Neo-Aether and the Medium of Space
in [Physics]
|
Prev: Very simple question on a REDOX reaction
Next: REDOX: How can this occur if an ion can't lose anymore electrons?
From: Timo Nieminen on 30 Jul 2010 01:39 On Thu, 29 Jul 2010, Paul Stowe wrote: > On Jul 29, 3:56 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Jul 29, 11:04 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > > > On Jul 28, 1:36 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > Well, yes, you didn't specify hard spheres. Other hard shapes won't > > > > give you superfluidity. I suspect that blobs won't either. Don't > > > > assume that they will give you superfluidity; show that they will, if > > > > you're going to depend on it. > > > > > What fundamental physical properties do you seem to think are missing > > > for superfluidity??? > > > > That's really the wrong question, on a par with showing a pile of > > scrap metal and asking why it isn't a watch. > > NOT even close... But I cannot address your objection(s) if you do > not articulate and quantify them. > > > Take two streams of gas flowing anti-parallel, next to each other. A > > bit like this: > > > > ----------------------------------------------------- > > flow ---> > > ----------------------------------------------------- > > <--- flow > > ----------------------------------------------------- > > OK... > > > For superfluidity, you must have no force acting between the flows. > > Yes... > > > That is, there must be no momentum transfer between the flows. > > Not so... There must be no net change in momentum directionality. Force _is_ momentum transfer. See Newton 2. > > For a kinetic gas at a finite temperature, there will be diffusion > > from one flow into the other. This will transfer momentum, so there > > will be a force. > > What you describe above is viscosity, honest, it is... Yes, it is. That's the point. From kinetic theory, we get viscosity. Specifically, we get viscosity, not superfluidity. > > If the particles making up the flows are little hard objects, little > > elastic objects, little blobby objects, with only contact forces, they > > will diffuse from one flow into the other, and the diffused particles > > will collide with the particles in their new flow. See Maxwell. > > Ummm, not so. By definition, these particles undergo center point > perfectly elastic collisions. Not by your original definition. First, note that you explicitly pointed out that you weren't restricting the shapes to spherical. Second, in general, the velocity will not be always parallel to the line joining the centres. Third, are you adding the requirement that the particles are all of identical mass? > While particles exchange directions the > momentum along the direction of travel NEVER! changes. These vector > lines are invariant. It does not matter if there is one or one > trillion, the magnitude of each vector line in the system is eternally > unchanging. Fourth, consider a collision of two frictionless hard elastic spherical particles of radius a, moving towards each other at initial speed v, along paths sqrt(2)*a apart. So, when they collide, the impulse is at 45 degrees to the initial paths. Final component of speed along the original path is v-dv, final component normal to this path is dv. From conservation of energy, dv=v. After the collision, there is no motion in the original direction of travel. For smaller or larger impact parameters (i.e., the distance between the original paths), this will be different. Explore the possibilities on a billiard table. You need to distinguish between the direction of travel, and the direction of a line joining the two centres at the moment of collision. > This leads to one of Helmholtz's theorem's, namely, > > "In the absence of rotational EXTERNAL forces, a fluid that is > initially irrotational remains irrotational." > > http://en.wikipedia.org/wiki/Helmholtz%27s_theorems > > Thus, the very definition OF inviscid... Fifth, it isn't the very definition of inviscid. Helmholtz's theorems (of fluid flow) only apply in inviscid fluids, but that's a long way from them being a definition, let alone the very definition, of inviscid. Sixth, are you saying that Maxwell was wrong? You say you have Maxwell kinetic theory paper on the viscosity of gases. Where was he wrong? Basically, your starting point, kinetic theory of bodies where no energy is transferred to internal energy, rotation, or other forms of energy - all of the energy remains as translations KE, as per the usual understanding of "elastic" in this context - is very closely approximated by a real mono-atomic gas when the temperature is such that the atoms are (very probably) all in their (electronic) ground state. For example, this is the case for helium, argon, etc., for temperatures that aren't too high. Note very well that such gases are not superfluid until quantum effects - deviation of the behaviour of the atoms from simple kinetic behaviour - become important. Before you waste any more time claiming repeatedly that such gases should be superfluids, sit down and read Maxwell's paper. If you think that Maxwell was right, then stop disagreeing with him. If you think that Maxwell was wrong, then go ahead and re-write kinetic theory. > > How to stop this? If you have the right kind of long range forces, you > > can. Not just any long-range forces, but special ones (don't expect > > superfluidity from Newton's gas theory!). Maybe if the particles don't > > collide with each other? > > In a 'perfect' inviscid fluid it is as if they don't, as you should > come to realized if you work through several actual collisions. See > also Feynman's lecture Vol II chapter 40. Superfluidity does not > require forces or fields, just a perfect fluid. A "perfect" fluid which doesn't result from kinetic theory. Again, see Maxwell. See experimental results for mono-atomic gases. Work through several actual collisions yourself. For identical particles, momenta along the line of collision (_not_ the line of travel) exchange, momenta normal to the line of collision remain unchanged. The one collision worked out above is a sufficient counterexample, really, but working through a sequence of collisions with varying directions of the line of collision relative to the direction of travel might be usefully instructive. Feynman II-40 isn't relevant. Feynman I-39 (on kinetic theory) and succeeding chapters are relevant, but iirc, Feynman doesn't do the microscopic origin of viscosity of gases. Feynman I-40-6 is very relevant to superfluidity. When Feynman's "freezing out" happens for linear motion at low temperature, then you get superfluid helium. Note well that such "freezing out" isn't a result of kinetic theory. -- Timo
From: Paul Stowe on 31 Jul 2010 12:49 On Jul 29, 10:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Thu, 29 Jul 2010, PaulStowewrote: > > On Jul 29, 3:56 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Jul 29, 11:04 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > On Jul 28, 1:36 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > Well, yes, you didn't specify hard spheres. Other hard shapes won't > > > > > give you superfluidity. I suspect that blobs won't either. Don't > > > > > assume that they will give you superfluidity; show that they will, if > > > > > you're going to depend on it. > > > > > What fundamental physical properties do you seem to think are missing > > > > for superfluidity??? > > > > That's really the wrong question, on a par with showing a pile of > > > scrap metal and asking why it isn't a watch. > > > NOT even close... But I cannot address your objection(s) if you do > > not articulate and quantify them. > > > > Take two streams of gas flowing anti-parallel, next to each other. A > > > bit like this: > > > > ----------------------------------------------------- > > > flow ---> > > > ----------------------------------------------------- > > > <--- flow > > > ----------------------------------------------------- > > > OK... > > > > For superfluidity, you must have no force acting between the flows. > > > Yes... > > > > That is, there must be no momentum transfer between the flows. > > > Not so... There must be no net change in momentum directionality. > > Force _is_ momentum transfer. See Newton 2. Yes, force is Newton's second law. However F = m(dv/dt) + v(dm/dt). Given the definition of c -> RMS and P is invariant (perfect elastisity) then for the kinetic population neither changes, thus, dv/ dt = 0 and dm/dt = 0. Even for individual centerpoint collisions regardless of the individual masses, the initial momenta will be exactly the same as the final along the vector line. No dP, no force. It really is hard to imagine a system with momentum and energy but no force. > > > For a kinetic gas at a finite temperature, there will be diffusion > > > from one flow into the other. This will transfer momentum, so there > > > will be a force. > > > What you describe above is viscosity, honest, it is... > > Yes, it is. That's the point. From kinetic theory, we get viscosity. > Specifically, we get viscosity, not superfluidity. Well, I told you earlier that it was my 'opinion' that the system was not an 'ideal' (or perfectly perfect [now how's that for an oxymorn]) Perfect fluid. I don't think some magic external something stirred the pot and created vorticity. And, given Helmholtz theorem that IS! what must have happen in a perfect linearized kinetic system. Morevover, perfect flids are static. That is to say, once established, unchanging in time. There probably IS! some viscosity, I can even quantify it in terms of this system. It is, x = cLz = 1.7E-10 kg/m-sec Where L is the mean Free path, c RMS, and z effective mass density. That's five orders of magnitude less that air... I doubt that laboratory superfluids are 'perfect'. Given sufficient time, they will devolve also. But, for modeling effects it is not uncommon to assume the idealized state. I even mentioned that what we term Voltage has the natural dimensions of kinematic viscosity in this system. The temperature unit (Kelvin) has dimensions of impact. force per unit time. However, generally, one never gets to this level of subtlety in newsgroup discussions. If there is some viscosity the system will evolve, slowly, yet still retain many features of superfluidity. Superfluid behavior is the key I think to understanding the larger scale behavior yet you cannot have a fluid at all without the kinetic underpinning. To get to the bottom of it all one has to start somewhere and that should lead to observed behaviors in a unified manner. > > > If the particles making up the flows are little hard objects, little > > > elastic objects, little blobby objects, with only contact forces, they > > > will diffuse from one flow into the other, and the diffused particles > > > will collide with the particles in their new flow. See Maxwell. > > > Ummm, not so. By definition, these particles undergo center point > > perfectly elastic collisions. > > Not by your original definition. Right, I did not quantify this then. You're right glancing collisions (like anti-parallel grazes) will alter directions but should even almost perfectly themselves on the grand scale. You're also right, it is here that the lack of perfection lies. > First, note that you explicitly pointed out that you weren't restricting > the shapes to spherical. > > Second, in general, the velocity will not be always parallel to the line > joining the centres. > > Third, are you adding the requirement that the particles are all of > identical mass? No, could be,, but I have no such restriction. Only strict conservation of momentum... > > While particles exchange directions the > > momentum along the direction of travel NEVER! changes. These vector > > lines are invariant. It does not matter if there is one or one > > trillion, the magnitude of each vector line in the system is eternally > > unchanging. > > Fourth, consider a collision of two frictionless hard elastic spherical > particles of radius a, moving towards each other at initial speed v, along > paths sqrt(2)*a apart. So, when they collide, the impulse is at 45 degrees > to the initial paths. > > Final component of speed along the original path is v-dv, final component > normal to this path is dv. From conservation of energy, dv=v. After the > collision, there is no motion in the original direction of travel. For > smaller or larger impact parameters (i.e., the distance between the > original paths), this will be different. Explore the possibilities on a > billiard table. You need to distinguish between the direction of travel, > and the direction of a line joining the two centres at the moment of > collision. > > > This leads to one of Helmholtz's theorem's, namely, > > > "In the absence of rotational EXTERNAL forces, a fluid that is > > initially irrotational remains irrotational." > > >http://en.wikipedia.org/wiki/Helmholtz%27s_theorems > > > Thus, the very definition OF inviscid... > > Fifth, it isn't the very definition of inviscid. Helmholtz's theorems > (of fluid flow) only apply in inviscid fluids, but that's a long way from > them being a definition, let alone the very definition, of inviscid. > > Sixth, are you saying that Maxwell was wrong? You say you have Maxwell > kinetic theory paper on the viscosity of gases. Where was he wrong? > > Basically, your starting point, kinetic theory of bodies where no energy > is transferred to internal energy, rotation, or other forms of energy - > all of the energy remains as translations KE, as per the usual > understanding of "elastic" in this context - is very closely approximated > by a real mono-atomic gas when the temperature is such that the atoms are > (very probably) all in their (electronic) ground state. For example, this > is the case for helium, argon, etc., for temperatures that aren't too > high. I'm saying I have not had time to read it yet. > Note very well that such gases are not superfluid until quantum effects - > deviation of the behaviour of the atoms from simple kinetic behaviour - > become important. I 'think' the quantum nature arises from Kelvin's instability and granularity. That, at some scale the vortices can no longer devolve into smaller elements and since vorticity is conserved, these vortex rings are the base quantum state and size. > Before you waste any more time claiming repeatedly that such gases should > be superfluids, sit down and read Maxwell's paper. If you think that > Maxwell was right, then stop disagreeing with him. If you think that > Maxwell was wrong, then go ahead and re-write kinetic theory. I shall waste no more time. I will ask you to propose an alternative way to attain the fluid condition of Continuum Mechanics. > > > How to stop this? If you have the right kind of long range forces, you > > > can. Not just any long-range forces, but special ones (don't expect > > > superfluidity from Newton's gas theory!). Maybe if the particles don't > > > collide with each other? > > > In a 'perfect' inviscid fluid it is as if they don't, as you should > > come to realized if you work through several actual collisions. See > > also Feynman's lecture Vol II chapter 40. Superfluidity does not > > require forces or fields, just a perfect fluid. > > A "perfect" fluid which doesn't result from kinetic theory. Again, see > Maxwell. See experimental results for mono-atomic gases. > > Work through several actual collisions yourself. For identical particles, > momenta along the line of collision (_not_ the line of travel) exchange, > momenta normal to the line of collision remain unchanged. The one > collision worked out above is a sufficient counterexample, really, but > working through a sequence of collisions with varying directions of the > line of collision relative to the direction of travel might be usefully > instructive. > > Feynman II-40 isn't relevant. Feynman I-39 (on kinetic theory) and > succeeding chapters are relevant, but iirc, Feynman doesn't do the > microscopic origin of viscosity of gases. > > Feynman I-40-6 is very relevant to superfluidity. When Feynman's "freezing > out" happens for linear motion at low temperature, then you get superfluid > helium. Note well that such "freezing out" isn't a result of kinetic > theory. I think your reference to I-40-6 is what I'm been trying to get across, that the very concept of temperature and, in turn, interal energy states at the kinetic level vanish (freeze out)... Paul Stowe > Timo
From: Timo Nieminen on 31 Jul 2010 17:40 On Aug 1, 2:49 am, Paul Stowe <theaether...(a)gmail.com> wrote: > On Jul 29, 10:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > For a kinetic gas at a finite temperature, there will be diffusion > > > > from one flow into the other. This will transfer momentum, so there > > > > will be a force. > > > > What you describe above is viscosity, honest, it is... > > > Yes, it is. That's the point. From kinetic theory, we get viscosity. > > Specifically, we get viscosity, not superfluidity. > > Well, I told you earlier that it was my 'opinion' that the system was > not an 'ideal' (or perfectly perfect [now how's that for an oxymorn]) > Perfect fluid. I don't think some magic external something stirred > the pot and created vorticity. And, given Helmholtz theorem that IS! > what must have happen in a perfect linearized kinetic system. > Morevover, perfect flids are static. That is to say, once > established, unchanging in time. There probably IS! some viscosity, I > can even quantify it in terms of this system. It is, > > x = cLz = 1.7E-10 kg/m-sec > > Where L is the mean Free path, c RMS, and z effective mass density. > > That's five orders of magnitude less that air... So, not superfluid. See below. [moved] > But, for modeling effects it is not uncommon to > assume the idealized state. Very common indeed. Thus, we will happily use Bernoulli to deal with fluid flow, and it works OK for many real flows, even for water which is certainly not really inviscid. 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. So, maybe you can assume that your kinetic aether is inviscid for some flows. Not for all. Why not actually check that it works OK as an assumption, rather than just assuming it? What is dangerous theoretically is to make use of very special behaviour for a a very special case, and assume it applies in your case, without checking the validity of this assumption. But the thing you're talking about, a kinetic theory gas, is a well- known thing. The main approximations made are the continuum approximation, approximating it as an ideal gas, and approximating it as inviscid. The limits of validity of these approximations are known. These are safe approximations to make, if due attention is paid to validity. 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. Given that we know about wave propagation in kinetic theory gases, in both the continuum limit (i.e., acoustics of gases) and out of it (e.g., high-speed flows in low-density gases when the mean free path can't be ignored), there's a good theoretical basis for maing use of what we know about wave propagation in kinetic theory gases. To close your eyes to all that, and to hope to be saved by the results for some special cases of superfluidity, doesn't look like a sound theoretical step. > I doubt that > laboratory superfluids are 'perfect'. Given sufficient time, they > will devolve also. Not perfect, but imperfect in a different way. That's the point of them being superfluid, rather than just fluids of very low viscosity. > > > > If the particles making up the flows are little hard objects, little > > > > elastic objects, little blobby objects, with only contact forces, they > > > > will diffuse from one flow into the other, and the diffused particles > > > > will collide with the particles in their new flow. See Maxwell. > > > > Ummm, not so. By definition, these particles undergo center point > > > perfectly elastic collisions. > > > Not by your original definition. > > Right, I did not quantify this then. You're right glancing collisions > (like anti-parallel grazes) will alter directions but should even > almost perfectly themselves on the grand scale. You're also right, it > is here that the lack of perfection lies. They don't even themselves out on the grand scale, in the sense of restoring the direction of this momentum back to the original. The subsequent collisions don't know what the original direction was. On the grand scale, they even themselves out to an isotropic distribution (in the zero total momentum frame). This means that the momentum of atoms entering a kinetic gas is transferred to the gas as a whole. > > First, note that you explicitly pointed out that you weren't restricting > > the shapes to spherical. > > > Second, in general, the velocity will not be always parallel to the line > > joining the centres. > > > Third, are you adding the requirement that the particles are all of > > identical mass? > > No, could be,, but I have no such restriction. Only strict > conservation of momentum... No, you are also using "While particles exchange directions the momentum along the direction of travel NEVER! changes" in your argument for superfluidity. This is not only strict conservation of momentum. Note that it requires spherical particles, no off-centre (glancing) collisions, and _equal masses_. > > Sixth, are you saying that Maxwell was wrong? You say you have Maxwell > > kinetic theory paper on the viscosity of gases. Where was he wrong? [...] > > I'm saying I have not had time to read it yet. Read it. > > Before you waste any more time claiming repeatedly that such gases should > > be superfluids, sit down and read Maxwell's paper. If you think that > > Maxwell was right, then stop disagreeing with him. If you think that > > Maxwell was wrong, then go ahead and re-write kinetic theory. > > 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?)? For liquids and solids, we need long-range forces. > > Feynman I-40-6 is very relevant to superfluidity. When Feynman's "freezing > > out" happens for linear motion at low temperature, then you get superfluid > > helium. Note well that such "freezing out" isn't a result of kinetic > > theory. > > I think your reference to I-40-6 is what I'm been trying to get > across, that the very concept of temperature and, in turn, interal > energy states at the kinetic level vanish (freeze out)... In the real-life examples we have, they don't vanish. However, they are discrete. So, we need a sufficiently large excitation to do anything. If that amount of momentum isn't available, no momentum is transferred. Thus, we see He superfluidity below a critical velocity. We don't usually see this for momentum, but we very commonly see it for internal (electronic) energy states (and for rotational and vibrational states, but this is harder to see). Even at 5000K, atomic hydrogen will mostly be in the ground state - collisions in an atomic hydrogen gas will usually not have enough energy to excite the 1st excited state. So no energy will go into internal energy. 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. -- Timo
From: Paul Stowe on 1 Aug 2010 15:13 On Jul 31, 2:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Aug 1, 2:49 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > > > On Jul 29, 10:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > For a kinetic gas at a finite temperature, there will be diffusion > > > > > from one flow into the other. This will transfer momentum, so there > > > > > will be a force. > > > > > What you describe above is viscosity, honest, it is... > > > > Yes, it is. That's the point. From kinetic theory, we get viscosity. > > > Specifically, we get viscosity, not superfluidity. > > > Well, I told you earlier that it was my 'opinion' that the system was > > not an 'ideal' (or perfectly perfect [now how's that for an oxymorn]) > > Perfect fluid. I don't think some magic external something stirred > > the pot and created vorticity. And, given Helmholtz theorem that IS! > > what must have happen in a perfect linearized kinetic system. > > Morevover, perfect fluids are static. That is to say, once > > established, unchanging in time. There probably IS! some viscosity, I > > can even quantify it in terms of this system. It is, > > > x = cLz = 1.7E-10 kg/m-sec > > > Where L is the mean Free path, c RMS, and z effective mass density. > > > That's five orders of magnitude less that air... > > So, not superfluid. See below. > > [moved] > > > But, for modeling effects it is not uncommon to > > assume the idealized state. > > Very common indeed. Thus, we will happily use Bernoulli to deal with > fluid flow, and it works OK for many real flows, even for water which > is certainly not really inviscid. > > 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? > So, maybe you can assume that your kinetic aether is inviscid for some > flows. Not for all. Why not actually check that it works OK as an > assumption, rather than just assuming it? > > What is dangerous theoretically is to make use of very special > behaviour for a a very special case, and assume it applies in your > case, without checking the validity of this assumption. Form follows from observed functional behavior. Maxwell's very successful inviscid (perfect fluid) model (his words, not mine) contains the vortex lattice structure seen in superfluids. Vortices are fluid structures, I know of no way to get them except through a kinetic theory model... > But the thing you're talking about, a kinetic theory gas, is a well- > known thing. The main approximations made are the continuum > approximation, approximating it as an ideal gas, and approximating it > as inviscid. The limits of validity of these approximations are known. > These are safe approximations to make, if due attention is paid to > validity. The system duality again... Yes there must be a discrete kinetic model underpinning (Maxwell assumed a perfect fluid) BUT! the 'super' behavior arises from the pseudo particle individual vortiex interactions. These DO! have circulation fields and long range effects. But, I reiterate, how else does the fluidic nature of these arise EXCEPT through the kinetic process. The only other option seems to be assume there just are magical 'fields', and they come out of nowhere. To me, that not a proper scientific approach. > 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. 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,,, > Given that we know about wave propagation in kinetic theory gases, in > both the continuum limit (i.e., acoustics of gases) and out of it > (e.g., high-speed flows in low-density gases when the mean free path > can't be ignored), there's a good theoretical basis for making use of > what we know about wave propagation in kinetic theory gases. To close > your eyes to all that, and to hope to be saved by the results for some > special cases of superfluidity, doesn't look like a sound theoretical > step. I have no idea what you're talking about here. > > I doubt that > > laboratory superfluids are 'perfect'. Given sufficient time, they > > will devolve also. > > Not perfect, but imperfect in a different way. That's the point of > them being superfluid, rather than just fluids of very low viscosity. Yes, and what, specifically and succinctly is it??? > > > > > If the particles making up the flows are little hard objects, little > > > > > elastic objects, little blobby objects, with only contact forces, they > > > > > will diffuse from one flow into the other, and the diffused particles > > > > > will collide with the particles in their new flow. See Maxwell. > > > > > Ummm, not so. By definition, these particles undergo center point > > > > perfectly elastic collisions. > > > > Not by your original definition. > > > Right, I did not quantify this then. You're right glancing collisions > > (like anti-parallel grazes) will alter directions but should even > > almost perfectly themselves on the grand scale. You're also right, it > > is here that the lack of perfection lies. > > They don't even themselves out on the grand scale, in the sense of > restoring the direction of this momentum back to the original. The > subsequent collisions don't know what the original direction was. On > the grand scale, they even themselves out to an isotropic distribution > (in the zero total momentum frame). 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. > > > First, note that you explicitly pointed out that you weren't restricting > > > the shapes to spherical. > > > > Second, in general, the velocity will not be always parallel to the line > > > joining the centres. > > > > Third, are you adding the requirement that the particles are all of > > > identical mass? > > > No, could be,, but I have no such restriction. Only strict > > conservation of momentum... > > No, you are also using "While particles exchange directions the > momentum along the direction of travel NEVER! changes" in your > argument for superfluidity. This is not only strict conservation of > momentum. Yes, a mean momentum is assumed. Thus the term P in my original post. Now for the individual entities they all could have the same mass or different masses, same speed or different speeds but the mean values of P & c are the only parameter I dealt with. They 'could have a Boltzmann distribution or something else like Fermi. I did not try to define this. The only initial assumption was force-free elastic collisions. > Note that it requires spherical particles, no off-centre (glancing) > collisions, and _equal masses_. I you want, fine. > > > Sixth, are you saying that Maxwell was wrong? You say you have Maxwell > > > kinetic theory paper on the viscosity of gases. Where was he wrong? > [...] > > > I'm saying I have not had time to read it yet. > > Read it. > > > > Before you waste any more time claiming repeatedly that such gases should > > > be superfluids, sit down and read Maxwell's paper. If you think that > > > Maxwell was right, then stop disagreeing with him. If you think that > > > Maxwell was wrong, then go ahead and re-write kinetic theory. > > > 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. 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&dmode=source P = ~5.15E-27 kg-m/sec L = ~6.43E-08 meters z = ~8.85E-12 kg/m^3 For the pseudo particle (vortex pair) model we can relate the kinetic energy of a vortex to the thermal energy content. Thus, KE = TE => (1/2)mc^2 = (3/2)kT => mc^2 = 3kT From my original definition then since mc^2 = hi we need the vibrational oscillation (i) for this base vortex. Assuming the most basic represents an electron, then the charge (q) to inertial mass ratio gives us a frequency value (using the Maxwellian model definitions provided in my original post) then i = q/m Therefore, h(q/m) = 3kT Thus, T = hq/3km All terms, h, q, k, and m are the standard physical constant for Plancks;, charge, Boltzmann's and electron mass in SI. Solve for the answer. You might be surprised. This coupled with such new relationships a k = h/qc, ...etc. suggest that the model is both useful and not coincidental. > For liquids and solids, we need long-range forces. Yes, the vortex pairing & lattice. > > > Feynman I-40-6 is very relevant to superfluidity. When Feynman's "freezing > > > out" happens for linear motion at low temperature, then you get superfluid > > > helium. Note well that such "freezing out" isn't a result of kinetic > > > theory. > > > I think your reference to I-40-6 is what I'm been trying to get > > across, that the very concept of temperature and, in turn, interal > > energy states at the kinetic level vanish (freeze out)... > > In the real-life examples we have, they don't vanish. However, they > are discrete. So, we need a sufficiently large excitation to do > anything. If that amount of momentum isn't available, no momentum is > transferred. Thus, we see He superfluidity below a critical velocity. > We don't usually see this for momentum, but we very commonly see it > for internal (electronic) energy states (and for rotational and > vibrational states, but this is harder to see). Even at 5000K, atomic > hydrogen will mostly be in the ground state - collisions in an atomic > hydrogen gas will usually not have enough energy to excite the 1st > excited state. So no energy will go into internal energy. > > 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. Perhaps I should ask you, what is you goal here? Paul Stowe
From: Timo Nieminen on 2 Aug 2010 03:19
On Sun, 1 Aug 2010, Paul Stowe wrote: > On Jul 31, 2:40 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Aug 1, 2:49 am, PaulStowe<theaether...(a)gmail.com> wrote: > > > On Jul 29, 10:39 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > > > For a kinetic gas at a finite temperature, there will be diffusion > > > > > > from one flow into the other. This will transfer momentum, so there > > > > > > will be a force. > > > > > > > What you describe above is viscosity, honest, it is... > > > > > > Yes, it is. That's the point. From kinetic theory, we get viscosity. > > > > Specifically, we get viscosity, not superfluidity. > > > > > Well, I told you earlier that it was my 'opinion' that the system was > > > not an 'ideal' (or perfectly perfect [now how's that for an oxymorn]) > > > Perfect fluid. I don't think some magic external something stirred > > > the pot and created vorticity. And, given Helmholtz theorem that IS! > > > what must have happen in a perfect linearized kinetic system. > > > Morevover, perfect fluids are static. That is to say, once > > > established, unchanging in time. There probably IS! some viscosity, I > > > can even quantify it in terms of this system. It is, > > > > > x = cLz = 1.7E-10 kg/m-sec > > > > > Where L is the mean Free path, c RMS, and z effective mass density. > > > > > That's five orders of magnitude less that air... > > > > So, not superfluid. See below. > > > > [moved] > > > > > But, for modeling effects it is not uncommon to > > > assume the idealized state. > > > > Very common indeed. Thus, we will happily use Bernoulli to deal with > > fluid flow, and it works OK for many real flows, even for water which > > is certainly not really inviscid. > > > > 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. 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 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. This is the bare-naked kinetic theory result, without appealing to any special results from elsewhere. Do your suggested numbers work? Is the attenuation of waves slow enough to be compatible with obervation? If not, back to the drawing board. > > 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. 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. So where does Maxwell clearly state that he is considering a gas of low vicosity? Rather than elastic bodies with molecular vortices? > 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"? > > 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. > 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. > 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? 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? > 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&dmode=source > > 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. > > 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. -- Timo |