GETTING RID OF EINSTEIN RELATIVITY
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From: Dr. Henri Wilson on 2 Oct 2007 17:10 On Tue, 02 Oct 2007 02:57:43 -0700, George Dishman <george(a)briar.demon.co.uk> wrote: >On 1 Oct, 21:57, HW@....(Dr. Henri Wilson) wrote: >> On Mon, 1 Oct 2007 21:41:16 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: >> >"Clueless Henri Wilson" <HW@....> wrote in message >> >> George, Have a look at:www.users.bigpond.com/hewn/ringgyro1.jpg > >It is OK as far as the calculation of time difference, >below that it is wrong because you used the wavelength >instead of the distance moved by the wave in a period. In the source frame, the distance move by the light in 'one period' is called the 'wavelength'.All lengths are absolute AND FRAME INDEPENDENT. Is that too hard for you?. >> It shows what a farce SR really is. > >It shows that you cannot do schoolboy algebra. George, a rod doesn't change when a moving observer goes past it.... >George Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 3 Oct 2007 05:37 On 2 Oct, 22:10, HW@....(Clueless Henri Wilson) wrote: > On Tue, 02 Oct 2007 02:57:43 -0700, George Dishman <geo...(a)briar.demon.co.uk> wrote: > >On 1 Oct, 21:57, HW@....(Clueless Henri Wilson) wrote: > >> On Mon, 1 Oct 2007 21:41:16 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: > >> >"Clueless Henri Wilson" <HW@....> wrote in message > > >> George, Have a look at:www.users.bigpond.com/hewn/ringgyro1.jpg > > >It is OK as far as the calculation of time difference, > >below that it is wrong because you used the wavelength > >instead of the distance moved by the wave in a period. > > In the source frame, the distance move by the light in 'one period' is called > the 'wavelength'. Good, you finally listened. The distance moved is not "called" the wavelength, that is the distance between wave crests, but since the source is at rest, they have the same value. In the SOURCE frame, the path length is the same for both beams hence there is no difference in the path lengths REGARDLESS of the (constant) speed of rotation, therefore the answer is: zero path difference / wavelength equals zero shift > All lengths are absolute AND FRAME INDEPENDENT. No, the wavelength is invariant in ballistic theory but the path length and distance moved are not, both change. > Is that too hard for you?. No, it's quite easy for me, you still haven't worked it out in enough detail. You have made progress though, you should now see why the answer is zero phase difference in the SOURCE frame which is consistent with the zero time difference you already agreed. You just need to translate the distance method into the lab frame and you will understand your error. > >It shows that you cannot do schoolboy algebra. > > George, a rod doesn't change when a moving observer goes past it.... 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
From: Dr. Henri Wilson on 3 Oct 2007 17:40 On Wed, 03 Oct 2007 02:37:30 -0700, George Dishman <george(a)briar.demon.co.uk> wrote: >On 2 Oct, 22:10, HW@....(Clueless Henri Wilson) wrote: >> On Tue, 02 Oct 2007 02:57:43 -0700, George Dishman <geo...(a)briar.demon.co.uk> wrote: >> >On 1 Oct, 21:57, HW@....(Clueless Henri Wilson) wrote: >> >> On Mon, 1 Oct 2007 21:41:16 +0100, "George Dishman" <geo...(a)briar.demon.co.uk> wrote: >> >> >"Clueless Henri Wilson" <HW@....> wrote in message >> >> >> George, Have a look at:www.users.bigpond.com/hewn/ringgyro1.jpg >> >> >It is OK as far as the calculation of time difference, >> >below that it is wrong because you used the wavelength >> >instead of the distance moved by the wave in a period. >> >> In the source frame, the distance move by the light in 'one period' is called >> the 'wavelength'. > >Good, you finally listened. The distance moved is >not "called" the wavelength, that is the distance >between wave crests, but since the source is at >rest, they have the same value. > >In the SOURCE frame, the path length is the same >for both beams hence there is no difference in the >path lengths REGARDLESS of the (constant) speed of >rotation, therefore the answer is: > > zero path difference / wavelength equals zero shift > >> All lengths are absolute AND FRAME INDEPENDENT.:) :) George, Sagnac doesn't rely on the source frame. It uses the non-rotating frame. Rotating frames can be very confusing and lead to errors like the one you just made. >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. >> Is that too hard for you?. > >No, it's quite easy for me, you still haven't >worked it out in enough detail. You have made >progress though, you should now see why the >answer is zero phase difference in the SOURCE >frame which is consistent with the zero time >difference you already agreed. You just need >to translate the distance method into the lab >frame and you will understand your error. George, I realise that this is very hard for you to take. you wil go down fighting...but really, you are just digging your hole deeper and deeper.... http://www.users.bigpond.com/hewn/ringgyro.htm >> >It shows that you cannot do schoolboy algebra. >> >> George, a rod doesn't change when a moving observer goes past it.... > >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. The wavelength is L. The number of wavelengths in the path is unequivocably (2pi.R+vt)/L., no matter what is the speed. >George Henri Wilson. ASTC,BSc,DSc(T) www.users.bigpond.com/hewn/index.htm
From: George Dishman on 3 Oct 2007 18:06 "Clueless Henri Wilson" <HW@....> wrote in message news:dh28g31kcobj79u7uprguto7o4l10eef88(a)4ax.com... > On Wed, 03 Oct 2007 02:37:30 -0700, George Dishman > <george(a)briar.demon.co.uk> wrote: >>On 2 Oct, 22:10, HW@....(Clueless Henri Wilson) wrote: >>> On Tue, 02 Oct 2007 02:57:43 -0700, George Dishman >>> <geo...(a)briar.demon.co.uk> wrote: >>> >On 1 Oct, 21:57, HW@....(Clueless Henri Wilson) wrote: >>> >> On Mon, 1 Oct 2007 21:41:16 +0100, "George Dishman" >>> >> <geo...(a)briar.demon.co.uk> wrote: >>> >> >"Clueless Henri Wilson" <HW@....> wrote in message >>> >>> >> George, Have a look at:www.users.bigpond.com/hewn/ringgyro1.jpg >>> >>> >It is OK as far as the calculation of time difference, >>> >below that it is wrong because you used the wavelength >>> >instead of the distance moved by the wave in a period. >>> >>> In the source frame, the distance move by the light in 'one period' is >>> called >>> the 'wavelength'. >> >>Good, you finally listened. The distance moved is >>not "called" the wavelength, that is the distance >>between wave crests, but since the source is at >>rest, they have the same value. >> >>In the SOURCE frame, the path length is the same >>for both beams hence there is no difference in the >>path lengths REGARDLESS of the (constant) speed of >>rotation, therefore the answer is: >> >> zero path difference / wavelength equals zero shift >> >>> 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. >>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. >>> Is that too hard for you?. >> >>No, it's quite easy for me, you still haven't >>worked it out in enough detail. You have made >>progress though, you should now see why the >>answer is zero phase difference in the SOURCE >>frame which is consistent with the zero time >>difference you already agreed. You just need >>to translate the distance method into the lab >>frame and you will understand your error. > > George, I realise that this is very hard for you to take. you wil go down > fighting...but really, you are just digging your hole deeper and > deeper.... I'll just keep correcting your error until you get bored parroting your assertions and actually think about what you are saying. > http://www.users.bigpond.com/hewn/ringgyro.htm Yep, same error. >>> >It shows that you cannot do schoolboy algebra. >>> >>> George, a rod doesn't change when a moving observer goes past it.... >> >>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. 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. George
From: George Dishman on 3 Oct 2007 18:27
"Clueless Henri Wilson" <HW@....> wrote in message news:6vjtf3d2q6sprg1t8e4kg5qlbc429cpj4i(a)4ax.com... > On Sat, 29 Sep 2007 14:23:51 +0100, "George Dishman" > <george(a)briar.demon.co.uk> wrote: >>"Clueless Henri Wilson" <HW@....> wrote in message >>news:5amdf3tiaiqgth380f9kmevr70ce8a62ss(a)4ax.com... > >>> The concept of 'field strength' in space is a rather nebulous thing. An >>> electric field is a mathematical construct that describes how unit >>> charge >>> would ACCELERATE if placed in that field at a certain point. >> >>The definition of a field is not nebulous in any sense >>at all. I already gave you the definition which you >>essentially just repeated. >> >>> ...but in what direction would the charge accelerate? Would it move >>> towards the >>> origin or towards the point of maximum opposite field. >> >>The field at any point is a vector, the charge >>accelerates in the direction specified by the value >>of the field. As you said, the value is how the >>charge would accelerate which is a vector divided >>by the charge which is scalar giving a vector >>result. > > George, a field exists whether or not it accelerates a charge. Of course, the definition of a "test particle" is one whose charge is so small it doesn't disturb the feld. > As a physicist, ... ROFL, you will never be a physicist Henry, you don't know the meaning of the word. >>> Would it move towards the >>> origin or towards the point of maximum opposite field. >>> The graph we normally see drawn with Maxwell's concept of EM is pretty >>> useless >>> if you think about it. >> >>For a philosopher perhaps, but if you understand the >>principles of physics, the value of the field tells >>you everything you need to know. > > Typical relativist...think he has all the answers. Just teaching you where science limits the field. >>>>If not there would be a beat and you would get a >>>>sine wave envelope. What I have described so far >>>>matches your "moving oboe" analogy. >>> >>> Except that the oboe will change length during an acceleration. >> >>Your diagram illustrates constant velocity and I was >>discussing that. Once you analyse that, you could >>apply an acceleration and it would make predictions. > > Just assume .. No Henry, we never assume, we calculate. >>> ...but the velocity of the intrinsic wave wrt the 'package' isn't >>> ncesssarily >>> related to the speed of the package itself...wrt anything... >> >>Right, it is most probable the speed of the underlying >>waves would need to be much higher and the observed >>speed is the smaller fractional difference between >>the speeds of the left and right propagating parts. >> >>> ...or maybe there IS some kind of not very obvious connection that needs >>> explaining.... >> >>It is close to a beat but more complex. > > That could be true. It is true, it is related to a "window function" in digital signal processing. Look it up if you want to learn the answer. >>>>No, in the frame in which it is drawn >>> >>> I'm not sure what you re getting at. >> >>I was agreeing. In the frame of the photon, the >>envelope doesn't propagate, it essentially _is_ >>the photon, however what I said is true in the >>frame from which you drew the illustration. >> >>> In the photon frame there are two waves traveling in oposite directions. >> >>That is also true in the frame of the illustration >>but there is an anisotropy, probably in the speeds, >>which results in the motion of the envelope. > > THis is indeedpossible.. A photon is capable of self-propagation...without > losing energy ...or not much anyway. Note that your equations do not allow for propagation in any sense, Maxwell's Equations do but as we discussed they are not usable in ballistic theory. > ...or maybe the photon just starts out at c wrt its source and would stay > around that speed except for extinction effects. .... >>>>Sure, but then you Fourier transform it to find the >>>>components. >>> >>> ...but that has nothing to do with he speed of the waves relative to >>> anything external. >> >>No, it has to do with frequencies. Your illustration >>doesn't have a single "absolute" frequency or wavelength >>as you described in other posts, it is a continuum of >>frequencies. > > 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 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. > 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. >>"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. > 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. 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. George |