CBR has the answer

Using only geometry and accurate fundamental physical constants, my model gives 2.1729 K, for the "lambda point" of He-4. According to RJ Donnelly, "Experimental Superfluidity" (U. of Chicago, 1967), the experimental value is 2.172.

According to ZM Galasiewicz, ed., "Helium 4" (Pergamon 1971), p. 67, the transformation heat for liquid He-4 I --> liquid He-4 II, is 0.13 cal/gram. This agrees with 0.5*k*2.1729=0.54 cal/mole.

Superfluid He-4's thermal conductivity peaks at about 1.885 K (ibid., p. 120). The shoulders are about 200 times as steep (i.e., d^2(log(f))/d(log(x))^2 ) as for a Planck curve. This was determined by finding second differences at three different intervals and extrapolating to zero on a parabola.

Let every other "deepfrozen" atom have its electron spin vectors reversed and compute the total energy change per side, including nonadjacent atoms. Let the neighbor side also reverse every other atom's spins; of the two ways to do this, choose the way that minimizes the energy change at the hexagon-hexagon bond. Add half that change in bond energy. Setting this equal to k*T, gives about 1.885 K, the temperature of peak thermal conductivity.

The Landau formula for the speed of sound in liquid He-4, together with experimental data, imply (ibid., pp. 179,183) that just above 1 K, the fraction of the helium's mass that is "normal", levels off at about 1/40. This could be because the hexagons are forming icosahedra of 20 surface hexagons each, plus about 20 more hexagons inside. Ultimately there would be only about 1/40 the number of particles, if the extra hexagons trapped between the icosahedra aren't effective.

"Spin down" occurs because the icosahedra acquire the angular momentum of the rotating fluid. So little vorticity remains, that its quantization becomes apparent. The icosahedra act as bearings like "rolamites" to allow superfluidity.

Joe,

You might have more luck getting feedback from us if you would <u>explicitly</u> compare your ideas to our ideas and talk about why your explanation for Phenomenon A is better than our explanation for Phenomenon A.

It will take some effort to learn our stuff well enough to do this, of course. But once you get started up the learning curve you will probably find it worth while.

Regards,
LB

Hi Larry!

Thanks for your input! I can't draw diagrams on this messageboard, but it's better than not publishing at all!

- Joe Keller

The superfluid heat conductivity curve resembles a backwards Planck curve starting at the lambda point and traveling right to left. Also, the Planck curve must be taken to the 4th power (approx. 4.3 at the peak) for the best fit to the data.

This suggests that heat conduction in the superfluid is a cyclic process involving four quanta, each of half, the electron spin reversal energy per bond.

When some hexagon sides deepfreeze near the lambda point, the energy of adjacent un-deepfrozen sides increases. So qualitatively, the helium tends to deepfreeze all at once. The rate of deepfreeze, w.r.t. temperature, becomes infinite at the lambda point, though with a finite nonzero integral. Inter-hexagon forces are stronger in the deepfrozen-but-disorganized state, so a discontinuous rise in viscosity occurs followed by a swift decrease in viscosity at lower temperatures.

The deepfrozen hexagons form polyhedra with triangular faces each containing a centered hexagon. The dihedral angle causes the bond energy in the icosahedron, for example, to be about 1.5% more than on a plane. Half the faces of each hexagon bond with hexagons of adjacent faces. If no more than one of these three bonds is broken for any face, the polyhedron will remain intact. These events are independent for every other face and statistically dependent for the faces in between. This gives a rational function of exp(-E/T) which is taken to the n/2 power, where n is the number of faces.

At the lambda point, 57% of tetrahedrons, 44% of hexahedrons, 33% of octahedrons and 6.3% of icosahedrons are intact. At 3/4 that temperature, 15% of icosahedrons are intact, and at 1/2 that temp, 53%. If the larger polyhedra are more effective at reducing viscosity, a smooth decrease in viscosity results. Packed spheres (face-centered cubic) would fill 74% of the volume. So, at 1/2 the lambda temp, the viscosity might be ((74-53)/(100-6))^2 times as much, and this is close to experiment.

Conclusion.

This shows that CMBR is emitted by intergalactic neutral atomic hydrogen, at an average distance of merely 9000 km/s redshift. The temperature is explained, to 0.5% accuracy, by a semi-relativistic ("ether drift") thermodynamic model based on FitzGerald contraction of the spinning electron. The deviation from blackbody shape is within the accuracy of present-day analyses.

Due to electron pairing, helium emits such radiation inefficiently. The horizontal components of the two electrons' spins fixate at right angles to each other, which quantitatively explains several properties (e.g., the lambda point to four significant figures) of cold helium.

One property of cold helium (the inflection points on the melting curve near 20 K) is explained, in temperature and magnitude (each to about 10% accuracy) by semi-relativistic ("ether drift" or "FitzGerald") addition of spin and orbital kinetic energies. Furthermore GPS accuracy does not disprove ether drift, because, as noted on the Geoengineering thread, the ether drift contributes a cross term to the time dilation factor, which cancels the ether drift effect on signal times and phases.