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From: Timo Nieminen on 4 Apr 2010 18:39 On Apr 5, 2:19 am, Paul Stowe <theaether...(a)gmail.com> wrote: > On Apr 3, 3:56 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Apr 4, 4:14 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > On Apr 1, 11:09 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Apr 2, 12:59 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > ... > > > > Since c is light speed it isn't free either! > > > v = c .... > > ... it looks at first glance like this assumption would break EM > > propagation. > > Not at all. Why not? If the mean speed of aether-atoms is c, and the speed of waves in the aether is c, then the wave moves faster than many of the atoms. Since you're also assuming that the only interactions between atoms are by collision, how does this work in practice? You have checked that this, mathematically? The best model we have of corpuscles weakly interacting by collision (i.e., cross-sections are very small) is an ideal gas. For sound waves in a monatomic ideal gas, we have c = sqrt(5kT/3m), or c = 0.75 v_mean. The number will change a little if different definitions of v_mean are used, but this is clearly different from c = v_mean. I don't think this is an important objection to the theory in itself, since EM waves are clearly not sound waves in an ideal gas medium, but it does indicate that one needs to check what is happening, not just hand-wave and analogise. > > 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? > > 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. > > See above... 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? > > > > G, drag, heating, and aberration are straightforward enough, > > > > physically. I don't remember anything about "fling/back-action" from > > > > Edwards - is this in there? (It's been some time since I read it.) A > > > > reference or quick explanation would be good. In particular, why would > > > > this "cancel" drag for something in an orbit. Also not obvious to me > > > > why drag would cancel aberration. > > > > Even in the GR analog of the 'rubber sheet' there is 'back-action'. > > > ... > > > Frame-dragging? Gravito-magnetism, which can be derived in the weak- > > field limit by assuming propagation at c and lack of aberration. So, > > naturally enough, cancels aberration. If le Sage back-action + drag > > cancel aberration, then le Sage back-action is not the same, > > quantitatively, as gravitomagnetism. How does it do with tests so far > > performed for rotational frame-dragging? > > See my article in Edwards book on the dynamic effects in LeSage > models. RFD is discussed. It should, again, be obvious that it would > occur in this type of model. "Obvious" is Cartesian hand-waving. How does it perform quantitatively? As in compare with experimental tests? How does it quantitatively differ from GR? What is obvious is that viscous fluid around a rotating object is dragged around with it, resulting in a drag torque on the rotating object. Velocity of the dragged fluid falls of as 1/r^3 away from a rotating sphere. This is for a viscous fluid, with negligibly short mean free path, unable to penetrate into the rotating body, with no- slip boundary conditions. These conditions are not satisfied by your le Sage model. I will look in Edwards. > > > In a closed loop path (like an orbit) > > > the field being pushed ahead of the moving mass (as a result of the > > > drag action) induces standard vorticity in the field (like a swizzle > > > stick in a coffe cup) causing the field to acquire a spin, which will > > > in turn, tend to reduces the effect of the future forward drag. (If > > > you let go of the swizzle stick after spinning it a few times the > > > stick will continue to circulate, carried along with the vorticity of > > > the field). That is the back-action I am referring to. > > > OK, sounds like this requires further assumptions about the > > interaction of le Sage corpuscles with each other. Perhaps a mean free > > path is sufficient. That this is suggested as occuring for orbits > > within the solar system is a big red STOP sign. The range of the > > inverse square law depends on the mean free path (since these > > interactions will affect the shadow, which gives Newton in the weak > > absorption, straight-line propagation limit), while this back-action > > looks like it needs a much smaller mean free path. > > Sigh, in perfectly elastic collisions, along any linear path, does it > matter to a particle's momentum if encounters no collision or a > billion? But it does matter however to the field's granularity. IOW, > the 1/r^2 is due to the attenuation mean free path NOT! the > interaction mean free path. It matters very much if the particle has collisions along its path. In particular, if it has collisions as it travels, elastic or otherwise, the path will generally not be a straight line. In perfectly elastic collisions, momentum is still transferred. See le Sage's comments on mean free path (i.e., interaction between corpuscles, and their collision frequency). The kind of back-action and swirling you described requires interaction between the corpuscles. For this to happen within the solar system, the mean free path can't be large compared to the solar system. > > > > > > What about the mean speed? The speed distribution? The mass of the > > > > > > corpuscles? Their mass distribution? > > > > > > Good questions. Mean speed, c... I doubt that a mass value is > > > > > relevant. Momentum, yes. Cross-sectional area, yes. Mean free path, > > > > > yes. All things need more work. But, who is willing to even look, > > > > > much less work on such? > > > > > Why isn't mass relevant? Is unknown physics required here? Hopefully > > > > not, since all that would be achieved would be to move the unknown > > > > from an open and acknowledged position to a hidden place. > > > > No, but I doubt that one can ever stop the aetheron to measure it rest > > > mass. > > > Mean speed c + temperature tells you mass. (Which would be rest mass, > > if we have Newtonian mechanics.) > > It's momentum divided by c would give you a calculated value also. > But, like the photon, if you can't stop it and actually measure it do > you want to count it??? If the mean speed is known, and the temperature is known (or if these are assumed), then the mass can be determined. Does this mass have any observable consequences? This is a potential route to falsification of the theory, and is thus a possible test for the theory. Even if you're not interested in knowing the mass for its own sake (surely this would be really interesting to know!), it's methodologically important.It's a possible source of behaviour differing from conventional gravitational theory, and, in the best of worlds as far as the theory is concerned, would lead to an experiment supporting it and falsifying conventional theory. The rest mass of the photon is important as well, since it can also lead to falsification of theories (classical electrodynamics and QED, for example). So, it's been measured, not by stopping and weighing one, of course, but there are other ways. > > > > Limits on mean free path result from the distance over which the > > > > inverse square law is seen to work well. This gives a limit of size, > > > > assuming Newtonian mechanics. Not fixed, only restricted by a lower > > > > bound (mean free path) and upper bound (size, for some number density > > > > of corpuscles). > > > > There is a difference between an attenuation mean free path and a > > > interaction mean free path. > > > Yes. Attenuation length tells you about shielding, MFP tells you about > > range of the inverse square law. One is about interaction between > > corpuscles and matter, the other is about interaction between > > corpuscles. > > Not in an inviscous media. MFP doesn't tell you about interactions between corpuscles in an inviscid medium? Consider the definition of mean free path - the distance travelled between interactions between corpuscles. > > > They probably are in equilibrium. This is where unification concepts > > > become important. In my model expanded upon Maxwell's, charge is the > > > fundamental divergence of the medium. This is an oscillation around > > > the mean, a linear harmonic effect. Thus it has a base frequency > > > equal to the value of charge (e) divded by the mass (m) of the > > > oscillator. Thus, in the case of a lowly electron, that is e/m = 1.76E > > > +11 Hz (nu). Then, > > > > E = h(nu) > > > > Now what is the thermal equivalence to this energy? > > > > n(nu) = 3kT > > > > Where k is Boltzmann's contant > > > > Thus, > > > > T = h(nu)/3k = 2.8 Degrees K > > > Why 3kT? Why not (1/2)kT per degree of freedom, as usual? > > Typically it's 3/2kT but, 1/2mv^2 = 3/2kT => mv^2 = 3kT... You wrote energy = h(nu) = 3kT, not KE = (3/2) kT. > > > > > > Also look at what _isn't_ predicted. Gravitational redshift. Gravitational > > > > > > deflection of light. > > > > > > Wrong... > > > > > Le Sage predicts gravitational redshift? Details, please. Deflection > > > > of light? Details, please. A reference will suffice. If it predicts > > > > them, give details. "Wrong" is insufficient. > > > > First off, LeSage's model yield's Newton's equation and that is known > > > to properly account for red-shift. Second, given the very nature of > > > the LeSagian field light bending should be obvious. No-one, not in > > > articles or herein (Including Feynman or Steve Carlip) has questioned > > > this aspect. > > > How is Newtonian gravitation known to account for gravitational > > redshift? Given a wave originating within a gravity well, and climbing > > out, where do the "extra" oscillations go, as the frequency falls as > > the wave climbs? Unlike Doppler shift due to motion away from the > > source, where the extra oscillations go into the increasing distance > > between source and reception point, the source and receiver remain at > > the same positions, but there is an observed frequency change. > > In the LeSage process gravity is the result of pressure gradient. > Pressure/density gradients results in changes in wave speed. Just > like GR. So, why does the frequency change? Where do the extra oscillations go? Why do pressure/density gradients result in changes in wave speed? (See what kinetic theory says about this!) Is the predicted change quantitatively correct? > > "Should be obvious" isn't a very good explanation. It should come out > > of the basic assumptions, quantitatively, or at least come out of the > > basic theory + a new special parameter that's introduced to model it. > > If it's just hand-waved, without quantitification, as "obvious", it > > isn't very convincing. > > But it does, that what's meant by 'should be obvious' from the very > process that is being modeled and described. All you gave above was more hand-waving. Where is the quantitative story? Where is a quantitative calculation of le Sage gravitational redshift? > > (Where did Feynman write about le Sage-like theories?) > > "In 1965 Richard Feynman examined the Fatio/Lesage mechanism, > primarily as an example of an attempt to explain a "complicated" > physical law (in this case, Newton's inverse-square law of > gravity) > in terms of simpler primitive operations without the use of > complex > mathematics, and also as an example of a failed theory. He notes > that the mechanism of "bouncing particles" reproduces the > inverse-square force law and that "the strangeness of the > mathematical relation will be very much reduced", but then notes > that the scheme "does not work", because of the drag it predicts > would be experienced by moving bodies, "so that is the end of > that > theory".[59]" > > See:http://en.wikipedia.org/wiki/Le_Sage's_theory_of_gravitation > > and, Feynman, R. P. (1995), Feynman Lectures on Gravitation, Addison- > Wesley, pp. 2328 > Feynman, R. P. (1967), The Character of Physical Law, The 1964 > Messenger Lectures, Cambridge, Mass.: Massachusetts Institute of > Technology, pp. 3739, ISBN 0-262-56003-8 In both cases, essentially an illustration of methodology. OK, Feynman shows it as an attempt that was made, rather than Feynman makes such an attempt. Thanks for refs. > > > > How? This doesn't arise in kinetic theory without long-range forces > > > > between molecules. You're not talking about ideal gas behaviour here, > > > > or even a hard-shere gas. > > > > Well this is the very first I've heard of this. > > > The viscosity of a dilute hard-sphere gas is 2 * sqrt(mkT/pi)/ > > (3*pi*cross-section). This goes back to Maxwell (I think in the form > > viscosity is proportional to sqrt(mT)/cross-section). > > > Elastic collisions means that no energy is transferred to the internal > > energy of each atom/molecule; you still have energy transfer to the > > internal energy of the gas. > > http://openlibrary.org/b/OL14662971M/second-order_accurate_kinetic-th... And? A perfect fluid can be assumed as an idealisation, useful due to mathematical simplicity. From this, a Lagrangian, and from this, quantisation, and from this (in this case), a computational method. This doesn't have anything to do with the viscosity of gasses from kinetic theory. > > You want superfluid behaviour from a kinetic theory? You need to put > > in more than just elastic contact-collisions; you need long-range > > interactions. > > Can you provide a reference? For explicitly this, not off-hand. For what you get from elastic contact-collisions, either try Maxwell (his Bakerian lecture in 1865 I believe, which will be in Phil Trans or Proc RS), or look in old kinetic theory books, as in the 100 year old kind you can find on google books or archive.org. Basically, you get viscosity as given above, which is not the viscosity of a superfluid. http://en.wikipedia.org/wiki/Superfluid#Background has the very basics - it's the result of interaction. The details should be available in many of the works specifically on superfluids (or BEC?); I don't do this stuff myself, so I'm not familiar with this literature and can't recommend anything in particular. > > If this is the first you've heard of this, time to study kinetic > > theory. (This particular topic isn't covered very well in typical > > courses. It's beyond ideal gas, so it's too complex for the general > > introductory courses, and stat mech courses focus on thermodynamic > > properties, not physical properties, and condensed matter courses > > focus on condensed matter, not dilute gasses. Perhaps a good start > > would be with Maxwell's original papers? He doesn't give the details > > in "Theory of heat", only a brief qualitative discussion.) > > Maxwell's model was an inviscid fluid model. Maxwell's model of the EM aether was an inviscid fluid model. What Maxwell got for a gas of particles interacting by collisions was not an inviscid fluid. Put the right kind of long-range interactions into a microscopic model, even a simple mechanical one, and you can start getting the right kind of behaviour. E.g., Fitzgerald's pulley and spring model. > > > > More than that, we don't have rigid billiard-balls anywhere else to > > > > use as a model - this is only an approximation of the behaviour of > > > > real observed particles. > > > > So? > > > So in the days when much less was known about matter, when a common > > working model of matter as rigid bodies was not only used, but assumed > > to perhaps even be true, a rigid billiard-ball model of atoms made > > sense. After all, that would make atoms just like little versions of > > bulk macroscopic matter. > > But, we're not talking about matter, we're talking about physical > models. We talking about something that's meant to model reality in a meaningful way. An artificial model, based on known-to-be unphysical "objects", designed to merely generate a known set of differential equations, explains nothing. (Such a model can be pedagogically or computationally useful, but that isn't the same as explanation.) > > Now, when we know the limitations and approximations inherent in rigid > > body models of macroscopic matter, it's clear that assuming billiard- > > ball models for microscopic matter involves hypothesising the > > existence of objects with properties completely unlike the properties > > of anything we observe. > > > It's a HUGE assumption. It's an un-natural assumption (in that we > > _don't_ see this type of thing in nature). > > Yes, we do... We do? Where do we see billiard-ball-like rigid objects in nature? > > > > > > 3. What is the density of aether-atoms? > > > > > > 8.854E-12 kg/m^3 (z) > > > > > A remarkably coincident number, given that the other place this occurs > > > > in electromagnetics, in conventional classical or quantum > > > > electrodynamics, the numerical value depends on the choice of units > > > > that have nothing to do with mass or distance. > > > > > Can you justify this assumption? > > > > Seehttp://en.wikipedia.org/wiki/Speed_of_sound > > > > The propagation speed in any medium is related directly to it > > > compressibility (u, the inverse of which is called [M]odulus) and mass > > > density (z). Thus, > > > > c^2 = 1/uz = M/z > > > > For the aether medium in SI density (z) is called permittivity. The > > > compressibility (u) permeability. Then standard equation applies, > > > > c^2 = 1/uz > > > > Ochkam's Razor, KIS and all that jazz... > > > All you're doing is drawing an analogy between v = sqrt(k/rho) for > > sound, and c = sqrt((1/mu)/epsilon) for light. To claim that the KIS > > solution is to claim that epsilon _is_ mu is nonsensical - the units > > are wrong. > > Not in Maxwell's model. > > > In Gaussian units, we have e0 = 1, so does this mean that > > the mass density of aether is 1 (1 what? g/cm^3?) is Gaussian units, > > but something else if we choose to use SI units? > > The Gaussian system 'assumes' that the constant of proportionality of > Coulomb's law is unitless. It then pushes any units that it might > have artifically into the charges. This, as is well known, results in > irrational units of charge in this system. This should have been a > BIG RED FLAG that the system was flawed. But, since we had no clue > what charge was, anything could be assigned to it as long as the > resulting system was internally self consistent. But we're talking about the numerical value of something that doesn't involve the units of charge - a mass density. Why should the choice of units of charge affect a mass density, numerically? To bring this back to relevance, consider that you've given a value for the mean speed of corpuscles in the le Sage aether, which you say is identical with the EM aether. Here, you give a mass density. Do these, together, give the correct momentum flux? Since you're confident that the density is 8.85e-12 kg/m^3, and the speed is c, it's easy to calculate. Compare with your Q. > > > > (I didn't ask this one clearly enough - what is the bulk average > > > > velocity of the aether-atoms?) > > > > c, light speed as with any medium. > > > Not "as with any medium". In what "any" medium do we see the mean > > particle speed being the same as the speed of waves in the medium? > > Well, IIRC the mean speed = propagation velocity since the only way > for such to occur IS by transmission by said particles, but I'll > review my references and perhaps so third party wants to chime in. Transmission of <something> by waves going from A to B, versus transmission by particles going from A to B. Different processes. These speeds are not the same in a dilute gas (or ideal gas). In some media, wave speed is lower than particle speed, and in other media, faster. It isn't usual for them to be the same. -- Timo
From: mpc755 on 4 Apr 2010 19:59 On Apr 4, 6:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Apr 5, 2:19 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > On Apr 3, 3:56 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > On Apr 4, 4:14 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > On Apr 1, 11:09 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Apr 2, 12:59 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > ... > > > > > Since c is light speed it isn't free either! > > > > v = c > ... > > > ... it looks at first glance like this assumption would break EM > > > propagation. > > > Not at all. > > Why not? If the mean speed of aether-atoms is c, and the speed of > waves in the aether is c, then the wave moves faster than many of the > atoms. Since you're also assuming that the only interactions between > atoms are by collision, how does this work in practice? You have > checked that this, mathematically? > > The best model we have of corpuscles weakly interacting by collision > (i.e., cross-sections are very small) is an ideal gas. For sound waves > in a monatomic ideal gas, we have c = sqrt(5kT/3m), or c = 0.75 > v_mean. The number will change a little if different definitions of > v_mean are used, but this is clearly different from c = v_mean. I > don't think this is an important objection to the theory in itself, > since EM waves are clearly not sound waves in an ideal gas medium, but > it does indicate that one needs to check what is happening, not just > hand-wave and analogise. > > > > > > 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? > > > > 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. > > > See above... > > 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? > > > > > > > > G, drag, heating, and aberration are straightforward enough, > > > > > physically. I don't remember anything about "fling/back-action" from > > > > > Edwards - is this in there? (It's been some time since I read it.) A > > > > > reference or quick explanation would be good. In particular, why would > > > > > this "cancel" drag for something in an orbit. Also not obvious to me > > > > > why drag would cancel aberration. > > > > > Even in the GR analog of the 'rubber sheet' there is 'back-action'. > > > > ... > > > > Frame-dragging? Gravito-magnetism, which can be derived in the weak- > > > field limit by assuming propagation at c and lack of aberration. So, > > > naturally enough, cancels aberration. If le Sage back-action + drag > > > cancel aberration, then le Sage back-action is not the same, > > > quantitatively, as gravitomagnetism. How does it do with tests so far > > > performed for rotational frame-dragging? > > > See my article in Edwards book on the dynamic effects in LeSage > > models. RFD is discussed. It should, again, be obvious that it would > > occur in this type of model. > > "Obvious" is Cartesian hand-waving. How does it perform > quantitatively? As in compare with experimental tests? How does it > quantitatively differ from GR? > > What is obvious is that viscous fluid around a rotating object is > dragged around with it, resulting in a drag torque on the rotating > object. Velocity of the dragged fluid falls of as 1/r^3 away from a > rotating sphere. This is for a viscous fluid, with negligibly short > mean free path, unable to penetrate into the rotating body, with no- > slip boundary conditions. These conditions are not satisfied by your > le Sage model. > This is a complete misunderstanding of aether as a frictionless super fluid/solid. Aether penetrates all matter except for a black hole. The 'viscous' fluid rotates around an object because of friction. Hence the 'frictionless' part of a frictionless superfluid. A particle moving through a frictionless superfluid displaces the frictionless superfluid. The frictionless superfluid returns to its previous state once it is done interacting with the particle. If you want to try and undermine aether as a frictionless super fluid/ solid go for it. You should at least make an attempt to understand how matter interacts with (i.e. connects to) a frictionless superfluid. "the state of the [ether] is at every place determined by connections with the matter and the state of the ether in neighbouring places" - Albert Einstein The state of the aether's displacement is determined by its connections with the matter and the state of the aether in neighboring places. When a particle is dropped into a frictionless superfluid what occurs to the superfluid? It is displaced. What occurs to the superfluid when the particle is removed? It displaces back. The pressure exerted by aether displaced by matter is gravity.
From: Paul Stowe on 4 Apr 2010 21:28 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: {snip... I'll focus on this one issue here} > > > OK, sounds like this requires further assumptions about the > > > interaction of le Sage corpuscles with each other. Perhaps a mean free > > > path is sufficient. That this is suggested as occuring for orbits > > > within the solar system is a big red STOP sign. The range of the > > > inverse square law depends on the mean free path (since these > > > interactions will affect the shadow, which gives Newton in the weak > > > absorption, straight-line propagation limit), while this back-action > > > looks like it needs a much smaller mean free path. > > > Sigh, in perfectly elastic collisions, along any linear path, does it > > matter to a particle's momentum if encounters no collision or a > > billion? But it does matter however to the field's granularity. IOW, > > the 1/r^2 is due to the attenuation mean free path NOT! the > > interaction mean free path. > > It matters very much if the particle has collisions along its path. In > particular, if it has collisions as it travels, elastic or otherwise, > the path will generally not be a straight line. In perfectly elastic > collisions, momentum is still transferred. > > See le Sage's comments on mean free path (i.e., interaction between > corpuscles, and their collision frequency). > > The kind of back-action and swirling you described requires > interaction between the corpuscles. For this to happen within the > solar system, the mean free path can't be large compared to the solar > system. OK, let's have three particles with different velocities thus momenta p1, p2, p3 Particle p1 is moving in the x direction, all other particles move randomly.... The particle are frictionless, and by definition, perfectly elastic. They undergo frictionless center of mass collisions. I'll try to graphically portray this, /(p2 m1v2) (p3 m3v3) / \ (p1 m1v1) / (p1 m2v1) \ (p1 m3v1) o---------x----->--o-----------------------x-----o => / \ / \ o (p2 m2v2) o (p3 m2v3) In the end, the momentum along the x axis is unchanged as if no collisions had ever occurred The momentum lines in a perfect fluid of this type are invariant, it does not matter HOW MANY collisions occur. Attenuation is a different animal. This why I said interaction mean free path is not the same. More on the other issues later. Paul Stowe
From: Timo Nieminen on 4 Apr 2010 23:03 On Apr 5, 11:28 am, Paul Stowe <theaether...(a)gmail.com> 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: > > {snip... I'll focus on this one issue here} > > > > > > > OK, sounds like this requires further assumptions about the > > > > interaction of le Sage corpuscles with each other. Perhaps a mean free > > > > path is sufficient. That this is suggested as occuring for orbits > > > > within the solar system is a big red STOP sign. The range of the > > > > inverse square law depends on the mean free path (since these > > > > interactions will affect the shadow, which gives Newton in the weak > > > > absorption, straight-line propagation limit), while this back-action > > > > looks like it needs a much smaller mean free path. > > > > Sigh, in perfectly elastic collisions, along any linear path, does it > > > matter to a particle's momentum if encounters no collision or a > > > billion? But it does matter however to the field's granularity. IOW, > > > the 1/r^2 is due to the attenuation mean free path NOT! the > > > interaction mean free path. > > > It matters very much if the particle has collisions along its path. In > > particular, if it has collisions as it travels, elastic or otherwise, > > the path will generally not be a straight line. In perfectly elastic > > collisions, momentum is still transferred. > > > See le Sage's comments on mean free path (i.e., interaction between > > corpuscles, and their collision frequency). > > > The kind of back-action and swirling you described requires > > interaction between the corpuscles. For this to happen within the > > solar system, the mean free path can't be large compared to the solar > > system. > > OK, let's have three particles with different velocities thus momenta > p1, p2, p3 Particle p1 is moving in the x direction, all other > particles move randomly.... The particle are frictionless, and by > definition, perfectly elastic. They undergo frictionless center of > mass collisions. I'll try to graphically portray this, > > /(p2 m1v2) (p3 m3v3) > / \ > (p1 m1v1) / (p1 m2v1) \ (p1 m3v1) > o---------x----->--o-----------------------x-----o => > / \ > / \ > o (p2 m2v2) o (p3 m2v3) > > In the end, the momentum along the x axis is unchanged as if no > collisions had ever occurred The momentum lines in a perfect fluid of > this type are invariant, it does not matter HOW MANY collisions > occur. Attenuation is a different animal. This why I said > interaction mean free path is not the same. For frictionless collisions, the force during the collision is along the line joining the centres of the particles. For identical particles, this means that the component of momentum along this line is swapped between the particles, and the component normal to this line remains the same. For anything other than a straight in-line collision (impact parameter of zero, if you like), this swapped component of the momentum is only part of the momentum p1 at the first collision, and so on for the subsequent collisions. Your diagram is the special case when all particle centres are lined up along the line of p1 at the moments of collision, and what is shown will not be the general result. You've played billiards/snooker/pool, yes? Hit a ball dead-centre with the cue ball, no extra spin, and the cue ball stops, and the hit ball goes off at the same speed that the cue ball hit with. Hit off-centre, a glancing hit, and neither ball continues on the original path. You know that with real particles, it matters whether or not the collision is dead-centre or off-centre. Off-centre, and both balls are deflected away from the original direction of travel. Yes, you can have friction with the real balls, and real balls won't be perfectly elastic, but you should be able to see that an off-centre collision with ideal elastic frictionless balls will still give deflection away from the original line of travel. If not, do the algebra. This is related to the viscosity of a hard-sphere gas! Fire an atom into a hard sphere gas, and its original momentum is spread amongst many particles by such collisions, and spread sideways. It's usual to say that this spread is by diffusion, but the mechanism of that diffusion is a sequence of collisions like this. -- Timo
From: mpc755 on 4 Apr 2010 23:46
On Apr 4, 7:59 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Apr 4, 6:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Apr 5, 2:19 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > On Apr 3, 3:56 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > On Apr 4, 4:14 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > > On Apr 1, 11:09 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > On Apr 2, 12:59 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > ... > > > > > > Since c is light speed it isn't free either! > > > > > v = c > > ... > > > > ... it looks at first glance like this assumption would break EM > > > > propagation. > > > > Not at all. > > > Why not? If the mean speed of aether-atoms is c, and the speed of > > waves in the aether is c, then the wave moves faster than many of the > > atoms. Since you're also assuming that the only interactions between > > atoms are by collision, how does this work in practice? You have > > checked that this, mathematically? > > > The best model we have of corpuscles weakly interacting by collision > > (i.e., cross-sections are very small) is an ideal gas. For sound waves > > in a monatomic ideal gas, we have c = sqrt(5kT/3m), or c = 0.75 > > v_mean. The number will change a little if different definitions of > > v_mean are used, but this is clearly different from c = v_mean. I > > don't think this is an important objection to the theory in itself, > > since EM waves are clearly not sound waves in an ideal gas medium, but > > it does indicate that one needs to check what is happening, not just > > hand-wave and analogise. > > > > > 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? > > > > > 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. > > > > See above... > > > 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? > > > > > > > G, drag, heating, and aberration are straightforward enough, > > > > > > physically. I don't remember anything about "fling/back-action" from > > > > > > Edwards - is this in there? (It's been some time since I read it.) A > > > > > > reference or quick explanation would be good. In particular, why would > > > > > > this "cancel" drag for something in an orbit. Also not obvious to me > > > > > > why drag would cancel aberration. > > > > > > Even in the GR analog of the 'rubber sheet' there is 'back-action'. > > > > > ... > > > > > Frame-dragging? Gravito-magnetism, which can be derived in the weak- > > > > field limit by assuming propagation at c and lack of aberration. So, > > > > naturally enough, cancels aberration. If le Sage back-action + drag > > > > cancel aberration, then le Sage back-action is not the same, > > > > quantitatively, as gravitomagnetism. How does it do with tests so far > > > > performed for rotational frame-dragging? > > > > See my article in Edwards book on the dynamic effects in LeSage > > > models. RFD is discussed. It should, again, be obvious that it would > > > occur in this type of model. > > > "Obvious" is Cartesian hand-waving. How does it perform > > quantitatively? As in compare with experimental tests? How does it > > quantitatively differ from GR? > > > What is obvious is that viscous fluid around a rotating object is > > dragged around with it, resulting in a drag torque on the rotating > > object. Velocity of the dragged fluid falls of as 1/r^3 away from a > > rotating sphere. This is for a viscous fluid, with negligibly short > > mean free path, unable to penetrate into the rotating body, with no- > > slip boundary conditions. These conditions are not satisfied by your > > le Sage model. > > This is a complete misunderstanding of aether as a frictionless super > fluid/solid. > > Aether penetrates all matter except for a black hole. > > The 'viscous' fluid rotates around an object because of friction. > > Hence the 'frictionless' part of a frictionless superfluid. > > A particle moving through a frictionless superfluid displaces the > frictionless superfluid. The frictionless superfluid returns to its > previous state once it is done interacting with the particle. > > If you want to try and undermine aether as a frictionless super fluid/ > solid go for it. You should at least make an attempt to understand how > matter interacts with (i.e. connects to) a frictionless superfluid. > > "the state of the [ether] is at every place determined by connections > with the matter and the state of the ether in neighbouring places" - > Albert Einstein > > The state of the aether's displacement is determined by its > connections with the matter and the state of the aether in neighboring > places. > > When a particle is dropped into a frictionless superfluid what occurs > to the superfluid? It is displaced. What occurs to the superfluid when > the particle is removed? It displaces back. > > The pressure exerted by aether displaced by matter is gravity. What happens when you place a mesh bag of marbles separated by springs into a super fluid? The super fluid is displaced by the marbles. The super fluid also penetrates the bag full of marbles. When you spin the bag of marbles the state of the super fluid is determined by its connections with the marbles. This state is the super fluids state of displacement. If your 'concept' of matter does not consist of nuclei separated by aether (i.e the matter is penetrated by the aether) then your 'understanding' of the connectedness between matter and aether is incorrect. |