Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I consider zero temperature and high lepton number chemical potential $\mu$. This results in a neutrino (or antineutrino, depending on the sign of the potential) "sea" filling a Fermi sphere in momentum space. There are profound effects on particle stability

In the $\mu << 0$ case, the neutron becomes stable. This is because it would produce an antineutrino upon decay but it would have to carry prohibitively large energy because of Pauli's exclusion principle. The negative pion becomes stable for the same reason

In the $\mu >> 0$ case, the positive pion becomes stable

It is then possible to consider bound states of pions and nucleons: neutrons and negative pions in the negative potential case, protons and positive pions in the positive potential case*. This is interesting since they have a mass ratio of about 1 : 7. In molecular physics, the high mass ratio between the electron and the nuclei is the reason for the immense structural richness**. Here the ratio is way smaller but maybe it leads to interesting effects anyway

Is this a correct analysis? What is known / can be said about physics in these conditions? In particular about particle and bound state spectra?

*You can't mix negative pions with protons, since it would produce neutrons. Similarily, mixing positive pions with neutrons yields protons

**At least I think it is. This mass ratio leads to the Born-Oppenheimer approximation which gives rise to a complicated effective potential for the nuclei which posses many local minima: the molecules

EDIT: Actually, I don't want the lepton number chemical potential to be too high, since then pairs of $e^- + \pi^+$ (or $e^+ + \pi^-$, depending on the potential sign) will start forming which introduces additional complications

EDIT: Let's make it a bit more quantitative. What is the potential needed to stabilize a charged pion? W.l.o.g. let's use a negative pion. Under normal condition it mostly decays into $\mu^- + \bar{\nu}$. If this decay is forbidden it still has the $e^-+ \bar{\nu}$ channel (albeit much slower). Since the electron mass is about 0.4% of the pion mass, the resulting electron is ultrarelativistic. Hence the energy splits roughly 50-50 between the electron and the antineutrino and each gets $m_\pi / 2 = 70 MeV$. Thus if the antineutrino Fermi level $-\mu$ is above 70 MeV, the pion is stabilized. The neutron is stabilized under much milder conditions, since $$m_n - m_p - m_e = 780 KeV$$ which is an upper bound on the required chemical potential. Now, to form $e^+ + \pi^-$ pairs we need the chemical potential to reach pion $m_\pi = 140 MeV$ (as above, position mass is relatively negligible). Thus the range of interest is 70 MeV - 140 MeV.

Trouble is, we can also have $2 \pi^- \rightarrow 2 e^- + 2 \bar{\nu}$ processes. Here momentum conservation doesn't constrain us hence to rule this out we are left with the very narrow range of 139 MeV - 140 MeV (the size of this range is $2m_e$). And we do need to rule this out to get multipion bound states

EDIT: There's another aspect to this thing. Sufficiently high negative chemical potential destabilizes the proton due to $p + \bar{\nu} \rightarrow n + e^+$ processes, where the excess energy comes from the antineutrino. Once this destabilization becomes greater than nuclear binding energy, protons cannot appear as constituents of nuclei. In similar fashion, high positive $\mu$ makes the neutron even less stable and at some point neutrons cannot appear as constituents of nuclei. For lower $\mu$ proton-neutron nuclei exist but the beta stability order might be modified

share|improve this question
add comment

1 Answer 1

I don't get this discussion. A finite lepton chemical potential leads to fermi spheres of electrons and neutrinos (there is an issue with electrons -- you need some kind of neutralizing back ground to avoid infinite Coulomb energies). All the weak decays you mention conserve lepton number, so a lepton chemical potential does not bias the decay in any direction.

share|improve this answer
    
Dear Thomas, this seems more like a comment than an answer. As you noted correctly, charged leptons cannot appear by themselves. They only appear in pairs with oppositely charged particles with vanishing lepton number. The lightest charged particle with vanishing lepton number is the pion. However, electron pion pairs will only appear for a potential above 140 GeV because these particles are massive. You can find it in the text of my question. Now, let me explain again why the potential does affect the decay. It's not a Le Chatelier bias, the decay actually becomes forbidden. –  Squark Dec 27 '11 at 8:28
    
Consider a neutron decaying into proton + electron + anti neutrino. In the presence of an anti neutrino Fermi sphere all energy less than mu antineutrino states are occupied. Hence the anti neutrino can only form with energy greater than mu. For sufficiently high mu this disables the decay process due to energy conservation –  Squark Dec 27 '11 at 8:32
    
A lepton chemical potential gives Fermi spheres for electrons and neutrinos, not anti-neutrinos. Do you have an isospin chemical potential in mind? –  Thomas Dec 27 '11 at 13:01
    
the lepton number chemical potential is a signed quantity (as is lepton number). For positive potential you get neutrinos, for negative potential you get antineutrinos. –  Squark Dec 27 '11 at 17:07
    
A neat way to think about it is that the chemical potential is the level of the neutrino Dirac sea. When the sea rises neutrinos appear. When the sea falls, holes = antineutrinos appear –  Squark Dec 27 '11 at 17:12
show 14 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.