# Behavior of a Chiral Condensate Around Astrophysical-Mass Schwarzschild and
Reissner-Nordström Black Holes

Ross DeMott, Alex Flournoy

(Department of Physics, Colorado School of Mines, Golden, CO 80401, USA.
April 2023)

###### Abstract

In this work, we develop a perturbative method to describe the behavior of a
chiral condensate around a spherical black hole whose mass is astrophysically
realistic. We use the inverse mass as the expansion parameter for our
perturbative series. We test this perturbative method in the case of a
Schwarzschild black hole, and we find that it agrees well with previous
numerical results. For an astrophysical-mass Schwarzschild black hole, the
leading order contribution to the condensate is much larger (in most of space)
than the next-to-leading order contribution, providing further evidence for
the validity of the perturbative approach. The size of the bubble of restored
chiral symmetry is directly proportional to the size of the black hole.
Next, we apply this perturbative method to a Reissner-Nordström (RN) black
hole. We find that, as the charge-to-mass ratio increases, the bubble of
restored chiral symmetry becomes larger relative to the black hole. This
effect is particularly pronounced for near-extremal RN black holes. The case
of an extremal RN black hole provides an interesting counterexample to the
standard thermal explanation for the formation of a bubble of restored chiral
symmetry around a black hole.

##  1 Introduction

In physics, spontaneous symmetry breaking is a critical tool, used to explain
how fundamentally symmetric theories can give rise to asymmetric low-energy
effective theories. Most of the familiar examples of spontaneous symmetry
breaking come from theories formulated in Minkowski spacetime, but it is also
possible for spontaneous symmetry breaking to occur in theories on curved
backgrounds [1, 2, 3, 4].
The QCD phase diagram is a matter of great interest to contemporary particle
physics. At high energies and/or high temperatures, QCD exhibits chiral
symmetry. At low energies and/or temperatures, QCD spontaneously breaks chiral
symmetry. If gravity is included in the theory, strong space-time curvature
can also cause QCD matter to transition from a phase with spontaneously broken
chiral symmetry to a phase with restored chiral symmetry [5, 6, 7, 8]. Such
strong curvature can be found in the vicinity of black holes.
In Ref. [8], the authors found that the chiral condensate approaches zero near
the event horizon of a Schwarzschild black hole. Thus, the event horizon is
surrounded by a “bubble” of approximately restored chiral symmetry. As the
mass of the black hole increases, the radius of this bubble (relative to the
radius of the black hole) decreases. At the boundary of the bubble, a second-
order phase transition takes place, in which the chiral-symmetric phase inside
the bubble meets the chiral-broken phase in the wider universe.
Throughout this article, we restrict our discussion to astrophysical-mass
black holes, whose masses are much larger than the Planck scale. Because of
this, we can use the inverse mass (in Planck units) as a small expansion
parameter. We use this perturbative approach to greatly simplify the equations
of motion for the condensate $\sigma$. We apply this method to Schwarzschild
black holes, and we find that the results thereby obtained are consistent with
the numerical results obtained in Ref. [8]. Having confirmed the validity of
the perturbative approach for Schwarzschild black holes, we apply the method
to Reissner-Nordström black holes (which possess electric charge but no
angular momentum). As far as we know, this is the first time that the behavior
of a chiral condensate around a charged black hole has been investigated.
In flat spacetime, there is a critical temperature $T_{c}$ that marks the
boundary between the phase with spontaneously broken chiral symmetry
($T<T_{c}$) and the phase with restored chiral symmetry ($T>T_{c}$) [9, 8]. At
any point, the Tolman temperature gives the effective local temperature of
spacetime [8]. As one approaches the event horizon of the black hole, the
Tolman temperature increases without bound. Assuming that the asymptotic
temperature is smaller than $T_{c}$, the Tolman temperature will cross $T_{c}$
at some finite radius outside the event horizon. Thus, one intuitively expects
that the chiral condensate will transition from a spontaneously broken phase
far away from the horizon to a restored phase close to the horizon. This is
the usual argument explaining why chiral symmetry is restored close to the
horizon [8]. However, we find that this intuitive argument fails for an
extremal Reissner-Nordström black hole.
Throughout this article, we use Planck units, so that $c=G=\hbar=4\pi\hskip
1.42262pt\epsilon_{0}=k_{B}=1$.

##  2 Effective Action for General Black Hole Space-Time

In a Minkowski background, a general, spherically-symmetric black hole metric
takes the form

$ds^{2}=-f\left(r\right)\hskip 1.42262ptdt^{2}+f\left(r\right)^{-1}\hskip
1.42262ptdr^{2}+r^{2}\hskip 1.42262ptd\Omega^{2},$ (1)

where $d\Omega^{2}$ is the line element on the two-sphere. Every black hole
has a temperature $T_{\textrm{BH}}$. To account for thermal effects, we work
in Euclidean time. The Euclidean time direction is periodic, and its period is
equal to the inverse temperature $\beta$. In Euclidean time, we may write the
metric as

$ds^{2}=f\left(r\right)\hskip 1.42262ptdt^{2}+f\left(r\right)^{-1}\hskip
1.42262ptdr^{2}+r^{2}\hskip 1.42262ptd\Omega^{2}.$ (2)

We model baryonic matter by a four-fermion interaction with interaction
strength $\lambda$. Let $N$ be the total number of different types of quarks,
counting both flavors and colors. We may write the classical action as [8]

$S=\int\mathrm{d}^{4}x\hskip 1.42262pt\left\\{\overline{\psi}\hskip
1.42262pti\gamma^{\mu}\nabla_{\mu}\psi+\frac{\lambda}{2N}\left(\overline{\psi}\psi\right)^{2}\right\\}.$
(3)

(The astute reader may notice that there is no factor of $\sqrt{g}$ beside the
volume element in Eqn. 3. This is because the determinant of the metric in
Schwarzschild spacetime is identical to that in Minkowski spacetime.) We
define the condensate $\sigma$ as

$\sigma=-\frac{\lambda}{N}\hskip 1.42262pt\langle\overline{\psi}\psi\rangle.$
(4)

Let us introduce the new variables $\Sigma_{\pm}$, which are defined by

$\Sigma_{\pm}=\sigma^{2}\pm\sqrt{f}\frac{d\sigma}{dr}.$ (5)

Note that $f$ is short for $f\left(r\right)$. Let us define
$\omega_{\nu}\left(u\right)$ as

$\omega_{\nu}\left(u\right)=\sum_{n=1}^{\infty}\left(-1\right)^{n}n^{-\nu}K_{\nu}\left(n\beta
u\right).$ (6)

To approximate the effective action, we use a resummed heat kernel expansion.
To second-order in the resummed heat kernel expansion, we may write the
effective action $\Gamma\left[\sigma\right]$ as [8, 5, 6]

$\Gamma\left[\sigma\right]=\beta\int\mathrm{d}^{3}\vec{x}\hskip
2.84526pt\mathcal{L},$ (7)

where $\vec{x}$ are the spatial coordinates. The Lagrangian density
$\mathcal{L}$ is given by [8, 5, 6]

$\mathcal{L}=-\frac{\sigma^{2}}{2\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(\mathcal{L}_{+}+\mathcal{L}_{-}\right)+\frac{1}{32\pi^{2}}\delta\mathcal{L}.$
(8)

The Lagrangian pieces $\mathcal{L}_{\pm}$ and $\delta\mathcal{L}$ are given by

$\mathcal{L}_{\pm}=\frac{3}{4}\Sigma_{\pm}^{2}-\left(\frac{1}{2}\Sigma_{\pm}^{2}+a_{\pm}\right)\ln\left(\frac{f\Sigma_{\pm}}{\ell^{2}}\right)+\frac{16\Sigma_{\pm}}{f\beta^{2}}\hskip
1.42262pt\omega_{2}\left(\sqrt{f\Sigma_{\pm}}\right)+4a_{\pm}\omega_{0}\left(\sqrt{f\Sigma_{\pm}}\right)$
(9)

and

$\delta\mathcal{L}=\left(\Sigma_{+}^{2}+\Sigma_{-}^{2}\right)\frac{\ln
f}{2}-\frac{\left(\Sigma_{+}+\Sigma_{-}\right)}{12f}\left\\{\left(\frac{df}{dr}\right)^{2}-2f\frac{d^{2}f}{dr^{2}}+\frac{4f}{r}\frac{df}{dr}\right\\}.$
(10)

With a little algebra, we may rewrite $\delta\mathcal{L}$ as

$\delta\mathcal{L}=\left(\sigma^{4}+f\left(\frac{d\sigma}{dr}\right)^{2}\right)\ln
f-\frac{\sigma^{2}}{6f}\left\\{\left(\frac{df}{dr}\right)^{2}-2f\frac{d^{2}f}{dr^{2}}+\frac{4f}{r}\frac{df}{dr}\right\\}.$
(11)

Let $\Delta_{\textrm{Eucl}}$ be the Laplacian for three-dimensional Euclidean
space. We may write $a_{\pm}$ as (see Appendix A for derivation)

$a_{\pm}=a_{\textrm{sp}}+\frac{f\left(r\right)^{2}}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{\pm}\right),$
(12)

where the pure space-time contribution $a_{\textrm{sp}}$ is given by

$\displaystyle\begin{split}a_{\textrm{sp}}=\hskip
2.84526pt&\frac{1}{180}\left\\{\frac{2f\left(r\right)^{2}}{r^{4}}-\frac{8f\left(r\right)^{3}}{r^{4}}+\frac{6f\left(r\right)^{4}}{r^{4}}+\frac{8f\left(r\right)^{2}}{r^{3}}f^{\prime}\left(r\right)-\frac{12f\left(r\right)^{3}}{r^{3}}f^{\prime}\left(r\right)+\frac{6f\left(r\right)^{2}}{r^{2}}f^{\prime}\left(r\right)^{2}-\frac{4f\left(r\right)^{2}}{r^{2}}f^{\prime\prime}\left(r\right)\right.\\\
&\left.+\frac{6f\left(r\right)^{3}}{r^{2}}f^{\prime\prime}\left(r\right)-\frac{6f\left(r\right)^{2}}{r}f^{\prime}\left(r\right)f^{\prime\prime}\left(r\right)+\frac{3}{2}f\left(r\right)^{2}f^{\prime\prime}\left(r\right)^{2}-\frac{6f\left(r\right)^{3}}{r}f^{\left(3\right)}\left(r\right)-f\left(r\right)^{2}f^{\prime}\left(r\right)f^{\left(3\right)}\left(r\right)\right.\\\
&\left.-2f\left(r\right)^{3}f^{\left(4\right)}\left(r\right)\right\\}.\end{split}$
(13)

We assume the black hole is of astrophysical size, which means that $\beta$ is
very large. Therefore, let us make the following assumptions:

$\beta\sqrt{f\Sigma_{+}}\gg 1,$ (14) $\beta\sqrt{f\Sigma_{-}}\gg 1.$ (15)

Later, we will verify that these assumptions are satisfied in most of space.
The Bessel functions $K_{\nu}\left(x\right)$ decay exponentially for large
values of $x$. Therefore, if Eqns. 14 and 15 are satisfied, the Lagrangian
terms involving $\omega_{0}\left(\sqrt{f\Sigma_{\pm}}\right)$ and
$\omega_{2}\left(\sqrt{f\Sigma_{\pm}}\right)$ are exponentially suppressed.
Hence, we may simplify the Lagrangian pieces $\mathcal{L}_{\pm}$ to

$\mathcal{L}_{\pm}=\frac{3}{4}\Sigma_{\pm}^{2}-\left(\frac{1}{2}\Sigma_{\pm}^{2}+a_{\pm}\right)\ln\left(\frac{f\Sigma_{\pm}}{\ell^{2}}\right).$
(16)

With some algebraic manipulation, we may write the sum
$\mathcal{L}_{+}+\mathcal{L}_{-}$ as

$\displaystyle\begin{split}\mathcal{L}_{+}+\mathcal{L}_{-}=&\left(\frac{3}{2}-\ln\left(\frac{f}{\ell^{2}}\right)\right)\left(\sigma^{4}+f\left(\frac{d\sigma}{dr}\right)^{2}\right)-\frac{1}{2}\left(\Sigma_{+}^{2}\ln\Sigma_{+}+\Sigma_{-}^{2}\ln\Sigma_{-}\right)-\frac{f\left(r\right)^{2}}{3}\Delta_{\textrm{Eucl}}\left(f\sigma^{2}\right)\ln\left(\frac{f}{\ell^{2}}\right)\\\
&-a_{\textrm{sp}}\ln\left(\sigma^{4}-f\left(\frac{d\sigma}{dr}\right)^{2}\right)-\frac{f\left(r\right)^{2}}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{+}\right)\ln\Sigma_{+}-\frac{f\left(r\right)^{2}}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{-}\right)\ln\Sigma_{-}.\end{split}$
(17)

(Note that we have omitted a term that does not depend on $\sigma$.) Recall
that the effective action $\Gamma\left[\sigma\right]$ is given by

$\Gamma\left[\sigma\right]=\beta\int\mathrm{d}^{3}\vec{x}\hskip
1.42262pt\mathcal{L}$ (18)

and that the total Lagrangian $\mathcal{L}$ is given by

$\mathcal{L}=-\frac{\sigma^{2}}{2\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(\mathcal{L}_{+}+\mathcal{L}_{-}\right)+\frac{1}{32\pi^{2}}\delta\mathcal{L}.$
(19)

For convenience, let us introduce the quantities $\mathcal{A}$ and
$\mathcal{B}$, which are defined as

$\mathcal{A}=\left(\frac{3}{2}-\ln\left(\frac{f}{\ell^{2}}\right)\right)\left(\sigma^{4}+f\left(\frac{d\sigma}{dr}\right)^{2}\right)-\frac{1}{2}\left(\Sigma_{+}^{2}\ln\Sigma_{+}+\Sigma_{-}^{2}\ln\Sigma_{-}\right)-a_{\textrm{sp}}\ln\left(\sigma^{4}-f\left(\frac{d\sigma}{dr}\right)^{2}\right),$
(20)
$\mathcal{B}=-\frac{1}{3}\Delta_{\textrm{Eucl}}\left(f\sigma^{2}\right)\ln\left(\frac{f}{\ell^{2}}\right)-\frac{1}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{+}\right)\ln\Sigma_{+}-\frac{1}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{-}\right)\ln\Sigma_{-}.$
(21)

From Eqns. 17 and 19, we see that the Lagrangian may be written as

$\mathcal{L}=-\frac{\sigma^{2}}{2\lambda}+\frac{f^{-2}}{32\pi^{2}}\mathcal{A}+\frac{1}{32\pi^{2}}\mathcal{B}+\frac{1}{32\pi^{2}}\delta\mathcal{L}.$
(22)

Combining Eqns. 17, 18, and 22, we may write the effective action as

$\Gamma\left[\sigma\right]=\frac{\beta}{32\pi^{2}}\int\mathrm{d}^{3}\vec{x}\left(-\frac{1}{3}\Delta_{\textrm{Eucl}}\left(f\sigma^{2}\right)\ln\left(\frac{f}{\ell^{2}}\right)-\frac{1}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{+}\right)\ln\Sigma_{+}-\frac{1}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{-}\right)\ln\Sigma_{-}\right)+\dots,$
(23)

where $\dots$ represents terms that do not depend involve the Laplacian
$\Delta_{\textrm{Eucl}}$. Using Green’s first identity, we may rewrite Eqn. 23
as

$\Gamma\left[\sigma\right]=\frac{\beta}{32\pi^{2}}\int\mathrm{d}^{3}\vec{x}\left(\frac{1}{3}\nabla_{\textrm{Eucl}}\left(f\sigma^{2}\right)\cdot\frac{\nabla_{\textrm{Eucl}}f}{f}+\frac{1}{6}\nabla_{\textrm{Eucl}}\left(f\Sigma_{+}\right)\cdot\frac{\nabla\Sigma_{+}}{\Sigma_{+}}+\frac{1}{6}\nabla_{\textrm{Eucl}}\left(f\Sigma_{-}\right)\cdot\frac{\nabla\Sigma_{-}}{\Sigma_{-}}\right)+\dots.$
(24)

Taking advantage of spherical symmetry, we may rewrite Eqn. 24 (after some
algebra) as

$\Gamma\left[\sigma\right]=\frac{\beta}{32\pi^{2}}\int\mathrm{d}^{3}\vec{x}\left(\frac{2}{3}\frac{df}{dr}\frac{d\left(\sigma^{2}\right)}{dr}+\frac{\sigma^{2}}{3f}\left(\frac{df}{dr}\right)^{2}+\frac{f}{6\Sigma_{+}}\left(\frac{d\Sigma_{+}}{dr}\right)^{2}+\frac{f}{6\Sigma_{-}}\left(\frac{d\Sigma_{-}}{dr}\right)^{2}\right)+\dots.$
(25)

Thus, we may rewrite the Lagrangian piece $\mathcal{B}$ as

$\mathcal{B}=\frac{2}{3}\frac{df}{dr}\frac{d\left(\sigma^{2}\right)}{dr}+\frac{\sigma^{2}}{3f}\left(\frac{df}{dr}\right)^{2}+\frac{f}{6\Sigma_{+}}\left(\frac{d\Sigma_{+}}{dr}\right)^{2}+\frac{f}{6\Sigma_{-}}\left(\frac{d\Sigma_{-}}{dr}\right)^{2}.$
(26)

##  3 Large Schwarzschild Black Hole

###  3.1 Expansion in Powers of $M^{-1}$

Let us consider a large Schwarzschild black hole with mass $M\gg 1$ in Planck
units. The metric function $f\left(r\right)$ is given by

$f\left(r\right)=1-\frac{2M}{r}.$ (27)

With this metric function, the quantity $a_{\textrm{sp}}$ takes the form [10]

$a_{\textrm{sp}}=\frac{4M^{2}}{5r^{6}}\left(1-\frac{2M}{r}\right)^{2}.$ (28)

So far, we have been working in Planck units. However, for an astrophysical
black hole, the mass $M$ should be at least a few solar masses. One solar mass
is around $10^{38}$ Planck masses. Therefore, the Schwarzschild radius of an
astrophysical black hole will be at least $\sim 10^{38}$ Planck lengths. Thus,
the metric tensor and its derivatives will hardly change at all if $r$ changes
by a Planck length. Since the profile of $\sigma$ as a function of $r$ depends
on the properties of space-time, it is reasonable to expect that
$\sigma^{\prime}\left(r\right)$ will be very small. To obtain a better
description, we rescale our unit of length by a factor of $M$. As a result, we
have a new radial coordinate $R$, which is given by

$R=\frac{r}{M}.$ (29)

Under this coordinate transformation, derivatives will transform as

$\frac{d^{n}}{dr^{n}}\longrightarrow\frac{1}{M^{n}}\frac{d^{n}}{dR^{n}}.$ (30)

We may write the metric function $f\left(R\right)$ as

$f\left(R\right)=1-\frac{2}{R}.$ (31)

Thus, in this new coordinate system, the event horizon is located at $R=2$.
Because Eqn. 31 does not contain any very large or very small coefficients, we
see that the coordinate $R$ is well-suited to capturing the spatial variation
in the properties of space-time. Therefore, it is also well-suited to
capturing the spatial variation in $\sigma$. Plugging Eqn. 29 into Eqn. 28, we
find that

$a_{\textrm{sp}}=\frac{4}{5M^{4}R^{6}}\left(1-\frac{2}{R}\right)^{2}.$ (32)

In Ref. [8], the authors employed non-perturbative, numerical methods to solve
for the condensate $\sigma\left(r\right)$. That work considered black holes
whose mass was within a few orders of magnitude of the Planck mass. Since the
black holes considered here are much larger, it is more convenient to expand
the Lagrangian $\mathcal{L}$ in powers of $M^{-1}$. Keeping terms up to
$M^{-2}$, we may write the Lagrangian pieces $\mathcal{A}$, $\mathcal{B}$, and
$\delta\mathcal{L}$ as [10]

$\mathcal{A}=\left(\frac{3}{2}-\ln\left(\frac{f}{\ell^{2}}\right)\right)\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)-\frac{1}{2}\left(\Sigma_{+}^{2}\ln\Sigma_{+}+\Sigma_{-}^{2}\ln\Sigma_{-}\right),$
(33)
$\mathcal{B}=\frac{8\sigma}{3M^{2}R^{2}}\frac{d\sigma}{dR}+\frac{4\sigma^{2}}{3M^{2}R^{4}f}+\frac{f}{6M^{2}\Sigma_{+}}\left(\frac{d\Sigma_{+}}{dR}\right)^{2}+\frac{f}{6M^{2}\Sigma_{-}}\left(\frac{d\Sigma_{-}}{dR}\right)^{2},$
(34)
$\delta\mathcal{L}=\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)\ln
f-\frac{\sigma^{2}}{3f}\left(\frac{8}{M^{2}R^{3}}-\frac{14}{M^{2}R^{4}}\right).$
(35)

To further simplify Eqns. 33 and 34, we must expand $\ln\Sigma_{\pm}$ and
$\Sigma_{\pm}^{-1}$ in powers of $M^{-1}$. We may write $\Sigma_{\pm}$ as

$\Sigma_{\pm}=\sigma^{2}\left(1\pm\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}\right).$
(36)

Thus, we may write $\ln\Sigma_{\pm}$ and $\Sigma_{\pm}^{-1}$ as

$\ln\Sigma_{\pm}=\ln\left(\sigma^{2}\right)+\ln\left(1\pm\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}\right),$
(37)
$\Sigma_{\pm}^{-1}=\frac{1}{\sigma^{2}}\left(1\pm\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}\right)^{-1}.$
(38)

Using Taylor series (up to second order in $M^{-1}$), we may re-express
$\ln\Sigma_{\pm}$ and $\Sigma_{\pm}^{-1}$ as

$\ln\Sigma_{\pm}=\ln\left(\sigma^{2}\right)\pm\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}-\frac{f}{2M^{2}\sigma^{4}}\left(\frac{d\sigma}{dR}\right)^{2},$
(39)
$\Sigma_{\pm}^{-1}=\frac{1}{\sigma^{2}}\left(1\mp\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}+\frac{f}{M^{2}\sigma^{4}}\left(\frac{d\sigma}{dR}\right)^{2}\right).$
(40)

Next, we plug Eqns. 39 and 40 into Eqns. 33 and 34 (again keeping terms only
up to $M^{-2}$) to obtain

$\mathcal{A}=-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)+\frac{3}{2}\sigma^{4},$
(41)
$\mathcal{B}=\frac{8\sigma}{3M^{2}R^{2}}\frac{d\sigma}{dR}+\frac{4\sigma^{2}}{3M^{2}R^{4}f}+\frac{4f}{3M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}.$
(42)

Thus, we may write the total Lagrangian as

$\displaystyle\begin{split}\mathcal{L}=\hskip
2.84526pt&-\frac{\sigma^{2}}{2\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)+\frac{3}{2}\sigma^{4}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\frac{8\sigma}{3M^{2}R^{2}}\frac{d\sigma}{dR}-\frac{8\sigma^{2}}{3M^{2}R^{3}f}+\frac{6\sigma^{2}}{M^{2}R^{4}f}+\frac{4f}{3M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}+\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)\ln
f\right)\end{split}$ (43)

Since $\mathcal{L}$ is spherically symmetric, we may write the effective
action as

$\Gamma\left[\sigma\right]=\beta\int\mathrm{d}R\hskip
1.42262ptR^{2}\mathcal{L}.$ (44)

Thus, we may write the Euler-Lagrange equation as

$\frac{\partial\mathcal{L}}{\partial\sigma}-\frac{d}{dR}\left(\frac{\partial\mathcal{L}}{\partial\sigma^{\prime}}\right)-\frac{2}{R}\frac{\partial\mathcal{L}}{\partial\sigma^{\prime}}=0.$
(45)

Plugging Eqn. 43 into Eqn. 45, we find that

$\displaystyle\begin{split}&-\frac{\sigma}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-4\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\sigma^{3}+4\sigma^{3}-\frac{2f}{M^{2}\sigma}\left(\frac{d\sigma}{dR}\right)^{2}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\frac{8}{3M^{2}R^{2}}\frac{d\sigma}{dR}-\frac{16\sigma}{3M^{2}R^{3}f}+\frac{12\sigma}{M^{2}R^{4}f}+4\sigma^{3}\ln
f\right)\\\
&-\frac{1}{32\pi^{2}}\frac{d}{dR}\left(-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\frac{2f^{-1}}{M^{2}}\frac{d\sigma}{dR}+\frac{8\sigma}{3M^{2}R^{2}}+\frac{8f}{3M^{2}}\frac{d\sigma}{dR}+\frac{2f}{M^{2}}\frac{d\sigma}{dR}\ln
f\right)\\\
&-\frac{1}{16\pi^{2}R}\left(-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\frac{2f^{-1}}{M^{2}}\frac{d\sigma}{dR}+\frac{8\sigma}{3M^{2}R^{2}}+\frac{8f}{3M^{2}}\frac{d\sigma}{dR}+\frac{2f}{M^{2}}\frac{d\sigma}{dR}\ln
f\right)=0.\end{split}$ (46)

###  3.2 Leading Order Condensate Profile

Let us expand the condensate $\sigma$ in powers of $M^{-2}$ as

$\sigma\left(R\right)=\sigma_{0}\left(R\right)+\frac{\sigma_{2}\left(R\right)}{M^{2}}+\dots$
(47)

At leading order (meaning without any $M$ dependence), we may write the
equation of motion (Eqn. 46) as

$-\frac{\sigma_{0}}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-4\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\sigma_{0}^{3}+4\sigma_{0}^{3}\right)+\frac{1}{32\pi^{2}}\left(4\sigma_{0}^{3}\ln
f\right)=0.$ (48)

Clearly, there is trivial solution $\sigma_{0}=0$ to Eqn. 48. However, we
expect $\sigma_{0}$ to be generally non-zero. Therefore, with some simple
algebra, we may rewrite Eqn. 48 as

$-\frac{1}{2\lambda}+\frac{f^{-2}}{16\pi^{2}}\left(\sigma_{0}^{2}-\sigma_{0}^{2}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+f^{2}\ln\left(f\right)\sigma_{0}^{2}\right)=0.$
(49)

We may solve this equation to obtain (see Appendix B)

$\sigma_{0}^{2}=-\frac{8\pi^{2}f^{2}}{\lambda}\left[W_{-1}\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right)\right]^{-1}.$
(50)

Eqn. 50 provides an expression for $\sigma_{0}^{2}$. Therefore, there are two
possible expressions for $\sigma_{0}$, which differ from each other by a sign.
Note that the Lagrangian $\mathcal{L}$ (Eqn. 43) and the equation of motion
(Eqn. 46) are both symmetric under the transformation $\sigma\to-\sigma$.
Therefore, it does not matter which expression for $\sigma_{0}$ we choose. For
simplicity, we will choose the positive sign. Thus, we may write $\sigma_{0}$
as

$\sigma_{0}=\sqrt{-\frac{8\pi^{2}f^{2}}{\lambda}\left[W_{-1}\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right)\right]^{-1}}.$
(51)

Below, we have graphed $\sigma_{0}\left(r\right)$ for several different
combinations of $\lambda$ and $\ell$ [10].

Figure 1: Graph of $\sigma_{0}\left(R\right)$ with respect to $R$. Note that
the event horizon is at $R=2$.

As we will demonstrate in the next subsection, $\sigma_{0}$ provides a very
close approximation to $\sigma$ almost everywhere outside the event horizon.
Therefore, we would like to compare the results obtained in Figure 1 to the
results obtained in Ref. [8]. As in Ref. [8], the condensate $\sigma$
asymptotically approaches a constant, non-zero value far from the event
horizon, indicating a phase of spontaneously broken chiral symmetry. Close to
the event horizon, $\sigma$ approaches zero, illustrating a phase of restored
chiral symmetry.
In Ref. [8], the authors work with black holes far smaller than the
astrophysical-mass black holes considered here. They found that, as the mass
of the black hole increases, the size of the bubble of restored chiral
symmetry decreases (relative to the Schwarzschild radius of the black hole).
For their black holes, the bubbles expanded out to $\sim 10-30$ Schwarzschild
radii. In Fig. 1, the bubble extends out to $\sim 2-3$ Schwarzschild radii.
Hence, we see that our results are consistent with the results of Ref. [8], in
that larger black holes create smaller bubbles of restored chiral symmetry
(relative to their Schwarzschild radii). However, for very large black holes
(like the astrophysical-mass black holes considered here), the size of the
bubble scales almost linearly with the mass of the black hole.

###  3.3 Next-to-Leading Order Condensate Profile

From Eqn. 43, we see that $\mathcal{L}$ contains terms proportional to
$\ln\sigma$. We wish to expand this term in powers of $M^{-2}$. To do this, we
must first rewrite Eqn. 47 as

$\sigma\left(r\right)=\sigma_{0}\left(R\right)\left(1+\frac{\sigma_{2}\left(R\right)}{M^{2}\hskip
1.42262pt\sigma_{0}\left(R\right)}+\dots\right).$ (52)

Using the Mercator series for the natural logarithm, we may write $\ln\sigma$
as

$\ln\sigma=\ln\sigma_{0}+\frac{\sigma_{2}}{M^{2}\hskip
1.42262pt\sigma_{0}}-\frac{1}{2}\left(\frac{\sigma_{2}}{M^{2}\hskip
1.42262pt\sigma_{0}}\right)^{2}+\dots$ (53)

At next-to-leading order (terms proportional to $M^{-2}$), we may write the
equation of motion (Eqn. 46) as

$\displaystyle\begin{split}&-\frac{\sigma_{2}}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-12\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\sigma_{0}^{2}\sigma_{2}+4\sigma_{0}^{2}\sigma_{2}-\frac{2f}{\sigma_{0}}\left(\frac{d\sigma_{0}}{dR}\right)^{2}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\frac{8}{3R^{2}}\frac{d\sigma_{0}}{dR}-\frac{16\sigma_{0}}{3R^{3}f}+\frac{12\sigma_{0}}{R^{4}f}+12\sigma_{0}^{2}\sigma_{2}\ln
f\right)\\\
&-\frac{1}{32\pi^{2}}\frac{d}{dR}\left(-2f^{-1}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right)\\\
&-\frac{1}{16\pi^{2}R}\left(-2f^{-1}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right)=0.\end{split}$
(54)

Solving Eqn. 54 for $\sigma_{2}$, we obtain

$\displaystyle\begin{split}\sigma_{2}=&\frac{1}{32\pi^{2}}\left\\{-\frac{8}{3R^{2}}\frac{d\sigma_{0}}{dR}+\frac{16\sigma_{0}}{3R^{3}f}-\frac{12\sigma_{0}}{R^{4}f}+\frac{2f^{-1}}{\sigma_{0}}\left(\frac{d\sigma_{0}}{dR}\right)^{2}\right.\\\
&\left.+\frac{d}{dR}\left(-2f^{-1}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right)\right.\\\
&\left.+\frac{2}{R}\left(-2f^{-1}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right)\right\\}\\\
&\times\left\\{-\frac{1}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-12\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\sigma_{0}^{2}+4\sigma_{0}^{2}+12\sigma_{0}^{2}\hskip
1.42262ptf^{2}\ln f\right)\right\\}^{-1}\end{split}$ (55)

Because the explicit expression for $\sigma_{2}\left(R\right)$ is extremely
complicated, we will not write it down here. Below, we have graphed
$\sigma_{2}\left(R\right)$ for several different combinations of $\lambda$ and
$\ell$ [10].

Figure 2: Graph of $\sigma_{2}\left(R\right)$ with respect to $R$.

In the perturbative expansion for $\sigma$, $\sigma_{2}$ has a coefficient of
$M^{-2}$. Since $\sigma_{2}$ is smaller than $\sigma_{0}$ almost everywhere
outside the event horizon, the profile of $\sigma$ outside the horizon is
dominated by the contribution from $\sigma_{0}$.

###  3.4 Verification of the Perturbative Approach

To obtain the above results for $\sigma_{0}$ and $\sigma_{2}$, we had to make
three assumptions. First, we assumed that $\beta\sqrt{f\Sigma_{\pm}}\gg 1$ for
both $\Sigma_{+}$ and $\Sigma_{-}$. Second, we assumed that $\sigma$ varies on
scales far larger than the Planck scale. Lastly, we assumed that $\sigma$ can
be expanded perturbatively in powers of $M^{-2}$. We wish to verify that these
assumptions hold almost everywhere outside the event horizon.
We start by verifying the assumption $\beta\sqrt{f\Sigma_{\pm}}\gg 1$. As
discussed earlier, an astrophysical black hole should have a mass of at least
$\sim 10^{38}$ Planck masses. The inverse temperature $\beta$ is given by [11,
12]

$\beta=8\pi M.$ (56)

Therefore, $\beta$ should be at least $\sim 10^{39}$. Hence, we see that the
assumption is satisfied if $f\Sigma_{\pm}\gg 10^{-78}$. We may write
$\Sigma_{\pm}$ as

$\Sigma_{\pm}=\sigma^{2}\pm\frac{\sqrt{f}}{M}\frac{d\sigma}{dR}.$ (57)

Next, we use Eqn. 47 to expand $\Sigma_{\pm}$ in powers of $M^{-1}$. Keeping
terms up to $M^{-2}$, we find that

$\Sigma_{\pm}=\sigma_{0}^{2}+\frac{2}{M^{2}}\sigma_{0}\sigma_{2}\pm\frac{\sqrt{f}}{M}\frac{d\sigma_{0}}{dR}.$
(58)

Fortunately, we already have expressions for $\sigma_{0}$ and $\sigma_{2}$.
Below, we graph the functions $f\left(R\right)\Sigma_{+}\left(R\right)$ and
$f\left(R\right)\Sigma_{-}\left(R\right)$ for $M=10^{39}$ Planck masses (a
stellar-mass black hole) and $M=10^{48}$ Planck masses (a very large
supermassive black hole) [10].

Figure 3: Graph of $f\left(R\right)\Sigma_{+}\left(R\right)$ with respect to
$\left(R-2\right)$, with $\lambda=10^{-2}$ and $\ell=10^{3}$. The curves for
$M=10^{39}$ Planck masses and $M=10^{48}$ Planck masses are indistinguishable
on the graph. Figure 4: Graph of $f\left(R\right)\Sigma_{-}\left(R\right)$
with respect to $\left(R-2\right)$, with $\lambda=10^{-2}$ and $\ell=10^{3}$.
The curves for $M=10^{39}$ Planck masses and $M=10^{48}$ Planck masses are
indistinguishable on the graph.

From Figures 3 and 4, we see that the assumption $\beta\sqrt{f\Sigma_{\pm}}\gg
1$ may be violated very close to the event horizon. However, the region where
this occurs only extends out to $\left(R-2\right)<10^{-3}$, or less than one
thousandth of a Schwarzschild radius outside the event horizon. Thus, the
assumption $\beta\sqrt{f\Sigma_{\pm}}\gg 1$ holds almost everywhere outside
the event horizon.
Next, we verify the assumption that $\sigma$ varies on scales far larger than
the Planck scale. In terms of the original Planck-scale coordinate $r$, we may
write this condition as

$\frac{d\sigma}{dr}\ll 1.$ (59)

Rewriting Eqn. 59 in terms of the rescaled coordinate $R$, we obtain

$\frac{d\sigma}{dR}\ll M.$ (60)

Up to second-order in $M^{-1}$, we may write $\sigma$ as

$\sigma\left(R\right)=\sigma_{0}\left(R\right)+\frac{\sigma_{2}\left(R\right)}{M^{2}}.$
(61)

Fortunately, we already have expressions for $\sigma_{0}$ and $\sigma_{2}$.
Below, we graph the derivative of $\sigma\left(R\right)$ for $M=10^{39}$
Planck masses and $M=10^{48}$ Planck masses [10]. We see that Eqn. 60 is
satisfied everywhere outside the event horizon.

Figure 5: Graph of $\sigma^{\prime}\left(R\right)$ with respect to
$\left(R-2\right)$, with $\lambda=10^{-2}$ and $\ell=10^{3}$. The curves for
$M=10^{39}$ Planck masses and $M=10^{48}$ Planck masses are indistinguishable
on the graph.

Finally, we verify the assumption that $\sigma$ can be expanded perturbatively
in powers of $M^{-2}$, as in Eqn. 47. This assumption holds if and only if
$\lvert\sigma_{2}/M^{2}\rvert\ll\sigma_{0}$, or equivalently,
$\lvert\sigma_{2}\rvert/\sigma_{0}\ll M^{2}$. From Figures 1 and 2, it is easy
to see that $\lvert\sigma_{2}\rvert<\sigma_{0}$ in most of the space outside
the event horizon. Thus, $\lvert\sigma_{2}\rvert/\sigma_{0}\ll M^{2}$ almost
everywhere outside the event horizon. Having verified all three assumptions,
we conclude that the perturbative analysis is valid almost everywhere outside
the event horizon.

##  4 Large Reissner-Nordström Black Hole

Let us consider a large Reissner-Nordström (RN) black hole with mass $M\gg 1$
in Planck units. We denote the charge of this black hole by $Q$. The event
horizon radius $r_{+}$ is given by [13, 12]

$r_{+}=M+\sqrt{M^{2}-Q^{2}}.$ (62)

The metric function $f\left(r\right)$ is given by [13]

$f\left(r\right)=1-\frac{2M}{r}+\frac{Q^{2}}{r^{2}}.$ (63)

We assume that the black hole is sub-extremal or extremal, so that $\lvert
Q\rvert\leq M$ [13]. As in the Schwarzschild case, we assume that the event
horizon radius sets the natural scale of variation for the condensate
$\sigma$. Thus, we define a new radial coordinate $R$, which is given by

$R=\frac{r}{M}.$ (64)

In this new coordinate system, we may write the metric function
$f\left(R\right)$ as

$f\left(R\right)=1-\frac{2}{R}+\frac{Q^{2}}{M^{2}R^{2}}.$ (65)

Let us expand the condensate $\sigma$ in powers of $M^{-2}$ as

$\sigma\left(R\right)=\sigma_{0}\left(R\right)+\frac{\sigma_{2}\left(R\right)}{M^{2}}+\dots$
(66)

We may write $\sigma_{0}$ and $\sigma_{2}$ as (see Appendix C)

$\sigma_{0}=\sqrt{-\frac{8\pi^{2}f^{2}}{\lambda}\left[W_{-1}\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right)\right]^{-1}},$
(67)
$\displaystyle\begin{split}\sigma_{2}=&\frac{1}{32\pi^{2}}\left\\{\frac{2f^{-1}}{\sigma_{0}}\left(\frac{d\sigma_{0}}{dR}\right)^{2}-\frac{8}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma_{0}}{dR}\right.\\\
&\left.-\frac{8\sigma_{0}}{3R^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+\frac{2\sigma_{0}}{3f}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right)\right.\\\
&\left.+\frac{d}{dR}\left[-2f^{-1}\hskip
1.42262pt\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right]\right.\\\
&\left.+\frac{2}{R}\left[-2f^{-1}\hskip
1.42262pt\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right]\right\\}\\\
&\times\left\\{-\frac{1}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-12\sigma_{0}^{2}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+4\sigma_{0}^{2}+12\sigma_{0}^{2}\hskip
1.42262ptf^{2}\ln f\right)\right\\}^{-1}.\end{split}$ (68)

Below, we graph $\sigma_{0}$ and $\sigma_{2}$ for several different values of
the charge-to-mass ratio $Q/M$ [10]. (In all the graphs, we set
$\lambda=10^{-2}$ and $\ell=10^{3}$.) Unlike the Schwarzschild case, the event
horizon for a Reissner-Nordstöm black hole will be located at different values
of $R$ depending on the charge-to-mass ratio $Q/M$. Therefore, the horizontal
axis will be in units of the dimensionless ratio $r/r_{+}$.

Figure 6: Graph of $\sigma_{0}$ with respect to $\frac{r}{r_{+}}$. Figure 7:
Graph of $\sigma_{2}$ with respect to $\frac{r}{r_{+}}$.

In the perturbative expansion for $\sigma$, $\sigma_{2}$ has a coefficient of
$M^{-2}$. Since $\sigma_{2}$ is smaller than $\sigma_{0}$ almost everywhere
outside the event horizon, the profile of $\sigma$ outside the horizon is
dominated by the contribution from $\sigma_{0}$. (In Appendix D, we
demonstrate that the perturbative approach is valid almost everywhere outside
the event horizon. Thus, higher-order effects should not significantly change
the condensate profile displayed in Figure 6.)
From Figure 6, we see that the bubble of restored chiral symmetry becomes
larger as $Q/M$ increases. This effect is particularly pronounced for near-
extremal black holes ($Q/M\approx 1$), as there is a much greater difference
between the $Q^{2}/M^{2}=0.9$ and $Q^{2}/M^{2}=1.0$ curves than between
$Q^{2}/M^{2}=0.75$ and $Q^{2}/M^{2}=0.9$ curves.
For an extremal black hole, the asymptotic temperature is zero [11, 12].
Therefore, the local Tolman temperature is zero everywhere (see Ref. [8] for
the definition of the Tolman temperature). However, from Figure 6, we see that
a bubble of restored chiral symmetry still forms, surrounded by a region of
spontaneously broken chiral symmetry that fills the rest of spacetime. This
contradicts the usual explanation for the formation of bubbles of restored
symmetry around black holes, which attributes this phenomenon to the increased
local (Tolman) temperature near the event horizon.

##  5 Conclusion

In this article, we have examined the behavior of a chiral condensate outside
a spherically symmetric, astrophysical-mass black hole. For a Schwarzschild
black hole, the behavior of the chiral condensate is consistent with the
results from Ref. [8]. Encouraged by this, we proceeded to analyze the
behavior of a chiral condensate in the presence of a charged (Reissner-
Nordström) black hole. Here, we find that radius of the bubble of restored
chiral symmetry increases as the charge-to-mass ratio increases. However, the
radius of the bubble never reaches infinity, even for an extremal Reissner-
Nordström black hole. Thus, the chiral condensate always exhibits a phase of
restored chiral symmetry close to the event horizon and a phase of
spontaneously broken chiral symmetry far away from the horizon.
Unfortunately, it may prove difficult to test any of the predictions made in
this article, since the bubble of restored chiral symmetry around an
astrophysical black hole would only extend out to a few Schwarzschild radii.
Therefore, in the future, we would like to investigate possible empirical
signatures of chiral symmetry restoration that could be detected from Earth.
For example, the restoration of chiral symmetry close to the black hole could
alter the behavior of matter swirling into it, which might produce detectable
radiation signatures.
In theory, the effect of curved spacetime on a chiral condensate would not be
confined to black holes. Much subtler effects could occur due to the curvature
of spacetime around the Earth. This could cause a very small difference in the
proton mass (or the mass of some other hadron) depending on altitude. With
sufficiently sensitive equipment, such a difference might be observable.
Although these possibilities for experimental verification remain highly
speculative, there is a tantalizing chance that gravity could provide an
entirely new experimental window on the QCD phase diagram.

##  6 Acknowledgements

We would like to thank Dr. Laith Haddad for many interesting conversations and
invaluable pieces of advice, as well as for his kind encouragement.

## References

  * [1] Stephen William Hawking  “Interacting quantum fields around a black hole”  In _Communications in Mathematical Physics_ 80 Springer, 1981, pp. 421–442
  * [2] I.. Moss  “Black-hole bubbles”  In _Phys. Rev. D_ 32 American Physical Society, 1985, pp. 1333–1344  DOI: 10.1103/PhysRevD.32.1333
  * [3] Sidney Coleman, John Preskill and Frank Wilczek  “Growing hair on black holes”  In _Phys. Rev. Lett._ 67 American Physical Society, 1991, pp. 1975–1978  DOI: 10.1103/PhysRevLett.67.1975
  * [4] Sidney Coleman, John Preskill and Frank Wilczek  “Quantum hair on black holes”  In _Nuclear Physics B_ 378.1, 1992, pp. 175–246  DOI: https://doi.org/10.1016/0550-3213(92)90008-Y
  * [5] Antonino Flachi and Takahiro Tanaka  “Chiral modulations in curved space I: formalism”  In _JHEP_ , 2011  DOI: 10.1007/JHEP02(2011)026
  * [6] Antonino Flachi  “Chiral modulations in curved space II: conifold geometries”  In _Journal of High Energy Physics_ 2012.1 Springer, 2012, pp. 1–18
  * [7] Antonino Flachi  “Dual fermion condensates in curved space”  In _Phys. Rev. D_ 88 American Physical Society, 2013, pp. 085011  DOI: 10.1103/PhysRevD.88.085011
  * [8] Antonino Flachi and Takahiro Tanaka  “Chiral phase transitions around black holes”  In _Phys. Rev. D_ 84 American Physical Society, 2011, pp. 061503  DOI: 10.1103/PhysRevD.84.061503
  * [9] Kenji Fukushima and Tetsuo Hatsuda  “The phase diagram of dense QCD”  In _Reports on Progress in Physics_ 74.1, 2010, pp. 014001  DOI: 10.1088/0034-4885/74/1/014001
  * [10] Wolfram Research Inc.  “Mathematica, Version 13.0.1.0” Champaign, IL, 2022  URL: https://www.wolfram.com/mathematica
  * [11] Stephen W Hawking  “Particle creation by black holes”  In _Communications in mathematical physics_ 43.3 Springer, 1975, pp. 199–220
  * [12] Harry W. Braden, J. Brown, Bernard F. Whiting and James W. York  “Charged black hole in a grand canonical ensemble”  In _Phys. Rev. D_ 42 American Physical Society, 1990, pp. 3376–3385  DOI: 10.1103/PhysRevD.42.3376
  * [13] Charles W. Misner, K.. Thorne and J.. Wheeler  “Gravitation”  San Francisco: W. H. Freeman, 1973
  * [14] Robert M Corless et al.  “On the LambertW function”  In _Advances in Computational mathematics_ 5.1 Springer, 1996, pp. 329–359

## Appendix A: Derivation of Eqn. 13

The conformally rescaled metric is given by [8]

$d\hat{s}^{2}=dt^{2}+f\left(r\right)^{-2}dr^{2}+f\left(r\right)^{-1}\hskip
1.42262ptr^{2}d\Omega^{2}.$ (69)

All quantities associated with this metric will have hats. The quantity
$a_{\pm}$ is defined by [8]

$a_{\pm}=\frac{1}{180}\left(\hat{R}_{\mu\nu\tau\rho}^{2}-\hat{R}_{\mu\nu}^{2}-\hat{\Delta}\hat{R}\right)+\frac{1}{6}\hat{\Delta}\left(f\Sigma_{\pm}\right).$
(70)

The Christoffel symbols are defined as (Wolfram)

$\hat{\Gamma}^{m}_{ij}=\frac{1}{2}\hat{g}^{mk}\left(\frac{\partial\hat{g}_{ik}}{\partial
x^{j}}+\frac{\partial\hat{g}_{jk}}{\partial
x^{i}}-\frac{\partial\hat{g}_{ij}}{\partial x^{k}}\right).$ (71)

Explicitly, the non-zero Christoffel symbols are given by

$\hat{\Gamma}^{r}_{rr}=-\frac{f^{\prime}\left(r\right)}{f\left(r\right)},$
(72) $\hat{\Gamma}^{r}_{\theta\theta}=\frac{1}{2}r^{2}\hskip
1.42262ptf^{\prime}\left(r\right)-r\hskip 1.42262ptf\left(r\right),$ (73)
$\hat{\Gamma}^{r}_{\phi\phi}=\sin^{2}\left(\theta\right)\left(\frac{1}{2}r^{2}\hskip
1.42262ptf^{\prime}\left(r\right)-rf\left(r\right)\right),$ (74)
$\hat{\Gamma}^{\theta}_{r\theta}=\Gamma^{\theta}_{\theta
r}=\frac{1}{r}-\frac{f^{\prime}\left(r\right)}{2f\left(r\right)},$ (75)
$\hat{\Gamma}^{\theta}_{\phi\phi}=-\sin\left(\theta\right)\cos\left(\theta\right),$
(76) $\hat{\Gamma}^{\phi}_{r\phi}=\Gamma^{\phi}_{\phi
r}=\frac{1}{r}-\frac{f^{\prime}\left(r\right)}{2f\left(r\right)},$ (77)
$\hat{\Gamma}^{\phi}_{\theta\phi}=\Gamma^{\phi}_{\phi\theta}=\cot\left(\theta\right).$
(78)

The Riemann curvature tensor is defined by (Wolfram)

$\hat{R}^{\alpha}_{\beta\gamma\delta}=\partial_{\gamma}\Gamma^{\alpha}_{\beta\delta}-\partial_{\delta}\Gamma^{\alpha}_{\beta\gamma}+\Gamma^{\mu}_{\beta\delta}\Gamma^{\alpha}_{\mu\gamma}-\Gamma^{\mu}_{\beta\gamma}\Gamma^{\alpha}_{\mu\delta}.$
(79)

The non-zero components of the Riemann curvature tensor are given by

$\hat{R}^{r}_{\theta r\theta}=-R^{r}_{\theta\theta
r}=\frac{1}{2}r^{2}f^{\prime\prime}\left(r\right)-\frac{r^{2}f^{\prime}\left(r\right)^{2}}{4f\left(r\right)},$
(80) $\hat{R}^{r}_{\phi r\phi}=-\hat{R}^{r}_{\phi\phi
r}=\sin^{2}\left(\theta\right)\left(\frac{1}{2}r^{2}f^{\prime\prime}\left(r\right)-\frac{r^{2}f^{\prime}\left(r\right)^{2}}{4f\left(r\right)}\right),$
(81) $\hat{R}^{\theta}_{rr\theta}=-\hat{R}^{\theta}_{r\theta
r}=\frac{f^{\prime}\left(r\right)^{2}}{4f\left(r\right)^{2}}-\frac{f^{\prime\prime}\left(r\right)}{2f\left(r\right)},$
(82)
$\hat{R}^{\theta}_{\phi\theta\phi}=-\hat{R}^{\theta}_{\phi\phi\theta}=\sin^{2}\left(\theta\right)\left(1+rf^{\prime}\left(r\right)-\frac{r^{2}f^{\prime}\left(r\right)^{2}}{4f\left(r\right)}-f\left(r\right)\right),$
(83) $\hat{R}^{\phi}_{rr\phi}=-\hat{R}^{\phi}_{r\phi
r}=\frac{f^{\prime}\left(r\right)^{2}}{4f\left(r\right)^{2}}-\frac{f^{\prime\prime}\left(r\right)}{2f\left(r\right)},$
(84)
$\hat{R}^{\phi}_{\theta\theta\phi}=-\hat{R}^{\phi}_{\theta\phi\theta}=-1-rf^{\prime}\left(r\right)+\frac{r^{2}f^{\prime}\left(r\right)^{2}}{4f\left(r\right)}+f\left(r\right).$
(85)

The Ricci curvature tensor is defined by (Wolfram)

$\hat{R}_{\mu\nu}=\hat{R}^{\lambda}_{\mu\lambda\nu}.$ (86)

The non-zero components of the Ricci curvature tensor are given by

$\hat{R}_{rr}=-\frac{f^{\prime}\left(r\right)^{2}}{2f\left(r\right)^{2}}+\frac{f^{\prime\prime}\left(r\right)}{f\left(r\right)},$
(87)
$\hat{R}_{\theta\theta}=\frac{1}{2}r^{2}f^{\prime\prime}\left(r\right)-\frac{r^{2}f^{\prime}\left(r\right)^{2}}{2f\left(r\right)}+rf^{\prime}\left(r\right)-f\left(r\right)+1,$
(88)
$\hat{R}_{\phi\phi}=\sin^{2}\left(\theta\right)\left(\frac{1}{2}r^{2}f^{\prime\prime}\left(r\right)-\frac{r^{2}f^{\prime}\left(r\right)^{2}}{2f\left(r\right)}+rf^{\prime}\left(r\right)-f\left(r\right)+1\right).$
(89)

The Ricci scalar $\hat{R}$ is defined by (Wolfram)

$\hat{R}=\hat{g}^{\mu\nu}\hat{R}_{\mu\nu}.$ (90)

Explicitly, we may write $\hat{R}$ as

$\hat{R}=-\frac{3}{2}f^{\prime}\left(r\right)^{2}+2f\left(r\right)f^{\prime\prime}\left(r\right)+\frac{2}{r}f\left(r\right)f^{\prime}\left(r\right)-\frac{2}{r^{2}}f\left(r\right)^{2}+\frac{2}{r^{2}}f\left(r\right).$
(91)

The scalars $\hat{R}^{2}_{\mu\nu\rho\tau}$ and $\hat{R}^{2}_{\mu\nu}$ are
given by [10]

$\displaystyle\begin{split}\hat{R}^{2}_{\mu\nu\rho\tau}=\hskip
1.42262pt&\frac{4f\left(r\right)^{4}}{r^{4}}+\frac{3}{4}f^{\prime}\left(r\right)^{4}-\frac{8f\left(r\right)^{3}}{r^{4}}-\frac{8f\left(r\right)^{3}}{r^{3}}f^{\prime}\left(r\right)+\frac{4f\left(r\right)^{2}}{r^{4}}+\frac{8f\left(r\right)^{2}}{r^{3}}f^{\prime}\left(r\right)+\frac{6f\left(r\right)^{2}}{r^{2}}f^{\prime}\left(r\right)^{2}\\\
&+2f\left(r\right)^{2}f^{\prime\prime}\left(r\right)^{2}-\frac{2f\left(r\right)}{r^{2}}f^{\prime}\left(r\right)^{2}-\frac{2f\left(r\right)}{r}f^{\prime}\left(r\right)^{3}-2f\left(r\right)f^{\prime}\left(r\right)^{2}f^{\prime\prime}\left(r\right),\end{split}$
(92) $\displaystyle\begin{split}\hat{R}^{2}_{\mu\nu}=\hskip
2.84526pt&\frac{2f\left(r\right)^{4}}{r^{4}}+\frac{3}{4}f^{\prime}\left(r\right)^{4}-\frac{4f\left(r\right)^{3}}{r^{4}}-\frac{4f\left(r\right)^{3}}{r^{3}}f^{\prime}\left(r\right)-\frac{2f\left(r\right)^{3}}{r^{2}}f^{\prime\prime}\left(r\right)+\frac{2f\left(r\right)^{2}}{r^{4}}+\frac{4f\left(r\right)^{2}}{r^{3}}f^{\prime}\left(r\right)\\\
&-\frac{2f\left(r\right)}{r^{2}}f^{\prime}\left(r\right)^{2}+\frac{4f\left(r\right)^{2}}{r^{2}}f^{\prime}\left(r\right)^{2}-\frac{2f\left(r\right)}{r}f^{\prime}\left(r\right)^{3}+\frac{2f\left(r\right)^{2}}{r^{2}}f^{\prime\prime}\left(r\right)+\frac{2f\left(r\right)^{2}}{r}f^{\prime}\left(r\right)f^{\prime\prime}\left(r\right)\\\
&-2f\left(r\right)f^{\prime}\left(r\right)^{2}f^{\prime\prime}\left(r\right)+\frac{3}{2}f\left(r\right)^{2}f^{\prime\prime}\left(r\right)^{2}\end{split}$
(93)

The operator $\hat{\Delta}$ represents the spatial Laplace-Beltrami operator
associated with the conformally rescaled metric. We only need to consider
expressions in which the operator $\hat{\Delta}$ acts on scalar quantities.
Thus, we may write $\hat{\Delta}$ as

$\hat{\Delta}=\partial_{i}\partial^{i}+\left(\partial_{i}\hat{g}^{ij}\right)\partial_{j}+\frac{\hat{g}^{ij}}{2\det\hat{g}}\left(\partial_{i}\det\hat{g}\right)\partial_{j},$
(94)

where $i$ and $j$ are spatial indices in the conformally rescaled space-time.
We only need to consider expressions in which the operator $\hat{\Delta}$ acts
on scalar functions that depend only on $r$. Thus, we may rewrite Eqn. 94 as

$\hat{\Delta}=g^{rr}\partial^{2}_{r}+\left(\partial_{i}\hat{g}^{ir}\right)\partial_{r}+\frac{\hat{g}^{ir}}{2\det\hat{g}}\left(\partial_{i}\det\hat{g}\right)\partial_{r}.$
(95)

Plugging Eqn. 69 into Eqn. 95, we find that

$\hat{\Delta}=f\left(r\right)^{2}\frac{d^{2}}{dr^{2}}+\frac{2f\left(r\right)^{2}}{r}\frac{d}{dr}=f\left(r\right)^{2}\Delta_{\textrm{Eucl}},$
(96)

where $\Delta_{\textrm{Eucl}}$ is the Laplacian for three-dimensional
Euclidean space. Finally, we may write $a_{\pm}$ as

$\displaystyle\begin{split}a_{\pm}=\hskip
2.84526pt&\frac{1}{180}\left\\{\frac{2f\left(r\right)^{2}}{r^{4}}-\frac{8f\left(r\right)^{3}}{r^{4}}+\frac{6f\left(r\right)^{4}}{r^{4}}+\frac{8f\left(r\right)^{2}}{r^{3}}f^{\prime}\left(r\right)-\frac{12f\left(r\right)^{3}}{r^{3}}f^{\prime}\left(r\right)+\frac{6f\left(r\right)^{2}}{r^{2}}f^{\prime}\left(r\right)^{2}-\frac{4f\left(r\right)^{2}}{r^{2}}f^{\prime\prime}\left(r\right)\right.\\\
&\left.+\frac{6f\left(r\right)^{3}}{r^{2}}f^{\prime\prime}\left(r\right)-\frac{6f\left(r\right)^{2}}{r}f^{\prime}\left(r\right)f^{\prime\prime}\left(r\right)+\frac{3}{2}f\left(r\right)^{2}f^{\prime\prime}\left(r\right)^{2}-\frac{6f\left(r\right)^{3}}{r}f^{\left(3\right)}\left(r\right)-f\left(r\right)^{2}f^{\prime}\left(r\right)f^{\left(3\right)}\left(r\right)\right.\\\
&\left.-2f\left(r\right)^{3}f^{\left(4\right)}\left(r\right)\right\\}+\frac{f\left(r\right)^{2}}{6}\Delta_{\textrm{Eucl}}\left(f\Sigma_{\pm}\right).\end{split}$
(97)

## Appendix B: Computation of $\sigma_{0}\left(r\right)$

At zeroth order in $M^{-2}$, we may write the Euler-Lagrange equations as

$-\frac{1}{2\lambda}+\frac{f^{-2}}{16\pi^{2}}\left(\sigma_{0}^{2}-\sigma_{0}^{2}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+f^{2}\ln\left(f\right)\sigma_{0}^{2}\right)=0.$
(98)

Through simple algebra, we may rewrite Eqn. 98 as

$\frac{8\pi^{2}f^{2}}{\lambda\sigma_{0}^{2}}+\ln\left(\sigma_{0}^{2}\right)=1-\ln\left(\frac{f}{\ell^{2}}\right)+f^{2}\ln\left(f\right).$
(99)

In Eqn. 99, we would like the argument of the logarithm on the LHS to be equal
to the first term on the LHS. With a bit more algebra, we may rewrite Eqn. 99
as

$-\frac{8\pi^{2}f^{2}}{\lambda\sigma_{0}^{2}}+\ln\left(-\frac{8\pi^{2}f^{2}}{\lambda\sigma_{0}^{2}}\right)=-1+\ln\left(\frac{f}{\ell^{2}}\right)-f^{2}\ln\left(f\right)+\ln\left(-\frac{8\pi^{2}f^{2}}{\lambda}\right).$
(100)

Next, we take the exponential of both sides of Eqn. 100 to obtain

$-\frac{8\pi^{2}f^{2}}{\lambda\sigma_{0}^{2}}\exp\left(-\frac{8\pi^{2}f^{2}}{\lambda\sigma_{0}^{2}}\right)=-\frac{8\pi^{2}}{e\hskip
1.42262pt\lambda\hskip 1.42262pt\ell^{2}}f^{3-f^{2}}.$ (101)

The Lambert W function is defined by [14]

$w\left(x\right)\exp\left(w\left(x\right)\right)=x.$ (102)

Thus, we may rewrite Eqn. 101 as

$-\frac{8\pi^{2}f^{2}}{\lambda\sigma_{0}^{2}}=w\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right),$
(103)

or equivalently,

$\sigma_{0}^{2}=-\frac{8\pi^{2}f^{2}}{\lambda}\left[w\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right)\right]^{-1}.$
(104)

Because we are working in Euclideanized space-time, $f$ is positive
everywhere. Therefore, the argument of $w\left(x\right)$ in Eqn. 104 will be
negative everywhere. For negative arguments, the Lambert W function has two
branches: $W_{0}\left(x\right)$ and $W_{-1}\left(x\right)$. From previous
numerical results, we expect that there exists a region of restored chiral
symmetry close to the event horizon [8]. Below, we graph both solutions for
$\sigma_{0}^{2}\left(R\right)$ (corresponding to both branches of the Lambert
W function) [10].

Figure 8: Comparison of the two possible solutions for
$\sigma_{0}^{2}\left(R\right)$

From Figure 8, we see that only the $W_{-1}$ solution restores chiral symmetry
close to the event horizon. Thus, we may rewrite Eqn. 104 as

$\sigma_{0}^{2}=-\frac{8\pi^{2}f^{2}}{\lambda}\left[W_{-1}\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right)\right]^{-1}.$
(105)

## Appendix C: Derivation of Eqns. 67 and 68

In the original coordinate system (with the original radial coordinate $r$),
the quantity $a_{\textrm{sp}}$ takes the form [10]

$a_{\textrm{sp}}=\frac{2f\left(r\right)^{2}}{5r^{8}}\left[\left(2M^{2}-Q^{2}\right)r^{2}-2MQ^{2}r+Q^{4}\right].$
(106)

In the new coordinate system (with the rescaled radial coordinate $R$), the
quantity $a_{\textrm{sp}}$ takes the form

$a_{\textrm{sp}}=\frac{2f\left(r\right)^{2}}{5M^{8}R^{8}}\left[\left(2M^{2}-Q^{2}\right)M^{2}R^{2}-2M^{2}Q^{2}R+Q^{4}\right].$
(107)

Since $Q$ can be no larger than $M$, $a_{\textrm{sp}}$ must be of order
$M^{-4}$ or smaller. We will approximate the Lagrangian to order $M^{-2}$.
Therefore, we will neglect the contribution of $a_{\textrm{sp}}$ to the
Lagrangian. Keeping terms up to $M^{-2}$, we may write the Lagrangian pieces
$\mathcal{A}$, $\mathcal{B}$, and $\delta\mathcal{L}$ as [10]

$\mathcal{A}=\left(\frac{3}{2}-\ln\left(\frac{f}{\ell^{2}}\right)\right)\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)-\frac{1}{2}\left(\Sigma_{+}^{2}\ln\Sigma_{+}+\Sigma_{-}^{2}\ln\Sigma_{-}\right),$
(108)
$\mathcal{B}=\frac{8\sigma}{3M^{2}R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma}{dR}+\frac{4\sigma^{2}}{3M^{2}\hskip
1.42262ptR^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+\frac{f}{6M^{2}\hskip
1.42262pt\Sigma_{+}}\left(\frac{d\Sigma_{+}}{dR}\right)^{2}+\frac{f}{6M^{2}\hskip
1.42262pt\Sigma_{-}}\left(\frac{d\Sigma_{-}}{dR}\right)^{2},$ (109)
$\delta\mathcal{L}=\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)\ln
f-\frac{\sigma^{2}}{3M^{2}\hskip
1.42262ptf}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right).$
(110)

In Eqns. 108-110, we have accounted for the fact that $Q$ and $M$ are at
similar scales for a near-extremal RN black hole. Even though we have
approximated the Lagrangian to the order of $M^{-2}$, some terms contain
factors of $M^{-4}$ or $M^{-6}$. This occurs because of the possibility that
factors of $Q$ could “raise” these terms up to order $M^{-2}$ or higher. As in
the Schwarzschild case, we may write $\Sigma_{\pm}$ as

$\Sigma_{\pm}=\sigma^{2}\left(1\pm\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}\right).$
(111)

In Eqns. 39 and 40, we derived Taylor series approximations for
$\ln\Sigma_{\pm}$ and $\Sigma_{\pm}^{-1}$. Fortunately, these approximations
are still valid in RN spacetime. Explicitly,

$\ln\Sigma_{\pm}=\ln\left(\sigma^{2}\right)\pm\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}-\frac{f}{2M^{2}\sigma^{4}}\left(\frac{d\sigma}{dR}\right)^{2},$
(112)
$\Sigma_{\pm}^{-1}=\frac{1}{\sigma^{2}}\left(1\mp\frac{\sqrt{f}}{M\sigma^{2}}\frac{d\sigma}{dR}+\frac{f}{M^{2}\sigma^{4}}\left(\frac{d\sigma}{dR}\right)^{2}\right).$
(113)

Thus, we may rewrite Eqns. 108-110 as

$\mathcal{A}=-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)+\frac{3}{2}\sigma^{4},$
(114)
$\mathcal{B}=\frac{8\sigma}{3M^{2}R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma}{dR}+\frac{4\sigma^{2}}{3M^{2}\hskip
1.42262ptR^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+\frac{4f}{3M^{2}}\left(\frac{d\sigma}{dR}\right)^{2},$
(115)
$\delta\mathcal{L}=\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)\ln
f-\frac{\sigma^{2}}{3M^{2}\hskip
1.42262ptf}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right).$
(116)

Using Eqn. 22, we may write the Lagrangian as

$\displaystyle\begin{split}\mathcal{L}=&-\frac{\sigma^{2}}{2\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)+\frac{3}{2}\sigma^{4}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\frac{8\sigma}{3M^{2}R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma}{dR}+\frac{4\sigma^{2}}{3M^{2}\hskip
1.42262ptR^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+\frac{4f}{3M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\left(\sigma^{4}+\frac{f}{M^{2}}\left(\frac{d\sigma}{dR}\right)^{2}\right)\ln
f-\frac{\sigma^{2}}{3M^{2}\hskip
1.42262ptf}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right)\right).\end{split}$
(117)

As in the Schwarzschild case, the Euler-Lagrange equation of motion is given
by

$\frac{\partial\mathcal{L}}{\partial\sigma}-\frac{d}{dR}\left(\frac{\partial\mathcal{L}}{\partial\sigma^{\prime}}\right)-\frac{2}{R}\frac{\partial\mathcal{L}}{\partial\sigma^{\prime}}=0.$
(118)

Plugging Eqn. 117 into Eqn. 118, we find that

$\displaystyle\begin{split}&-\frac{\sigma}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-4\sigma^{3}\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)+4\sigma^{3}-\frac{2f}{M^{2}\hskip
1.42262pt\sigma}\left(\frac{d\sigma}{dR}\right)^{2}+\frac{8f^{2}}{3M^{2}R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma}{dR}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\frac{8\sigma}{3M^{2}\hskip
1.42262ptR^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+4\sigma^{3}\ln
f-\frac{2\sigma}{3M^{2}\hskip
1.42262ptf}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right)\right)\\\
&-\frac{1}{32\pi^{2}}\frac{d}{dR}\left\\{-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\frac{2f^{-1}}{M^{2}}\frac{d\sigma}{dR}+\frac{8\sigma}{3M^{2}R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3M^{2}}\frac{d\sigma}{dR}+\frac{2f\ln
f}{M^{2}}\frac{d\sigma}{dR}\right\\}\\\
&-\frac{1}{16\pi^{2}R}\left\\{-\ln\left(\frac{f\sigma^{2}}{\ell^{2}}\right)\frac{2f^{-1}}{M^{2}}\frac{d\sigma}{dR}+\frac{8\sigma}{3M^{2}R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3M^{2}}\frac{d\sigma}{dR}+\frac{2f\ln
f}{M^{2}}\frac{d\sigma}{dR}\right\\}=0.\end{split}$ (119)

At zeroth order in $M^{-1}$, we may write the equation of motion as

$-\frac{\sigma_{0}}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-4\sigma_{0}^{3}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+4\sigma_{0}^{3}\right)+\frac{1}{32\pi^{2}}\left(4\sigma_{0}^{3}\ln
f\right)=0.$ (120)

We look for non-trival solutions, where $\sigma$ is not constant and
approaches a non-zero value at spatial infinity. Therefore, we may rewrite
Eqn. 120 as

$-\frac{1}{2\lambda}+\frac{f^{-2}}{16\pi^{2}}\left(\sigma_{0}^{2}-\sigma_{0}^{2}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+f^{2}\ln\left(f\right)\sigma_{0}^{2}\right)=0.$
(121)

It is easy to see that Eqn. 121 is equivalent to Eqn. 49, except that
$f\left(r\right)$ is now given by the Reissner-Nordström metric instead of the
Schwarzschild metric. Therefore, we may write $\sigma_{0}$ as

$\sigma_{0}=\sqrt{-\frac{8\pi^{2}f^{2}}{\lambda}\left[W_{-1}\left(-\frac{8\pi^{2}}{e\lambda\ell^{2}}f^{3-f^{2}}\right)\right]^{-1}}.$
(122)

Using the Mercator series, we may write $\ln\sigma$ as

$\ln\sigma=\ln\sigma_{0}+\frac{\sigma_{2}\left(r\right)}{M^{2}\hskip
1.42262pt\sigma_{0}\left(r\right)}-\frac{1}{2}\left(\frac{\sigma_{2}\left(r\right)}{M^{2}\hskip
1.42262pt\sigma_{0}\left(r\right)}\right)^{2}+\dots$ (123)

At second order in $M^{-1}$, we may write the equation of motion as

$\displaystyle\begin{split}&-\frac{\sigma_{2}}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-12\sigma_{0}^{2}\sigma_{2}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+4\sigma_{0}^{2}\sigma_{2}-\frac{2f}{\sigma_{0}}\left(\frac{d\sigma_{0}}{dR}\right)^{2}+\frac{8f^{2}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma_{0}}{dR}\right)\\\
&+\frac{1}{32\pi^{2}}\left(\frac{8\sigma_{0}}{3R^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+12\sigma_{0}^{2}\sigma_{2}\ln
f-\frac{2\sigma_{0}}{3f}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right)\right)\\\
&-\frac{1}{32\pi^{2}}\frac{d}{dR}\left\\{-2f^{-1}\hskip
1.42262pt\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right\\}\\\
&-\frac{1}{16\pi^{2}R}\left\\{-2f^{-1}\hskip
1.42262pt\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right\\}=0.\end{split}$
(124)

Using simple algebra, we may write the solution $\sigma_{2}$ as

$\displaystyle\begin{split}\sigma_{2}=&\frac{1}{32\pi^{2}}\left\\{\frac{2f^{-1}}{\sigma_{0}}\left(\frac{d\sigma_{0}}{dR}\right)^{2}-\frac{8}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)\frac{d\sigma_{0}}{dR}\right.\\\
&\left.-\frac{8\sigma_{0}}{3R^{4}f}\left(1-\frac{Q^{2}}{M^{2}R}\right)^{2}+\frac{2\sigma_{0}}{3f}\left(\frac{8}{R^{3}}-\frac{14}{R^{4}}+\frac{24Q^{2}}{M^{2}R^{5}}-\frac{10Q^{2}}{M^{2}R^{4}}-\frac{8Q^{4}}{M^{4}R^{6}}\right)\right.\\\
&\left.+\frac{d}{dR}\left[-2f^{-1}\hskip
1.42262pt\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right]\right.\\\
&\left.+\frac{2}{R}\left[-2f^{-1}\hskip
1.42262pt\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)\frac{d\sigma_{0}}{dR}+\frac{8\sigma_{0}}{3R^{2}}\left(1-\frac{Q^{2}}{M^{2}R}\right)+\frac{8f}{3}\frac{d\sigma_{0}}{dR}+2f\ln\left(f\right)\frac{d\sigma_{0}}{dR}\right]\right\\}\\\
&\times\left\\{-\frac{1}{\lambda}+\frac{f^{-2}}{32\pi^{2}}\left(-12\sigma_{0}^{2}\ln\left(\frac{f\sigma_{0}^{2}}{\ell^{2}}\right)+4\sigma_{0}^{2}+12\sigma_{0}^{2}\hskip
1.42262ptf^{2}\ln f\right)\right\\}^{-1}.\end{split}$ (125)

## Appendix D: Verification of the Perturbative Approach for a Large Reissner-
Nordström Black Hole

To obtain the results of Appendix C, we had to make three assumptions. First,
we assumed that $\beta\sqrt{f\Sigma_{\pm}}\gg 1$ for both $\Sigma_{+}$ and
$\Sigma_{-}$. Second, we assumed that $\sigma$ varies on scales far larger
than the Planck scale. Lastly, we assumed that $\sigma$ can be expanded
perturbatively in powers of $M^{-2}$. We wish to verify that these assumptions
hold almost everywhere outside the event horizon.
Let us first verify the assumption $\beta\sqrt{f\Sigma_{\pm}}\gg 1$. As
discussed earlier, an astrophysical black hole should have a mass of at least
$\sim 10^{38}$ Planck masses. The inverse temperature $\beta$ is given by [11,
12]

$\beta=4\pi r_{+}\hskip 1.42262pt\left(1-\frac{Q^{2}}{r_{+}^{2}}\right)^{-1}.$
(126)

From Eqns. 126 and 62, we see that $\beta\geq 4\pi M$. Therefore, $\beta$
should be at least $\sim 10^{39}$. Hence, we see that the assumption
$\beta\sqrt{f\Sigma_{\pm}}\gg 1$ is satisfied if $f\Sigma_{\pm}\gg 10^{-78}$.
We may write $\Sigma_{\pm}$ as

$\Sigma_{\pm}=\sigma^{2}\pm\frac{\sqrt{f}}{M}\frac{d\sigma}{dR}.$ (127)

Next, we use Eqn. 47 to expand $\Sigma_{\pm}$ in powers of $M^{-1}$. Keeping
terms up to $M^{-2}$, we find that

$\Sigma_{\pm}=\sigma_{0}^{2}+\frac{2}{M^{2}}\sigma_{0}\sigma_{2}\pm\frac{\sqrt{f}}{M}\frac{d\sigma_{0}}{dR}.$
(128)

Fortunately, we already have expressions for $\sigma_{0}$ and $\sigma_{2}$.
Below, we graph the functions $f\left(R\right)\Sigma_{+}\left(R\right)$ and
$f\left(R\right)\Sigma_{-}\left(R\right)$ for $M=10^{39}$ Planck masses and
several different values of the charge ratio $Q/M$ [10].

Figure 9: Graph of $f*\Sigma_{+}$ with respect to $r/r_{+}$, with
$\lambda=10^{-2}$ and $\ell=10^{3}$. Figure 10: Graph of $f*\Sigma_{-}$ with
respect to $r/r_{+}$, with $\lambda=10^{-2}$ and $\ell=10^{3}$.

From Figures 3 and 4, we see that the assumption $\beta\sqrt{f\Sigma_{\pm}}\gg
1$ may be violated very close to the event horizon. Nevertheless, it is also
clear from these graphs that the assumption $\beta\sqrt{f\Sigma_{\pm}}\gg 1$
holds almost everywhere outside the event horizon.
Next, let us verify the assumption that $\sigma$ varies on scales far larger
than the Planck scale. In terms of the original Planck-scale coordinate $r$,
we may write this condition as

$\frac{d\sigma}{dr}\ll 1.$ (129)

Rewriting Eqn. 129 in terms of the rescaled coordinate $R$, we obtain

$\frac{d\sigma}{dR}\ll M.$ (130)

Up to second-order in $M^{-1}$, we may write $\sigma$ as

$\sigma\left(R\right)=\sigma_{0}\left(R\right)+\frac{\sigma_{2}\left(R\right)}{M^{2}}.$
(131)

Fortunately, we already have expressions for $\sigma_{0}$ and $\sigma_{2}$.
Below, we graph $\sigma^{\prime}\left(R\right)$ for $M=10^{39}$ Planck masses
and several different values of $Q/M$ [10]. From this graph, we see that Eqn.
130 is satisfied everywhere outside the event horizon. (To account for the
different event horizon radii for different values of $Q/M$, we will plot all
the curves with respect to $r/r_{+}$. However, the curves will still represent
the derivative of $\sigma\left(R\right)$ with respect to $R$.)

Figure 11: Graph of $\sigma^{\prime}\left(R\right)$ (the derivative of
$\sigma\left(R\right)$ with respect to $R$) as a function of $r/r_{+}$, with
$\lambda=10^{-2}$ and $\ell=10^{3}$.

Finally, we verify the assumption that $\sigma$ can be expanded perturbatively
in powers of $M^{-2}$, as in Eqn. 47. This means that
$\lvert\sigma_{2}/M^{2}\rvert\ll\sigma_{0}$, or equivalently,
$\lvert\sigma_{2}\rvert/\sigma_{0}\ll M^{2}$. From Figures 6 and 7, it is easy
to see that $\sigma_{2}<\sigma_{0}$ in most of the space outside the event
horizon. Thus, $\sigma_{2}/\sigma_{0}\ll M^{2}$ almost everywhere outside the
event horizon. Having verified all three assumptions, we conclude that the
perturbative analysis is valid almost everywhere outside the event horizon.