Flaw in Stowe/LeSage Heating
in [Relativity]
|
Prev: Achieving Critical Mass. Anarchist Communism, EMC squared and the Arc of the Covenant. How to construct a home-made nuclear weapon.
Next: Joan-Claude van Dirk Helps to Trivialize Special Relativity
From: mingstb on 11 Mar 2005 01:50 TC wrote: > mingstb(a)sim-ss.com wrote: > > TC wrote: > > > Clarke: 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. > > > LOL! The only way to find that out is to actually *use the theory* > to > > make predictions. Using other theories (or mere hand-waving) is > > useless. > > Other theories do pretty well. Not for claims about aether theory, they don't. > <snip> > > There was only one post that attempted anything of substance: > > > http://groups-beta.google.com/group/sci.physics.relativity/msg/2080fab1f8cc1d21 > > > > > 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. > > > That's called weaseling. What *specific* point did I not address? > > Let's look at two points: The speed of the ultramundane particles: > ________________ > Shadows: the possibility and consequences of > > ultra-mundane particles moving at various speeds, including c or much > > greater than c, is addressed in detail; > > Mingst: "Shadows" simply makes the false claim that the speeds must > be vastly greater than 'c' (using hand-waving arguments, coupled with > hilariously erroneous mathematical manipulations). > _______________ > How is this a refutation? It is just as statement - without reason - > that shadows is wrong. LOL! What a dishonest, troll statement. Where in Shadow's statement is there anything specific to address? I was pointing out that Shadow (in his/her one attempt do distract from the substantive issues) refused to address my explicit refutation of his mixed-unit attempts at equations *in the webpage*. > And of course if you read > http://www.mathpages.com/home/kmath131/kmath131.htm > you find reasons for the estimates of ultramundance particle > velocities given. What *reasons* are those, Jim? The "equations" made up of mixed-up units? The ones that would result in failing high-school physics? And you'll find that the "reasons" were explicitly blown out of the water with the post that Shadow references -- but never addresses. > Reflection versus scattering: > ___________________ > Shadows: the effects of reflection versus absorption are discussed. > > Mingst: Acutally, all "Shadows" does is make a bald assertion that > reflection won't work: > "(Note that we can exclude from consideration all the reflected > particles, > because these contribute nothing to the net force on a body, e.g., with > > perfect reflection there would be no net force at all. Thus we need > consider > only the absorbed particles.)" > And the above claim is flatly untrue. > ___________________ > The above counterclaim is flatly untrue. > So this is hardly a refutation, either. > Enough for now. LOL! So how is the "counterclaim" flatly untrue? I see you refuse to address the specifics in the rebuttal to the webpage (which is where all of the detail is). The link above, is the pointer to all of that. Three rounds of unsubstantiated claims, attempted smears, and constantly-shifting arguments. I have better things to do that to feed your trolling. Bye in this thread..... greywolf42
From: Bilge on 11 Mar 2005 02:59 Paul Stowe: >On 10 Mar 2005 08:40:50 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: >> >> Is mu, in effect, the ratio of absorption to elastic scattering of >> the ultramundane particles? > > ý (mu) is the mass attenuation coefficient, look up the concept. You look up the concept. I(r) = I \exp(-ur) is the _unscattered_ flux at a distance r. [...] > I jumped back in just to clarify these points. You didn't succeed.
From: TC on 11 Mar 2005 09:18 mingstb(a)sim-ss.com wrote: > TC wrote: > > mingstb(a)sim-ss.com wrote: > > > TC wrote: > > <snipped to get to the points> > > > > > 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. > > I had provided. > > http://www.mazepath.com/uncleal/eotvos.htm > > bu this one seems even better for history > > One Hundred Years of the Eotvos Experiment > > http://www.kfki.hu/~tudtor/eotvos1/onehund.html > This is not a reference to the point under discussion. Your "creative" > snip alters the subject of your avoidance. I wasn't referring to > questions about the actual Eotvos experiment. > The quote under discussion was: "Eotvos type experiments have shown > that most things that make up matter gravitate. Including binding > energy." I searched the web a bit for a reference to more modern Eotvos experiments, like those done by Dicke. But I couldn't find a good one, so sue me. It's probably in his book: Dicke, R The Theoretical Significance of Experimental Relativity (Gordon and Breach: New York, 1965) > Where is your experimental support for the claim about binding energy > gravitating? All around you. As I said 6 moles of deuterium weights more than 1 mole of carbon-12. [This would actually be negative binding energy but it is a cleaner example (no proton/neutron conversion, than something like a mole of uranium weighing more than a mole of krypton and a mole of barium.] So if you want to balance some deuterium with some carbon-12, the carbon-12 side of the balance will have more particles than the deuterium side. The extra particles on the carbon side will be balanced by energy. <snip for brevity> > > > > Binding energy is observed. It is why nuclear weapons go boom. > > > I see you again have no experimental support to give. > > How do you account for the energy released by nuclear weapons - and > > by nuclear reactors? > I see you still have no experimental support for your assertion that > binding energy is a measurable quantity. The government needs people like you who think that nuclear weapons won't explode! <snip to issue> > > So does inertial mass include binding energy? Or does the "Period" > > exclude it? > Inertial mass is inertial mass. Whether it does or does not include > "binding energy" is irrelevant to LeSagian gravitation. Also to > Newtonian gravitation. All that matters is the inertial mass. Evasion. > > > > > 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. > > You make my point. To show equivalence of inertial and gravitational > > mass you need a reducto type argument. > There is no need for "equivalence" of a concept that does not exist in > the theory. So you reject all gravitational theory not in the Principia? > > The equivalence, the identity, of > > the two is built into GR. > And into Newtonian gravitation. And into Lesagian gravitation. > On this issue, there is no difference between theories. ........ > > Why do you think Eotvos started his experimentation to test the > > equivalence of the two types of mass in 1889, long before GR? > Because he wanted to see if he could disprove Newton's empirical and > LeSage's theoretical assumptions. It's called science. Fair enough, I suppose. But this implies that Newton's theory could persist in modified form if Eotvos had been successful. F=m'a=GMm/R^2 where m' is inertial mass and m,M are gravitational masses. > Still totally irrelevant. Good experimental technique, though. Had > Eotvos found a difference between feathers and lead, all three types of > theory (Newton, LeSage, and GR) would be knocked into a cocked hat. I don't know about LeSage but Newton would have survived just fine - see above - GR would have a problem since equivalence is at the core of the theory. > > > > 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. > No response, I see. I don't limit Newtonian gravity to the Principia. That would be like limiting evolutionary theory to "The Origin of Species" <snip stuff covered in post to Stowe> > > > 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? > > If the interaction is absorptive, which I had assumed it was, yes. > > But now you say it can be non-absorptive, scattering, in which > > case there might not be any heating. > Bingo. And the most likely real situation is somewhere inbetween > complete absorption and perfect elasticity. Only observations allow us > to quantify the amount. I understand the claim. <snip more covered in reply to Stowe> > > > > 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. > > You have a known. The force - the momentum flux attributed to > > gravity. > You fail elementary algebra. One cannot determine the momentum flux > (Phi_0) from the gravitational force -- because you don't know mu_0. Force has units ML/T^2 or (ML/T)/T or (momentum)/T > > You have a lower bound on the speed of the ultramundane > > particles set by lack of aberration. > You fail elementary logic. Aberration cannot be examined in isolation > from dyanmic drag. Lesagian theories *ALWAYS* have all four resulting > physical manifestations: gravitational force, flux heating, drag and > aberration. Dynamical drag opposes orbital aberration -- and can > balance same (i.e. Bode's law). So what is the speed of the ultramundane particles? Power or heating rate has units ML^2/T^2 or (ML/T^2)(L/T) or force-times-velocity or momentum-flux-times-velocity. ....... > Momentum flux is not force -- which was your prior claim. Depends on defintion of flux I suppose. 6. the rate of transfer of fluid, particles, or energy across a given surface This mentions "a given surface". That is probably our difference. You seem to take a unit area, I tend to take the entire surface of the gravitatiing body. > > > Even if heating is dependent upon flux, it is > > > also dependent upon the constant, mu. > > Is mu, in effect, the ratio of absorption to elastic scattering of > > the ultramundane particles? > No. But it *is* affected by that ratio. Mu is the mass attenuation > coefficient. As described in the mathematics and detailed explantions > that existed below -- which you snipped. > {replacing explanatory mathematics, snipped by TC} Resnip for brevity. Stowe put this back in amply and I await interlibrary loan of "Pushing Gravity". ........................................................... > > > > > 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? > > There is an omnidirectional rain of ultramundane particles. If these > > are scattered elastically then even the presence of a massive > > body will not modify the flux of the ultramundane near the body. > There are other types of scattering (non-absorption) than simple > billiard-ball collisions. To my knowledge, the only person who ever > postulated such a Lesagian variant was Darwin. And even Darwin > obtained a net force. What are these other types of scattering? > > Particles that do no penetrate the body are replaced by particles > > from the other direction that scatter from the body and wind up going > > in the same direction. So there is no shadowing for elastic > scattering. > But they aren't "scattered" back to the other body. Some are. My intuition - admittedly not backed up by calculation - is that an elastically scattering body in an isotropic field has no net affect on the field. > So, I'm afraid > that you'll have to back up your claim with real mathematics. But if > you'd rather not, that's OK. Because no one I know is postulating such > a situation. Does anyone in Pushing Gravity address this issue? > I'm afraid I don't have time to bother with one who snips every > explanatory equation as soon as it is made -- and as if it had never > been provided. > Enjoy yourself. Bye.... Bye. Tom
From: TC on 11 Mar 2005 10:01 mingstb(a)sim-ss.com wrote: > TC wrote: Snip pointless replies. > Bye in this thread..... Bye. Tom
From: TC on 12 Mar 2005 20:12
Paul Stowe wrote: > On 11 Mar 2005 05:46:03 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: <snip off-point speculation as to what Mingst meant> <snip down to substantive discussion> > >> Pay attention! the force due to two bodies is a function of µ > >> SQUARED! That where G come from..., G = ¿µ^2 > > Can you use a different font or something. These upside down > > question marks are not illuminating. Maybe just spell out the > > symbols, or use TeX notation. > The so-called upside down ?, ¿ is a symbol. Like all algebraic > symbols it represent a term, momentum flux, and it alway has as > I used it in our discussions. The dimension of ¿ are kg/m-sec^2 "Upside down question mark" is a bit wordy. Plus I don't have any idea how to make my keyboard produce same. So it is a bit tough to communicate with this symbol. I see from your web site that upside down question mark must be phi. ["fee" in some of the articles], so I'll use phi. > Let me be pedantic, Some people would use the term "precise". > ¿ ~= 6.740E+00 kg/m-sec^2 (LeSagian Momentum Flux) > µ ~= 3.147E-06 m^2/kg (LeSagian Mass Attenuation Coefficient) > G = ¿µ^2, you do the math & dimension check! This actually makes sense. Phi is momentum/area/time for force/area. [Presumably this is per steradian since LeSagian flux comes in from all directions] Mu is cross section per unit mass. So M times mu times phi would be force. [per steradian] Or dividing out M, mu times phi is accceleration. <an earlier equation? > >> Thus the one body drag equation, > >> a' = ¿µ(v/c) Why the term v/c ? Phi times mu is accleration. <snip insult> > >> Note that the term µM/R is a distance factor (in units of > >> length). > > Can you explain why this is a length? Is it the mean free path > > of the ultramundance particles in the body of mass M? > An 'effective' length. Well, do the analysis, <snip> I know its a length, dimensionally, I was asking for the physical significance of this length. > > We find that the Newtonian force is simply, > >> ¿[µM/R][µm/R] => GMm/R^2 > > Why is big "R" applied to little "m"? For that matter why is > > it applied to big "M"? R is center to center distance and > > thus is a property of the two body system, not of any > > single body. > It 'becomes' the distance between the center of masses. 'Becomes' in quotes? Why in quotes? > Each > mass M & m can be assumed to smeared out in a uniform density > of 3M/4piR^3 and 3m/4piR^3. That's the property of the weak, > linear limit of the attenuation process. The force is just M mu phi . (integrated over angles). There is no R, no density involved so far as I can tell. Density is determined by the physical properties of the bodies involved. This may then reenter though the integral over angles but I don't see the basis for assuming the above values. > > Is there an accidental equivocation of mathematical symbol > > "R" here? > No. There is a formal mathematical equivalency. > ¿[µM/R][µm/R] -- DOES EQUAL -- GMm/R^2 > Given the LeSagian relationship, G = ¿µ^2 That is just a decompostion of G, there is nothing in there involving density. > >> Again, we CLEARLY see the 1/R relationship for 'each' of the > >> attenuating bodies. > > Again, why is it not little "r" for little "m" body then? > Because it is a two body problem... I'll have to wait for the book I guess. <snip for brevity> > > Ignoring my questions about notation, then Q (power) would > > be F time v where "v" is the speed of the ultramundane particles. > > Right? > No, it would be Q = F · dv. No work or power is present when > Force is applied statically. We are talking about a different velocity. You are talking about the velocity of the moving body, I am talking about the speed of the ultramundane particles. <snip for brevity what ws in part another misintepretation of a Mingst statement> <snip misunderstanding of Feynman lectures> <Left for reference> > > So what is the speed of ultramundance particles in your theory? > Sqrt(3)c <snip for brevity> > >> Note, from the heating expression we had, > >> q' = kM/R > > That is your heating expression. You need to supply > > a reason to not use the heating expression from elementary > > physics. > Sheez... The character Scar of "The Lion King" comes to mind So why is not Q=F Sqrt(3)c = mu phi M Sqrt(3)c ? <snip sparing> > > You should know by now that I don't think LeSagian theory works! > Then shut up. Don't misrepresent the concept. So fars as I can tell the supporters of the concept make many errors. Tom |