Flaw in Stowe/LeSage Heating
in [Relativity]
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From: TC on 1 Mar 2005 13:12 Paul Stowe wrote: > On Tue, 01 Mar 2005 17:26:58 GMT, Paul Stowe <ps(a)acompletelyjunkaddress.net> > wrote: > [Snip...] > See "Pushing Gravity" I've put in another interlibrary loan request. Last one was during the holidays and the ILL office was closed. Then I forgot to resubmit. > >> It doesn't address the core issues of the enormous ultramundane > >> velocities required to avoid tangential acceleration > One way for this is TVF's "Gravity" pages 93-122 of above, I don't look forward to this. I don't like TVF's arguments against GR. I think they are bogus. > >> and the enormous energy depostion - heating rates - that that > >> would require. > Another is Slablinski's "Force, Heat and Drag in a Graviton Model" > pages 123-128 in same... Does he mean gravitions in QFT sense? I guess I'll find out. > > It is addressed, see references given... > As I said... We will see. Tom
From: Bilge on 2 Mar 2005 06:04 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.
From: TC on 2 Mar 2005 08:02 Bilge wrote: > TC: > >But I remain troubled. Stowe's formulas just do not ring true > >for me. > What formalism? Physical common sense, I guess. E.G. M/r is escape velocity or potential in gravitational context. To make it an areal flux, would require some more r's, I would think. Tom
From: TC on 2 Mar 2005 21:46 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. > 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. > 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. Energy? Is this the same as your "fluence"? > 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. 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. {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. 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, so the energy flux (which is not signed) could actually be much larger than (net momentum flux) times velocity/2. 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. Tom
From: Aetherist on 3 Mar 2005 01:24
On Wed, 02 Mar 2005 16:17:35 GMT, dubious(a)radioactivex.lebesque-al.net (Bilge) wrote: [Snip of pure whining...] Bilge's obvious problem... http://www.mazepath.com/uncleal/sunshine.jpg. Paul Stowe |