Flaw in Stowe/LeSage Heating
in [Relativity]
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From: Bilge on 5 Mar 2005 09:44 Aetherist: > >On Wed, 02 Mar 2005 16:17:35 GMT, dubious(a)radioactivex.lebesque-al.net >(Bilge) wrote: > > [Snip of pure whining...] Does that mean you didn't understand the part about the taylor series expansion of \exp(x)?
From: Paul Stowe on 5 Mar 2005 11:02 On Sat, 05 Mar 2005 14:44:06 GMT, dubious(a)radioactivex.lebesque-al.net (Bilge) wrote: > Aetherist: > > > >On Wed, 02 Mar 2005 16:17:35 GMT, dubious(a)radioactivex.lebesque-al.net > >(Bilge) wrote: > > > > [Snip of pure whining...] > > Does that mean you didn't understand the part about the > taylor series expansion of \exp(x)? http://groups-beta.google.com/group/sci.physics/msg/4d6965b484d9d081?dmode=source http://groups-beta.google.com/group/sci.physics.relativity/msg/e9a6ecf588d540c0?dmode=source Showing all your lack of attention again I see... Paul Stowe
From: Bilge on 6 Mar 2005 18:10 Paul Stowe: >On Sat, 05 Mar 2005 14:44:06 GMT, dubious(a)radioactivex.lebesque-al.net: >wrote: > >> Aetherist: >> > >> >On Wed, 02 Mar 2005 16:17:35 GMT, dubious(a)radioactivex.lebesque-al.net >> >(Bilge) wrote: >> > >> > [Snip of pure whining...] >> >> Does that mean you didn't understand the part about the >> taylor series expansion of \exp(x)? > >http://groups-beta.google.com/group/sci.physics/msg/4d6965b484d9d081? >dmode=source >http://groups-beta.google.com/group/sci.physics.relativity/msg/e9a6ec >f588d540c0?dmode=source > > Showing all your lack of attention again I see... What you wrote made it clear that you didn't understand what a taylor series is and I assume that is why you snipped what you wrote.
From: mingstb on 7 Mar 2005 21:38 Bilge wrote: > TC: > >Bilge wrote: > >> TC: > >> >I was looking at Paul Stowe's > >> >"An Overview of the Concept of Attenuation [Pushing] Gravity" > >> >http://www.mountainman.com.au/index_ps.htm > > > >> >He derives from the LeSage theory that the heat output > >> >of a body resulting from gravity should be > >> >q = kM/R where k is a constant, M the mass and R the radius > >> >of the body. > > > >Stowe has clarified that q is heat flux/unit area, not total > >heat flux as I had misread. > > > >> >It occured to me to ask what the temperature of a laboratory > >> >scale body would be from this mechanism > > > >It seems that there will be a small effect using Stowe's formulas. > >A 1 meter radius sphere of iron would have a minimum temperature > >of 0.19K absent other heating or cooling. > > > >But I remain troubled. Stowe's formulas just do not ring true > >for me. > > What formalism? > > >Also Bilge supplies: > > > >> http://www.mathpages.com/home/kmath131/kmath131.htm > > > >Which derives very different conclusions from the LeSage model at > >great odds with Stowe's results. > > > >Has Stowe ever refuted these arguments? > > No. LOL! But that pathetic and ignorant site was completely refuted by myself. Paul didn't have to. greywolf42
From: mingstb on 7 Mar 2005 21:56
TC wrote: > Paul Stowe wrote: > > On 1 Mar 2005 09:54:18 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: > > ......Snip since it seems a new start is beginning at this point. > > > > Come to think of it why is there an "r" at all in the expression > > > for heating? Wouldn't heating just depend on the mass? > > > Again, let's go through this from the beginning. In equilibrium, > > we know that input should equal output. For LeSage's process we > > can define this precisely. That is, flux [i] integrated over the > > area upon which it impinges defined how much CAN interact in a > > region enclosed by said area. > > At this point I interject to ask why not think of the interaction > as the sum over individual small masses? That is the Earth > is 10^whatever particles. Each particle has a LeSagian heating > so the Earth would heat as the sum of all these particles > so total mass is all that is involved. As long as you are making > the weak absorption approximation, there will be little reduction > in heat flux at the inner particles compared to the outer particles. Indeed, you cut directly to the essence of the situation! Nucleons (specifically, neutrons) have been experimentally determined to gravitate. LeSagian gravitation is not based on the macroscopic radius of matter. If you have 'N' nucleons, you will have the same gravitational force at distance 'R', outside the body. This is simply Gauss' flux law. Since density goes as the cube of volume, and area goes as the square of volume, the radial parameter arises in the above calculation. > > For a sphere this is, > > > T = (4pir^2)i > > > Now, let r -> 0, then T -> 0. Thus the total of all types of > > fluence is a function of area (r^2) > > Thus? You assumed the flux is "integrated over the area" > and you concluded that the total is a function of the area. > Seems tautological. It's called a definition. One integrates over a solid angle of 4 pi steradians to get the contribution from all directions. > > Now, the material mass [m] within the enclosed region is simply > > its mean density [p] multiplied by the volume of that region, > > right? That is, > > > m = p([4pir^3]/3) > > > > We see then that the density changes as r^3 and energy as r^2. > > Density changes? To first approximation density does not depend > on size. Horsefeathers. If mass is held constant, and volume changes, then the density changes. > Energy? Is this the same as your "fluence"? No. Fluence is fluence. Specifically, the momentum flux. > > There is a 1/r (r^2/r^3) relationship between the mass, m and > > the total of i. This means, the lowest interacting state is > > the one with the highest density for any given given radius. > > Increase the radius and hold mass constant, and increase the > > total interaction that will occur within the region. It is > > simple math. The key is the realization that it is the flux > > not the total that is the invariant property. > > If you make your (apparently tautological) assumptions > it follows. Definitions are not tautologies. > But I don't seem why area is the thing. Gravity is universal > so it effects small particles in such a way that the sum > gives the gravitational effect on a large body made from > the small particles. Surface area has nothing to do with > it. Precisely! GRAVITY is not affected by surface area. But energy deposition (for a constant mass) *IS* affected by surface area. Because gravity is a two-body effect, and gravitational flux heating is a one-body effect. (The mass will heat up even if there is nothing around it with which to gravitate.) > {snip unchanged stuff} > > > > I'll await how you avoid the problem of energy flux=momentum flux > > > times velocity/2. > > > What problem? > > As you say, input equals output, so in the LeSage theory > the net momentum flux of ultramundane particles absorbed by > the body equals the gravitational force on a body. Not quite. But you are on the right track... > The energy flux (power) of the ultramundane particles > is given by the above. Since momentum is a signed quantity > it is the net absorded that gives rise to gravitational force, Close. It is the net transfer of momentum flux. Not the net absorption. > so the energy flux (which is not signed) could actually > be much larger than (net momentum flux) times velocity/2. Speaking purely mathematically, without incorporating any real-world observations, yes. Because heating is a one-body problem. And graviation is always a two-body problem. And we don't know -- a priori -- what the scattering/ absorption relationship is. > Put in speed of light for speed of ultramundane particles > and the power input to a kg mass in the Earth's > gravity field will be something greater a gigawatt. In this, you have made a fundamental error. You cannot make such a claim, if you actually use Le Sagian mathematics. In short, you cannot get any heating values solely from gravitational calculations. You are attempting to resolve two unknowns with only one equation. greywolf42 |