Flaw in Stowe/LeSage Heating
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From: mingstb on 8 Mar 2005 18:28 > 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. The usual Bilge reply. Empty of content and thought. Full only of invective and fear of failure. greywolf42
From: TC on 8 Mar 2005 22:47 mingstb(a)sim-ss.com wrote: > TC wrote: > > mingstb(a)sim-ss.com wrote: > > > TC wrote: > > > > Paul Stowe wrote: > > > > Snip....as Mingst has tagged in mingstb(a)sim-ss.com wrote: > 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: Sigh. Why? It just clutters up the post. The context is clear enough. Snip for clarity. > > > 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? Are you denying that 6 moles of neutrons and 6 moles of protons in the form of a mole of carbon weighs less than 6 moles of neutrons and 6 moles of protons in the form of 6 moles of deuterium? [6 moles of electrons are also needed in each case] > > > 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. Binding energy is observed. It is why nuclear weapons go boom. > 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." So LeSagian interaction couples to binding energy as in GR? > Only GR postulates a distinction between "gravitational" mass and > "inertial" mass. Where do you get that from. GR is based on principle of equivalence which equates the two. It is Newton's theory of gravity that distinguishes them. > Such efforts are irrelevant to non-GR theory. There > is no reason for non-GR theories to see a difference. You have it absolutely backwards. .......... [snip irrelevant quibble] > Again, replacing some context: > =================== [snip for clarity] > Again, totally irrelevant to the issue. The phase of matter is > immaterial to LeSagian gravitational theory (also to Newtonian theory). Then why is there an "r" in the expression. If the phas is irrelevant then it follows that the density does not matter. > 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. There you go. So why the "r" in the expression for heating by LeSagian gravity? [snip some more for clarity] > > > > 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. If you don't like potential, then think of terminal velocity of something falling from infinity or the equivalent escape velocity. These are observable and have the same dependence on m and r as the potential. > > > 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. OK. Where does this go? > 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. It will do no such thing. Under LeSagian theory it shadows the second body. > 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. No it doesn't. It is shadowed quite a bit by the larger body reducing the flux it sees. > 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. You can bound the heating from the orbital situation. The imbalance of flux that causes F cannot be larger than the total flux absent a second body. This provides a lower bound on the heating. The heating has to be at least as large as the momentum flux (the force F) times the speed of the ultramundane particles. > > > > {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. You mean to say that LeSagian theory has action at a distance or that it takes the view of GR that motion is along geodesics of space time? You build up this apparatus of ultramundane particles and then say its not them that cause the gravitational force after all? > 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. Oh. But doesn't scattering spoil the shadowing affect that leads to the LeSagian explanation of gravity? [snip reiteration of the one body/two body discussion] > > > 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.) I'm awaiting interlibrary loan. But I have grave doubts. > > > > 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. Fantasy is yet another, I suspect. Tom > > 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: TC on 8 Mar 2005 22:56 mingstb(a)sim-ss.com wrote: > > 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. Of course not. The issue is whether LeSagian theory is a consistent physical theory that agrees with observation. > 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. Are you Mingst or Stowe? It seems to me Stowe first mentioned the one/two body thing. No I didn't read the entire thread. > Perhaps you'd care to be a bit more specific. Please point out a > specific argument of mine that you felt "failed." I can't find the URL pointing to that thread right now. Can you resupply? But it wasn't that your argument failed, but that it missed the point. Tom
From: Dirk Van de moortel on 9 Mar 2005 07:34 "TC" <tclarke(a)ist.ucf.edu> wrote in message news:1110340036.670638.58520(a)f14g2000cwb.googlegroups.com... > > mingstb(a)sim-ss.com wrote: > > TC wrote: > > > mingstb(a)sim-ss.com wrote: > > > > TC wrote: > > > > > Paul Stowe wrote: > > > > > > Snip....as Mingst has tagged in > mingstb(a)sim-ss.com wrote: > > 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: > > Sigh. Why? It just clutters up the post. > The context is clear enough. The idea is to put up a smokescreen so you give up. Previous examples of this particular Mingst tactic. #14 http://groups.google.co.uk/groups?&as_umsgid=Q_hAd.35825$Bw5.29564(a)news.flashnewsgroups.com #13 http://groups.google.co.uk/groups?&as_umsgid=1090870ihgt6p5f(a)corp.supernews.com #12 http://groups.google.co.uk/groups?&as_umsgid=vuh3v912ilk51f(a)corp.supernews.com #11 http://groups.google.co.uk/groups?&as_umsgid=vm1fb4d1hqc897(a)corp.supernews.com #10 http://groups.google.co.uk/groups?&as_umsgid=6ffd15bd.0308191548.79f09a30(a)posting.google.com #9 http://groups.google.co.uk/groups?&as_umsgid=vibm42507ohma5(a)corp.supernews.com #8 http://groups.google.co.uk/groups?&as_umsgid=vhtma4hh0cs0ac(a)corp.supernews.com #7 http://groups.google.co.uk/groups?&as_umsgid=vhm4f6rucptia4(a)corp.supernews.com #6 http://groups.google.co.uk/groups?&as_umsgid=vg0ndom3o5ihcd(a)corp.supernews.com #5 http://groups.google.co.uk/groups?&as_umsgid=vfrn4g1rmj1gbd(a)corp.supernews.com #4 http://groups.google.co.uk/groups?&as_umsgid=vfrlt15m8qus68(a)corp.supernews.com #3 http://groups.google.co.uk/groups?&as_umsgid=vfgtonnqejtf29(a)corp.supernews.com #2 http://groups.google.co.uk/groups?&as_umsgid=vfem054fm2dk78(a)corp.supernews.com #1 http://groups.google.co.uk/groups?&as_umsgid=vemlos4174kffa(a)corp.supernews.com Enjoy :-) Dirk Vdm
From: mingstb on 10 Mar 2005 01:42
TC wrote: > mingstb(a)sim-ss.com wrote: > > TC wrote: > > > mingstb(a)sim-ss.com wrote: > > > > TC wrote: > > > > > Paul Stowe wrote: > > > > > > Snip....as Mingst has tagged in > mingstb(a)sim-ss.com wrote: > > 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: > > Sigh. Why? It just clutters up the post. > The context is clear enough. > > Snip for clarity. I didn't think that it was clear enough in the particluar post -- that's why I put it back. Because the question under discussion was with regards to your apparent confusion about macroscopic versus microscopic gravitation. Kudos to you for marking the snip, though. Many here don't properly mark the snip. {snip higher levels} > > > 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. I see that you didn't provide an actual reference. > > 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? > > Are you denying that 6 moles of neutrons and 6 moles of protons > in the form of a mole of carbon weighs less than 6 moles of neutrons > and 6 moles of protons in the form of 6 moles of deuterium? > [6 moles of electrons are also needed in each case] No. Again, are you alluding to theoretical claims about 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. > > > Please, be specific. Which experiment? I believe you will find that > > the calculation of energy is completely circular. > > Binding energy is observed. It is why nuclear weapons go boom. I see you again have no experimental support to give. And I don't disagree with the conversion between mass and energy. It is irrelevant to this discussion, however. > > 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." > > So LeSagian interaction couples to binding energy as in GR? The LeSagian interaction couples to inertial mass. Period. Whether a different theory uses "binding energy" to calculate that mass doesn't matter (pun intended). > > Only GR postulates a distinction between "gravitational" mass and > > "inertial" mass. > > Where do you get that from. Newton's Principia, for one. Take a look at Newton's empirical relationship. The mass in F = - G mM/r^2 is inertial mass. Otherwise Kepler's orbital equations (from which Newton derived the above) don't work. > GR is based on principle of equivalence > which equates the two. GR *calls* it the 'Principle of Equivalence'. But only GR makes any claims that there are two kinds of mass. > It is Newton's theory of gravity that > distinguishes them. Your assertion is incorrect and unsupported. Feel free to cite section number in the Principia to support your position. > > Such efforts are irrelevant to non-GR theory. There > > is no reason for non-GR theories to see a difference. > > You have it absolutely backwards. Feel free to cite section number in the Principia to support your position. > ......... > [snip irrelevant quibble] > > > Again, replacing some context: > > =================== > [snip for clarity] > > > Again, totally irrelevant to the issue. The phase of matter is > > immaterial to LeSagian gravitational theory (also to Newtonian > > theory). > > Then why is there an "r" in the expression. If the phas is irrelevant > then it follows that the density does not matter. Matter phase is not density. One can have liquid, gas or solid all at the same density. Phase is irrelevant. Mass density is often relevant. > > 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. > > There you go. So why the "r" in the expression for heating by > LeSagian gravity? Because gravitational heating is not gravitational force. > [snip some more for clarity] > > > > 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. > > If you don't like potential, then think of terminal velocity of > something > falling from infinity or the equivalent escape velocity. These are > observable and have the same dependence on m and r as the > potential. None of these concepts (potential, terminal velocity or escape velocity) enter into the calculation. > > > > 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. > > OK. Where does this go? Where does *what* go? Do you understand the concept that if body A interacts with the corpuscle flux, that it will heat even if no other body is near it? > > 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. > > It will do no such thing. Under LeSagian theory it shadows the second > body. Momentum shadowing *IS* the direct cause gravitation -- in LeSagian theory. > > 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. > > No it doesn't. It is shadowed quite a bit by the larger body reducing > the flux it sees. Neither body removes any significant fraction of the momentum. A gravitational field of 1 g is a variation on the order of 1 part in 10^9 of the incoming momentum flux. > > 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. > > You can bound the heating from the orbital situation. No, you cannot. Mathematics will not allow you to determine two unknowns (Phi and mu) from only one equation. > The imbalance of flux that causes F cannot be larger than > the total flux absent a second body. This provides a lower > bound on the heating. The heating has to be at least as large > as the momentum flux (the force F) times the speed of the > ultramundane particles. Heating is not flux. Even if heating is dependent upon flux, it is also dependent upon the constant, mu. Your simple assertion is simply wrong. As noted in the only substantive discussion in my rebuttal (identified below): ==================== My argument is based on only two things. 1) "Shadows" did not use Le Sagian theory to determine the predictions of Le Sagian theory -- it simply asserted what the equations should be. 2) "Shadows" never justified it's hand-waving assertions. The latter is what Shadow now claims are "perfectly general arguments". But (s)he still provides no basis for ignoring actual Le Sagian equations. Gravitational Force Law: p 188 (one of several ways) F_g = Phi_0 mu_s^2 m M / r2 = G m M / r^2 Drag from Inertial Motion: p 197 F_d = sqrt(3) Phi_0 mu_s m v / vg In the above equations, Phi_0 is the momentum flux of the Le Sagian aether (kg/m-sec2), and mu_s is the mass attenuation coefficient (m2/kg). The ratio between the two forces is therefore given by the equation: Fd/Fg = [sqrt(3)/mu_s M] (v/vg) The overall form of the above equations does not change from one Le Sagian to the next. There is always a "mu_s" term (or variant) that prevents one from determining the drag-to-gravity ratio from purely gravitational permutations. {In Darwin's theory, it is (v a b).} Shadow simply ignores the little problem. ==================== {snip higher levels} > > > How is momentum transfered without absorption of ultramundane > > > particles? > > > However it is done in the real universe, of course. > > You mean to say that LeSagian theory has action at a distance > or that it takes the view of GR that motion is along geodesics of > space time? Not at all, of course. Your strawman is quite silly. > You build up this apparatus of ultramundane particles and then > say its not them that cause the gravitational force after all? Never said any such thing. Your inability to respond substantively is noted. > > 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. > > Oh. But doesn't scattering spoil the shadowing affect that leads > to the LeSagian explanation of gravity? No. Why would you expect it to? > [snip reiteration of the one body/two body discussion] > > > > 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.) > > I'm awaiting interlibrary loan. But I have grave doubts. The true mark of a believer. But not of science. > > > > > 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. > > Fantasy is yet another, I suspect. The last refuge of a losing argument. I'm terribly sorry that a real theory is more difficult to blow off than disconnected pieces of strawman. {snip the rest... uncommented by Jim.) But you may try reading the only substantive response was on the "Shadows" sub-thread: http://groups-beta.google.com/group/sci.physics.relativity/msg/2080fab1f8cc1d21 greywolf42 |