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Natural Science Forum / Physics / Relativity / May 2005Slabinski and Mingst/Stowe disagree in Pushing Gravity
A while back I was discussing with Paul Stowe the expression on his web site for the heating caused by LeSagian gravity. Stowe derives a formua for heat emission that is of the form: (constant)x(mass)/(radius). I did not understand why there was a radius involved and was referred to _Pushing Gravity_, espcially the article by Slabinski. Well, I finally got Pushing Gravity through interlibraty loan. On page 191, equation (26) in the article by Mingst and Stowe, agrees with the above formula that was under discussion. But on page 127, equation (19) in the article by Slabinski has the form (constant)x(mass). The radius of the body is not involved. Thus Slabinski and Mingst/Stowe disagree. What gives? Tom > A while back I was discussing with Paul Stowe the > expression on his web site for the heating caused by [quoted text clipped - 23 lines] > > What gives? So they have been playing with your feet to keep you busy. That's what trolls do :-) Dirk Vdm Dirk Vdm wrote: > "TC" <tclarke@ist.ucf.edu> wrote in message [snip crap] > > What gives? > > So they have been playing with your feet to keep you busy. > That's what trolls do :-) Well yeah, but don't expect your comment to influence Tom Clarke. I can think of no one in this newsgroup who has attempted to give more legitimacy to the extreme crank contingent than Clarke. I suggest he approach the department chair or other faculty in the UCF physics department, and check their level of interest toward the "researchers" who are published by Apeiron. Who knows, maybe Clarke can be the proud host of a colloquium of academia-bashers. ---Tim Shuba--- > Dirk Vdm wrote: > > "TC" <tclarke@ist.ucf.edu> wrote in message > [snip crap] > > > What gives? > > So they have been playing with your feet to keep you busy. > > That's what trolls do :-) > Well yeah, but don't expect your comment to influence Tom Clarke. > I can think of no one in this newsgroup who has attempted to give [quoted text clipped - 3 lines] > the "researchers" who are published by Apeiron. Who knows, maybe > Clarke can be the proud host of a colloquium of academia-bashers. That hurts a bit. Why do you think I got the book through interlibrary loan instead of purchasing it? I don't want to monetarily support them. The scary thing is that the book came, not from a university library, but from the Ocala public library where impressionable youngsters interested in science might check it out and think it real science. I remember picking up Velikowsky's Worlds in Collision at the library as a child and thinking - briefly - that it might be legitimate. But I guess even then I had decent BS filters as I didn't even finish the book. Do you really think my attempts to convince the cranks of the errors of their ways by talking to them as if they weren't cranks is harmful? That it gives them legitimacy? I really thought it was at worst just a harmless hobby. Tom > > Dirk Vdm wrote: > [quoted text clipped - 8 lines] > > > Well yeah, but don't expect your comment to influence Tom Clarke. My comment was not really intended to influence Tom, but rather to tease the trolls. > > I can think of no one in this newsgroup who has attempted to give > > more legitimacy to the extreme crank contingent than Clarke. I think Tom did a remarkably fine job at it :-) > > I > > suggest he approach the department chair or other faculty in the [quoted text clipped - 22 lines] > > I really thought it was at worst just a harmless hobby. Keep up the good work, Tom :-) Dirk Vdm [..] > > I can think of no one in this newsgroup who has attempted to give > > more legitimacy to the extreme crank contingent than Clarke. I [quoted text clipped - 8 lines] > interlibrary loan instead of purchasing it? I don't want > to monetarily support them. Giving your time is much more valuable. > The scary thing is that the > book came, not from a university library, but from the > Ocala public library where impressionable youngsters > interested in science might check it out and think it > real science. I just came across a book at one of my local public libraries that is potentially far more damaging in that respect. My first reaction was similar to yours, but then I decided that having it on the shelf isn't a problem. There's plenty of lousy information on the internet, but anyone wishing to learn a subject in depth isn't likely to be fooled for long. That's one reason I have much respect for our academic system. When individuals come out with nonsense, the free exchange of ideas is a self-correcting feature. > I remember picking up Velikowsky's Worlds in Collision > at the library as a child and thinking - briefly - that > it might be legitimate. But I guess even then I had decent > BS filters as I didn't even finish the book. I had a very similar experience with that book. > Do you really think my attempts to convince the cranks of the > errors of their ways by talking to them as if they weren't cranks > is harmful? That it gives them legitimacy? > > I really thought it was at worst just a harmless hobby. I do think it's potentially harmful, but I'm not sure if I can make a strong case. The particular people you are dealing with have shown utter contempt for anyone involved in the educational system. Their ignorance of modern physics is surpassed only by their arrogance. Since no serious research team would want anything to do with these people, they seek the appearance of legitimacy by trolling knowledgeable people into a discussion. One needs only look at the abuse directed at physicists by these cranks to see where that leads. On the other hand, nobody should take usenet seriously. Watching (and helping) people to make fools out of themselves is in itself harmless, and can be entertaining. The "upside down question mark" is a classic. Here we have someone trying to promote a "theory" (for years!) who has no idea how to even present an equation in a suitable electronic form. In general, there's no way to give legitimacy to cranks, but treating them as non-cranks might do injustice to the many alternative researchers who are not cranks. I just think that people who are so rabidly opposed to the workings of the academic world don't deserve any pretense to legitimacy. ---Tim Shuba--- > .... That's one > reason I have much respect for our academic system. When > individuals come out with nonsense, the free exchange of ideas is > a self-correcting feature. Yes, well said. Unfortunately cranks advocate ideas that were rejected long ago in a free exchnage of ideas. The "maintream" has no interest in conducting an exchange with cranks, so unfortunately all too often their ideas go unchallenged (on the net anyway) unless die-hard's like me respond and comment. Many cranks are self-evidently cranks [See Baez's Crackpot Index: http://math.ucr.edu/home/baez/crackpot.html] But some - the ones I respond to - manage to sound legitimate. Hence my hobby of refuting their ideas. .... [cranks] they seek the appearance of > legitimacy by trolling knowledgeable people into a discussion. > One needs only look at the abuse directed at physicists by these > cranks to see where that leads. I don't bother with the really abusive ones, they are self evidently cranks. I like to think [perhaps erroneously] that I demonstrate the errors of those who attempt to pass. > On the other hand, nobody should take usenet seriously. Oh, I've gotten questions answered in a useful way on usenet, so it's not all bad. > In general, there's no way to give legitimacy to cranks, but > treating them as non-cranks might do injustice to the many > alternative researchers who are not cranks. If I were to just drop in on a thread by van Flandern/McCarthy/Mingst/Stowe/etc and call them a crank, I don't think it would have an effect. They would continue in their pseudo-scientific jargon perhaps misleading young lurkers astray. I think it better to point out the errors in their pseudo-theories. Usually they give up after a while, but of course, they are not convinced of the error of their pseudo-theories. But maybe some lurker is benefited. And - god help me- I find the verbal jousting amusing. Tom > Unfortunately cranks advocate ideas that were rejected long > ago in a free exchnage of ideas. The "maintream" has no [quoted text clipped - 6 lines] > But some - the ones I respond to - manage to sound > legitimate. Hence my hobby of refuting their ideas. [..] > If I were to just drop in on a thread by > van Flandern/McCarthy/Mingst/Stowe/etc and call them a crank, [quoted text clipped - 5 lines] > error of their pseudo-theories. But maybe some lurker is > benefited. Okay, that's reasonable. I retract my comment that you have attempted to give legitimacy to cranks. Clearly, you have not. ---Tim Shuba--- > > .... I think it better to point out the > > errors in their pseudo-theories. Usually they give up > > after a while, but of course, they are not convinced of the > > error of their pseudo-theories. But maybe some lurker is > > benefited. > Okay, that's reasonable. I retract my comment that you have > attempted to give legitimacy to cranks. Clearly, you have not. I posted a response to Gerald L. O'Barr yesterday. His response may well cure me of posting in response to cranks on s.p.r. Thanks for your comments. Tom TC: >A while back I was discussing with Paul Stowe the >expression on his web site for the heating caused by [quoted text clipped - 23 lines] > >What gives? You've been duped into wasting your time trying to come up with an argument against a well discredited idea for the sake of trying to persudae two kooks who don't even grasp basic classical physics why it's unphysical. You'd have a better shot trying to convince the pope to become a rabbi at a jewish temple and his argument for not doing so would be more plausible, even to an athiest. > A while back I was discussing with Paul Stowe the > expression on his web site for the heating caused by [quoted text clipped - 6 lines] > was referred to _Pushing Gravity_, espcially the article > by Slabinski. Ummm, actually that was to help you with the question of 'excess' heating. However, you have a good question, and I will address it this weekend (when I have more time to devote). Note however Slabinski's equation is not one of specific heat (watts/m^2), as is ours, but of total heat (watts). > Well, I finally got Pushing Gravity through interlibraty > loan. [quoted text clipped - 9 lines] > The radius of the body is not involved. Thus Slabinski and > Mingst/Stowe disagree. Actually, it is. But it's buried in the solid angle. More on the weekend. However, note the total lack of any technical content OR attempt to address your actual question but the cynics around here... Paul Stowe > What gives? How does one determine the the solid angle??? Paul Stowe Mingst/Stowe > > On page 191, equation (26) has form > > (constant)x(mass)/(radius). page 127, equation (19) has form Slabinksi (constant)x(mass) > ... Note however Slabinski's equation is not one of > specific heat (watts/m^2), as is ours, but of total heat > (watts). Then if Slabinski is correct, your equation should take the form. (constant)x(mass)/(radius^2) .... > > The radius of the body is not involved. Thus Slabinski and > > Mingst/Stowe disagree. > Actually, it is. But it's buried in the solid angle. Slabinski works out a different functional form for heating. Is he wrong then? Does he not account for angles correctly? > More on the weekend. I await your explanation. Tom > Mingst/Stowe > [quoted text clipped - 7 lines] > Then if Slabinski is correct, your equation should take the form. > (constant)x(mass)/(radius^2) Fine, point to the specific step in the following that is either, - mathematically wrong, or, - logically inconsistent We start will conservation and say, at equilibrium, q_in = q_out Where q is the power flux per unit area of the graviton field. Note that q is the current, not omni-directional flux. For ease of notation we'll set q_in = q & q_out = q'... Then for an attenuating mass we say, -ß -ß q = q' = qe + q(1 - e ) The ß term is the total attenuation parameter. Clearly, if ß -> 0 then we have, q' = q + (q - q) = q And, if ß -> oo then, q' = 0 + (q - 0) = q In he first case, nothing interacts, in the second, all interacts & is ultimately re-emitted as a secondary flux. Either way, in equilibrium is strictly maintained! Thus, the 'delta' or interacting component of q is (q''), -ß q'' = q(1 - e ) Therefore, when ß << unity the Taylor series shows us that this can be quantified (to a very high precision) by simply writing q'' = qß The ß term is an expression of the departure from equilibrium of the graviton fluence. This can be found on page 190 Eq(22) and is given as 2GM/rc^2. Now follow this through q'' = q(2GM/rc^2) Regrouping this we get, q'' = (q2G/c^2)M/r Finally, we know that within LeSage's model, G = ¿µ^2 and q = ¿c/4pi Thus, (q2G/c^2) => ([¿µ]^2/[2pi]c) => k q'' = kM/r Now, to Slabinski version. ... In our analysis above we've taken a 'big picture' or macroscopic continuum approach to the issue. In Slabinki's work he take the microscopic or kinetic theory type approach. You could say ours was a top down and Slabinki's a bottom up analysis. We must also quantify Slabinki's terms and map them to their counterparts in our approach. Slabinki defines, N = graviton particle density (particles per unit volume) A = test area (length squared) A' = absorption cross-sectional area for the smallest possible interacting material particle... (length squared) A''= scattering cross-sectional area for the smallest possible interacting material particle... (length squared) ç = Solid angle sutended by A (Radians) K = mass absorption coefficient (length squared per unit mass) K' = mass scattering coefficient (length squared per unit mass) m = m, m' test particles of gravitating mass r = distance of A from m R = Rates (R, R', R'') net, direct, scatter £ = net decrease in graviton flux density c = graviton mean speed w = graviton mass Mapping into our version. ¿ = Nwc^2 µ = Sqrt[K(K + K'[1 - Cos æ])] Thus mapping Slabinki's Eq 19 we get, H = (¿2piKc)m Dimensionally this is, kg | m^2 | kg | m kg-m^2 -------+-----+----+--- => ------ m-sec^2| kg | |sec sec^3 Converting to a per unit area (4piL^2)we get, (¿Kc/2)m/L^2 => kg/sec^3 Note that Slabinki is evaluating the test area of a single interacting differential particle of matter. A, A', A'' as well as ç and æ are set by this. HE IS NOT! evaluating a macroscopic body consisting of multiple test particles. The 'solid' angles ç and æ are affected by size. However, for his analysis size doesn't change. Note, area is proportional to r^2 and density to r^3, a 1/r differential. Paul Stowe > > Mingst/Stowe > > [quoted text clipped - 16 lines] > > q_in = q_out > Where q is the power flux per unit area of the graviton field. Note > that q is the current, not omni-directional flux. For ease of > notation we'll set q_in = q & q_out = q'... Then for an attenuating > mass we say, > -ß -ß > q = q' = qe + q(1 - e ) > The ß term is the total attenuation parameter. Clearly, if ß -> 0 > then we have, Actually this expression is true for any value of beta, you have just added and substracted teh same thing, q e to the minus beta. > q' = q + (q - q) = q > And, if ß -> oo then, > q' = 0 + (q - 0) = q True also for any value of beta, as noted. > In he first case, nothing interacts, in the second, all interacts & > is ultimately re-emitted as a secondary flux. Either way, in > equilibrium is strictly maintained! It's just adding and subtracting the same thing. > Thus, the 'delta' or interacting component of q is (q''), > -ß > q'' = q(1 - e ) Why "thus"? I would tend to think that this is the non-interacting component. e to the minus beta would be the term describing absorption and thus interaction. > Therefore, when ß << unity the Taylor series shows us that this > can be quantified (to a very high precision) by simply writing > q'' = qß We all know how Taylor series work. > The ß term is an expression of the departure from equilibrium of > the graviton fluence. This can be found on page 190 Eq(22) and > is given as 2GM/rc^2. Now follow this through Unfortunately, I left the book at my desk. But that doesn't sound correct. The absorbtion should just be determined by M. r should not have anything to do with it. Probably this is where the r term comes in in your expressions that Slabinsky does not have in his. > q'' = q(2GM/rc^2) > Regrouping this we get, > q'' = (q2G/c^2)M/r > Finally, we know that within LeSage's model, > G = ¿µ^2 > and > q = ¿c/4pi > Thus, > (q2G/c^2) => ([¿µ]^2/[2pi]c) => k > q'' = kM/r > Now, to Slabinski version. ... > [quoted text clipped - 4 lines] > quantify Slabinki's terms and map them to their counterparts in > our approach. Slabinki defines, > N = graviton particle density (particles per unit volume) > A = test area (length squared) [quoted text clipped - 11 lines] > c = graviton mean speed > w = graviton mass > Mapping into our version. > > ¿ = Nwc^2 > µ = Sqrt[K(K + K'[1 - Cos æ])] I really wish you wouldn't use that upside down question mark. If you mean psi, write psi, please. > Thus mapping Slabinki's Eq 19 we get, > H = (¿2piKc)m Heat absorbed and thus re-emitted is a constant times the mass. > Dimensionally this is, > > kg | m^2 | kg | m kg-m^2 > -------+-----+----+--- => ------ > m-sec^2| kg | |sec sec^3 > Converting to a per unit area (4piL^2)we get, > (¿Kc/2)m/L^2 => kg/sec^3 > Note that Slabinki is evaluating the test area of a single > interacting differential particle of matter. A, A', A'' > as well as ç and æ are set by this. HE IS NOT! evaluating > a macroscopic body consisting of multiple test particles. A macroscopic body is comprised of test particles. The sum of the heat input to test particles should give the value for the macroscopic body. I guess is beta >> 1 there might be shielding of the interior that would make the sum not proportional to total mass, but you explicitly assume beta <<1. > The 'solid' angles ç and æ are affected by size. The sum of the solid angles is the sum of the solid angles. > However, > for his analysis size doesn't change. Note, area is > proportional to r^2 and density to r^3, a 1/r differential. This makes no sense. Tom >>> Mingst/Stowe >>> [quoted text clipped - 29 lines] > Actually this expression is true for any value of beta, you have > just added and substracted teh same thing, q e to the minus beta. Yes, that was the point of showing both boundaries... >> q' = q + (q - q) = q > [quoted text clipped - 9 lines] > > It's just adding and subtracting the same thing. That's the whole of conservation, don't ya'think? >> Thus, the 'delta' or interacting component of q is (q''), > [quoted text clipped - 4 lines] > component. e to the minus beta would be the term describing > absorption and thus interaction. Nope, you're confused. In the expression, -ß I' = Ie -ß I' is that amount the 'actually makes it to be measured, I(1 - e ) is the amount attenuated... >> Therefore, when ß << unity the Taylor series shows us that this >> can be quantified (to a very high precision) by simply writing >> >> q'' = qß > > We all know how Taylor series work. You'd be surprised... But, given you have a degree in applied math I'd think you would... >> The ß term is an expression of the departure from equilibrium of >> the graviton fluence. This can be found on page 190 Eq(22) and [quoted text clipped - 94 lines] > > This makes no sense. I'm make one last try. The LeSagian flux in any quantity (particle, momentum, ..., etc.) IS a property of its density and mean speed. Given a relatively uniform densiy and mean speed this is pretty consistent and can be treated as a constant (i). Thus 'per unit area' it has the ability to transfer up to that maximum amount of power from itself to an intervening attenuator. So, let now have an attenuator that, for sake argument, transfer some fraction which is a function of its mass across an area A. Thus for this part let consider only the mono-directional fluence shown by the => travresing the slab indicated by |<--t--|, => | | => | | => | | i => |<---- t------>| => | | => | | => | {rho} | We say, -µ{rho}t H = iA(1 - e ) Crush t to 0.5t but keep the same mass, still => | | => | | => | | i => |<-t'->| => | | => | | => |{rho'}| -µ{rho'}t' H = iA(1 - e ) And, rho' = 2{rho}, t' = t/2 No change in H... In this case H is simply a function of mass, right? I agree! Any questions about this so far??? Paul Stowe > > Why "thus"? I would tend to think that this is the non-interacting > > component. e to the minus beta would be the term describing > > absorption and thus interaction. > Nope, you're confused. In the expression, > -ß > I' = Ie -ß > I' is that amount the 'actually makes it to be measured, I(1 - e ) > is the amount attenuated... I understand your terminology now. ................. > > A macroscopic body is comprised of test particles. > > The sum of the heat input to test particles should give the [quoted text clipped - 15 lines] > (particle, momentum, ..., etc.) IS a property of its density > and mean speed. What does the pronoun "its" refer to? The quantity or the LeSagian flux? > Given a relatively uniform densiy and mean > speed this is pretty consistent and can be treated as a > constant (i). Thus 'per unit area' it has the ability to > transfer up to that maximum amount of power from itself to > an intervening attenuator. >From this I conclude it is the LeSagian flux. > So, let now have an attenuator > that, for sake argument, transfer some fraction which is a > function of its mass across an area A. This area would be the cross section of the attenuator due to the attenuator's mass. > Thus for this part > let consider only the mono-directional fluence shown by the [quoted text clipped - 7 lines] > => | | > => | {rho} | > We say, > -µ{rho}t > H = iA(1 - e ) > Crush t to 0.5t but keep the same mass, still > => | | > => | | [quoted text clipped - 7 lines] > H = iA(1 - e ) > And, > rho' = 2{rho}, t' = t/2 > No change in H... In this case H is simply a function > of mass, right? I agree! Yes indeed. So why does your expression differ from Slabinski's? > Any questions about this so far??? The analysis is much easier when the concept of cross sectional area/unit mass is used. If I recall correctly that is how Slabinski does it. Tom >The analysis is much easier when the concept of cross sectional >area/unit mass is used. If I recall correctly that is how Slabinski >does it. The rate of energy absorption of a material body of mass M in a Lesagean theory is well known to be 6pi G A v_g M, where G is Newton's gravitational constant, A is the opaque cross-sectional area of one unit of mass, and v_g is the speed of the ultramundane corpuscles. This is based on the assumption that Newton's third law is exactly satisfied by the gravitational force, i.e., that the mutual forces that bodies exert on each other are exactly equal and opposite. If we relax this assumption, and allow some violation of Newton's third law (within empirical limits), the rate of energy absorption can be reduced by postulating a certain degree of reflection along with absorption. However, even with this, we still find that the Earth must be absorbing energy at a rate that is many orders of magnitude greater than the Sun's total rate of energy output, which would be sufficient to vaporize the Earth in a fraction of a second. > >The analysis is much easier when the concept of cross sectional > >area/unit mass is used. If I recall correctly that is how Slabinski [quoted text clipped - 4 lines] > gravitational constant, A is the opaque cross-sectional area of one > unit of mass, and v_g is the speed of the ultramundane corpuscles. A complete falsehood. Provided simply by hand-waving. No reference given, of course. From what reference, exactly, did you get this "well known" equation? Since you refer to an "opaque cross-sectional area of one unit of mass" you immediately identify that you aren't using "LeSage's" theory at all. For LeSage's matter was 99.999...% transparent. I suspect you are trying to use either Darwin's variant or Slabinski's variant -- but failing anyway. You fail in both of the specific theoretical variations, because the gravitational heating equation in a LeSage theory does not contain "G" at all! In a Lesagian theory, "G" is a product of the momentum flux and mass attenuation coefficient to the second power (or some variant thereof). But the heating equation only includes mass attenuation coefficient to the first power. In Slabinski (p 127 "Pushing Gravity"), for example, the equation is: H = 2 pi N m_g v_g^3 K_abs M > This is based on the assumption that Newton's third law is exactly > satisfied by the gravitational force, i.e., that the mutual forces [quoted text clipped - 6 lines] > than the Sun's total rate of energy output, which would be sufficient > to vaporize the Earth in a fraction of a second. Ah! Another classic myth. Provided merely as proof-by-repetition and assertion. Again, to what reference are you specifically asserting? Or are you again just repeating the myth? I can e-mail you the old chestnut of Poincare, if you like. -- greywolf42 ubi dubium ibi libertas {remove planet for e-mail} >Since you refer to an "opaque cross-sectional area of one unit of mass" >you immediately identify that you aren't using "LeSage's" theory at all. >For LeSage's matter was 99.999...% transparent. The opaque cross-sectional area of one kilogram of matter in any Fatio-LeSage theory of gravity can be no greater than about 1E-10 square meters. In other words, macroscopic bodies are almost totally transparent to the ultramundane flux. This is just Fatio 101. But those bodies cannot be TOTALLY transparent, or there would be no interaction. A given quantity of matter must have a non-zero opaque cross-sectional area, which, for ordinary bodies must equal the sum of the areas of the constituent microscopic entities, e.g., Lesages "bars" or, equivalently, Darwin's particles, both of which are perfectly opaque. Understand? > ...the gravitational heating equation in a LeSage theory does not > contain "G" at all! The empirical value of G is a measure of the coupling strength of gravity. If G was zero, there would be no interaction, and therefore, no heat absorption at all. So it makes no sense to claim that the heat absorption does not depend on G. Understand? >From what reference, exactly, did you get this "well known" equation? You can find a derivation on the web at http://www.mathpages.com/home/kmath426/kmath426.htm > >Since you refer to an "opaque cross-sectional area of one unit of mass" > >you immediately identify that you aren't using "LeSage's" theory at all. [quoted text clipped - 23 lines] > You can find a derivation on the web at > http://www.mathpages.com/home/kmath426/kmath426.htm Ouch! Pointing Mingst to the mathpages is a Definite Nono. Prepare for a bucket of something with a nasty smell ;-) Dirk Vdm >> Since you refer to an "opaque cross-sectional area of one unit of >> mass" you immediately identify that you aren't using "LeSage's" [quoted text clipped - 18 lines] > no heat absorption at all. So it makes no sense to claim that the > heat absorption does not depend on G. Understand? And G is 'in fact' the coupling constant for the interaction of two bodies and a 'resulting' force between them. Thus it is, Fatio/LeSage momentum flux (¿) AND that attenuation coefficent (µ) SQUARED or, G = ¿µ^2! For energy deposition only ONE body & the LeSagian field are involved or just ¿µ NOT ¿µ^2 'Understand'??? Trying to relate the magnitude of force generated between two bodies to the energy deposited in one is mixing apples & oranges... Understand! >> From what reference, exactly, did you get this "well known" equation? > > You can find a derivation on the web at > http://www.mathpages.com/home/kmath426/kmath426.htm OK, where in, -µ{rho}t q' = ¿(1 - e ) Or for the weak limit, q' = ¿[µ(rho)t)] Thus, as can be clearly seen, q' = (¿µ)[rho)t] Now, if you want, desire, or require that this be cast in a form that contains G, then of course, q' = G[rho)t/µ and how does one determine the value for µ??? Paul Stowe > ...how does one determine the value for µ??? See the referenced web page, which gives the complete derivation in detail. In brief, one determines the ultramundane flux and the absorption (and reflection if you're willing to accept small violations of Newton's third law for gravity) necessary to account for the observed force of gravity, consistent with the absence of any appreciable shielding. This amount of flux and absorption corresponds to a specific amount of energy absorption, depending on the speed of the supposed ultramundane particles. Lower bounds on this speed are then determined, giving lower bounds on the rate of energy absorption. >> ...how does one determine the value for µ??? > > See the referenced web page, which gives the complete derivation in > detail. Really? I musta' missed the equation that quantified the value of µ. Care to point it out? > In brief, one determines the ultramundane flux and the absorption > (and reflection if you're willing to accept small violations of > Newton's third law for gravity) necessary to account for the observed > force of gravity, consistent with the absence of any appreciable > shielding. This is quite vague and I presume that you're talking about the weak limit, linearized cases. > This amount of flux and absorption corresponds to a specific amount > of energy absorption, ... Duh! > ... depending on the speed of the supposed ultramundane particles. Ah, do you mean for the ultra-mundane particles of mass m' and mean speed c that dp = m'(dc) and dE = m'c(dc) > Lower bounds on this speed are then determined, ... Not by lack of aberration you can't! There are many specific detail problems in the page you referenced. Barry took the time & effort to point many of these out. http://groups-beta.google.com/group/sci.physics.relativity/msg/982030b9f3d78380? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/3db4a663f4a27799? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/53e6d3fafce5c1ad? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/ab1f374a72cab477? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/6fdf4a19783e5397? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/7ba762e40f7fb88c? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/440347db69626a4c? dmode=source&hl=en http://groups-beta.google.com/group/sci.physics.relativity/msg/3bfa252862d7d98c? dmode=source&hl=en (Comment for the peanut gallery [this means YOU Dinky!]: Trying to distract by references to irrelvant postings is the mark of someone that has no capability for any other relevant technical input.) And this page is a moving target and is modified often. In the time these were posted we have, http://groups-beta.google.com/group/sci.physics.relativity/msg/a5a08ded256cc736? dmode=source&hl=en and it has continued to be changed... Nature herself are not bound by assumptions and constraints placed on model by those unwilling or too ignorant question them! > .. giving lower bounds on the rate of energy absorption. Oh, BTW the correct specific power equation is, q' = ¥[rho)t q' = (¿c/4pi)µ[rho)t Since power flux ¥ is related to momentum flux by, ¥ = ¿c/4pi You failed one implicit test of your knowledge of details of a LeSagian model!!! Hopefully you are NOT the author of that reference :) Paul Stowe > There are many specific detail problems in the page you referenced. Barry > took the time & effort to point many of these out. The series of (rather demented and, one notices, thoroughly refuted) posts you referenced by "Barry" (greywolf42?) do not even pertain to the web page I gave you, they refer to a different page on the same site, called Lesages Shadows. The page I referred you to is called Omni-Directional Flux. >> There are many specific detail problems in the page you referenced. >> Barry took the time & effort to point many of these out. > > The series of (rather demented and, one notices, thoroughly refuted) Great! Now provide the references to the post(s) that contain any technical content of refutation. Again, I musta' missed'em. All I ever saw was narrative & ad hominems attacks, like this... > posts you referenced by "Barry" (greywolf42?) do not even pertain > to the web page I gave you, they refer to a different page on the > same site, called Lesages Shadows. The page I referred you to is > called Omni-Directional Flux. I know, that was posted AFTER the rebutals were posted. As I said in the previous post, this page has, & is, ungoing rather frequent revisions. However, Slabinki does a much more concise job than that page. And then again, they say, the proof of the pudding is in the eatting!. The equation derived empirically for heating is, -(UA/MC)t q' = kM/r(1 - e ) Where U is the overall heat transfer coefficient A is the radiating surface area M is the mass r is a spherical body's radius C is the heat capacity t is time since creation k is an emperically determined LeSagian constant ~2.4E-19 q' is Power per unit area As t -> oo this becomes simply, q' = kM/r Now the challenge to you is, go figure out if it fits the excess thermal emmissions of Planetary bodies. Want'a guess??? Paul Stowe > Great! Now provide the references to the post(s) that contain any > technical content of refutation. http://groups-beta.google.com/group/sci.physics.relativity/msg/c7bca3bd38b03b69 >> The page I referred you to is called Omni-Directional Flux. > I know.... Slabinki does a much more concise job than that page. So now your objection to the referenced page is that it's not as concise as it could be? That's certainly possible, but from the standpoint of someone like yourself (i.e., someone who has no clue about how to even begin thinking about the subject), one would think you would actually benefit more from a less concise presentation, one that explains the reasoning behind each step. > And then again, they say, the proof of the pudding is in the eatting! >The equation derived empirically for heating is... q' = kM/r. Now the > challenge to you is, go figure out if it fits the excess thermal > emmissions of Planetary bodies. Want'a guess??? There's no need for guessing. The point of the exercise is to infer from the Fatio-Lesage model of gravity (along with the observed properties of gravity, and Newton's three laws of motion) how much kinetic energy must be absorbed by material bodies. The answer is entirely unambiguous: A body with the mass of the Earth must be absorbing energy at a rate that would be sufficient to vaporize the Earth in a fraction of a second. You've been pointed to a detailed and rigorous proof of this. What more is there to say? Strael Nosduj: >> Great! Now provide the references to the post(s) that contain any >> technical content of refutation. [quoted text clipped - 11 lines] >you would actually benefit more from a less concise presentation, one >that explains the reasoning behind each step. I think paul's idea of concise means something that doesn't contain any math he can't follow -- especially equations accompanied by explanations that justify them. Unfortuntely, that seems to be anything more complex than basic algebra. The only way to make something concise enough for paul is to remove all of the physics so he can work on a semantic argument. > ... The equation derived empirically for heating is, empirically? No theory? > -(UA/MC)t > q' = kM/r(1 - e ) [quoted text clipped - 7 lines] > k is an emperically determined LeSagian constant ~2.4E-19 > q' is Power per unit area > As t -> oo this becomes simply, > q' = kM/r Once again the problem is that this is incompatible with Slabinski or any other LeSage/Fato theory in which provides heat on a per unit mass basis. Mass is M - no dependence on r. Area is 4pi r^2 so to be compatible with Slabinski et al your expression should read q' = KM/r^2 [pi etc goes into little k to big K switch] Tom >> ... The equation derived empirically for heating is, > > empirically? No theory? Come'on Tom, every significant step is spelled out in our article (Eq. 22-26, pages 190-191). Within the assumptions specified there is, as far as I'm aware, NO mathematical inconsistency there. >> -(UA/MC)t >> q' = kM/r(1 - e ) [quoted text clipped - 15 lines] > Slabinski or any other LeSage/Fato theory in which provides > heat on a per unit mass basis. Yes, on a 'per mass basis'. > Mass is M - no dependence on r. > Area is 4pi r^2 [quoted text clipped - 4 lines] > > [pi etc goes into little k to big K switch] OK Tom, do a simple fit test. Let Jupiter be the base and we'll say that the net heat is proprtional to either the mass divided by radius (s) OR mass (m). Now, we know that the net heat flux for Jupiter is ~ 6.6 Watts/m^2. Thus, either, x = 6.6(m/m') or x = 6.6(s/s') Where s' & m' are the mass & mass per radius of our basis, Jupiter. Go through the Planetary data & see which empirically fits best... The Moon net is 0.02 Watts/m^2 & is two times higher than expected. Neptune is ~1 Watt/m^2, Saturn 2.3 Watts/m^2, Earth is 0.06 Watts, ... etc. Now the Earth is an interesting study. We know that, __ q' = U \/T __ And, assuming that \/T is ~ 10,000° C and q is 0.06, then one can solve for U... Then, C (heat capacity) is roughly density dependent such that, C ~ 0.13(rho) Where rho is density in kg/m^3. Thus, with Earth's overall density @ 5525 kg/m^3 C-> ~718. Thus the 'Lumped Heat' thermal response coefficient is, UA -- mC Given A = 4pir^2 and m = (rho)4pir^3/3 we get, 3U ------- C(rho)r (There's that thar area to volume 1/r effect showing up in the exponential response term) and, U is ~6E-06. Thus, -1 1/t = [3(6E-06)]/[718(5525)6.374E+06] = 7.12E-19 sec this is 44.5 Billion years... The Earth cannot have reached any thermal equilibrium status EVEN IF! the was just a one time induction of thermal input a t = 0 (time of creation). So the 0.06 MUST be significantly lower than what will be expected at equilibrium. Paul Stowe > >> ... The equation derived empirically for heating is, > > empirically? No theory? > Come'on Tom, every significant step is spelled out in our article > (Eq. 22-26, pages 190-191). Within the assumptions specified > there is, as far as I'm aware, NO mathematical inconsistency there. Then why did you use the word "empirically"? [But there must be a mistake somewhere in your paper or in Slabinski's or else your results would agree with Slabinski.] Snip definitions for brevity. > >> q' = kM/r > > Once again the problem is that this is incompatible with > > Slabinski or any other LeSage/Fato theory in which provides > > heat on a per unit mass basis. > Yes, on a 'per mass basis'. > > Mass is M - no dependence on r. > > Area is 4pi r^2 > > so to be compatible with Slabinski et al your expression > > should read > > q' = KM/r^2 > > > > [pi etc goes into little k to big K switch] > OK Tom, do a simple fit test. Let Jupiter be the base and > we'll say that the net heat is proprtional to either the > mass divided by radius (s) OR mass (m). Don't forget that m/r can also be accounted for by classical gravitational potential energy of collapse when the planets formed. > ... Now, we know that > the net heat flux for Jupiter is ~ 6.6 Watts/m^2. Thus, > either, > x = 6.6(m/m') > or > x = 6.6(s/s') > Where s' & m' are the mass & mass per radius of our basis, > Jupiter. [Paul argues that m/r is the best fit by comparing some heat fluxes from various planets] Thereby he disproves LeSagian gravity (which predicts m dependence) and supports classical theory wherein m/r is deposited in the body from its gravitational collapse and is currently escaping from the body as relict heat. Tom >> >> ... The equation derived empirically for heating is, > [quoted text clipped - 5 lines] > > Then why did you use the word "empirically"? Because, like Newton & his equation, it is tied directly to observation. It is one thing to say, q' = (¥2G/c^2)m/r and another to solve for ¥, by 'linking it' to an observed value. Like Newton when he said, I observe that, F o< Mm/r^2 And the solves for the magnitude of the proportionality constant BY linking it back to measured observations... That's the 'empirical' part! > [But there must be a mistake somewhere in your paper or in Slabinski's > or else your results would agree with Slabinski.] As I told you earlier, now that we've been made aware of this, it IS being worked on. My gut says, we're both right. We'll see. > Snip definitions for brevity. > [quoted text clipped - 42 lines] > is deposited in the body from its gravitational collapse > and is currently escaping from the body as relict heat. Sigh, why do you deliberately distort? You know damned well that I've said no such thing. No,r do I think so. You also know damned well that if what you are claiming is true there would have never been and issue of 'anomalous excess heat' for the gas giant planets. Further, you should also know that that very same ¥ term is utilized to decompose G into its constitute parts of ¿ & µ. These in turn, are used in the classic derivation of the drag equation, which, just amazingly, matches the precise magnitude of that observed in Pioneer & Ulysses spacecraft. Two from one, an amazing series of coincidences, eh? Paul Stowe > > Then why did you use the word "empirically"? > > Because, like Newton & his equation, it is tied directly to > observation. It is one thing to say, > q' = (¥2G/c^2)m/r > and another to solve for ¥, by 'linking it' to an observed value. You can always make one observation fit a theory by adjusting a constant [well almost always] > Like Newton when he said, I observe that, > F o< Mm/r^2 > And the solves for the magnitude of the proportionality constant > BY linking it back to measured observations... Newton had a whole lot, enormously more support, than you can claim. > That's the 'empirical' part! But the form of the equation is based on a theory. It is not a purely empirical formula like a regression equation. Better to term it a theoretical formula. > > [But there must be a mistake somewhere in your paper or in Slabinski's > > or else your results would agree with Slabinski.] > As I told you earlier, now that we've been made aware of this, > it IS being worked on. My gut says, we're both right. We'll > see. OK. .....snip for brevity..... > > [Paul argues that m/r is the best fit by comparing some > > heat fluxes from various planets] > > Thereby he disproves LeSagian gravity (which predicts m > > dependence) and supports classical theory wherein m/r > > is deposited in the body from its gravitational collapse > > and is currently escaping from the body as relict heat. > Sigh, why do you deliberately distort? You know damned well > that I've said no such thing. I know you didn't say that your results support classical gravity theory. You say excess heat fits an equation of the form Km/r. _I_ pointed out that this can be seen as support for classical gravity theory. > No,r do I think so. You also > know damned well that if what you are claiming is true there > would have never been and issue of 'anomalous excess heat' > for the gas giant planets. I just point out that potential energy has form m/r. I'm really not that familiar with all the possible explanations people have posited. But any one having an m/r dependence would fit the same data. By the way the fit is not all that good. > Further, you should also know that that very same ¥ term is > utilized to decompose G into its constitute parts of ¿ & µ. What does this have to do with the alternative explanation provided by classical gravity theory? The decomposition is only part of your theory. Also, Slabinski does a decomposition and derives an upper bound from it. > These in turn, are used in the classic derivation of the drag > equation, which, just amazingly, matches the precise magnitude > of that observed in Pioneer & Ulysses spacecraft. > Two from one, an amazing series of coincidences, eh? Now you have to tackle the binary pulsar, deflection of light by the sun, gravitational redshift ... Tom >Now you have to tackle the binary pulsar, deflection of >light by the sun, gravitational redshift ... ... the gravitational attraction between material bodies... >>Now you have to tackle the binary pulsar, deflection of >>light by the sun, gravitational redshift ... > > ... the gravitational attraction between material bodies... The LeSagian force equation for the weak limit is, (µm)(µM) F = ¿ -------- d^2 And strong limit as, (r^2pi)(R^2pi) F = ¿ ------------ d^2 Where, ¿ = momentum flux (kg/m-sec^2) µ = total mass attenuation coefficient (m^2/kg) d = mean distance between centers (m) m,M = mass of bodies (kg) r,R = radius of bodies (m) F = force (kg-m/sec^2) G = ¿µ^2 (m^3/kg-sec^2) This has been known since the time of LeSage and is not in dispute except, maybe by you... Paul Stowe > The LeSagian force equation for the weak limit is, > F = ¿(µm)(µM)/d^2 where [quoted text clipped - 6 lines] > This has been known since the time of LeSage and is not in > dispute except, maybe by you... The problem is not with that equation, it's with your failure to realize how much energy absorption is implied by that equation in order to account for the observed strength of gravity. Let me try to explain this in simple words: Gravity exerts known forces on objects of known masses, in accord with Newton's law and the empirical value of the gravitational constant G, which in your model equals phi mu^2. Furthermore, the observed absence of any appreciable shielding or saturation effects - even for objects as large as the planets - implies a very low upper bound on the value of mu. Knowing the value of G, and taking the largest possible value of mu consistent with observation, we can compute the minimum possible value of phi as G/mu^2. This is the minimum amount of momentum flux that must be absorbed by material bodies in order for them to be attracted with the observed force of gravity. (It's not hard to show that mixing reflection with absorption results in unequal mutual forces between bodies, i.e., it violates Newton's third law. We could allow a certain amount of this, within empirical limits, but it doesn't change the overall conclusion, so you should first try to understand the absorption case, and then if you're really interested you can learn about the effects of including reflection.) For a given amount of absorbed momentum flux there is a definite corresponding amount of kinetic energy, depending on the velocity of the momentum flux, because the ratio of the kinetic energy to the momentum of a particle is simply v/2. If the ultramundane corpuscles are slow and heavy, the kinetic energy they deliver for a given amount of momentum is small, but if they are fast and light, the kinetic energy for a given amount of momentum is large, directly proportional to the velocity. This is just elementary physics. So the question is, how fast are the corpuscles moving? We can place lower bounds on this speed in several different ways, and so we have everything we need to compute a lower bound on how much energy must be absorbed in order to account for the observed force of gravity and lack of shielding, and lack of aberration and lack of drag. When we make this calculation we find that matter must be absorbing energy at a stupendous rate - fast enough to vaporize any material object in a fraction of a second. This is true even if we assume the relativiely slow value of c for the speed of the corpuscles. Two factors were glossed over in the above discussion. One of them decreases the required amount of energy absorption, and the other increases it, so the conclusion doesn't change when they are taken into account... they would only serve to confuse you, so I will leave them out. (One is the effect of reflection combined with absorption, and the other is to account for the heating effect of the entire amount of absorbed flux, rather than just the net unbalanced amount.) These conclusions can't be avoided by postulating some occult mechanism for isotropically re-radiating the stupendous amounts of absorbed energy in a form (ultra-ultra-mundane corpuscles?) that is totally invisible to and non-interacting with ordinary matter, because any such mechanism would violate the second law of thermodynamics. >Knowing the value of G, and taking the largest possible value >of mu consistent with observation, we can compute the minimum possible >value of phi as G/mu^2. This is the minimum amount of momentum flux >that must be absorbed by material bodies in order for them to be >attracted with the observed force of gravity. Ooppss.. I meant to say the minimum absorbed flux is phi mu, which equals G/mu, which gets smaller as mu gets larger, and since we have an upper bound on mu, we have a lower bound on the absorbed flux (phi mu). Of course, this is just the net unbalanced flux. The total absorbed flux is many orders of magnitude greater. >> Knowing the value of G, and taking the largest possible value >> of mu consistent with observation, we can compute the minimum possible [quoted text clipped - 7 lines] > mu). Of course, this is just the net unbalanced flux. The total > absorbed flux is many orders of magnitude greater. Let's go with this... The flux [intensity (¿)] is momentum flux which is that amount of momentum traversing a unit of area per unit of time. In MKS, that's kg/m-sec^2. Given µ is a standard total attenuation cross-section it is the 'effective' area per unit of mass, i.e., m^2/kg... So, ¿µ is, (kg/m-sec^2)(m^2/kg) -> m/sec^2 So, ¿µ is the field's effective deceleration term. However, rather than speculating & guessing we should try quantifing the problem with what we do know. One should expect that the directionalized velocity 'defect' due to a LeSagian process would have to be opposed to overcome to pull away (to infinity) from such a body. Thus, logically, the escape velocity should represent this magnitude. This is quite quantifiable and is, of course, 2GM/r. Ah!, but why G? Because to solve for this value in this case we need another mass body to calculatethe value. Therefore dv = Sqrt(2GM/r) And given a mean speed of the LeSage gravitons' of some magnitude c, the energy defect due to attenuation is simply proportional to (dv/c)^2, or, dE o< 2GM/rc^2 And, assuming dE is directly proportional to the attenuation, which, in turn is directly related to the dv traversing the body, we can say the power flux [in Watts/m^2 (¥')] absorbed is related to the uncollided (or its full strength value) as, ¥' = ¥(2GM/rc^2) Now, if we only knew ¥' & c we have one equation & one unknown (¥) we could solve for this, ¥ = ¥'/(2GM/rc^2) If we then 'assume' that c is the same as that of light, and use ¥' as the value of some measured excess heat emission, we can quantify the full strength value ¥. We do this in our paper, and obtain the value for ¥ of ~1.6E+08 Watts/m^2. Now that we have this value we can return to our original expression, ¥' = ¥(2GM/rc^2) Grouping the constants such that, ¥' = (¥2G/c^2)M/r and designating the terms in the parentheses as k, thus we get back, ¥' = kM/r However, ¥ is not ¿ but, they are relatable, and we can thus now decouple ¿ and µ from G (which we do in our published papers) we then find, ¿ ~= 6.740 kg/m-sec^2 µ ~= 3.147E-06 m^2/kg We can now resolve the value for ¿µ. It is ~ 2.12E-05 m/sec^2. At the weak attenuation limit, any body in linear motion relative to the LeSagian omni-directional fluence experiences a net drag deceleration (w) equal to, w = ¿µ(v/c) Thus we come to the Pioneer spacecraft. Given Pioneer's net linear velocity is ~12 kps, and given the assumption for deriving ¥ was that c = light speed, we now solve for w, -> ~8.5E-10 m/sec^2... So, we've taken an actual observation for net heat emission, utilized that in LeSage's model to derive the full strength power flux ¥. Then, using that derived value, we then solved for ¿ & µ and quantifed their values. Using the values, of ¿ & µ in a direct, straight forward manner, we next found that the drag predicted matches the unexpected observation of drag on non-orbital spacecraft. Hmmm, an amazing set of coincidental conjunctions here. And, through it all, no magical vaporization of the Planets seemed to have been required... Paul Stowe >> the minimum absorbed flux [per unit mass] is phi mu, which >> equals G/mu, [where mu is the maximum possible value >> consistent with the observed absence of shielding and >> saturation]... > Let's go with this... Given µ is a standard total attenuation > cross-section it is the 'effective' area per unit of mass... Right, it is the opaque cross-sectional area of a unit mass, which is the reciprical of the quantity called rho_0 in the referenced web page. > However, rather than speculating & guessing we should try > quantifing the problem with what we do know. We know there is no appreciable shielding or saturation of gravity, even for objects as large as the Earth. This place VERY stringent upper bounds on the possible values of mu. In other words, mu must be extremely small, i.e., ordinary material bodies must be nearly transparent to the flux. This is not guessing or speculating, it is honestly confronting the implications of the model. If you cannot find within yourself the intellectual integrity to face the fact that lack of shielding and saturation places limits on the value of mu, then so be it. Since G/mu [= G rho_0] is the minimum possible rate of momentum absorption (per unit mass), and we multiply this by v_g / 2 to get the corresponding absorbed kinetic energy per unit mass, it follows that the minimum possible absorbed energy for a body of mass M is G rho_0 v_g M / 2, which is identical to what is derived on the referenced web page, except for the constant factor, which is due to the fact that we haven't really correctly accounted for the fully 3D flux. But that is insignificant for just an order-of-magnitude calculation of the minimum possible energy absorption necessary to account for the force of gravity with no shielding or saturation. Plug in the values, and you find that the energy is sufficient to vaporize any material body in a fraction of a second. >>> the minimum absorbed flux [per unit mass] is phi mu, which >>> equals G/mu, [where mu is the maximum possible value [quoted text clipped - 7 lines] > which is the reciprical of the quantity called rho_0 in the > referenced web page. So what? >> However, rather than speculating & guessing we should try >> quantifing the problem with what we do know. > > We know there is no appreciable shielding or saturation of gravity, > even for objects as large as the Earth. We do? How? Please tell us all how to determine the mass of a Planet WUITHOUT invoking the gravitational force equation. If you do, then F = ¿Aa/d^2 and F = ¿(µM)(µm)/d^2 are BOTH! proportional to 1/d^2. You cannot infer anything about the magnitude of saturation from the force of interaction based on these equations... Unless of course, you can provide another, independent means of determining the mass of the bodies... Now THAT, I'd love to see! > This place VERY stringent upper bounds on the possible values of > mu. Sorry, but no, it does not. > In other words, mu must be extremely small, i.e., ordinary material > bodies must be nearly transparent to the flux. Only if you can answer the question posed above... > This is not guessing or speculating, it is honestly confronting the > implications of the model. If you cannot find within yourself the > intellectual integrity to face the fact that lack of shielding and > saturation places limits on the value of mu, then so be it. I wouldn't be braying about 'lack of intellectual integrity' if I were you buddy. I'm not the whining that 'it can't be'. You want to play in the same league fine, pony up to the bar & show how it can't possibly be. I just showed how it could possibly be... And, before yo go off spouting about how the lack of aberration MUST require a graviton mean speed of >> c I'll remind you that EM 'retarded' potentials have the very same problem. And the force law of the the very same form. So then, EM cannot possible have a speed of c, we all know this because aberration's retarded potential wouldn't allow nature to be as we observe :) > Since G/mu [= G rho_0] is the minimum possible rate of momentum > absorption (per unit mass), and we multiply this by v_g / 2 to > get the corresponding absorbed kinetic energy per unit mass, it > follows that the minimum possible absorbed energy for a body of > mass M is G rho_0 v_g M / 2, So, you assume that the gravitons loses all of their energy? Also, It's would seem that your expression is one of power (not energy, you haven't even gotten your dimensions correct), Last I heard, energy had dimensions of kg-m^2/sec^2. Your expression is, m^3 | kg | m | kg kg-m^2 ---------+-----+----+---- => ------ kg-sec^2 | m^2 | sec sec^3 which, of course, are units of power, not 'energy'. > which is identical to what is derived on the referenced web page, > except for the constant factor, which is due to the fact that we [quoted text clipped - 4 lines] > you find that the energy is sufficient to vaporize any material body > in a fraction of a second. I cannot help it that you've invalid assumptions, that's solely your problem... Paul Stowe >> Right, it is the opaque cross-sectional area of a unit mass, >> which is the reciprical of the quantity called rho_0 in the >> referenced web page. > So what? I was given to understand that you and greywolf42 were of the same mind on this subject. He stridently insisted in this thread that matter has NO opaque cross-sectional area in a Lesage theory. In fact, the very suggestion that matter might have a non-zero opaque cross-sectional area causes him to laugh out loud. I congratulate you for dis-associating yourself from his views. >> We know there is no appreciable shielding or saturation of gravity, >> even for objects as large as the Earth. > We do? How? Lesage himself estimated roughly that the opaque cross-sectional area of ordinary matter, in comparison with it's bulk cross-section, must be so small that "all the particles included in the terrrestrial globe intercept not the ten-thousandth part of the corpuscles which present themselves to traverse it". Laplace showed that at most only one ten-millionth of the ultra-mundane corpuscles incident on the earth are intercepted. The absence of shielding and saturation for gravity was well known to Huygens, Newton, Fatio, Lesage, Laplace, Maxwell, Kelvin, Darwin, Poincare, Feynman, and every other (sane) person who has ever given a moment's thought to the idea of "pushing gravity". In fact, it is known even to some insane people... in this very thread your formerly esteemed greywolf42 said "LeSage's matter was 99.999...% transparent." You see, if mu is too large, we do not have gravity directly proportional to mass, and this would be apparent in many different observable phenomena, from Evotos-type experiments (which are among the most precise ever performed) to the movements of the planets. Newton himself performed very careful pendulum experiments to demonstrate the exact proportionality of gravity to inertial mass for a wide range of subtances of differing densities and compositions, and was able to establish fairly precise bounds on this equality, giving the basis of Lesage's rough estimate for an upper bound on mu. The precision of these measurements, along with astronomical observations, has improved ever since. In addition, measurements of the gravitational attraction between metal balls near the surface of the Earth (such as in freshman physics lecture halls) are sufficient to show that the gravitational "flux" is not appreciably diminished even in the shadow of an object as large as the Earth. This too places very strict upper limits on mu. From all these, it is perfectly clear that mu cannot be greater than about 1E-10 meters/kilogram. > Unless of course, you can provide another, independent means of > determining the mass of the bodies... Now THAT, I'd love to see! You might enjoy a grade school physics class then (except for the tests). Mass is defined independently of the phenomenon of gravitation. It is the proportionality between applied force and resulting acceleration. We take a standard object, and see how much it accelerates when subjected to a certain external influence (like a spring at a certain extension). Then we apply that same influence to other objects and see how much they accelerate, and to each object we assign a value of mass inversely proportional to the observed acceleration when subjected to that standard external influence. (This is done on smooth horizontal surfaces so as to be unaffected by gravity.) Now, if the force of gravity really is exactly proportional to the inertial mass of an object, we can infer a great many things about how astronomical bodies will behave, and also how they would behave if the force of gravity was not exactly proportional to mass, such as with a bit of saturation and/or shielding. (For example, Kepler's third law would give inconsistent results when applied to the Sun-Jupiter system and the Jupiter-Ganymede system.) We find in all cases, in all circumstances, that there is no appreciable shielding or saturation, and therefore we have very strict limits on the empirically viable value of mu in any Lesagian theory. There's not much point in addressing the remainder of your misconceptions until you understand and acknowledge the fundamental requirement of an extremely small value of mu for any Lesagian theory of gravity. >that mu cannot be greater than about 1E-10 meters/kilogram. The should be meters^2/kilogram. >>> Right, it is the opaque cross-sectional area of a unit mass, >>> which is the reciprical of the quantity called rho_0 in the [quoted text clipped - 26 lines] > other (sane) person who has ever given a moment's thought to the idea > of "pushing gravity". See pages 219-237 of "Pushing Gravity" for a discussion on this topic. You do have an all or nothing mentality typical of arrogant cynics. > In fact, it is known even to some insane people... in this very > thread your formerly esteemed greywolf42 said "LeSage's matter was > 99.999...% transparent." I've never said otherwise, I simply asked you a question! One that was targeted to try to make you think (although I kinda' knew that THAT was a lost cause...) > You see, if mu is too large, we do not have gravity directly > proportional to mass, Look Dickhead, if µ is some value (and that's ANY non-zero value) it then depends upon the density & thickness as to whether or not we retain the weak limiting situation. A mountain might fall within that limit, the Moon 'might' still be but the Earth COULD not. Now note the word "could', I am NOT claiming it is. > and this would be apparent in many different observable phenomena, > from Evotos-type experiments (which are among the most precise ever > performed) to the movements of the planets. Your argument here is simply silly, indicative of lack of thought & sheer arrogance. So, then answer the question posed in the previous post, quantitatively. I claim that orbits would exist even IF the was full 'absorption' & the equation ¿Aa/d^2 is used. Second, one doesn't just suddenly go from the weak state to a strong state and certainly small mass (asteroid sized or smaller) are found to be in the weak limiting state. The fact is Bozo, if the was some divergence from the weak limit it would NOT show up in a simple two body orbital situation observed from afar. If you think otherwise, fine, get quantative & show it!!! > Newton himself performed very careful pendulum experiments to > demonstrate the exact proportionality of gravity to inertial > mass for a wide range of subtances of differing densities and > compositions, ROTFLMAO. So what? Did he do this with a mountain??? See the reference to Majorana... > and was able to establish fairly precise bounds on this equality, > giving the basis of Lesage's rough estimate for an upper bound on [quoted text clipped - 7 lines] > perfectly clear that mu cannot be greater than about > is 1E-10 meters/kilogram. >> Unless of course, you can provide another, independent means of >> determining the mass of the bodies... Now THAT, I'd love to see! [quoted text clipped - 18 lines] > would give inconsistent results when applied to the Sun-Jupiter system > and the Jupiter-Ganymede system.)... Given the expression, F = iAa/d^2 Or F = GMm/d^2 And assuming that M & m, A & a are fixed quantities thus their product is constant k & k' then F = ik'/d^2 Or F = Gk/d^2 Further that ¿ & G are constants, then ik' & Gk -> constant x' & x, the forece exerted is strictly proprtional to 1/d^2 for both situations. Thus the so-called saturated masses of A & a can be related to an 'effective' mass M & m by saying, GMm = ¿Aa Now tell me HOW to know if we have a weak limit situation or not strictly for this??? (Hint, Allasis Effect) > We find in all cases, in all circumstances, that there is no > appreciable shielding or saturation, and therefore we have very > strict limits on the empirically viable value of mu in any > Lesagian theory. Me'thinks you're ignoring solid angles... > There's not much point in addressing the remainder of your > misconceptions until you understand and acknowledge the fundamental > requirement of an extremely small value of mu for any Lesagian theory > of gravity. You're right, your misconceptions are simply too gross... Paul Stowe >> Laplace showed that at most only one ten-millionth of the ultra-mundane >> corpuscles incident on the earth are intercepted. The absence of shielding [quoted text clipped - 4 lines] > > See pages 219-237 of "Pushing Gravity" for a discussion on this topic. I said every other _sane_ person. Also it must be pointed out that you< |
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