Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: George Dishman on 6 Nov 2005 05:31 "Henri Wilson" <HW@..> wrote in message news:h3rfm11mk9sladl6qt8le2093f8l7ik3qe(a)4ax.com... > On Tue, 1 Nov 2005 19:18:26 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:vq3dm19b3tv3uoie6l6etdlcl2sbacl0uc(a)4ax.com... >>> On Mon, 31 Oct 2005 20:20:03 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> > >>>> >>>>Two cars driving along the road, the first moving at 30m/s >>>>and the second moving at 40m/s, closing speed 10m/s. The >>>>first has a head start of 200m. The second car catches the >>>>first when the first car has moved 600m and the second has >>>>travelled 800m. >>>> >>>>To find the time, you divide the separation at the start >>>>by the closing speed, 200 / 10 = 20s. "Closing speed" is >>>>the rate at which the separation is decreasing. What you >>>>try to do below is divide the 600m by 10m/s which gives >>>>you the wrong answer. >>>> >>>>> The beam travels in a straight line from the source. The >>>>> travel length is greater than the source/mirror distance. >>>>> For the opposite beam, the distance is shorter. >>>> >>>>No. To get the time taken, you use the separation between >>>>the source and mirror at the instant the light is emitted. >>>>That is simply the distance measured when the table is not >>>>rotating. That distance is unchanged by the rotation and >>>>as you said the closing speed is always c so the time is >>>>also unchanged by rotation. >>> >>> No. 'c' is the closing speed of the beam and the next mirror, >>> the direction of which is 45 degrees. >> >>So far we agree. >> >>> The beam is not the horizontal one. It is angled slightly >>> downwards to meet the mirror in its new position. >> >>No, the "new position" is the 600m figure in the car >>example. You have to use the distance to the mirror >>at the instant when the light is emitted, not when >>the light will be reflected, if you choose to use >>closing speed. > > George, now give your leading car a sideways speed component as well as > the 30 > m/s instantaneous speed component in the direction of the second car. The > second car has to point diagonally to reach the first. It takes longer > than 20 > seconds to catch it and moves more than 800 metres. This is perhaps getting away from the point a bit but it might help to clear this up. 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 can measure the distance not along the straight line between cars but along the path. Think of two cars each holding constant speed but on a winding mountain road. Closing speed is then the rate at which that length is decreasing. If you take the initial separation and divide by the closing speed, you get the time for the second car to catch the first. > Thus the travel distance of the light beam is the hyponenuse of the > triangle. > the closing speed is c. ...and the path lengths are different for the > opposing > beams. Henri, the mirror is moving in a circle relative to the source so the length is no greater even though there is a transverse component. >>> The distance is greater than >>> in the non-rotating case. >>> >>>> >>>>You have again proved that Ritz predicts a null result. >>> >>> George, at what speed (in the table frame) is the light reflected from >>> the >>> first mirror towards the next mirror? >> >>I thought we had agreed that several times before. Perhaps >>I wasn't clear so I'll go through it again in more detail. >>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. I agree what you say next, do you agree what I say above? > I would say that in the mirror frame, the light arrives at c and is > therefore > reflected at c....which makes it the same situation as light from the > source > towards the first mirror. I agree but your previous question was "(in the table frame)". >>>>I am happy to accept what you have again proved, Ritz >>>>predicts a null result for Sagnac. >>> >>> It does not...and stop repeating something you would like to be true but >>> isn't. >> >>I will if you will. I am just clarifying the consequences >>of the corrected version of what you wrote. > > You car analogy was plainly wrong. > You have to move the leading car sideways as well as forward. You missed that the motion of the target mirror is circular hence there is no increase of distance in the source frame. There is an increase in the table frame of course. >>> I realize you are trying desperately to convince yourself but you aren't >>> impressing me. >>> >>> I now think it might predict a fringe shift of 1/root2 times the >>> classical >>> one. >>> That would be near enough to the same. >> >>Not near enough, you need to get it within about one >>part per million IIRC from one of the web pages we >>discussed some time ago. I'll try to find the reference >>again if I can. > > I should imagine ring gyros are calibrated empirically anyway. The maximum range of calibration would be orders of magnitude smaller though, there is no way you can get away with such a large error. >>>>> In the table frame, the closing speed between the beam and the moving >>>>> mirrors >>>>> is always c. >>>>> that is all that matters. >>>> >>>>Right, and that gives a null prediction. >>> >>> No the path lengths are distinctly different in either direction. >>> >>>>> The paths lengths of the opposite beams are different. >>>> >>>>No, see the example of the cars above to learn how to use >>>>closing speed correctly. >>> >>> That is a linear example. Sagnac involves 45 degrees. >> >>The distance is to the point of reflection at the instant >>the light is emitted, the consequence of the motion of the >>mirror is included in using closing speed so if you adjust >>the distance too, you double count its effect. > > No your car example was wrong. Try again. The motion of the mirror in the source frame is at a fixed radius so the distance doesn't change. George
From: George Dishman on 6 Nov 2005 07:00 "Henri Wilson" <HW@..> wrote in message news:nknfm1pp6cufk08o6ije9uqtcpd29a2q8j(a)4ax.com... > On Tue, 1 Nov 2005 20:04:10 -0000, "George Dishman" > <george(a)briar.demon.co.uk> > wrote: > >> >>"Henri Wilson" <HW@..> wrote in message >>news:k8bam191oeqaaidmdeqccad4pl12pci3i2(a)4ax.com... >>> On Sun, 30 Oct 2005 10:56:41 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>> >>>> >>>>"Henri Wilson" <HW@..> wrote in message >>>>news:fnpvl152mh7h3rhvc4oi4rad3lne51lrp2(a)4ax.com... >>>>> On Wed, 26 Oct 2005 20:50:41 +0100, "George Dishman" >>>>> <george(a)briar.demon.co.uk> >>>>> wrote: >>> >>>>> It should be plainly obvious to anyone with even half a brain that no >>>>> light >>>>> BEAM moves along any one of the infinite number of infinitesimally >>>>> thin >>>>> diagonal lines that represent the paths if infinitesimal elements of >>>>> the >>>>> light >>>>> beam that is vertical in the rest frame. >>>>> >>>>> What moves up a diagonal is an infinitesimal point. It is not a >>>>> physical >>>>> entity. >>>> >>>>It is the smallest, indivisible piece of light. Your >>>>diagram is like tracking the path of a water molecule >>>>in the jet from a hose. >>> >>> No it is smaller than that. It is infinitesimal. >> >>"infinitesimal" means of negligible size but finite. A >>photon like electrons and other fundamental particles >>is of zero size so smaller than infinitesimal, unless >>you want to speculate about them being of the Planck >>size and the consequences of string theory. > > Talk science, not religion. In science, infinitesimal is not zero. >>> It is the same size as a line on a graph. >>> Didn't you ever study maths George? >> >>Enough to highlight the errors in your program Henri. >>Ad hominems won't get you anyhere Henri, shall we drop >>them and stick to the science? > > How thick is a line on a graph George. > Talk science, not religion. In physics the line is representative of the motion of something measurable, such as the centre of momentum of a ball bearing, or some trackable location on a wavefront, or a photon. >>>>The thing whose path is being plotted exists and >>>>the line is a plot of its path. >>> >>> The point where the moving blades of scissors cross is also real is it >>> George? >> >>You can plot the path of something that exists or >>something that is not. The light from the laser is >>real regardless of what frame you choose. > > You just said it has zero size. > Do you know what you mean? > If it has zero size it also has zero roperties. OBVIOUSLY. Talk science, not religion Henri. Your assertion that something of zero size cannot have properties is religious. Scientifically, a photon has no measurable size but does have measurable properties including energy and momentum. >>>>>>Because your program shows that the diagonal path >>>>>>of each photon or flash or wavefront on the top >>>>>>diagram is longer than the vertical line showing >>>>>>the corresponding path on the bottom diagram. >>>>> >>>>> IT IS NOT THE DIAGONAL PATH OF A PHOTON. >>>>> >>>>> It is a diagonal line showing the path of an infinitesimal point of >>>>> the >>>>> vertical beam. >>>> >>>>And an "infinitesimal point of the ... beam" is what >>>>we call "a photon". >>> >>> You seem to believe that photons are like ball bearings. >> >>In some cases, except that they have zero size and zero >>mass, in other cases they act like waves. > > In other words you haven't a clue but will stick rigidly to your religion. No, in other words I follow the scientific method, I accept what is repeatably measured. >>> 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. > 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 '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. >>>>No, he didn't. The source obviously moves while, if >>>>the beam went along the diagonal, the source would >>>>need to be static. >>> >>> My animation shows the difference between a diagonal beam (purple) and >>> the >>> diagonally moving 'infinitesimal point'. >>> How can you compare the two? >> >>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. >>> 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. >>>>No, I'm saying you are misquoting Einstein and, >>>>perhaps unintentionally, creating a strawman. >>> >>> Sorry George, that's the essence of SR. >> >>No Henri it isn't. As you said before, you have deliberately >>avoided learning SR so it is to be expected that you will >>have these misunderstandings. > > Sorry George, the 'light clock' idea summarizes the principle behind SR > and > also shows why the whole theory is nonsense from the first chapter. The 'light clock' is a consequence, not the essence. >>> It works in LET, it fails in SR. >>> >>>>>>> BUT ALL THE WAVECRESTS REMAIN >>>>>>> VERTICALLY ALIGNED in the moving frame. >>>>>> >>>>>>Yes, that is also true. Nobody is disputing it. >>>>> >>>>> Well can you not see tat hte line of vertical wavecrests constitute >>>>> the >>>>> light beam. IT REMAINS VERTICAL IN BOTH FRAMES. >>>> >>>>As I said, nobody is disputing that. >>> >>> Well does it move at 'c' vertically? >> >>No, it doesn't move at all. The "beam" always has its >>base at the aperture of the laser. > > The elements that make up the beam remain vertically aligned in all > frames. True but irrelevant, Einstein's explanation relates to the path of an individual element. > 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. > 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. >>> Does it take the same time to reach the top no matter who goes past? >> >>It doesn't take any time to "reach the top", one end of the >>beam IS the top (of the beam). How much time does it take >>a piece of string to reach the end of the piece of string? > > Talk science, not religion. I'm talking basic English here, just as "the blade" is the whole object while "each link" is an item that moves along the blade. Stick the physics Henri, you never win these silly word games. >>>>> 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? > 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). >>>>> Now do the same with the bullets. >>>>> In the moving frame, as before, the centre of each bullet moves along >>>>> its >>>>> own unique diagonal. Only one bullet centre moves along any >>>>> particular >>>>> diagonal. >>>> >>>>Let's consider what you said of the shot: "However each shot moves >>>>along a different diagonal path." >>>> >>>>That remains true of the bullets, each bullet moves >>>>along a diagonal path and that is the basis of the >>>>argument. >>> >>> But its axis is not diagonal. It remains vertical. >>> Similarly, if a hole was drilled through the lead shot, the holes would >>> remain >>> lined up vertically.. >> >>Einstein's gedanken deals with the length of the path >>the photon took. It doesn't matter what a photon looks >>like, only the length of the path it takes is relevant >>to his illustration of the consequences of the speed >>being c in both frames. > > He wrongly assumed that photons were like ball bearings He didn't even know about photons, Planck quantised light some years after Einstein published SR. > and he wrongly > postulated that these imaginary objects travel at c in both frames. He derived his theory from the observation, encapsulated in Maxwell's Equations, that light wavefronts propagate at c. >>>>> However, NOW, the axis of each bullet is angled wrt the >>>>> diagonal...because >>>>> each >>>>> element of the bullet emerges from the barrel at a slightly different >>>>> time. It >>>>> remains vertical..so that all the bullet axes are still lined up >>>>> vertically. >>>> >>>>Yes, motion of the source can influence polarisation. >>>>See for example how that is used on the CMBR. That >>>>doesn't change the fact that each photon travels a >>>>diagonal path just as for the shot and bullets. >>> >>> This is bloody stupid. Can you not get it into your head that the >>> diagonal >>> 'path' is just a line on a graph. >> >>Can you not get it into your head that you have to compare >>like with like? The length of the line on that graph is >>being compared with length of the equivalent line for the >>same "infinitesimal element", if that's what you want to >>call it, on the laser-frame graph. > > It IS an infinitesimal element...in other words its size approaches zero. 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. > ,,,and since the time taken to reach the top is unaffected by observer > motion, > the speed along the diagonal must be aqrt(c^2+v^2) Talk science, not religion Henri. You are assuming that the universe obeys Galilean relativity, that is not what is measured. >>> It is the path taken by an infinitesimal part of something. >>> Do you understand what 'infinitesimal' means George? >>> I gather you don't. >> >>Whatever you think it means, you must still compare like >>with like. The path of the 'infinitesimal' in one frame >>with the path of the same 'infinitesimal' in the other. > > We aren't really interested in the infinitesimal points. We want to know > what > the light beam is doing. No, we are interested in the length of the path of a single wavefront. That is what can be related to the speed defined by Maxwell's equations. Where later wavefronts lie in relation to that one is of no interest. > It remains vertical in all frames and it is only a > true 'light beam' in the 'vertical line' that appears to move sideways in > the > moving frame.. >> >>http://www.google.co.uk/search?hl=en&q=define%3A+infinitesimal >> >>"a variable that has zero as its limit" >> >>"In mathematics, an infinitesimal ... is a number that is >>greater in absolute value than zero yet smaller than any >>positive real number." >> >>"an abstract description of nearly nothing." >> >>"a very small quantity, approaching zero" >> >>> A photon might be small but it is certainly not infinitesimal. >> >>Right, it has zero size so is smaller than "infinitesimal", >>unless string theory turns out to be correct. > > If it has zero size, George, please tell the world how it differs from > 'nothing'. Just learn what the words mean Henri. >>> Whatever moves along the graphed path of an infinitesimal point is not >>> light >>> and it moves at sqrt(c^2+v^2) >> >>It is pointless to repeat assertions unless you have >>something to back them up. Experimentally the measured >>speed is c. > > It has never been measured and you know it. Roemer, etc. Why do you say such idiotic things Henri, I know you are smarter than that. >>>>> Of course...which means that the infintesimal point that follows each >>>>> diagonal >>>>> path moves at sqrt(c^2+v^2) and NOT at 'c', as Einstein stupidly >>>>> believed. >>>> >>>>It moves at c as measured, you therefore have to revise >>>>your belief that the times must be the same. >>> >>> George, don't talk nonsense. >>> You are starting to look like a complete fool. >>> You are proving that you have no understanding of physics at all. >> >>Your inability to offer anyhing other than insults is telling. > > I have explained over and over, in the simplest of terms why Einstein is > wrong. And in return I have explained repeatedly why you are not comparing like with, you considering the path of a single wavefront in the moving frame so you must compare that to the path of the same wavefront in the laser frame. You have simple ignored my corretion, endlessly repeating your original erroneous statements. Change the record if you want me to resond differently. > You refuse to accept, saying only that 'PHOTONS HAVE ZERO SIZE'. That is > meaningless drivel. That is the reason I gave for why your illustration fails to disprove Einstein, as an aside I was correcting yet another of your errors. SR was published before Planck quantised radiation, and even at that time Planck still thought it was a quantisation of the amount of energy in a wave rather than a particulate theory. >>>>> 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 vacuuum then using the centre of momentum of the drop would be reasonable. Nothing "infinitesimal" is involved at all. >>> 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 ;-) > Are you about to show you know even less about physics than I thought? > >> >>> The drops take the same time to reach the ground no matter how fast >>> Einstein >>> moves. >> >>Almost, but not quite ;-) > > Why 'not quite' George. Because you forgot the extra factor involving gamma from SR. "the same time" is a very good approximation though. > Are you about to show you know even less about physics than I thought? > >>>>OK, you might replace your short line for your >>>>"infinitesimal element" by a single pixel of a >>>>different colour in that case. It would make more >>>>sense anyway if it is supposed to be "infinitesimal". >>> >>> You should know what 'infinitesimal' means. >> >>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". >>>>>>> Not so. It is a point on a graph.....nothing physical. >>>>>> >>>>>>It is a mrker on the light beam, for example a wavefront >>>>>>as you suggested. >>>>> >>>>> A 'point' >>>> >>>>The location of "an infinitesimal element of >>>>light", a photon. >>> >>> Is this a magic photon designed by the fairies, George? >>> >>> Infinitesimal points don't have physical properties George. >> >>Yet more baseless assertions Henri? > > An infinitesimal point IS INDISTINGUISHABLE FROM NOTHING!!!!!!! But a wavefront is quite different and that's what the mathematical point on the graph describes. > 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. >>>>>>The vertical line is a plot of the history of one >>>>>>wavefront in the rest frame of the laser. In fact >>>>>>it is the plot of all the wavefronts which lay on >>>>>>top of each other. In the moving frame the lines >>>>>>are equivalent plots of the same wavefronts but >>>>>>the motion means they are no longer superimposed. >>>>> >>>>> George, I think you should spend a litle more time thinking about >>>>> this. >>>>> You seem utterly confused. >>>>> Come back when you have worked out what is really happening. >>>> >>>>Read it again Henri but this time think about >>>>it, I am correct. >>> >>> You 'correction' of my output is wrong. >> >>What is wrong with your argument is that you aren't >>comparing like with like. You would be better to show >>a single diagonal green line and the same element >>moving vertically in the laser frame. However you >>choose to show it, you need to compare like with like. >> >>The diagram was to illustrate how consistent graphics >>would look. If you are using a line element for the >>path, the other way to get consistency is to correct >>the symbol for the element. That was a minor point >>intended to be helpful. > > 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. >>>>No, that's your confused alternative, "the beam" is >>>>always attached to the laser, it is the "infinitesimal >>>>elements" that are moving at c, either vertically in >>>>one frame or diagonally in the other. >>> >>> HoHohahahaha!!! Whoever measured that? >> >>Well I could start with Roemer, but that's beside >>the point. The gedanken illustrates a consequence by >>comparing the length of the paths in the two frames, >>not the directions of the beam in the two frames. Your >>diagram confirms the path in one is vertical but >>diagonal in the other. > > 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? >>> 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. George
From: Henri Wilson on 6 Nov 2005 17:12 On Sun, 6 Nov 2005 12:00:19 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >Talk science, not religion Henri. Your assertion that >something of zero size cannot have properties is >religious. Scientifically, a photon has no measurable >size but does have measurable properties including >energy and momentum. > continued... >>>> >>>> You seem to believe that photons are like ball bearings. >>> >>>In some cases, except that they have zero size and zero >>>mass, in other cases they act like waves. >> >> In other words you haven't a clue but will stick rigidly to your religion. > >No, in other words I follow the scientific method, I accept >what is repeatably measured. > >>>> 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. >> 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. 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 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. The whole beam remains aligned vertically in all frames. >>>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) > >>>> 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. You have somehow turned one arrow into an infinite number. Brilliant!!!! >> >> Sorry George, the 'light clock' idea summarizes the principle behind SR >> and >> also shows why the whole theory is nonsense from the first chapter. > >The 'light clock' is a consequence, not the essence. It is the very basis of SR. >>>> It works in LET, it fails in SR. >>>> Well does it move at 'c' vertically? >>> >>>No, it doesn't move at all. The "beam" always has its >>>base at the aperture of the laser. >> >> The elements that make up the beam remain vertically aligned in all >> frames. > >True but irrelevant, Einstein's explanation relates to >the path of an individual element. Einstein's 'explanation' is plain nonsense. >> 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. It obviously moves at sqrt(u^2+v^2) > >> 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. It has never been proved. >>>> Does it take the same time to reach the top no matter who goes past? >>> >>>It doesn't take any time to "reach the top", one end of the >>>beam IS the top (of the beam). How much time does it take >>>a piece of string to reach the end of the piece of string? >> >> Talk science, not religion. > >I'm talking basic English here, just as "the blade" is the >whole object while "each link" is an item that moves along >the blade. Stick the physics Henri, you never win these >silly word games. George, it is obvious that the time taken for each link to reach the top is not dependent on the movement of any stray observer. It is a direct aether concept that the moving observer will MEASURE a different time than the rest observer. SR merely uses the same principle in a disguised fashion. >> >> 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'. > >> 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? >>>Einstein's gedanken deals with the length of the path >>>the photon took. It doesn't matter what a photon looks >>>like, only the length of the path it takes is relevant >>>to his illustration of the consequences of the speed >>>being c in both frames. >> >> He wrongly assumed that photons were like ball bearings > >He didn't even know about photons, Planck quantised light >some years after Einstein published SR. He didn't even get that far. He got the whole picture wrong when he 'deduced' that the raindrops moving diagonally past his train window took longer to reach the ground BECAUSE the train was moving. > >> and he wrongly >> postulated that these imaginary objects travel at c in both frames. > >He derived his theory from the observation, encapsulated in >Maxwell's Equations, that light wavefronts propagate at c. wrt an 'aether'. >>>Can you not get it into your head that you have to compare >>>like with like? The length of the line on that graph is >>>being compared with length of the equivalent line for the >>>same "infinitesimal element", if that's what you want to >>>call it, on the laser-frame graph. >> >> It IS an infinitesimal element...in other words its size approaches zero. > >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. > >> ,,,and since the time taken to reach the top is unaffected by observer >> motion, >> the speed along the diagonal must be aqrt(c^2+v^2) > >Talk science, not religion Henri. You are assuming that >the universe obeys Galilean relativity, that is not what >is measured. You are assuming that 'absolute' states of matter are observer dependent. This is a direct consequence of aether theory belief. >>>> It is the path taken by an infinitesimal part of something. >>>> Do you understand what 'infinitesimal' means George? >>>> I gather you don't. >>> >>>Whatever you think it means, you must still compare like >>>with like. The path of the 'infinitesimal' in one frame >>>with the path of the same 'infinitesimal' in the other. >> >> We aren't really interested in the infinitesimal points. We want to know >> what >> the light beam is doing. > >No, we are interested in the length of the path of a >single wavefront. That is what can be related to the >speed defined by Maxwell's equations. Where later >wavefronts lie in relation to that one is of no >interest. > >> It remains vertical in all frames and it is only a >> true 'light beam' in the 'vertical line' that appears to move sideways in >> the >> moving frame.. >>> >>>http://www.google.co.uk/search?hl=en&q=define%3A+infinitesimal >>> >>>"a variable that has zero as its limit" >>> >>>"In mathematics, an infinitesimal ... is a number that is >>>greater in absolute value than zero yet smaller than any >>>positive real number." >>> >>>"an abstract description of nearly nothing." >>> >>>"a very small quantity, approaching zero" >>> >>>> A photon might be small but it is certainly not infinitesimal. >>> >>>Right, it has zero size so is smaller than "infinitesimal", >>>unless string theory turns out to be correct. >> >> If it has zero size, George, please tell the world how it differs from >> 'nothing'. > >Just learn what the words mean Henri. Learning any number of words wont turn 'nothing' into 'something' George. >>>It is pointless to repeat assertions unless you have >>>something to back them up. Experimentally the measured >>>speed is c. >> >> It has never been measured and you know it. > >Roemer, etc. Why do you say such idiotic things Henri, I know >you are smarter than that. .....and did Roemer get 2.997E8 m/sec George? >>> >>>Your inability to offer anyhing other than insults is telling. >> >> I have explained over and over, in the simplest of terms why Einstein is >> wrong. > >And in return I have explained repeatedly why you are not >comparing like with, you considering the path of a single >wavefront in the moving frame so you must compare that to >the path of the same wavefront in the laser frame. You >have simple ignored my corretion, endlessly repeating your >original erroneous statements. Change the record if you >want me to resond differently. Says George, the magician who can turn ONE arrow into an infinite number just by driving past in his car. > >> You refuse to accept, saying only that 'PHOTONS HAVE ZERO SIZE'. That is >> meaningless drivel. > >That is the reason I gave for why your illustration fails >to disprove Einstein, as an aside I was correcting yet >another of your errors. SR was published before Planck >quantised radiation, and even at that time Planck still >thought it was a quantisation of the amount of energy >in a wave rather than a particulate theory. Well I agree that we have to ultimately regard all matter as quantized waves in fields, whatever that means. >>>>>> 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 vacuuum >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) ....so let's stick with arrows. They are more like photons than are raindrops. >>>> 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. > >> Are you about to show you know even less about physics than I thought? >> >>> >>>> The drops take the same time to reach the ground no matter how fast >>>> Einstein >>>> moves. >>> >>>Almost, but not quite ;-) >> >> Why 'not quite' George. > >Because you forgot the extra factor involving gamma from >SR. "the same time" is a very good approximation though. George, you apparently want to use a prediction of your theory to prove your theory. >>>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)? since Einstein's postulate that they move at 'c' has never been even remotely supported by ezperiment, one must regard his whole theory as a complete joke. >>>> Infinitesimal points don't have physical properties George. >>> >>>Yet more baseless assertions Henri? >> >> An infinitesimal point IS INDISTINGUISHABLE FROM NOTHING!!!!!!! > >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? In my demo, the purple beam represents such a continuous wave. It is obviously quite different from the 'green element'. >> 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 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. If the beam was perfectly narrow and parallel, you wouldn't see it at all. You would only detect a minute flash. >> >> 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. >> >> 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. >>>> 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. > >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 6 Nov 2005 17:41 On Sun, 6 Nov 2005 10:31:41 -0000, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@..> wrote in message >news:h3rfm11mk9sladl6qt8le2093f8l7ik3qe(a)4ax.com... >> On Tue, 1 Nov 2005 19:18:26 -0000, "George Dishman" >> <george(a)briar.demon.co.uk> >>>> No. 'c' is the closing speed of the beam and the next mirror, >>>> the direction of which is 45 degrees. >>> >>>So far we agree. >>> >>>> The beam is not the horizontal one. It is angled slightly >>>> downwards to meet the mirror in its new position. >>> >>>No, the "new position" is the 600m figure in the car >>>example. You have to use the distance to the mirror >>>at the instant when the light is emitted, not when >>>the light will be reflected, if you choose to use >>>closing speed. >> >> George, now give your leading car a sideways speed component as well as >> the 30 >> m/s instantaneous speed component in the direction of the second car. The >> second car has to point diagonally to reach the first. It takes longer >> than 20 >> seconds to catch it and moves more than 800 metres. > >This is perhaps getting away from the point a bit but it >might help to clear this up. 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 can measure the distance not along the straight line >between cars but along the path. Think of two cars each >holding constant speed but on a winding mountain road. >Closing speed is then the rate at which that length is >decreasing. If you take the initial separation and divide >by the closing speed, you get the time for the second car >to catch the first. Don't try to wriggle out by changing the subject. Your initial example only considered the speed component along the line joining the two cars. If you introduce a sideways components as well, as in the sagnac, you get an entirely difffernet result. > >> Thus the travel distance of the light beam is the hyponenuse of the >> triangle. >> the closing speed is c. ...and the path lengths are different for the >> opposing >> beams. > >Henri, the mirror is moving in a circle relative to the >source so the length is no greater even though there is >a transverse component. Let's go back to the non-rotating table frame. Here, the beam moves at very nearly c+v/root2v towards the mirror''s new position when the beam arrives. After each reflection, it continues at c+v/root2 wrt the table. The second beam moves at c-v/root2 towards the third mirror's new position. This approach gives a similar answer to your classical analysis except that the fringe shift should be about 35% of the 'aether prediction', (I think). Since the gyros are calibrated empirically, I don't think the difference would be noticed. >>> >>>I thought we had agreed that several times before. Perhaps >>>I wasn't clear so I'll go through it again in more detail. >>>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. > >I agree what you say next, do you agree what I say above? > >> I would say that in the mirror frame, the light arrives at c and is >> therefore >> reflected at c....which makes it the same situation as light from the >> source >> towards the first mirror. > >I agree but your previous question was "(in the table frame)". > >>>>>I am happy to accept what you have again proved, Ritz >>>>>predicts a null result for Sagnac. >>>> >>>> It does not...and stop repeating something you would like to be true but >>>> isn't. >>> >>>I will if you will. I am just clarifying the consequences >>>of the corrected version of what you wrote. >> >> You car analogy was plainly wrong. >> You have to move the leading car sideways as well as forward. > >You missed that the motion of the target mirror is circular >hence there is no increase of distance in the source frame. >There is an increase in the table frame of course. Yes. I am reverting to the table frame now. We should be looking at travel times around the loop rather than distance traveled. >> I should imagine ring gyros are calibrated empirically anyway. > >The maximum range of calibration would be orders of >magnitude smaller though, there is no way you can get >away with such a large error. the 'error' is about 65%, quite easy to live with. It isn't really an error. >>>The distance is to the point of reflection at the instant >>>the light is emitted, the consequence of the motion of the >>>mirror is included in using closing speed so if you adjust >>>the distance too, you double count its effect. >> >> No your car example was wrong. Try again. > >The motion of the mirror in the source frame is at a fixed >radius so the distance doesn't change. your car example was still wrong. > >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 6 Nov 2005 17:45
On 5 Nov 2005 14:55:10 -0800, "Eric Gisse" <jowr.pi(a)gmail.com> wrote: > >Henri Wilson wrote: >> On 3 Nov 2005 16:01:58 -0800, "Eric Gisse" <jowr.pi(a)gmail.com> wrote: >> >> > >> Geese, you are totally confused with regard to my theories. > >That is what happens when you refuse to talk about the details of your >theory. > >I am flexible, it doesn't have to be c+v - even though that is the only >assumption you have given me. You can derive it any way you want, as >long as you derive it using your theory. > >If I say "there is no such thing as a reverse field bubble in the >'BaTh' theory", you have no way of proving me wrong. geese, the WRFB is not related to the BaTh. > >If you say "there is no length contraction in special relativity", I >can easily show you to be incorrect because I have a theory that is >mathematical in nature and as such makes concrete predictions that are >unable to be influenced by anyone's sophistry. > >> >> >> >If you have something to say that hasn't been said before, have at it. I >> >> >think you are just plowing soil that has already been over plowed, over >> >> >planted and over grazed. Watch out for the gnomes. >> >> >> >> Stop trolling . >> >> >> >> Learn something. >> > >> >When are you going to learn SR so you stop making so many basic >> >mistakes? >> >> Ther is nothing much to learn about complete bullshit. > >You spend an amazing amount of time talking about the details of a >"bullshit" theory. > >Just because you think it is bullshit does not mean either that it is >or that you can say whatever you want about it and have it be true. > >> >> >> >> >http://newton.ex.ac.uk/aip/physnews.211.html >> >> >[quote] These jets appear to be traveling at greater- than-light speeds. >> >> >This is actually an optical illusion owing to the alignment of the object >> >> >relative to us,.... [unquote] >> >> >http://www.npl.washington.edu/AV/altvw71.html >> >> >> >> They ARE traveling at >c wrt Earth. >> >> Don't believe the nonsense relativists dish out. >> > >> >This problem is an exercise in both MTW and "Spacetime and Geometry". >> > >> >Just because you don't understand it does not mean it is nonsense. >> >> It's bullshit propaganda. >> ...part of the brainwashing process. > >An utterly irrational and emotional response to something you do not >understand. > >> >> If you don't give the 'right' answer according to Einsteiniana, you fail your >> course. > >It is mathematics, not sophistry. > >Just because you are incapable of learning a mathematical theory and >applying it correctly does not mean either that it is bullshit or that >the people who have learned it are "brainwashed". They are. They end up incapable of using their own brains. >> >> >Explain it to me, I don't know it well at all. >> >> >> >> study the physiology and psychology of the sensory system. >> > >> >Why are you incapable of answering a direct question? >> >> It would take me a few days. > >Then don't give examples you are incapable of explaining. Poor boy.... 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". |