CBR has the answer

Thanks for posting this information, JMB! Maybe the solar apex motion and heliopause asymmetry cause Dayton Miller's ether drift. The CMB dipole is roughly 120 degrees away from the solar apex motion.

The galactic ether drift causing Oort's law, might cause "high velocity clouds". Also it is roughly antiparallel to the solar apex motion. It would cause a second harmonic (in the polar angle phi) in proper motions, about 1/1000 the amplitude of the first harmonic.

For Gould Belt stars, the random part of the proper motion might be 1/30 as great. They are moving roughly radially away from us. Using radial velocity data alone (corrected for Oort's law), their point of origin and proper motion could be estimated. Then a sample of a mere 1000 stars could give enough precision to detect the second harmonic.

Because the hypothetical 270 km/s galactic tangential ether drift is not perpendicular to the plane of the ecliptic, the apparent perihelion of Venus (orbital eccentricity 0.007) would vary by a large angle, depending on the Earth-season of observation. This assumes that the effect penetrates the heliosphere.

Nor is the drift parallel to the ecliptic. So the apparent ascending node of Mars (inclination 1.8 degrees) would show an analogous variation. However a 2 minute of arc elevation of the entire ecliptic circle is not seen. Sunlight might be protected by its intensity or by the sun's proximity. In the latter case, these effects on apparent orbits might be detectable only for the outer planets.

As 2.8016 K is to atomic hydrogen, so should 2.8016*1.5^4=14.18 K be to helium. The pairing of helium's S1 electrons apparently prevents noticeable blackbody radiation of this temperature from occurring. However, several experimental values in the physical chemistry of helium (see Dugdale JS & FE Simon, "Thermodynamic properties and melting of solid helium", Proceedings of the Royal Society of London, Series A, vol. 218, no. 1134, 7 Jul 1953, pp. 298, 303,304) might be explained.

The "alpha solid" a.k.a. "hexagonal close packed" state of He-4, and its "beta solid" a.k.a. "face centered cubic" state, meet its liquid state in a kind of triple point at 14.9 K. The line dividing the alpha and beta states curves slightly; extrapolation to zero pressure would give about 14 K.

The molar volumes of solid or liquid He-4, solid He-3, or the solid minus liquid difference for He-3 or He-4 (for He-3 see GA Cook, ed., "Argon, Helium & the Rare Gases", vol. 1, Interscience 1961, p. 351), on the melting curve in the temperature-pressure plane, all have, roughly, double inflection points w.r.t. temperature (i.e., zero 2nd & 3rd derivatives) around 14 to 16 K. Using the smoothed data in Dugdale, I averaged the solid and liquid molar volumes, and applied 8th order (lower order at the ends of the interval) Bessel numerical differentiation, to find inflection points near 16 and again near 20 K. Parabolic interpolation found the first zero of d^2V/dT^2 at 14.9 K, and the first zero of d^3V/dT^3 at 15.9 K.

Using a trapezoidal rule (accurate, due to cancellation for smoother interpolants) I found that the deficit in dV/dT due to the aberrant value of d^2V/dT^2 near 16 K, was 0.00115/(molar ideal gas vol/std temp). Solid He-4 gets about half its volume from its quantum gas, so the ideal gas constant, R, becomes effectively 2*(3/2)*k*T. The energy put into the ether by FitzGerald contraction of the spinning electrons, is 2/3*k*T. Treating this like light, would require multiplication by beta/2 to find the equivalent electron kinetic energy for momentum purposes. This predicts a change in dV/dT, of 0.00122 (within 10% of experiment) if the "dumbells" become uncoupled and that vibration energy in the ether gets loose for temperatures above 14 K.

The internal energy at 0 K, of He-4, is experimentally -11.9 cal/mol. The 2/3*k*T ether energy above, plus 2*1/2*k*T for photon energy in the orbital plane, assuming T=14.18 remains applicable to that subsystem, equals 11.7 cal/mol.

The spin dumbells also add their kinetic energy relativistically to the orbital kinetic energy. These kinetic energies should be equal by equipartition. Two-thirds of the dumbells are rotating in an appropriate plane. Taylor expansion of square roots in the addition formula, shows 5/4 the average energy deficit, for this spin-orbit addition phenomenon, as for the spin-orbit contraction phenomenon discussed above. So T=14.18*5/4=17.7. Internal energy at 0 K is unaffected because the two S1 electrons cannot "freeze" at complementary orientations as with spin. The predicted change, near 18 K, in dV/dT, is 5/4*0.00122/(molar ideal gas vol/std temp)=0.001525. The observed deficit in dV/dT, estimated as above from Dugan's He-4 data, is 0.00166.

The trapezoidal, or rather triangular rule, used a trough 6 K wide this time, instead of 4 K, because both 20 K and 22 K seemed aberrant. The last data point had to be estimated from the percent contraction of solid alone, because liquid was not given. By parabolic interpolation the zero of d^3V/dT^3 was 20.4 K (predicted: 15.9*5/4=19.9), and the first zero of d^2V/dT^2 was 18.9 K (predicted: 14.9*5/4=18.6).

Suppose six He-4 atoms form a hexagon. For each atom, let the two electron spins, projected onto the plane of the hexagon, be frozen 90 degrees apart (according to my theory, this should be the case, below 14 K). Now let the spins all precess at the same frequency, with a 60 degree frozen phase difference from each atom to the next. Let the interatomic distance = (cube rt 2; i.e., the experimental value)*(Bohr radius for helium; i.e., Bohr radius for hydrogen / 1.5). Then the average energy of spin alignment, counting only nearest neighbors, is -k*(T=14.18)/96=-0.073 cal/mole of solid alpha He-4, close to the experimental value of -0.08 cal/mole for beta-->alpha crystalline transition.

Tile the plane with these hexagons. The sense of progession of the phase difference cannot reverse from each hexagon to the next. Stabilization energy, comes from stacking hexagons, in phase (analogous stacking occurs in beta crystals). This interatomic bond has four times the energy of the hexagon bond. A stack of hexagons has energy -(5/16)*k*(T=14.18) per hexagon.

Liquid hexagons resemble graphite. Superfluidity occurs when the (horizontal components of) the spins stop precessing. Let one side of the hexagon have spins (vector sum of the atom's electrons) deepfrozen at +/- 30 degree angles to the side. Align this side with another similar hexagon, at the usual interatomic distance. Summing the interactions of the four adjacent dipoles, subtracting the pre-deepfreeze self-interaction of one hexagon-side, and setting the result equal to 0.5*k*Tsuperfluid, gives Tsuperfluid=2.17 K (experimental: 2.18 K). Below this temperature, liquid He-4 forms ribbons. By contrast, He-3 spins cannot deepfreeze, because of the influence of the nuclear dipole.

At zero pressure, He-3 is said to become superfluid at 0.93 mK. The electron spins, whose projections on the orbital plane are, theoretically, fixed at right angles, give 1836*sqrt(2) times more dipole, in that plane, than would the nucleus, if the nuclear moment were one nuclear magneton. The nuclear moment of He-3 is 2.1276 nuclear magnetons. There are two degrees of freedom, for the nuclear spin. There was only one degree of freedom, for the relative phase of the electron spins in the adjacent hexagons. So the predicted temperature of superfluidity, assuming the same interatomic distance at zero pressure for He-3 and He-4, would be 2.17/2*(2.1276/1836/sqrt(2))=0.89 mK for He-3.