Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 17 Nov 2005 08:30 Black Knight wrote: > "Paul B. Andersen" <paul.b.andersen(a)hiadeletethis > > Ok. Anything to oblige. > Andersen, you have convinced me. > Your stupidity IS so gigantic that you do not > understand why your statement is nonsense. > Please don't forgive me for not having doubted that. > I'm sure it will happen again. > > Androcles. > > Ouch! You really got me there, Androcles. Since you say so, I am sure my statement must have been nonsense. But which statement are you referring to? Would you quote it please, so we can laugh at it together? Paul
From: bz on 17 Nov 2005 08:24 HW@..(Henri Wilson) wrote in news:5vifn1hpot3i2kr8doc3daq19qnhdd4jcm(a)4ax.com: > On Sun, 13 Nov 2005 02:15:52 +0000 (UTC), bz > <bz+sp(a)ch100-5.chem.lsu.edu> wrote: > >>HW@..(Henri Wilson) wrote in >>news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com: >> ..... You said it. >>> Actually, you have shown that the BaTh does what it should do. It >>> expects NO fringe shifts under constant rotation. IOW, no position change (back to zero). >>> >>> BUT!!!! >>> The standard SR explanation says that there WILL BE a continuous >>> fringe shift during steady rotation. IOW, a constant position change. (offset remains constant). >>> >>> Sagnac proves SR to be wrong!!! >> >>So, you are saying that BaT predicts the fringes will shift during a- >>acceleration and return to original position when contant a-velocity is >>reached while SR predicts the fringes will move during acceleration and >>maintain a constant position when a constant velocity is reached? >> >>In otherwords BaT predicts return to original position upon ceasation of >>acceleration while SR predicts return to original position upon >>ceasation of rotation. >> >>If I understand the implications, it should be easy to tell the >>difference. >> >>Also, a light ring gyro should measure angular acceleration rather than >>angular position. > > Idiot. Learn the facts. tiodI. Facts are that in a BaT universe, the sagnac device would only show fringe shift when accelerating. Fringes would go back to zero when velocity was constant. YOU said it correctly. Now you are backpeddling. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: George Dishman on 17 Nov 2005 08:51 Henri Wilson wrote: > On 16 Nov 2005 06:10:53 -0800, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > >Henri Wilson wrote: > >> On 15 Nov 2005 06:55:28 -0800, "George Dishman" <george(a)briar.demon.co.uk> > >> wrote: > snip <snip uncommented parts> > >> >> >>>As you said of Maxwell's > >> >> >>>Equations, "a solution involves a wave moving at c" > >> >> >>>and the magnetic fields still exist and are still > >> >> >>>governed by Maxwell's Equations in the moving frame. > >> >> >> > >> >> >> Maxwell's equation applies to a wave that is symmetrical around an axis. > >> >> > > >> >> >No, Maxwell's Equations apply to the interactions of > >> >> >electric and magnetic fields regardless of symmetry. > >> >> >You can define a set of boundary conditions and they > >> >> >will tell you how the fields evolve thereafter. For > >> >> >example the waves produced by applying a sine wave > >> >> >voltage to a metal sphere will differ from those > >> >> >produced by a flat plate or a long wire. Maxwell's > >> >> >Equations apply to the fields regardless of the shape > >> >> >or motion of the source. > >> >> > >> >> That's rubbish George. > >> > > >> >I suggest you open a textbook and find out ho Maxwell's > >> >equations are used. > >> > >> George, an infinitesimal point cannot contain any wave of the Maxwellian type. > > > >The wavefront above for the laser was 1.1mm in diameter. > >Open a basic textbook and see if you can find out how > >Maxwell's Equations would predict that would evolve. > >Breaking it into infinitesimal parts was your choice. > > george, I'l try to make this even more simple. You are still have trouble > understanding it. Below is a laser beam with a infinitesimally wide diagonal > element drawn through it. > > | | > | | > | | > / > / > | | > | | So tell me, how does an element of infinitesimal width stretch all the way across a beam which is 1.1mm wide? What you don't seem to follow is that if you are going to analyse the evolution of the wavefront using infinitesimal elements, you have to break the wavefront into an infinite number of such elements and then integrate over the whole set. > Tell me George, does that element show any tendency to shoot off in the > diagonal direction? The element doesn't move at all, it produces a contribution to the fields in the space around it and you integrate the contributions from all such elements to find the field at some new position. You really should open a textbook sometime. > Of course not! Why? Because it is not a light beam. It is not governed by > Maxwell's equations. It is infinitesimal. It is nothing. > > >> > >> and its axis remains vertical. > >> There are no diagonal ants crawling diagonally.. > > > >Their path length is that of the diagonal, that's all that matters. > > There are no diagonal ants crawling diagonally.. We are only interested in the direction of their motion, not their orientation. That direction is diagonal as Einstein said and you illustrated proving him right. > >> >Right, but they do apply to electromagnetic fields so > >> >how are you going to resolve that ;-) > >> > >> George, an infinitesimal point cannot contain any wave of the Maxwellian type. > > > >Henri, the 1.1mm diameter wavefront from the laser is part > >of such a wave. If you cannot handle that using infinitesimal > >elements, find another way to do your math. > > See above. > > According to you, a laser beam would disperse in all directions, in the source > frame. I think you are being deliberately difficult. I think you are deliberately avoiding addressing what I am saying. If you deal with ONLY one point source then it disperses. If you shine a laser on a pinhole much smaller than the wavelengh, then diffraction at the edges gives you spherical waves emanating from the hole. In order to keep the narrow beam, you need a pinhole larger than the wavelength and you have to model the wavefront as a large number of small elements and integrate. > You know I am right. I can see you don't have any clue how to use Maxwell's Equations and you don't seem to be aware of the relationship between aperture and dispersion. > >> George, an infinitesimal point cannot contain any wave of the Maxwellian type. > > > >Henri, the 1.1mm diameter wavefront from the laser is part > >of such a wave. If you cannot handle that using infinitesimal > >elements, find another way to do your math. > > The wavefront is horizontal. In the laser frame that is correct, so your "infinitesimal elements" have to be aligned horizontally and you have to integrate over them. > According to you, a laser beam would disperse in all directions, in the source > frame. No, according to me, the wavefront from a point source (such as ONE infinitesimal element) would disperse in all directions in the ALL frames. To model a laser you MUST include all the elements across the finite width of the beam in some way. > >> >> George, hold a pen vertically. Now move you hand sideways. > >> > > >> >It draws a straight line. > >> > > >> >> Does the pen lean over? > >> > > >> >I don't care, now repeat the sideways motion but this > >> >time also move it away from you at the same time. It > >> >moves diagonally and for the same amount of sideways > >> >motion the line is longer than the first time, that fact > >> >is all that Einstein uses. The line isn't a pen, but then > >> >it wasn't the pen the first time either so your stuff > >> >about "the light isn't light anymore" is just > >> >meaningless nonsense. > >> > >> George, hold the pen vertical then move it sideways and upwards. > >> That's what Einstein tried to do.... > > > >No Henri, you have utterly failed to understand what he did. > >What he siad is that the length of the line drawn by the pen > >along the diagonal is longer than that drawn vertically. He > >is of course entirely correct as you have illustrated. > > What's this 'the diagonal'? The diagonal line being drawn by this single pen you introduced into the discussion. > There is an infinite number of diagonals involved. > ...and of course each one is longer than the vertical. > Plain Pythagoras. Exactly, and plain Pythogoras is what Einstein was using. I can't believe how much effort you are putting into trying to hide that simple fact. <snip> > >> >Of course the speed of light has been measured Henri, > >> >what are you raving about. > >> > >> The speed of the diagonally moving indinfitesimal points has never been > >> measured George. > > > >I said the speed of light, not of "indinfitesimal points" (sic). > > You claimed above somewhere that the diagonal speed of the points had been > measured Where? I said the speed of light had been measured. > and found to be c. That is not true and you know it. TWLS has been measured and found to be c, always, and you know it, and that is what I said. > >> >> There is NO wave moving along any diagonal. > >> > > >> >The ant is still an ant. > >> > >> a vertical ant....in all horiziontally moving frames, at any instant. > >> > >> That's the point you are missing.,,the 'instant' bit. > > > >I'm not missing it at all, I agreed it the first time you said it. > >I am waiting for you to explain why you think the direction > >of the axis affects the length of the path. > > > >> If you take a movie shot of ants crawling up a pole while you flash past in > >> your car, do their bodies appear vertical or diagonal on each frame? > > > >I don't care, it doesn't affect the length of the diagonal path. > > It affects what is supposed to be moving along each diagonal. Nope, what comes out of a laser it is still light whether you move your hand or not. I am still waiting for you to explain why it affects the length of the path because if it doesn't it is irrelevant. > >> George, an infinitesimal point cannot contain any wave of the Maxwellian type. > > > >The beam (wavefront) is 1.1mm wide for the laser example > >above. All beams have a finite width Henri. > > I don't care what you laser is. > Its beam is made up of an infinitte number of infinitesimally thin vertical > lines. The field is continuous over the width. Your approach of breaking it into a large number of finite segments is giving you problems. > Work out what happens to each one of those. By your own argument, none of them exists since they are infinitesimal. That is nonsense of course but if you want to use that argument in the moving frame, it also applies in the laser frame. > According to you, a laser beam would disperse in all directions, in the source > frame. No, according to Maxwell's Equations, a point source produces spherical wavefronts. > >> Nobody has tested the polarization of a sideways moving, vertical light beam. > >> There is no actual diagonal beam to test it on. > > > >I'll see if I can find the WMAP map. The details of the > >interaction of moving matter and light are complex, and > >your ants don't explain it well, but it remains a powerful tool. > > You are trying to change the subject. Just pointing out that your assertion is nonsense. > >> >Points aren't, no. Not in either the laser or moving > >> >frames, but the light whose location is represented > >> >by those points is still light, the ants are still ants. > >> > >> and they are still vertical at any instant in all frames. > > > >How does that affect the path length of the 1.1mm disc > >shaped wavefronts? > > George, I think you are trying to tell me that the wavefronts are lined up like > this in the moving frame: > > _ > _ > _ > _ > _ > > or this: > > \ > \ > \ > \ > \ > > They do neither, > > They remain like this: > > _ > _ > _ -> > _ > _ > > In all frames. No, what I am saying is that you can integrate over all the inifinitesimal elements to apply Maxwell's Equations (ME) and the result tells you the direction power will flow. If you start with a horizontal wavefront and apply ME then the wavefront will move vertically like this which is obviously wrong: > _ > _ > _ ^ > _ | > _ > > In all frames. What you will find is that the result in the moving frame becomes like this with each wavefront moving diagonally towards the top right. > \ > \ > \ -> > \ > \ Try to work out what the wavefronts look like inside the laser as they bounce between the mirrors and you should finally understand. > >> >> >>>> It obviously moves at sqrt(u^2+v^2) > >> > >> >> >> You know there has never been such a measurement, George. > >> >> > > >> >> >The speed of light has been measured many times > >> >> >Henri, I have no idea why you are denying that. > >> >> > >> >> HoHohahaha! > >> >> > >> >> George, TWLS has been measured and found to be consistent and precise > >> > > >> >Exactly what I have been saying throughout. If you are > >> >now admitting I was right, I fail to see why you keep > >> >claiming it has never been measured. > >> > >> George, what you are refering to is not light and its speed has certainly never > >> been measured, either OW or TW.. > > > >What comes out of a green laser is light Henri, even if > >you move your hand while holding it. > > It is only light in the vertical direction, in all frames. It is light in all frames. The beam is also vertical in all frames. The difference is that the wavefronts are propagating in the diagonal direction. > >> >Which part of "constant" escapes you? > >> > >> None. > >> The fact that both observers calculate the same value for c from the constants > >> escapes me...since the beam DOES NOT approach them at the same speed. > > > >The fact that they find the same values means the speed must be > >the same. That was exactly Einstein's route from the equations to > >the postulate. > > But we know they are not thsame. > The beam approaches the two observers at different speeds. Do we? That is YOUR competing postulate and the point is that it conflicts with Maxwell's Equations where the speed is defined by the constants. > >> >Energy and momentum for example. Been here before? > >> > >> Energy and momentum are properties of energy and momentum, not of photons. > > > >I'm not even going to waste my time on that one. > > Why not? Because properties are like adjectives that apply to nouns, you are trying to apply an adjective to an adjective. If I say a brick is hot, then temperature is a property of the brick. > You have wasted a lot more time worshipping the hoaxer Einstein. > > > > ><snip nonsense> > > Snip what you don't want to hear. Snip what has no scientific content, like the comment above. If all you can do is toss insults around, you are obviously unable just justify your assertions, you reduce them to religious claims, and there is no point in taling about it any farther. > >> >> >Your short lines moving up the screen are vertical > >> >> >while the wavefronts should be horizontal. > > >> Consider the laser beam to be infinitesimal in width....or at least much > >> smaller than a wavelength of the light used. > > > >Then it would produce spherical wavefronts and wouldn't > >be a laser at all. The width must be many wavelengths for > >the beam to have a small dispersion so let's consider the > >beam to be 1.1mm in diameter. > > It matters not how wide the beam is. It still doesn't spontaneously disperse in > all diagonal directions as you seem to think it does. Think about shining a laser onto a pinhole. The width is very important in a classical analysis. > >> >> I did...and it makes no sense at all. Infinitesimal points on a graph do not > >> >> constitute 'wavefronts'. > >> > > >> >The points on the graph aren't infinitesimal, they are purely > >> >mathematical points of zero size. They represent the "very > >> >very small pieces" of the actual wavefront, or isn't that what > >> >you mean, it is what you have been saying. > >> > >> Not 'very small' but 'infinitesimal'. > > > >Same thing Henri. > > Not the same George. Exactly the same Henri, open a textbook on basic calculus. > >> Even if the laser beam is wide, only an infinitesimally thin section will move > >> up any diagonal line....far too thin to be a light wave. > > > >Each circular disc wavefront sweeps out a tube with > >elliptical cross section with a major axis of 1.1mm. > >Maxwell's Equations must describe the evolution of > >that wavefront or they are invalid. > > The wavefront is nothing more than a line on a graph. > It is infinitesimally thin and has no light-like properties. Let me remind you of what you said above: > The wavefront is horizontal. That horizontal line is merely a mathematical trick, it marks the highest field strength in a region but the field is actually a sine wave filling the volume of the beam. That propagating sine wave is what is called light (both fields of course but I'm keeping it simple here). > >> >Then you have to throw Maxwell's Equations in the bin. > >> >That was what was worrying them. > >> > >> Maxwell's equations apply to waves moving in one direction. > > > >Nope, they apply to the evolution of the fields whether there are > >waves involved or not and they apply in all frames. Think of a > >mexican wave in a stadium. You can describe the overall effect > >using a wave equation but you can also define the behaviour of > >one person solely in terms of the movement of those around him. > >You can then produce a wave by defining the starting condition > >for everyone at some instant and then using the individual > >behaviour to evolve the next state and so on. That's how they > >work. A plane wave is just one solution. > > According to you, George, a laser beam would spontaneously disperse in all > diagonal directions, in the source frame. No, according to Maxwell's Equations, the field from any single one of your "infinitesimal elements" would disperse in ALL frames. It is the interference between the infinite number of elements covering the horizontal surface of the wavefront that allows the beam to avoid dispersing. George
From: Black Knight on 17 Nov 2005 09:13 "Paul B. Andersen" <paul.b.andersen(a)hiadeletethis Ok. Anything to oblige. Andersen, you have convinced me. Your stupidity IS so gigantic that you do not understand why your statement is nonsense. Please don't forgive me for not having doubted that. I'm sure it will happen again. Androcles.
From: Paul B. Andersen on 17 Nov 2005 10:27
Henri Wilson wrote: > On Wed, 16 Nov 2005 23:34:14 +0100, "Paul B. Andersen" > <paul.b.andersen(a)hiadeletethis.no> wrote: > > >>Henri Wilson wrote: >> >>>I don't htink we will discuss ring lasers becasue they might operate on an >>>entirely different principle to the four mirror type. >> >>I don't think you will discuss ring lasers because they so >>obviously falsifies the BaT. > > We are discussing sagnac. > Your statement above shows these aren't basd on sagnac. > > >>I have explained why before, and you have fled the discussion before >>because you were unable to refute my arguments. Which is the real reason you don't like to discuss ring lasers. >>In fact any gas laser falsifies the BaT. Coherent light means all >>the light is going at the same speed. In a gas laser, the gas atoms >>which are the sources of the light, are moving fast relative to >>each other. BaT falsified. > > > It is YOUR theory that each atom is a source. > Other don't believe you. Don't be ridiculous, Henri. :-) What in a gas laSER do you think is Stimulated to Emit Radiation, if not the gas atoms? > Even if it were, the effect would be too small to worry about. Really, Henri? The light from a laser can go to the Moon and back, and still be coherent light. Do you think that could happen if the speeds of the photons were as different as the speeds of the atoms in the gas? The mere existence of gas lasers falsifies the BaT. >>>FoGs are similar but effectively have an infinitie number of mirrors which >>>reflect at infinitesimal angle. >>>We aren't going to get anywhere multiplying zero by infinity. >> >>Not unless you know some math, of course. >>Hint: limits. >>What is sin(x)/x for x = 0? > > > Irelevant. It is an example where "multiplying zero by infinity" get you somewhere. >>And light in a mono-mode fibre is never reflected. >>It's a wave guide. >>BaT falsified. > > > There is constant internal reflection at grazing angles. You don't know what a mono-mode fibre is, do you? :-) >>>So let's just stick with the four mirror sagnac eh? >>> >>>I think by now you will have realised that it fully supports the BaTh and >>>probably refutes SR. >> >>Any Sagnac ring falsifies the BaT. >>No question about it. > > > My diagram clearly show the opposite. > Path lengths change during acceleration. Wavelength is absolutely constant > according to the BaTh. Therefore fringes will MOVE during angular acceleration > and will NOT move during constant rotation. > > Sagnac disproves SR. Inventing new laws of nature proves nothing. Any Sagnac ring falsifies the BaT. No question about it. > > >>>Poor old George has spent years proving that according to the BaTh, fringes >>>will not move during constant rotation. That is of course what happens. >> >>Of course it is. >>But I have in a much shorter time proved that according >>to the BaT, the phase relationship between the two waves >>will be the same regardless of the rate of a constant rotation. >>That is NOT what happens. >>The BaT falsified. > > > You haven't proved that at all. You have a very selective memory. Paul B. Andersen wrote January 2005: | To a first order approximation, (that is, ignoring | terms containing higher than first order of | the tangential mirror speed v) the light will use | the same time in both directions. | The math isn't very hard, but it isn't trivial either. | I won't bother to go through all the math in this awkward | medium, but I will write the first order terms: | The length of one chord of the light path will be: | d = srt(2)*r + v*t/sqrt(2) | where r is the radius of the circle tangenting the mirrors, | and t is the time the light uses to traverse the chord. | The speed of the light will be: | c' = c + v/sqrt(2) | Note that these equations are valid for both direction, | v being negative for the beam going in the opposite direction. | So we have: | c'*t = d | c*t + v*t/sqrt(2) = sqrt(2)*r + v*t/sqrt(2) | t = sqrt(2)*r/c | The ballistic theory predicts that the time | has no first order dependency on the speed! | | The sagnac effect IS a first order effect! | | You are proven wrong. Henri Wilson responded: | I did that calculation a long time ago. So Henri Wilson agree to the statement: " The ballistic theory predicts that the time has no first order dependency on the speed!" Which obviously implies that the phase relationship between the two waves will be the same regardless of the rate of a constant rotation. > You haven't even understood the significance of acceleration in all of this. > You haven't a clue. > Study my diagram again. I understand that acceleration have no significance in this. >>And you never refuted my proof. Quite the contrary, you have confirmed it. >>You fled it by claiming that interferometers works >>in an entirely different manner than physicists think. >>You are unable to explain _how_ you think it works, thought. >>All you know is that it works in some mysterious way which >>make the fringes appear at different positions even when >>the phase relationship between the two waves are the same. Henri's explanation of why the fringes change position despite the fact that BaT predicts no change in the phase relationship: Henri Wilson wrote: | I have told you many times. The sagnac effect is caused by the angular change | in the mirrors during the travel time of the light between them. The change in | opposite ways for the two beams and cause opposite sideways displacements when | reunited. That causes sideways fringe movements. Henri Wilson wrote: | Actually, the sagnac effect is completely unrelated to SR, the BaT or aether | theories. | | The fringes shift because the mirrors rotate slightly during lights travel time | between them. | | The clockwise beam ends up displaced one way, the anti-clockwise one the other. See? The fringes do not change position because the phase relashionship between the waves changes, interferometers doesn't measure phase differences. They work in an entirely different manner. How is not clear. This is unbelivable stupid, Henri. :-) >>A typical Wilsonian escape by inventing new laws of nature >>for every new phenomenon that must be explained away. > > > The sagnac effect occurs when the apparatus is experiencing angular > acceleration. The reason is that, during light transit time, successive > components are displaced by a little more than they would be under constant > rotation. D = kw when dw/dt = 0 But D is caused by dw/dt, not w, because dD/dt <> 0 only when dw/dt <> 0. How incredible stupid! > The amount is virtually the same whether c or c+v is used. > Path length difference alone determines the position of the fringes. Small > variations in light speed do not make any significant difference to the result. How incredible stupid! The Sagnac effect IS "the small variations in light speed". delta_n = n+ - n- where n+ and n- are the number of wavelengths around the ring for the two beams, and delta_n is the difference. n+ = (S/lambda)(c+v)/c S = circumference of ring n- = (S/lambda)(c-v)/c delta_n = (S/lambda)2v/c S = 2*pi*r, A = pi*r^2, v = w*r, lambda = c/f, T = 1/f delta_n = f*4*pi*r^2*w/c^2 = f4Aw/c^2 delta_t = delta_n*T = delta_n/f = 4Aw/c^2 Seen this before? If this is the Sagnac effect, then it is caused by "the small variations in light speed". Paul |