GETTING RID OF EINSTEIN RELATIVITY
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From: Henri Wilson on 31 Aug 2007 19:30 On Fri, 31 Aug 2007 11:40:27 +0100, "George Dishman" <george(a)briar.demon.co.uk> wrote: > >"Henri Wilson" <HW@....> wrote in message >> That was true before the advent of computers. Time you brought yourself up >> to >> date george. > >The computer will do the calculation, you still >need the equations. For example there are programs >as part of PCB design suites that will predict the >EMI for a board, you can get the router to place >tracks to minimise crosstalk or have specific >impewdances and so on, but they all work by solving >Maxwell's Equations and without those the prgorams >couldn't be written. George, as an example, I produce elliptical orbits without using Kepler's equations. The program actually generates those equations from basic Newton then uses them to poduce graphed results.. Equations are not much good on their own. >>>>>Wrong, SR predicts the difference in speed should >>>>>be c+/-v so it is compatible with SR. Ballistic >>>>>theory predicts no fringe shift so is falsified. >>>> >>>> Ah! You finally accept that the speeds are c+/-v.... >>> >>>No, the "difference in speed" is what I said. >> >> That's what I thought you said. >> You finally accept the fact. > >I have always accepted the fact that you >didn't know the difference between the >algebraic difference between speeds in one >frame (closing speed as it is sometimes >called) and the speed in another, or at least >that you pretend to not know to facilitate >your silly word joke. See my previous post... >>>It doesn't, it relies on the difference being c >>>and that isn't the value measured by Sagnac, >>>your theory is measured to be wrong. >> >> Your a losing it George. Definitely time for a holiday.... > >More petty attempts at abuse Henry, you always >make it easy to see where you know I am right, >ballistic theory depends on the 'closing speed' >being c, not c+v and you know that. Completely wrong George. ..no wonder you're confused about the finer points of BaTh... Have a good holiday George... >>>> That's how physicists try to model mechanical behavior. >>> >>>That's how 19th century physicists worked. Modern >>>physicists know it is a valid approximation for >>>macroscopic behaviour but most cutting edge work >>>is pushing the quantum level. >> >> ..and getting nowhere... > >Other than the invention of the transistor and >everything that came from, lasers, laser and >fibre gyros, and on and on. The whole of modern >technology has come from relativity and QM. Transistors hardly came from QM George...and certainly relativity has nothing whatsoever to do with them.... >>>> George, PHOTONS are described by FIELDS, a la Maxwell. >>> >>>Photons are not dsescribed by fields, they are >>>described by Schroedinger's Wave Equation and >>>quantum mechanics. >> >> Schroedinger doesn't describe an individual photon. > >I suggest you have a look at this though the >topic is beyond the scope of our conversation: > > http://galileo.phys.virginia.edu/classes/252/wave_equations.html > >Hit "next" a couple of times to here > > http://galileo.phys.virginia.edu/classes/252/FiniteSquareWell/FiniteSquareWell.htmland look at the graphic about two thirds down.Think of the potential curve as the cross-sectionof an optical fibre. Does the tail remind you ofthe evanescent wave?>>> How can FIELDS be>>> described by PHOTONS?>>>> E = rP/q>> Where's the physics?You snipped it earlier:>>>>... The>>>>definition of the electric field for example>>>>is:>>>>>>>> E = f / q>>>>>>>>and the definition of force is:>>>>>>>> f = dp/dt>>>>>>>>hence>>>>>>>> E = 1/q * dp/dt>>>>>>>>so if a stream of similar photons each carrying>>>>momentum P hit charge q at a mean rate of r>>>>photons per second, the field is:>>>>>>>> E = rP/qThat is the physics. Did you mean "where isthe philosophical interpretation?"? What thephysics (i.e. the equation) shows is that thefield E is a name given to the mean effect ofa flux of photons, it has no other physicalinterpretation.>>> Admit that nobody has a clue....except for me of course.>>>>ROFL, Henry >you are only slightly less clueless>>than the other cranks around here, you cannot>>even cope with calculus never mind physics,>>and you only think you understand that because>>you confuse it with philosophy.>> George I probalby know a lot more about maths than you do.ROFL, get a life Henry, you couldn't evenwork out the ADoppler equation, I knew morethen you do now when I was 13.>>> Physicists can be engineers but engineers can't be physicists.>>>>Lucky I got my degree first then.>> All of mine were largely physics. I tended towards engineering later."All" of your are a faked photo of somebody elses.>>Nonsense, as I have told you before, look at how>>the slope of a pulsar's dispersion can be used to>>determine the electron column density. It will>>tell you more than you think about speeds.>> ..and it certainly backs up my unification idea.Yep, it shows SR is correct.> It also shows why DeSitter's 'disproof' of the BaTh was wrong.Nope, there is no unification in Ritz's theoryto which >De Sitter's disproof was valid.>>> This is>>> not true and shows how primative and ignorant most humans are...even the>>> scientists.>>>>>> When relative light speed is included, as it should be, the wholepicture>>> changes.>>>>Sure, you get predictions of multiple images of>>binaries and then have to invoke speed changing>>fairies to give yourself an excuse for why we>>don't see that phenomenon.>> The conditions required for multiple imagery to occur are never reached> .....rarely even approached.Never even approached, hence ADoppler is alwaysundetectable. Reasonable really since it doesn'texist.>>>>> Willusions don't occur in the lab...in fact they are virtually>>>>> negligible>>>>> o>>>>>> George, theories are usually considered 'right' if they come up withthe>>>>> previously known answer....even if they are completely wrong.>>>>>>>>If they give the right answer, they are "right" by>>>>definition.>>>>>> George, I can easily 'fiddle' my above statistical diffraction theory to>>> >match>>> the QM results, by introducing a few other likely factors. Would thatmake>>> it>>> correct in your mind?>>>>Nope, like your program you would prove you had>>nothing more than a curve fitting program.>> That's your last ditch argument....It's the first ditch, and valid you proved itby proving that the theme from Close Encounterswas a Keplerian orbit ;-)> when all else fails make stupid statements> like that one. My program can only produce a very stringent range ofcurves.> It so happens that most star curve fit into this range...And flutes (or whatever the synth was simulating).> Coincidence George?No, an indication of too many free parameters.>>The way to do it is to take the measured>>temperature curve and _calculate_ the luminosity>>variation that it produces using Planck's Law>>and then you have no factors to fiddle. That is>>physics, not your string of fudge factors.>> Move a planck curve sideway 1% and see what it does to the V band.0.01% at most Henry, usually less, >and the answeris ... no more than 0.01% or less than 0.15K for atemperature of 6000K (remember the Stefan-BoltzmannLaw), completely irrelevant.George www.users.bigpond.com/hewn/index.htm The difference between a preacher and a used car salesman is that the latter at least has a product to sell.
From: YBM on 31 Aug 2007 20:09 Henri Wilson a �crit : > On Fri, 31 Aug 2007 11:40:27 +0100, "George Dishman" <george(a)briar.demon.co.uk> > wrote: > >> "Henri Wilson" <HW@....> wrote in message > >>> That was true before the advent of computers. Time you brought yourself up >>> to >>> date george. >> The computer will do the calculation, you still >> need the equations. For example there are programs >> as part of PCB design suites that will predict the >> EMI for a board, you can get the router to place >> tracks to minimise crosstalk or have specific >> impewdances and so on, but they all work by solving >> Maxwell's Equations and without those the prgorams >> couldn't be written. > > George, as an example, I produce elliptical orbits without using Kepler's > equations. > The program actually generates those equations from basic Newton then uses them > to poduce graphed results.. > Equations are not much good on their own. ;-))))))))))))))))))) Gosh ! How can you be soooo stupid Henri ?
From: T.M. Sommers on 6 Sep 2007 00:52 T.M. Sommers wrote: > Henri Wilson wrote: > >> ADoppler DOES exist. George merely wants it to go quietly away so his >> faith >> wont be threatened. I'll try to explain briefly. >> >> According to BaTh, (ballistic theory) > > Wasn't that theory shown to be false about 95 years ago? > >> ADoppler occurs like this. >> As light is emitted by an orbiting star, it moves at c wrt the star and >> c+v.sin(ft) wrt planet Earth. Over time, the fast photons catch up >> with the >> slower ones and vice versa causing a bunching or separation pattern >> across >> space. > > I assume f is the frequency of the light. t is presumably a time, but > what time? It appears possible to change the results just by changing > the zero of time. v is presumably the orbital speed. In what frame is > it measured? Is it the total linear speed, or the component in the > direction of the observer? > > How does your theory account for the period-luminosity relationship for > Cepheids? > > The A in ADoppler evidently means acceleration, but there is no > acceleration dependence in your equation. So why don't all stars look > like Cepheids? > >> The process does not continue forever because the speed of all light >> moving in a particular direction tends to become unified. > > That sounds fishy. What accounts for it? > >> The arrival of the >> 'photon density' pattern over time causes the star to APPEAR to vary in >> brightness when in fact it is perfecty stable. > > You seem to be saying that Cepheids are not really variable, but appear > variable because they are part of binary systems. So why don't all > binaries look like Cepheids? Why don't opposite spiral arms of galaxies > look different? Why don't the stars in close orbits around the black > hole at the center of our galaxy look like Cepheids? > >> Since the 'ends' of INDIVIDUAL photons are also emitted by the >> accelerating >> source, it was pointed out by George that these should also behave >> like the >> macroscopic bunching that occurs BETWEEN photons. This is just the >> classical >> wave theory, in which individual photons are regarded as being just small >> snippets of the whole wave. That is possible...and it answered a big >> question >> about the phase differences between the observed brightness and >> velocity curves >> of Cepheids. However, it was obvious that the photon 'length changes' >> could not >> be of the same order as the observed brightness variations. >> My theory says that photons DO contract and extend due to this source >> acceleration and these effects are roughly in phase with the macroscopic >> bunching...... BUT the changes are reduced markedly by a factor 'K', >> which may >> be of the order 10^-5. > > That sounds like a fudge factor to me. What is the theoretical > justification for K? > >> This appears logical because, whilst there appears to be >> NO restriction on the way aligned photons might move wrt each other, >> photons >> are likely to resist 'compression ' and extension like an conventional >> elastic >> material. > > Are you saying that photons are not points but have extent and some > internal structure, or are you just talking about wavelength? Are you going to answer these questions, Henri? -- Thomas M. Sommers -- tms(a)nj.net -- AB2SB
From: sean on 6 Sep 2007 10:33 On 30 Aug, 23:55, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: > sean <jaymose...(a)hotmail.com> wrote in news:1188512224.511353.237820 > @i13g2000prf.googlegroups.com: > > > If you bothered analysing your sean planets sim, youd > > see that its the only way to have light propagating away > > from any source at c. > > One of the most fundamental laws of physics is that an object in motion > continues in that motion unless acted upon by an outside force. You ignore several things here, First of all if emmision theory predicts that light always is at c relative to a source then it would be impossible for light to be effected by gravity under these terms. Secondly you ignore what I say repeatedly in the post youve just replied to. That emmision theory cannot have light as a particle with mass. You ignore the parameter that in emmision theory light isnt a particle and therefore has to be distinct from mass. Not least because its also clear classicl theory says light is a wave only. And finally you ignore the fact that SR whether it admits this or not has light as not obeying gravitational pull sometimes. Look at the classic lorentz transformation.. What does it do to explain SR ?? It adds or subtracts velocity when it puts the lambda (sometimes called gamma) symbol in the formula. This is an example of where SR has light breaking Newtons laws. So if SR can have light special. Why cant emmision theory say light is special in that it isnt a particle but a wave in a non particulate medium..? You are a hypocrite > The light emitted consists of 'objects'. Once emitted, they must travel in > straight lines at a constant velocity unless acted upon by an outside > force. > > What force would 'tie' the light, once emitted, to the source so that it > must follow the source? > > There is no evidence of any such 'tie' between light, once emitted, and its > source. You jump to conclusions here. What "evidence" do you have that light doesnt travel at c in any direction from a source? My evidence that they do is MMx. Notice here that the MMx source is rotating around the earths axis yet we also know that the light is travelling at c on both paths. This would not be possible if light travelled away from a rotating source at different speeds in different directions as you erroneously suggest.above > In a cyclotron, particles traveling around a circular path emit beams of > xrays. > Those xrays travel in straight lines from the point of emission. They do > NOT follow, in any way, the subsequent motion of the emitting particle. Two things here.. Firstly,.. whats your proof these xrays travel in straight lines?? You have none. Show me the part of an experimental setup that checks the direction relative to the known source and the speed of the xray. You have neither nor do both tests exist. This is a fabrication on your part. Secondly,.. you make the assumption that the xrays are produced by moving particles/ie ...rotating sources like the MMx or sagnac source. But what proof have you that the sources in a cyclotron is rotating?? NONE You can only assume the xray `sources` are rotating theoretically. IM not familar enough with the technical setup of a cyclotron but my reference refers to the `particles` not as atoms but as electrons As far as Im aware electrons are emr. And emr in a wave only model does not consist of particles. In fact the existence of electrons is hypothetical and can easily be explained as electro magnetic waves travelling through a magnetic medium in emmision theory. In which case the xray source isnt rotating as its static particles in the cyclotron that are being energized by wave only electromagnetic radiation which in itself even isnt a particle in wave theory. In other words the xrays are generated by static particles exited by a emr current. To show otherwise youd have to prove that the so called roating particles in teh cyclotron were not electromagnetic. I think this would be impossible for you to prove. Like George you make up your evidence for almost all your arguments Lets see if for starters you can supply the lab setup in a cyclotron that measures the speed and dierection of the observed xrays.. You cant, because no such test was ever made. > >> > And one way to test this would be to put a small MMx setup > >> > onto a rotating table in a lab. And rotate that setup in > >> > a circle in the lab. If this was tried then either > >> > the small MMx setup wouldnt give a null result, in which case > >> > you could say that yes light doesnt travel at c relative to a source > >> > OR... > >> > The rotating MMx setup WOULD still give a null result. In which case > >> > I could say to you... light does always travel at c relative to a > >> > source. > >> > I bet the results would confirm my argument and refute yours. > > >> The result would be that the centrifugal force > >> would distort the arms and wreck the experiment, > >> you need to find a more practical method. > > Do you really think a setup like this > > rotating around a table at lets say once per 5 seconds would > > break apart??? > > He did NOT say 'break apart', he said 'distort'. > In the MMX, all it takes is moving part of the apparatus by a small > fraction of the wavelength of light in order to create a detectable > indication. > > THAT is why he said 'distort'. Spinning the apparatus would cause changes > in the instrument that would make it useless for the purpose you propose to > use it for. I assume by this that you and george imply that a rotating MM experiment physically stretches one arm moreso than another and doesnt give a null result? If so please supply a reference for such an observation, because I dont believe you have any proof to back up this claim.(notice that will not be able to supply this as none exists. Youve just fabricated this evidence) And secondly assuming the remote chance that it does distort such an apparatus and not allow us to check what I propose may be observed (ie whether or not a rotating MMx that isnt distorted does still have light at c on both paths)...THen how is it that you or george can say with certainty that a rotating MMx source does not allow light to propagate down both arms at c? After all, youve just claimed that any experiment to test this is impossible. Id also like to point out Georges failure to respond to a simple request I make ,.. that he supply or help simulate the path of an emmision wave in a sagnac setup. Notice how he cannot do such a thing as he knows that it will prove that a light path at c relative to a source in a moving sagnac setup will,if simulated correctly, show that an emmision model CAN explain the fringe shift detected in sagnac. Furthermore Id like to point out that an emmision model where the light travels at c in straight lines relative to a source does NOT allow light to travel in straight lines in a lab. THis is an erroneous assumption made by all relativistas when they try to argue that emmision theory cannnot model sagnac. The point here is that relativistas when trying to simulate a emmision version of sagnac,... INCORRECTLY show light travelling in straight lines in the lab frame. This is incorrect . In fact if one models emmision theory correctly for sagnac in the lab frame the light paths have to be curved. It is precisely because dogmatic relativistas incorrectly model emmision theory in straight lines that they arrive at the wrong conclusions.Were these morons to redo sagnac correctly for emmision theory they would have to model light in curved lines in the lab frame. If they did, like myself, they would find that mathematically,.. emmision theory predicts a fringe shift. Not the erroneous no fringe shift relativistas pretend to predict for emmision theory. See my correct sagnac simulations at... http://www.youtube.com/profile?user=jaymoseleygrb Sean www.gammarayburst.com
From: bz on 6 Sep 2007 12:33
sean <jaymoseley(a)hotmail.com> wrote in news:1189089230.764471.53320(a)22g2000hsm.googlegroups.com: > On 30 Aug, 23:55, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> sean <jaymose...(a)hotmail.com> wrote in news:1188512224.511353.237820 >> @i13g2000prf.googlegroups.com: >> >> > If you bothered analysing your sean planets sim, youd >> > see that its the only way to have light propagating away >> > from any source at c. >> >> One of the most fundamental laws of physics is that an object in motion >> continues in that motion unless acted upon by an outside force. > You ignore several things here, First of all if emmision theory > predicts that light always is at c relative to a source That only applies at the moment of emission. If the source changes its motion after emission, the light does not know or care. > then it would be impossible for light to be effected by gravity under > these terms. Once light has left the source, gravity, mirrors, lenses, etc can effect its motion. The 'leaves its source at c' only applies at the moment of emission and then only when measured with equipment that are co-moving with the source and hence in the same inertial frame of reference. If the source suddenly accelerates at right angles and goes flying off, into the night (you have just thrown your flashlight away) the light that has been already emitted from that flashlight does NOT suddenly change its motion in order to maintain a velocity of c wrt the source that it has left far behind. > Secondly you ignore what I say repeatedly in the post > youve just replied to. I ignore many things because I am trying to focus your attention on one small point. > That emmision theory cannot have light as a particle with mass. I don't care if light is a wave or a particle at this moment. It doesn't matter. The light, once emitted, goes its merry way. Unlike the teenager, leaving home for the first time, the light does not 'write home' and ask for more money. The light doesn't care what happens to the source after it has left home. [snipping the rest of the stuff that is unrelated to the point I am trying to make and snipping your insults. If you want to continue our discussion, you will avoid insults. I don't bother with those that lack enough self respect to treat others with respect.] ..... >> The light emitted consists of 'objects'. Once emitted, they must travel >> in straight lines at a constant velocity unless acted upon by an >> outside force. >> >> What force would 'tie' the light, once emitted, to the source so that >> it must follow the source? >> >> There is no evidence of any such 'tie' between light, once emitted, and >> its source. > You jump to conclusions here. What "evidence" do you have that light > doesnt > travel at c in any direction from a source? I made no such claim. It DOES travel at c in all directions from where the source was at the moment of emission. But once emitted, it doesn't care where the source goes afterwards. > My evidence that they do is MMx. Notice here that the MMx source is > rotating > around the earths axis yet we also know that the light is travelling > at c on > both paths. This would not be possible if light travelled away from a > rotating > source at different speeds in different directions as you erroneously > suggest.above You misread my suggestion. >> In a cyclotron, particles traveling around a circular path emit beams >> of xrays. >> Those xrays travel in straight lines from the point of emission. They >> do NOT follow, in any way, the subsequent motion of the emitting >> particle. > Two things here.. Firstly,.. whats your proof these xrays > travel in straight lines?? You have none. You put three lead plates with small holes in them along a straight line leading away from the source. The xrays only reach the detector when all three plates line up with the holes in a straight line. If the xrays were traveling along a curved path, the holes would need to line up along a curved path. > Show me the part > of an experimental setup that checks the direction relative to the > known > source and the speed of the xray. You have neither nor do both tests > exist. You have never worked with an xray beam line at a synchrotron. visit a facility such as http://www.camd.lsu.edu/ and you will see that they use the synchrotron to generate xray beams. BEAMS of xrays traveling in straight lines. If your idea about the emission following the movements of the source were correct, CAMD would not work as it does. > This is a fabrication on your part. I don't need to fabricate anything because I know what I am talking about. > Secondly,.. you make the assumption that the xrays are produced by > moving particles/ie ...rotating sources like the MMx or sagnac > source. But what proof have you that the sources in a cyclotron > is rotating?? NONE The xrays are produced when a beam of particles is forced to CHANGE directions. When this is done, they emit photons. This is known as cyclotron radiation. > You can only assume the xray `sources` are rotating theoretically. > IM not familar enough with the technical setup of a cyclotron I am. > but my reference refers to the `particles` not as atoms but as > electrons Electrons or positrons or protons or heavier particles can be accelerated. When the beams are bent(forced to change direction of motion) photons are emitted. > As far as Im aware electrons are emr. You are quite mistaken. The motion of electrons generates electro magnetic radiation but electrons are NOT electro magnetic radiation. But, all of this is beside the point. The point being that once emitted, light doesn't care what happens to the source. [snipping the rest of the side issues] ..... > Lets see if for starters you can supply the lab setup in a cyclotron > that measures the speed and dierection of the observed xrays.. > You cant, because no such test was ever made. Wrong. The direction is easily determined by three lead plates with holes in them. The speed is easily determined by measuring the time between the moment that the batch of particles gets forced to change direction of motion and the time the xray pulse hits a target along the beam line. Also, the beam line can contain two detectors, a known distance apart and the time for the pulse to travel between those detectors can be measured. ..... >> He did NOT say 'break apart', he said 'distort'. >> In the MMX, all it takes is moving part of the apparatus by a small >> fraction of the wavelength of light in order to create a detectable >> indication. >> >> THAT is why he said 'distort'. Spinning the apparatus would cause >> changes in the instrument that would make it useless for the purpose >> you propose to use it for. > I assume by this that you and george imply that a rotating MM > experiment > physically stretches one arm moreso than another and doesnt give a > null > result? If so please supply a reference for such an observation, > because > I dont believe you have any proof to back up this claim.(notice that > will not be able to supply this as none exists. Youve just fabricated > this evidence) If you ever spin anything (a car tire for example) you will notice that getting everything 'in dynamic balance' is NOT trivial. Any experimental apparatus consists of many different parts, each with their own mass, density and, in this case, distance from the point about which the apparatus is to be rotated. The MMX apparatus must maintain all dimensions within fractions of the wavelength of light. ANY slight imbalance or difference in the distances from the center would produce different torques and invalidate the experiment. Have you ever seen an MMX apparatus? And one final point. The MMX apparatus IS spinning at a slow rate of spin, 1 revolution per day. > > And secondly assuming the remote chance that it does distort such an > apparatus and not allow us to check what I propose may be observed > (ie whether or not a rotating MMx that isnt distorted does still > have light at c on both paths)...THen how is it that you or george > can say with certainty that a rotating MMx source does not allow > light > to propagate down both arms at c? After all, youve just claimed > that any experiment to test this is impossible. The math is not subject to the physical limitations of the equipment. [snipping the rest because it has nothing to do with **the point** that once light is emitted, subsequent motion of the source can not and does not effect the already emitted light ] -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap |