GETTING RID OF EINSTEIN RELATIVITY
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From: George Dishman on 7 Oct 2007 07:31 "Dr. Henri Wilson" <HW@....> wrote in message news:hn0gg3le7a72fm1mqmsmn16fi6lgh4fd2r(a)4ax.com... > 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: .... >>> 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. This isn't SR Henry, in a snapshot at any instant, it is the same in _all_ frames, inertial or not, in ballistic theory. >>> ,,. 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. Yes, but also it is zero. You cannot see that from the diagram but you can in Jerry's simulaton because the two waves arriving at the detector hit the radial line at exactly the same point at all times. >>> 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. All frames Henry, and you didn't make that distinction previously. >>> ...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. No, it is correct, check the code for yourself. >>>>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. It is 2piR/lambda regardless of which frame you calculate in, that is your error. >>>>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. No, there is none as the maths shows when you do it correctly and Jerry's simulation confirms. > Your model is wrong. > In the rotating frame, wavelength is NOT absolute and constant. > The moral is, "Don't try to use rotating frames". The simulation is in the inertial frame and makes no assumption about the wavelength, it only propagates the waves at the correct speed and the phase can be seen directly. >>>>> 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. Wrong, try writing you own program if you cannot understand Jerry's. > 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... No, a displacement is observed which means they must fail to align by an amount proportional to the speed. >>>>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.. You are wrong, you have only shown cannot even apply your own model nor understand it when it is done for you. George
From: Dr. Henri Wilson on 7 Oct 2007 19:52 On Sun, 7 Oct 2007 12:31:28 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Dr. Henri Wilson" <HW@....> wrote in message >news:hn0gg3le7a72fm1mqmsmn16fi6lgh4fd2r(a)4ax.com... >> On Sat, 6 Oct 2007 12:34:59 +0100, "George Dishman" >>>> 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. > >Yes, but also it is zero. You cannot see that from the >diagram but you can in Jerry's simulaton because the >two waves arriving at the detector hit the radial line >at exactly the same point at all times. :) It isn't zero George, It depends solely on the relative numbers of wavelengths in the two paths. >>>> 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. > >All frames Henry, and you didn't make that >distinction previously. It isn't absolute OR constant in a rotating frame. Straight lines become curved for one thing.. ...best not to use rotating frames George. >>>> ...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. > >No, it is correct, check the code for yourself. > >>>>>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. > >It is 2piR/lambda regardless of which frame >you calculate in, that is your error. Just have another look at: http://www.users.bigpond.com/hewn/ringgyro.htm where's the error there, George? >ust 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. > >No, there is none as the maths shows when you do it >correctly and Jerry's simulation confirms. > >> Your model is wrong. >> In the rotating frame, wavelength is NOT absolute and constant. >> The moral is, "Don't try to use rotating frames". > >The simulation is in the inertial frame and makes >no assumption about the wavelength, it only >propagates the waves at the correct speed and the >phase can be seen directly. > >>>>>> 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. > >Wrong, try writing you own program if you cannot >understand Jerry's. > >> 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... > >No, a displacement is observed which means they >must fail to align by an amount proportional to >the speed. > >>>>>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.. > >You are wrong, you have only shown cannot even apply >your own model nor understand it when it is done for >you. > >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 8 Oct 2007 14:47 "Clueless Henri Wilson" <HW@....> wrote in message news:ftrig3h6nri7ra2t6d2hm7sehuvc7iaqnd(a)4ax.com... > On Sun, 7 Oct 2007 12:31:28 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:hn0gg3le7a72fm1mqmsmn16fi6lgh4fd2r(a)4ax.com... >>> On Sat, 6 Oct 2007 12:34:59 +0100, "George Dishman" > >>>>> 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. >> >>Yes, but also it is zero. You cannot see that from the >>diagram but you can in Jerry's simulaton because the >>two waves arriving at the detector hit the radial line >>at exactly the same point at all times. > > :) > It isn't zero George, It depends solely on the relative numbers of > wavelengths > in the two paths. The numbers are equal. >>>>> 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. >> >>All frames Henry, and you didn't make that >>distinction previously. > > It isn't absolute OR constant in a rotating frame. > Straight lines become curved for one thing.. Yes Henry, so you measure along the curved path since your speed of c+v is likewise along that path. The distance moved by the wave is (c+v)/f measured along whatever path it takes (including reflecting from mirrors) while the source moves v/f so they are separated by c/f when the next wave is emitted, hence that is the wavelength. > ..best not to use rotating frames George. Jerry's animation is in the inertial frame, your web page has a division that is half in each! >>>>> 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. >> >>It is 2piR/lambda regardless of which frame >>you calculate in, that is your error. > > Just have another look at: http://www.users.bigpond.com/hewn/ringgyro.htm > > where's the error there, George? You divide the path length by lambda when it should be (c+v)/f or (c-v)/f for the opposing beam. The result is 2piR/lambda in the inertial frame. George
From: Dr. Henri Wilson on 8 Oct 2007 18:08 On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Clueless Henri Wilson" <HW@....> wrote in message >news:ftrig3h6nri7ra2t6d2hm7sehuvc7iaqnd(a)4ax.com... >> On Sun, 7 Oct 2007 12:31:28 +0100, "George Dishman" >> <george(a)briar.demon.co.uk> wrote: >>>"Clueless Henri Wilson" <HW@....> wrote in message >>>news:hn0gg3le7a72fm1mqmsmn16fi6lgh4fd2r(a)4ax.com... >>>> On Sat, 6 Oct 2007 12:34:59 +0100, "George Dishman" >> >>>>>> 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. >>> >>>Yes, but also it is zero. You cannot see that from the >>>diagram but you can in Jerry's simulaton because the >>>two waves arriving at the detector hit the radial line >>>at exactly the same point at all times. >> >> :) >> It isn't zero George, It depends solely on the relative numbers of >> wavelengths >> in the two paths. > >The numbers are equal. George, I know this is difficult. 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. 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. >>>>>> 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. >>> >>>All frames Henry, and you didn't make that >>>distinction previously. >> >> It isn't absolute OR constant in a rotating frame. >> Straight lines become curved for one thing.. > >Yes Henry, so you measure along the curved path >since your speed of c+v is likewise along that >path. The distance moved by the wave is (c+v)/f >measured along whatever path it takes (including >reflecting from mirrors) while the source moves >v/f so they are separated by c/f when the next >wave is emitted, hence that is the wavelength. > >> ..best not to use rotating frames George. > >Jerry's animation is in the inertial frame, your >web page has a division that is half in each! > >>>>>> 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. >>> >>>It is 2piR/lambda regardless of which frame >>>you calculate in, that is your error. >> >> Just have another look at: http://www.users.bigpond.com/hewn/ringgyro.htm >> >> where's the error there, George? > >You divide the path length by lambda when it should >be (c+v)/f or (c-v)/f for the opposing beam. The >result is 2piR/lambda in the inertial frame. > >George > Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 8 Oct 2007 18:44
"Clueless Henri Wilson" <HW@....> wrote in message news:ih9lg3da9pc4jt1dg32t9041s25nfbs72f(a)4ax.com... > On Mon, 8 Oct 2007 19:47:09 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:ftrig3h6nri7ra2t6d2hm7sehuvc7iaqnd(a)4ax.com... >>> On Sun, 7 Oct 2007 12:31:28 +0100, "George Dishman" >>> <george(a)briar.demon.co.uk> wrote: >>>>"Clueless Henri Wilson" <HW@....> wrote in message >>>>news:hn0gg3le7a72fm1mqmsmn16fi6lgh4fd2r(a)4ax.com... .... >>>>> It is obvious what happens to phase. >>>>> At constant rotation speed, the phase relationship between the two is >>>>> constant. >>>> >>>>Yes, but also it is zero. You cannot see that from the >>>>diagram but you can in Jerry's simulaton because the >>>>two waves arriving at the detector hit the radial line >>>>at exactly the same point at all times. >>> >>> :) >>> It isn't zero George, It depends solely on the relative numbers of >>> wavelengths >>> in the two paths. >> >>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. > 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. 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. 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. George |