What is the Aether?
in [Relativity]
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From: Greg Neill on 19 Jun 2007 16:04 "Florian" <firstname(a)lastname.net> wrote in message news:1hzyjoi.thrw70pivls0N%firstname(a)lastname.net... > Greg Neill <gneillREM(a)OVEsympatico.ca> wrote: > > > > Could be a swirl or a stationnary wave, anything that is not random. > > > > To the limits of experimental precision, electrons are > > point particles with no structure. > > Electrons are not point particles. They are certainly stationary waves. Strange, because experiments designed to measure their diameter show them to be indistinguishible from point particles. Note that this is not the same as determining their position in, say, an atomic orbital, or their "fuzzy" position vs velocity relationship due to the uncertainty principle. > > > > They don't fragment > > into smaller bits when they are collided (although their > > energy of motion can spawn new particle pairs). How > > does charge conservation work? Which aetheron carries > > the electric charge of the electron (or proton)? > > Etherons would not have charge or mass or spin. Those properties would > be macroscopic properties derived from the motion of etherons that form > the particle. How are these properties then quantized and conserved? > > > > > If electrons are a collection of swirly bits, they should > > be decomposable into smaller bits. And you should be able > > to add more swirly bits; there should be a whole > > continuum of masses of particles, but there isn't. > > There would be a continuum only if intermediate state are stable. > Obviously, experiments tell us that intermediate state are not stable. > That's why there are quanta. Quanta are bit of energy that can be > absorbed by a particle to jump from one stable level to the next one. Nah. If aether is absorbed continuously then there must be a continuum of masses for every particle. Otherwise you'd have to propose that the aether simply disappears from existence as it is absorbed. And if you did that, you'd have no way to make new particles during collisions. I think your theory has a problem. Also, we can measure events and reactions down to time intervals of about a femtosecond with pulsed laser technology. Plenty of very short unstable states have been "seen" in atomic processes. > > > > > No force. Collisions like any wave. Right? > > > > There no such thing as a true collision -- all interactions > > are mediated by forces. Atoms don't touch, their fields > > do. > > At a particle scale, you can certainly consider that etherons make > collisions. That's an approximation of course. But there is no need to > complicate the picture trying to figure out what etherons are. Let's > proceed sequentially. You need a theory of their interaction in order to proceed. What could it mean for particles to actually touch? We have no examples from experimental or theoretical physics. > > > Waves carrying energy usually transmit their energy via > > the electromagnetic force. > > Right. How does a wave transmit energy to particle though > electromagnetic force? > In other words: what is that energy? What is an electromagnetic force? > What is a particle? How do they interact? ;-) Maxwell's equations. Photons. It's all in the standard models. Photons carry momentum and spin. > > > Why would I avoid relativity? what's wrong with relativity? > > > > With relativity there is no need for an aether. > > There is no need for ether but at the same time it does not rule it out. Well, invisible massless pink elephants are also not ruled out by theory. Do we need them? > > Sometimes, the simpler model is not the best one. How so? If the simple model makes accurate predictions under all circumstances, why would a more complex model be better? It wouldn't predict anything different. > > > > > There is repulsion in GR? > > > > Pressure causes repulsion. Also, the cosmological > > constant of General Relativity introduces a repulsive > > component -- > > Which is a mathematical trick to make it work. Otherwise, GR would imply > that the universe would collapse. It's just mathematics. Calling it a "trick" doesn't mean anything. The form of the equations of General Relativity are in fact the simplest solutions of the differential equations, and that constant is part of the general solution (which is why Einstein originally included it in the initial derivation). > > > you've heard of the recently discovered > > acceleration of the expansion of the universe? The > > so-called "Dark Energy" is its manifestation. > > I know. we need to invent more trick to make it work. At some point > there would be so many tricks (MOND theory, dark matter, dark energy) > that it will certainly be time to find a better model. Right? > I wonder if the physicist who will find a better model is born yet? It's not "inventing more trick" if the mathematics is already in place in the current theory. MOND is an attempt to ignore theory and employ a curve-fit model instead. Dark matter is not a "trick" in that it is an empirical necessity. > > A side question. You mentioned the expansion of the universe. Do you > think that the Big Bang theory is definitively proven? In science no theory is ever definitively proven. All we can do is keep testing our current best theories against reality to see if and where they are deficient and change or modify them as required. Suppose it were shown that General Relativity or The Big Bang were incorrect in some way. A new theory to replace them would have to not only fix the 'wrong' bit, but also make all the currently verified predictions of the theory it replaces. Now, GR and BB have been testied in a *lot* of scenarios. GR in particular makes certain predictions that have been experimentally verified to about 14 decimal places. So any replacement for GR is going to look an *awful* lot like GR -- so don't expect its tensor calculus to be suddenly replaced with high school algebra. > > > > Of course it does! We're discussing concepts, are we? Nothing is frozen. > > > I remark that you don't make no suggestion yourself? Don't you have > > > ideas for a concept? > > > > I'm happy with (most of) the standard model, including > > Special and General Relativity and no aether. I also > > think that the standard model of particle physics is > > doing pretty well. > > I'm sure you don't like status quo and wouldn't mind making some > progress? How do you define progress in a scientific model? What if it were to turn out that GR is correct? (It's probably not, but it may be damned close). What kind of progress would you like to see? > > > > > Because the density of the flow would depend on the mass. If the mass is > > > large enough, most etherons will continue to wander. > > > > I don't understand that. Aren't they attracted to mass? > > Otherwise why would you say that their flow is radial to > > mass? > > They would be globally attracted to a mass like the particle of a gas > would be attracted by a low pressure center. A lower pressure attract a > higher amount of particles. You can't say that the particle of the gas > all move toward the low pressure, right? What maintains a low pressure of aether around a mass? In a gas, a pressure imbalance is quickly equilibrated unless some work is being doe to maintain it. > > > > "It" means all etherons? Because the strength of the flow is > > > insufficient? > > > > I should think that the "strength of the flow", whatever > > that means, would be irrelevant for a finite volume if > > the particles move at c. > > If the etherons move at c, would that imply that ether moves at c? It would mean that, like a photon bath in a closed container, it would soon be absorbed by the container and "darkness" would prevail in a very, very short time. > > > > > How could a box be impermeable to etherons? > > > > You said that they move radially into mass. The > > box is composed of matter (atoms) to which all > > aetherons must therefore flow radially. One can > > easily posit a box where the sides are thus > > opaque to aetherons, simply by assuring that the > > thickness of the sides is several times the mean > > free path length of the aetherons given the > > cross sectional area of the atoms comprising it and > > their lattice spacing. > > You assume that all etherons would flow toward the atoms. See above. I need only assume that they are absorbed by matter. Then, any path from outside to inside could be blocked by an intervening atom. > > Second, can you make a box made of water that can prevent water to go > inside that box? ;-) Sure. Ice. > > Third, let's imagine that you find a way to block etherons from going > inside the box. If the global movement of etherons is dictated by low > pressure, would an equilibrium soon be reached when the amount of > etheron inside the box decreases. Is that a question or a statment? If aetherons move at c and are absorbed by matter then they cannot exist for long as free particles in a closed matter container. What is your model for how pressure, well let's call it density, dictates the flow of particles in a gas? Are you imagining some kind of force pulling the particles into the low density region?
From: FrediFizzx on 19 Jun 2007 17:51 "Richard Schultz" <schultr(a)mail.biu.ack.il> wrote in message news:f58eq3$bra$1(a)news.iucc.ac.il... > In sci.physics.particle FrediFizzx <fredifizzx(a)hotmail.com> wrote: > : "Richard Schultz" <schultr(a)mail.biu.ack.il> wrote in message > : news:f55he2$t0e$2(a)news.iucc.ac.il... > > :> Or, as the author of our textbook put it, > :> > :> For a treatment of the fundamental physics of fields and > :> matter, [the MKS system] has one basic defect. Maxwell's > :> equations for the vacuum fields, in this system, are > :> symmetrical in the electric and magnetic field only if H, > :> not B, appears in the role of the magnetic field. . . . > :> On the other hand. . . B, not H, is the fundamental > :> magnetic field inside matter. This is not a matter of > :> definition or of units, but a fact of nature, reflecting > :> the absence of magnetic charge. Thus the MKS system, as > :> it has been constructed, tends to obscure either the > :> fundamental electromagnetic symmetry of the vacuum, or > :> the essential asymmetry of the sources. > : > : This is perhaps as bad as John thinks only from the other direction. > : Any consistent system of units CANNOT change physics and CANNOT have > : "defects". SI and CGS are fully consistent systems of units. I > really > : doubt that there is any problem in physics that couldn't be solved > : properly using either one. > > This is what the author of the textbook was saying. He was defending > his > choice to use CGS units. He felt, as he explained, that when CGS > units are > used, it is easier to see the symmetries of the underlying physics, > not > that MKS units cannot be used to solve physics problems -- in fact, in > the same appendix that I quoted above, he explicitly says that MKS > units > are more convenient for solving engineering problems. He didn't need to "defend" his choice; he only needed to state that he was using CGS. Yes, a particular system of units has it advantages and disadvantages depending on what kind of problem is being solved. > :> But then again, the author of the above was only a Nobel Prize > winner > :> in physics, and hence can't really be expected to know what he was > :> talking about. > : > : Well, who? Nobel Prize winners are not always right. > > I would guess that they are right about elementary aspects of their > fields more often than not. In this case, the textbook from which I > quoted was written by the appropriately named Edward M. "EM" Purcell. Haven't read it but I hear that Purcell's textbook is a decent one. Parts of what he says about this is OK. I suppose the main argument is whether or not the "vacuum" is polarizable. SI'ers are convinced that CGS buries "vacuum" polarization when it doesn't necessarily do so. If the "vacuum" is polarizable, then we will have "vacuum" capacitance and "vacuum" inductance. In order to have those, there must be "vacuum" charge. Which can be expressed as "Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be all bound. SI's statement c^2 = 1/(eps0 mu0), is merely shorthand for c^2 = lambda^2/(4pi^2 Cvac Lvac), which would be true in any system of units if the "vacuum" is polarizable. So it is pretty easy to see that it doesn't really matter which unit system is used. Best, Fred Diether Moderator sci.physics.foundations
From: John C. Polasek on 19 Jun 2007 21:37 On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx" <fredifizzx(a)hotmail.com> wrote: >"Richard Schultz" <schultr(a)mail.biu.ack.il> wrote in message >news:f58eq3$bra$1(a)news.iucc.ac.il... >> In sci.physics.particle FrediFizzx <fredifizzx(a)hotmail.com> wrote: >> : "Richard Schultz" <schultr(a)mail.biu.ack.il> wrote in message >> : news:f55he2$t0e$2(a)news.iucc.ac.il... >> >> :> Or, as the author of our textbook put it, >> :> >> :> For a treatment of the fundamental physics of fields and >> :> matter, [the MKS system] has one basic defect. Maxwell's >> :> equations for the vacuum fields, in this system, are >> :> symmetrical in the electric and magnetic field only if H, >> :> not B, appears in the role of the magnetic field. . . . >> :> On the other hand. . . B, not H, is the fundamental >> :> magnetic field inside matter. This is not a matter of >> :> definition or of units, but a fact of nature, reflecting >> :> the absence of magnetic charge. Thus the MKS system, as >> :> it has been constructed, tends to obscure either the >> :> fundamental electromagnetic symmetry of the vacuum, or >> :> the essential asymmetry of the sources. >> : >> : This is perhaps as bad as John thinks only from the other direction. >> : Any consistent system of units CANNOT change physics and CANNOT have >> : "defects". SI and CGS are fully consistent systems of units. I >> really >> : doubt that there is any problem in physics that couldn't be solved >> : properly using either one. >> >> This is what the author of the textbook was saying. He was defending >> his >> choice to use CGS units. He felt, as he explained, that when CGS >> units are >> used, it is easier to see the symmetries of the underlying physics, >> not >> that MKS units cannot be used to solve physics problems -- in fact, in >> the same appendix that I quoted above, he explicitly says that MKS >> units >> are more convenient for solving engineering problems. > >He didn't need to "defend" his choice; he only needed to state that he >was using CGS. Yes, a particular system of units has it advantages and >disadvantages depending on what kind of problem is being solved. > >> :> But then again, the author of the above was only a Nobel Prize >> winner >> :> in physics, and hence can't really be expected to know what he was >> :> talking about. >> : >> : Well, who? Nobel Prize winners are not always right. >> >> I would guess that they are right about elementary aspects of their >> fields more often than not. In this case, the textbook from which I >> quoted was written by the appropriately named Edward M. "EM" Purcell. > >Haven't read it but I hear that Purcell's textbook is a decent one. >Parts of what he says about this is OK. > >I suppose the main argument is whether or not the "vacuum" is >polarizable. SI'ers are convinced that CGS buries "vacuum" polarization >when it doesn't necessarily do so. If the "vacuum" is polarizable, then >we will have "vacuum" capacitance and "vacuum" inductance. In order to >have those, there must be "vacuum" charge. Which can be expressed as >"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be all >bound. SI's statement > >c^2 = 1/(eps0 mu0), > >is merely shorthand for > >c^2 = lambda^2/(4pi^2 Cvac Lvac), It looks like you are assigning lump constants Cvac and Lvac to a continuum. I have lost track of your derivation. Is lambda the Compton wavelength? >which would be true in any system of units if the "vacuum" is >polarizable. > >So it is pretty easy to see that it doesn't really matter which unit >system is used. > >Best, > >Fred Diether >Moderator sci.physics.foundations Well, I'm distressed to see you chose not to include any of my arguments, but I will repeat one of them. With cgs ignoring eps0 and mu0, cgs is reduced to such critical equations as D = E and B = H which contain absolutely no information of any value and in fact are highly misleading. As a practical test, given a field intensity of 1 million volts/meter calculate D and H D = E*eps0 = 10^6V/m*eps0 = 8.857e-6 coulombs/meter^2 Similarly in free space the same electric field can determine H: H = E/Z = 10^6V/m/377 ohms = 2.6e3 Amp/m How would you write these equations in cgs? John Polasek
From: FrediFizzx on 20 Jun 2007 01:15 "John C. Polasek" <jpolasek(a)cfl.rr.com> wrote in message news:l20h7359ctfot73g0qeng6mg9d1h3e28te(a)4ax.com... > On Tue, 19 Jun 2007 14:51:53 -0700, "FrediFizzx" > <fredifizzx(a)hotmail.com> wrote: >>I suppose the main argument is whether or not the "vacuum" is >>polarizable. SI'ers are convinced that CGS buries "vacuum" >>polarization >>when it doesn't necessarily do so. If the "vacuum" is polarizable, >>then >>we will have "vacuum" capacitance and "vacuum" inductance. In order >>to >>have those, there must be "vacuum" charge. Which can be expressed as >>"Quantum Vacuum Charge" = +,- sqrt(hbar*c) in CGS and which must be >>all >>bound. SI's statement >> >>c^2 = 1/(eps0 mu0), >> >>is merely shorthand for >> >>c^2 = lambda^2/(4pi^2 Cvac Lvac), > > It looks like you are assigning lump constants Cvac and Lvac to a > continuum. I have lost track of your derivation. Is lambda the Compton > wavelength? What does c = 1/sqrt(eps0 mu0) represent? Lambda is the wavelength of the EM radiation of interest relative to a detector. >>which would be true in any system of units if the "vacuum" is >>polarizable. >> >>So it is pretty easy to see that it doesn't really matter which unit >>system is used. > Well, I'm distressed to see you chose not to include any of my > arguments, but I will repeat one of them. With cgs ignoring eps0 and > mu0, cgs is reduced to such critical equations as > D = E and B = H > which contain absolutely no information of any value and in fact are > highly misleading. > As a practical test, given a field intensity of 1 million volts/meter > calculate D and H > > D = E*eps0 = 10^6V/m*eps0 = 8.857e-6 coulombs/meter^2 > Similarly in free space the same electric field can determine H: > H = E/Z = 10^6V/m/377 ohms = 2.6e3 Amp/m > How would you write these equations in cgs? Why do you think it is important to distinguish D and E or B and H for free space EM radiation? I don't think it really matters. I would rather use Cvac and Lvac anywise since they would be valid for any system of units. Best, Fred Diether Moderator sci.physics.foundations
From: Florian on 20 Jun 2007 06:08
Greg Neill <gneillREM(a)OVEsympatico.ca> wrote: > Strange, because experiments designed to measure their diameter show them > to be indistinguishible from point particles. Note that this is not the > same as determining their position in, say, an atomic orbital, or their > "fuzzy" position vs velocity relationship due to the uncertainty > principle. Measure the diameter of a stationary wave? What a strange idea. ing the Measurwavelength is certainly a much better idea. > > Etherons would not have charge or mass or spin. Those properties would > > be macroscopic properties derived from the motion of etherons that form > > the particle. > > How are these properties then quantized and conserved? Quantized means there are stable level for the stationary wave. Conserved means that collisions of etherons are elastic. > Nah. If aether is absorbed continuously then there must be a continuum of > masses for every particle. Nope. Particles couldn't absorbed etherons in motion, unless the absorbtion is sufficient to reach the next stable level of the particle. > Otherwise you'd have to propose that the aether simply disappears from > existence as it is absorbed. And if you did that, you'd have no way to > make new particles during collisions. Still looking for good ideas here. You're welcome to share yours. A possibilty would be that particles absorbes ether in quanta, i. e., get excited, and at some point, they become unstable so that they disintegrate into more particles? > I think your theory has a problem. Sure, that's not even a theory, just thoughts. > Also, we can measure events and reactions down to time intervals of about > a femtosecond with pulsed laser technology. Plenty of very short unstable > states have been "seen" in atomic processes. femtosecond? Light has the time to travel 300 nm during 1 femtosecond. I'm afraid that's way too slow. > > At a particle scale, you can certainly consider that etherons make > > collisions. That's an approximation of course. But there is no need to > > complicate the picture trying to figure out what etherons are. Let's > > proceed sequentially. > > You need a theory of their interaction in order to proceed. > What could it mean for particles to actually touch? We have no examples > from experimental or theoretical physics. The use you make of the term "particle" is confusing. I reserve the term particle for matter, i. e., stationary wave in ether = etherons in organized motion. > > > Waves carrying energy usually transmit their energy via the > > > electromagnetic force. > > > > Right. How does a wave transmit energy to particle though > > electromagnetic force? In other words: what is that energy? What is an > > electromagnetic force? What is a particle? How do they interact? ;-) > > Maxwell's equations. Photons. It's all in the standard models. Photons > carry momentum and spin. That's the macroscopic description. Tell me about the microscopic description. A photon or an electromagnetic field, what is it made of? > Well, invisible massless pink elephants are also not ruled out by theory. > Do we need them? You can perfectly describe a wave in the ocean without considering the existence of water molecules. But would those wave exists without water molecules? Of course not. See, it's not different for ether. > > Sometimes, the simpler model is not the best one. > > How so? If the simple model makes accurate predictions under all > circumstances, why would a more complex model be better? It wouldn't > predict anything different. GR does not make accurate predictions under all circumstances. e. g., what are the prediction of GR inside a black hole? Why doesn't it predict the rotation curve of galaxies? > It's not "inventing more trick" if the mathematics is already in place in > the current theory. MOND is an attempt to ignore theory and employ a > curve-fit model instead. MOND is the demonstration that the current theory is innaccurate in some circumstances, much like newton gravitation law was in some circumstances. > Dark matter is not a "trick" in that it is an empirical necessity. Dark matter is an empirical necessity to support the theory, but has not been empirically identified. Unless it is found, it means that the theory does not fit the observations. Time to get a more accurate one. > > A side question. You mentioned the expansion of the universe. Do you > > think that the Big Bang theory is definitively proven? > > In science no theory is ever definitively proven. All we can do is keep > testing our current best theories against reality to see if and where they > are deficient and change or modify them as required. > > Suppose it were shown that General Relativity or The Big Bang were > incorrect in some way. Not incorrect, but inaccurate in some context. > A new theory to replace them would have to not only fix the 'wrong' bit, > but also make all the currently verified predictions of the theory it > replaces. Now, GR and BB have been testied in a *lot* of scenarios. GR > in particular makes certain predictions that have been experimentally > verified to about 14 decimal places. So any replacement for GR is going to > look an *awful* lot like GR -- so don't expect its tensor calculus to be > suddenly replaced with high school algebra. Of course not. But you're conscient that the model of the new theory has to me a little bit more complex than the previous one. Ether could certainly be that complication. > > I'm sure you don't like status quo and wouldn't mind making some > > progress? > > How do you define progress in a scientific model? improving accuracy. > What if it were to turn out that GR is correct? (It's probably not, but > it may be damned close). GR is correct in a specific context. It may not be correct in all contexts. > What kind of progress would you like to see? A theory that would provide a more accurate description of nature :-) > > They would be globally attracted to a mass like the particle of a gas > > would be attracted by a low pressure center. A lower pressure attract a > > higher amount of particles. You can't say that the particle of the gas > > all move toward the low pressure, right? > > What maintains a low pressure of aether around a mass? In a gas, a > pressure imbalance is quickly equilibrated unless some work is being doe > to maintain it. That's were I'm struggling. I can't figure out why the stationary wave would correspond to a depression in ether. Eventually, the etherons in the stationary wave occupy less space? > It would mean that, like a photon bath in a closed container, it would > soon be absorbed by the container and "darkness" would prevail in a very, > very short time. Photons are wave in a medium. Can you compare the behaviour of the wave and the medium ? > > Second, can you make a box made of water that can prevent water to go > > inside that box? ;-) > > Sure. Ice. Ice is a different phase of water. What about liquid water? Can you make a box made of liquid water that can prevent liquid water to go inside that box? > > > > > Third, let's imagine that you find a way to block etherons from going > > inside the box. If the global movement of etherons is dictated by low > > pressure, would an equilibrium soon be reached when the amount of > > etheron inside the box decreases. > > Is that a question or a statment? Question, see below. > If aetherons move at c and are absorbed by matter then > they cannot exist for long as free particles in a > closed matter container. If all etherons were absorbed by matter, that would mean that matter correspond to a zero pressure of ether. If matter is constituted by ether, is it possible that it does correspond to zero pressure of ether? > What is your model for how pressure, well let's call it density, dictates > the flow of particles in a gas? Are you imagining some kind of force > pulling the particles into the low density region? I imagine nothing special. The particles of gas will tend to fill the whole volume of a close container because they collide and propagate to all direction. But you know that. sidenote: There are plenty of simulation on the web that show this. For example: http://www.falstad.com/gas/ Set gravity to zero and volume to minimum. Then increase the volume to maximum. Funny isn't it ;-) -- Florian "Tout est au mieux dans le meilleur des mondes possibles" Voltaire vs Leibniz (1-0) |