Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Black Knight on 11 Nov 2005 23:49 "Henri Wilson" <HW@..> wrote in message news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... > On Fri, 11 Nov 2005 14:17:16 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:6ua7n15d9rn5rld9rco5lrfa752igqbdvs(a)4ax.com... >>> >>>>> On 9 Nov 2005 13:25:38 -0800, "George Dishman" >>>>> <george(a)briar.demon.co.uk> >>>>> wrote: >>>>>>> > >>>In Ritzian theory, the light is emitted at some speed greater >>>>>>> > >>>than c from the source. The speed can be found by taking the >>>>>>> > >>>magnitude of the vector sum of the mirror velocity and a >>>>>>> > >>>vector of magnitude c whose direction is such that the light >>>>>>> > >>>eventually reaches the detector. >>> >>> Consider this. >>> >>> You are driving along a road at v and your passenger wants to shoot a >>> duck >>> that >>> is sitting on the ground 500 metres to your left. Where does he aim? >>> Where does he aim if the duck appears at 45 degree forward? >>> >>> Where does he aim if the duck flies off at v and exactly 90 degrees away >>> from >>> the road ? >>> >>> ____________v<-car >>> /gun >>> >>> >>> >>> >>> >>> | >>> | >>> | >>> \/ v >>> Duck >> >>You have it in a nutshell Henri. The only thing you missed >>is that the speed of the bullet is known relative to the gun >>but I'm sure you realise that. Now look at what I said. I'll >>change some terms to make it appropriate to your analogy: >> >> The speed [of the bullet across the ground] can be found >> by taking the magnitude of the vector sum of the [gun] >> velocity [over the ground] and a vector of magnitude >> [bullet speed relative to the gun] whose direction is >> such that the [bullet] eventually reaches the [duck]. >> >>I think that is pretty good agreement, don't you? > > That is right.... and according to the BaTh, the time taken for a photon > to > reach the mirror is constant even though the path length changes. > > But sagnac does not refute the Bath for this reason. > > Under Bath, wavelength does not change with source speed. One can imagine > the > wavecrests being connected by tiny rods. > > Since the path lengths are different in both directions of sagnac, (the > duck > example shows that) the number of wavelengths between the source and the > mirror > is not the same in each beam... > The number varies with table rotation speed. > > That'll make you think... > > >> >>George >> > > > HW. Message rating several casks. Androcles.
From: George Dishman on 12 Nov 2005 05:35 "Henri Wilson" <HW@..> wrote in message news:9r5an1pjg0a2vce62k32juf9345b93kp9b(a)4ax.com... > On Fri, 11 Nov 2005 14:17:16 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:6ua7n15d9rn5rld9rco5lrfa752igqbdvs(a)4ax.com... >>> >>>>> On 9 Nov 2005 13:25:38 -0800, "George Dishman" >>>>> <george(a)briar.demon.co.uk> >>>>> wrote: >>>>>>> > >>>In Ritzian theory, the light is emitted at some speed greater >>>>>>> > >>>than c from the source. The speed can be found by taking the >>>>>>> > >>>magnitude of the vector sum of the mirror velocity and a >>>>>>> > >>>vector of magnitude c whose direction is such that the light >>>>>>> > >>>eventually reaches the detector. >>> >>> Consider this. >>> >>> You are driving along a road at v and your passenger wants to shoot a >>> duck >>> that >>> is sitting on the ground 500 metres to your left. Where does he aim? >>> Where does he aim if the duck appears at 45 degree forward? >>> >>> Where does he aim if the duck flies off at v and exactly 90 degrees away >>> from >>> the road ? >>> >>> ____________v<-car >>> /gun >>> >>> >>> >>> >>> >>> | >>> | >>> | >>> \/ v >>> Duck >> >>You have it in a nutshell Henri. The only thing you missed >>is that the speed of the bullet is known relative to the gun >>but I'm sure you realise that. Now look at what I said. I'll >>change some terms to make it appropriate to your analogy: >> >> The speed [of the bullet across the ground] can be found >> by taking the magnitude of the vector sum of the [gun] >> velocity [over the ground] and a vector of magnitude >> [bullet speed relative to the gun] whose direction is >> such that the [bullet] eventually reaches the [duck]. >> >>I think that is pretty good agreement, don't you? > > That is right.... and according to the BaTh, the time taken for a photon > to > reach the mirror is constant even though the path length changes. Actually the path length doesn't change. Consider that the car is chasing the duck round a roundabout. If their motion is at radius R from the centre, you could tie a string of length R*sqrt(2) between the duck and the car. > But sagnac does not refute the Bath for this reason. > > Under Bath, wavelength does not change with source speed. One can imagine > the > wavecrests being connected by tiny rods. Possibly but I'm going to lay aside thinking about that for the moment and use your first statement: "the time taken for a photon to reach the mirror is constant". Consider the situation when the table is not rotating. A wavecrest is emitted from the source, split into two equal parts, and they then proceed round the mirrors in opposed directions. The path lengths are obviously equal so the travel times are equal and the wavecrests arrive at the detector simultaneously. That means the peaks add and you get maximum brightness on the detector. You get minimum when the wavecrest along one path coincides with the arrival of a trough of along the other path and the signals always cancel. Now start the table turning. Since the travel times are unchanged, the wavecrests still arrive simultaneously and you still get the same brightness. The same argument applies to the next pair of wavecrests which arrive a "tiny rod-length" later. > Since the path lengths are different in both directions of sagnac, (the > duck > example shows that) the number of wavelengths between the source and the > mirror > is not the same in each beam... > The number varies with table rotation speed. > > That'll make you think... I'll give it some thought later (I have too much to do and still owe you several replies) but it doesn't change the phase relationship of the arriving crests so doesn't affect the detector output. You still have a null result but I think your understanding of why that is the case is much better. George
From: George Dishman on 12 Nov 2005 05:51 "Henri Wilson" <HW@..> wrote in message news:1ft4n1d2j2esvvn9th4gg4iiefpvvk9rv9(a)4ax.com... > On 9 Nov 2005 13:25:38 -0800, "George Dishman" <george(a)briar.demon.co.uk> > wrote: >>Henri Wilson wrote: >>> On 7 Nov 2005 05:59:38 -0800, "george(a)briar.demon.co.uk" >>> <george(a)briar.demon.co.uk> wrote: <snip stuff now agreed> >>> In the table frame, the whole beam appears to move at 45 degrees, like >>> the >>> green beam in my demo. >> >>We aren't interested in "the beam" but individual wavefronts. >>Think of the links on the chainsaw in your other analogy. >> >>> I'll have to think about this a bit more before I enlarge on it. >>> Can't draw it here. >> >>You need to think about the path of an individual >>representative wavefront. Each subsequent one >>just follows the same relative path but from a starting >>point farther round the circumference. > > Nah! OK, think about the car chasing the duck round the roundabout only this time the hunter has an automatic. The first bullet travels as you drew: v<-car /gun | / v Duck The second is fired 90 degrees later: car |\gun | \ v \ --> Duck Can you see that the same analysis applies, the drawing has just been turned a bit? The same is true for intermediate angles so once you do the maths for one bullet, the calculated time taken for the bullet to reach the duck will also apply to each subsequent bullet. <snip> >> Closing speed is meaningful when one object is following >> the same path as another but is some distance behind, for >> example cars on a race track. You measure the distance >> not along the straight line between the cars but along the >> path they have to travel along the track. Or think of the two >> cars each holding constant speed but on a winding mountain >> road. Closing speed is then the rate at which that distance >> between them is decreasing, the numerical difference between >> their individual ground speeds, not the vector difference. If you >> take the initial separation and divide by the closing speed, you >> get the time for the second car to catch the first. > >>> You have to move the leading car diagonally at 45 deg. >>> Then your figures will be different. >> >>No, if the first car is moving at 30m/s and the second is >>following the same path at 40m/s then their separation >>measured _along_the_track_ decreases by 10m each >>second. This applies OK to the circular path in an iFOG >>but isn't really applicable to the discrete mirror case. > > I wouldn't agree it applies to an iFOG because that really involves an > infinite > number of 'mirrors'. The reson it applies is that the path is a series of chords. If you calculate the sum of the lengths of the N chords of a regular polygon within a circle and then let N tend to infinity, the length of each tends to zero (infinitesimal) while the sum is asymptotic to the circumference. As N tends to infinity, the direction of the light is also asymptotic to the tangential at the point of reflection. The the path of the light is asymptotic to that of the mirrors, your use of closing speed is then reasonable. > You can still use it in the four mirror system but you have to accomodate > the > sideways movement. Closing speed does not deal with sideways, it is one dimensional by definition. George
From: George Dishman on 12 Nov 2005 10:12 "Henri Wilson" <HW@..> wrote in message news:k5rsm1lubih0v1k2b0dpm5lt4c2i9jrfjf(a)4ax.com... > On Sun, 6 Nov 2005 12:00:19 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > > continued... > >>>>> In that case, their diagonal speed is sqrt(c^2+v^2). >>>> >>>>That's what any self-respecting 18th century physicist >>>>would have believed. Maxwell's equations suggested it >>>>was wrong and observation has confirmed that, they move >>>>at c in the moving frame too. >>> >>> Maxwell's equation are not relevant. >> >>Of course they are Henri. > > Are you losing your marbles George? > Do you have the faintest idea what Maxwell's equations describe? > They relate to a lossless resonance between an electric field and a > magnetic > one in a dielectric material. It turns out that a solution involves a wave > moving at c where the value of c is found via the two measured constants. > > I fail to see how an infinitesimal point can represent a wavelike process > moving at c. Easy, I gave you that answer lower in the post: >>The size of the point on the graph does not 'approach zero' >>Henri, it is _precisely_ zero as is true for all mathematical >>points. To relate it to the real world, you need a definition, >>for example you might identify the point as the intersection of >>a particular wavefront with a ray, which is itself defined as a >>mathematical line passing down the central axis of the laser. >>> There is no wavelike structure moving along each diagonal. There is ONE >>> infinitesimal point on a graph. >> >>There is infinitesimal point on your graph which indicates >>the motion of the wavefront. When a wave near the top of >>your display was emitted, the laser was near the left edge >>of the screen so the diagonal line marks the path it took. >>How you relate your point on the graph to that wavefront is >>for you to define. > > The only wavelike structure in any frame follows a vertical path in that > frame. 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. > 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. 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. >>> The only 'wave' remains in the vertical direction of all frames. >> >>Try drawing the motion of an actual wavefront and you'll >>learn some interesting aspects that aren't apparent to >>you yet. > > 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 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. >>>>That's the key Henri, you have to compare like with like. >>>>If you want to compare one of the diagonal green lines, >>>>which is the historical path of one photon, with something >>>>in the laser frame, it has to be the historical path of the >>>>same photon, not the "beam" as you are trying to do. >>> >>> The diagonal green line is NOT the path of ONE PHOTON. >>> It is the path of an infinitesimal length of one photon. >> >>It is a mathematical line of zero width which tracks >>some point related to the physical phenomenon being >>plotted, be that a photon or a wavefront or whatever. >> >>> ..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 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). >>>>> 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. 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 > 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. I'm going to snip some less relevant stuff, this is much too long. >>> Do you remember my chainsaw experiment? >>> Fasten a chainsaw in your car roof with the blade vertical. Start the >>> saw then drive the car along the road. >>> Do you really believe the chain leans over diaginally in the road frame? >>> Each link in the chain appears to move diagonally but on close >>> inspection it >>> can be seen that the axis of the link remains vertical adn that each >>> infinitesimal section of each link moves along a different diagonal. >> >>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. >>> 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. >>>>>>> Your problem George is that you think photons are like ball >>>>>>> bearings. >>>>>> >>>>>>Sort of, they are like other sub-atomic particles. >>>>> >>>>> So you claim to know something about the structure of photons eh >>>>> George? >>>>> Tell us more..... >>>> >>>>I made no such claim. >>> >>> You said they have zero size. >> >>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. >>> 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'. <big snip, all repeating what has been said above or just agreed> >>>>>>> I'm afraid he spent too much time misinterpreting the way in which >>>>>>> raindrops moved past train windows. >>>>>> >>>>>>Your diagram shows he was right, the paths are diagonal. >>>>> >>>>> Each droplet APPEARS to move diagonally. >>>> >>>>The PATH of each droplet IS diagonal in the train frame >>>>while the path of the SAME droplet was vertical in the >>>>trackside frame. Compare like with like Henri. >>> >>> George, if you were able to magnify each droplet a billion times as it >>> moved >>> 'diagonally' past your train window, would you see each molecule moving >>> along >>> the same diagonal? Would the nucleus of each atom move along the same >>> diagonal >>> as the electron shell? >>> >>> No, of course not. >> >>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. > ...so let's stick with arrows. They are more like photons than are > raindrops. Actually they aren't, rigid ball bearings are probably closest. >>>>> In fact, the paths of each >>>>> infinitesimal point inside each water molecule travels along a >>>>> different >>>>> diagonal at the speed sqrt(u^2+v^2). >>>> >>>>Almost, but not quite ;-) >>> >>> Why 'not quite' George. >> >>Because you forgot the extra factor involving gamma from >>SR. sqrt(u^2+v^2) is a very good approximation though ;-) > > George, you apparently want to use a prediction of your theory to prove > your theory. No, you made a statement and I just corrected the error. I added a " ;-) " because the effect you missed is so tiny that it is immeasurable in practice for raindrops. >>>>I do, better than you it appears. You are highlighting >>>>the finite-sized pixel within which the 'infinitesimal >>>>element' can be found at the current time and pducing a >>>>path of all such pixels where the 'infinitesimal element' >>>>was located at previous times. Showing a single pixel of >>>>a different colour is a more accurate representation of >>>>an 'infinitesimal element' than your multiple pixels. >>> >>> My representation should be obvious to anyone with inltelligence. >> >>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. >>But a wavefront is quite different and that's what >>the mathematical point on the graph describes. > > But George, can you not see that no continuous WAVE follows the > 'wavefront' up its unique diagonal? Try drawing wavefronts and see what happens. > 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. >>> George, here is an experiment for you. >>> >>> Set up a vertically pointing laser beam. >>> No move past that beam at c/2. >>> How would you detect the presence of the CONTINUOUS beam if you had a >>> photodetector that was only on micron wide? >>> >>> Would you angle the detector at 60 degrees somewhere offset wrt your own >>> vertical or would you place it above you and then move it sideways >>> at -c/2 >>> when >>> the beam was detected? >>> >>> For the former, there would be a very brief and probably unnoticed >>> flash. >>> >>> In other words, the continuous beam exists only in the source frame. >>> Maxwell >>> only applies in the source frame. >> >>Suppose the detector is at the end of a narrow tube to >>exclude ambient light. The detector would be vertical >>if it was at rest wrt the laser but if I try to use >>it when moving past I have to angle it to compensate >>for aberration, and Maxwell's Equations apply to the >>propagation of the wavefront along the tube. > > this is not a good example, however... You chose it Henri. > You angle it so as to pick up the different part of the beam that ends up > in > your new direction. The distance traveled by the beam you detect is longer > than > the vertical. 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. >>> Each diagonal green line represents the path of a 'wavecrest' of the >>> vertical >>> beam. That should be obvious enough. >> >>It would be equally obvious if you used a distinct colour >>and that would also highlight why you are not comparing >>like with like, the moving frame would show a diagonal >>of one colour with your moving element at the tip in >>another. See if you can imagine the corresponding display >>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. >>> There is a vertical beam of light in one frame. There is NO diagonal >>> beam of light in the other. Do you want to disagree with that? >> >>"The beam" is of no relevance, Einstein's gedanken is about >>the speed of one particular wavefront of a propagating wave >>as defined by Maxwell's Equations. The path of EACH wavefront >>is vertical in the laser frame but diagonal in the moving >>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? >>>>> You are quoting a postulate not a proven fact. >>>> >>>>All science is based on postulates, it is a fact >>>>that no measurement has ever falsified the postulate >>>>and any possible deviation is tightly constrained, >>>>which is as good as any science can ever get. >>> >>> No measurement has ever supported the postulate George. >> >>Since no measurement has ever falsified it, every >>measurement ever made relating to the speed of light >>has supported it, including Michelson-Morely, Fizeau, >>aberration, Sagnac, Kennedy-Thorndike and so on. You >>know better than to write nonsense like that Henri. > > Give up George. OK, my point is made. George
From: George Dishman on 12 Nov 2005 11:48
"George Dishman" <george(a)briar.demon.co.uk> wrote in message news:dl51a0$td0$1(a)news.freedom2surf.net... > > "Henri Wilson" <HW@..> wrote in message > news:mg5an1dg95m92cg1ji7l4pf2cj9cunpr1k(a)4ax.com... >> >> Scientifically speaking, the size of the infinitie number of >> infinitesimal >> elements is 1/infinity > > Nope, it is infinity/infinity. Do you really want to go > off down another sidetrack? That could have been clearer, the total of elements is infinity/infinity which is finite. I could perhaps also have pointed out that each wavefront is perpendicular to the direction of the beam. It is not clear which way your are breaking it into elements. George |