Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 15 Oct 2005 18:27 On Sat, 15 Oct 2005 10:04:39 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:1u60l1t5qumtkfhqh5g9d45oqo4u64h3of(a)4ax.com... > >>>> As far as I can see, there is no radial velocity between any two >>>> components in ANY frame. >>> >>>Right, but there are a couple of complex points you >>>are missing even neglecting the complication of using >>>different frames for each leg. The origin of the >>>mirror frame is not inertial, it is moving in a >>>circle so you get an apparent centrifugal force >>>between the source and the mirror. That means you >>>also get "gravitational redshift" and Shapiro delay. >> >> you can believe that if you want to. I wont. > >I'm not talking SR here, I think just applying >Galilean relativity to Newtonian mechanics will >give you an equivalent result. You might want to >call it something like "centrifugal force redshift" >to distinguish them Hmm. Using rotating frames always introduces some confusion. >>>Also the speed of the light varies with the radius >>>because the frame is accelerating. There are lots >>>of tricky aspects to consider that come from using >>>an accelerating origin. >> >> The whole thing is very tricky....particularly when nobody has a real clue >> as >> to what a 'photon' actually is. > >You may not. You do not either. >>>It enters too. If you take all the above into account, >>>you will still get different answers if you compare a >>>model where the light is moving at c in the lab frame >>>or at c+kv where k depends on the number of mirrors. >> >> I maintain that each mirror is moving normally in the frame of the next. > >I agree, the motion is purely transverse. > >> This >> means that source speed makes no difference to the fringe shifts. > >No it doesn't, it means the radial distance is >constant. That in turn means the time taken is >inversely proportional to the speed. Working >out the speed becomes complex in the mirror >frame. I thnk you have gotten that wrong. I think the speed remains c but the path length changes. >>>> We know why the MMX predicts a null result. >>>> That's a very simple application of the BaTh. >>> >>>Yes, but Ritz predicts null for both MMX and Sagnac. >> >> Only in YOUR imagination. > >I've shown you the maths repeatedly or you can >find it on several pages on the web if you don't >trust me. It is a simple statement of fact that >Ritz produces a null prediction for Sagnac. I have now told you why Ritz was wrong then. >>>If you introduce a new feature like a "photon gyro" >>>to produce a non-null result for Sagnac, it may also >>>produce a non-null prediction for the MMX. >> >> It doesn't. > >You don't know that, you don't have any equations >for your "photon gyro" concept so you can't do the >calculation to find out. George, you SR 'explanation' is really just the LET explanation. As always, SR reverts to LET when it tries to go physical instead of plain mathematical. > >> The MMX result is straight forward. > >Without your "photon gyro" it is, you have no >idea how that might affect it. But you know that the fringes become displaced only when movement is actually occuring. They return to their original positions when the rotation ceases. I doubt if Michelson watched while he rotated the apparatus ,...but if he had, it would have done what any sagnac would do. > >> If the MMX apparatus is rotated, it becomes a sagnac. > >That depends on the details but generally it >won't. The delay in Sagnac is propotional to the >area enclosed by the path but in MMX the light >usually returns back along the same path on each >leg so the ennclosed area is zero. Not when the MMX is rotating. A sagnac interferometer is just a Michelson one set up so that the fringe offset can be observed during rotation. > >>>That's >>>why you can't just guess, you have to write out the >>>equations and work each of the experiments. If you >>>can't provide the equations, you don't have a theory, >>>just a speculation. >> >> One must begin with speculation. > >Sure, you have a speculation but not a theory. well forget the 'photon gyro' business. I think that the path length difference is not affected by source speed because that is always normal to the next mirror (in that mirror's frame), not 45 degrees as you previously thought. > >George > HW. www.users.bigpond.com/hewn/index.htm see: www.users.bigpond.com/hewn/variablestars.exe "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong".
From: Henri Wilson on 15 Oct 2005 18:33 On Sat, 15 Oct 2005 10:09:48 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <H@..> wrote in message >news:hp70l1d2jeh8nki5qk2si9fro9lkbulhf6(a)4ax.com... >> On Fri, 14 Oct 2005 00:29:18 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> >> wrote: >... >>>The point is that even though there is a finite >>>propagation time, there is still no "transverse >>>Doppler". >> >> But that only applies at the very centre. >> >> To an observer at any point between the centre and the whistle, there IS >> doppler. The length between the whistle and observer is then constantly >> changing. > >Yes and that is because the radial distance is changing. >If you want to calculate the shift, you only need to >know the rate of change of radial distance, there is no >component dependent on the transverse speed, hence no >"transverse Doppler". > >> It reaches maximum and minimum when the three points are in line. > >Should you have said zero at those points? You talk of >the maximum again next. Yes, sorry, I changed my line of thought half way throught the sentence. > >> I haven't worked out the position where maximum doppler occurs.....but >> that >> shouldn't be too difficult. > >Indeed but it depends on the paths and speeds of both >objects. Interestingly, if detectors are placed at two different radii, I think it should be possible to calculate the true radius of the whistle path and its speed, if they weren't already known. > >George > HW. www.users.bigpond.com/hewn/index.htm see: www.users.bigpond.com/hewn/variablestars.exe "Sometimes I feel like a complete failure. The most useful thing I have ever done is prove Einstein wrong".
From: Eric Gisse on 15 Oct 2005 20:52 Henri Wilson wrote: [snip] I find it interesting that you completely ignored every point George made about you not having any equations so you can make actual predictions.
From: George Dishman on 16 Oct 2005 03:51 "Henri Wilson" <H@..> wrote in message news:tj03l11qdo4p2femd24onrf8f7gjqtd007(a)4ax.com... > On Sat, 15 Oct 2005 10:09:48 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Henri Wilson" <H@..> wrote in message >>news:hp70l1d2jeh8nki5qk2si9fro9lkbulhf6(a)4ax.com... >>> >>> I haven't worked out the position where maximum doppler occurs.....but >>> that shouldn't be too difficult. >> >>Indeed but it depends on the paths and speeds of both >>objects. > > Interestingly, if detectors are placed at two different radii, I think it > should be possible to calculate the true radius of the whistle path and > its > speed, if they weren't already known. I think for one detector, you only get the angular velocity but with two, you can effectively use the two angles to locate the source. George
From: George Dishman on 16 Oct 2005 04:23
"Henri Wilson" <H@..> wrote in message news:bqv2l1lcficiap82dlub1bfn07nna0f0pf(a)4ax.com... > On Sat, 15 Oct 2005 10:04:39 +0100, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <H@..> wrote in message >>news:1u60l1t5qumtkfhqh5g9d45oqo4u64h3of(a)4ax.com... > >> >>>>> As far as I can see, there is no radial velocity between any two >>>>> components in ANY frame. >>>> >>>>Right, but there are a couple of complex points you >>>>are missing even neglecting the complication of using >>>>different frames for each leg. The origin of the >>>>mirror frame is not inertial, it is moving in a >>>>circle so you get an apparent centrifugal force >>>>between the source and the mirror. That means you >>>>also get "gravitational redshift" and Shapiro delay. >>> >>> you can believe that if you want to. I wont. >> >>I'm not talking SR here, I think just applying >>Galilean relativity to Newtonian mechanics will >>give you an equivalent result. You might want to >>call it something like "centrifugal force redshift" >>to distinguish them > > Hmm. Using rotating frames always introduces some confusion. For the iFOG where the path is circular, it is easier. With mirrors where the path is chords, the lab frame is easier. However, isn't it an interesting that Sagnac gives a hand-waving way to predict the equivalence of acceleration and gravity? >>>>Also the speed of the light varies with the radius >>>>because the frame is accelerating. There are lots >>>>of tricky aspects to consider that come from using >>>>an accelerating origin. >>> >>> The whole thing is very tricky....particularly when nobody has >>> a real clue as to what a 'photon' actually is. >> >>You may not. > > You do not either. No, but the people who have studied QED do. >>>>It enters too. If you take all the above into account, >>>>you will still get different answers if you compare a >>>>model where the light is moving at c in the lab frame >>>>or at c+kv where k depends on the number of mirrors. >>> >>> I maintain that each mirror is moving normally in the frame of the next. >> >>I agree, the motion is purely transverse. >> >>> This >>> means that source speed makes no difference to the fringe shifts. >> >>No it doesn't, it means the radial distance is >>constant. That in turn means the time taken is >>inversely proportional to the speed. Working >>out the speed becomes complex in the mirror >>frame. > > I thnk you have gotten that wrong. > I think the speed remains c but the path length changes. I think you misunderstood me. The general result is that in calculating the time taken, the speed still matters if the distance is constant. I also think that the speed of emission would remain equal to c irrespective of the angular velocity according to Ritz. However, the path length does not change because "each mirror is moving normally in the frame of the next". If there is no distance change and Ritz predicts no speed change, then it also predicts no change to the time taken, hence a null output. >>>>> We know why the MMX predicts a null result. >>>>> That's a very simple application of the BaTh. >>>> >>>>Yes, but Ritz predicts null for both MMX and Sagnac. >>> >>> Only in YOUR imagination. >> >>I've shown you the maths repeatedly or you can >>find it on several pages on the web if you don't >>trust me. It is a simple statement of fact that >>Ritz produces a null prediction for Sagnac. > > I have now told you why Ritz was wrong then. Yes, we are no in agreement :-) >>>>If you introduce a new feature like a "photon gyro" >>>>to produce a non-null result for Sagnac, it may also >>>>produce a non-null prediction for the MMX. >>> >>> It doesn't. >> >>You don't know that, you don't have any equations >>for your "photon gyro" concept so you can't do the >>calculation to find out. > > George, you SR 'explanation' is really just the LET explanation. No Henri it isn't even vaguely close. I tried to explain some of the differences and you seemed to be getting an inkling of how SR works but you stopped replying to that thread. See message news:dfaarq$9ag$1(a)news.freedom2surf.net > As always, SR reverts to LET when it tries to go physical instead of plain > mathematical. You do that, SR uses geometry alone to derive the LTs and is then mathematical. >>> The MMX result is straight forward. >> >>Without your "photon gyro" it is, you have no >>idea how that might affect it. > > But you know that the fringes become displaced only when movement is > actually > occuring. > They return to their original positions when the rotation ceases. > I doubt if Michelson watched while he rotated the apparatus ,...but if he > had, > it would have done what any sagnac would do. No, there would have been no shift or only a tiny shift if they used multiple passes along the legs to increase the effective distance, the reflection points need to be slightly offset. >>> If the MMX apparatus is rotated, it becomes a sagnac. >> >>That depends on the details but generally it >>won't. The delay in Sagnac is propotional to the >>area enclosed by the path but in MMX the light >>usually returns back along the same path on each >>leg so the ennclosed area is zero. > > Not when the MMX is rotating. A sagnac interferometer is just a Michelson > one > set up so that the fringe offset can be observed during rotation. For Sagnac the first leg is A->B->C->D->A A--->---B | | | v ^ | | | D---<---C That encloses an area and the empirical result is that the shift is proportional to the area. In MMX the first leg goes from A to B and back: A-<--->-B | | | | D Neither leg encloses any area so empirically we expect no shift when rotating. Your ideas on a "photon gyro" would suggest to me that the act of rotating should produce a shift even in MMX but that is hand-waving. Until you produce the equations, we can only guess ... as I said: >>>>That's >>>>why you can't just guess, you have to write out the >>>>equations and work each of the experiments. If you >>>>can't provide the equations, you don't have a theory, >>>>just a speculation. >>> >>> One must begin with speculation. >> >>Sure, you have a speculation but not a theory. > > well forget the 'photon gyro' business. I think we have to until you can write down the equations so we can both apply them. Degrees of freedom for a photon are also intimately tied to thermodynamic considerations and you would need to do a pile of work to show your idea didn't fail those experiments. It's not something we can do here. It's also somewhat academic since part of the design of iFOGs is to apply a modulation to the light and it also exhibits the same time delay. You might find a way to use a "photon gyro" to explain the carrier shift but not the modulation. > I think that the path length difference is not affected by source speed Indeed, the path length isn't affected, but the time taken to traverse that path depends on the emission speed. I think that's obvious. Constant speed means constant time taken, but the experiment shows variable time hence Ritz is wrong. > because > that is always normal to the next mirror (in that mirror's frame), not 45 > degrees as you previously thought. I never thought that Henri, I said the _light_ path is at 45 degrees to the mirror hence the speed in the _lab_ frame is c + v / sqrt(2) according to Ritz. In the non-rotating mirror-centred frame, the source is moving in a circle but the light moves on a curved path at variable speed, it all gets very complicated! George |