Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Henri Wilson on 20 Nov 2005 23:41 On Sat, 19 Nov 2005 12:43:51 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:c2fsn1t9elj93qhgdn2bp0ok8m14r5dp8s(a)4ax.com... >That is because it is just a convenience we invent >to help keep track of the motion of the wave. In >reality, the field varies as the sine along the >beam, remember this: > >> >... I'll add a plot of field versus >> >distance to the next bit: >> > >> >> > \ >> >> > - ) >> >> > / >> >> > ( >> >> > \ >> >> > - ) >> >> > | | >> >> > |-| >> >> > | | >> > >> >Does that help at all? Each hyphen is a cross >> >section of a plane of maximum voltage. >> >> clever drawing... > >So what you have is a 1.5mm ellipse with a thickness >of half a wavelength which is still an arbitrary >cut. The thin plane merely identifies the location >of the _peak_ of the field. Energy is distributed >throughout the volume. > >> It is not light. It is nothing. > >Right, it is a mathematical plane we use only to aid >description. The light (E and B fields) fills the >volume of the beam. > >>>>>TWLS has been measured and found to be c, always, and >>>>>you know it, and that is what I said. >>>> >>>> You claimed the diagonal speed of the elements was also c. >>> >>>Light moving diagonally is just light and you have >>>accepted that the speed of light has been measured >>>using two-way techniques. >> >> Very funny George. >> >> How would you apply a TWLS technique to measuring the speed of ONE >> infinitesimal thin plane along an infinitesimally thin diagonal line? > >Oops, what was "really an elliptical plane about 1.5mm >long." just became "an infinitesimally thin diagonal >line" again. No George. I was refereingn toi the movement of the infinitesimally thin plane along the infinitesimally thin diagonal line, which describes its path in the moving frame. >The beam isn't "infinitesimally thin" in >the moving frame Henri, it is close to 1.1mm wide and >the volume of the beam is filled with light obeying >Maxwell's Equations. The beam is the same width in all frames. It merely APPEARS to move sideways in the moving frame. What moves diagonally is the infinitesimally thin elliptical plane that does not constitute light or anything else. >>>> I'm afraid you are so hopelessly indoctrinated, your brain has ceased to >>>> function freely. >>> >>>I see you cannot refute my statement. >> >> You are acting dumb George. I think by now you must realise that I'm >> right. > >You had to switch above from "elliptical plane" >to "an infinitesimally thin diagonal line" to >avoid the problem, and for the second time you >had to snip what I said. You didn't read what I said. You are totally confused. >You wouldn't have had >to do that if you were right. The beam has finite >width in both frames and the volume is filled with >light. Mathematical lines tracing some feature like >the centre of the beam are mathematical lines in >both frames so you can't compare the beam in one >frame with the line in the other frame to prove one >is light and the other isn't. OK, let's use a beam of infinitesimal width. >> So where are all these diagonal light beams you imagine are continuously >> emerging from the sides of a vertical laser beam? That's what you are >> claiming... > >No it isn't Henri. I've corrected you on that several >times now. Telling lies won't solve your problem. > >> according to Huygens and Maxwell, a laser beam will emit wavefronts >> in all directions. >> >> I don't think that actually happens do you George? >> I think you are babbling hopelessly in an attempt to prop up your faith.. > >I think you are inventing babble that I never said >because you cannot address what I really said. My >statement was nothing more than the Huygens method. Well it doesn't work with a coherent, parallel laser beam. >> You (and Einstein) claim that a vertical beam in one frame becomes a >> diagonal >> one in another. > >Nope, here is what I drew again, you can see it above: > >>>>>What you will find is that the result in the moving >>>>>frame becomes like this with each wavefront moving >>>>>diagonally towards the top right. >>>>> >>>>>> \ >>>>>> \ >>>>>> \ -> >>>>>> \ >>>>>> \ > >I'll say it yet again to see if it can penetrate: >the beam is vertical in the moving frame. ...and I repeat, if there were diagonal wavefronts like these, they would be seen in the source frame as well as the movingnframe. There are not. >Aside: the ASCII exaggerates the slope, the wavefronts >would be almost horizontal. Yes I know what you mean. > ><snip insults - no content to reply to> They weren't insults, they were plain truths. >> >> even SR says their closing speeds with the light are different. > >From a different frame, correct. > >> If that is the case, why should they get the answer c from Maxwell's >> equation. > >Because Maxwell's Equations apply to EM disturbances >so the light moves at c. The other object is moving >at some speed v and "closing speed" is defined as the >difference in whatever other frame you are using. This >is simply a question of definitions Henri. So you agree that the light beam is 'approaching' the two differently moving observers at different speeds even though your theory says they will both MEASURE the OW speed of that beam to be in their own respective frames? >>>> That is unless you want to tear up the whole of physics as it stands. >>> >>>Physics as it stands has been based on SR for nearly >>>a century. You are going back to physics as it stood >>>before Maxwell's Equations. >> >> No I'm not. > >Yes you are, you are postulating Galilean Relativity >and trying to explain light within that framework >using a ballistic model for propagation. Maxwell's >Equations are like a classical wave theory except >that the speed of the medium doesn't appear. They rely on a medium. >> ,,and any SRian should be very interested in my question. >> Your theory has to explain why. > >"the whole of physics as it stands" is currently based >on Lorentz invariance so the question of tearing up >physics doesn't arise. You are the one trying to replace >it with Galilean Relativity and a Ritzian light model. It matters not if you assume invariancce of MEASURED light speed. (it has never been done of course) You still have to explain why both observers should derive the same value of c from Maxwell's equation. >>>> I'm giving up on you George. >>>> You are a hopeless case. >>> >>>That's it Henri, run away. >> >> I'm not running George, just making better use of my time. >> there is no point in arguing with the hopelessly indoctrinated. > >I am not "indoctrinated" in 18th century physics, we have >long left this stuff behind, but it's interesting to see >whether Ritz might have fitted into it. Ritz was 100% correct. If he hadn't died, Einstein would never have become the world's greatest hoaxer. >>>>>> >> Not 'very small' but 'infinitesimal'. >>>>>> > >>>>>> >Same thing Henri. >>>>>> >>>>>> Not the same George. >>>>> >>>>>Exactly the same Henri, open a textbook on basic calculus. >>>> >>>> Calculus doesn't work to well with 'very small' increments, George. >>> >>>Open a book on calculus Henri, I'm not going to teach you. >> >> Why do you think it is called 'infititesimal calculus'? > >Because it uses limits of quantites that are not zero >as they tend towards zero. If y=f(x) then dy/dx is >undefined if dy and dx are identically zero but is >defined (with certain limitations on function f(x)) >provided they are finite. They are called infititesimal >because we work with the asymptotic limits, not the >actual finite values. Close George. they are not 'very small' , they are infinitesimal. >>>> George, when I move past a vertical laser beam, I don't see it suddenly >>>> dispersing in all directions. >>> >>>No Henri, perhaps that's because it is "the interference >>>between the infinite number of elements covering the >>>horizontal surface of the wavefront that allows the beam >>>to avoid dispersing." >>> >>>As you said above, I am just explaining Huygens method to >>>you. >> >> But you claim that ONE diagonal element replaces the vertical beam. >> What are you actually trying to say George? > >Go back to what you said at the top: > >> By 'width' I was refering to the thickness of what is really an elliptical >> plane about 1.5mm long. It has infinitesimal thickness. >... >> How would you apply a TWLS technique to measuring the speed of ONE >> infinitesimal thin plane along an infinitesimally thin diagonal line? > >What I am saying is that you get a beam that doesn't >disperse if you take the "elliptical plane about >1.5mm long" and apply Huygens. ...you don't get a beam. You get an infinitie number of different such planes all moving along different diagonal lines. That would make the original beam infinitely bright in the moving frame George.....very clever... >If instead you use just one "infinitesimal element" you >talked of previously which is where the plane of >"infinitesimal thickness" intersects the "infinitesimally >thin diagonal line" then you have a point source which >falsely suggests dispersion. > >I am saying that, to analyse the plane as "infinitesimal >elements", you have to consider the sum of all the elements >that form the plane. I think you have become totally confused George. You are ignoring plain logic. Indoctrination can do that. >>>Not if you can't even understand Huygens method, I tend to >>>agree. >> >> Then the laser beam should be dispersing in all directions in its own >> frame.. >> It doesn't. > >Right, and you will find that if you consider >_all_ the "infinitesimal elements", not just >one, which is what I said. The beam doesn't change every time an observer moves past it. > >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 20 Nov 2005 23:54 On Sat, 19 Nov 2005 10:18:10 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:7tgsn11flp2iggll8is15lrdpqck37voaj(a)4ax.com... >>>In that case you are wrong if I understand your diagram. >>>When the rotation is at constant speed, the path length >>>is also constant at the length of line AD or roughly: >>> >>> L' = L + v * t / sqrt(2) >>> >>>When the rotation is at constant acceleration, the path >>>length is also constant at the length of line AE or >>>roughly: >>> >>> L' = L + (v * t + a * t^2 / 2) / sqrt(2) >>> >>> _________ >>> / >>> / ^ >>> __________/ | speed >>> | >>> ______________________ >>> >>> ------> >>> time >>> >>>Your diagram predicts there would be an output like >>>this: >>> >>> __ >>> | | ^ >>> | | | output >>> | | | >>> | | >>> __________|__|________ >>> >>> ------> >>> time >> >> No it doesn't. >> >> Assume that fringe movement is +ve for +ve acceleration and -ve for -ve >> acceleration. >> There is no -ve acceleration in the above diagram showing a +ve speed >> change.. > >> Therefore fringe displaement moves in only one direction during the >> acceleration period. > >The fringes ARE DISPLACED in only one direction during >theacceleration period as shown by your diagram. When >the acceleration is zero, the factor a t^2 / 2 is zero >so there is zero displacement, not zero rate-of-change >of displacement. > >> It doesn't suddenly revert ot zero when acceleration ceases. > >Your diagram says it does, if 'a' is zero then the >length difference is zero. > >> SO: >> _________ >> | ^ >> | | output >> | | >> | >> __________|__________ >> >> ------> >> time >> >> Is the correct result. > >Let's add a negative acceleration period: > > _________ > / \ > / \ ^ > __________/ \______ | speed > | > _______________________________ > > ------> > time > >Your diagram predicts there would be an output like >this: > > __ >+ve | | ^ > | | | output > | | | > 0 ________|__|__________________ > | | > ------> | | > time | | >-ve |__| > >This is basically a graph of the "a t^2 / 2" factor. How are you defining 'output'. I'm using 'output' to mean 'fringe displacement' from the central (non-rotating) position. Maybe you are using it as rotation angle. In either case, I think I disagree with you. Fringes move only during acceleration periods. If a +ve acceleration is followed by an identical -ve one, the rotation speed is back to where it was....and so is fringe displacement. >>>There is no method to provide physical integration >>>because the number of wavelengths in the path does >>>not affect the time difference between wavefront >>>arrivals in the two beams which is what produces >>>the output. >> >> The 'change in fringe displacement' is effectively an integration of the >> path >> length increase during the acceleration period. > >No, the fringe displacement is a t^2 / 2 as you show >in your diagram. Yes, I suppose that its right. It is proportional to the path length difference of the two beams.. > >>>Actually I think your diagram is oversimplified but >>>we can go with it for the moment, it is close enough. >> >> It shows what happens during a constant acceleration. In practice, >> acceleration >> would vary with time. > >Indeed. I can't show a quadratic start and end to >each period of acceleration but if I could the >resulting output would look like this: > > +ve __ > / \ > / \ > / \ > 0 ___/ \_______ _____ > \ / > \ / > \ / > -ve \__/ > >Since the acceleration is changing, you now have to >understand the 'a' in your diagram to be the mean >acceleration during the flight time. yes, something like that. > >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 21 Nov 2005 00:49 On Sun, 20 Nov 2005 10:23:12 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:nfhsn1p5qtslgismtsq4tr1akram78c641(a)4ax.com... >> On Fri, 18 Nov 2005 15:31:03 -0000, "George Dishman" >> I should imagine that in a four mirror sagnac, fringe movement would be >> quite >> small. In a FoG with many turns, a much longer path length is provided and >> hence much greater fringe movement. > >That's true, you can use more turns in the fibre >coil, but it is still quite small. Are you sure? > >> This is detected by focussing the fringe >> pattern onto a narrow slit and counting the alternate dark/light >> fluctuations. > >There is no slit or counter. A photodiode is used >but the displacement is only a fraction of a fringe. >The photodiode measures the brightness of the light >which depends on the phase shift between the beams >or to put it another way, which part of the pattern, >a light or dark band, falls on the detector. that's what I meant. It counts the light and dark pulses. It probably also senses 'fractions of fringe.' >> I'm not sure how fringe direction is monitored but it would be. > >That's actually the clever bit, the source is >modulated so that the output is the product of a sine >wave with the cosine produced by the interference. It >means there is a sign change at the zero rotation point >but gives the sensitivity of an interferometer. That's >slightly different from the lab experiment version >where fringes can be seen and counted as you suggest >but the physical principles are the same, it's an >engineering solution to the question of directional >ambiguity. I don't quite see how the beam is modulated but I'll believe you. >Given that the displacement is a fraction of a fringe >and generally has a sine curve, the current in the >photodiode is roughly proportional to the fringe >displacement, or technically to the phase difference >between the two beams. > >Phase difference is 2 * pi times time difference >over the period or 2 * pi * frequency * delta_t. > >The shift is actually much smaller but: > >> Now, during an acceleration period, 50 such pulses might be counted. > >The displacement is proportional to the acceleration >so "As the acceleration smoothly increases, 50 such >pulses might be counted." > >> When the >> acceleration ceases, ... > >"As the acceleration smoothly decreases, 50 such >pulses would be counted but in the opposite >direction." so that, once the speed returned to >zero, the counter would also read zero. It would read what it read before the acceleration, not zero. >> ... the visible fringe might not be exactly in the centre of >> the slit >> and I assume there is a way of determining that offset...because it >> is essential for accuracy..... > >The offset would have been present before the >acceleration and could be calibrated out in >the factory. Only long term drift or temperature >sensitivity causes a problem. By 'offset' I meant 'fringe displacement'. > >> since the offset is integrated electronically >> with time to obtain total rotation angle. > >No, the output from what you say above if integrated >would give speed since it is proportional to >acceleration. Wait I think we are talking about different things here George. The integration of fringe displacement with time gives total angle moved during that time. >> I think you and Andersen are omiting a factor of (2-Root2) when you claim >> that >> the displacement should return to zero during constant rotation, according >> to >> the BaTh...In fact, the travel time around each path is not the same. > >A factor of root two applied to zero is still zero >but I am prepared to consider there might be a root >two scale factor difference, I haven't looked at it >in detail. There is a fundamental difference between >an output proportional to speed as predicted by SR >and an output proportional to acceleration as >predicted by Ritz which makes a factor of 0.7 of >lesser significance at the moment. The output, as in 'fringe displacement' is still proportional to rotation speed according to Ritz. >>>Correct, by a constant amount while the acceleration is >>>present, and since it is proportional to the acceleration, >>>it goes back to zero if the speed subsequently becomes >>>constant again. >> >> Haha. >> So you apply an acceleration that slowly approaches zero. During that >> time, the >> fringes move in one direction only. > >Yes, they slowly approach zero because their >displacement is proportional to the acceleration. No george. Their CHANGE of displacement occurs DURING acceleration. They slowly approach their new position. > >> Do you really think they suddenly flip right back to the starting point >> when >> the acceleration ceases? >> >> That's nonsense George. > >It is indeed. Good. > >>>> My diagram shows why. The path length change only during acceleration, >>>> not >>>> during constant rotation. >>> >>>Careful with your wording Herni, your diagram shows the >>>path length ARE CHANGED during constant acceleration from >>>their values during constant velocity. >> >> The path lengths remain constant during constant rotation. > >And the speeds remain constant at c+v and c-v so the times >are equal. what about the (2-root2) factor? >> During a period of acceleration, I think the path length is CONTINUALLY >> increasing. > >That's not what your diagram shows. For CONSTANT >acceleration, the path lengths are constant and >changed from the non-accelerating length by an >amount you show as 1/2 a t^2. > >> I too have been working on that point. It's not all that simple. > >Trust your diagram then. You are probably right. ....but you seem to believe the path lengths revert to their old values after the acceleration ceases. They don't. Each rotation speed has a different path length. You are ignotring the 'vt' term. >>>> The fringe 'displacement' at any instant is the integrated effect of all >>>> previous ACCELERATIONS. >>> >>>There is no physical mechanism involved that could integrate >>>the difference in arrival times of wavecrests. The actual >>>path times don't directly produce an output. >> >> During an acceleration, the two beams continuosly 'beat'. Even Androcles >> get's >> it right occasionally. > >During a _change_ of acceleration, they beat. When >the acceleration becomes constant, the stop beating >at take a stable value of phase shift at the value >they had at that time. That's right...meanwhile the number of fringes moved (including the partial bit at the end) has been counted. > >> The beats are counted. > >That would be an electronic integration which would >give speed. It gives change in speed. A second continuous integration of instantaneous displacement with time gives the rotation angle from zero. >You could imagine that this is how commercial units >work (it isn't) but that wouldn't apply to the lab >experiment. Remember in the original experiment of >Sagnac, he saw a shift of 7% of a fringe while the >table was turning at 120rpm. What you are describing >would be that he counted 0.07 fringes during the >acceleration phase and the displacement returned to >zero once constant speed was achieved. I dont think that would be accurate enough for any practical purpose. I'm sure fringes move a lot more in multi turn FoGs. >> The count IS really an integration of the path length >> change (or, more correctly, the number of wavelengths in each path) > >Yes but counting would have to be done as part of the >electronics, it is not inherent in the physical process. The path length change effectively integrates acceleration. No need for electronics. > >> The count itself (fringe diplacement) is also electronically integrated to >> give >> total rotation angle. > >That would be a second integration. yes. > >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 21 Nov 2005 00:51 On 18 Nov 2005 18:02:49 -0800, "Jeff Root" <jeff5(a)freemars.org> wrote: >George, > >Would you explain to me why you are posting in this thread? >What are you trying to accomplish? >Are you succeeding, or making progress? If you cannot see the connection, why are you bothering to post at all. You know nothing ! > > -- Jeff, in Minneapolis 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: Paul B. Andersen on 21 Nov 2005 05:51
Henri Wilson wrote: > On Fri, 18 Nov 2005 14:08:21 +0100, "Paul B. Andersen" > <paul.b.andersen(a)hiadeletethis.no> wrote: > > >>Henri Wilson wrote: >> >>>On Thu, 17 Nov 2005 16:27:07 +0100, "Paul B. Andersen" >>><paul.b.andersen(a)hiadeletethis.no> wrote: >>> > > >>>Wouldn't you like that to be true eh? >>> >>>At what tempertatures do they run? >> >>Depend on the cooling. >>Say 50 - 150C. >> >> >>>What determines the direction of emission of a photon from an atom Paul? >>>Come on, you are the expert..... >> >>I am a little reluctant to use much time to explain >>to you what you will ignore anyway. > > > I take that to mean you don't know. > ..which isn't surprising because I doubt if anyone has the faintest idea. I didn't say I couldn't or wouldn't explain it. I said I was reluctant do to so, because you would probably not care to read it anyway. Which you have demonstrated below. >>But in a laSEr, we are talking about Stimulated Emission. >>That means that a photon passing close by an atom may stimulate >>it to emit a new photon with the same direction and with the same >>phase. > > > Well now, that IS interesting. > It appears to support my W-aether theory. > > >>>Why should gas lasers work at all? >> >>Why indeed. :-) >>I am not going to give a thorough description, you can look it up >>yourself if you really want to know. >>But you never really want to know, you will rather invent it yourself. >> >>But I will mention a few points of special interest to this discussion. But Henri didn't read it, and made no attempt to understand it. >>Let us consider a HeNe laser. Let us assume the gas temp. is 350k. >>Let us assume the length of the tube is ca. 0.75 m. >>The rms speed of the He atoms will be: >>v^2 = 3kT/m where m is 4 proton masses. >>v = 1.47 km/s. (rms) >>v/c = 5*10^-6. >>This means that the frequency emitted by the atoms >>will be Doppler shifted, so the frequency will be >>distributed like a Gauss function with relative width ca. 5*10^-6. >>The central frequency fo = ca. 0.5*10^15 Hz >> >>However, the laser tube is a resonator, and only the frequencies >>given by f = m*c/2L can exist in the resonator. >>The distance between the possible frequencies is: >>delta_f = c/2L = 2*10^8 Hz. >>delta_f/fo = 4*10^-7 >> >>Note that a laser do not emit strictly monochromatic >>light, but a number of close spectral lines. >>We can see that in the order of 10 spectral lines can exist >>within the frequency distribution above. >> >>We KNOW this is happening in a laser. >>Each spectral line is emitted by atoms with >>a specific longitudinal velocity component. >> >>According to the BaT, the light from these spectral >>lines should travel at different speeds. >>They don't. >>BaT falsified. > > > Yes, very funny Paul. > > I think it is a little more complicated than that. So what is too simple to you, Henri? :-) > I think the photon fields combone somehow to create the resonance. Of course they do. They combine to a number of standing waves in the resonator. That WAS what I explained above, wasn't it? > I don't think molecular speeds would make much difference. You are babbling, Henri. The distribution of atomic speeds is the very reason for why there are several close spectral lines in a laser beam. And the number of spectral lines is exactly as predicted by the speed distribution. Read the explanation above again, please. The light in one of these of spectral lines can only come from He atoms with exactly the right speed to Doppler shift the light by exactly the correct amount. So we know what the speed of the atoms emitting the light in each spectral lines is. To sum it up: 1. We know that the speed of the atoms emitting the light in the different spectral lines are different. 2. We know that the speed of the light in the different spectral lines is the same. The BaT is falsified. Again. > Remember it was YOU > who claimed that electric fields act instantaneously. I claimed that a force acts on a charged particle at the same instant it enters a static electric field. You claimed this could be used to instant communication! :-) The fact that you mention this utterly irrelevant matter show your inability to give any sensible respons to the issue at hand. Paul |