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18 years 4 months ago #16907 by Joe Keller
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A wall clock, could be compared to a gravitational/inertial clock based on comparing the directions of a gyroscope and of a plumb line. The two semicircles of daily travel might require wall times differing by 3 microsec.

Also, regarding my Chernobyl post above, a large error could come from the asphericity of the uranium nucleus (proportionally ep^4/12, where ep is the eccentricity, according to my estimate). I dealt with the question of whether the field is inside or outside, by multiplying it by (1-w), where w is the ratio of charge density, to charge density at the origin.

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18 years 4 months ago #16856 by Joe Keller
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I think that the apparent kinematic viscosity of ether determined experimentally by Yuri Galaev, is about equal to the kinematic viscosity of the totality of the electrons, in his (I guess) brass or aluminum tubes. If the tubes are copper, the usual assumption of one free electron per atom gives too large a kinematic viscosity. Brass, with typically 28% the conductivity of copper (therefore, according to the Drude formula, having perhaps the equivalent of 28% as many totally free electrons) would give a kinematic viscosity for the electron gas, about equal to what Galaev observed for the ether.

The peak of the Maxwellian speed distribution for a free electron at room temperature is 90 km/s, vs. 0.4 km/s for a nitrogen molecule. So, the evacuated metal tubes shielding "null" ether drift experiments, contain plenty of free electrons which are at rest relative to the 20 to 40 km/s ether drift (Earth orbital + solar apex motions); these might nullify the ether drift result. On the other hand, the electrons in or near the light path of Miller's apparatus were bound to atoms of air, glass, paper or wood, with practically the lab velocity relative to the ether.

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18 years 4 months ago #16863 by Joe Keller
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In the above, the kinematic viscosity of a metal's electron gas is calculated as follows. Thermal conductivity is divided by density, then divided by specific heat (at constant volume). This gives a measure of diffusivity, for metals or for air. Diffusivity is essentially the same thing as kinematic viscosity. For the electron gas of a metal, this measure is multiplied by 2+1=3 to account for the energy soaked up by the whole atom (which has 3 kinetic and 3 potential degrees of freedom, vs. only 3 kinetic d.o.f. for the free conduction electron). The ratio of these measures, for a metal and for air, multiplied by the kinematic viscosity of air, gives the kinematic viscosity of the electron gas. Thus the electron gas of brass has about 6/7 of the value found by Galaev for the kinematic viscosity of the ether. His article doesn't state what kind of metal he used for his tubes. Was it brass?

This doesn't contradict the Debye theory of specific heat. Suppose there are only a small fraction, say 1/100, as many free electrons as there are atoms. This makes the electron contribution to specific heat negligible, as in the Debye theory. Instead of multiplying by 3 to account for energy soaked up by the whole atoms, we must multiply by 300. Also we must divide by 100 because this new larger kinematic viscosity reflects only the activity of the free electrons, while we assume 99% of the conduction electrons (one per atom for copper) are unfree at any given moment. So the kinematic viscosity calculated remains the same.

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18 years 4 months ago #16912 by Joe Keller
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Correction: Galaev's article does state that the tubes were copper. Also, his article uses an assumed atomic diameter of 1 Angstrom instead of the value 2.6 typically given for copper. The main difference between copper and brass is that the thermal conductivity (and according to some standards the electrical conductivity also) of brass is only 28% as great. So copper's electron gas has a kinematic viscosity about equal to (6/7)/0.28=3.1x Galaev's value for the ether, but Galaev's value must be multiplied by 2.6 to account for the correct diameter of a copper atom, and there is still agreement.

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18 years 4 months ago #16914 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Joe Keller</i>
<br />clock desynchronization masks the ether drift, to first order. ... the lab rest clock is slow by a fraction 0.5*v^2, and the moving clock is slow by 0.5*((v+w)^2+y^2)=0.5*(v^2+2vw+u^2).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">These small effects are irrelevant to GPS. The GPS system simply measures the time interval for a light signal to travel from a circular orbit to the ground, and does so on both sides of the Earth at once. If there is aether drift, signals on one side will be retarded by the drift, and signals on the opposite side will be sped up by the drift.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If GPS were done with 2-way times ...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Ordinary receivers have no atomic clocks and no transmission capability. But the Air Force Master Stations have both, and do send 2-way signals measured separately at both ends.

Study up on GPS if you want to avoid years of wasted investment in a falsified possibility. For example, see metaresearch.org/solar%20system/gps/absolute-gps-1meter.ASP -|Tom|-

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18 years 4 months ago #16917 by Joe Keller
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Using the thermal expansion coefficients for copper at all temperatures, from Kroeger's 1974 Physics Dissertation at Iowa State Univ., I used the density near absolute zero to calculate the atomic diameter assuming a face-centered cubic lattice. Putting that diameter into Galaev's formula implies a kinematic viscosity for the ether of &lt; 17.89 * 10^(-5), in cgs units. The kinematic viscosity of the electron gas for copper, calculated by the above method, is 17.27 * 10^(-5) at 20 Centigrade, 17.39 @ 0 C., and 17.14 @ 40 C. This conforms to Galaev's idea that it should be just less than the limiting value (of 17.89).

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