# Phases of QCD at nonzero isospin and strangeness chemical potentials with
application to pion stars

Jens O. Andersen<EMAIL_ADDRESS>Martin Kjøllesdal Johnsrud
<EMAIL_ADDRESS>Department of Physics, Norwegian University of Science
and Technology, Høgskoleringen 5, Trondheim, N-7491, Norway

###### Abstract

We study pion and kaon condensation using three-flavor chiral perturbation
theory at finite isospin and strangeness quark chemical potentials $\mu_{I}$
and $\mu_{S}$. The phase diagram consists of a vacuum phase and three distinct
Bose condensed phases with condensates of charged pions as well as charged and
neutral kaons. Adding electromagnetic interactions, a phase with a charged
condensate becomes a Higgs phase and the resulting phase diagram is modified
due to the electromagnetic mass splittings of the mesons. The results for the
pion-condensed phase are applied to calculate mass-radius relation of pion
stars. Local electric charge neutrality is imposed by adding electrons, muons
together with their neutrionos. Finally, we compare our results for the mass-
radius relations with those from recent lattice simulations, and find very
good agreement.

###### keywords:

QCD phase diagram, chiral perturbation theory, meson condensation, compact
stars

††journal: Physics Letters B

##  1 Introduction

The phase diagram of quantum chromodynamics (QCD) has received a lot of
attention since the first sketchy version of it appeared in the 1970s. At the
time, it was expected that it consists of two phases, a hadronic phase at low
temperature and small baryon chemical potential $\mu_{B}$, and a deconfined
phase at high temperature and/or large baryon chemical potential. In 1984,
Bailin and Love [1] suggested that QCD at large baryon chemical potential is a
color superconductor. The result is based on the fact that there is an
attracting channel in the one-gluon exchange diagram in QCD and that
perturbation QCD is valid for large $\mu_{B}$ due to asymptotic freedom. This
renders the Fermi surface unstable and in analogy with BCS theory leads to the
formation of Cooper pairs. Since then a number of colors-superconducting
phases have been found whose exact location is model dependent. See Refs. [2,
3, 4] for reviews.

Conventionally, the QCD phase diagram has been drawn in the $\mu_{B}$–$T$
plane, but we can add more axes to it, for example by allowing for an
independent quark chemical potential $\mu_{f}$ for each flavor $f$. For two
flavors, this implies that we introduce an isospin chemical potential
$\mu_{I}$ and for three flavors an additional strangeness chemical potential
$\mu_{S}$. The new $\mu_{I}$–$T$ plane has received particular attention, in
part by the fact that fermion determinant in QCD at finite isospin density and
zero baryon and strangeness density is real, which implies that the system is
free of the sign problem. Thus one can carry out lattice simulations using
standard importance sampling techniques and study pion condensation and the
transition line in the $\mu_{I}$–$T$ plane. The first lattice simulations
carried out more than 20 years ago [5, 6, 7, 8, 9] suggested that the phase
transition is second order with critical exponent from the $O(2)$ universality
class. At zero temperature the transition takes place at $\mu_{I}=m_{\pi}$,
which is expected for a second-order phase transition. In recent years, high
precision lattice simulations have been carried out confirming the second-
order nature of the transition at $T=0$. Moreover, quark and pion condensates
as functions of $\mu_{I}$ were obtained [10, 11, 12, 13].

Condensation of pions, kaons, and diquarks in QCD and QCD-like theories has
also been studied using chiral perturbation theory ($\chi$PT) [19, 20, 21, 22,
23, 24, 25, 26, 27]. $\chi$PT is an effective low-energy theory based on the
global symmetries and relevant degrees of freedom of the the fundamental
theory (QCD or QCD-like) [28, 29]. The second-order nature of the transition
at low temperatures, where $\chi$PT is valid, as well as the critical chemical
potential being equal to the mass of the condensing particle (at $T=0$) were
confirmed. 111This has been shown at order ${\cal O}(p^{2})$ and ${\cal
O}(p^{4})$ in the low-energy expansion of $\chi$PT and is expected to hold to
all orders. Finally, excellent agreement between lattice and ${\cal O}(p^{4})$
result in chiral perturbation for the quark and pion condensates at $T=0$ was
found. A recent review of meson condensation can be found in Ref. [30].

In Ref. [31], the authors considered the problem of pion condensation in a
dense neutrino gas. At $T=0$, it was shown that for sufficiently large
neutrino chemical potentials and therefore neutrino densities, a charged pion
condensate is formed in electrically neutral matter in chemical equilibrium.
The electron and muon provide the background electric charge that neutralizes
the charged pion condensate. This opens up the possibility to probe the QCD
phase diagram if these conditions are met, for example in the early universe.
This question was addressed in Refs. [15, 16, 17]. In Ref. [17], the authors
use an HRG model where the pion contribution is replaced by an interacting
pion gas based on lattice simulations [18], to allow for pion condensation.
The resulting pressure is parameterized by the temperature $T$ and five
different chemical potentials, corresponding to the conserved charges $B$,
$Q$, and the three lepton numbers. Given that the baryon and electric charge
per entropy ratios are tightly constrained, the cosmic trajectories can be
parameterized by the lepton asymmetries $l_{e,\mu,\tau}$. While the sum of
them is tightly constrained, the individual contributions are not. By varying
$l_{e}+l_{\mu}$ with $l_{e}=l_{\mu}$ and $l_{e}+l_{\mu}+l_{\tau}=0$, different
cosmic trajectories are obtained. For lepton asymmetry $l_{e}+l_{\mu}=0.1$ or
larger, the trajectories enter the pion condensed phase in the $\mu_{Q}$–$T$
plane, where the transition line was obtained from lattice calculations [18].
The idea of a pion star formed by a Bose condensate of charged pions was first
suggested in Ref. [24]. A detailed study of their properties, with and without
charge neutrality constraint, based on an EoS from lattice simulations of QCD
at finite isospin density, can be found in Ref. [14].

In this paper, we continue the study of pion and kaon condensation in the
context of $\chi$PT, including electromagnetic effects. We apply the results
for pion condensation to pion stars. A detailed analysis will be presented
elsewhere [32], see also [33]. The letter is organized as follows. In Sec. 2,
we briefly discuss the chiral Lagrangian at finite chemical potentials and map
out the phase diagram at $T=0$ in the $\mu_{I}$–$\mu_{S}$ plane including
electromagnetic effects. In Sec. 3, we discuss pion stars at finite isospin
chemical potential. We discuss the nonrelativistic and Newtonian limits as
well as the effects of imposing local charge neutrality. The mass-radius
relation is calculated and we compare our results with those of recent lattice
simulations.

##  2 Phases of QCD at nonzero $\mu_{I}$ and $\mu_{S}$

The original formulation of chiral perturbation theory by Gasser and Leutwyler
was in the strong sector [28, 29]. Later, electromagnetic interactions were
included by Ecker et al. [34] at ${\cal O}(p^{2})$, and by Urech [35, 36] and
by Meissner et al. [37] at order ${\cal O}(p^{4})$. In the presence of
electromagnetic interactions, the ${\cal O}(p^{2})$ chiral Lagrangian for
three-flavor QCD is

$\displaystyle{\cal L}_{2}$ $\displaystyle=-{1\over
4}F_{\mu\nu}F^{\mu\nu}+{1\over
4}f^{2}\langle\nabla_{\mu}\Sigma\nabla^{\mu}\Sigma^{\dagger}\rangle$
$\displaystyle\quad+{1\over
4}f^{2}\langle\chi^{\dagger}\Sigma+\Sigma^{\dagger}\chi\rangle+C\langle
Q\Sigma Q\Sigma^{\dagger}\rangle$ $\displaystyle\quad+{\cal L}_{\rm gf}+{\cal
L}_{\rm ghost}\;,$ (1)

where $\langle A\rangle$ means the trace of the matrix $A$, $F_{\mu\nu}$ is
the electromagnetic field tensor,
$\nabla_{\mu}\Sigma=\partial_{\mu}\Sigma-i[v_{\mu}-A_{\mu}Q,\Sigma]$ is the
covariant derivative, $\chi=2B_{0}{\rm diag}(m_{u},m_{d},m_{s})$, $Q=e\,{\rm
diag}({2\over 3},-{1\over 3},-{1\over 3})$ is the charge matrix, $f$ is the
bare pion (or kaon) decay constant, and $C$ is a constant that determines the
electromagnetic mass splittings at tree level. ${\cal L}_{\rm gf}$ and ${\cal
L}_{\rm ghost}$ are the ghost and gauge-fixing terms in the chiral Lagrangian,
but their expressions are not needed in this letter. Finally, $\Sigma$ is the
standard parameterization of the Goldstone fields,

$\displaystyle\Sigma=\exp\left[i{\lambda_{a}\phi_{a}}\over f\right],$ (2)

where $\lambda_{a}$ are the Gell-Mann matrices and $v_{\mu}=v_{0}\delta_{\mu
0}$ with

$\displaystyle v_{0}$ $\displaystyle=\mbox{$1\over
3$}(\mu_{B}-\mu_{S})\mathbb{1}+\mbox{$1\over
2$}\mu_{I}\lambda_{3}+\mbox{$1\over\sqrt{3}$}\mu_{S}\lambda_{8}\;,$ (3)

where the baryon, isospin, and strangeness chemical potentials expressed in
terms of the three quark chemical potentials $\mu_{f}$ are

$\displaystyle\mu_{B}$ $\displaystyle={3\over 2}(\mu_{u}+\mu_{d})\;,$ (4)
$\displaystyle\mu_{I}$ $\displaystyle=(\mu_{u}-\mu_{d})\;,$ (5)
$\displaystyle\mu_{S}$ $\displaystyle={1\over 2}(\mu_{u}+\mu_{d}-2\mu_{s})\;.$
(6)

Alternatively, if we introduce the chemical potentials $\mu_{K^{\pm}}={1\over
2}\mu_{I}+\mu_{S}$ and $\mu_{K^{0}}=-{1\over 2}\mu_{I}+\mu_{S}$, we can write

$\displaystyle v_{0}$ $\displaystyle=\mbox{$1\over
3$}(\mu_{B}-\mu_{S})\mathbb{1}+{1\over 2}\mu_{K^{\pm}}\lambda_{Q}+{1\over
2}\mu_{K^{0}}\lambda_{K}\;,$ (7)

where $\lambda_{Q}=\lambda_{3}+{1\over\sqrt{3}}\lambda_{8}$ and
$\lambda_{K}=-\lambda_{3}+{1\over\sqrt{3}}\lambda_{8}$. Note that
$\\{\lambda_{4},\lambda_{5},\lambda_{Q}\\}$ and
$\\{\lambda_{6},\lambda_{7},\lambda_{K}\\}$ form the remaining two $SU(2)$
algebras of $SU(3)$. After having introduced the chemical potentials, the
symmetry is reduced to $U(1)_{I}\times U(1)_{Y}$ or equivalently
$U(1)_{Q}\times U(1)_{K}$. In the absence of electromagnetic interaction,
these are global symmetries. Including them, the $U(1)_{Q}$ symmetry is being
gauged and is thus local. Since the unit operator commutes with all the
generators of $SU(3)$, the results will be independent of $\mu_{B}$, which
simply reflects that the mesons have zero baryon number and that $\Sigma$ is a
singlet under $U(1)_{B}$ transformations.

By expanding the Lagrangian Eq. (1) with vanishing chemical potentials to
second order in the fields $\phi_{a}$, we can read off the tree-level meson
masses in the vacuum. Considering first the case without EM contributions, we
have

$\displaystyle m_{\pi^{0}}^{2}$
$\displaystyle=m_{\pi^{\pm},0}^{2}=B_{0}(m_{u}+m_{d})\;,$ (8) $\displaystyle
m_{K^{\pm},0}^{2}$ $\displaystyle=B_{0}(m_{u}+m_{s})\;,$ (9) $\displaystyle
m_{K^{0}{/\bar{K}^{0}}}^{2}$ $\displaystyle=B_{0}(m_{d}+m_{s})\;,$ (10)
$\displaystyle m_{\eta}^{2}$
$\displaystyle=\frac{1}{3}B_{0}(m_{u}+m_{d}+4m_{s})\;.$ (11)

Including EM contributions, the masses of the charged particles become

$\displaystyle m_{\pi^{\pm}}^{2}$ $\displaystyle=m_{\pi^{\pm},0}^{2}+\Delta
m_{\text{EM}}^{2}\;,$ (12) $\displaystyle m_{K^{\pm}}^{2}$
$\displaystyle=m_{K^{\pm},0}^{2}+\Delta m_{\text{EM}}^{2}\;,$ (13)

where $\Delta m_{\rm EM}^{2}=2C{{e^{2}}\over f^{2}}$.

In order to understand the phase diagram in three-flavor QCD, it may be useful
to first take a look at the two-flavor case and pion condensation. In this
case $v_{0}={1\over 3}\mu_{B}\mathbb{1}+{1\over 2}\mu_{I}\tau_{3}$, where
$\tau_{a}$ are the Pauli matrices. For $\mu_{I}=0$, the vacuum state
$\Sigma_{0}$ is given by the $2\times 2$ unit matrix, $\Sigma_{0}=\mathbb{1}$.
For finite $\mu_{I}$, the most general ansatz for the ground state is [19, 20]

$\displaystyle\Sigma_{\alpha}^{\pi^{\pm}}$
$\displaystyle=\cos\alpha\mathbb{1}+i\hat{\phi}_{a}\tau_{a}\sin\alpha\;,$ (14)

where $\hat{\phi}_{a}$ are variational parameters and $\alpha$ can be thought
of as a tilt or rotation angle of the ground state. The variational parameters
satisfy $\hat{\phi}_{1}^{2}+\hat{\phi}_{2}^{2}+\hat{\phi}_{3}^{2}=1$ to ensure
that the ground state is normalized, i.e
$(\Sigma_{\alpha}^{\pi^{\pm}})^{\dagger}\Sigma_{\alpha}^{\pi^{\pm}}=\mathbb{1}$.
The thermodynamic potential $\Omega_{0}$ is a function of the parameters
$\hat{\phi}_{a}$ and the tilt angle $\alpha$. After a straightforward
calculation using the ansatz Eq. (14), $\Omega_{0}$ becomes

$\displaystyle\Omega_{0}$ $\displaystyle=-f^{2}m_{\pi^{0}}^{2}\cos\alpha$
$\displaystyle\quad-{1\over 2}f^{2}\left(\mu_{I}^{2}-\Delta m^{2}_{\rm
EM}\right)\sin^{2}\alpha(\hat{\phi}_{1}^{2}+\hat{\phi}_{2}^{2})\;.$ (15)

We notice that the energy is minimized for $\hat{\phi_{3}}=0$, so that neutral
pions do not condense (the third component of the isospin is zero for a
neutral pion). Thus $(\hat{\phi}_{1}^{2}+\hat{\phi}_{2}^{2})=1$ and the energy
is independent of the values of the parameters $\hat{\phi}_{1}$ and
$\hat{\phi}_{2}$, this degeneracy reflects the breaking of the $U(1)_{I}$
symmetry in the pion-condensed phase. We can therefore choose
$\hat{\phi}_{2}=1$ without loss of generality, which yields

$\displaystyle\Sigma_{\alpha}^{\pi^{\pm}}$
$\displaystyle=\cos\alpha\mathbb{1}+i\tau_{2}\sin\alpha=e^{i\alpha\tau_{2}}\;,$
(16) $\displaystyle\Omega_{0}$
$\displaystyle=-f^{2}m_{\pi^{0}}^{2}\cos\alpha-{1\over 2}f^{2}\mu_{I,\rm
eff}^{2}\sin^{2}\alpha\;,$ (17)

where $\mu_{I,\rm eff}^{2}=\mu_{I}^{2}-\Delta m^{2}_{\rm EM}$. The first term
arises from the second term in the Lagrangian (1), while the second term comes
from the second and third term. In minimizing the thermodynamic potential
there is a competition between two terms in Eq. (17), where the first term
prefers $\alpha=0$ and the second term $\alpha={\pi\over 2}$. The optimum
value is found by minimizing $\Omega_{0}$, giving

$\displaystyle\cos\alpha^{*}=\left\\{\begin{array}[]{ll}1\;,&\mu_{I}^{2}\leq
m_{\pi^{\pm}}^{2},\\\ {m_{\pi^{0}}^{2}}\over{\mu_{I,{\rm
eff}}^{2}}\;,&\mu_{I}^{2}\geq m_{\pi^{\pm}}^{2}.\\\ \end{array}\right.\;$ (20)

Thus there is a transition from the vacuum phase to a pion-condensed phase at
$\mu_{I}^{2}=m_{\pi^{\pm}}^{2}$. In the chiral limit, $\alpha^{*}={\pi\over
2}$ and the quark condensate is rotated into a pion condensate for any nonzero
$\mu_{I}$.

We next consider three flavors. For $\mu_{S}=0$, the two-flavor results and
the fact that $\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}$ form an $SU(2)$
algebra suggest that there is a transition to a pion-condensed phase at
$\mu_{I}^{2}=m_{\pi^{\pm}}^{2}$. Similarly, for $\mu_{K}^{0}=0$, we expect a
transition to a phase with a charged kaon condensate at
$\mu_{K^{\pm}}^{2}=m_{K^{\pm}}^{2}$. Finally, for $\mu_{K^{\pm}}=0$, a
transition to a phase with a neutral kaon condensate takes place at
$\mu_{K^{0}}^{2}=m_{K^{0}}^{2}$. In the pion-condensed phase the $U(1)_{I}$
(and therefore also $U(1)_{Q}$) is broken. In the kaon condensed phases,
either the $U(1)_{Q}$ or the $U(1)_{K}$ symmetry is broken. Without
electromagnetic effects, these phases are superfluid. There is a single
Goldstone boson associated with this breaking. If one includes them, the
charged pion and kaon condensates are superconductors since the gauged
$U(1)_{Q}$ is broken. There is no Goldstone boson in these phases as the Higgs
mechanisms eats this degree of freedom giving rise to massive photons with
three polarizations. The form of the ground states is therefore

$\displaystyle\Sigma_{\alpha}^{\pi^{\pm}}=e^{i\alpha\lambda_{2}}\;,\hskip
8.5359pt\Sigma_{\beta}^{K^{\pm}}=e^{i\beta\lambda_{5}}\;,\hskip
8.5359pt\Sigma_{\gamma}^{K^{0}/\bar{K}^{0}}=e^{i\gamma\lambda_{7}}\;,$ (21)

where $\alpha$, $\beta$, and $\gamma$ are the tilt angles in the different
cases and the superscript indicates the condensing meson. Generally, one
should allow for the simultaneous condensation of two or more mesons. However,
this is not supported by actual calculations. The relevant term from the
static Lagrangian, $-{1\over
4}f^{2}\langle[v_{0},\Sigma_{\alpha}][v_{0},\Sigma_{\alpha}^{\dagger}]\rangle$
does not prefer multiple condensates [32].

In analogy with the two-flavor results, the transition points between the
vacuum and Bose-condensed phases are determined by

$\displaystyle\mu_{I}^{2}=m_{\pi^{\pm}}^{2}\;,\hskip
8.5359pt\mu_{K^{\pm}}^{2}=m_{K^{\pm}}^{2}\;,\hskip
8.5359pt\mu_{K^{0}}^{2}=m_{K^{0}/\bar{K}^{0}}^{2}\;.$ (22)

These equations define lines in the $\mu_{I}$–$\mu_{S}$ plane and their
intersection define the region of the vacuum phase. Outside this region, the
Bose-condensed phase with the higher pressure wins.

In the pion-condensed phase, the pressure and isospin density are given by

$\displaystyle p$
$\displaystyle=-\Omega_{0}\Big{|}_{\alpha=\alpha^{*}}={1\over
2}f^{2}\mu_{I,\rm eff}^{2}\left[1-{m_{\pi^{\pm},0}^{2}\over\mu_{I,\rm
eff}^{2}}\right]^{2}\;,$ (23) $\displaystyle n_{I}$
$\displaystyle=-{\partial\Omega_{0}\over\partial\mu_{I}}\bigg{|}_{\alpha=\alpha^{*}}=f^{2}\mu_{I}\left[1-{m_{\pi^{\pm},0}^{4}\over\mu_{I,\rm
eff}^{4}}\right]\;,$ (24)

where we here and in the following have subtracted a constant such that $p=0$
in the vacuum phase. The strangeness density
$n_{S}=-{\partial\Omega_{0}\over\partial_{\mu_{S}}}\big{|}_{\alpha=\alpha^{*}}$
vanishes. As $\mu_{I}\rightarrow m_{\pi^{\pm}}$ from above, the isospin
density goes to zero continuously, which suggests that the transition to the
vacuum state is of second order. In order to find the critical exponent, we
construct a Landau-Ginzburg energy functional by expanding the thermodynamic
potential around $\alpha=0$. This yields, after omitting a constant term,

$\displaystyle\Omega_{0}$ $\displaystyle=-{1\over
2}f^{2}[\mu_{I}^{2}-m_{\pi^{\pm}}^{2}]\alpha^{2}$ $\displaystyle\quad+{1\over
24}f^{2}[4\mu_{I}^{2}-m_{\pi^{\pm}}^{2}-3\Delta m_{\rm
EM}^{2}]\alpha^{4}+\cdots\;.$ (25)

We define a critical chemical potential $\mu_{I}^{c}$ by the vanishing of the
coefficient of the quadratic term. This yields $\mu_{I}^{c}=m_{\pi^{\pm}}$.
Since the quartic term is positive when evaluated at the critical chemical
potential, the transition is of second order as expected. Since
$\alpha^{*}\propto[\mu_{I}^{2}-{\mu_{I}^{c}}^{2}]^{1\over 2}$, for small
$\mu_{I}$, the mean-field critical exponent is ${1\over 2}$.

In the $K^{\pm}$-condensed phase, the isospin and strangeness densities are
given by

$\displaystyle p$ $\displaystyle={1\over 2}f^{2}\mu_{K^{\pm}{,\rm
eff}}^{2}\left[1-{m_{K^{\pm}{,0}}^{2}\over\mu_{K^{\pm},\rm
eff}^{2}}\right]^{2}\;,$ (26) $\displaystyle n_{I}$ $\displaystyle={1\over
2}f^{2}\mu_{K^{\pm}}\left[1-{m_{K^{\pm}{,0}}^{4}\over\mu^{4}_{K^{\pm},\rm
eff}}\right]\;,$ (27) $\displaystyle n_{S}$
$\displaystyle=f^{2}\mu_{K^{\pm}}\left[1-{m_{K^{\pm}{,0}}^{4}\over\mu^{4}_{K^{\pm},\rm
eff}}\right]\;,$ (28)

where $\mu^{2}_{K^{\pm},\rm eff}=\mu^{2}_{K^{\pm}}-\Delta m_{\rm EM}^{2}$. The
result for the $K^{0}/\bar{K}^{0}$ condensed phases is obtained by making the
substitutions $\mu^{2}_{K^{\pm}}\rightarrow\mu^{2}_{K^{0}}$,
$\mu^{2}_{K^{\pm},\rm eff}\rightarrow\mu^{2}_{K^{0}}$, and
$m^{2}_{K^{\pm},0}\rightarrow m^{2}_{K^{0}}$ in Eqs. (26)–(28).

Comparing the pressure of the phases with charged condensates, the transition
line, on which the pressures (23) and (26) are equal, is determined by

$\displaystyle\mu_{K^{\pm}{\rm,eff}}$
$\displaystyle=\frac{1}{2\mu_{I{\rm,eff}}}\bigg{(}\mu_{I,{\rm
eff}}^{2}-m_{\pi^{\pm}{,0}}^{2}$ $\displaystyle\quad+\sqrt{(\mu_{I,{\rm
eff}}^{2}-m_{\pi^{\pm}{,0}}^{2})^{2}+4\mu^{2}_{I{\rm,eff}}m^{2}_{K^{\pm}{,0}}}\bigg{)}\;.$
(29)

A similar equation for the transition line between the neutral and charged
kaon condensates can be obtained.

So far we have said nothing about $f$ appearing in the various expressions for
the pressure and the densities. In an ${\cal O}(p^{4})$ analysis of the phase
transition, similar to Eq. (25), $f$ would be replaced by $f_{\pi}$ if the
transition is to a pion-condensed phase and $f_{K}$ if the transition is to a
kaon-condensed phase. It is therefore natural to identify $f$ with the
physical value for $f_{\pi}$ or $f_{K}$ depending on which phase we are
studying. Identifying $f$ with $f_{\pi}$ also guarantees that the two and
three-flavor results for the pion-condensed phase at ${\cal O}(p^{2})$ are the
same. In the next section, we therefore set $f=f_{\pi}$, whose experimental
value is [38]

$\displaystyle f_{\pi}$ $\displaystyle=92.2,{\rm MeV}\;.$ (30)

In order to generate the phase diagram, we need the physical values for the
pion and kaon masses taken from the Particle Data Group [38],

$\displaystyle m_{\pi^{0}}$ $\displaystyle=134.98\,{\rm MeV}\;,$
$\displaystyle m_{\pi^{\pm}}$ $\displaystyle=139.57\,{\rm MeV}\;,$
$\displaystyle m_{K^{\pm}}$ $\displaystyle=493.68\,{\rm MeV}\;,$
$\displaystyle m_{K^{0}}$ $\displaystyle=497.61\,{\rm MeV}\;.$

Figure 1: Phase diagram as predicted by $\chi$PT in the
$\mu_{I}$–$\mu_{S}$-plane. The black lines are results without electromagnetic
interactions, while red lines are results including them. To the right, we
have zoomed in on two of the triple points. Upper panel shows the intersection
of the normal, neutral kaon condensed and charged kaon condensed phases, while
the lower shows the intersection of the normal, pion condensed, and charged
kaon condensed phases.

In Fig. 1, we show the phase diagram at $T=0$ in the $\mu_{I}$–$\mu_{S}$
plane. The black lines are the transition lines without electromagnetic
interactions and the red lines are with electromagnetic interactions. The
former was first obtained by Kogut and Toublan [22] with $m_{u}=m_{d}$. The
transitions from the normal phase to a meson-condensed phase is always second
order with mean-field exponents. The transitions are in the $O(2)$
universality class, as discussed above. The transitions between the various
condensed phases are always first order and involves the competition between
the order parameters between the different phases. As we cross the transition
lines, the order parameters as well as the densities $n_{I}$ and $n_{S}$ jump
discontinuously. For example, coming from the $\pi^{+}$-condensed phase and
crossing the the transition line to the $K^{+}$-condensed phase, the
strangeness density $n_{S}$ jumps from zero to the value given by Eq. (28).
The small offset of the dashed vertical lines is due to the mass difference
between the charged and neutral kaons, which is both due to $\Delta
m_{\text{EM}}\neq 0$, and $m_{u}\neq m_{d}$. These contributions, however,
pull in opposite directions, as we see in the phase diagram. The contribution
due to the difference in quark masses adds to the mass of the
$K^{0}/\bar{K}^{0}$ meson, which is why the black transition line between the
kaon condensate is to the left of the $\mu_{I}=0$ line, while the
electromagnetic contribution adds to the mass of the charged kaon, which is
why the red line is between these two lines. The partition function in the
normal phase is independent of the two chemical potentials $\mu_{I}$ and
$\mu_{S}$, which is the Silver Blaze property [44]. It implies that all the
thermodynamic functions are constant, for example that the pressure and the
charge densities vanish. Similarly, the partition function in the condensed
phases is independent of one of the chemical potential $\mu_{I}$ or
$\pm{1\over 2}\mu_{I}+\mu_{S}$, implying that $n_{S}=0$, $n_{K^{\pm}}=0$, or
$n_{K^{0}/\bar{K}^{0}}=0$, respectively.

The phase diagram has been mapped out under the assumption that the chemical
potentials are small so that $\chi$PT is inside its region of validity. The
expansion parameter is $M/4\pi f$, where $M$ is a mass or a chemical
potential. An expansion parameter of ${1\over 2}$ corresponds to $M\approx
565$ MeV. Another constraint on the region of validity of $\chi$PT is the
appearance of new resonances not in the model. The first new resonance to
appear is the $\rho$-meson, at $M\approx 770$ MeV. A natural question to ask
is: what happens at larger values of $\mu_{I}$ and $\mu_{S}$. For
asymptotically large values of $\mu_{I}$, asymptotic freedom implies that
quarks are weakly interacting and a description in terms of these degrees of
freedom applies [19, 20]. One-gluon exchange gives rise to an attractive
channel, rendering the Fermi surface unstable to the formation of Cooper pairs
of $\bar{u}$ and $d$ quarks or $\bar{d}$ and $u$ depending on the sign of
$\mu_{I}$. The resulting BCS phase is characterized by the same order
parameter as the pion-condensed phase [19, 20].

##  3 Pion stars

Describing a pion star using the perfect fluid approximation as done for
neutron stars, requires an equation of state. For an ideal Bose gas at zero
temperature, all the bosons are in the zero-momentum state, the pressure is
vanishing, and there is no equation of state [39]. In contrast to neutron
stars, where the Pauli principle gives rise to a quantum pressure at $T=0$, we
need interactions in the bosonic case. In the pion-condensed phase, the mean-
field pressure is given in Eq. (23), which vanishes for $f=f_{\pi}=0$. Using
$p$ and $\epsilon$ to describe matter, the Tolman-Oppenheimer-Volkov (TOV)
equation for a spherically symmetric star follows directly from Einstein’s
fields equation. The continuity equation and the TOV equation for a
spherically symmetric star are

$\displaystyle{dm\over dr}$ $\displaystyle=4\pi r^{2}\epsilon(r)\;,$ (31)
$\displaystyle{dp\over dr}$ $\displaystyle=-{{G\epsilon(r)m(r)}\over
r^{2}}\left[1+{p(r)\over\epsilon(r)}\right]$
$\displaystyle\quad\times\left[1+{4\pi r^{2}p(r)\over
m(r)}\right]\left[1-{2Gm(r)\over r^{2}}\right]^{-1}\;.$ (32)

The TOV equation is the general-relativistic generalization of the
corresponding equation in Newtonian physics, which is derived from imposing
hydrostatic equilibrium throughout the star. The two equations above are
coupled and involve three unknowns, namely the pressure $p(r)$, the energy
density $\epsilon(r)$, and the gravitational mass inside a shell of radius
$r$, $m(r)$. We therefore need the equation of state $p(\epsilon)$ to close
the system. The pressure is a function of one or more chemical potentials
depending on the composition of the star. Imposing electric charge neutrality
and assuming chemical equilibrium yield a number of a constraints among them
and one ends up with a single free chemical potential, for example $\mu_{I}$.
The pressure and the energy density can then be parameterized uniquely by it.

The energy density is given by a Legendre transform of the pressure,

$\displaystyle\epsilon$ $\displaystyle=-p+\sum_{i}n_{i}\mu_{i}\;,$ (33)

where the sum $i$ is over all relevant chemical potentials.

We first discuss a pure pion star with the thermodynamic quantities at ${\cal
O}(p^{2})$ and set for simplicity $\Delta m^{2}_{\rm EM}=0$ in Eqs. (34)–(37)
below. For the pion-condensed phase, we can find the energy density using Eqs.
(23) and (24). Expressing $\epsilon$ as a function of $p$, we obtain the EoS

$\displaystyle\epsilon(p)$
$\displaystyle=-p+2\sqrt{p(p+2f_{\pi}^{2}m_{\pi^{\pm},0}^{2})}\;.$ (34)

In the ultrarelativistic limit $\mu_{I}\rightarrow\infty$, the EoS reduces to
$\epsilon=p$, which is a polytrope with index $n=\infty$ or $\gamma=1$. In the
opposite limit, we write the isospin chemical potential as
$\mu_{I}=m_{\pi^{\pm}{,0}}+\mu_{\rm NR}$, where $\mu_{\rm NR}\ll
m_{\pi^{\pm}{,0}}$ is the non-relativistic chemical potential. One can then
show that

$\displaystyle p$ $\displaystyle=2f_{\pi}^{2}\mu_{\rm NR}^{2}\;,$ (35)
$\displaystyle n_{I}$ $\displaystyle=4f_{\pi}^{2}\mu_{\rm NR}\;,$ (36)
$\displaystyle\epsilon$ $\displaystyle=m_{\pi^{\pm}{,0}}n_{I}+{1\over
8f_{\pi}^{2}}n_{I}^{2}\;.$ (37)

This yields the equation of state is $p=K\epsilon^{2}$, i.e. it is a polytrope
with polytropic index $n=1$ and constant
$K^{-1}=8m_{\pi^{\pm}}^{2}f_{\pi}^{2}$. It is also a polytrope with $n=1$ for
$\Delta m_{\rm EM}^{2}\neq 0$, but now with
$K^{-1}=8m_{\pi^{\pm}}^{4}f_{\pi}^{2}/m_{\pi^{0}}^{2}$. The pressure and
energy density are those of a weakly interacting nonrelativistic Bose gas with
an $s$-wave scattering length $a={m\over 16\pi f_{\pi}^{2}}$ in the mean-field
approximation, first obtained by Bogoliubov [43]. 222To obtain the energy
density of a dilute Bose gas, we subtract the leading term
$m_{\pi^{\pm},0}n_{I}$ associated with the rest mass energy. For stars with a
sufficiently low central pressure, we expect that the EoS of a dilute Bose gas
is a good approximation. Moreover, if $\epsilon(r)\gg p(r)$ and $r\gg 2Gm(r)$,
the TOV equation reduces to the Newtonian equation for hydrostatic
equilibrium, namely the Lane-Emden equation. It is expected that this is also
a good approximation for sufficiently low-mass stars. Solving the Lane-Emden
equation for $\gamma=2$, one finds that the radius $R$ is independent of its
mass, i.e. it corresponds to a vertical line in the mass-radius diagram.
333This limit is analogous to the Chandrasekhar limit for white dwarfs, where
its radius is independent if its mass attains its maximum value, approximately
$1.4M_{\odot}$. The radius in the Newtonian limit is given by

$\displaystyle R$ $\displaystyle=\pi\sqrt{\frac{K}{2\pi G}}.$ (38)

This is shown in Fig. 2, where we plot the mass-radius relation in different
approximations for a pure pion star without electromagnetic interactions.

Figure 2: Mass-radius relation for a pure pion star without electromagnetic
interactions in different approximations. See main text for details. Figure 3:
Mass-radius relation for a pure pion star. See main text for details.

In Fig. 3, we show the mass-radius relation for a pure pion star. The solution
to the TOV equation without electromagnetic interactions is given by the solid
line, where the stable stars are to right of the maximum mass indicated by a
cross. The maximum mass is $M_{\rm max}=10.47M_{\odot}$ and the corresponding
radius is $R=55.4$ km. The mass-radius relation including electromagnetic
interactions is given by the black dashed line in Fig. 3. The maximum mass is
now $10.1M_{\odot}$ with a radius of $53.6$ km. The maximum radius for
vanishing mass is $R_{\rm max}=86$ km. If we include electromagnetic effects,
the maximum radius is reduced to $R_{\rm max}=80$ km. These numerical results
are in excellent agreement with the limit Eq. (38).

We next discuss electric charge neutrality. Since condensed pions are
electrically charged, a pure pion star has a macroscopic electric charge.
However, due to the Coulomb repulsion among the pions, there is an enormous
energy cost of having bulk matter that is not electrically neutral [42]. We
will therefore impose local electric charge neutrality by adding a background
charge of leptons. This amounts to requiring

$\displaystyle n_{I}-\sum_{l}n_{l}$ $\displaystyle=0\;,$ (39)

where the sum is over lepton species $l$. Chiral perturbation theory can be
extended to include the light leptons as dynamical degrees of freedom [45].
The ${\cal O}(p^{2})$-term one should add to the Lagrangian Eq. (1) is

$\displaystyle{\cal L}_{2}=\sum_{l}$
$\displaystyle\left[\bar{l}(i\partial\\!\\!\\!/-\mu_{l}\gamma_{0}+eA\\!\\!\\!/-m_{l})l\right.$
$\displaystyle\left.+\bar{\nu}_{lL}(i\partial\\!\\!\\!/-\mu_{\nu_{l}}\gamma_{0})\nu_{lL}\right]\;,$
(40)

where we include the electron and the muon in the sum, $m_{l}$ is the lepton
mass, $\mu_{l}$ and $\mu_{\nu_{l}}$ are the corresponding chemical potentials.
The neutrinos are massless, $m_{l_{\nu}}=0$, and the subscript $L$ indicates
that there are only left-handed neutrinos. In the power counting scheme
adopted in Ref. [45], a fermion bilinear and lepton mass both count as ${\cal
O}(p)$. Below we include the leading contributions to the pressure and energy
density, which according to the above counting scheme are ${\cal O}(p^{4})$.
It is therefore not consistent to include strong and electromagnetic
interactions among the pions to ${\cal O}(p^{2})$ and the leptons to ${\cal
O}(p^{4})$. We therefore include the ${\cal O}(p^{4})$ corrections from the
pions in the pressure, isospin density, and energy density. Details of the
calculations can be found in Ref. [32] or in a forthcoming paper [33].
Moreover, the ${\cal O}(p^{4})$ electromagnetic effects are prohibitively
difficult to calculate, but since we are mainly interested in the comparison
with the lattice simulations, we do not need them anyway.

A lepton with chemical potential $\mu_{l}>m_{l}$ contributes to the pressure
and energy density as

$\displaystyle p_{l}$ $\displaystyle={2m_{l}^{4}\over
3(4\pi)^{2}}\left[(2x_{l}^{3}-3x_{l})\sqrt{1+x_{l}^{2}}+3{\rm
arcsinh}\,x_{l}\right]\;,$ (41) $\displaystyle\epsilon_{l}$
$\displaystyle={2m_{l}^{4}\over(4\pi)^{2}}\left[(2x_{l}^{3}+x_{l})\sqrt{1+x_{l}^{2}}-{\rm
arcsinh}\,x_{l}\right]\;,$ (42)

where $x_{l}={\sqrt{\mu_{l}^{2}-m_{l}^{2}}\over m_{l}}$ is the dimensionless
ratio of the Fermi momentum and the mass of the lepton. The contribution to
the charge density from a lepton is

$\displaystyle n_{l}$ $\displaystyle={\partial p\over\partial\mu_{l}}={16\over
3(4\pi)^{2}}(\mu_{l}^{2}-m_{l}^{2})^{3\over 2}\;.$ (43)

Considering again the nonrelativistic limit for a fermion, the standard
results for the lepton pressure and energy density are

$\displaystyle p_{l}$ $\displaystyle={16m_{l}^{4}\over
15(4\pi)^{2}}(\mu_{l}^{2}-m_{l}^{2})^{5\over 2}\;,\hskip
14.22636pt\epsilon_{l}=m_{\pi^{\pm}}n_{I}\;,$ (44)

which leads to the equation of state $p_{l}=K\epsilon_{l}^{5\over 3}$. The
neutrinos are massless so $p$ and $\epsilon$ are given by the their
ultrarelativistic limit, including a factor of ${1\over 2}$ due to the absence
of right-handed neutrinos and left-handed antineutrinos,

$\displaystyle p_{\nu_{l}}={2\mu_{\nu_{l}}^{4}\over 3(4\pi)^{2}}\;,\hskip
28.45274pt\epsilon_{\nu_{l}}$
$\displaystyle={2\mu_{\nu_{l}}^{4}\over(4\pi)^{2}}\;.$ (45)

Pions are unstable particles that mainly decay via weak interactions. The
dominant decay mode is $\pi\rightarrow\mu+\nu_{\mu}$. In chemical equilibrium,
the reactions rates of $\pi^{+}\rightarrow l^{+}+\nu_{\mu}$ and
$\pi^{-}\rightarrow l^{-}+\bar{\nu}_{\mu}$ are the same, implying the relation

$\displaystyle\mu_{I}$ $\displaystyle=\mu_{l}-\mu_{\nu_{l}}\;,$ (46)

where $l$ is either an electron or a muon. Although we ignore the neutrino
masses in the analysis, the nonzero masses imply a relation between their
chemical potentials via neutrino oscillations. Chemical equilibrium among the
two species yields

$\displaystyle\mu_{\nu_{e}}$ $\displaystyle=\mu_{\nu_{\mu}}\;,$ (47)

or equivalently $\mu_{e}=\mu_{\mu}$. We have six chemical potentials,
$\mu_{I}$, $\mu_{Q}$, $\mu_{e}$, $\mu_{\mu}$, $\mu_{\nu_{e}}$, and
$\mu_{\nu_{\mu}}$. Chemical equilibrium and charge neutrality reduce this
number to a single independent chemical potential.

We note that at the transition when $\mu_{I}=m_{\pi^{\pm}}$, the lepton
density also vanishes due to the charge neutrality constraint (39). However,
chemical equilibrium, Eq. (46), implies that the neutrino chemical potential
and consequently the neutrino density is nonzero at the transition point, i.e.
on the surface of the pion star. The star therefore has a neutrino atmosphere
with an EoS $\epsilon=3p$.

Figure 4: Mass-radius relations for a pion star for the different cases as
indicated by the legends. Electromagnetic interaction are not included. The
dashed lines are results using the EoS obtained from lattice calculations
[12].

In the absence of electromagnetic interactions, we take the physical mass of
the neutral pion and kaon as our common meson masses. The masses of the
leptons are also taken from Ref. [38]

$\displaystyle m_{e}$ $\displaystyle=0.51\;{\rm MeV}\;,$ $\displaystyle
m_{\mu}$ $\displaystyle=105.6\;{\rm MeV}\;.$ (48)

In order to compare with lattice simulations, we also use the set in Ref.
[14]. The central values are given by

$\displaystyle m_{\pi}=131\,{\rm MeV}\;,\;m_{K}=481\,{\rm
MeV}\;,\;f_{\pi}=90.8\,{\rm MeV}\;.$ (49)

In Fig. 4, we show the mass-radius for pion stars including leptons and
neutrinos. The blue band is obtained from lattice simulations including the
statistical and systematic errors in the EoS. The black solid and dashed lines
are obtained from the ${\cal O}(p^{2})$ EoS from $\chi$PT using the PDG and
lattice values for the meson masses, respectively. The red solid and dashed
lines are obtained from the ${\cal O}(p^{4})$ EoS and the same sets of meson
masses. The results show that the mass-radius relations are rather sensitive
to the masses and decay constant. This change can mostly be understood as a
scaling of the variables in the TOV equation. As we will discuss further, the
EoS is dominated by the neutrinos, so the most important mass scale is
$p_{\text{min}}\propto f_{\pi}^{2}m_{\pi}^{2}$, where $p_{\rm min}$ is the
pressure on the surface of the star. Scaling the pressure and energy density
in the TOV equation by $t^{2}$ leads to a scaling of $M,R$ by $t^{-1}$, so we
expect both mass and radius to be proportional to $1/m_{\pi}f_{\pi}$. Using
the lattice constants, this quantity is around $5\%$ larger than using the PDG
constants, which accounts for most of the difference in 3.

Moreover, the results for the ${\cal O}(p^{2})$ and ${\cal O}(p^{4})$ curves
are very close. The quantity ${\mu_{\pi}\over m_{\pi}}-1$ is equal to
$2.0\times 10^{-2}$ at ${\cal O}(p^{2})$ and $1.8\times 10^{-2}$ at ${\cal
O}(p^{4})$ at the center for the heaviest star, denoted by the crosses. This
corresponds to the largest values of $\mu_{I}$ on the entire branch. In this
region the corrections to the thermodynamic quantities are small suggesting
that the we can reliably use $\chi$PT to calculate the properties of pion
stars. As noted above, at the surface of the star, $\epsilon=3p$. Chemical
equilibrium and charge neutrality imply that $\mu_{e}=\mu_{\mu}=m_{e}$ and
therefore $\mu_{\mu_{e}}=\mu_{\mu_{\nu}}=m_{\pi^{\pm{,0}}}+m_{e}$. This yields
the pressure on the surface $p_{\rm min}={4\over
3(4\pi)^{2}}(m_{\pi^{\pm}{,0}}+m_{e})^{4}$.

The equation of state in the charge neutral pion condensate is well
approximated by $\epsilon=3p$, especially for low pressures, as the neutrino
contribution dominates. In Fig 4, we have included the results of solving the
TOV equation with this EoS, and a surface pressure $p=p_{\rm min}$. The result
is the grey dashed curve in Fig. 4, demonstrating that the EoS is dominated by
the neutrinos if the star is not too compact.

A further investigation of pion stars requires a more complete understanding
of their evolution and instabilities. A necessary requirement for the
stability of stars against spherical perturbations is the increase of their
mass as the central pressure increases [40],

${dM\over dp_{c}}\geq 0\;.$ (50)

Brandt et al. showed, as expected, that pion stars including leptons and
neutrinos are stable for central pressures below that of the maximum mass
configuration [14]. In addition to perturbations, an important component of
the stability of pion stars is the decay rate of its constituent particles. As
the pion star is modeled at zero temperature, the decay of the pion condensate
into neutrinos and charged leptons will be Pauli blocked. The neutrino
atmosphere, however, is not gravitationally bound as we consider neutrinos
massless. This will lead to the depletion of the lepton number at the edge of
the star.

##  4 Acknowledgements

The authors would like to thank B. Brandt, G. Endrődi, and S. Schmalzbauer for
useful discussions as well as their lattice data for the mass-radius relation
of pion stars [41].

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