Wednesday, 18 May 2011

The Axial Force and the neutrino sea.

Regularly readers will know that my blog is many about the possibility of a fifth force, which acts primary between neutrinos. The strength of the force is unknown, but
can be comparatively large and still not noticeable by traditionally experimental because it doesn't interact with electrons at all. The simplest case a long force is a fully conversed charges with a massless force carrier, the axi-photon. I have not yet
investigate the case of massive force carrier, but the massive case would show up to easierly in accelerator experiments for masses ranges 137 MeV (more massive than a Pion) to 100 GeV.

If neutrinos carry this conversed charge so much protons and or neutrons, specifically the charge on the Neutron must equal the charge on a proton plus the charge on a neutrino, so that beta decay can occur. Further since both protons and neutrons can flip the spins easierly. The axial charge on a spin up nucleon is the same as that on a spin down nucleon. This is the opposite to neutrinos where by definition a right handed neutrino or anti-neutrino has the opposite axial charge to the left handed neutrino.

In any material with a axial net charge its nuclei, the axial force will collect a sea of neutrinos in order to the cancel out it the net axial charge. Since neutrinos are light this will effective screen out the axial force, and make it very difficult to observe, hence why the axial force has not yet been observed. However neutrinos are fermions can as such there is a limit to how many can be placed in a given volume of material. I make a terrible error, in my first paper. I used the wrong formule of the Fermi Energy of the neutrino sea, I used the non-relativistic, when formule when most of the neutrinos are clearly travelling near the speed of light.
The correct formula is.

$$E_F = {\h c}/{2} ({3 ρ}/{π} )^{1/3} $$

Thats the Energy of the most energetic neutrino assuming a neutrino number density of ρ The total Energy is just

$$E_T = {3/4}N E_F = {3/4} {ρ}V E_F $$

We don't need to know the strength of the axial force to calculate the Fermi Energy due to neutrinos (we assume 3 kinds as so far known). If the force isn't strong enough to bind neutrinos of the maximum (fermi) energy to the substance, then the substance will be left with an overall charge. This might lead to a detectable long range forces between substances, which might well have been observed already.

In fact the Fermi energy for most substance is rather large, assuming the axial charge is -1/2 on a proton and +1/2 on a neutron. We have.

MaterialNeutrino Density $cm^{-3}$Fermi Energy
Day Air $1 * 10^{13} $ 1.9 eV
Water$3.34 * 10^{22} $ 1.3 KeV
Uranium-238 $3.5 * 10^{24}$ 4.68 KeV
Bismith $1.78 * 10^{24} $ 3.5 KeV
NaCl (Salt)$2.7 * 10^{22} $ 1.27 KeV
Pyrex Glass$ 1.16 *10^{21} $ 0.4 KeV
Palladium$4.89* 10^{23} $ 3.3 KeV
Copper$ 2.34* 10^{23} $ 2.6 KeV
Zinc$ 1.76 * 10^{23} $ 2.4 KeV
Barium Chloride$ 2.9*10^{23} $ 2.24 KeV

The total energies are thus just to high to be practicle, some 2.7 Mega Joules in one cubic centimeter of tap water.

In order to save the axial force we need to add either several sterlie neutrinos in the 1eV to 1KeV ranges, or more add scalar sneutrinos in the mass range 0.1 eV to 100eV. Scalar neutrinos would not generate any Fermi Energy at all, and would allow any density of matter. Similar arguments apply for any long ranges force that need to cancelled inside matter. For instance a chameleonic B-L (baryon number minus Lepton number) force,
again needs a light charged scalar particle in order to solve the problem of Fermi-Energy.

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