Flaw in Stowe/LeSage Heating
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From: TC on 28 Feb 2005 15:00 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. It occured to me to ask what the temperature of a laboratory scale body would be from this mechanism so I equated it to the Stephan Boltzman law q = sigma.Area.T^4, where sigma is a constant, T the temperature. Since area = 4pi.R^2, the result is T^4 = A.M/R^3 for some constant A. But since M/R^3 is some constant times the density, the result is T^4 = B.density for some constant B. Thus a body of a given density with no other source of heat will come to an equilibrium temperature that depends only on its density if LeSage theory as expounded by Paul Stowe is true. It's beyond me calculate what B is right now, and if I tried I'd probably get it wrong, but in Stowe's paper the emission of Jupiter is claimed to be predicted by LeSage theory. The temperature of Jupiter is 170 K and its density 1.34 times that of water. So it seems any body of density 1.34 would reach a temperature of 170K if well insulated. It seems to me that this phenonmena would have been noted in laboratory cryogenic experiments as it would presented an additional difficulty in cooling to temperatures near 0K. Tom
From: Paul Stowe on 28 Feb 2005 20:36 On 28 Feb 2005 12:00:16 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: >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. > > It occured to me to ask what the temperature of a > laboratory scale body would be from this mechanism > so I equated it to the Stephan Boltzman law > > q = sigma.Area.T^4, where > sigma is a constant, T the temperature. Let ý = Stephen-Boltzmann's constant 5.67E-08 W/m^2-K^4 Then equate them exactly, kM/R = ýT^4 Thus, _____ / kM T = 4 / -- (Note: 4 mean fourth root) \/ ýR Since M is simply, M = pV Where p = density in kg/m^3 V = Volume in m^3 and for a sphere of radius R V = 4piR^3/3 Then __________ / 4pikpR^2 T = 4 / -------- \/ 3ý We're relating apples to apples (radiative heat flux in Watts per square meter). Now, for example, iron has a density of 7800 kg/m^3 and let's, for simplicity, use a radius of 1 meter. Then, 4pi(2.4E-19)(7.8E+03)(1)^2 -------------------------- = 1.383E-07 3(5.67E-08) And then, __________ / T = 4 / 1.383E-07 => 0.0193 ýK \/ So, you're right. To get below this value you must remove heat at a rate of > 7.84E-15 Watts/m^2... That's quite possible. However, the 'prediction' based on this indicates that it is physically impossible to get to absolute zero :) Snip of flawed analysis, nice thinking though, you're on the ball :) > It seems to me that this phenonmena would have been noted > in laboratory cryogenic experiments as it would presented an > additional difficulty in cooling to temperatures near 0K. Have they ever achieved absolute zero? http://www.infoplease.com/ce6/sci/A0802223.html Paul Stowe
From: TC on 28 Feb 2005 22:07 Paul Stowe wrote: > On 28 Feb 2005 12:00:16 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: > >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. > > It occured to me to ask what the temperature of a > > laboratory scale body would be from this mechanism > > so I equated it to the Stephan Boltzman law Led me to conclude T^4 = B.density for some constant B. Paul Stowe writes >Let ñ = Stephen-Boltzmann's constant 5.67E-08 W/m^2-K^4 >Then equate them exactly, >kM/R = ñT^4 I don't thinkk this is right. There needs to be an R^2 on the right side to cancel the 1/m^2 in ñ/ kM/R is an energy flux - watts - right? Thus > __________ > / 4pikpR^2 > T = 4 / -------- > \/ 3ñ Should not have the R^2 term inside. With this correction it agrees with my expression when B is equated to all the stuff other than p. > We're relating apples to apples (radiative heat flux in > Watts per square meter). Radiative heat flux from a planet as in your paper would just be watts, I think. The per area would be integrated out since it is flux for entire planet. > Now, for example, iron has a density of 7800 kg/m^3 > and let's, for simplicity, use a radius of 1 meter. > Then, > 4pi(2.4E-19)(7.8E+03)(1)^2 > -------------------------- = 1.383E-07 > 3(5.67E-08) __________ > / > T = 4 / 1.383E-07 => 0.0193 °K > \/ Since this has R=1 meter, the root-R factor won't make a different. By your calculation then the contribution of gravity to a planet's (total) radiated heat is << 1K. So I don't see how the table in your paper that shows a significant contribution can be right if k is this small. ............ > > It seems to me that this phenonmena would have been noted > > in laboratory cryogenic experiments as it would presented an > > additional difficulty in cooling to temperatures near 0K. > Have they ever achieved absolute zero? It's impossible to achieve. It can only be approached asymptotically. Tom
From: Paul Stowe on 28 Feb 2005 23:08 On 28 Feb 2005 19:07:59 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: >Paul Stowe wrote: >> On 28 Feb 2005 12:00:16 -0800, "TC" <tclarke(a)ist.ucf.edu> wrote: > > >>> 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. > >>> It occured to me to ask what the temperature of a >>> laboratory scale body would be from this mechanism >>> so I equated it to the Stephan Boltzman law > > Led me to conclude T^4 = B.density for some constant B. > > Paul Stowe writes >> Let ý = Stephen-Boltzmann's constant 5.67E-08 W/m^2-K^4 >> Then equate them exactly, > > kM/R = ýT^4 > > I don't think this is right. There needs to be an R^2 on the > right side to cancel the 1/m^2 in ý/ > > kM/R is an energy flux - watts - right? No, Watts per square meter... Watts is 4pikMR >> Thus >> __________ >> / 4pikpR^2 >> T = 4 / -------- >> \/ 3ý >> > > Should not have the R^2 term inside. With this correction it > agrees with my expression when B is equated to all the stuff > other than p. > >> We're relating apples to apples (radiative heat flux in >> Watts per square meter). > > Radiative heat flux from a planet as in your paper would > just be watts, I think. The per area would be integrated out > since it is flux for entire planet. No, it was on a per unit area basis... I'll reference you to the full & detailed evaluation from our paper in "Pushing Gravity" on pages 189-192. Note that k had/has dimensions of m/sec^3. Thus, m | kg | kg Watts -----+------+--- -> ----- = ----- sec^3| | m sec^3 m^2 >> Now, for example, iron has a density of 7800 kg/m^3 >> and let's, for simplicity, use a radius of 1 meter. > >> Then, > >> 4pi(2.4E-19)(7.8E+03)(1)^2 >> -------------------------- = 1.383E-07 >> 3(5.67E-08) >> __________ >> / >> T = 4 / 1.383E-07 => 0.0193 ýK >> \/ > > Since this has R=1 meter, the root-R factor won't make a different. > By your calculation then the contribution of gravity to a planet's > (total) radiated heat is << 1K. Take Jupiter its mass is ~1.9E+27 kg and its radius is ~7E+07 meters. Thus, (2.4E-19)(1.9E+27)/(7E+07) ~ 6.5 Watt/m^2 This multiplied by 4piR^2 yields it total thermal output which is, ~4E+17 Watts. > So I don't see how the table in your paper that shows a significant > contribution can be right if k is this small. Hopefully this clears up the confusion. >........... >>> It seems to me that this phenonmena would have been noted >>> in laboratory cryogenic experiments as it would presented an >>> additional difficulty in cooling to temperatures near 0K. > >> Have they ever achieved absolute zero? > > It's impossible to achieve. It can only be approached asymptotically. Not if there is no energy input into the region. Paul Stowe
From: Bilge on 1 Mar 2005 00:23
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. > >It occured to me to ask what the temperature of a laboratory >scale body would be from this mechanism so I equated it >to the Stephan Boltzman law q = sigma.Area.T^4, where >sigma is a constant, T the temperature. >Since area = 4pi.R^2, the result is T^4 = A.M/R^3 for >some constant A. > >But since M/R^3 is some constant times the density, the >result is T^4 = B.density for some constant B. > >Thus a body of a given density with no other source of heat >will come to an equilibrium temperature that depends only on >its density if LeSage theory as expounded by Paul Stowe is true. > >It's beyond me calculate what B is right now, and if I tried >I'd probably get it wrong, but in Stowe's paper the emission >of Jupiter is claimed to be predicted by LeSage theory. >The temperature of Jupiter is 170 K and its density 1.34 times >that of water. So it seems any body of density 1.34 would >reach a temperature of 170K if well insulated. > >It seems to me that this phenonmena would have been noted >in laboratory cryogenic experiments as it would presented an >additional difficulty in cooling to temperatures near 0K. See: http://www.mathpages.com/home/kmath131/kmath131.htm |