GETTING RID OF EINSTEIN RELATIVITY
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From: Dr. Henri Wilson on 5 Oct 2007 17:13 On Fri, 5 Oct 2007 15:14:24 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"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. I attempted to show a moving 'standing wave'. The main point was that an oscillating 'field' is not just a sqiggly line drawn on paper. It has physical propeties. > >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 6 Oct 2007 07:27 "Clueless Henri Wilson" <HW@....> wrote in message news:v4adg3pdknssk3dsh4rvfaa7072vhdcuaj(a)4ax.com... > On Fri, 5 Oct 2007 15:14:24 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"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. > > I attempted to show a moving 'standing wave'. :) I love that phrase. A standing wave is a combination of two propagating waves of the same frequency moving in opposite directions, each of which is as illustrated on the sites cited above. It is called "standing" because it doesn't move :) > The main point was that an oscillating 'field' is not just a sqiggly line > drawn > on paper. It has physical propeties. Sure, it has the property that it accelerates a charge, but all you have done is draw another squiggly line which _isn't_ a solution to Maxwell's Equations whereas the sites above show the correct solution and all other solutions are linear combinations of that form. George
From: George Dishman on 6 Oct 2007 07:34 "Dr. Henri Wilson" <HW@....> wrote in message news:k49dg39a5s7uh2j15a1acrbbn526mhtiiq(a)4ax.com... > 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. Henry, you keep telling me 'wavelength' is the same in _all_ frames, that's what you mean by "absolute" isn't it? >>> >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. The phase isn't even shown. >>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. You told me it was absolute, and it is. > ...best not to use rotating frames george. I'll use it when I like, it doesn't matter that you can't cope with it. >>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 Nope, I am well aware of that, I told you your error lies elsewhere. >>> 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. Agreed but that's not the point, there is also no displacement. >>> 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 question is do they? If they always align regardless of speed then there is no fringe displacement. If the two fast-wheel dots pass somewhere other than as they pass the dot on the slow wheel then you get a displacement. > the two at the detector have to remain in constant relationship,,,,which > they > do. If they do, ballistic theory is wrong. >>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. Everything you need to know is already in what I said, you just didn't read it. Try again or look at jerry's animation which shows it perfectly. George
From: Dr. Henri Wilson on 6 Oct 2007 17:59 On Sat, 6 Oct 2007 12:34:59 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:k49dg39a5s7uh2j15a1acrbbn526mhtiiq(a)4ax.com... >> On Fri, 05 Oct 2007 01:06:25 -0700, George Dishman >> <george(a)briar.demon.co.uk> >> wrote: >>>> >>>> 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. > >Henry, you keep telling me 'wavelength' is the same >in _all_ frames, that's what you mean by "absolute" >isn't it? Don't act dumb... It is the same in all INERTIAL frames George. >>>> 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. > >The phase isn't even shown. It is obvious what happens to phase. At constant rotation speed, the phase relationship between the two is constant. It only changes during an acceleration. >>>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. > >You told me it was absolute, and it is. ...in inertial frames. >> ...best not to use rotating frames george. > >I'll use it when I like, it doesn't matter >that you can't cope with it. You made an error. >>>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 > >Nope, I am well aware of that, I told you your >error lies elsewhere. George, I don't have an error. The number of wavelengths around the paths is (2piR+/-vt)/lambda. I've done the rest of the maths for you. >>>> 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. > >Agreed but that's not the point, there is also >no displacement. The point George, is that THERE IS DISPPLACEMENT. Your model is wrong. In the rotating frame, wavelength is NOT absolute and constant. The moral is, "Don't try to use rotating frames". >>>> 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 question is do they? If they always align >regardless of speed then there is no fringe >displacement. If the two fast-wheel dots pass >somewhere other than as they pass the dot on >the slow wheel then you get a displacement. George, the way the dots align depends on rotation speed. Alignment changes only during an acceleration. > >> the two at the detector have to remain in constant relationship,,,,which >> they >> do. > >If they do, ballistic theory is wrong. George, they only do so at constant speed..which is what is observed... >>>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. > >Everything you need to know is already in >what I said, you just didn't read it. Try >again or look at jerry's animation which >shows it perfectly. I am right George..get used to it. Sagnac is purely a ballistic phenomenon. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: Dr. Henri Wilson on 6 Oct 2007 18:02
On Sat, 6 Oct 2007 12:27:43 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:v4adg3pdknssk3dsh4rvfaa7072vhdcuaj(a)4ax.com... >> On Fri, 5 Oct 2007 15:14:24 +0100, "George Dishman" >>>>>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. >> >> I attempted to show a moving 'standing wave'. > >:) I love that phrase. > >A standing wave is a combination of two propagating >waves of the same frequency moving in opposite >directions, each of which is as illustrated on the >sites cited above. > >It is called "standing" because it doesn't move :) It appears to move if you run past it.... Do you know what 'frames' are george? >> The main point was that an oscillating 'field' is not just a sqiggly line >> drawn >> on paper. It has physical propeties. > >Sure, it has the property that it accelerates a charge, >but all you have done is draw another squiggly line >which _isn't_ a solution to Maxwell's Equations whereas >the sites above show the correct solution and all other >solutions are linear combinations of that form. No, I have shown the 'field strength' in colour. >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm |