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From: Paul Stowe on 10 Apr 2010 18:17 On Apr 6, 9:57 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Wed, 7 Apr 2010, Timo Nieminen wrote: > > On Tue, 6 Apr 2010, Paul Stowe wrote: > > > > Yup, that an interesting problem, isn't it... The exposure of more > > > attenuating area ican increase the effect per unit volume because its > > > a 4 pi omni-directional flux and, being constant per unit volume has > > > some counter intuitive aspects. > > > Problem 2: Not that counter-intuitive. 6.74N is what you'd have if the > > flux were unidirectional. Not being unidirectional will reduce the maximum > > force possible, and make things worse. Not much worse, but it certainly > > doesn't help. > > In Edwards (pg 187), you have max flux = Q * pi * (r/R)^2 for a spherical > body that absorbs completely. Given that Q is defined as the incident flux > per unit area, it is odd that shielding from one side gives a net flux > larger than this, but I assume this is in the definition of "net flux" as > a different kind of flux. The maximum attenuation is terms of field flux is given in equation 12 (mentioned above) BUT! R^2 must, at all times but physically equal or greater than pir^2 (if r' is the physical radius of a spherical body) and and R = r' then the 'visible' area pir^2 is less than piR^2, meaning, the sphere is physically blocking the view or horizon of some physical area. IOW, you cannot lose more than you have to begin with. This was assumed to be understood. > OK, with this, you get a maximum force, next to a completely shielding > body, of 6.74 * pi = 21 N per kg. Qnet = Q but this is the maximum impinging flux, not a force, or an acceleration. The 'force' for this would be, F = QAa/R^2 Where A and a are the physical shadowing areas and R the physical distance between them. So, given the constraint mentioned above, QA/ R^2 must be simply Q, as a maximum (A/R^2 <= 1). Thus, given the case where a is unity, the maximum mutual 'force' between them would be 6.74 Nt on the surface. But, this is an 'idealized' black body interaction case, not likely to be seen in any real physical situation. Further, the concept of 'mass' or 'masiveness' is not applicable under such conditions. Note that mass does not appear, and has no bearing on, the outcome in situation. The concept of mass or massiveness is an 'emergent' quantity arrising from the weakly attenuating situations in the LeSage model. > (Eqn (19), on pg 188, is wrong; this should be F = Qu m2 pi (r/R)^2, not > F = Qu^2 m2 pi (r/R)^2. Just repeat the weak limit calculation, replacing > the weak limit net flux with the strong limit net flux, and this is what > you get. Equation 19 is the mixed bag, where a perfectly 'black body' is interacting with a normal (weakly attenuating) object. So, OK, the constraint for the strong limit applies, and A/R^2 <= 1, thus a black body exerts a maximum force of QuM between them. But, again, the 'strong' attenuator does not have mass in any definable sense. What does this say about LeSage's model? Black holes are gravitationally extremely weak objects... and would tend to NOT! influence their surrounding much at'tall... > In Edwards, pg 188, it's clearly stated (in words, rather than as a > mathematical expression) that F = Q_net u m. So why have you been trying > to say that Qu * mass isn't a force? Where? It states that the force between two bodies in the LeSage field of magnitude Q will behave as described under the conditions described. For mutually weak attenuators, that is, F = Q(uM)(um)/R^2 => (Qu^2)Mm/R^2 => GMm/R^2 Where G = Qu^2 And, for "Black bodies' it is, F = QAa/R^2 Where A/R^2 <= 1 etc... There is nothing either logically, or mathematically, inconsistent here. Believe me, if there was, Matt Edwards, TVF, and other reviewers would have been all over it like a cheap suit. Paul Stowe
From: Timo Nieminen on 10 Apr 2010 19:01 On Apr 11, 8:17 am, Paul Stowe <theaether...(a)gmail.com> wrote: > On Apr 6, 9:57 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On Wed, 7 Apr 2010, Timo Nieminen wrote: > > > On Tue, 6 Apr 2010, Paul Stowe wrote: > > > > > Yup, that an interesting problem, isn't it... The exposure of more > > > > attenuating area ican increase the effect per unit volume because its > > > > a 4 pi omni-directional flux and, being constant per unit volume has > > > > some counter intuitive aspects. > > > > Problem 2: Not that counter-intuitive. 6.74N is what you'd have if the > > > flux were unidirectional. Not being unidirectional will reduce the maximum > > > force possible, and make things worse. Not much worse, but it certainly > > > doesn't help. > > > In Edwards (pg 187), you have max flux = Q * pi * (r/R)^2 for a spherical > > body that absorbs completely. Given that Q is defined as the incident flux > > per unit area, it is odd that shielding from one side gives a net flux > > larger than this, but I assume this is in the definition of "net flux" as > > a different kind of flux. > > The maximum attenuation is terms of field flux is given in equation > 12 (mentioned above) BUT! R^2 must, at all times but physically equal > or greater than pir^2 (if r' is the physical radius of a spherical > body) and and R = r' then the 'visible' area pir^2 is less than piR^2, > meaning, the sphere is physically blocking the view or horizon of some > physical area. IOW, you cannot lose more than you have to begin > with. This was assumed to be understood. Yes, the _maximum_ possible flux is Q*pi (for R=r). (It's explicitly stated in Edwards that r<=R.) > > OK, with this, you get a maximum force, next to a completely shielding > > body, of 6.74 * pi = 21 N per kg. > > Qnet = Q but this is the maximum impinging flux, not a force, or an > acceleration. The 'force' for this would be, > > F = QAa/R^2 > > Where A and a are the physical shadowing areas and R the physical > distance between them. So, given the constraint mentioned above, QA/ > R^2 must be simply Q, as a maximum (A/R^2 <= 1). Thus, given the case > where a is unity, the maximum mutual 'force' between them would be > 6.74 Nt on the surface. But, this is an 'idealized' black body > interaction case, not likely to be seen in any real physical > situation. And for the case where only one body completely shields, and the 2nd body (the small one) is weakly absorbing, then a = mass * u, since a is the absorption cross-section for the body. For weak absorption by the 2nd body, the maximum possible gravitational force is 6.74N/kg. For complete absorption by the 2nd body, the maximum possible is 6.74kg/m^2. Yes, we're not likely to see this, since it is an extreme limiting case. That's the problem, it's the maximum force possible, and all gravitational forces we observe are likely to be less - much less - than these values. Measuring g here I find that the observed force is 9.78N/kg (uncorrected for centripetal acceleration, but that doesn't make enough difference), in excess of the strong-weak limit. Since I can make a plate of area 1m^2 with a mass of 1kg, I can also easily observe forces in excess of the strong-strong limit. Since I observe forces in excess of the _maximum possible_ gravitational force predicted using your values of Q and u, either the theory is wrong, or your values of Q and u are wrong. > > (Eqn (19), on pg 188, is wrong; this should be F = Qu m2 pi (r/R)^2, not > > F = Qu^2 m2 pi (r/R)^2. Just repeat the weak limit calculation, replacing > > the weak limit net flux with the strong limit net flux, and this is what > > you get. > > Equation 19 is the mixed bag, where a perfectly 'black body' is > interacting with a normal (weakly attenuating) object. So, OK, the > constraint for the strong limit applies, and A/R^2 <= 1, thus a black > body exerts a maximum force of QuM between them. But, again, the > 'strong' attenuator does not have mass in any definable sense. What > does this say about LeSage's model? Black holes are gravitationally > extremely weak objects... and would tend to NOT! influence their > surrounding much at'tall... No, "black holes" have the maximum possible gravitational force. If they don't influence their surroundings much at all, nothing does, gravitationally. They have less gravitational force per unit mass, due to shielding, but adding mass never reduces the gravitational force in a le Sage theory. > > In Edwards, pg 188, it's clearly stated (in words, rather than as a > > mathematical expression) that F = Q_net u m. So why have you been trying > > to say that Qu * mass isn't a force? > > Where? It states that the force between two bodies in the LeSage > field of magnitude Q will behave as described under the conditions > described. For mutually weak attenuators, that is, > > F = Q(uM)(um)/R^2 => (Qu^2)Mm/R^2 => GMm/R^2 Where G = Qu^2 Yes, and since Q_net = Q(uM)/R^2, this means that F = Q_net u m. > There is nothing either logically, or mathematically, inconsistent > here. Believe me, if there was, Matt Edwards, TVF, and other > reviewers would have been all over it like a cheap suit. It's a rare error that's caught by peer review, so I'm not surprised if there's a mistake there. Check the numbers for yourself. Using your value of u, and typical densities, the linear absorption coefficient (i.e., what is usually called the linear absorption coefficient, not your "linear attenutation coefficient" which is something else altogether) is lambda = u*density = approx 0.01 m^-1. That is, going through ordinary terrestrial matter, we expect a unidirectional beam of le Sage corpuscles to fall in intensity as exp(-lambda*distance) = exp(-0.01*distance in metres). This means that most le Sage corpuscles would be absorbed after travelling a few 100m into the Earth, which is only a tiny fraction of the thickness of the Earth - almost none would make it through. Meanwhile, F = Q(uM)(um)/R^2 => (Qu^2)Mm/R^2 => GMm/ R^2 assumes weak absorption - contradicted by this value of u*density. The strong absorption limit meanwhile gives a maximum gravitational force of 6.74N per kg or per m^2 (depending on whether the 2nd body is weakly or strongly absorbing), both of which are observed to be exceeded. -- Timo
From: Ken S. Tucker on 10 Apr 2010 23:50 Hi Timo and all. On Apr 10, 1:48 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Apr 10, 10:58 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > On Apr 9, 2:38 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Apr 10, 12:31 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > Something I enjoyed was using a little electric powered > > > > toy car, allow me to provide the 'menu'. > > > [cut] > > > > Bright students appreciate the unit conversations > > > > and the scaling, if ya can get a bit of time in a > > > > gymnasium it's good fun. > > > > Nice exercise. Very good for engineering students in particular. Do > > > you include drag (i.e., air resistance) in the scaling? > > > Ha-ha, scaling Drag Coefficient (?), no. I've worked with > > scaling aerodynamic models, but that's complicated, if your > > serious I'm happy to discuss it though. > > It is complicated, but it's the done thing when doing any scale work > in fluid flow, and more generally, when using scale models to > investigate real systems.Should be compulsory for all engineers. You're tough, scaling is a specialization, I'd agree though that an introduction for engineers would be ok. Engineering is highly specialized. I (we) do a bit of hobby scaling for model aircraft that we launch using a catapult, http://www.flickr.com/photos/35156618(a)N03/page2/ just for fun. > Physicists work a little differently. A physicist will simplify the > model until something can be said from basic physics. Hopefully the > simplified model still behaves enough like the real system so as to be > meaningful. The engineer, OTOH, will measure a scale model (or a > different-scale real system) and extrapolate. Two different answers to > two different questions; the physicist answers "why", and the engineer > answers "how much". The engineer's approach works well, as long as the > scaling is understood. Not knowing the "why" can make this dangerous. > There are some spectacular roof collapses that resulted from naive > scaling, and other disasters too. It was the fundamental problem with > the Tacoma Narrows bridge, although stupidity contributed too (i.e., > not fixing it as soon as its behaviour was seen). Yes Tacoma was built cheap and oscillation was not quite appreciated as it is now. > > IMO, a big weakness among phyicists is sliding through units. > > That's why an intro to Volts x Current = Power (Watts) to HP > > and to Force x velocity , (1 HP = 550 foot pounds/sec) and so > > on is worth knowing, (ok a yawner), but later on is dimensional > > analysis. > > Mostly people (scientists) focus on energy and momentum, and > > power and action are sidelined. > > In mechanics, yes, power is sidelined. Friction is largely ignored, > and while P=Fv will appear as a brief note for motion at constant > speed with friction, it's minor. For circuits, of course, energy and > momentum are ignored, and power is king. But these aren't connected, > even appearing in different semesters in the usual courses. > > The newest generation of students are very metrified. Foot-pounds > would be traumatic to them, let alone HP. Well in the real world Outboard Motors are still measured by HP, but that's a whole nuther topic, best to go bi. > > > I'd always do the Bernoulli bottle, plastic drink bottle with hole in > > > side. From Bernoulli, predict the speed of the outflow. From the > > > projectile motion of the water squirting out onto the table, find the > > > flow speed from measurement of the horizontal range. That's the basic > > > part of the exercise. The advanced part is to ask why these two values > > > aren't the same. > > > I've always found that one tough. > > In the proper context, it shouldn't be. As a mystery rule introduced > in a hydrodynamics section of a cource, divorced from the rest of the > course, it is, well, a mystery. > > I'd introduce it in mechanics, as an example of the conservation of > energy. With no pressure difference (e.g., take the two reference > points at the surface of the water and at the hole, where the pressure > is atmospheric in both cases), Bernoulli just says "KE + gravitational > PE = constant". (Well, KE per unit volume and PE per unit volume.) If > there is a pressure difference, deltaP is the work done per unit > volume to move from point 1 to point 2, so the complete Bernoulli is > "work done = change in (KE + PE)". And you can do the measurement with > garbage and a ruler, with the most complicated maths being a > quadratic. There's a Maxima-Minima problem in calculus, that involved moving the exit orfice vertically to max the stream, (how far out it would go squirt horizontally). The column height was fixed. Maybe I got spooked, but I found it tough. > Timo Regards Ken
From: NoEinstein on 11 Apr 2010 00:20 On Apr 10, 4:48 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > As I expected, persona non grata, Timo, got his non-understanding about mechanics from textbooks written by those who don't understand mechanics, either. The correct formula for kinetic energy is: KE = a/ g (m) + v /32.174 (m). The latter, my own formula, is a variant of the momentum formula: F = mv. The "textbook" formula is Coriolis's KE = 1/2 mv^2 (sic). Note that the KE increases exponentially while the only variable (assuming a unit weight) is v. All near-Earth falling objects increase in velocity UNIFORMLY, or linearly with respect to the time of fall. So Coriolis's formula violates the Law of the Conservation of Energy. Interestingly, Einstein based his E = mc^2 on Coriolis's formula. Both violate the LCE. NoEinstein > > On Apr 10, 10:58 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > > > On Apr 9, 2:38 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Apr 10, 12:31 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > Something I enjoyed was using a little electric powered > > > > toy car, allow me to provide the 'menu'. > > > [cut] > > > > Bright students appreciate the unit conversations > > > > and the scaling, if ya can get a bit of time in a > > > > gymnasium it's good fun. > > > > Nice exercise. Very good for engineering students in particular. Do > > > you include drag (i.e., air resistance) in the scaling? > > > Ha-ha, scaling Drag Coefficient (?), no. I've worked with > > scaling aerodynamic models, but that's complicated, if your > > serious I'm happy to discuss it though. > > It is complicated, but it's the done thing when doing any scale work > in fluid flow, and more generally, when using scale models to > investigate real systems.Should be compulsory for all engineers. > > Physicists work a little differently. A physicist will simplify the > model until something can be said from basic physics. Hopefully the > simplified model still behaves enough like the real system so as to be > meaningful. The engineer, OTOH, will measure a scale model (or a > different-scale real system) and extrapolate. Two different answers to > two different questions; the physicist answers "why", and the engineer > answers "how much". The engineer's approach works well, as long as the > scaling is understood. Not knowing the "why" can make this dangerous. > There are some spectacular roof collapses that resulted from naive > scaling, and other disasters too. It was the fundamental problem with > the Tacoma Narrows bridge, although stupidity contributed too (i.e., > not fixing it as soon as its behaviour was seen). > > > IMO, a big weakness among phyicists is sliding through units. > > That's why an intro to Volts x Current = Power (Watts) to HP > > and to Force x velocity , (1 HP = 550 foot pounds/sec) and so > > on is worth knowing, (ok a yawner), but later on is dimensional > > analysis. > > Mostly people (scientists) focus on energy and momentum, and > > power and action are sidelined. > > In mechanics, yes, power is sidelined. Friction is largely ignored, > and while P=Fv will appear as a brief note for motion at constant > speed with friction, it's minor. For circuits, of course, energy and > momentum are ignored, and power is king. But these aren't connected, > even appearing in different semesters in the usual courses. > > The newest generation of students are very metrified. Foot-pounds > would be traumatic to them, let alone HP. > > > > I'd always do the Bernoulli bottle, plastic drink bottle with hole in > > > side. From Bernoulli, predict the speed of the outflow. From the > > > projectile motion of the water squirting out onto the table, find the > > > flow speed from measurement of the horizontal range. That's the basic > > > part of the exercise. The advanced part is to ask why these two values > > > aren't the same. > > > I've always found that one tough. > > In the proper context, it shouldn't be. As a mystery rule introduced > in a hydrodynamics section of a cource, divorced from the rest of the > course, it is, well, a mystery. > > I'd introduce it in mechanics, as an example of the conservation of > energy. With no pressure difference (e.g., take the two reference > points at the surface of the water and at the hole, where the pressure > is atmospheric in both cases), Bernoulli just says "KE + gravitational > PE = constant". (Well, KE per unit volume and PE per unit volume.) If > there is a pressure difference, deltaP is the work done per unit > volume to move from point 1 to point 2, so the complete Bernoulli is > "work done = change in (KE + PE)". And you can do the measurement with > garbage and a ruler, with the most complicated maths being a > quadratic. > > -- > Timo- Hide quoted text - > > - Show quoted text -
From: Paul Stowe on 11 Apr 2010 13:26
On Apr 10, 4:01 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On Apr 11, 8:17 am, Paul Stowe <theaether...(a)gmail.com> wrote: > > > On Apr 6, 9:57 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On Wed, 7 Apr 2010, Timo Nieminen wrote: > > > > On Tue, 6 Apr 2010, PaulStowewrote: > > > > > > Yup, that an interesting problem, isn't it... The exposure of more > > > > > attenuating area ican increase the effect per unit volume because its > > > > > a 4 pi omni-directional flux and, being constant per unit volume has > > > > > some counter intuitive aspects. > > > > > Problem 2: Not that counter-intuitive. 6.74N is what you'd have if the > > > > flux were unidirectional. Not being unidirectional will reduce the maximum > > > > force possible, and make things worse. Not much worse, but it certainly > > > > doesn't help. > > > > In Edwards (pg 187), you have max flux = Q * pi * (r/R)^2 for a spherical > > > body that absorbs completely. Given that Q is defined as the incident flux > > > per unit area, it is odd that shielding from one side gives a net flux > > > larger than this, but I assume this is in the definition of "net flux" as > > > a different kind of flux. > > > The maximum attenuation is terms of field flux is given in equation > > 12 (mentioned above) BUT! R^2 must, at all times but physically equal > > or greater than pir^2 (if r' is the physical radius of a spherical > > body) and and R = r' then the 'visible' area pir^2 is less than piR^2, > > meaning, the sphere is physically blocking the view or horizon of some > > physical area. IOW, you cannot lose more than you have to begin > > with. This was assumed to be understood. > > Yes, the _maximum_ possible flux is Q*pi (for R=r). (It's explicitly > stated in Edwards that r<=R.) > > > > OK, with this, you get a maximum force, next to a completely shielding > > > body, of 6.74 * pi = 21 N per kg. > > > Qnet = Q but this is the maximum impinging flux, not a force, or an > > acceleration. The 'force' for this would be, > > > F = QAa/R^2 > > > Where A and a are the physical shadowing areas and R the physical > > distance between them. So, given the constraint mentioned above, QA/ > > R^2 must be simply Q, as a maximum (A/R^2 <= 1). Thus, given the case > > where a is unity, the maximum mutual 'force' between them would be > > 6.74 Nt on the surface. But, this is an 'idealized' black body > > interaction case, not likely to be seen in any real physical > > situation. > > And for the case where only one body completely shields, and the 2nd > body (the small one) is weakly absorbing, then a = mass * u, since a > is the absorption cross-section for the body. For weak absorption by > the 2nd body, the maximum possible gravitational force is 6.74N/kg. > For complete absorption by the 2nd body, the maximum possible is > 6.74kg/m^2. > > Yes, we're not likely to see this, since it is an extreme limiting > case. That's the problem, it's the maximum force possible, and all > gravitational forces we observe are likely to be less - much less - > than these values. Measuring g here I find that the observed force is > 9.78N/kg (uncorrected for centripetal acceleration, but that doesn't > make enough difference), in excess of the strong-weak limit. Since I > can make a plate of area 1m^2 with a mass of 1kg, I can also easily > observe forces in excess of the strong-strong limit. THe real question is, what, in terms of LeSage's model is inertial/ gravitational mass? We measure this by Newton's second law, i.e. by the inertial response... Thus the strong equivilence principle. But, as you should clearly see in the above expressions, mass (M) is NOT! the primary player in the process, interaction area is. As I've said, massiveness is an emergent quantity. Thus, in terms of LeSage's process, mass is NOT! a fundamental property. It is in inertia, not gravity, that the answer to this is found. And, inertia is not gravity. To answer this question you do need some sort of unification. > Since I observe forces in excess of the _maximum possible_ > gravitational force predicted using your values of Q and u, either the > theory is wrong, or your values of Q and u are wrong. Not if you throw out M as an invariant property. Also Q is the LeSagian the momentum interaction parameter. Nothing says it cannot be field density (z) multiplied by a differntial change in c per event squared, i.e., Q = z(dc)^2 I am NOT! saying this is, in fact, the case BUT! now we get down into the nitty-gritty underlying foundational questions of possible different variants of the LeSage models. Matt Edwards for example perfers a Casmir type effect for the apparent momentum attenuation in the field. > > > (Eqn (19), on pg 188, is wrong; this should be F = Qu m2 pi (r/R)^2, not > > > F = Qu^2 m2 pi (r/R)^2. Just repeat the weak limit calculation, replacing > > > the weak limit net flux with the strong limit net flux, and this is what > > > you get. > > > Equation 19 is the mixed bag, where a perfectly 'black body' is > > interacting with a normal (weakly attenuating) object. So, OK, the > > constraint for the strong limit applies, and A/R^2 <= 1, thus a black > > body exerts a maximum force of QuM between them. But, again, the > > 'strong' attenuator does not have mass in any definable sense. What > > does this say about LeSage's model? Black holes are gravitationally > > extremely weak objects... and would tend to NOT! influence their > > surrounding much at'tall... > > No, "black holes" have the maximum possible gravitational force. If > they don't influence their surroundings much at all, nothing does, > gravitationally. They have less gravitational force per unit mass, due > to shielding, but adding mass never reduces the gravitational force in > a le Sage theory. Maybe I should have said 'black bodies' to indicate this model's version. > > > In Edwards, pg 188, it's clearly stated (in words, rather than as a > > > mathematical expression) that F = Q_net u m. So why have you been trying > > > to say that Qu * mass isn't a force? > > > Where? It states that the force between two bodies in the LeSage > > field of magnitude Q will behave as described under the conditions > > described. For mutually weak attenuators, that is, > > > F = Q(uM)(um)/R^2 => (Qu^2)Mm/R^2 => GMm/R^2 Where G = Qu^2 > > Yes, and since Q_net = Q(uM)/R^2, this means that F = Q_net u m. > > > There is nothing either logically, or mathematically, inconsistent > > here. Believe me, if there was, Matt Edwards, TVF, and other > > reviewers would have been all over it like a cheap suit. > > It's a rare error that's caught by peer review, so I'm not surprised > if there's a mistake there. > > Check the numbers for yourself. Using your value of u, and typical > densities, the linear absorption coefficient (i.e., what is usually > called the linear absorption coefficient, not your "linear > attenutation coefficient" which is something else altogether) is > lambda = u*density = approx 0.01 m^-1. That is, going through ordinary > terrestrial matter, we expect a unidirectional beam of le Sage > corpuscles to fall in intensity as exp(-lambda*distance) = > exp(-0.01*distance in metres). This means that most le Sage corpuscles > would be absorbed after travelling a few 100m into the Earth, which is > only a tiny fraction of the thickness of the Earth - almost none would > make it through. Meanwhile, F = Q(uM)(um)/R^2 => (Qu^2)Mm/R^2 => GMm/ > R^2 assumes weak absorption - contradicted by this value of u*density. > The strong absorption limit meanwhile gives a maximum gravitational > force of 6.74N per kg or per m^2 (depending on whether the 2nd body is > weakly or strongly absorbing), both of which are observed to be > exceeded. I have been a radiation transport specialist since 1980 and am the primary author of both ProShield and SmartShield (discrect Ordinate transport model [QADCGGP] based) and am very well aware of fundamentals. Thus I have always said u is a mass 'attenuation' coefficient NOT! an absorption coefficient since the principle underlying processes remain undefined! Further, yes, u is, at 3.146E-6, incongruent with a mass density of 5525 kg/m^3 Earth's bulk density for example. It is not for a field density of the order of 8.854E-12 kg/m^3 however (which is the EM density). In fact, 3E-6 x 9E-12 => 2.7E-17 1/m (linear 'attenuation' coefficient). However, this raises the big question as to 'what, exactly, is matter?'. This is a question that goes to the very heart of unification. We know that matter is, fundamentally, quantum mechanical, thus, hu = mc^2. Then hd(nu) = mc(dc) + dmc^2 It is my take that both the inertia response and gravity itself is the result from a second order effect of changes in velocity. First order changes in velocity carry the sign (+/-) and can result in opposing effects, however, effects of the second order are decidedly non linear AND unlike first order effects, cannot have a change in sign (directionality). I speculate that, in fact, both inertia and gravity are the second order effects of EM processes. Thus, again, this leads back to the question where does matter's apparent density come from? That is a very interesting question. FYI, there are other striking correlations (coincidences) such as (in SI), k = h/ec Where k is Boltzmann's constant, e elemental charge, c light speed. And, before you go off on saying the units don't match, they doe using Maxwell's definition where e has units of mass devided by time. This leads to temperature (in Kelvins) of have mlt units of 'impact' of kg- m/sec^3. Also expressions with physical units can yield coincidential correlations BUT NEVER numerological ones. IMO the Planck units are really just silly exercises in blind man's numerology that are not even coincidental or correlated. I try very hard NOT! to intrduce baseless and unitless scaling factors which WOULD BE numerology... Paul Stowe |