# Physics in high lepton chemical potential

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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

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asked Dec 24, 2011

## 1 Answer

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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.

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answered Dec 27, 2011 by (720 points)
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.

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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

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A lepton chemical potential gives Fermi spheres for electrons and neutrinos, not anti-neutrinos. Do you have an isospin chemical potential in mind?

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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.

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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

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Yes, but you either get electrons and neutrinos, or anti-electrons and anti-neutrinos, so there is no direct effect on, for example, p->n + e^- + \bar{\nu}.

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At low chemical potential there are no electrons since they are massive and have to be complemented by pions which makes it even worse (i.e. makes the threshold potential for their appearance even higher).

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But the electron mass is much less than the pion mass. So we can either consider large lepton chemical potential, which gives equal Fermi spheres of electrons and neutrinos, or very small lepton chemical potential, which gives a small background density of neutrinos (essentially neutrino dark matter), but no effect on anything else.

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No Thomas pls reread the question. The neutron becomes stable somewhere between 0 and -780 KeV much before the appearance of positrons. Between 70 MeV and 140 MeV with either sign we have a stable pion no electrons or positrons yet. Two pions can decay upon meeting though. Between 139 MeV and 140 MeV there can be multipion bound states. This range is so narrow precisely because the electron mass is small compared to the pion mass

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But my original argument still applies: The lepton chemical potential is a positive (negative) shift in the energy of all leptons (anti-leptons). This means that there is no effect on p + e^- -> n + \nu, or any other reaction that conserves lepton number (any reaction that has the same number of leptons in the initial and final state).

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the reaction n -> p + e^- + \bar{\nu} is affected not because of the formal energy associated with lepton number but because of the presence of antineutrinos in the ground state. This forces the new antineutrino produced to have high energy due to Pauli's exclusion principle. If you wish, you can treat chemical potential as a formal addition of energy to reaction equations. Then the original equation is not affected in vacuum, however the vacuum is no longer the ground state. Transitions 0 -> \bar{\nu} occur until we get a Fermi sphere of antineutrinos.

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In their presence, the new equation is n + N \bar{\nu} -> p + e^- + (N+1) \bar{\nu} which is different. As a simple analogy, consider a chemical reaction X -> Y which goes different ways depending on the presence of a solvent Z. The solvent doesn't participate in the reaction hence its chemical potential is conserved. Nevertheless, turning on Z chemical potential reverses the direction of the reaction.

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For example consider the reaction Na+ + Cl- <-> NaCl. Under pressure 1atm and temperature 300K this reaction would be exothermic to the right, if it happens inside gas He, for example. However, if placed in H20 it becomes exothermic to the left

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I don't think you can shift the equilibrium via the density (that's why the chemical potential is called the chemical potential), but you can suppress rates by Pauli blocking. In the present case I am not sure how much of a difference it makes: If you make the (anti) lepton chemical potential bigger than the final anti-neutrino momentum in free space, you end up making a Fermi sphere of anti-electrons, and you can get a decay n+e^+-> p+\bar{\nu}, where both the e^+ and the \bar{nu} sit near the Fermi surface.

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you are neglecting the numbers here. 780 KeV is sufficient to block the anti neutrino from forming but very far from the 140 MeV required to form positron pion pairs

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And of course I shift the equilibrium. The stable state of zero electric charge and unit baryon number is the state of minimal energy + chemical potential * lepton number with these properties. The neutron wins under my conditions since a very high energy antineutrino is removed, the lepton charge doesn't change, and the only loss is the small mass difference between the neutron and the proton electron pair

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Also I dont understand how your logic accounts for my example from chemistry

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1) If $\mu$ is the control parameter, then you work in the grandcanonical ensemble. There is a reservoir, and for $|\mu|>m_e$ there is a Fermi surface of (anti) electrons. 2) I don't claim to understand physical chemistry, but I think what is typically done is to assign independent chemical potentials to each of the reactands (i.e. not a chemical potential for atom number, but one for the number of Na, one for Cl, one for NaCl, even though atom number, not molecule number is conserved).

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Thomas, you are wrong since positrons have to come paired with pions hence it is the sum of the positron and pion mass that appears here. Regarding 2 I am using molecular chemical potentials. The point is that the chemical potentials of water and Helium in my example affect the equilibrium of a reaction not involving these substances.

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