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

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 $c=G=\hbar=4\pi\hskip 1.42262pt\epsilon_{0}=k_{B}=1$.

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)

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 $r/r_{+}$.