Flaw in Stowe/LeSage Heating
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From: TC on 7 Mar 2005 22:54 mingstb(a)sim-ss.com wrote: > Bilge wrote: > > TC: > > >> http://www.mathpages.com/home/kmath131/kmath131.htm > > >Which derives very different conclusions from the LeSage model at > > >great odds with Stowe's results. > > >Has Stowe ever refuted these arguments? > > No. > > LOL! > But that pathetic and ignorant site was completely refuted by myself. > Paul didn't have to. If your refutation is in the threads to which Stowe provided the URLs then you didn't refute very much. You pointed out how the terminology in the mathpages article differs from that used by "LeSagians" and made some other minor corrections. However, you did not demonstrate how a LeSagian theory can avoid making the absurd physical predictions that the mathpages article points out. Tom
From: TC on 7 Mar 2005 23:11 mingstb(a)sim-ss.com wrote: > TC wrote: > > Paul Stowe wrote: Snip....as Mingst has tagged in > > At this point I interject to ask why not think of the interaction > > as the sum over individual small masses? That is the Earth > > is 10^whatever particles. Each particle has a LeSagian heating > > so the Earth would heat as the sum of all these particles > > so total mass is all that is involved. As long as you are making > > the weak absorption approximation, there will be little reduction > > in heat flux at the inner particles compared to the outer particles. > Indeed, you cut directly to the essence of the situation! I do my best. > Nucleons > (specifically, neutrons) have been experimentally determined to > gravitate. Eotvos type experiments have shown that most things that make up matter gravitate. Including binding energy. > LeSagian gravitation is not based on the macroscopic radius > of matter. If you have 'N' nucleons, you will have the same > gravitational force at distance 'R', outside the body. This is simply > Gauss' flux law. How do you account for Eotvos experiments then? These experiments show that the same number of nucleons can have differing gravitational force at distance R because of variations in binding energy. Compare sphere of gold to one of silicon. This seems to be a discussion: http://www.mazepath.com/uncleal/eotvos.htm > Since density goes as the cube of volume, and area goes as the square > of volume, the radial parameter arises in the above calculation. Density is constant for the most part. Area goes as the square of radius. ....... > > Density changes? To first approximation density does not depend > > on size. > Horsefeathers. If mass is held constant, and volume changes, then the > density changes. If you are talking about a gas, volume is a free parameter for the most part. But many planets are solid, and those that are not have a complicated density/radius relation determined by the total mass, internal energy sources, equation of state etc. > > Energy? Is this the same as your "fluence"? > No. Fluence is fluence. Specifically, the momentum flux. Nonstandard usage. > > > There is a 1/r (r^2/r^3) relationship between the mass, m and > > > the total of i. This means, the lowest interacting state is > > > the one with the highest density for any given given radius. > > > Increase the radius and hold mass constant, and increase the > > > total interaction that will occur within the region. It is > > > simple math. The key is the realization that it is the flux > > > not the total that is the invariant property. > > If you make your (apparently tautological) assumptions > > it follows. > Definitions are not tautologies. If you define A. You cannot say that A has been derived from A. > > But I don't seem why area is the thing. Gravity is universal > > so it effects small particles in such a way that the sum > > gives the gravitational effect on a large body made from > > the small particles. Surface area has nothing to do with > > it. > Precisely! GRAVITY is not affected by surface area. But energy > deposition (for a constant mass) *IS* affected by surface area. I would tend to think it would be affected by radius (or square root of surface area) since the potential is affected by inverse radius. > Because gravity is a two-body effect, and gravitational flux heating is > a one-body effect. (The mass will heat up even if there is nothing > around it with which to gravitate.) Stowe said this too. It really doesn't make much sense. > > {snip unchanged stuff} > > As you say, input equals output, so in the LeSage theory > > the net momentum flux of ultramundane particles absorbed by > > the body equals the gravitational force on a body. > Not quite. But you are on the right track... > > The energy flux (power) of the ultramundane particles > > is given by the above. Since momentum is a signed quantity > > it is the net absorded that gives rise to gravitational force, > Close. It is the net transfer of momentum flux. Not the net > absorption. How is momentum transfered without absorption of ultramundane particles? > > so the energy flux (which is not signed) could actually > > be much larger than (net momentum flux) times velocity/2. > Speaking purely mathematically, without incorporating any real-world > observations, yes. Because heating is a one-body problem. And > graviation is always a two-body problem. Again that one/two body statement. > And we don't know -- a priori -- what the scattering/ absorption > relationship is. You know what it must do to account for attraction of gravity. > > Put in speed of light for speed of ultramundane particles > > and the power input to a kg mass in the Earth's > > gravity field will be something greater a gigawatt. > In this, you have made a fundamental error. You cannot make such a > claim, if you actually use Le Sagian mathematics. In short, you cannot > get any heating values solely from gravitational calculations. You are > attempting to resolve two unknowns with only one equation. So LeSagian calculations are not gravitational calculations? Tom
From: Bilge on 8 Mar 2005 02:31 TC: >mingstb(a)sim-ss.com wrote: >> Bilge wrote: >> > TC: > >> > >> http://www.mathpages.com/home/kmath131/kmath131.htm > >> > >Which derives very different conclusions from the LeSage model at >> > >great odds with Stowe's results. > >> > >Has Stowe ever refuted these arguments? > >> > No. >> >> LOL! > >> But that pathetic and ignorant site was completely refuted by myself. >> Paul didn't have to. > >If your refutation is in the threads to which Stowe provided the He didn't refute anything either. In fact, he only made himself appear even less competent. Read the thread for yourself. You'll be hard pressed to find the quantitative objections made on that url refuted with anything but a long-winded soliloquy of content-free bullshit.
From: mingstb on 8 Mar 2005 18:22 TC wrote: > mingstb(a)sim-ss.com wrote: > > TC wrote: > > > Paul Stowe wrote: > > Snip....as Mingst has tagged in Putting back the context of the discussion: > > > > > Come to think of it why is there an "r" at all in the expression > > > > > for heating? Wouldn't heating just depend on the mass? > > > > > > Again, let's go through this from the beginning. In equilibrium, > > > > we know that input should equal output. For LeSage's process we > > > > can define this precisely. That is, flux [i] integrated over the > > > > area upon which it impinges defined how much CAN interact in a > > > > region enclosed by said area. > > > > > > At this point I interject to ask why not think of the interaction > > > as the sum over individual small masses? That is the Earth > > > is 10^whatever particles. Each particle has a LeSagian heating > > > so the Earth would heat as the sum of all these particles > > > so total mass is all that is involved. As long as you are making > > > the weak absorption approximation, there will be little reduction > > > in heat flux at the inner particles compared to the outer > > > particles. > > > Indeed, you cut directly to the essence of the situation! > > I do my best. > > > Nucleons > > (specifically, neutrons) have been experimentally determined to > > gravitate. > > Eotvos type experiments have shown that most things that make > up matter gravitate. Including binding energy. To what experiments are you referring? The Eotvos experiment is a measure of two units of mass of different chemical compositions. There are no "binding energy" experiments within the Eotvos type. Because energy (of any form) is not an observable property. It is always calculated. Are you alluding to theoretical claims? > > LeSagian gravitation is not based on the macroscopic radius > > of matter. If you have 'N' nucleons, you will have the same > > gravitational force at distance 'R', outside the body. This is > > simply Gauss' flux law. > > How do you account for Eotvos experiments then? These experiments > show that the same number of nucleons can have differing gravitational > force at distance R because of variations in binding energy. Please, be specific. Which experiment? I believe you will find that the calculation of energy is completely circular. Such claims always start with a given mass. > Compare sphere of gold to one of silicon. > This seems to be a discussion: > http://www.mazepath.com/uncleal/eotvos.htm Totally irrelevant to the issue under discussion. You may note that I never claimed that nucleons always share the same value of mass in different circumstances. The LeSagian interaction is coupled to the *mass* of the nucleon (or nucleus), not to it's "number" or "type." Only GR postulates a distinction between "gravitational" mass and "inertial" mass. Such efforts are irrelevant to non-GR theory. There is no reason for non-GR theories to see a difference. > > Since density goes as the cube of volume, and area goes as the square > > of volume, the radial parameter arises in the above calculation. > > Density is constant for the most part. Not when one varies the volume of a constant mass (or changes the mass within a given radius/volume) -- which is the issue under discussion. > Area goes as the square of radius. Oops. Thanks for catching the typo. > ....... Again, replacing some context: =================== > > > Thus? You assumed the flux is "integrated over the area" > > > and you concluded that the total is a function of the area. > > > Seems tautological. > > > >It's called a definition. One integrates over a solid angle of 4 pi > >steradians to get the contribution from all directions. > > > > > > Now, the material mass [m] within the enclosed region is simply > > > > its mean density [p] multiplied by the volume of that region, > > > > right? That is, > > > > > > m = p([4pir^3]/3) > > > > > > We see then that the density changes as r^3 and energy as r^2. =================== > > > > > Density changes? To first approximation density does not depend > > > on size. > > > Horsefeathers. If mass is held constant, and volume changes, then > > the density changes. > > If you are talking about a gas, volume is a free parameter for the most > part. > But many planets are solid, and those that are not have a complicated > density/radius relation determined by the total mass, internal energy > sources, equation of state etc. Again, totally irrelevant to the issue. The phase of matter is immaterial to LeSagian gravitational theory (also to Newtonian theory). Gauss' law assures us that the distribution within a given mass has no effect on a body outside the mass. It simply becomes a result of the total mass of the body. > > > Energy? Is this the same as your "fluence"? > > > No. Fluence is fluence. Specifically, the momentum flux. > > Nonstandard usage. It's *standard usage* in LeSagian theory. Flux is a generic term meaning "something" per unit area. You have to specify what that "something" is. > > > > There is a 1/r (r^2/r^3) relationship between the mass, m and > > > > the total of i. This means, the lowest interacting state is > > > > the one with the highest density for any given given radius. > > > > Increase the radius and hold mass constant, and increase the > > > > total interaction that will occur within the region. It is > > > > simple math. The key is the realization that it is the flux > > > > not the total that is the invariant property. > > > > If you make your (apparently tautological) assumptions > > > it follows. > > > Definitions are not tautologies. > > If you define A. You cannot say that A has been derived from A. And no one did. > > > But I don't seem why area is the thing. Gravity is universal > > > so it effects small particles in such a way that the sum > > > gives the gravitational effect on a large body made from > > > the small particles. Surface area has nothing to do with > > > it. > > > Precisely! GRAVITY is not affected by surface area. But energy > > deposition (for a constant mass) *IS* affected by surface area. > > I would tend to think it would be affected by radius (or square root > of surface area) since the potential is affected by inverse radius. Your thought is wrong, because "potential" (another unobservable, mathematical concept) doesn't enter into it at all. I pointed out the derivation, above. > > Because gravity is a two-body effect, and gravitational flux heating > > is a one-body effect. (The mass will heat up even if there is nothing > > around it with which to gravitate.) > > Stowe said this too. It really doesn't make much sense. Sure it does. Postulate a single matter body in the universe filled with LeSagian corpuscles. It will be subject to heating from the interaction of said corpuscles with itself. It has nothing whatsoever to do with any gravitational attraction with any other body. Now place a second body of equal mass a distance 'd' away from the first body. The first body will continue to absorb just as much energy as it did before. But it will begin to gravitate. Now make the second body 1000000 times the mass of the first body. The first body *still* absorbs the same energy from the LeSagian aether as it did before. The deposition of energy into the first matter body is a function of the product of momentum flux density (Phi_0) and the mass interaction coefficient of that body (mu_s). The gravitaional force is given by: F = -Phi_0 (mu_s)^2 m M / d^2 or, conventionally: F = -G m M / d^2 No gravitation calculation can give you aught else but the factor G. It cannot give you the product (Phi_0) (mu_s). Hence, you cannot determine the heating of a body from any orbital situation. > > > {snip unchanged stuff} > > > > As you say, input equals output, so in the LeSage theory > > > the net momentum flux of ultramundane particles absorbed by > > > the body equals the gravitational force on a body. > > > Not quite. But you are on the right track... > > > > The energy flux (power) of the ultramundane particles > > > is given by the above. Since momentum is a signed quantity > > > it is the net absorded that gives rise to gravitational force, > > > Close. It is the net transfer of momentum flux. Not the net > > absorption. > > How is momentum transfered without absorption of ultramundane > particles? However it is done in the real universe, of course. One can easily work out the macroscopic effects, without having to first know the microscopic detail. Scattering is just as valid a Newtoninan method for transfer of momentum as total absorption. > > > so the energy flux (which is not signed) could actually > > > be much larger than (net momentum flux) times velocity/2. > > > Speaking purely mathematically, without incorporating any real-world > > observations, yes. Because heating is a one-body problem. And > > graviation is always a two-body problem. > > Again that one/two body statement. And for the same reason. It is not possible to determine heating from "G", alone. > > And we don't know -- a priori -- what the scattering/ absorption > > relationship is. > > You know what it must do to account for attraction of gravity. No, we do not. A priori, the attraction of gravity can result from any combination of scattering and absorption (full or partial). It requires actual observation (i.e. experiment) to determine the value of mu_s. (See the prior reference in "Pushing Gravity" for the full detail.) > > > Put in speed of light for speed of ultramundane particles > > > and the power input to a kg mass in the Earth's > > > gravity field will be something greater a gigawatt. > > > In this, you have made a fundamental error. You cannot make such a > > claim, if you actually use Le Sagian mathematics. In short, you > > cannot get any heating values solely from gravitational calculations. > > You are attempting to resolve two unknowns with only one equation. > > So LeSagian calculations are not gravitational calculations? Some LeSagian calculations are gravitation (i.e. they include the derived parameter "G"). Some LeSagian calculations do not use "G". Gravitation is only one aspect of the LeSagian model. Heating is another. Drag is another. Orbital dynamics is another. greywolf42
From: mingstb on 8 Mar 2005 18:26
> ming...(a)sim-ss.com wrote: > > Bilge wrote: > > > TC: > > > >> http://www.mathpages.com/home/kmath131/kmath131.htm > > > >Which derives very different conclusions from the LeSage model > at > > > >great odds with Stowe's results. > > > >Has Stowe ever refuted these arguments? > > > No. > > > LOL! > > But that pathetic and ignorant site was completely refuted by myself. > > Paul didn't have to. > > If your refutation is in the threads to which Stowe provided the > URLs then you didn't refute very much. You pointed out how the > terminology in the mathpages article differs from that used by > "LeSagians" and made some other minor corrections. > However, you did not demonstrate how a LeSagian theory can > avoid making the absurd physical predictions that the mathpages > article points out. You mean the "predictions" that "Shadow" made? LOL! Those aren't LeSagian predictions. That's why "Shadow" and Bilge "both" cut and ran. ;) And it is obvious that you didn't actually read the replies. Otherwise you would have understood the reason for my references to one and two-body effects. Perhaps you'd care to be a bit more specific. Please point out a specific argument of mine that you felt "failed." greywolf42 |