There is no "pull" of gravity, only the PUSH of flowing ether!
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From: spudnik on 5 Apr 2010 21:56 see; you didn't even read your own, massively "quoted" ****, at the very front of your epistle. there are classes, for free, at your local community college, called ESL, dood. but, like I said, below, you also get the last word, here -- try not to make another turd -- unless you've got a special sauce. (I mean, what if I have other things to do, than spend time with other, latter-day Einsteins?) > Incorrect. Beam splitters do not cause a photon to 'split' into 'two > photons' of half the energy. What you are mistaking for two photons is > the associated aether wave propagating the available paths while the > photon 'particle' travels a single path. > > If you actually read the experiment associated with the experiment which > will provide evidence of Aether Displacement you would know this. > "a beta barium borate crystal (labeled as BBO) causes spontaneous > parametric down conversion (SPDC), converting the photon (from either > slit) into two identical entangled photons with 1/2 the frequency of the > original photon." thus: please, stop abusing fulerenes.... excusez-moi; *what* is incorrect -- you mean, "incorrect in the interpretation of my [language- challenged] theory?..." your theory, that adds not one thing to any observation, except perhaps spme sort of "metaphysical" say-so? get with it, dood; read some elementary, old textbooks on wave-theory/electromagnetism, in your mother tongue, if you can't find the original discoverers write-ups.... ah: the brachistochrone of Leibniz and Bernoulli (in French) !! it's bad enough, that people habitually refer to photons, when all they manifestly are is waves!... now, if you want to retool teh wave-theory into a theory of aether, it's going to take a Hell of a lot more than mere, illinguistic assertions. and, please, stop quoting that poor, old man, Einstien; he was probably very far from perfect, at any time, in almost any writing (possible exception: the paper on the photo-electrical effect, which is after all what they gave him a Nobel, for; or, the paper on Brownian motion; or, the patent with Dr. Strangelove for an acoustic refrigerator .-) [I mean, what if it was really a political thing, strictly to unbury Newton's corpuscles?] > Incorrect. Beam splitters do not cause a photon to 'split' into 'two > photons' of half the energy. What you are mistaking for two photons is > the associated aether wave propagating the available paths while the > photon 'particle' travels a single path. > is everywhere the same. The ether of the general theory of relativity > is transmuted conceptually into the ether of Lorentz if we substitute > constants for the functions of space which describe the former, > disregarding the causes which condition its state."http://www-groups.dcs.st-and.ac.uk/~history/Extras/Einstein_ether.html --Light: A History! http://21stcenturysciencetech.com --NASCAR rules on rotary engines! http://white-smoke.wetpaint.com thus: that's why I always suggest Shakespeare, becuase *no* one can *begin* to comprehend English, til he *tries* to read the bard. (he also had a hand in translating the KJV of the Bible .-) dear editor; The "cap & trade" omnibus bill -- what Waxman-Markey should be known as, being so fundamental to the Stupid, economy -- is at least as old as Waxman's '91 bill to ameliorate acid rain. One must really stop and consider, just who really opposes this "last hurrah" for Wall Street (like-wise, the healthcare bill, also under Waxman's House committee, and which, after all, is geared toward funding a smaller aspect of the S-- the economy, already tremendously leveraged by the "voluntary" cap & trade, which the bill would essentially mandate, a la the much-larger, market-making EU scheme). Not so long ago, there was a guest-editorial in the WSJ, which mentioned that a carbon tax would achieve the same thing, more or less, as the total "free" trade approach of cap & trade; oh, but, there're certain, so-called Republicans, who refer to the bill as "cap & tax!" Well, before any "reform" of the financial system, why would one put all of one's eggs into such a casino -- especially considering that the oil companies have not bothered to release the carbon-dating "fingerprints" that they use, to determine whether two wells are connected, underground; so, guys & gals, how old is the stuff, on average, anyway? Surely, the green-niks who lobby for "renewable" energy, do not think that oil comes only from dinosaurs, and their associated flora -- all, from before the asteroid supposedly offed them (I refer them to the recent issue of Nature -- several articles that may be related!) Finally, note that, in a sense, the whole world is going a) nuclear, and b) into space, while we are essentially frozen into '50s and '60s techniques in these crucial frontiers. (While some folks dither about Iran's nuke-weapons policy, they are rapidly achieving a full-scale nuke-e and process-heat capbility for industry & infrastructure.) --yr humble servant, the Voting Rights Act o'65 (deadletter since March 27, 2000, when Supreme Court refuzed appeal in LaRouche v. Fowler ('96))
From: NoEinstein on 5 Apr 2010 21:58 On Apr 5, 8:52 pm, spudnik <Space...(a)hotmail.com> wrote: > > Dear spudnik: Light doesn't need a medium to travel in, because light is photons, only! The ether is discontinuous between most galaxies. That's what accounts for the Swiss Cheese voids observed in the heavens. Yet, the light can travel perfectly well through an "ether vacuum". The notion that light is... waves necessitates having an ether, solely, for the purpose of passing light. If an ether medium were required for the passage of light, the ether would have to be close to homogenious across the Universe, otherwise the light "quality" would vary all over the place. If ether must be dedicated to just the passage of light, then ether couldn't be the BUILDING BLOCK of the Universewhich it most definitely is! The main mistakes most... science (non) thinkers have made is to lock- in one notion about an issue, and not consider all of the observations in nature which must be accounted for. The one great unifying component of everything observed in nature is: Varying ether flow and density. Ether is the ENERGY from which you and I and everything that is is made. And I, NoEinstein, discovered that fact! NE > > nah; we should blame Pascal for discovering, > experimentally, his "plenum," which he thought was perfect. I mean, > it's always good to have a French v. English dichotomy, > with a German thrown-in for "triality." > > > of Newton's "action at a distance" of gravity, > > via the re-adumbration of his dead-as- > > a-doornail-or-Schroedinger's-cat corpuscle, > > "the photon." well, and/or "the aether," > > necessitated by "the vacuum." > > --Light: A History!http://21stcenturysciencetech.com > > --NASCAR rules on rotary engines!http://white-smoke.wetpaint.com > > thus: > Death to the lightcone -- > long-live Minkowski!... yeah; and, > the photon is *still* dead, > no matter what herr Albert said about it! > > > > <<pseudoscientists rarely revise. The first edition of > > > Principia Mathematica, a product of a committee, > > > the Royal Society, after "the MS burnt in an alchemical > > > process that set the trunk in which it was resting, afire," > > > has had several editions, the latter of which take pains > > > to omit mention of Robert Hooke. The sole calculus is > > > is a rectangle, dxdy, in Book 2, Section 2, Paragraph 2.>> > > thus: > as a student of Bucky Fuller -- an army of one, I say -- you've bit- > off > more than you should want to chew, with the n-hole spin on fullerenes; > and that is my clue, because a fullerene should have a very large > manifestation of polarization, not unlike in a game of futbol. I > mean, > just becaus the ball went through only one slit, why wouldn't it be > affected by the total symmetry of the instrumentation?... > all of it, down to teh electronics etc. > > my main thing was, though, that you should at least *try* > to consider the theory of light using only waves, > which can still be pieced-together from almost any "undergrad" > textbook, post-Copenhagen, especially older ones. > > or, just stick with Einstein's refurbishment of Newton's crappy > "theory," > nothing of which is needed for relativity & so on. anyway, > one simply does not need to analyze a phenomenon > by *both* its wavey & bullety aspects -- at the same time; > once you have proven a theorem in projective geometry e.g., > you do not have to give the "2nd column proof," unless > you're just learning it, for the first time! > > >http://en.wikipedia.org/wiki/Louis_de_Broglie > > any moving particle or object had an associated wave." > > thus: > a-ha, I was correct: > say "half," with respect to the beamsplitters, please (as > I comprehend, they generally split the "photon" > into "two photons" of half the energy, I think > of a different frequency, not amplitude -- although > the "photon" is really more akin to a phonon, > such as the audible "click" of the geiger-counter. the *proviso* > with these experiments is that the waves are highly modified > in the LASER apparatus, so that some folks more easily think > of them as "rocks o'light." > > it could have been worse; > lots of more-or-less literate folks use "of" > in the place of "have" -- to be or not to be owned, > that is this particualr question! > > >http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser#The_experi... > > thus: > if you let go of the empty notion of "photon," > there isn't any difficulty, at all, with a geometrical picture. > Death to the lightcone -- long-live the lightcone-heads (because, > Minkowski was only one of them, by haphazard default/death). > yes, I know, that *photonics* is a whole field of engineering; > thank you, herr doktor-professor E., > for unburying Newton's bogus corpuscle and attendant "theory," > that Young had successfully popped! > > thus: > on the wayside, if > you are really going to set so much store in a two-hole procedure > for fullerenes, maybe you shouold read the original article, and > try to question its purpose. as it is, I'd guess that > English is not your mother-tongue, > which can sometimes prove difficult in *using* it; so, > that's why I always suggest Shakespeare, becuase > *no* one can *begin* to comprehend English, > til he *tries* to read the bard. (he also had a hand > in translating the KJV of the Bible .-) > > thus: > NB, quaternions are not "quadrays" (for an amateur attempt > at homogenous co-ordination), but you can "do" special rel. > with them (according to Lanczos .-) > > thus: > The "cap & trade" omnibus bill -- what Waxman-Markey should > be known as, being so fundamental to the Stupid, economy -- is > at least as old as Waxman's '91 bill to ameliorate acid rain. One > must really stop and consider, just who really opposes this "last > hurrah" for Wall Street (like-wise, the healthcare bill, also > under Waxman's House committee, and which, > after all, is geared toward funding a smaller aspect of the S-- > the economy, already tremendously leveraged by the "voluntary" > cap & trade, which the bill would essentially mandate, > a la the much-larger, market-making EU scheme). > > Not so long ago, there was a guest-editorial in the WSJ, > which mentioned that a carbon tax would achieve the same thing, > more or less, as the total "free" trade approach of cap & trade; oh, > but, there're certain, so-called Republicans, who refer to the bill > as "cap & tax!" > > Well, before any "reform" of the financial system, why > would one put all of one's eggs into such a casino -- especially > considering that the oil companies have not bothered > to release the carbon-dating "fingerprints" that they use, > to determine whether two wells are connected, underground; so, > guys & gals, how old is the stuff, on average, anyway? > > Surely, the green-niks who lobby for "renewable" energy, do not think > that oil comes only from dinosaurs, and their associated flora -- > all, from before the asteroid supposedly offed them (I refer them > to the recent issue of Nature -- several articles that may be > related!) > > Finally, note that, in a sense, the whole world is going a) > nuclear, and b) into space, while we are essentially frozen > into '50s and '60s techniques in these crucial frontiers. (While some > folks dither about Iran's nuke-weapons policy, they are rapidly > achieving a full-scale nuke-e and process-heat capbility > for industry & infrastructure.) > > --yr humble servant, the Voting Rights Act o'65 > (deadletter since March 27, 2000, > when Supreme Court refuzed appeal in LaRouche v. Fowler ('96))
From: mpc755 on 5 Apr 2010 22:08 On Apr 5, 9:56 pm, spudnik <Space...(a)hotmail.com> wrote: > see; you didn't even read your own, massively "quoted" ****, > at the very front of your epistle. there are classes, for free, > at your local community college, called ESL, dood. > > but, like I said, below, > you also get the last word, here -- > try not to make another turd -- unless you've got a special sauce. > (I mean, what if I have other things to do, > than spend time with other, latter-day Einsteins?) > > > Incorrect. Beam splitters do not cause a photon to 'split' into 'two > > photons' of half the energy. What you are mistaking for two photons is > > the associated aether wave propagating the available paths while the > > photon 'particle' travels a single path. > > > If you actually read the experiment associated with the experiment which > > will provide evidence of Aether Displacement you would know this. > > "a beta barium borate crystal (labeled as BBO) causes spontaneous > > parametric down conversion (SPDC), converting the photon (from either > > slit) into two identical entangled photons with 1/2 the frequency of the > > original photon." > > thus: > please, stop abusing fulerenes.... > excusez-moi; *what* is incorrect -- you mean, > "incorrect in the interpretation of my [language- > challenged] theory?..." your theory, > that adds not one thing to any observation, except > perhaps spme sort of "metaphysical" say-so? > A C-60 molecule is in the slit(s). While the C-60 molecule is in the slit(s) detectors are placed at the exits to the slits. When there are detectors at the exits to the slits the C-60 molecule is always detected exiting a single slit. If the detectors are placed and removed from the exits to the slits while the C-60 molecule is in the slit(s) the C-60 molecule creates an interference pattern. Explain how this is possible without aether. http://en.wikipedia.org/wiki/Louis_de_Broglie "This research culminated in the de Broglie hypothesis stating that any moving particle or object had an associated wave." 'Interpretation of quantum mechanics by the double solution theory Louis de BROGLIE' http://www.ensmp.fr/aflb/AFLB-classiques/aflb124p001.pdf "I called this relation, which determines the particle's motion in the wave, "the guidance formula". It may easily be generalized to the case of an external field acting on the particle." "This result may be interpreted by noticing that, in the present theory, the particle is defined as a very small region of the wave where the amplitude is very large, and it therefore seems quite natural that the internal motion rythm of the particle should always be the same as that of the wave at the point where the particle is located." de Broglie's definition of wave-particle duality is of a physical wave and a physical particle. The particle occupies a very small region of the wave. In AD, the external field is the aether. In a double slit experiment the particle occupies a very small region of the wave and enters and exits a single slit. The wave enters and exits the available slits. In AD, the C-60 molecule has an associated aether displacement wave. The C-60 molecule always enters and exits a single slit while the associated aether displacement wave enters and exits the available slits. The displacement wave creates interference upon exiting the slits which alters the direction the C-60 molecule travels. Detecting the C-60 molecule causes decoherence of the associated aether displacement wave (i.e. turns it into chop) and there is no interference.
From: Paul Stowe on 5 Apr 2010 22:21 On Apr 4, 3:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Apr 5, 2:19 am, Paul Stowe <theaether...(a)gmail.com> wrote: ... > > > I wanted to do a quick calculation of heating, so looked at your > > > numbers: > > > > "the mass attenuation coefficient [u] (m^2/kg) is ~3.147E-06. The > > > "momentum flux (Q) is ~6.74E+00 kg/m-sec^2." > > > > Are these correct? Given the typical density of matter around here, is > > > this compatible with > > > > > > > The 'linear attenuation coefficient is on the order of 1E-20 1/m, > > > > As I understand it, the intercepted momentum flux, if completely > > > transferred to some object, gives the maximum force possible, if there > > > was complete shielding on the other side. So, the maximum > > > gravitational force on an object (presumably super-dense) of some > > > given cross-sectional area. Meanwhile, an object of "normal" mass and > > > density, completely shielded from one side, experiences a force of > > > Qu*mass. For gravitational attraction, the fluxes from each side, > > > accounting for shadowing are Q(1-u*term dependent of mass and geometry > > > of shading object) from the shady side, and Q from the other side, for > > > a net transfer of momentum of Qu^2 * f(m1,r)*m2. > > > In the case of heating it's power flux, not momentum flux. The power > > flux (W) is Qc/4pi => ~1.6E+08 Watts/m^2. Then for any body w = W(2GM/ > > rc^2), conbining all 'constants' we have (2WG/c^2)(M/r). Thus, > > > k = (2WG/c^2) ~= 2.38E-19 m/sec^3 and, > > > w = kM/r > > > Where M & r are the mass and radius of the body. > > > Then, > > > w = W - W' > > > where W is as defined above and W' the amount that makes it out. > > Then, > > > w = W(1 - e^-2lr) > > > Now, solve for l... (the linear attenuation coefficient) > > So, w = W(1 - exp(-2lr)) is approximately w = 2lrW for weak > attenuation, or k M/r = 2lrW. > > We have M = 4*pi*r^3/3 * rho where rho is the mean density.With your > numbers, this gives > > l = 2e-20 * rho * r, > > for l = 1e-16 m^-1 using rho, r for the Earth. Why does this value > depend on r? The flux (Q) is constant on a per area basis. Thus, the total momentum/power impinging upon a mass is QA - WA. As area goes to zero the intercepted flux goes to zero. If the mass is constant the density increase as a function of r^3 and the linear attenuation remains constant (in the weak region) BUT! the total area is decreasing as function of r^2. This differential results in a 1/r reduction in the total attenuated quantities. Thus the form M/r... > > > What have I gotten wrong here? From the above, I have maximum force > > > possible per kg is only Qu = 2e-5N, which is unrealistically low. > > > Meanwhile, for a block a steel, the linear attenuation coefficient > > > would be 0.02. > > You don't comment at all on the calculation of force. Using the values > you give for Q and u, I get a maximum force per kg of 2e-5N. This is > rather less than we observe terrestrially, and in turn, this must be > far less than the maximum possible since shielding is negligible. > Since you have repeated the correctness of your values for Q and u, > why does the above calculation give a value of l very different from > rho*u? As for the force of gravity, this is why big G is Qu^2, NOT! qu. The Newtonian is, technically in LeSage's weak region, F = Q(uM)(um)/r^2 The total 'potential', or what we call acceleration due to gravity, is, a = Qu(Mu/r^2) BUT! you MUST have that second mass m to realize this 'potential'. Given that u is independent of m this 'potential' is considered mass indenpendent. This is WHY! big G is only truly valid for multibody problems. Trying to get the right answer with only Qu just doesn't cut it. Look at Qu (kg/m-sec^2)(m^2/kg) => m/sec^2, it's not a force, it's the field's attenuation potential. In fact, its the key factor, NOT G! but for Qu(Mu/r^2) we have this potential (Qu) occuring throughout M as (Mu/r^2) resulting in the dimensionless degree of magnitude of how these individual interactions are creating the total cummulative effect. Paul Stowe
From: Timo Nieminen on 5 Apr 2010 23:03
On Mon, 5 Apr 2010, Paul Stowe wrote: > On Apr 4, 3:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Apr 5, 2:19 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > ... > > > > > I wanted to do a quick calculation of heating, so looked at your > > > > numbers: > > > > > > "the mass attenuation coefficient [u] (m^2/kg) is ~3.147E-06. The > > > > "momentum flux (Q) is ~6.74E+00 kg/m-sec^2." > > > > > > Are these correct? Given the typical density of matter around here, is > > > > this compatible with > > > > > > > > > The 'linear attenuation coefficient is on the order of 1E-20 1/m, > > > > > > As I understand it, the intercepted momentum flux, if completely > > > > transferred to some object, gives the maximum force possible, if there > > > > was complete shielding on the other side. So, the maximum > > > > gravitational force on an object (presumably super-dense) of some > > > > given cross-sectional area. Meanwhile, an object of "normal" mass and > > > > density, completely shielded from one side, experiences a force of > > > > Qu*mass. For gravitational attraction, the fluxes from each side, > > > > accounting for shadowing are Q(1-u*term dependent of mass and geometry > > > > of shading object) from the shady side, and Q from the other side, for > > > > a net transfer of momentum of Qu^2 * f(m1,r)*m2. > > > > > In the case of heating it's power flux, not momentum flux. The power > > > flux (W) is Qc/4pi => ~1.6E+08 Watts/m^2. Then for any body w = W(2GM/ > > > rc^2), conbining all 'constants' we have (2WG/c^2)(M/r). Thus, > > > > > k = (2WG/c^2) ~= 2.38E-19 m/sec^3 and, > > > > > w = kM/r > > > > > Where M & r are the mass and radius of the body. > > > > > Then, > > > > > w = W - W' > > > > > where W is as defined above and W' the amount that makes it out. > > > Then, > > > > > w = W(1 - e^-2lr) > > > > > Now, solve for l... (the linear attenuation coefficient) > > > > So, w = W(1 - exp(-2lr)) is approximately w = 2lrW for weak > > attenuation, or k M/r = 2lrW. > > > > We have M = 4*pi*r^3/3 * rho where rho is the mean density.With your > > numbers, this gives > > > > l = 2e-20 * rho * r, > > > > for l = 1e-16 m^-1 using rho, r for the Earth. Why does this value > > depend on r? > > The flux (Q) is constant on a per area basis. Thus, the total > momentum/power impinging upon a mass is QA - WA. As area goes to zero > the intercepted flux goes to zero. If the mass is constant the > density increase as a function of r^3 and the linear attenuation > remains constant (in the weak region) BUT! the total area is > decreasing as function of r^2. This differential results in a 1/r > reduction in the total attenuated quantities. Thus the form M/r... OK, so this is the decrease in flux as it moves into a spherical body? I'd suggest calling it something different; this is not what is usually meant by a "linear attenuation coefficient". > > > > What have I gotten wrong here? From the above, I have maximum force > > > > possible per kg is only Qu = 2e-5N, which is unrealistically low. > > > > Meanwhile, for a block a steel, the linear attenuation coefficient > > > > would be 0.02. > > > > You don't comment at all on the calculation of force. Using the values > > you give for Q and u, I get a maximum force per kg of 2e-5N. This is > > rather less than we observe terrestrially, and in turn, this must be > > far less than the maximum possible since shielding is negligible. > > Since you have repeated the correctness of your values for Q and u, > > why does the above calculation give a value of l very different from > > rho*u? > > As for the force of gravity, this is why big G is Qu^2, NOT! qu. The > Newtonian is, technically in LeSage's weak region, > > F = Q(uM)(um)/r^2 > > The total 'potential', or what we call acceleration due to gravity, > is, > > a = Qu(Mu/r^2) > > BUT! you MUST have that second mass m to realize this 'potential'. > Given that u is independent of m this 'potential' is considered mass > indenpendent. This is WHY! big G is only truly valid for multibody > problems. Trying to get the right answer with only Qu just doesn't > cut it. > > Look at Qu (kg/m-sec^2)(m^2/kg) => m/sec^2, it's not a force, it's the > field's attenuation potential. I didn't say that Qu was a force. I said that Qu*mass is a force, the force you'd get on a mass if you had complete screening, with the momentum flux Q incident from one side only. If you want to use a cube of some size, of some density, you get the same result, as long as you're still in the weak attenuation limit. Isn't saying that the momentum flux Q is 6.74 (kg.m/s)/m^2/s the same as saying that the force that would act on a perfectly absorbing square 1m by 1m in area, completely shielded from one side, would be 6.74N? If not, why not? Surely the momentum flux of the corpuscles must be much greater than the gravitational forces we see, since a tiny reduction in the flux from one side, resulting in an equally tine anisotropy in the flux, with a tiny absorption, is meant to lead to the gravitational forces we see. If the maximum force we could get on an object of 1m^2 in cross-section were only 6.74N, how can we see forces larger than this, in real life, on objects smaller than this? -- Timo |