions), then in a given time, the same number of >>pulses arrive (in both directions). > > Try thinking for once. I know this is difficult...but you should be able > to > understand. > The path lengths are different. The distance between pulses is 1 metre in > all > frames. The number of pulses in each path differs by 124. When a pulse is > split > into two and sent down both fibres, it takes the same time to get to the > detector in both directions. Both halves arrive together. > So 124 more pulses have arrived from one direction than the other. > IN OTHER WORDS, the pulse arrival rate is different from both paths. > > >>> Convert that to 'wavecrests', in the case of light, and you get a beat >>> frequency that is somehow related to fringe displacement. Don't ask me >>> how >>> at >>> the moment but the reason will reveal quite a lot about the nature of a >>> photon. >> >>Because it is nonnsese. There is no beat frequency. >> >>> The fact >> >>Lie >> >>> that the frequencies in each direction are different >> >>They are not. >> >>> annihilates the >>> Roberts, Jerry, Andersen 'rotating frame' argument as well as your own. >> >>Nope .. you've just proven your own argument wrong. Well done. > > Use your brain... I did .. and you just refuted your own theory. Saying 'it predicts something that doesn't happen and somehow instead what is observed does happen anyway and my theory cannot explain why' means your theory is refuted.
From: Henry Wilson DSc on 14 Feb 2010 21:45 On Mon, 15 Feb 2010 13:32:23 +1100, "Inertial" <relatively(a)rest.com> wrote: > >"Henry Wilson DSc" <..@..> wrote in message >news:0fbhn59irun7u51r5r9u2dnbtv4ehbi11d(a)4ax.com... >> On Mon, 15 Feb 2010 11:50:45 +1100, "Inertial" <relatively(a)rest.com> >> wrote: >>> >>>> annihilates the >>>> Roberts, Jerry, Andersen 'rotating frame' argument as well as your own. >>> >>>Nope .. you've just proven your own argument wrong. Well done. >> >> Use your brain... > >I did .. and you just refuted your own theory. Saying 'it predicts >something that doesn't happen and somehow instead what is observed does >happen anyway and my theory cannot explain why' means your theory is >refuted. Don't kid yourself. By counting the pulses arriving from the two directions, I can calculate the earth's rotation speed. I don't need an interferometer. A clock will do just as well. Sagnac fully supports BaTh. Henry Wilson... ........provider of free physics lessons
From: Inertial on 14 Feb 2010 21:52
"Henry Wilson DSc" <..@..> wrote in message news:0fbhn59irun7u51r5r9u2dnbtv4ehbi11d(a)4ax.com... > On Mon, 15 Feb 2010 11:50:45 +1100, "Inertial" <relatively(a)rest.com> > wrote: > >> >>"Henry Wilson DSc" <..@..> wrote in message >>news:a05hn55udgqmbu5rs41ap1shceh27dki2d(a)4ax.com... >>> On Mon, 15 Feb 2010 10:28:50 +1100, "Inertial" <relatively(a)rest.com> >>> wrote: >>> >>>>"Henry Wilson DSc" <..@..> wrote in message >>>>news:1d0hn59n8ofm6um5qk7l1nds3bgv32q3rs(a)4ax.com... >>>>> Two different arrival rates produces a beat frequency at the detector. >>>> >>>>There is no beat frequency in a Sagnac device. The two beams have the >>>>same >>>>frequency at the detector. This is the case in both emission theory and >>>>SR. >>>>SR, however, predicts differing arrival TIMES for pulses emitted at the >>>>same >>>>time .. hence the shift. Emission theory says they arrive at the same >>>>time >>>>(as you agree) .. hence there is no shift predicted (this refuting the >>>>theory). >>>> >>>>That your theory is even worse and somehow predicts differing arrival >>>>rates >>>>and a beat frequency that does not occur is enough proof to refute it. >>> >>> You still don't get it. >> >>I get it just fine >> >>> The two beams DO NOT have the same frequency at the detector. >> >>Yes, they do in reality. That your theory predicts otherwise shows it to >>be >>wrong. >> >>> In the clock example I gave, 40000062 pulses arrive at the detector from >>> one >>> direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME >>> of >>> any >>> particular split pulse. Those numbers change during a speed CHANGE. >> >>And that is nonsense .. as the time taken for each pulse to make the trip >>is >>the same (in both directions), and the time between pulses being emitted >>is >>the same (in both directions), then in a given time, the same number of >>pulses arrive (in both directions). > > Try thinking for once. I know this is difficult... It is for you > but you should be able to > understand. I do > The path lengths are different. > The distance between pulses is 1 metre in all > frames. The number of pulses in each path differs by 124. When a pulse is > split > into two and sent down both fibres, it takes the same time to get to the > detector in both directions. Both halves arrive together. > So 124 more pulses have arrived from one direction than the other. > IN OTHER WORDS, the pulse arrival rate is different from both paths. So the following is equivalent to what you have here: You have two paths from (say) A to B. You have n objects leaving A over a period of 1 second down one path and taking time T to get to B. The first object arrives at time T, the last at time T+1. You have another n objects leaving A over a period of 1 second down the other path and taking the same time T to get to B. The first object arrives at time T, the last at time T+1. Yet by your logic, you would claim that more would arrive per second at B from one path than the other. That is just pure impossible nonsense. You also claim that even though in any given time you emit the same number of pulses in both directions, and that they are all in the tube for the same time, that there are more pulses in the tube going in one direction than their are in the other. Where do the extra pulses come from? That is also just pure impossible nonsense. At any given time the tube has the same length in both directions .. the length is the circumference of the tube, it does not change. And as you say that the pulses are the same distance apart, there must be the same number of them in the tube at a time. But you claim that at any given time there are more pulses in the tube going in one direction than that other. That is also just pure impossible nonsense. You really have just totally refuted your 'thoery'. It is all based on pure impossible nonsense. |