Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,741 comments
1,470 users with positive rep
818 active unimported users
More ...

  Physics in high lepton chemical potential

+ 4 like - 0 dislike
2866 views

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

This post has been migrated from (A51.SE)
asked Dec 24, 2011 in Theoretical Physics by Squark (1,725 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Dec 27, 2011 by tmschaefer (720 points) [ no revision ]
Most voted comments show all comments
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

This post has been migrated from (A51.SE)
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

This post has been migrated from (A51.SE)
Also I dont understand how your logic accounts for my example from chemistry

This post has been migrated from (A51.SE)
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).

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
Most recent comments show all comments
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.

This post has been migrated from (A51.SE)
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

This post has been migrated from (A51.SE)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...