Ballistic Theory, Progress report...Suitable for 5yo Kids
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From: Paul B. Andersen on 15 Nov 2005 16:17 Paul B. Andersen wrote: > Henri Wilson wrote: > >> According to the BaTh, the path length difference between both beams is >> dependent solely on rotation speed. > > > Of course. > How could it be otherwise? > All theories agree on that. > >> Therefore fringe displacement remains constant at constant angular >> velocity. > > > Does not follow. Sorry, that DOES follow. What does NOT follow is that the displacement depend on the rotation, because: > The BaT predicts that the wave going the longest path > will go faster than the wave going the shorter path, > so the phase difference will remain the same. So according to the BaT, there will be no displacement. But there is. The BaT is falsified. Paul
From: George Dishman on 15 Nov 2005 17:13 "Henri Wilson" <HW@..> wrote in message news:djuhn1l5gcdaijqlsiahpqfhp8f5tcd6fu(a)4ax.com... > On 14 Nov 2005 05:19:05 -0800, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > >>Henri Wilson wrote: >>> On Sun, 13 Nov 2005 11:52:57 -0000, "George Dishman" >>> <george(a)briar.demon.co.uk> >>> wrote: >>> >>> > >>> >"Henri Wilson" <HW@..> wrote in message >>> >news:omncn19cle09dml5jjtgdc7ib6bcsuvh0l(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. >>> >>> George, a few months ago, you (or maybe it was Paul) went to great >>> trouble >>> to explain to me that fringes shift only during angular acceleration. >>> Gyros >>> indicate total rotation by continuously integrating the rate of fringe >>> shift >>> with time. >>> >>> Are you now saying this is wrong? >> >>I am sticking to what we both agreed. There had >>been some talk earlier but we revised this around >>the 11th October. See your message, ID: >> >> a4cok15ekbocc95d6ahg68foum9hm082hn(a)4ax.com >> >>[Henri wrote:] >>>[George wrote:] >>>>[Henri wrote:] >>>>> George, if a sagnac is rotating at constant angular speed, Do the >>>>> fringes move continuously or remain steady but offset? >>> >>>>The latter, steady but with an offset proportional >>>>to the speed of rotation. >>> >>> (Yes I thought that's what came up before. Just checking). >> >>The formula for the Sagnac effect is: >> >> delta_t = 4Aw/(c^2-(wR)^2) >> >>where where w is the angular speed. For wR << c >>that simplifies to: >> >> delta_t = 4Aw / c^2 >> >>so the output is proportional to the angular speed, >>not the acceleration. > > That's correct. > But the fringes only move during acceleration. Right. >>> >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. >>> >>> When acceleration occurs, path lengths vary. >>> The number of wavelengths in each path changes. >>> Fringes MOVE. >>> >>> It matters not whether the beam moves at c or c+v. >>> >>> During constant rotation, including zero, there is no fringe movement. >> >>>From page 6 of the comprehensive reference Jerry >>posted: >> >> "Sagnac's polygonal interferometer was mounted >> on a turntable. It had an area of 0.0860 m^2, a >> rotation rate of order 2 Hz, and the resulting >> fractional fringe shift 0.07 +/- 0.01." >> >>> >> Actually, you have shown that the BaTh does what it should do. >>> >> It expects NO fringe shifts under constant rotation. >> >>Exactly. >> >>> >> 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. >>> >>> The output is given in degrees rotation from zero angle. That is >>> calculated via >>> a time integral during acceleration. >> >>Angle turned is the integral of angular speed. There is >>a single integration to get the heading from the raw >>signal which is proportional to speed, not acceleration. > > Yes George. The fringe shift is proportional to speed BUT IT OCCURS ONLY > DURING > ACCELERATION. That's wrong but from other messages I think it was more a question of wording, there is a shift (displacement) at all speeds but the displacement only changes during acceleration. Barring that confusion between shift as a noun and a verb, I believe we are in agreement so far. >>> I think you and your colleagues should learn a few facts. >> >>Facts: >> >>1) Ritz predicts no fringe shift for constant angular >> speed. >> >>2) Sagnac measured a fringe shift of 0.07 at constant >> rate of 2Hz (120rpm) compared to non-rotating. >> >>3) The experiment measures the speed of light from the >> moving source and it turns out to be unchanged from >> that in the non-rotating situation. >> >>George > > Sorry George, you have it all wrong. Well so far you have not said anything that differs from my view. > The fringes DO NOT move during constant rotation. They are displaced by a > constant amount. That is correct. Item 2 above says the same, the amount the fringes have moved, or shifted or been displaced, however you want to put it, was constant at 7% of the fringe spacing in Sagnac's experiment at 120rpm. > You said this yourself on for instance,12th October. > > """"" >>> George, if a sagnac is rotating at constant angular speed, Do the >>> fringes >>> move continuously or remain steady but offset? >> >>The latter, steady but with an offset proportional >>to the speed of rotation." > """"" Yep, and you seem to agree with that above. As I say, no difference in views so far. <I'll swap the next two sentences of your reply.> > The amount of shift signifies a rotation rate. Integrate that over very > short > time intervals and you have a fairly accurate measurement of the total > angle of > rotation from zero. Yep, again that is entirely correct and exactly what I said earlier, still no difference in views. However ... > The displacement arises from the path length change that occurs DURING > ACCELERATION. You need to explain that Henri. If the displacement occurs only during acceleration then there should be no displacement whenrunning at constant speed, but you just agreed there was a constant (non-zero) displacement at constant speed. > The number of 'wavelengths' in each path changes ONLY during > acceleration. I'm less concerned about your ideas on the mechanism at this stage, we might get to that later. > You should be able to see now, (from the duck shoot experiment) that > virtually > the same path length difference occurs whether you use c or c + v.....it > differs by only c/(c+v) > > With a FoG, the path length is greatly increased for higher accuracy. > > So Sagnac DOES NOT refute the BaTh.
From: Paul B. Andersen on 15 Nov 2005 17:07 Henri Wilson wrote: > On Sun, 13 Nov 2005 15:48:41 +0100, "Paul B. Andersen" > <paul.b.andersen(a)hiadeletethis.no> wrote: > > >>Henri Wilson wrote: >> >>>Wavefronts really exist only in the source frame. >> >>Quite. >>When I go out in my boat, the wavefronts disappear >>as soon as I start moving. Whatever is hitting my boat so hard >>are not wavefronts, they are only infinitesimal points. >> >>Paul > > > A fine laser beam is not like the ocean surface Paul. The light from a laser and a water surface wave have that in common that they both have wave fronts which are physical entities that exist independent of frames of reference. It is incredible stupid to claim that a physical entity - any physical entity - exists in one frame of reference but not in another. And what is even more remarkable is that I have to point out to you that the statement: "Wavefronts really exist only in the source frame." is mindless babble. It should be self evident to any sane person. > However, in case you are correct, I shall look through my telescope to see if > star light is really coming to us via cosmic ships. If you from the cosmic ship Tellus ever look through a telescope, be sure to notice the stellar aberration. Having done so, you can kick and shout: WHATEVER IS COMING FROM THAT STAR, MOVING WITH CHANGING DIRECTION, IS NOT LIGHT, IT IS ONLY INFINITESIMAL POINTS. That will make you look exactly as intelligent as you are. Paul
From: Henri Wilson on 15 Nov 2005 17:22 On 15 Nov 2005 06:55:28 -0800, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >Henri Wilson wrote: >> On Sun, 13 Nov 2005 11:44:10 -0000, "George Dishman" <george(a)briar.demon.co.uk> >> wrote: >> >> > >> >"Henri Wilson" <HW@..> wrote in message >> >news:0mlcn197uu5d7383ug28ct9oav3t2ij5kq(a)4ax.com... >> >> On Sat, 12 Nov 2005 15:12:11 -0000, "George Dishman" >> >> >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: >> >> You are still drawing only a few of them. > >No, they are all there. Perhaps we have a different >understanding of 'wavecrest' or 'wavefront' but it >seems unlikely. I'll add a plot of field versus >distance to the next bit: > >> > \ >> > - ) >> > / >> > ( >> > \ >> > - ) >> > | | >> > |-| >> > | | > >Does that help at all? Each hyphen is a cross >section of a plane of maximum voltage. clever drawing... but the 'hyphens' are really just points on a graph. > >> >>>> 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. >> >> It would fool some people. >> >> How about using spinning spheres eh George.... > >Better. > >> Does their spin axis lean over diagonally? Of course not. > >Right, that is why polarisation is affected. but there is no continuous diagonal beam which would enable you to check that. > >> >> 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. >> >> George, there are NO diagonal ants. > >The ants move diagonally in the moving frame and they are >still ants. The ants bodies remain vertical. The vertical axis of their bodies moves sideways. Did you try moving a vertically held pen sideways George, like I suggested? Even if you move it upwards at the same time, the axis of the pen remains vertical at any instant. That should be obvious by now. > >> >> 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. >> >> That's his unproven postulate. > >That reply makes no sense in the context Henri. >Deriving time wehen speed and distance are >known is not a postulate. Diagonal speed is sqrt(c^2+v^2) >> >>>As you said of Maxwell's >> >>>Equations, "a solution involves a wave moving at c" >> >>>and the magnetic fields still exist and are still >> >>>governed by Maxwell's Equations in the moving frame. >> >> >> >> Maxwell's equation applies to a wave that is symmetrical around an axis. >> > >> >No, Maxwell's Equations apply to the interactions of >> >electric and magnetic fields regardless of symmetry. >> >You can define a set of boundary conditions and they >> >will tell you how the fields evolve thereafter. For >> >example the waves produced by applying a sine wave >> >voltage to a metal sphere will differ from those >> >produced by a flat plate or a long wire. Maxwell's >> >Equations apply to the fields regardless of the shape >> >or motion of the source. >> >> That's rubbish George. > >I suggest you open a textbook and find out ho Maxwell's >equations are used. George, an infinitesimal point cannot contain any wave of the Maxwellian type. > >> Both waves are symmetrical about an axis which also defines the direction of >> movement. >> The whole concept depend on that movement. >> >> >For our purposes, just note that they apply equally >> >well in both frames. >> >> They don't. There is no wave moving diagonally. there is just a dimesnionless >> point. > >The ant is still an ant. and its axis remains vertical. There are no diagonal ants crawling diagonally.. > >> >> You 'diagonal field' is skewed. >> > >> >Yes, and Maxwell's Equations must still apply or >> >they would be invalid. >> >> they don't apply to any skewed wave. > >Right, but they do apply to electromagnetic fields so >how are you going to resolve that ;-) George, an infinitesimal point cannot contain any wave of the Maxwellian type. > >> If they did, the speed of the wave would be sqrt(c^2+v^2) > >Nope, the speed is the product of the constants in the >equations. George, an infinitesimal point cannot contain any wave of the Maxwellian type. > >> >> 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. >> >> There is only one wave. It is vertical in the source frame and it is vertical >> in the moving frame but moving sideways. >> >> George, hold a pen vertically. Now move you hand sideways. > >It draws a straight line. > >> Does the pen lean over? > >I don't care, now repeat the sideways motion but this >time also move it away from you at the same time. It >moves diagonally and for the same amount of sideways >motion the line is longer than the first time, that fact >is all that Einstein uses. The line isn't a pen, but then >it wasn't the pen the first time either so your stuff >about "the light isn't light anymore" is just >meaningless nonsense. George, hold the pen vertical then move it sideways and upwards. That's what Einstein tried to do.... but he stuffed up because he used spherical raindrops instead of something with an identifiable 'axis'. > >> >> ...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. >> >> What do you think the green dashes represent in my program? > >You said they were just a way of distinguishing the point >representing the wavefront form the diagonal line so they >don't mean anything. The vertical dashed COULD signify individual photons....which have some kind of longitudinal axis about which the maxwellian wave is symmetrical. ..Isn't that called the Poynting Vector? > >> >>>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. >> >> Well it certainly has NOT been measured. > >Of course the speed of light has been measured Henri, >what are you raving about. The speed of the diagonally moving indinfitesimal points has never been measured George. >> >>>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. >> >> There is NO wave moving along any diagonal. > >The ant is still an ant. a vertical ant....in all horiziontally moving frames, at any instant. That's the point you are missing.,,the 'instant' bit. If you take a movie shot of ants crawling up a pole while you flash past in your car, do their bodies appear vertical or diagonal on each frame? >> >> Physical processes are not observer dependent. > >Right, but some values are, so while Maxwell's Equations >must apply in all frames but the specific values of electric >and magnetic field might differ from frame to frame. George, an infinitesimal point cannot contain any wave of the Maxwellian type. >> >> Do the pen experiment again and you will see that vertical objects can move >> sideways without leaning over. > >Einstein didn't say it wouldn't affect the polarisation of the >light Henri, that's what your lean represents. Nobody has tested the polarization of a sideways moving, vertical light beam. There is no actual diagonal beam to test it on. >> >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). >> >> The points that 'move diagonally' are not physical entities at all. > >Points aren't, no. Not in either the laser or moving >frames, but the light whose location is represented >by those points is still light, the ants are still ants. and they are still vertical at any instant in all frames. > >> >>>> It obviously moves at sqrt(u^2+v^2) >> >> You know there has never been such a measurement, George. >> > >> >The speed of light has been measured many times >> >Henri, I have no idea why you are denying that. >> >> HoHohahaha! >> >> George, TWLS has been measured and found to be consistent and precise. > >Exactly what I have been saying throughout. If you are >now admitting I was right, I fail to see why you keep >claiming it has never been measured. George, what you are refering to is not light and its speed has certainly never been measured, either OW or TW.. >> >It wasn't, it was based on Maxwell's Equations, the speed >> >is the product of two measured constants, remember? >> >> Then why should differently moving observers produce the same value for c, via >> Maxwell, > >Because the constants are, ... well ..., constant? > >> when any light beam will approach them at different speeds? >> ...you seem very confused. > >Which part of "constant" escapes you? None. The fact that both observers calculate the same value for c from the constants escapes me...since the beam DOES NOT approach them at the same speed. > >> >> Anything that has zero size and no properties must be 'nothing'. >> > >> >But photons don't have "no properties" so your argument >> >fails. >> >> Well please give us a physical description of their properties. > >Energy and momentum for example. Been here before? Energy and momentum are properties of energy and momentum, not of photons. The kinetic energy of a car is not a property of the car. It is a contribution to the total ENERGY of the frame in which you measure it. >> >> Define photon properties George. > >Energy and momentum for example. Been here before? see above >> >> well it has never been adequately tested. >> > >> >In your opinion. >> >> George, if a vertical laser beam is moved sideways, there is no way to measure >> any part of it that is moving diagonally. >> >> Any detector positioned to sense the diagonally moving element would merely >> detect an infinitesimal blip. > >Two detectors on the diagonal path a known distance apart >each with a partly silvered mirror and photodetector. Use >the photodectors to measure the time the light takes hence >speed. What's your problem? It wouldn't get past the first one because it is infinitesimal. >> I suppose an approximate answer could be obtained using adjacent diagonals of >> finite size. But I can tell you the answer now. It is sqrt(c^2+v^2). > >That's not what is measured. It's the only way it could be done. >> >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. >> >> Clocks at rest can be absolutely synched using Einstein's method. >> Just make tAB=tBA. > >Fine, use that for the test above. it wont work. infinitesimal points involve infinitesimal energy. >> >Your short lines moving up the screen are vertical >> >while the wavefronts should be horizontal. >> >> Oh for christ's sake George, there is no continuous wave moving up each >> infinitesimally wide diagonal. Can't you see that? > >The wavefront is roughly the same width as the laser >aperture. Your concept of describing that as an infinite >number of adjacent infinitesimal parts seems to be >confusing you. Consider the laser beam to be infinitesimal in width....or at least much smaller than a wavelength of the light used. >> >> I did...and it makes no sense at all. Infinitesimal points on a graph do not >> constitute 'wavefronts'. > >The points on the graph aren't infinitesimal, they are purely >mathematical points of zero size. They represent the "very >very small pieces" of the actual wavefront, or isn't that what >you mean, it is what you have been saying. Not 'very small' but 'infinitesimal'. Even if the laser beam is wide, only an infinitesimally thin section will move up any diagonal line....far too thin to be a light wave. >> >> If the green beam was diagonal, the detector would pick up a continuous signal. > >And you would then have to modulate it to measure its >speed, the simplest way being to switch is on for a very >short time and measure the flash. This is becoming quite ridiculous. >> >> I suppose I can show dots moving up the green beam in the lower diagram. >> I will do that. > >Cool, that would be a major improvement. I have done that, have a look. >> >> I can see those answer to that problem. The fields a skewed in all frames other >> than that of the source. they don't move at c. > >Then you have to throw Maxwell's Equations in the bin. >That was what was worrying them. Maxwell's equations apply to waves moving in one direction. The waves are symmetrical only in that direction. > >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 15 Nov 2005 17:37
On Tue, 15 Nov 2005 15:00:06 GMT, The Ghost In The Machine <ewill(a)sirius.tg00suus7038.net> wrote: >In sci.physics, donstockbauer(a)hotmail.com ><donstockbauer(a)hotmail.com> > wrote >on 15 Nov 2005 04:49:09 -0800 ><1132058949.576411.92810(a)f14g2000cwb.googlegroups.com>: >>> Not a ghost of a chance of understanding sticks in water are not really >>> bent just because they look bent, I suppose? >>> Didn't think so. >> >> ****************** >> >> They are bent if you stick an actually bent stick in the water. >> > >He is referring to the common illusion that when a whole >straight stick stuck into a tank of water, when viewed >from above the surface of the water, the stick appears >bent, if not broken, into two parts. > >This is similar to, apparently, his view that we are >looking at an illusion of the Universe, which only appears >to work as though GTR is true. In this view, c'=c+v, which >is after all consistent with all Earthly observations >at low v, holds sway. Stars such as Aql1493 appear to >confirm this view. (Other stars confirm GR.) > >Why he and H. Wilson disagree is not clear to me; both hold >this general view, though Black Knight suggests he wants to >be neutral and hold no preferred theory, while suggesting >(if I'm reading him correctly) that we are all holding a >massively corrupt variant of physics theory by using SR/GTR >in such things as particle acceleration, GPS calibration, >and astronomical observation. The first two of course >require large sums of money; the last isn't quite as >expensive and the average Joe can easily procure a >telescope -- and wouldn't have too much trouble procuring >a water jug, some electronics, and a photodetector as well, >for measurement of incoming muons, superluminal or otherwise. >Another possibility is energizing the plates of an old >airgap capacitor until it sparks on occasion -- a primitive >Geiger counter. > >This view is also consistent with MMX, which was only >really designed to show the absolute velocity in a rigid >luminiferous aether, where light therein travels at c >relative to the aether. MMX cannot tell the difference >between > > - lightspeed is c relative to everyone regardless of motion > (SR/GTR) > - lightspeed is c relative to everyone if motion is confined > to non-accelerated motion only > - lightspeed is c relative to the source only (Galilean relativity) > - lightspeed is always vertical relative to its propagation > and therefore does something rather peculiar (Seto) > >though in the last case one might ask whether MMX will show >anything at all, in light of the motion of gas molecules in >the light source. > >In any event, he appears to follow in the footsteps of >Sekerin, and will also suggest superluminal muons, which is >perfectly logical in this view though as yet unconfirmed >by Earthly experiments. However, H. Wilson has a simple >method around this: particles in such experiments are >pushed around by EM radiation and therefore can't travel >faster than light when pushed thereby. > >The main question here would be whether one can tell >the diff between two Earth-accelerated 500 MeV protons >colliding and two Earth-accelerated 7 TeV protons >colliding. Regrettably, I have no data on either one. >If one *can* tell the difference the EM "particle energy >speed limit" is effectively ruined. One can also work >around this using various target experiments -- e.g., throw >subluminal protons at a wax target and measure the speed of >electrons knocked out thereby, somewhat a la Rutherford. >The curvature of the path in a cloud chamber should be >more than sufficient, and one might not even need the wax; >a proton could collide with a water molecule instead. > >A quick Google coughed up this reference of an experiment >apparently done in 1935, and published in 1935-03-18: > >http://prola.aps.org/abstract/PR/v47/i7/p573_1 >DOI: 10.1103/PhysRev.47.573 >URL: http://link.aps.org/abstract/PR/v47/p573 >Collisions of High Energy Protons in Hydrogen >by Milton G. White, Department of Physics, University of California > >Not horribly surprising, really; it's an obvious experiment. >However, I can't get at the results -- they require subscription >to the American Physical Society journal. > >It is not clear whether the Black Knight shares this workaround, >or not. > >Particles such as Oh My God, which had 51 J of energy >(for a proton, muon, or pi meson that's a whallop!) suggest >that the Universe isn't limited by the "EM particle energy >speed limit". Such particles are rare but are easily >measured in detectors such as SuperKamiokande, which is >basically a very large tank of ultrapure water surrounded >by light detectors. > >SR has no "EM energy limit", of course; all particles simply >cannot travel faster than light, no matter how much energy >is thrown at the problem. I actually read all that Ghost. I don't know why! Here's a thought. Take a thick block of very clear glass (or prefereably crystal) cool it to 100K degrees or less and shine a beam of light through it . According to your friends, light speed remains c between atoms but takes time to be absorbed and re-emitted by each one. This gives the impression that light is being slowed... and hence 'refractive index'. Now heat the crystal to as fairly high temprerature. One would expects the thermal atomic speeds to disperse the beam by a considerable amount. On might expect strange doppler effects too. Do you know if this has been tried? .. 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". |