Flaw in Stowe/LeSage Heating
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From: Paul Stowe on 13 Mar 2005 13:42 On 12 Mar 2005 17:12:20 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: > >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 ý >>>> SQUARED! That where G come from..., G = ýý^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 ?, ý 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 ý 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. Jeez, all you had to do is say that... These characters are simple on a Windows based system... Hold down the ALT key & type in the numeric keypad the following, ALT 135 -> ý ALT 145 -> ý ALT 146 -> ý ALT 156 -> ý ALT 157 -> ý ALT 159 -> ý ALT 166 -> ý ALT-168 -> ý ALT 171 -> ý ALT 172 -> ý ALT 174 -> ý ALT 175 -> ý ALT 225 -> ý ALT 230 -> ý ALT 241 -> ý ALT 246 -> ý ALT 248 -> ý (Degrees symbol) ALT 249 -> ý (Dot operator) ALT 253 -> ý (2 as in squared) Thus for example, you can write, xý or, 3ý2 = 1.5, or 273 ýK, ... etc. > 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. yep... >> Let me be pedantic, > > Some people would use the term "precise". > >> ý ~= 6.740E+00 kg/m-sec^2 (LeSagian Momentum Flux) >> ý ~= 3.147E-06 m^2/kg (LeSagian Mass Attenuation Coefficient) If it were the 'first' time you were told, it'd be being precise, after many such, its being pedantic... >> G = ýý^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. ý = the total flux (as opposed to directional current) > So M times mu times phi would be force. [per steradian] Nope, total... > Or dividing out M, mu times phi is accceleration. Yes! > <an earlier equation? > >>>> Thus the one body drag equation, > >>>> a' = ýý(v/c) > > Why the term v/c ? > Phi times mu is accleration. http://groups-beta.google.com/group/sci.physics/msg/5ee4d6b55d80917d?dmode=source > <snip insult> > >>>> Note that the term ýM/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. Since ý is the LeSagian pressure, the apparent force due to to this pressure is dependent upon the area to which it is applied. It take TWO! bodies to realize an interaction of this type. Each contributes an 'effective' length to cross-sectional area of the problem. Thus the multiplication of [ýM/R][ým/R]. We note that this is WHY! mu squared is required by this type of interaction! >>> We find that the Newtonian force is simply, > >>>> ý[ýM/R][ým/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? Because its a modeling concept... >> 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. Then you have not understood the basic derivation of the LeSagian form. >>> Is there an accidental equivocation of mathematical symbol >>> "R" here? > >> No. There is a formal mathematical equivalency. > >> ý[ýM/R][ým/R] -- DOES EQUAL -- GMm/R^2 > >> Given the LeSagian relationship, G = ýý^2 > > That is just a decompostion of G, there is nothing in there > involving density. Only the masses... >>>> 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. That would help... If you really read with comprehension. Note, as I've said before, you don't have to buy into a concept to understand it in detail. > <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 ý 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. I guess it the difference between gross flux & net current. > <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> It would. But that is a dimensionless scalar and factors out in most problems. The reason, we're limited to measurable verses model values. >>> 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. I think you'll find most of those errors lie in the misunderstanding and lack of expertise on the part of detractors. Such as with the case of aberration... Paul Stowe
From: TC on 13 Mar 2005 20:51
Paul Stowe wrote: > On 12 Mar 2005 17:12:20 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: > > "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. > Jeez, all you had to do is say that... > These characters are simple on a Windows based system... > Hold down the ALT key & type in the numeric keypad the > following, Oh, yeah, the IBM specific stuff. But the net isn't limited to IBM compatiable computers so I'll stick to "phi". <snip examples> > > 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. ............... > >> ¿ ~= 6.740E+00 kg/m-sec^2 (LeSagian Momentum Flux) > >> µ ~= 3.147E-06 m^2/kg (LeSagian Mass Attenuation Coefficient) > If it were the 'first' time you were told, it'd be being > precise, after many such, its being pedantic... You really should consider the use of "phi" upside down question mark just does not stick to the neurons. > >> 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. > ¿ = the total flux (as opposed to directional current) Probably that is a constant factor different (4 pi?) except when another body is present. > > So M times mu times phi would be force. [per steradian] > Nope, total... Same comment. But I see you agree as to units. > > Or dividing out M, mu times phi is accceleration. > Yes! This is exciting? > > <an earlier equation? > >>>> Thus the one body drag equation, > >>>> a' = ¿µ(v/c) > > Why the term v/c ? > > Phi times mu is accleration. > http://groups-beta.google.com/group/sci.physics/msg/5ee4d6b55d80917d?dmode=source So this is a velocity dependent drag force that your theory predicts? > > <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. > Since ¿ is the LeSagian pressure, the apparent force due to > to this pressure is dependent upon the area to which it is > applied. It take TWO! bodies to realize an interaction > of this type. Interaction requires two bodies, of course, but by above URL uniform motion of one body leads to force accoding to LeSagian theory. > Each contributes an 'effective' length to > cross-sectional area of the problem. Thus the > multiplication of [µM/R][µm/R]. We note that this is WHY! > mu squared is required by this type of interaction! Sound too much like handwaving to me. To amplify, gravity equation is F=GMm/R^2, so there needs to be an R^2 from some where. This language about 'effective' lenght is a not very precise way to get two factors of R into force due to gravity it seems to me. The right way to do it, IMHO, is to determine the total cross section for M, A=mu M, and for m, a=mu M, Then the asymmetry produced by absorption is proportional to mu A/(4 pi R^2) and mu a/(4 pi R^2). That is proportioanl to the fraction of the matter cross section to the area of the sphere of radius R. So the force on m is [phi a mu A/(4 pi R^2)] or GMm/R^2 using the expression of G in terms of mu and phi and ignoring the 4pi business. The same result follows for the force on M. ..................... > >>> 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. > Only the masses... There is nothing about the radius of the bodies. ....................................... > > <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. > I guess it the difference between gross flux & net current. No. If the mass absorbs mu phi of ultramundane flux, then the energy absorbed is mu phi v-ultramnundane. Basic physics. .................................. > > So why is not Q=F Sqrt(3)c = mu phi M Sqrt(3)c ? > > <snip sparing> > It would. But that is a dimensionless scalar and factors out > in most problems. The reason, we're limited to measurable > verses model values. What is a dimensionless scalar? > >>> 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. > I think you'll find most of those errors lie in the misunderstanding > and lack of expertise on the part of detractors. Such as with > the case of aberration... I think the advocates make serious errors. Tom |