Phase-transition Theory of Kerr Black Holes in the Electromagnetic Field

Black hole thermodynamics has been an intriguing subject of discussion for decades. The analogy between space–time with black hole horizon and thermodynamics have been extensively investigated (Bardeen et al. 1973). The four laws of black hole thermodynamics were discovered by Carter, Hawking, and Bardeen (Bardeen et al. 1973). These laws are physical properties that black holes are believed to satisfy. A black hole behaves as a black body with the Hawking temperature ${T}_{{\rm{H}}}=\tfrac{{\rm{\hslash }}\kappa }{2\pi {k}_{{\rm{B}}}c}$, where κ, c, k_B, and ${\rm{\hslash }}$ denote the surface gravity of the event horizon (EH), light velocity, Boltzmann constant, and Planck constant respectively (Carr & Hawking 1974; Hawking 1975). Since thermodynamic quantities correspond to microscopic structures, some researchers have attempted to explain black hole properties through exploring phase transition of the small-large charged Schwarzschild black hole (SBH/LBH; Hawking & Page 1983; Chamblin et al. 1999a, 1999b; Wei & Liu 2015). This kind of phase transition is similar to the liquid–gas transitions, which are first-order (Liu et al. 2014). Compared with the Schwarzschild black hole, there might exist more complex phenomena in the Kerr black hole (KBH) since it has additional spin J.

Blandford & Znajek (1977) gave the nonlinear Grad–Shafranov (G–S) equation for a force-free magnetosphere (FFM) of the curved Kerr space–time, which can describe the energy extraction process (Blandford & Znajek 1977). In this progress, KBH can be heated or cooled to reach a phase transition. One recalls Boltzmanns insight: "If you can heat it, it has microscopic structure." The rotational energy of the KBH can be converted into the thermal and kinetic energy of the surrounding plasma. The electron would emit many photons, which in turn can produce a plentiful supply of the electron–positron pairs. When a KBH is immersed in a steady electromagnetic field, for the Wald vacuum solution (WVS; Wald 1974), King et al. (1975) found that all magnetic fields are expelled out of the EH of the extremal KBH (King et al. 1975). The WVS is time-independent and it describes a KBH being immersed in a uniform magnetic field aligned with the black hole spin axis. Bičák & Dvořák (1976) and Bičák & Janis (1985) generalized this result (Bičák & Dvořák 1976; Bičák & Janis 1985). They showed that if the black hole is extremal (namely M/a = 1, where M denotes the mass of KBH and a = J/M), the non-monopole component of the magnetic flux could not penetrate the EH for all steady axisymmetric vacuum solutions. The interesting phenomenon is called Meissner effect (ME), which is the expulsion of magnetic field lines out of the EH and the quenching of jet power for the KBH (Bičák & Janis 1985; Penna 2014). However, WVS does not consider the Hawking radiation (HR) but treats KBH as an absolute black body (Carr & Hawking 1974; Wald 1974). If the HR is considered, researchers believe that the condition is not so restricted and it is possible for ME to occur in less-extreme case (Penna 2014). Some observations about astrophysical black holes with high spins would support this argument (McClintock et al. 2006; Gou et al. 2011; McClintock et al. 2014). The spin parameters of the near-extreme KBHs in Cyg X-1 and GRS 1915+105 have been measured to be M²/J greater than 1. Jets from these black holes could be quenched by the ME.

Recently, for zero or non-zero sources, the ME in the steady axisymmetric FFM attracts lots of attention, where HRs have been considered. Many physicists want to know whether the Blandford–Znajek (BZ) process would be quenched if the ME exists in the real astrophysical environment. However, this effect was never seen in the previous general relativistic magnetohydrodynamic simulations. In our previous work (Gong et al. 2016), we discuss the external-field condition for ME to occur, which is ${H}_{\phi }=0,\omega ={{\rm{\Omega }}}_{0}\equiv \tfrac{a}{{a}^{2}+{r}_{+}^{2}}$ on the external EH. Here Ω₀ is the angular velocity of KBH-EH, H_ϕ corresponds to twice the poloidal electric current, and $\omega =-{A}_{0,r}/{A}_{\phi ,r}=-{A}_{0,\theta }/{A}_{\phi ,\theta }$, which satisfies ${\boldsymbol{E}}=-{\boldsymbol{\omega }}\times {\boldsymbol{B}}$, with A_i and ${r}_{+}$ being the ith component of vector potential ${\boldsymbol{A}}$ and the radius of external EH (Blandford & Znajek 1977; Znajek 1977), we denote $\partial X/\partial y$ as X_,y. Besides, Znajek (1978) suggested that the effective electric resistance of the KBH-EH surface is a non-zero constant indicated by G–S equation if the BZ mechanism works (Blandford & Znajek 1977; Znajek 1978). In Gong et al. (2016), we prove that the BZ mechanism only works when the angular velocity ω of the field is strictly slower than the angular velocity Ω₀ of KBH. If the BZ mechanism does not work, the resistance may vanish. It could be regarded as a superconducting phase transition since the electric resistance suddenly vanishes. At the EH, there does not exist any magnetic field, this means ME will occur. Furthermore, the resistance could be vanishing at the EH. These facts mean that the EH has superconductivity. In condensed matter physics (CMP) research, due to flux quantization, the ME is closely related to zero-resistance phenomena. Generally speaking, when ME occurs, the superconductivity happens.

In this paper, we study the connection between the superconductivity of KBH-EH and the existence of Weyl fermion (WF). We also give the thermodynamic KBH state equation and the inherent-parameter condition for the existence/non-existence of the ME (ME/NME) phase transition, where ${M}^{2}/J={\epsilon }_{c}=1.5$. The divergence of the specific heat at constant J tells us that this is a second-order phase transition. Therefore, we also calculate the critical exponents and find the critical points, further we draw the phase diagrams in (M, J) and (M, Ω) coordinates. Moreover, we provide a possible explanation that the parameters of ME phase exactly correspond to the positive specific heat condition. It is a necessary condition to reach holographic superconductivity (Papantonopoulos 2011).

For simplicity, the constant ${\rm{\hslash }}$, k_B, c, and the gravitational constant G equals a unit of 1.

The Kerr metric in Boyer–Lindquist coordinates is

$$\begin{eqnarray}{{ds}}^{2} & = & {g}_{\mu \nu }{{dx}}^{\mu }{{dx}}^{\nu }\\ & = & -\left(1-\displaystyle \frac{2{Mr}}{{\rm{\Sigma }}}\right){{dt}}^{2}-\displaystyle \frac{4{Mar}\,{\sin }^{2}\theta }{{\rm{\Sigma }}}{dtd}\phi +\displaystyle \frac{{\rm{\Sigma }}}{{\rm{\Delta }}}{{dr}}^{2}\\ & & +{\rm{\Sigma }}d{\theta }^{2}+\displaystyle \frac{A\,{\sin }^{2}\theta }{{\rm{\Sigma }}}d{\phi }^{2}.\end{eqnarray} \tag{ 1 }$$

Here ${x}^{\nu }=(t,r,\theta ,\phi )$, $\nu =(0,1,2,3)$, ${\rm{\Delta }}={r}^{2}-2{Mr}+{a}^{2}$, ${\rm{\Sigma }}={r}^{2}+{a}^{2}{\cos }^{2}\theta$, and $A=({r}^{2}+{a}^{2}){\rm{\Sigma }}\,$ $+\,2{{Mra}}^{2}{\sin }^{2}\theta \,$ $=\,{({r}^{2}+{a}^{2})}^{2}-{\rm{\Delta }}{a}^{2}{\sin }^{2}\theta$.

For solving the problem about a KBH with spin J and mass M in a steady electromagnetic field, one needs to know some information about the electromagnetic field, such as the electromagnetic tensor F_μν or the vector potential A_ν. A special WVS has been extensively researched.

2.1. Wald Vacuum Solution

In the WVS, the field reads (Wald 1974)

$$\begin{eqnarray}{F}_{\mu \nu } & = & 2{B}_{0}({m}_{[\mu ,\nu ]}+2{{ak}}_{[\mu ,\nu ]}),\\ {A}_{\nu } & = & {B}_{0}(2{{ag}}_{{tt}}+{g}_{t\phi },0,0,2{{ag}}_{t\phi }+{g}_{\phi \phi }),\end{eqnarray} \tag{ 2 }$$

where the square bracket $[\mu ,\nu ]$ denotes antisymmetrization of the indices, ${k}^{\nu }={\partial }_{t}$ and ${m}^{\nu }={\partial }_{\phi }$. The Carter orthonormal vector fields are in Boyer–Lindquist components as follows,

$$\begin{eqnarray}{e}_{0}^{c} & = & \left(\displaystyle \frac{{r}^{2}+{a}^{2}}{\sqrt{{\rm{\Sigma }}{\rm{\Delta }}}},0,0,\displaystyle \frac{a}{\sqrt{{\rm{\Sigma }}{\rm{\Delta }}}}\right),\\ {e}_{r}^{c} & = & \left(0,\sqrt{\displaystyle \frac{{\rm{\Delta }}}{{\rm{\Sigma }}}},0,0\right),\\ {e}_{\theta }^{c} & = & \left(0,\displaystyle \frac{1}{\sqrt{{\rm{\Sigma }}}},0,0\right),\\ {e}_{\phi }^{c} & = & \left(\displaystyle \frac{a\sin \theta }{\sqrt{{\rm{\Sigma }}}},0,0,\displaystyle \frac{1}{\sin \theta \sqrt{{\rm{\Sigma }}}}\right).\end{eqnarray} \tag{ 3 }$$

The electromagnetic field in the Carter frames reads

$$\begin{eqnarray}{E}_{r} & = & [{{aA}}_{\phi ,r}+({r}^{2}+{a}^{2}){A}_{t,r}]/{\rm{\Sigma }},\\ {B}_{r} & = & ({A}_{\phi ,\theta }+a{\sin }^{2}\theta {A}_{t,\theta })/({\rm{\Sigma }}\sin \theta ),\\ {E}_{\theta } & = & [{{aA}}_{\phi ,\theta }+({r}^{2}+{a}^{2}){A}_{t,\theta }]/({\rm{\Sigma }}\sqrt{{\rm{\Delta }}}),\\ {B}_{\theta } & = & -\,\sqrt{{\rm{\Delta }}}({A}_{\phi ,r}+a{\sin }^{2}\theta {A}_{t,r})/({\rm{\Sigma }}\sin \theta ),\\ {E}_{\phi } & = & {B}_{\phi }=0.\end{eqnarray} \tag{ 4 }$$

Based on these following relations

$$\begin{eqnarray}{{ag}}_{t\phi }+({r}^{2}+{a}^{2}){g}_{{tt}} & = & -\,{\rm{\Delta }},\\ {{ag}}_{\phi \phi }+({r}^{2}+{a}^{2}){g}_{t\phi } & = & {\rm{\Delta }}a{\sin }^{2}\theta ,\\ {g}_{t\phi }+a{\sin }^{2}\theta {g}_{{tt}} & = & -\,a{\sin }^{2}\theta ,\\ {g}_{\phi \phi }+a{\sin }^{2}\theta {g}_{t\phi } & = & ({r}^{2}+{a}^{2}){\sin }^{2}\theta ,\end{eqnarray} \tag{ 5 }$$

one can get

$$\begin{eqnarray}{E}_{r} & = & -\,2[{B}_{0}a(r-M)(1+{\cos }^{2}\theta )+{{rA}}_{t}]/{\rm{\Sigma }},\\ {B}_{r} & = & 2\cos \theta [{B}_{0}({r}^{2}-{a}^{2})-{{aA}}_{t}]/{\rm{\Sigma }},\\ {E}_{\theta } & = & {B}_{0}a\sqrt{{\rm{\Delta }}}\sin 2\theta /{\rm{\Sigma }},\\ {B}_{\theta } & = & -\,2{B}_{0}\sqrt{{\rm{\Delta }}}r\sin \theta /{\rm{\Sigma }},\\ {E}_{\phi } & = & {B}_{\phi }=0.\end{eqnarray} \tag{ 6 }$$

In the extreme KBH, where a = M, it is easy to know that at the EH, ${A}_{t}={A}_{\phi }={E}_{r}={B}_{r}={E}_{\theta }={B}_{\theta }={E}_{\phi }={B}_{\phi }=0$. In other words, there does not exist any electromagnetic field at the EH for the extreme KBHs.

2.2. BZ Mechanism

Wald finished his work before the pioneering literature about HR was published (Carr & Hawking 1974; Wald 1974). So WVS does not consider the HR but treats KBH as an absolutely black body. Where the particle number is small enough, WVS is a good approximation. Later, a more general nonlinear equation, called the G–S equation, was discovered (Blandford & Znajek 1977). It is primarily used to analyze the energy extraction process of a force-free magnetosphere in the curved Kerr space–time. In the G–S equation, the field induced by HR particles can be added, which makes it very hard to solve.

The constraint differential equation for the FFM around KBHs is given by Menon & Dermer (2005) as

$$\begin{eqnarray}\displaystyle \frac{\sqrt{\gamma }}{2\alpha {\gamma }_{\phi \phi }}\displaystyle \frac{{{dH}}_{\phi }^{2}}{{{dA}}_{\phi }} & = & \omega {\partial }_{r}\left[\displaystyle \frac{{\gamma }_{\theta \theta }}{\alpha \sqrt{\gamma }}({\gamma }_{\phi \phi }\omega +{\beta }_{\phi }){A}_{\phi ,r}\right]\\ & & +\omega {\partial }_{\theta }\left[\displaystyle \frac{{\gamma }_{{rr}}}{\alpha \sqrt{\gamma }}({\gamma }_{\phi \phi }\omega +{\beta }_{\phi }){A}_{\phi ,\theta }\right]\\ & & +{\partial }_{r}\left[\displaystyle \frac{{\gamma }_{\theta \theta }}{\alpha \sqrt{\gamma }}({\beta }^{2}-{\alpha }^{2}+{\beta }_{\phi }\omega ){A}_{\phi ,r}\right]\\ & & +{\partial }_{\theta }\left[\displaystyle \frac{{\gamma }_{{rr}}}{\alpha \sqrt{\gamma }}({\beta }^{2}-{\alpha }^{2}+{\beta }_{\phi }\omega ){A}_{\phi ,\theta }\right],\end{eqnarray} \tag{ 7 }$$

where $\alpha =\sqrt{\tfrac{{\rm{\Delta }}{\rm{\Sigma }}}{A}}$, ${\beta }_{\phi }=-\tfrac{2{Mra}}{{\rm{\Sigma }}}{\sin }^{2}\theta$, $\sqrt{\gamma }=\sqrt{\tfrac{A{\rm{\Sigma }}}{{\rm{\Delta }}}}\sin \theta$, ${\gamma }_{\phi \phi }=\tfrac{A{\sin }^{2}\theta }{{\rm{\Sigma }}}$, ${\gamma }_{\theta \theta }={\rm{\Sigma }}$, ${\gamma }_{{rr}}=\tfrac{{\rm{\Sigma }}}{{\rm{\Delta }}}$, ${\beta }^{2}-{\alpha }^{2}=\tfrac{2{Mr}}{{\rm{\Sigma }}}-1$. The angular velocity $\omega ({A}_{\phi })$ is a function of A_ϕ. Considering the Euler–Lagrange equation, which is ${\partial }_{\nu }\tfrac{\partial {\mathscr{L}}}{\partial {A}_{\phi ,\nu }}=\tfrac{\partial {\mathscr{L}}}{\partial {A}_{\phi }}$, the Lagrangian ${\mathscr{L}}$ for Equation (7) can be expressed as (Gong et al. 2016)

$$\begin{eqnarray}{\mathscr{L}} & = & \displaystyle \frac{1}{\sin \theta }({g}_{33}{\omega }^{2}+2{g}_{03}\omega +{g}_{00})\left({A}_{\phi ,r}^{2}+\displaystyle \frac{1}{{\rm{\Delta }}}{A}_{\phi ,\theta }^{2}\right)\\ & & +\displaystyle \frac{1}{\sin \theta }{g}_{11}{H}_{\phi }^{2}.\end{eqnarray} \tag{ 8 }$$

One can get the external and inner light surfaces for the given ω by solving

$$\begin{eqnarray}&&L\equiv {g}_{33}{\omega }^{2}+2{g}_{03}\omega +{g}_{00}=0.\end{eqnarray} \tag{ 9 }$$

Namely $(\omega -{\rm{\Omega }}^{\prime} )\varpi =\pm \alpha$, where ${\rm{\Omega }}^{\prime} =\tfrac{2{aMr}}{A}$, $\varpi =\sqrt{\tfrac{A}{{\rm{\Sigma }}}}\sin \theta$. Let $\mu =-\cos \theta$, and we assume that $L({A}_{\phi },r,\mu )$ is a function of ${A}_{\phi },r,\mu$. Then Equation (7) becomes G–S equation,which reads

$$\begin{eqnarray}&&L\left(\displaystyle \frac{{A}_{\phi ,{rr}}}{1-{\mu }^{2}}+\displaystyle \frac{{A}_{\phi ,\mu \mu }}{{\rm{\Delta }}}\right)+\displaystyle \frac{{L}_{,r}{A}_{\phi ,r}}{1-{\mu }^{2}}+\displaystyle \frac{{L}_{,\mu }{A}_{\phi ,\mu }}{{\rm{\Delta }}}\\ &&\quad +\displaystyle \frac{1}{2}{L}_{,{A}_{\phi }}\left(\displaystyle \frac{{A}_{\phi ,r}^{2}}{1-{\mu }^{2}}+\displaystyle \frac{{A}_{\phi ,\mu }^{2}}{{\rm{\Delta }}}\right)-\displaystyle \frac{{g}_{11}{H}_{\phi }}{1-{\mu }^{2}}\displaystyle \frac{{{dH}}_{\phi }}{{{dA}}_{\phi }}=0.\end{eqnarray} \tag{ 10 }$$

Here, i = r or μ, ${L}_{,i}={g}_{33,i}{\omega }^{2}+2{g}_{03,i}\omega +{g}_{00,i}$, and ${L}_{,{A}_{\phi }}=2\left({g}_{33}\omega \tfrac{d\omega }{{{dA}}_{\phi }}+{g}_{03}\tfrac{d\omega }{{{dA}}_{\phi }}\right)$.

In Gong et al. (2016), for the G–S equation in the case where $\tfrac{d\omega }{{{dA}}_{\phi }}\leqslant 0$ and ${H}_{\phi }\left(\tfrac{{{dH}}_{\phi }}{{{dA}}_{\phi }}\right)\geqslant 0$, we prove that the ME expels magnetic fields out of the external EH and it also expels magnetic fields out of the inner light surface. The magnetic fields outside the inner light surface could not go to the region between the external EH and the inner light surface. For a stationary, axisymmetric, force-free and magnetically dominated field, its configuration cannot possess a closed loop of poloidal field line (Gralla & Jacobson 2014). Thus the inner light surface must meet the EH when ME occurs. In other words, the angular velocity ω must equal Ω₀ at the EH. If the inner light surface does not meet the EH, ME would expel magnetic fields out of the external EH and the inner light surface. Since the inner light surface, which is defined by L = 0, must be located outside the external EH, the configuration of magnetic field between the external EH and the inner light surface will possess a closed loop of poloidal field line. It will contradict to the result of Gralla & Jacobson (2014).

The Lorentz invariant $\tfrac{1}{2}{{\mathscr{F}}}^{2}=\tfrac{1}{2}{F}_{\mu \nu }{F}^{\mu \nu }$ to the Carter observers is ${B}^{2}-{E}^{2}$, where the Carter field components given by Znajek (1977) are

$$\begin{eqnarray}{E}_{r} & = & [a-\omega ({r}^{2}+{a}^{2})]{A}_{\phi ,r}/{\rm{\Sigma }},\\ {B}_{r} & = & [(1-a\omega {\sin }^{2}\theta )]{A}_{\phi ,\theta }/({\rm{\Sigma }}\sin \theta ),\\ {E}_{\theta } & = & [a-\omega ({r}^{2}+{a}^{2})]{A}_{\phi ,\theta }/({\rm{\Sigma }}\sqrt{{\rm{\Delta }}}),\\ {B}_{\theta } & = & -[\sqrt{{\rm{\Delta }}}(1-a\omega {\sin }^{2}\theta )]{A}_{\phi ,r}/({\rm{\Sigma }}\sin \theta ),\\ {E}_{\phi } & = & 0,\\ {B}_{\phi } & = & {H}_{\phi }/(\sqrt{{\rm{\Delta }}}\sin \theta ).\end{eqnarray} \tag{ 11 }$$

Then we can get $\tfrac{1}{2}{{\mathscr{F}}}^{2}$ $=\,({B}_{r}^{2}+{B}_{\theta }^{2}+\,{B}_{\phi }^{2})$ $-\,({E}_{r}^{2}$ $+\,{E}_{\theta }^{2}+{E}_{\phi }^{2})$ $=\,[1-{({\nu }_{c}^{\phi })}^{2}][{B}_{r}^{2}+{B}_{\theta }^{2}]\,$ $+\,{B}_{\phi }^{2}$, where ${\nu }_{c}^{\phi }$ is the velocity of a observer rotating around the KBH with angular velocity ω with respect to the Carter observers (${\nu }_{c}^{\phi }=\tfrac{\sin \theta }{\sqrt{{\rm{\Delta }}}}\tfrac{\omega ({r}^{2}+{a}^{2})-a}{1-a\omega {\sin }^{2}\theta }$). The final result of the Lorentz invariant is

$$\begin{eqnarray}\tfrac{1}{2}{{\mathscr{F}}}^{2} & = & \displaystyle \frac{-1}{{\rm{\Sigma }}{\sin }^{2}\theta }\left(\displaystyle \frac{A\,{\sin }^{2}\theta }{{\rm{\Sigma }}}{\omega }^{2}-\displaystyle \frac{4{Mar}\,{\sin }^{2}\theta }{{\rm{\Sigma }}}\omega \right.\\ & & \left.-1+\displaystyle \frac{2{Mr}}{{\rm{\Sigma }}}\right)\left({A}_{\phi ,r}^{2}+\displaystyle \frac{1}{{\rm{\Delta }}}{A}_{\phi ,\theta }^{2}\right)+\displaystyle \frac{1}{{\sin }^{2}\theta }\displaystyle \frac{1}{{\rm{\Delta }}}{H}_{\phi }^{2}\\ & = & \displaystyle \frac{-L}{{\rm{\Sigma }}{\sin }^{2}\theta }\left({A}_{\phi ,r}^{2}+\displaystyle \frac{1}{{\rm{\Delta }}}{A}_{\phi ,\theta }^{2}\right)+\displaystyle \frac{1}{{\sin }^{2}\theta }\displaystyle \frac{1}{{\rm{\Delta }}}{H}_{\phi }^{2}.\end{eqnarray} \tag{ 12 }$$

Then we can get the current density ${j}_{\mu }\equiv {g}_{\mu \nu }{j}^{\nu }={F}_{\mu \nu ;\nu }$ (; denotes the covariant derivative). Specially, for the vacuum case, which means j_μ ≡ 0, the electromagnetic field vanishes at the EH if ME occurs. There is no charged particle produced by HR.

For the G–S equation, one can get the electric current density in the $(t,r,\theta ,\phi )$ coordinates from Equation (11), which reads (Blandford & Znajek 1977)

$$\begin{eqnarray}{j}^{0} & = & -\,\displaystyle \frac{1}{{\rm{\Sigma }}}\left\{{\left[\displaystyle \frac{({g}_{33}\omega +{g}_{03}){A}_{\phi ,r}}{1-{\mu }^{2}}\right]}_{,r}+{\left[\displaystyle \frac{({g}_{33}\omega +{g}_{03}){A}_{\phi ,\mu }}{{\rm{\Delta }}}\right]}_{,\mu }\right\},\\ {j}^{r} & = & {g}^{-\tfrac{1}{2}}\displaystyle \frac{{{dH}}_{\phi }}{{{dA}}_{\phi }}{A}_{\phi ,\theta },\\ {j}^{\theta } & = & -\,{g}^{-\tfrac{1}{2}}\displaystyle \frac{{{dH}}_{\phi }}{{{dA}}_{\phi }}{A}_{\phi ,r},\\ {j}^{\phi } & = & \omega {j}^{0}+{g}^{-\tfrac{1}{2}}\displaystyle \frac{{{dH}}_{\phi }}{{{dA}}_{\phi }}\displaystyle \frac{{H}_{\phi }{\rm{\Sigma }}}{{\rm{\Delta }}\sin \theta }.\end{eqnarray} \tag{ 13 }$$

Here, g is the determinant of the metric tensor g_μν. When ω = Ω₀, the boundary condition ${H}_{\phi }=\tfrac{\sin \theta [\omega ({r}_{+}^{2}+{a}^{2})-a]}{{r}_{+}^{2}+{a}^{2}{\cos }^{2}\theta }{A}_{\phi ,\theta }$ at the EH leads to H_ϕ = 0.

The angular velocity ω is a function of A_ϕ and ω = Ω₀ at the EH, then ${A}_{\phi ,\theta }=\tfrac{{{dA}}_{\phi }}{d\omega }{\omega }_{,\theta }=0$, Δ = 0 and L = 0 at the EH. Since ${A}_{\phi ,r\theta }$ is continuous, if we further assume it is bounded (namely, $| {A}_{\phi ,r\theta }| \lt \infty$), we can deduce that ${A}_{\phi ,\theta }=O(\omega -{{\rm{\Omega }}}_{0})=O(\sqrt{{\rm{\Delta }}}),L=O({\rm{\Delta }})=O({H}_{\phi }),$ so that $\tfrac{1}{2}{{\mathscr{F}}}^{2}=0$. Here $y=O(x)$ means that infinitesimal varibles x and y have the same order. From Equation (13), one can know that the electromagnetic field vanishes at the EH. Based on Equation (13), we get

$$\begin{eqnarray}{j}_{\mu }{j}^{\mu } & = & {{Lj}}^{0}{j}^{0}+2({g}_{03}+{g}_{33}\omega ){j}^{0}{j}^{s}+{g}_{33}{j}^{s}{j}^{s}\\ & & +{g}_{{rr}}{j}^{r}{j}^{r}+{g}_{\theta \theta }{j}^{\theta }{j}^{\theta },\end{eqnarray} \tag{ 14 }$$

where ${j}^{s}\equiv {j}^{\phi }-\omega {j}^{0}={g}^{-\tfrac{1}{2}}\,\tfrac{{{dH}}_{\phi }}{{{dA}}_{\phi }}\tfrac{{H}_{\phi }{\rm{\Sigma }}}{{\rm{\Delta }}\sin \theta }$. Moreover, at the EH,

$$\begin{eqnarray}&&{g}_{33}{j}^{s}{j}^{s}+{g}_{{rr}}{j}^{r}{j}^{r}+{g}_{\theta \theta }{j}^{\theta }{j}^{\theta }\geqslant 0,L=0,{g}_{03}+{g}_{33}\omega =0.\end{eqnarray} \tag{ 15 }$$

Besides, due to the fact that the current density ${\boldsymbol{j}}$ is a physical observable that cannot move faster than light, ${\boldsymbol{j}}$ is not a space-like vector (e.g., ${j}_{\mu }{j}^{\mu }\leqslant 0$). In order to satisfy ${j}_{\mu }{j}^{\mu }\leqslant 0$, $\tfrac{{{dH}}_{\phi }}{{{dA}}_{\phi }}$ and ${j}_{\mu }{j}^{\mu }$ need to be 0 at the EH. From Equation (13), one can know that j^r = 0 at the EH when the ME occurs. The charged particles produced by HR cannot drop into the KBH. We think the charged particle with light velocity can stay on the EH. If the current density ${\boldsymbol{j}}$ is a null-vector (namely ${j}_{\mu }{j}^{\mu }=0$) and ${\boldsymbol{j}}\ne {\boldsymbol{0}}$, the current carrier is a charged particle with light velocity, whose mass should be nought. When the interaction between them is very weak, the resistance should be vanishing (Punsly 2008). The most possible candidate serving as the carrier is the mysterious WF, which is massless and permitted to carry charges. If the carrier is an ordinary charged particle with mass, the null-vector ${\boldsymbol{j}}$ must satisfy ${j}_{\mu }\equiv 0$. In other words, the electric current at the external EH vanishes. Thus, in theory, we can test the existence of WF by measuring if there is a nonvanshing current on the EH surface when ω = Ω₀. Moreover, because the resistance of particles with mass hardly vanishes (Punsly 2008), these charge carriers should be massless. Then, WF is a more possible candiate than some so-called supersymmetric particles. It should be pointed out that the above arguments are not explicitly based on the quantum effect. However, since the thermal behaviors of the black hole are essentially quantum mechanical, so are its electric behaviors. The quantum fluctuations might change this result, the thorough answer to this problem may require a better self-consistent theory of quantum gravity (Natsuume 2015).

For KBH, the external and inner EH exists at ${r}_{\pm }=M\pm \sqrt{{M}^{2}-{a}^{2}}$. Furthermore, we notice that the surface gravity κ of KBH reads $\kappa =\tfrac{{r}_{+}-{r}_{-}}{2({a}^{2}+{r}_{+}^{2})}=\tfrac{\sqrt{{M}^{2}-{a}^{2}}}{{a}^{2}+{r}_{+}^{2}}$. Thus we have the equation of the mass function $M(\kappa ,J)$ satisfying

$$\begin{eqnarray}&&\kappa =\displaystyle \frac{\sqrt{{M}^{4}-{J}^{2}}}{2M({M}^{2}+\sqrt{{M}^{4}-{J}^{2}})}.\end{eqnarray} \tag{ 16 }$$

At constant J, when M = J, κ = 0; when $M=+\infty$, κ = 0. Thus, for the same surface gravity κ, there exists a different mass M. Meanwhile, we have the first law of KBH thermodynamics in the electromagnetic field, which is ${dM}={dW}+{\rm{\Omega }}{dJ}+\kappa {{dA}}_{+}$, with ${A}_{+}$ the external surface area of EH. Here dW is the rotation energy extracted by the field. Punsly (2008) has shown that axisymmetic vacuum fields could not extract energy from a black hole (Punsly & Coroniti 1989; Punsly 2008), namely dW = 0. For the G–S equation, ${dW}=\dot{W}{dt}$, the effective power $\dot{W}$ of electromagnetic field affecting KBH satisfies (Blandford & Znajek 1977)

$$\begin{eqnarray}\dot{W} & = & \displaystyle \int {\rm{\Sigma }}{U}^{r}\sin \theta d\theta d\phi ,{U}^{r}\\ & = & \omega ({\rm{\Omega }}-\omega ){\left(\displaystyle \frac{{A}_{\phi ,\theta }}{{r}_{+}^{2}+{a}^{2}\cos {\theta }^{2}}\right)}^{2}({r}_{+}^{2}+{a}^{2}).\end{eqnarray} \tag{ 17 }$$

When ω = Ω, $\dot{W}=0$. Therefore, we have the first law of KBH thermodynamics that reads

$$\begin{eqnarray}&&{dM}={\rm{\Omega }}{dJ}+\kappa {{dA}}_{+}.\end{eqnarray} \tag{ 18 }$$

Here, M, $-{\rm{\Omega }}$, J, and A₊ is analogous to the internal energy, pressure, volume, and entropy of gas (Ho et al. 1997). The field does not affect the state of KBH, so we have ${\rm{\Omega }}={{\rm{\Omega }}}_{0}=\tfrac{a}{{a}^{2}+{r}_{+}^{2}}=\tfrac{a}{2{{Mr}}_{+}}.$ Therefore, ${(2M{\rm{\Omega }})}^{2}+(4M\kappa )\,$ $=\,\tfrac{{a}^{2}}{{r}_{+}^{2}}+\tfrac{2\sqrt{{M}^{2}-{a}^{2}}}{{r}_{+}}\,$ $=\,\tfrac{{a}^{2}+2({r}_{+}-M){r}_{+}}{{r}_{+}^{2}}=\tfrac{2{r}_{+}^{2}-2{{Mr}}_{+}+{a}^{2}}{{r}_{+}^{2}}=1.$ In the thermodynamic state equation, ${\rm{\Omega }}(\kappa ,J)$ satisfies

$$\begin{eqnarray}&&{(2M{\rm{\Omega }})}^{2}+(4M\kappa )=1;\kappa =\displaystyle \frac{\sqrt{{M}^{4}-{J}^{2}}}{2M({M}^{2}+\sqrt{{M}^{4}-{J}^{2}})}.\end{eqnarray} \tag{ 19 }$$

Let $\epsilon \equiv \tfrac{{M}^{2}}{J},M=\sqrt{J\epsilon }$, $\epsilon \geqslant 1$, we have the variable satisfying

$$\begin{eqnarray}\mathrm{ln}\kappa & + & \displaystyle \frac{1}{2}\mathrm{ln}J+\mathrm{ln}2\\ & & =\displaystyle \frac{1}{2}[\mathrm{ln}({\epsilon }^{2}-1)-\mathrm{ln}\epsilon )]-\mathrm{arccosh}\epsilon .\end{eqnarray} \tag{ 20 }$$

The specific heat at constant spin J is

$$\begin{eqnarray}&&{C}_{J}\equiv {\left(\displaystyle \frac{\partial M}{\partial \kappa }\right)}_{J}=\displaystyle \frac{J}{2M}{\left(\displaystyle \frac{\partial \epsilon }{\partial \kappa }\right)}_{J}.\end{eqnarray} \tag{ 21 }$$

Differentiating Equation (20) with respect to κ, we have ${\left(\tfrac{\partial \epsilon }{\partial \kappa }\right)}_{J}^{-1}=\kappa \left[\left(\tfrac{\epsilon }{{\epsilon }^{2}-1}\right)-\left(\tfrac{1}{2\epsilon }+\tfrac{1}{\sqrt{{\epsilon }^{2}-1}}\right)\right].$ Apart from the trivial diverging points (κ = 0 and $\epsilon =+\infty$), the singular points of C_J correspond to the roots of the function f() which can be written as

$$\begin{eqnarray}&&f(\epsilon )\equiv \left(\displaystyle \frac{\epsilon }{{\epsilon }^{2}-1}\right)-\left(\displaystyle \frac{1}{2\epsilon }+\displaystyle \frac{1}{\sqrt{{\epsilon }^{2}-1}}\right)=0.\end{eqnarray} \tag{ 22 }$$

The reciprocal of specific heat satisfies

$$\begin{eqnarray}&&{C}_{J}^{-1}=\displaystyle \frac{2M\kappa }{J}f\left(\displaystyle \frac{{M}^{2}}{J}\right).\end{eqnarray} \tag{ 23 }$$

As shown in Figure 1, the zero point is _c ≈ 1.5 by numerical calculation.

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Substituting ${M}_{c}^{2}/J={\epsilon }_{c}$ into Equation (19), $4{M}_{c}{\kappa }_{c}=\tfrac{2\sqrt{{M}_{c}^{4}-{J}^{2}}}{({M}_{c}^{2}+\sqrt{{M}_{c}^{4}-{J}^{2}})}=\tfrac{2\sqrt{{\epsilon }_{c}^{2}-1}}{({\epsilon }_{c}+\sqrt{{\epsilon }_{c}^{2}-1})}$. The critical point $({M}_{c},{{\rm{\Omega }}}_{c})$ at constant J satisfies

$$\begin{eqnarray}{M}_{c} & = & \sqrt{{\epsilon }_{c}J}=1.22\sqrt{J},{{\rm{\Omega }}}_{c}\\ & = & \displaystyle \frac{1}{2}\sqrt{\displaystyle \frac{{\epsilon }_{c}-\sqrt{{\epsilon }_{c}^{2}-1}}{{\epsilon }_{c}(\epsilon +\sqrt{{\epsilon }_{c}^{2}-1})}}\displaystyle \frac{1}{\sqrt{J}}=\displaystyle \frac{0.16}{\sqrt{J}}.\end{eqnarray} \tag{ 24 }$$

The critical surface gravity κ_c reads

$$\begin{eqnarray}&&{\kappa }_{c}=\displaystyle \frac{\sqrt{{\epsilon }_{c}^{2}-1}}{2\sqrt{{\epsilon }_{c}}({\epsilon }_{c}+\sqrt{{\epsilon }_{c}^{2}-1})}\displaystyle \frac{1}{\sqrt{J}}=\displaystyle \frac{0.174}{\sqrt{J}}=\displaystyle \frac{0.213}{{M}_{c}}.\end{eqnarray} \tag{ 25 }$$

The critical exponent δ and η is defined by

$$\begin{eqnarray}&&| {C}_{J}^{-1}(\tilde{M})| \sim | \tilde{M}{| }^{\delta },| {C}_{J}^{-1}(\tilde{\kappa })| \sim | \tilde{\kappa }{| }^{\eta }.\end{eqnarray} \tag{ 26 }$$

Here, the reduced surface gravity $\tilde{\kappa }$ and mass $\tilde{M}$ satisfies $\kappa \equiv {\kappa }_{c}(1+\tilde{\kappa })$, $M\equiv {M}_{c}(1+\tilde{M})$. Then we let J = 1, so  = M², M_c = 1.224, and κ_c = 0.174. From Equation (23), the reciprocal of specific heat C₁ satisfies

$$\begin{eqnarray}{C}_{1}^{-1}(\tilde{M}) & = & 2.45\kappa (1+\tilde{M})f(1.5{(1+\tilde{M})}^{2})\equiv {C}_{1}^{-1}(\tilde{\kappa })\\ & = & 0.35(1+\tilde{\kappa }){Mf}({M}^{2}).\end{eqnarray} \tag{ 27 }$$

When $\tilde{M}\ll 1$, near critical point M_c, we can re-write it as ${C}_{1}^{-1}(\tilde{M})$ $\approx \,0.7[1.2(1-5\tilde{M})-0.3(1-2\tilde{M})-0.9(1-3.6\tilde{M})]$ $\approx \,-1.5\tilde{M}\sim \tilde{M}$. Meanwhile, we have $\tilde{\kappa }\approx \tfrac{1+3.6\tilde{M}-1.1{\tilde{M}}^{2}}{1+3.6\tilde{M}+2.7{\tilde{M}}^{2}}\,$ $-\,1\approx -3.8{\tilde{M}}^{2}\sim {\tilde{M}}^{2}$. So, $| {C}_{1}^{-1}(\tilde{\kappa })| \sim | \tilde{\kappa }{| }^{1/2}$ near point κ_c. Therefore, the exponent $\delta =1$ and $\eta =\tfrac{1}{2}$.

As shown in Figure 2, in the ME/NME phase diagram, the critical curve $\partial C$ in $(M,J)$ coordinates and the corresponding curve $\partial C^{\prime}$ in $(M,{\rm{\Omega }})$ coordinates reads

$$\begin{eqnarray}&&\partial C\,:J-0.67{M}^{2}=0.\mathrm{and}\,\partial C^{\prime} \,:{\rm{\Omega }}-\displaystyle \frac{0.19}{M}=0.\end{eqnarray} \tag{ 28 }$$

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When the Equation (28) is established, the spectific heat is unbounded. It means this is a second-order phase transition, which is the same as the superconducting transition in CMP. A recent work suggests that due to the absence of an adequate free paremeter in black holes, all kinds of phase transitons and critical phenomena share the same critical exponents and scaling powers (Mandal et al. 2016). It is well-known that for an extreme KBH, the ME must occur. Furthermore, there are not any other critical points for the Kerr sequence. Based on the continuity of the phase zone, one could believe that in the region sandwiched by J ≤ M² and J ≥ 0.67 M², there exists the ME phase. Due to the nonlinearity and complexity, more elaborate theoretical proof and numerial calcuations need to be done.

One can notice that the parameters of the ME phase exactly correspond to the positive specific heat condition. It is a necessary condition to reach holographic superconductivity. In the superconductor, we need condensation of the matter field produced by HR. A non-zero condensate corresponds to a static non-zero field outside the black hole. This viewpoint is helpful to understand the physical cause of the phase transition. Furthermore, the near-horizon extreme KBH is an anti-de Sitter(AdS) space (Guica et al. 2009). AdS space is another necessary condition to reach holographic superconductivity. Besides, similar to AdS/CFT correspondence (Maldacena 1999), the extreme KBH has the Kerr/CFT correspondence (Guica et al. 2009). Here, conformal field theory is abbreviated as CFT. One may develop a kind of perturbation theory for the less-extreme KBH case. It is a possible way to resolve this problem. We will focus on the proof of superconductivity in our future work.

We give the thermodynamic state equation of KBH, calculate the condition of the ME/NME phase transition, and draw the phase diagrams. We also get the critical exponents δ = 1 and $\eta =\tfrac{1}{2}$. It must be pointed out that the ME/NME phase transition is only a terminology, like SBH/LBH phase transition. Lacking the knowledge about microscopic structure of KBH, it is hard to tell whether ME is the essential criterion for the phase transition or not. Comparing with Bardeen–Cooper–Schrieffer superconductivity, there may exist a more natural phenomenon where the electric resistance vanishes. However, the phase transition suggests that there exists a kind of massless and charged particle, possibly WF, which waits to be confirmed in the future. If ME and zero-resistance occurs simultaneously when the BZ mechanism does not work, one can study if the critical magnetic field exists for the linear vacuum case or more complex cases. Since ME/NME phase transition is second order, one could find the order parameter and construct the corresponding Landau–Ginzburg equation. Besides, AdS KBH with negative cosmological constant Λ or Kerr–Newman black hole with charge Q may exhibit richer phase transitions. It is very difficult to find the condition for ME to occur in these cases. The present calculation for phase transition of KBH serves as the first step to explore the above mentioned problems, which will be studied in our future work. We are looking forward to understanding these phenomena in future research.

This work was supported in part by startup funding of the South University of Science and Technology of China, the Shenzhen Peacock Plan and Shenzhen Fundamental Research Foundation (grant No. JCYJ20150630145302225) and National Natural Science Foundation of China (grant No. 11681240276 and grant No. 11674152).