GETTING RID OF EINSTEIN RELATIVITY
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From: Dr. Henri Wilson on 4 Oct 2007 16:49 On Wed, 3 Oct 2007 23:06:49 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:dh28g31kcobj79u7uprguto7o4l10eef88(a)4ax.com... >>>> All lengths are absolute AND FRAME INDEPENDENT.:) >> >> :) >> George, Sagnac doesn't rely on the source frame. It uses the non-rotating >> frame. > >Rubbish, the frame is just YOUR choice of a coordinate >system in which you are going to do your calculations. >Nature doesn't know about coordinate systems. > >> Rotating frames can be very confusing and lead to errors like the one >> you just made. > >The error is your Henry, you have divided lengths >from two different frames. You can choose either >frame but having made the choice you have to stick >with values from that one frame, you cannot mix them. George, I am using one frame, the inertial one. Lengths are absolute and the same in all frames. >>>No, the wavelength is invariant in ballistic theory >>>but the path length and distance moved are not, both >>>change. >> >> It is the path length differences in the non rotating fame that determines >> fringe movement and displacement. > >You can choose that if you like, and then you need >to divide by the distance moved by the wave in that >frame, not the distance moved in the rotating frame >as you have done. I'm NOT using the rotating frame...you are.. wavelength...like all lengths.... is absolute and has the same value in ALL frames. >> http://www.users.bigpond.com/hewn/ringgyro.htm > >Yep, same error. No error George.. The blade of a chainsaw has the same number of teeth no matter how fast it runs. >>>True but irrelevant, the unchanging length is >>>the wavelength, not the distance moved. The >>>distance from an object to a point which is at >>>fixed coordinates in a frame depends on the >>>choice of frame. >> >> George, the distance moved by the ray is 2pi.R+vt. > >That's the total distance moved, yes. > >> The wavelength is L. > >The distance moved by the wave is longer than the >wavelength in the inertial frame, you failed to >take that into account. > >> The number of wavelengths in the path is unequivocably (2pi.R+vt)/L., no >> matter >> what is the speed. > >The number between the source and detector at any >instant is unequivocably 2.pi.R/L no matter what >the speed. This is true and highlights the error one can make when trying to use rotating frames. >Rotating frame: N = (2.pi.R) / L > >Inertial frame: N = (2pi.R+vt) / ((1+v/c).L) > >You get the same number whichever frame you choose >and the number is same for both directions. What you are saying is that if you mark 1000 equally spaced points around an wheel, then there will be 1000 marks around it no matter how fast it rotates....and the absolute distance bewteen them is always the same. That's very clever...and it can even be used as a model here. What we really want to know is the number of marks around wheel between the points that are marked say, 999 and 1000. The answer is '999 one way and 1001 the other'...and again that number does not change with rotation speed. Let's use the wheel to simulate sagnac. We actually need three such wheels, each with say 10,000,000 marks around the edges. Two of the wheels will spin around the same axis at high speeds (c) and in opposite directions, to simulate the frame of the light. The other will rotate very slowly (at v) or not at all around that axis, simulating the rotation of the ring. Let's say the slow wheel moves ten marks (from points 1 to 10) for every complete rotation of the other two. The numbers of marks on the fast wheels that lie between the points 1 and 10 on the slow wheel are 10,000,010 and 9,999,990 in the two directions. We already established that this number is not dependent on the speed of the fast wheels...so we can speed them up to c+v or c-v and still get the same result.....which is what hapens in the ring gyro. So you can see that the two numbers of marks varies when and only when the ratio of v/c changes, c being a universal constant. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 4 Oct 2007 17:09 On Wed, 3 Oct 2007 23:27:34 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:6vjtf3d2q6sprg1t8e4kg5qlbc429cpj4i(a)4ax.com... >> >> It has 'wavelength'...the absolute distance between the wave peaks. > >No, to get the envelope, you need a continuum >of overalapping sine waves hence there is a >continuum of wavelengths too. Treat the window >function as AM applied to a CW signal and you >should realise the transform gives a carrier >plus sidebands. > >> Frequency, as we know it, is the rate at which those peaks arrive. > >Your illustration requires multiple sine waves >with different frequencies. > >>>>>It is the mirror that moves at c+v, the pretty pattern >>>>>behind it is irrelevant, and without another mirror >>>>>following behind, you need continuous power from the >>>>>source. The whole thing is nuts, but you could learn >>>>>a lot about Fourier analysis by ignoring that flaw and >>>>>working through the maths. Most of it you can find done >>>>>for you on the web. >>>> >>>> I don't know what you are talking about. >>>> >>>> What mirror? 'c+v' wrt what? >>> >>>Your pattern needs waves moving in both directions. >>>The "mirror" is whatever you propose reflects the >>>the wave going away from the source back towards it. >>>You haven't said anything about that yet. >> >> The intention was to illustrate Maxwell's model. > >Your model does not do that, this does: > > http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35 That's just a graph of field intensity. Mine is supposed to show the field intensity itself, varying in space...and moving along. >This also shows the differential equation for the >electric field which your solution must satisfy: > > http://www.analyzemath.com/antenna_tutorials/wave_propagation.html > >There are several more you can find with a cursory >search. No need, the concept is simple. >> I still have a pretty open mind about the nature of photons. I still >> reckon >> they might just be a spinning pair of charges....in which case the field >> strength tapers off with distance away. After all, there are lots of >> charges >> around. > >Clueless, "the field" is the effect of photons. That's a typically meaningless Dishman statement.. >>>"c+v" is the normal sense of the phrase in ballistic >>>theory, the speed at which the measurable energy >>>conveyed by the packet travels. The envelope that >>>you have drawn is what moves at c+v in ballistic >>>theory, not the underlying sine waves from which >>>it is constructed. >> >> Sure. > >I thought you should agree that, perhaps my first >statement was just unclear. George, what you don't seem to understand is that the idea of 'field strength' in space is not just a graph drawn on a piece of paper. It has a physical structure. A quality of that space changes with the field. Have you ever considered what a 'field gradient' might physically involve? Of course you haven't. You are incapable. >> I say there is no connection between the intrinsic wave and say, an RF >> signal. > >That's a different matter. What we know is that >the detectable wave aspect of a single photon >has exactly the same frequency as the macroscopic >signal of which it is part from the single photon >grating experiment. ....that does not follow. You would have to perform the experiment using photons that make up an RF or similar signal, not a light beam.. >The high speed underlying waves >required for your model are unrelated to either. > >> Mind you, it would really matter if there was a connection. It would >> throw some light on the problem. > >You have misunderstood your model entirely. You have misunderstood what I was modelling. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 5 Oct 2007 04:06 On 4 Oct, 21:49, HW@....(Clueless Henri Wilson) wrote: > On Wed, 3 Oct 2007 23:06:49 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message news:dh28g31kcobj79u7uprguto7o4l10eef88(a)4ax.com... <big snip> I've trimmed most because I think the next bit explains all of the rest as well and it might get lost in a morasse of minor points. Your wheels analogy is a good idea. In fact it would be better if you read on down to that before replying and come back to the earlier part afterwards, or even snip it if you prefer, but it explains again why phase controls the fringes. > >The distance moved by the wave is longer than the > >wavelength in the inertial frame, you failed to > >take that into account. > > >> The number of wavelengths in the path is unequivocably (2pi.R+vt)/L., no > >> matter what is the speed. > > >The number between the source and detector at any > >instant is unequivocably 2.pi.R/L no matter what > >the speed. > > This is true and highlights the error one can make when trying to use rotating > frames. There can be no error, all frames _must_ give the same answer. > >Rotating frame: N = (2.pi.R) / L > > >Inertial frame: N = (2pi.R+vt) / ((1+v/c).L) > > >You get the same number whichever frame you choose > >and the number is same for both directions. > > What you are saying is that if you mark 1000 equally spaced points around an > wheel, then there will be 1000 marks around it no matter how fast it > rotates....and the absolute distance bewteen them is always the same. > That's very clever...and it can even be used as a model here. > > What we really want to know is the number of marks around wheel between the > points that are marked say, 999 and 1000. NO, that's the fundamental point. Suppose at some instant one positive peak is just being emitted from the source and both path lengths are exactly 1000 wavelengths. A positive peak emitted 1000 cycles earlier is just arriving at the detector and you get constructive interference and a bright fringe. On the other hand, if one path is half a wavelength different at 999.75 and 1000.25 wavelengths, you get destructive interference and a dark fringe. I'll do the next bit in both frames so read both before replying please: What you need to know therefore, working in the rotating frame, is how many cycles fit between the source and the detector along these slightly curved paths: http://www.briar.demon.co.uk/Henri/paths.gif It should be obvious that the number of waves along each path will be the same. Now imagine you drew that out but with a superimposed sine wave stretched along the paths like Androcles' attempts but without his errors. Translate that static picture into the inertial frame and the path length now is the distance from source at the moment the first cycle was emitted but the waves also get stretched along that path by the same factor. That is *NOT* the sme as the wavelength in the inertial frame! If you took a snapshot of the current situation, the waves in flight at any instant must start from the _present_ location of the source and the wavelength then is the invariant value. > The answer is '999 one way and 1001 the other'...and again that number does not > change with rotation speed. No the numbers cannot depend on the choice of frame. They are obviously 1000 on each path as seen in the rotating frame so they must be 1000 on each path in the inertial frame too. When you realise the total length you used is along the flight path of the light from the location that is reaching the detector, from S in your diagram and not the present position S', you should also realise that the distance you must divide by isn't the wavelength but the distance moved by a wave in a cycle period. > Let's use the wheel to simulate sagnac. > > We actually need three such wheels, each with say 10,000,000 marks around the > edges. > Two of the wheels will spin around the same axis at high speeds (c) and in No, one at c+v and the other c-v, but you pick that up later. > opposite directions, to simulate the frame of the light. The other will rotate > very slowly (at v) or not at all around that axis, simulating the rotation of > the ring. > > Let's say the slow wheel moves ten marks (from points 1 to 10) for every > complete rotation of the other two. OK so far. > The numbers of marks on the fast wheels > that lie between the points 1 and 10 on the slow wheel are 10,000,010 and > 9,999,990 in the two directions. Hold on, it's not that simple. Just for ease of tracking, put dots of paint on one tooth on each wheel. The dot on the slow wheel identifies the point on the beam splitter which acts as source and detector. The dots on the fast wheels are the halves of a positive half-cycle emitted at some instant. At the instant, all three dots momentarily coincide at the same angle, say zero degrees. The wheels then turn until the c+v wheel has made slightly more than one turn, the c-v wheel slightly less and the slow wheel has made the fraction (v/c) of a revolution. At that time the dots come together. The question is, are the teeth on the fast wheel exactly aligned when the match up with each other or is there a phase difference, and if so how much? > We already established that this number is not dependent on the speed of the > fast wheels...so we can speed them up to c+v or c-v and still get the same > result.....which is what hapens in the ring gyro. > > So you can see that the two numbers of marks varies when and only when the > ratio of v/c changes, c being a universal constant. The question is do the dots all align if the speed is constant. We agree there will be odd effects when there is acceleration because the speed of the fast wheels is defined as c+v and c-v at the moment they pass the slow wheel. After that you can change the speed of th slow wheel because it doesn't affect the speed of any light already in motion so you can get the dot on the slow wheel to be anywhere you like (within limits) when the fast dots next align, certainly you can produce any phase difference you like. George
From: George Dishman on 5 Oct 2007 10:14 "Clueless Henri Wilson" <HW@....> wrote in message news:3ukag351olgfcemn9opmcas5g9a0i3dg1k(a)4ax.com... > On Wed, 3 Oct 2007 23:27:34 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:6vjtf3d2q6sprg1t8e4kg5qlbc429cpj4i(a)4ax.com... .... >>> The intention was to illustrate Maxwell's model. >> >>Your model does not do that, this does: >> >> http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35 > > That's just a graph of field intensity. Mine is supposed to show the field > intensity itself, varying in space...and moving along. Check your java installation, the waves move away from the source at the origin on my machine. These waves are a solution to Maxwell's Equations, yours are not. >>This also shows the differential equation for the >>electric field which your solution must satisfy: >> >> http://www.analyzemath.com/antenna_tutorials/wave_propagation.html >> >>There are several more you can find with a cursory >>search. > > No need, the concept is simple. Very, but your illustration of it is wrong if you intended it to accord with Maxwell's Equations. Of course you can merge numerous such solutions by Fourier analysis to get a propagating wave packet and use dispersion to get the envelope shape to change but you can't get the pattern you have shown because of the other complexities I explained before. George
From: Dr. Henri Wilson on 5 Oct 2007 17:11
On Fri, 05 Oct 2007 01:06:25 -0700, George Dishman <george(a)briar.demon.co.uk> wrote: >On 4 Oct, 21:49, HW@....(Clueless Henri Wilson) wrote: >> On Wed, 3 Oct 2007 23:06:49 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: >> >"Clueless Henri Wilson" <HW@....> wrote in message news:dh28g31kcobj79u7uprguto7o4l10eef88(a)4ax.com... > ><big snip> > >I've trimmed most because I think the next bit >explains all of the rest as well and it might >get lost in a morasse of minor points. Your >wheels analogy is a good idea. In fact it would >be better if you read on down to that before >replying and come back to the earlier part >afterwards, or even snip it if you prefer, but >it explains again why phase controls the fringes. > >> >The distance moved by the wave is longer than the >> >wavelength in the inertial frame, you failed to >> >take that into account. >> >> >> The number of wavelengths in the path is unequivocably (2pi.R+vt)/L., no >> >> matter what is the speed. >> >> >The number between the source and detector at any >> >instant is unequivocably 2.pi.R/L no matter what >> >the speed. >> >> This is true and highlights the error one can make when trying to use rotating >> frames. > >There can be no error, all frames _must_ give >the same answer. George, in the rotating frame, 'wavelength' is neither absolute nor constant. >> >Rotating frame: N = (2.pi.R) / L >> >> >Inertial frame: N = (2pi.R+vt) / ((1+v/c).L) >> >> >You get the same number whichever frame you choose >> >and the number is same for both directions. >> >> What you are saying is that if you mark 1000 equally spaced points around an >> wheel, then there will be 1000 marks around it no matter how fast it >> rotates....and the absolute distance bewteen them is always the same. >> That's very clever...and it can even be used as a model here. >> >> What we really want to know is the number of marks around wheel between the >> points that are marked say, 999 and 1000. > >NO, that's the fundamental point. Suppose at some >instant one positive peak is just being emitted from >the source and both path lengths are exactly 1000 >wavelengths. A positive peak emitted 1000 cycles >earlier is just arriving at the detector and you get >constructive interference and a bright fringe. On the >other hand, if one path is half a wavelength different >at 999.75 and 1000.25 wavelengths, you get destructive >interference and a dark fringe. forget it George. see: www.users.bigpond.com/hewn/ringgyro.htm Phase is taken care of. >I'll do the next bit in both frames so read both >before replying please: > >What you need to know therefore, working in the rotating >frame, is how many cycles fit between the source and the >detector along these slightly curved paths: > > http://www.briar.demon.co.uk/Henri/paths.gif > >It should be obvious that the number of waves along >each path will be the same. > >Now imagine you drew that out but with a superimposed >sine wave stretched along the paths like Androcles' >attempts but without his errors. Translate that static >picture into the inertial frame and the path length >now is the distance from source at the moment the first >cycle was emitted but the waves also get stretched >along that path by the same factor. That is *NOT* the >sme as the wavelength in the inertial frame! If you >took a snapshot of the current situation, the waves in >flight at any instant must start from the _present_ >location of the source and the wavelength then is the >invariant value. > >> The answer is '999 one way and 1001 the other'...and again that number does not >> change with rotation speed. > >No the numbers cannot depend on the choice of >frame. They are obviously 1000 on each path as >seen in the rotating frame so they must be 1000 >on each path in the inertial frame too. I told you, the error lies in the fact that wavelength is not constant in the rotating frame. ....best not to use rotating frames george. >When >you realise the total length you used is along >the flight path of the light from the location >that is reaching the detector, from S in your >diagram and not the present position S', you >should also realise that the distance you must >divide by isn't the wavelength but the distance >moved by a wave in a cycle period. You are missing the fact that the path lengths are different as shown in www.users.bigpond.com/hewn/ringgyro.htm >> Let's use the wheel to simulate sagnac. >> >> We actually need three such wheels, each with say 10,000,000 marks around the >> edges. >> Two of the wheels will spin around the same axis at high speeds (c) and in > >No, one at c+v and the other c-v, but you pick >that up later. > >> opposite directions, to simulate the frame of the light. The other will rotate >> very slowly (at v) or not at all around that axis, simulating the rotation of >> the ring. >> >> Let's say the slow wheel moves ten marks (from points 1 to 10) for every >> complete rotation of the other two. > >OK so far. > >> The numbers of marks on the fast wheels >> that lie between the points 1 and 10 on the slow wheel are 10,000,010 and >> 9,999,990 in the two directions. > >Hold on, it's not that simple. Just for ease of >tracking, put dots of paint on one tooth on each >wheel. The dot on the slow wheel identifies the >point on the beam splitter which acts as source >and detector. The dots on the fast wheels are >the halves of a positive half-cycle emitted at >some instant. At the instant, all three dots >momentarily coincide at the same angle, say zero >degrees. The wheels then turn until the c+v wheel >has made slightly more than one turn, the c-v >wheel slightly less and the slow wheel has made >the fraction (v/c) of a revolution. At that time >the dots come together. The question is, are the >teeth on the fast wheel exactly aligned when the >match up with each other or is there a phase >difference, and if so how much? see the bottom diagram on www.users.bigpond.com/hewn/ringgyro.htm at consant rotation, the phase relationships are unchangng and no fringe movement occurs. >> We already established that this number is not dependent on the speed of the >> fast wheels...so we can speed them up to c+v or c-v and still get the same >> result.....which is what hapens in the ring gyro. >> >> So you can see that the two numbers of marks varies when and only when the >> ratio of v/c changes, c being a universal constant. > >The question is do the dots all align if the speed >is constant. They don't have to align. the two at the detector have to remain in constant relationship,,,,which they do. >We agree there will be odd effects >when there is acceleration because the speed of >the fast wheels is defined as c+v and c-v at the >moment they pass the slow wheel. After that you >can change the speed of th slow wheel because it >doesn't affect the speed of any light already in >motion so you can get the dot on the slow wheel >to be anywhere you like (within limits) when the >fast dots next align, certainly you can produce >any phase difference you like. I'll give you some more time to think about it George. >George Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |