CBR has the answer

A higher-order ether drift effect for atomic hydrogen, is the rotation of the above-mentioned dumbell, to and fro about the vertical axis as it FitzGerald contracts, traveling along its orbit. The temperature of this would correspond to the popular shortwave frequency 80m, but the power, (80,000)^(-4) x the CMB, is too low.

The atoms above hydrogen all seem to emit much less "spin contraction" radiation than their temperature would imply. Apparently the CMB radiation can pass through a billion hydrogen atoms, so it does not seem that outer atomic electrons block it; indeed in helium there are none. Maybe paired electrons confine the ether disturbance, by maintaining the orbital-plane components of their spins, 90 degrees out of phase with each other as they revolve on opposite sides of the nucleus.

This amounts to a very low emissivity (high reflectivity, or high transparency) in the Stefan-Boltzmann formula. For helium, one might think the power would be 12%/87% abundance * 2 electrons * 1.5^(-2) S orbital surface area * (1.5^4)^4 = 80 times the CMB, instead of the far CIRB power of 5 or 6% times the CMB. The calculation for atomic oxygen is 0.06%/87% * 2 * 7.5^14 = 2.5 billion times the CMB. Because of this un-blackness, even helium only effectively comprises 12%/ 1.5^2 * 0.055/80=0.004% of the area of the blackbody cavity.

Many atoms have unpaired outer electrons; atomic nitrogen might emit P-orbital electron radiation commensurate with an effective Z of roughly 1.5 to 2; the power would be roughly 8 * 10^(-5)/87% abundance * 1.75^14 = 2% of the CMB power, with a peak at 5 to 16x higher frequency; perhaps this is the far CIRB, not helium.

Chlorine might be convenient experimentally. The effective Z for its unpaired P electron might be 2 to 3; using photochemically dissociated dilute chlorine vapor in helium, one might observe blackbody radiation at roughly 2.80*2.5^4=109 K, observable if the mix is under 100 K.

For all but the hard CXB, replacement of the T^4 Stefan-Boltzmann law with a T^1 law, fits observation within an order of magnitude. Therefore the abundance of heavy elements in the universe might be much larger than now believed.

On the Hertzsprung-Russell diagram (Phillips, The Physics of Stars, John Wiley, 1999), the geometric mean, of the two temperatures where the curvature of the line is greatest (i.e., of the knee temperatures) is 8610 K. The center of the "Cepheid pulsation channel" (or "strip") (Hwang & Yu, Stellar Astrophysics, Springer, 1998) crosses the Hertzsprung-Russell line at 9020 K. It is the "dwarf Cepheids" which overlie the H-R line; where they lie on it, they average 8710 K. The above FitzGerald agitation temperature calculated for atomic oxygen, 8841 K, therefore equals the "main sequence" - Cepheid intersection temperature to within 2% accuracy.

Thermodynamics might require the maximum temperature of the solar corona to equal the true maximum temperature of the solar interior. The sun's corona reaches 1 to 2 million K. The corona of Sirius is only 40% hotter than the sun's, though its surface is 71% hotter. The INTERIOR temperatures of main sequence stars might be stabilized near my calculated agitation temperature of IRON (1.15 million K), just as ice cubes stabilize the temperature of water.

According to ideal gas laws and gravity, interior temperature is proportional to radius regardless of density (the radius of Sirius is 68% more than the sun's). In the midrange of the main H-R sequence, the luminosity varies as the fourth power of surface temperature, showing that radius is independent of surface temperature. Thus interior temperature is independent of surface temperature, for main sequence stars.

If the density of a star magically doubles, then the reaction rate, at a given temperature, quadruples. Assuming negligible change in the interior temperature profile, the density increase causes each adiabatic convection packet to contain twice as much heat; the viscosity is twice as much, but the packet rises twice as fast because the bouyant force is four times as much, due to density and gravity. So, the quadrupled heat can be delivered, with negligible change in interior temperature.

If instead, the SURFACE temperature is stabilized near the agitation temperature of atomic OXYGEN (8841 K), one has the Cepheid variables. If both stabilizations fail (all the ice melts) one has a nova.

Radiotelescope data from Table 2, "H1 Redshift Surveys and Large-Scale Structure" by Salzer & Haynes (2001), show the "convergence depth" at which concentric shells of galaxies, attain the same average velocity (relative to our Local Group of galaxies) as have the sources of the CMB. Only "vpar", the component, of a shell's velocity, parallel to the CMB velocity, showed a trend. The best fit, giving no more than about 1% error for any of the four shells, was vpar = 266 + 9.5*(r/1000)^2. Distance, r, is in "km/s redshift"; speed, vpar, is in km/s. The characteristic distance of CMB absorption satisfies 627 = 266 + 9.5*2*(r/1000)^2, i.e., 1% redshift.

"Virgo infall" is a term for the Local Group's velocity of roughly 250 km/s, relative to the shell of galaxies at roughly 1250 km/s redshift distance, which includes the Virgo cluster. It is roughly parallel to the CMB velocity, so it agrees well with the above formula's prediction, 280 km/s.

In a second-order Taylor expansion of the velocity of the extragalactic matrix, only unmixed second partial derivative terms fail to cancel over spheres, thus giving the quadratic term; its coefficient is 3/c, to 5% accuracy. The constant term is the component, parallel to the CMB velocity, of the Local Group's random velocity relative to the extragalactic matrix. I used the midpoint of each distance interval, because the authors had roughly equal numbers of galaxies at each distance.

Thanks to my brother Ed, for starting all the above, by sending me Hatch's article suggesting that FitzGerald contraction causes "Thomas precession" radiation. When I modelled that, I found CMB radiation instead.

Hello all

I have been reading all your comments.
Sounds like fun.

Unable to add anything. My Brain is half dead,,,,,,,too much work.


Only that the universe is not expanding.

Harry

Hi Harry!

Thanks for tuning in. I've never been south of the equator. Have you ever been north of it?

- Joe