Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: bz on 13 Nov 2005 03:58 "Eric Gisse" <jowr.pi(a)gmail.com> wrote in news:1131855605.794683.277520(a)g43g2000cwa.googlegroups.com: > > bz wrote: > > [snip] > >> >> If I understand the implications, it should be easy to tell the >> difference. > > You should have seen by now that Henri has zero interest in testing his > theory. > Negative. -- 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: mmeron on 13 Nov 2005 04:06 In article <Xns970D1E95F2849WQAHBGMXSZHVspammote(a)130.39.198.139>, bz <bz+sp(a)ch100-5.chem.lsu.edu> writes: >"Eric Gisse" <jowr.pi(a)gmail.com> wrote in >news:1131855605.794683.277520(a)g43g2000cwa.googlegroups.com: > >> >> bz wrote: >> >> [snip] >> >>> >>> If I understand the implications, it should be easy to tell the >>> difference. >> >> You should have seen by now that Henri has zero interest in testing his >> theory. >> > >Negative. > Imaginary. Mati Meron | "When you argue with a fool, meron(a)cars.uchicago.edu | chances are he is doing just the same"
From: George Dishman on 13 Nov 2005 06:44 "Henri Wilson" <HW@..> wrote in message news:0mlcn197uu5d7383ug28ct9oav3t2ij5kq(a)4ax.com... > On Sat, 12 Nov 2005 15:12:11 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: >> >>Try illustrating the wavefronts as short lines >>perpendicular to their direction of motion. In >>the laser frame, it might look like this (drawn >>side by side as I can't animate in ASCII): >> >> - >> - >> - >> | | | | | | |-| | | | | | | >> |-| | | |-| | | |-| | | |-| >> | | |-| | | | | | | |-| | | >> >> >>So you get wavefronts coming out the top of the >>laser tube at regular intervals. I can't show the >>mirrors at the ends. >> >>See how that works in the moving frame. Each point >>on your graph is the centre of one of those wavefront >>segments show as "-" above. > > That's right. > ..but you have only drawn one point of the wavefront. No, each hyphen is a finite width, it is the entire wavefront across the width of the tube, or as close as I can get in ASCII. > There is an infintie number, each one moving along a different diagonal. > If you represented them all, you would have a broad diagonal line. This is the best I can do in reasonable width: - - - - - - - - - - - - - - - - - - - - - - - - | | |-| | | | | | | |-| | | | | | | |-| | | | | | | |-| | | | | |-| | | |-| | | |-| | | |-| | | |-| | | |-| | | |-| | | |-| | | | | | | | | |-| | | | | | | |-| | | | | | | |-| | | | | | | |-| That's harder to read, at any time the picture would be typically this showing two wavefronts moving up the screen having left the tube and the wavefront inside the laser moving down towards the rear mirror: - - | | |-| | | >>> The whole path is moving sideways in the moving frame. >>> George, when you drive past a light pole, does it lean over diagonally? >>> If ants are crawling up it, do their bodies point diagonally? >>> >>> Of course not!! >> >>The path of each ant is diagonal which is what we >>are discussing, but if you want to go into more >>detail, consider what the above diagram would look >>like in the moving frame. > > The ant bodies remain aligned vertically in the moving frame. Indeed but that is where your analogy breaks down and a spherical object would be more appropriate. > Each infinitesimal element of an ant (infinitely smaller than a molecule) > follows a diagonal path. The mathematical point locating the centre of momentum of the ant moves diagonally. Your infinitesimal elements only clouds the issue but yes their paths would also be diagonal. > The ants take the same time to reach the top no matter who moves past. In your religion it does. I Einstein's gedanken, the aim is to derive the time from the speed. >>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. For our purposes, just note that they apply equally well in both frames. > You 'diagonal field' is skewed. Yes, and Maxwell's Equations must still apply or they would be invalid. > The only true wave remains vertical in all frames. > > >>> You keep thinking in terms of the paths of infinitesimal elements. >> >>That's what your simulation depicts, according to what you >>have told me. Actually I am thinking in terms of wavefronts >>moving in accordance with Maxwell's Equations. > > The purple laser beam in my demo represents that. > The green elements are completely different. Unfortunately your wiggly lines are easy to understand but don't explain the behaviour of the wavefronts. >>> The whole beam remains aligned vertically in all frames. >> >>Yes, we have agreed that several times. The line of >>narrow segments of wavefronts remains vertically >>aligned while each segment moves diagonally. > > ...and the axis of wave symmetry remains vertical in all frames. Truye, but I don't think you yet realise the consequence of that, or at least you have avoided illustrating it. >>>>> ..and whatever moves up it has a velocity sqrt(c^2+v^2) >>>> >>>>Talk science, not religion Henri. >>> >>> I take it from that, you are starting to realize your error. >>> ..It obviously has a velocity sqrt(c^2+v^2) >> >>No, scientifically speaking, if we measure it we find >>it has a speed of c whether we understand why or not. >>That is an empirical statement. > > You are simply preaching the unproven second postulate. No, I am stating what is measured regardless of postulates. > That is hardly a scientific statement George. If I was stating the postulate, you would have a point. >>You have a religious conviction that Galilean relativity >>should hold and therefore you expect on faith that the >>speed will be sqrt(c^2+v^2). > > It obviously is. "obvious" being a statement of faith. > It isn't light. It is a point on a graph. It can move at any speed. The point is representative of the wavefront which must be a solution to Maxwell's equations in _both_ frames. >>>>>>> Are you being deliberately stubborn? >>>>>> >>>>>>No, I'm being intelligent. Can we drop the insults? >>>>> >>>>> Well if you cannot understand the difference between an 'infinitesimal >>>>> point on >>>>> a graph' and a whole 'photon' then you leave yourself open to >>>>> ridicule. >>>> >>>>If you don't know that a plotted point on a graph is not >>>>'infinitesimal' but of zero size then I'm not the one >>>>who will be ridiculed. Similarly if you don't know that >>>>the mathematical point must be related to the object >>>>being described by some definition (which you have so >>>>far omitted) to relate the maths to the physics. >>> >>> George, I will try to make my point even MORE clear. >>> >>> You drive your car past an archer who fires an arrow vertically into the >>> air. >>> In your frame, each minute element of that arrow moves along a diagonal >>> path. >>> >>> ACCORDING TO YOUR LOGIC, EVERY SUCH INFINITESIMAL DIAGONAL PATH HAS AN >>> ARROW GOING UP IT. >> >>No. In my interpretation of _your_ world, if the archer >>fires a series of arrows, each arrow would be seen to >>be still vertical but it would be flying along a diagonal >>path relative to the car. > > Of course. Thank you. No need to shout then. > but each part of the arrow moves along a different diagonal.... Yes. > and the arrow > takes the same time to reach the top no matter how fast the car moves. Your religious assertion again. Don't state it, derive it. >>The series of arrows would also still be in line vertically >>because, although each is moving diagonally, the archer is >>also moving sideways (relative to the car) and stays directly >>beneath the line as he fires. >> >>What Maxwell's equations say is that arrows always move in >>the direction of the shaft. :-o > > But you claim they are moving diagonally. As you said, all the arrows move diagonally, and the time evolution of wavefronts must be a solution to Maxwell's Equations. Keep going Henri, we have 2 and 2 but you haven't tried to add them yet. > You have now identified your own mistake. > >>> You have somehow turned one arrow into an infinite number. Brilliant!!!! >> >>You did that when you talked of a beam, which is a >>collection of wavefronts. > > It is a collection of infinitesimal elements. If you wish to view it that way, that's your choice though I don't see how it will help (unless you intend to use Huygens method). >>>>And it is that diagonal path that is being described. You >>>>simply keep proving Einstein was right. >>> >>> He said each link moves at the same speed vertically and diagonally. >> >>He took as a postulate that it would be the same because >>that is what Maxwell's Equations tell us as you said >>above, "a wave moving at c where the value of c is found >>via the two measured constants." >> >>> It obviously moves at sqrt(u^2+v^2) >> >>If you prefer faith to measurement. > > 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. >>>>> Do you really think the blade takes any longer to reach the top if >>>>> different >>>>> observers time it? >>>> >>>>"the blade" doesn't take any time to "reach the top", >>>>it is "each link" that moves from bottom to top, and yes >>>>scientific measurements show that the time _does_ depend >>>>on the motion of the instrument measuring that time. >>> >>> Yes I meant each 'link'. >>> >>> Bull. >>> That is just the postulate. >> >>And that is the whole point of the gedanken, he >>is demonstrating a consequence of the postulate. >> >>> It has never been proved. >> >>A thought experiment can never prove a postulate. > > ...a postulate can never prove itself either... That's what I just said >:-| > particularly when that postulate > was based on a thought experiment anyway. It wasn't, it was based on Maxwell's Equations, the speed is the product of two measured constants, remember? >>>>You said I claimed they had structure. How could something >>>>of zero size have structure Henri? >>> >>> That's what I asked you. >>> If it doesn't have structure, then it is no different from 'nothing'. >> >>That is your faith. I am saying photons are as much >>like ball-bearings as electrons or quarks, I made no >>other claim to have knowledge of any structure more >>fundamental than any of those particles. > > Anything that has zero size and no properties must be 'nothing'. But photons don't have "no properties" so your argument fails. >>>>> In that case you must surely have some idea of what makes them >>>>> different >>>>> from 'nothing'. Please tell us. >>>> >>>>Momenergy (hate that word but that's what they call it). >>> >>> Preaching your religion again George? >>> How can something that has no volume, no properties and is >>> indistinguishable >>> from 'nothing' suddenly possess energy and momentum? >> >>That it possesses energy and momentum is what distinguishes >>it from 'nothing'. > > How does your 'nothing' possess energy George? Easy, it isn't 'nothing' because your requirement that it be 'nothing' is that it has no properties. Your argument is circular. >>>>Very good Henri, you're starting to think. So how do >>>>you define the mathematical point on the graph that >>>>represents the raindrop? If it is falling in a vacuum >>>>then using the centre of momentum of the drop would be >>>>reasonable. Nothing "infinitesimal" is involved at all. >>> >>> You are moving into a differnet area here. If you want to consider >>> raindrops as >>> rigid spheres then you will find that the momentum of each in your >>> moving >>> frame >>> increases to p*sqrt(u^2+v^2) >> >>Momentum is frame dependent. For your laser, note that >>moving the laser will give the usual Doppler effect >>for an observer and the change in frequency gives a >>change in momentum. > > Standard NM. Yes, to first order. >>> ...so let's stick with arrows. They are more like photons than are >>> raindrops. >> >>Actually they aren't, rigid ball bearings are probably >>closest. > > So you DO have a model for a photon, eh George? > Tell us about it. No model, just properties. >>>>It is obvious. With knowledge of science as well as >>>>intelligence, the error is also obvious, you are not >>>>comparing like with like, just playing word games >>>>comparing "the beam" versus an "infinitesimal element". >>> >>> George, the question is, do the links of the 'photon chainsaw' move >>> diagonally at c or at sqrt(v^2+c^2)? >> >>No Henri, that isn't the question. Einstein is illustrating >>a consequence of the postulate so the question is "Assuming >>the links move at c, what other conclusions can we draw?". >>Finding differences between that and the conclusions if >>we assume the speed is sqrt(v^2+c^2) then provides a means >>to test the postulate. > > well it has never been adequately tested. In your opinion. > No direct measurement of OWLS from a moving source has ever been achieved. Nor can it ever be measured since a one-way measure requires synchronisation of clocks at the ends and the means of sync completes the loop turning it into a two-way measure. As I said, the two-way measure plus a measure of anisotropy gives the one-way value indirectly and both have been done many times. >>> But George, can you not see that no continuous WAVE follows the >>> 'wavefront' up its unique diagonal? >> >>Try drawing wavefronts and see what happens. > > I have. That's what my program shows. Your short lines moving up the screen are vertical while the wavefronts should be horizontal. > They all move up diffrent diagonals. Yes, that part is fine. > Surely you have enough intelligence to see that. > Even Paul Andersen can. Yes but I don't think you realised what I was suggesting. >>> In my demo, the purple beam represents such a continuous wave. It is >>> obviously quite different from the 'green element'. >> >>Since the wavefront has to be a solution of Maxwell's >>Equations, that would indicate you have another error >>that we haven't discussed yet. Drawing the wavefronts >>will help you find it. You'll probably learn more that >>way than if I just tell you. > > You are raving. > Wavefronts have nothing to do with this. Much more than you think. >>Try adding such a detector to your animation when you do >>the wavefronts and see how they are related to the tube. >> >>> If the beam was perfectly narrow and parallel, you wouldn't see it at >>> all. >>> You would only detect a minute flash. >> >>Well obviously, the same is true in both frames as your >>detector is passing through the beam briefly, but the >>angle that has to change if you want to see anything at >>all. > > You didn't understand my point. I did but again you aren't comparing like with like. In one case you have the detector passing through the beam so it sees a flash while in the other it is permanently in the beam and sees steady illumination so there is no comparison. >>>>in the laser frame, you have choices to make. >>> >>> George, I have shown only a small number of representative elements. >>> Obviouslky >>> I cannot draw an infinite number. >>> If I used your idea, I would end up with one wide rainbow instead of >>> lots >>> of narrow green ones. >> >>My suggestion is that you should draw just one as >>it can be representative of all. > > By drawing a number of them, I show how the beam remains vertically > aligned in > the moving frame. Understood, but in the lower diagram their paths lie on top of each other so you cannot see what is happening. Perhaps you could highlight the first on the moving frame and show it on lower then show the subsequent wavefronts only on the upper where they don't overlap. That shows both aspects. >>>>frame. Do you want to disagree with that, because that's >>>>what you have to do to prove him wrong? >>> >>> Yes I certainly do want to disagree with that. >>> What moves diagonally is NOT the wavefront of a light beam moving at c. >> >>Well in what direction does an individual wavefront >>move? > > A 'wavefront' is perpendicular to the wave axis. Wavefronts really exist > only > in the source frame. The electric and magnetic fields exist in both, and disturbances propagate in accordance with Maxwell's Equations which must apply equally in both. I think you are now starting to see one of the problems that faced science at the end of the nineteenth century. George
From: George Dishman on 13 Nov 2005 06:52 "Henri Wilson" <HW@..> wrote in message news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(a)4ax.com... > On Sat, 12 Nov 2005 10:35:04 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >>"Henri Wilson" <HW@..> wrote in message >>news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... > > George, George, George. > > I have finally woken up to your (and MY) complete misinterpretation of the > problem. > > We have both been arguing about whether or not the fringes will move > during > constant angular rotation...and of course they don't. Actually they do, that's exactly how fibre gyros work. The output is proportional to the angular velocity. That is why these devices are such important evidence and precisely why I have taken the trouble of bringing them to your attention. > We should only be considering what happens during angular ACCELERATION > !!!!!! > > That is when the two path lengths change. > That is when more 'wavelengths' fit into one path than the other. > That is when fringes move. > > Path lengths chaneg because each mirror accelerates slightly as light from > the > previous one is in flight. Small second order effect, you say. No way! It > is > the whole basis of operation. Acceleration is a more complex subject, involving Doppler at the source, Doppler at the receiver and the temporary lack of cancellation due to the flight time delay between. > Actually, you have shown that the BaTh does what it should do. It expects > NO > fringe shifts under constant rotation. > > BUT!!!! > The standard SR explanation says that there WILL BE a continuous fringe > shift > during steady rotation. > > Sagnac proves SR to be wrong!!! If the output was proportional to acceleration then you would be right. In fact the output is exactly what SR predicts, it is proportional to the angular velocity in actual devices and experiments. George
From: Jerry on 13 Nov 2005 07:58
Henri Wilson wrote: > On Sat, 12 Nov 2005 10:35:04 -0000, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > > >"Henri Wilson" <HW@..> wrote in message > >news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... > > George, George, George. > > I have finally woken up to your (and MY) complete misinterpretation of the > problem. > > We have both been arguing about whether or not the fringes will move during > constant angular rotation...and of course they don't. Huh? http://www.physics.berkeley.edu/research/packard/Competition/Gyros/LaserRingGyro/Steadman/StedmanReview1997.pdf Jerry |