Neo-Aether and the Medium of Space
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From: Paul Stowe on 7 Aug 2010 15:13 On Aug 2, 12:19 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > What matters is the Reynolds number, which measures the relative > > > importance of inertia to viscosity. Re > 1, maybe you can assume that > > > the flow is inviscid. Re << 1, you can ignore inertia. For the same > > > fluid, just different flows. > > > Does not Reynolds apply to the boundaries of material systems > > containing a fluid/gas in a flow situation? > > It does so apply, and it also applies to other situations. A ratio of > inertial effects to viscous effects is a general thing. > > We have > > Re = rho V L / mu = V L / nu, > > where rho is the density, V is the speed, mu is the dynamic viscosity, nu > is the kinematic viscosity, and L is a characteristic length. Yes, V is the bulk 'current' speed of the fluid, not the mean speed of its constitutes... > For wave propagation, L = lambda/2 is a good choice, since the speed of > the fluid will vary from zero to max to zero over this distance. We can > also choose V = c, the speed of the wave. So, we have not really. But, let use the kinetic properties, V = c L = L mu = cL Then R = cL/cL => Unity... Useless... > Re = c lambda / 2 nu. > > This tells us, more or less, the number of wavelengths a wave will travel > before being reduced to 1/e of the original intensity. > > Don't like this rough-and-ready calculation? Get the real deal from Lamb, > Hydrodynamic, art 359, from which we get for the decay distance (d) in > wavelengths: > > d/lambda = (3 / 8 pi^2) c lambda / nu. > > This is what we have for wave propagation in simple gases. If d/lambda is > large, and d is large compared to relevant lengths, then we can ignore > viscosity. If not, we can't. I don't know about you, but I have never advocated a 'simple gas'. I have always define the aether as a superfluidic vortex lattice of the Maxwellian variety. > This is the bare-naked kinetic theory result, without appealing to any > special results from elsewhere. All known substance are kinetic in nature, the only differences are how their 'fields' interact & behave. I don't think you really understand the duality. Material substance would be made up of fluidic interacting Skyrmions and vortices. These circulation fields either attract or repel depending upon the velocity gradient at the shear boundary between them. These structure, in turn, are ALSO! made up of kinetic entities. It is not, nor have I ever claimed it was, a simple 'gas'! > Do your suggested numbers work? Is the attenuation of waves slow enough to > be compatible with obervation? If not, back to the drawing board. Well there would be a non-zero attenuation coefficient for acoustic propagation and my estimate is that this attenuation coefficient (s) is on the order of the mid 10^-26. In other words, i = Ie^-sd and d is distance in meters... This attenuation coefficient IS NOT related to any kinetic mean free path. > > > To take results for, e.g., a superfluid liquid, and to assume that > > > they apply to a gas of low viscosity is a Bad Thing. > > > Go look a Maxwell's "On Physical Lines of Force", it is clearly stated > > there. > > Where? Maxwell starts off with the mathematical description of the > behaviour of an elastic solid, and goes with that for a while. No, he does not... Read it again, http://www.vacuum-physics.com/Maxwell/maxwell_oplf.pdf He starts WITH a vortex interaction model, period. And he does explicitly state that he assumes a perfect incompressible fluid. > He > introduces his molecular vortices, not because they naturally arise from > kinetic theory, but because his differential equation suggest them. He has > difficulty with the interpretation of the vortices. If vortices could not naturally arise in a gas we'd never see or known tornados or hurricanes. Vorticity is a fundamental property of a continuum state. > So where does Maxwell clearly state that he is considering a gas of low > vicosity? Rather than elastic bodies with molecular vortices? He states a "perfect fluid" and never defines gas, liquid, or solid, just quantifies behavior.... > > It's not superfluid liquid applies TO a gas, It is that the > > so-called gas gives rise to quantum vortices and those, acting in a > > particle like manner, forms the copper pair lattice in a superfluidic > > or, possibly, a supersolid manner. Your statement above is totally > > backwards... As for being a Bad Thing, without that model OPLF would > > not ever have been written,,, > > Where does Maxwell mention Cooper pairs? Why call the vortices "quantum"? Because if LOOK! at figure 2 of his illustrations after page 348 you'll clearly see the regular 'quantized' vortex lattice structure just like Bernard Cells... > > > This means that the momentum of > > > atoms entering a kinetic gas is transferred to the gas as a whole. > > > Entering? From where? In the universe is a kinetic medium model the > > whole system would be conservative. There is no entering or exiting, > > just distribution. > > Entering from the adjacent stream of flow. Remember, the two adjacent > flows, in opposite directions. Molecule will diffuse from one stream of > flow to the other (a basic result of kinetic theory). If they carry, on > average, the momentum of their original flow, and transfer it to the other > flow, that's a transfer of momentum from one flow to the other. I.e., a > force. Viscosity, straight from kinetic theory, as per Maxwell. I am not discussing, nor have I ever proposed, a simple gas. > > They 'could have > > a Boltzmann distribution or something else like Fermi. I did not try > > to define this. > > We know the kinetic theory distribution - the Maxwell distribution. Why > would it be something else? This is, after all, what results from kinetic > theory of particles only interacting through collision. You're still not comprehending the duality if the kinetic forming fluid structures giving rise to more complex fluidic behavior. > > The only initial assumption was force-free elastic > > collisions. > > Not force-free. You mean loss-free? Which would mean no energy transferred > to internal energy of the particles, all energy remaining as > translational KE. > > > > > I shall waste no more time. I will ask you to propose an alternative > > > > way to attain the fluid condition of Continuum Mechanics. > > > > Define "the fluid condition of Continuum Mechanics". What we do know a > > > lot about is when we can use continuum mechanics. For a gas, the most > > > important condition is usually whether or not the mean free path is > > > small. You mentioned mean free path in your "x = cLz = 1.7E-10 kg/m- > > > sec" above. Do you have a value for it (or just the product Lz?)? > > > No, I asked you to propose an alternative method to GET! Continuum > > Mechanics other than a kinetic particulate interacting population. > > Kinetic theory of contact-collision particles get us the continuum > mechanics of simple gases. Why do we need an alternative method? I want you to give me another path to continuum mechanics behavior. > If we want a different continuum mechanics, try a different starting > point. If you want to know how continuum mechanics of liquids and solids > relates to the statistical mechanics of such, there are plenty of good > books available. Do you want some reading suggestions? What I WANT is another foundation FOR Continuum Mechanics. > > As > > for where the values of L, and z come from see my original post, > > >http://groups.google.com/group/sci.physics/msg/515110642eab5644?hl=en.... > > > P = ~5.15E-27 kg-m/sec > > L = ~6.43E-08 meters > > z = ~8.85E-12 kg/m^3 > > Problematic. Note that your suggested L is longer than the wavelength of > waves for which we observe clear wave behaviour. Ah yes, with the kinetic basis the so-called ultra-(sonic/violet) catastrophe is, simply NEVER an option. But gien that for center of mass collisions MFP is meaningless, the the L term most likely does not represent the actual physical spacing (which could be much tighter) but, instead, the viscous mean free path. > > > But this isn't classical, and it isn't what we get with kinetic theory > > > of hard particles, or kinetic theory of classical elastic blobs. > > > Classically, there is no amount of momentum that is too small to be > > > transferred - our momentum states are form a continuum. Further note > > > that for a free quantum particle, our momentum states are also > > > continuous. As far as we know, the only way to get the kind of > > > discrete states we need for superfluidity through unavailability of > > > momentum states is long-range interaction. > > > As I see it, It all in the vortex pairing and quantization of these. > > I said as much in my original posting here. > > Don't assume it; show it! Then you might have something. Without it, it > just looks like unsubstantiated wishful thinking. > > > Perhaps I should ask you, > > what is you goal here? > > I initially commented to correct a factual error. If I'm going to read > some aether theory stuff, I'd rather spend my time on something that's at > least plausible, rather than something wrong at the beginning. That's fine BUT physics is far from complete. Paul Stowe
From: Timo Nieminen on 8 Aug 2010 06:13 On Aug 8, 5:13 am, Paul Stowe <theaether...(a)gmail.com> wrote: > On Aug 2, 12:19 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > What matters is the Reynolds number, which measures the relative > > > > importance of inertia to viscosity. Re > 1, maybe you can assume that > > > > the flow is inviscid. Re << 1, you can ignore inertia. For the same > > > > fluid, just different flows. > > > > Does not Reynolds apply to the boundaries of material systems > > > containing a fluid/gas in a flow situation? > > > It does so apply, and it also applies to other situations. A ratio of > > inertial effects to viscous effects is a general thing. > > > We have > > > Re = rho V L / mu = V L / nu, > > > where rho is the density, V is the speed, mu is the dynamic viscosity, nu > > is the kinematic viscosity, and L is a characteristic length. > > Yes, V is the bulk 'current' speed of the fluid, not the mean speed of > its constitutes... > > > For wave propagation, L = lambda/2 is a good choice, since the speed of > > the fluid will vary from zero to max to zero over this distance. We can > > also choose V = c, the speed of the wave. So, we have > > not really. But, let use the kinetic properties, > > V = c > L = L > mu = cL > > Then R = cL/cL => Unity... Useless... Recall that the Reynolds number compares two macroscopic properties, the momentum density and the effect of viscosity. Why do you expect to get anything meaningful from plugging in microscopic kinetic theory parameters into a formula where they don't apply? > > Re = c lambda / 2 nu. > > > This tells us, more or less, the number of wavelengths a wave will travel > > before being reduced to 1/e of the original intensity. > > > Don't like this rough-and-ready calculation? Get the real deal from Lamb, > > Hydrodynamic, art 359, from which we get for the decay distance (d) in > > wavelengths: > > > d/lambda = (3 / 8 pi^2) c lambda / nu. > > > This is what we have for wave propagation in simple gases. If d/lambda is > > large, and d is large compared to relevant lengths, then we can ignore > > viscosity. If not, we can't. > > I don't know about you, but I have never advocated a 'simple gas'. You started with: "Thus, for such a model we will need quantum entities (axeons) which have the following characteristics, - Of finite size - has momentum (P) and thus, a speed (c) - cannot occupy the same volume at the same time - interacts with each other solely by elastic collisions, period!" which is a simple gas. > I > have always define the aether as a superfluidic vortex lattice of the > Maxwellian variety. Fine. Then don't start with a simple gas from which these properties don't follow. This is important for two main reasons. First, if you start with assumption A, and claim properties B, and we know that properties B do not follow from A, then it all looks like junk, with gross errors in the very first step. This is what you are doing when you start with a simple gas and try to magic up some incompressible superfluid from it. Second, it's misleading as to what the real initial assumptions are. If you really assuming a set of macroscopic behaviours - e.g., an incompressible superfluid - you should do so explicitly. If you're assuming a kinetic theory where the particles have just the right long range interactions to give, e.g., an incompressible superfluid, then you should say so. And you should show that your assumption do in fact give you the properties you desire. Show it, don't wish it! > > This is the bare-naked kinetic theory result, without appealing to any > > special results from elsewhere. > > All known substance are kinetic in nature, the only differences are > how their 'fields' interact & behave. I don't think you really > understand the duality. Material substance would be made up of > fluidic interacting Skyrmions and vortices. These circulation fields > either attract or repel depending upon the velocity gradient at the > shear boundary between them. These structure, in turn, are ALSO! made > up of kinetic entities. It is not, nor have I ever claimed it was, a > simple 'gas'! A simple gas was your starting point. I don't think you really understand how a simple gas behaves. Start reading with Maxwell. > > Do your suggested numbers work? Is the attenuation of waves slow enough to > > be compatible with obervation? If not, back to the drawing board. > > Well there would be a non-zero attenuation coefficient for acoustic > propagation and my estimate is that this attenuation coefficient (s) > is on the order of the mid 10^-26. In other words, > > i = Ie^-sd > > and d is distance in meters... > > This attenuation coefficient IS NOT related to any kinetic mean free > path. Derivation? Is d = (3 / 8 pi^2) c lambda^2 / nu wrong, or somehow not applicable? Why is it wrong or why is it inapplicable? > > > > To take results for, e.g., a superfluid liquid, and to assume that > > > > they apply to a gas of low viscosity is a Bad Thing. > > > > Go look a Maxwell's "On Physical Lines of Force", it is clearly stated > > > there. > > > Where? Maxwell starts off with the mathematical description of the > > behaviour of an elastic solid, and goes with that for a while. > > No, he does not... Read it again, > > http://www.vacuum-physics.com/Maxwell/maxwell_oplf.pdf > > He starts WITH a vortex interaction model, period. He starts off (briefly) discussing (Kelvin's) elastic solid aether, then discusses an aether at the most general level, that of stresses. Then introduces a fluid aether, and considers the flow needed to provide said stresses (as opposed to in an elastic solid aether, where one considers the displacements (i.e., strains) required to provide said stresses). What he doesn't do is claim that the properties of a superfluid liquid apply to a simple kinetic gas. I said he didn't do that, you said it is "clearly stated there." > And he does > explicitly state that he assumes a perfect incompressible fluid. Where? He goes to a great deal of trouble to introduce and discuss his "idler wheels" to avoid frictional/viscous losses. See, e.g., pg 347. Where does he explicitly state the above? > > Where does Maxwell mention Cooper pairs? Why call the vortices "quantum"? > > Because if LOOK! at figure 2 of his illustrations after page 348 > you'll clearly see the regular 'quantized' vortex lattice structure > just like Bernard Cells... Where does Maxwell mention Cooper pairs? Also, that still doesn't explain why you call the vortices "quantum". In particular, "quantum" is used as a technical term in physics, and it can be misleading to apply it willy-nilly. If it's just like Bernard cells, it isn't "quantum". > > > > > I shall waste no more time. I will ask you to propose an alternative > > > > > way to attain the fluid condition of Continuum Mechanics. > > > > > Define "the fluid condition of Continuum Mechanics". What we do know a > > > > lot about is when we can use continuum mechanics. For a gas, the most > > > > important condition is usually whether or not the mean free path is > > > > small. You mentioned mean free path in your "x = cLz = 1.7E-10 kg/m- > > > > sec" above. Do you have a value for it (or just the product Lz?)? > > > > No, I asked you to propose an alternative method to GET! Continuum > > > Mechanics other than a kinetic particulate interacting population. > > > Kinetic theory of contact-collision particles get us the continuum > > mechanics of simple gases. Why do we need an alternative method? > > I want you to give me another path to continuum mechanics behavior. Why? What's wrong with statistical mechanics? It appears to work as a description of reality, and agrees with what we know about both microscopic and macroscopic behaviour. Why do you want me to give you another path? > > If we want a different continuum mechanics, try a different starting > > point. If you want to know how continuum mechanics of liquids and solids > > relates to the statistical mechanics of such, there are plenty of good > > books available. Do you want some reading suggestions? > > What I WANT is another foundation FOR Continuum Mechanics. Again, why? It's been done. The original foundation was to assume that the "continua" we see are really continuous, all the way down. Can start with basic properties such as stress, strain, flow, and density. Connect these with constitutive relations (historically often assumed to linear, but this isn't necessary). Measure experimentally. Done! > > > As > > > for where the values of L, and z come from see my original post, > > > >http://groups.google.com/group/sci.physics/msg/515110642eab5644?hl=en... > > > > P = ~5.15E-27 kg-m/sec > > > L = ~6.43E-08 meters > > > z = ~8.85E-12 kg/m^3 > > > Problematic. Note that your suggested L is longer than the wavelength of > > waves for which we observe clear wave behaviour. > > Ah yes, with the kinetic basis the so-called ultra-(sonic/violet) > catastrophe is, simply NEVER an option. But it is a serious problem for modelling the real world if the high frequency cutoff is lower than frequencies that we see. > But gien that for center of > mass collisions MFP is meaningless, the the L term most likely does > not represent the actual physical spacing (which could be much > tighter) but, instead, the viscous mean free path. MFP is not meaningless for collisions in a real gas. THE COLLISIONS WILL NOT ALWAYS BE LINED UP ALONG THE CENTRES! THERE WILL BE GLANCING COLLISIONS! READ MAXWELL! PLAY ON A POOL TABLE! STOP MAKING THIS GHASTLY ERROR OVER AND OVER AGAIN! > > > > But this isn't classical, and it isn't what we get with kinetic theory > > > > of hard particles, or kinetic theory of classical elastic blobs. > > > > Classically, there is no amount of momentum that is too small to be > > > > transferred - our momentum states are form a continuum. Further note > > > > that for a free quantum particle, our momentum states are also > > > > continuous. As far as we know, the only way to get the kind of > > > > discrete states we need for superfluidity through unavailability of > > > > momentum states is long-range interaction. > > > > As I see it, It all in the vortex pairing and quantization of these. > > > I said as much in my original posting here. > > > Don't assume it; show it! Then you might have something. Without it, it > > just looks like unsubstantiated wishful thinking. > > > > Perhaps I should ask you, > > > what is you goal here? > > > I initially commented to correct a factual error. If I'm going to read > > some aether theory stuff, I'd rather spend my time on something that's at > > least plausible, rather than something wrong at the beginning. > > That's fine BUT physics is far from complete. -- Timo
From: harald on 8 Aug 2010 13:44 On Jul 29, 2:29 am, Paul Stowe <theaether...(a)gmail.com> wrote: > On Jul 28, 1:41 am, harald <h...(a)swissonline.ch> wrote: > > > > > On Jul 25, 9:59 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > OVERVIEW > > > > What is Neo-Aether Theory? I identify Neo-Aether ss the so-called > > > classic aether model adapted and integrated to accomidate the > > > observations and experimental evidence garnered over the last > > > century. In other words, the model is explicitly demonstrated to be > > > compatible with and, in many cases, leads to, such concepts as Local > > > Lorentz Invariance, Planck's Constant, quntum elemental charge, > > > Newton's laws of motion, basic quantum nature, the uncertainty > > > principle, ... etc. Aether theory, especially this modern > > > interpretation is a 'bottoms up' approach to science, that is to say, > > > on starts with the a basic kinetic quantum entity model and builds up > > > all else from that. It truly is, the ultimate in simplicity... at its > > > base. See, > > > >http://www.archive.org/details/historyoftheorie00whitrich > > > for an excellent detailed presentation of the development of the > > > theory through circa ~ 1910. > > > Thanks for the link! > > > > So, now let's start at the bottom and build a universe... First let's > > > define the necessary fundamentals of this type of model. Aether is a > > > energetic substance, fluidic in nature. To my knowledge, there is > > > only one way to get such a medium, by kinetic theory. Thus, for such > > > a model we will need quantum entities (axeons) which have the > > > following characteristics, > > > > - Of finite size > > > - has momentum (P) > > > Sorry but I find such a model unsatisfying: IMHO inertia (with the > > resulting momentum) should be *caused* by a good ether model as an > > emerging property. Else you are adding a probably useless layer of > > complexity, for it implies a substratum under your ether that gives it > > the property of inertia. > > > Regards, > > Harald > > I'm confused Harald. For our universe momentum must be a > fundamental. The model proposed has this and, does not have inertia > at the level of the axeons. I then, specifically, show how inertia is > emergent from a -dE/dt response of speed changes. Inertial mass > (ponderable if you will) is a direct result of this, and this only. > The physical properties of momentum and energy cannot appear from > nothing however. > > Regards, > > Paul Stowe Dear Paul, OK... does your ether model constitute a Newtonian absolute space, so that "straight" motion is defined as relative to the ether? Does it work without such a background for the motions of your ether particles to relate to? Harald
From: Paul Stowe on 8 Aug 2010 16:12 On Aug 8, 10:44 am, harald <h...(a)swissonline.ch> wrote: > On Jul 29, 2:29 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > On Jul 28, 1:41 am, harald <h...(a)swissonline.ch> wrote: > > > > On Jul 25, 9:59 pm, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > OVERVIEW > > > > > What is Neo-Aether Theory? I identify Neo-Aether ss the so-called > > > > classic aether model adapted and integrated to accomidate the > > > > observations and experimental evidence garnered over the last > > > > century. In other words, the model is explicitly demonstrated to be > > > > compatible with and, in many cases, leads to, such concepts as Local > > > > Lorentz Invariance, Planck's Constant, quntum elemental charge, > > > > Newton's laws of motion, basic quantum nature, the uncertainty > > > > principle, ... etc. Aether theory, especially this modern > > > > interpretation is a 'bottoms up' approach to science, that is to say, > > > > on starts with the a basic kinetic quantum entity model and builds up > > > > all else from that. It truly is, the ultimate in simplicity... at its > > > > base. See, > > > > >http://www.archive.org/details/historyoftheorie00whitrich > > > > for an excellent detailed presentation of the development of the > > > > theory through circa ~ 1910. > > > > Thanks for the link! > > > > > So, now let's start at the bottom and build a universe... First let's > > > > define the necessary fundamentals of this type of model. Aether is a > > > > energetic substance, fluidic in nature. To my knowledge, there is > > > > only one way to get such a medium, by kinetic theory. Thus, for such > > > > a model we will need quantum entities (axeons) which have the > > > > following characteristics, > > > > > - Of finite size > > > > - has momentum (P) > > > > Sorry but I find such a model unsatisfying: IMHO inertia (with the > > > resulting momentum) should be *caused* by a good ether model as an > > > emerging property. Else you are adding a probably useless layer of > > > complexity, for it implies a substratum under your ether that gives it > > > the property of inertia. > > > > Regards, > > > Harald > > > I'm confused Harald. For our universe momentum must be a > > fundamental. The model proposed has this and, does not have inertia > > at the level of the axeons. I then, specifically, show how inertia is > > emergent from a -dE/dt response of speed changes. Inertial mass > > (ponderable if you will) is a direct result of this, and this only. > > The physical properties of momentum and energy cannot appear from > > nothing however. > > > Regards, > > > Paul Stowe > > Dear Paul, > > OK... does your ether model constitute a Newtonian absolute space, so > that "straight" motion is defined as relative to the ether? Does it > work without such a background for the motions of your ether particles > to relate to? > > Harald The void containing the axeons (and thus, our universe) can be considered Newton's concept of absolute space. It has only the property of volume (3-D). No OTHER physical qualities apply to it. One can impose upon the void any 'geometry' they wish to conceive as long as that geometry is compatible with a 3-D void and mathematically internally consistent. In the simplest kinetic model the axeons within the void only move in straight lines thus Euclidean geometry seems appropriate at this level If the axeons have no fields or surface 'roughness' (friction) they cannot attain any spin (English) during self interactions. So, relative to the void only straight linear motion applies. However, Helmholtz's Theorems applies to the bulk 'macroscopic' behavior of the overall population (aetherial medium). That carries with it vorticity, pressure, kinematic viscosity, ... etc. Therefore the geometry which emerges from the bulk behavior of a vortex dominated system is not simply linear, thus describing it physical behavior will not be simply Eucidean. There is a duality, the kinetic media (or field) and fluidic structures that arise from it. These structures interact with each other, forming a secondary pseudo-particulate medium (the vortex lattice medium). It is at this level to which we ascribe, observe, and measure the physical behaviors. Regards, Paul Stowe
From: franklinhu on 10 Aug 2010 16:38
On Jul 28, 1:41 am, harald <h...(a)swissonline.ch> wrote: > On Jul 25, 9:59 pm, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > > > OVERVIEW > > > What is Neo-Aether Theory? I identify Neo-Aether ss the so-called > > classic aether model adapted and integrated to accomidate the > > observations and experimental evidence garnered over the last > > century. In other words, the model is explicitly demonstrated to be > > compatible with and, in many cases, leads to, such concepts as Local > > Lorentz Invariance, Planck's Constant, quntum elemental charge, > > Newton's laws of motion, basic quantum nature, the uncertainty > > principle, ... etc. Aether theory, especially this modern > > interpretation is a 'bottoms up' approach to science, that is to say, > > on starts with the a basic kinetic quantum entity model and builds up > > all else from that. It truly is, the ultimate in simplicity... at its > > base. See, > > >http://www.archive.org/details/historyoftheorie00whitrich > > for an excellent detailed presentation of the development of the > > theory through circa ~ 1910. > > Thanks for the link! > > > So, now let's start at the bottom and build a universe... First let's > > define the necessary fundamentals of this type of model. Aether is a > > energetic substance, fluidic in nature. To my knowledge, there is > > only one way to get such a medium, by kinetic theory. Thus, for such > > a model we will need quantum entities (axeons) which have the > > following characteristics, > > > - Of finite size > > - has momentum (P) > > Sorry but I find such a model unsatisfying: IMHO inertia (with the > resulting momentum) should be *caused* by a good ether model as an > emerging property. Else you are adding a probably useless layer of > complexity, for it implies a substratum under your ether that gives it > the property of inertia. > > Regards, > Harald > > [..]- Hide quoted text - > > - Show quoted text - Absolutely correct, inertia should be explained in terms of the aether. I consider the aether to be composed of "poselectrons" which are nothing more than positron/electron dipole pairs. It is the attraction of these dipoles that create the inertia effect by storing and releasing the energy given to objects moving through the field. My aether also explains how the magnetic force is mediated and how "charge" creates attractive forces through phased wave interactions and how objects get "mass" by interacting with the electrostatic fields of the aehter. A "good" aether theory ought to explain all of these phenomenon under the same conceptual framework. See the bottom of this page: http://franklinhu.com/theory.html |