-
The Kerr-AdS black hole solution of the Einstein equations in the Boyer-Lindquist coordinates reads
$ \begin{aligned}[b] \mathrm{d}s^{2} =\;& - \frac{\Delta}{\rho^2}\big( \mathrm{d}t-\frac{a\sin^2\theta}{\Xi}\mathrm{d}\phi \big)^2+\frac{\rho^2}{\Delta}\mathrm{d}r^{2}\\ &+ \frac{\rho^2}{\Sigma}\mathrm{d}\theta^{2}+\frac{\Xi\sin^2\theta}{\rho^2}\big( a\mathrm{d}t-\frac{r^2+a^2}{\Xi}\mathrm{d}\phi \big)^2, \end{aligned} $
(1) with
$ \begin{aligned}[b] & \rho^2 = r^2+a^2\cos^2\theta,\\ & \Xi = 1-\frac{a^2}{l^2},\\ & \Sigma = 1-\frac{a^2}{l^2}\cos^2\theta,\\ & \Delta = (r^2+a^2)(1+\frac{r^2}{l^2})-2mr. \end{aligned} $
(2) Here
$ \Lambda = -3l^{-2} $ is the cosmological constant,$ m, a $ are the mass parameter and angular momentum parameter. The associated thermodynamic quantities are [16]:$ T = \frac{3r_+^4+(a^2+l^2)r_+^2-l^2a^2}{4\pi l^2r_+(r_+^2+a^2)}, \quad S = \frac{\pi(r_+^2+a^2)}{\Xi}, \quad \Omega_{H} = \frac{a\Xi}{r_+^2+a^2}, $
(3) where
$ r_{+} $ represents the horizon radius satisfying$ \Delta(r_{+}) = 0 $ , T, S and$ \Omega_{H} $ are defined as the Hawking temperature, the Bekenstein-Hawking entropy and the angular velocity respectively. The energy M and the angular momentum J are$ M = \dfrac{m}{\Xi^2} $ and$ J = \dfrac{am}{\Xi^2} $ respectively.In 2012, S. Gunasekaran, D. Kubiz
$ \check{n} $ ï¿œk, and R. B.Mann introduced the extended phase space for AdS black hole [23], where the cosmological constant Λ can be interpreted as a pressure term via the following relation$ P = -\frac{\Lambda}{8\pi}. $
(4) The first law of the black hole thermodynamics and the Smarr formula are
$ \delta M = T\delta S+\Omega_{H}\delta J+V\delta P, $
(5) $ \frac{M}{2} = TS+\Omega_{H}J-VP, $
(6) where the thermodynamic volume V is
$ \dfrac{4\pi l^2 r_{+}(a^2+r_{+}^2)}{3(-a^2+l^2)} $ . Followed Gunasekaran et al.'s perspective, many works had been down to explore the thermodynamic properties of AdS black holes on the extended phase space [24−27]. -
It is known that there is ambiguity on energy notion of the Kerr-AdS black holes [20, 21]. This ambiguity has been a long standing issue. In the very recent article [22], the authors proposed a natural criteria to justify the notion of energy. In particular, they examined whether the associated first law of the black hole thermodynamics exists. Within the Iyer-Wald formalism, two versions of the first law and the Smarr formula were established for different energies. The difference originated from the choice of the Killing vectors. The standard energy notion
$ m/\Xi^{2} $ is associated with the Killing vector$ \dfrac{\partial}{\partial T} = \dfrac{\partial}{\partial t}+\dfrac{1}{3}a\Lambda \frac{\partial}{\partial \phi} $ . The relevant thermodynamic quantities, the first law and the standard results are presented in section II. The other energy notion$ \hat{M} = m/\Xi^{3/2} $ is related to the Killing vector$ \dfrac{1}{\sqrt{\Xi}}\dfrac{\partial}{\partial t} $ , using the notation from [22], the corresponding thermodynamic quantities are$ \hat{\Omega}_{H} = \frac{\Omega_{H}}{\sqrt{\Xi}}, \quad \hat{T} = \frac{T}{\sqrt{\Xi}}, \quad \hat{V} = \frac{V}{\sqrt{\Xi}}. $
(7) The modified first law of black hole thermodynamics is presented as
$ \delta\hat{M} = \hat{T}\delta S+ \hat{\Omega}_{H}\delta J+\hat{V}\delta P. $
(8) Based on the concrete expressions of the Killing vectors, there exists a relative rotation between two kinds of observers. Unlike a naive observation,
$ \dfrac{\partial}{\partial T} $ coincides with the generator of the conformal boundary, corresponding to non-rotating observers, whereas rotating observers are related to$ \dfrac{\partial}{\partial t} $ . -
Using eq.(7), the functions
$ \hat{\Omega}_{H} $ ,J and$ \hat{T} $ can be expressed as$ J = \frac{3a(a^2+r_+^2)(3+8P\pi r_+^2 )}{2(3-8a^2P \pi)^2r_+}, $
(9) $ \hat{\Omega}_{H} = \frac{2a\sqrt{P(-a^2+\dfrac{3}{8P\pi})}\sqrt{\dfrac{2\pi}{3}}}{a^2+r_+^2}, $
(10) $ \hat{T} = \frac{\sqrt{\dfrac{1}{9-24a^2P\pi}}(a^2(-3+8P\pi r_+^2)+3(r_+^2+8P\pi r_+^4))}{4\pi r_+(a^2+r_+^2) }. $
(11) The above equations can be viewed as equation of state of the Kerr-AdS black hole.
In this section, we focus on the phase structure of the modified black hole thermodynamics on the
$ (\hat{\Omega}_{H},J) $ plane. Previous work [16] had considered similar problems for the standard black hole thermodynamics on the$ (J,\Omega_{H}) $ section, without considering the extended phase space.According to the equation of state (9)-(11), the isotherms for various temperatures on
$ \hat{\Omega}_{H}-J $ plane are presented in Fig. 1. Qualitatively, Fig. 1 is similar to the liquid/gas PVT diagram [28]. Under the following correspondence,Figure 1. (color online) We draw isotherms of Kerr-AdS black hole with
$ l = 1 $ in$ \hat{\Omega}_{H}-J $ plane. Qualitatively, the isotherms near the critical temperature$ \hat{T}_{c} $ have similar behavior of van der Waals system. The black solid line is the critical isotherm.$ \hat{\Omega}_{H}\rightarrow P, J\rightarrow V, $
(12) there is a van der Waals-like phase structure. By contrast, the correspondence in the previous article [16] was
$ J\rightarrow P, \Omega_{H}\rightarrow V $ .As the temperature decreases until
$ \hat{T}_{c} $ , an inflection point is formed. So, the temperature$ \hat{T}_{c} $ and the angular velocity$ \hat{\Omega}_{H} $ at the critical point satisfy [28]$ \left(\frac{\partial \hat{\Omega}_{H} }{\partial J}\right)_{\hat{T}_{c}} = 0, \quad\left(\frac{\partial^2 \hat{\Omega}_{H} }{\partial J^2}\right)_{\hat{T}_{c}} = 0. $
(13) Combining with the equation of state (9)-(11), the critical point is
$ J_{c} = \frac{3(4+\sqrt{2})(1+2\sqrt{2})^{3/2}\sqrt{-5+4\sqrt{2}}(-6+5\sqrt{2})}{448(-3+\sqrt{2})^2P\pi}, $
(14) $ \hat{\Omega}_{Hc} = \frac{2\sqrt{\dfrac{2}{3}(-23+17\sqrt{2})\pi P}}{(-2+3\sqrt{2})}, $
(15) $ \hat T_{c} = \frac{7(-2+3\sqrt{2})\sqrt{\frac{2P}{9\pi -3\sqrt{2}\pi}}}{(1+2\sqrt{2})^{3/2}(-6+5\sqrt{2})}. $
(16) -
In this subsection, we study firstly the coexistence condition between two states, from which the law of equal area will be obtained. For two stable states
$ \hat{a} $ and$ \hat{b} $ , it is well-known that coexistence condition is the equal free energy [28], i.e.$ G(\hat{a}) = G(\hat{b}) $ . Compared with the PTV system, the free energy for the black hole system should be$ {\rm{d}}G = -S{\rm{d}}\hat{T}+J{\rm{d}}\hat{\Omega}_{H}. $
(17) Fixing the temperature, one can get the free energy by integrating above equation along the isothermal curve. The difference of free energy between two states should be
$ \int_{\hat{a}}^{\hat{b}}J {\rm{d}}\hat{\Omega}_{H} $ (this integral is along the isothermal curve which connects the two states). From Fig. 2(a), since$ \hat{a} $ and$ \hat{b} $ are coexistent states, it meansFigure 2. (color online) (a) The brown solid line represents a isotherm with temperature
$ \hat{T} = 0.248291 $ and$ l = 1 $ ,$ \hat{a} $ and$ \hat{b} $ are coexistence states. Because of equal free energy requirement of coexistence states$ G(\hat{a}) = G(\hat{b}) $ , area(A) = area(B). (b) The red dashed line is the coexistence curve of phase transition above$ \hat{T}_{c} $ . The black solid line represents the critical isotherm with$ l = 1 $ .$ G(\hat{a})-G(\hat{b}) = \int_{\hat{\Omega}_{H\hat{a}}}^{\hat{\Omega}_{H\hat{b}}}J {\rm{d}}\hat{\Omega}_{H} = 0, $
(18) which implies the areas of region A and region B in Fig. 2(a) are equal, this is just the Maxwell equal area law.
In order to investigate the critical behavior near the critical point, we need to define the order parameter. Analogous to the van der Waals system,
$ \eta = \dfrac{J_{b}-J_{a}}{2} $ is defined as the order parameter. The coexistence curve is the red dashed line in Fig. 2(b). -
For van der Waals system, near the critical point, one can get the critical exponents [28]. For the correspondence (12), the analogous critical exponents can be obtained as following.
$ \bullet $ Degree of critical isotherm:$ \hat{\Omega}_{H}-\hat{\Omega}_{Hc} = A_{\delta}\lvert J-J_{c}\rvert^\delta sign(J-J_{c}),\qquad \hat{T} = \hat{T}_{c}. $
(19) $ \bullet $ Degree of coexistence curve:$ \eta = -A_{\beta}(\hat{T}-\hat{T}_{c})^\beta,\qquad \hat{T}>\hat{T}_{c}. $
(20) $ \bullet $ Degree of heat capacity($ J = J_{c} $ ):$ C_{J} = \left\{ \begin{array}{l} A_{\alpha'}\left\{-(\hat{T}-\hat{T}_{c})\right\}^{-\alpha'}\quad \hat{T}<\hat{T}_{c}\\ A_{\alpha}\left\{+(\hat{T}-\hat{T}_{c})\right\}^{-\alpha} \quad \hat{T}>\hat{T}_{c}. \end{array} \right . $
(21) $ \bullet $ Degree of isothermal compressibility:$ \kappa{_{T}} = -\frac{1}{J}\left(\frac{\partial J }{\partial \hat{\Omega}_{H}}\right)_{\hat T} = \left\{ \begin{array}{l} A_{\gamma'}\left\{-(\hat{T}-\hat{T}_{c})\right\}^{-\gamma'}\quad \hat{T}<\hat{T}_{c}\\ A_{\gamma}\left\{+(\hat{T}-\hat{T}_{c})\right\}^{-\gamma} \quad \hat{T}>\hat{T}_{c}. \end{array} \right . $
(22) In the following of this subseciton, we calculate explicitly the critical exponents for the modified black hole thermodynamics.
-
At critical point, the first and second derivatives of
$ \hat{\Omega}_{H} $ with respect to J satisfy$ \big(\frac{\partial \hat{\Omega}_{H} }{\partial J}\big)_{\hat T_{c}} = 0, \quad\big(\frac{\partial^2 \hat{\Omega}_{H} }{\partial J^2}\big)_{\hat T_{c}} = 0. $
(23) The third derivative can be calculated as
$ \big(\frac{\partial^3 \hat{\Omega}_{H} }{\partial J^3}\big)_{\hat T_{c}} = -\frac{16384 (\pi P) ^{7/2}}{9 \sqrt{87 \sqrt{2}+123} }\neq 0, $
(24) hence
$ \delta = 3 $ by the definition of δ in eq.(19). -
In Fig. 2(b), we plot the curve of the coexisting states using Maxwell's equal-area law. Along this curve, all the thermodynamical quantities only depend on the temperature. To get the value of the degree of coexistence curve, β, we need to obtain the relations between η and
$ \hat T $ near the critical point. For this purpose, we expand$ \hat{\Omega}_{H} $ in terms of J and$ \hat{T} $ to the third order as$ \begin{aligned}[b] \hat{\Omega}_{H}-\hat{\Omega}_{Hc}\approx \;& (\partial_{\hat{T}}\hat{\Omega}_{H})_{J}\vert_{c}(\hat{T}-\hat{T}_{c})+\frac{1}{2}(\partial^2_{\hat{T}}\hat{\Omega}_{H})_{J}\vert_{c}(\hat{T}-\hat{T}_{c})^2\\&+ \big(\partial_ {\hat{T}}(\partial _{J}\hat{\Omega}_{H})_{\hat{T}}\big)_{J}\vert_{c}(\hat{T}-\hat{T}_{c})(J-J_{c})\\ &+\frac{1}{6}(\partial^3_{\hat{T}}\hat{\Omega}_{H})_{J}\vert_{c}(\hat{T}-\hat{T}_{c})^3+ \frac{1}{6}(\partial^3_{J}\hat{\Omega}_{H})_{\hat{T}}\vert_{c}(J-J_{c})^3 \end{aligned}$
$ \begin{aligned}[b] &+\frac{1}{2}\big(\partial^2_ {\hat{T}}(\partial _{J}\hat{\Omega}_{H})_{\hat{T}}\big)_{J}\vert_{c}(\hat{T}-\hat{T}_{c})^2(J-J_{c})\\ &+ \frac{1}{2}\big(\partial_ {\hat{T}}(\partial^2 _{J}\hat{\Omega}_{H})_{\hat{T}}\big)_{J}\vert_{c}(\hat{T}-\hat{T}_{c})(J-J_{c})^2. \end{aligned}$
(25) For simplicity, let us introduce
$ \omega = \hat{\Omega}_{H}-\hat{\Omega}_{Hc},\qquad \overline{t} = \hat{T}-\hat{T}_{c},\qquad j = J-J_{c}, $
(26) then eq.(25) becomes
$ \omega = c_{10}\overline{t}+c_{20}\overline{t}^2+c_{11}\overline{t}j+c_{30}\overline{t}^3+c_{03}j^3+c_{21}\overline{t}^2j+c_{12}\overline{t}j^2. $
(27) According to the equation of state (9)-(11), one can calculate all coefficients in eq.(27),
$ \begin{aligned}[b]& c_{10} = -\sqrt{2} \pi, \quad c_{20} = \sqrt{6 \left(29 \sqrt{2}-41\right)} \pi ^{3/2} \sqrt{\frac{1}{P}}, \\& c_{11} = \frac{64}{3} \left(\sqrt{2}-1\right) \pi ^2 P,\quad \ c_{30} = \frac{9 \left(11 \sqrt{2}-15\right) \pi ^2}{P}, \\& c_{03} = -\frac{16384 (\pi P) ^{7/2}}{9 \sqrt{87 \sqrt{2}+123} },\\& c_{21} = -{544 \sqrt{\frac{2P}{5331 \sqrt{2}+7539}} \pi ^{5/2}}, \\& c_{12} = \frac{2048}{3} \left(4-3 \sqrt{2}\right) \pi ^3 P^2. \end{aligned} $
(28) The equilibrium condition
$ \hat{\Omega}_a = \hat{\Omega}_b $ implies$ \omega(j_{a},t) = \omega(j_{b},t) $ , which in turn yields$ \begin{aligned}[b] \omega_b-\omega_a =\;& (j_{b}-j_{a})\big[c_{11}\overline{t}+c_{21}\overline{t}^2+c_{12}\overline{t}(j_{a}+j_{b})\\&+c_{03}(j_{a}^2+j_{a}j_{b}+j_{b}^2)\big] = 0. \end{aligned} $
(29) By employing the Maxwell's equal area law, we know the area of the shaded region in Fig. 3 is equal to the area of the rectangle
$ cd\hat{b}\hat{a} $ in Fig. 3. So we haveFigure 3. (color online) According to the equal area law area(A) = area(B), the shaded area should equal to the area of the rectangle
$ cd\hat{b}\hat{a} $ .$ \int^{J_b}_{J_a}{\hat\Omega}dJ = (J_b-J_a)\frac{1}{2}(\Omega_b+\Omega_a) $
(30) (The left side of eq.(30) is the area of the shaded region in Fig. 3 and the right side is the area of the rectangle cdba in Fig. 3. The factor
$ \dfrac{1}{2}(\Omega_b+\Omega_a) $ comes from the equilibrium condition$ {\hat\Omega}_a = {\hat\Omega}_b $ .) By using eq. (26), the eq. (30) becomes$ \begin{aligned}[b] \int^{J_b}_{J_a}{\hat\Omega}dJ =\;& \left[\int^{J_b}_{J_a}({\hat\Omega}-{\hat\Omega}_c)d(J-J_c)\right]+{\hat\Omega}_c[(J_b-J_c)-(J_a-J_c)]\\ =\;& \left[\int^{j_b}_{j_a}({\hat\Omega}-{\hat\Omega}_c)dj\right]+{\hat\Omega}_c(j_b-j_a)\\ =\;& \left[\int^{j_b}_{j_a}\omega dj\right]+{\hat\Omega}_c(j_b-j_a)\\ =\;& (J_b-J_a)\frac{1}{2}({\hat\Omega}_b+{\hat\Omega}_a)\\ =\;& (j_b-j_a)\frac{1}{2}([{\hat\Omega}_b-{\hat\Omega}_c]+[{\hat\Omega}_a-{\hat\Omega}_c])+(j_b-j_a){\hat\Omega}_c\\ =\;& (j_b-j_a)\frac{1}{2}(\omega_b+\omega_a)+(j_b-j_a){\hat\Omega}_c. \end{aligned} $
(31) Above result implies
$ \int^{j_b}_{j_a}\omega dj = (j_b-j_a)\frac{1}{2}(\omega_b+\omega_a). $
(32) Submitting eq.(27) into eq.(32), one can obtain
$ \begin{aligned}[b]& \int_{j_{a}}^{j_{b}}(c_{10}\overline{t}+c_{20}\overline{t}^2+c_{11}\overline{t}j+c_{30}\overline{t}^3+c_{03}j^3+c_{21}\overline{t}^2j+c_{12}\overline{t}j^2) {\rm{d}}j\\ =\;& c_{10}\overline{t}(j_{b}-j_{a})+c_{20}\overline{t}^2(j_{b}-j_{a})+\frac{1}{2}c_{11}\overline{t}(j_{b}^2-j_{a}^2)+c_{30}\overline{t}^3(j_{b}-j_{a})\\& +\frac{1}{4}c_{03}(j_{b}^4 -j_{a}^4)+\frac{1}{2}c_{21}\overline{t}^2(j_{b}^2-j_{a}^2)+\frac{1}{3}c_{12}\overline{t}(j_{b}^3-j_{a}^3)\\ =\;& (j_{b}-j_{a})[c_{10}\overline{t}+c_{20}\overline{t}^2+\frac{1}{2}c_{11}\overline{t}(j_{b}+j_{a}) +c_{30}\overline{t}^3 \end{aligned} $
$ \begin{aligned}[b]\qquad & +\frac{1}{4}c_{03}(j_{b}+j_{a})(j_{b}^2+j_{a}^2)\\&+\frac{1}{2}c_{21}\overline{t}^2(j_{b}+j_{a})+\frac{1}{3}c_{12}\overline{t}(j_{b}^2+j_{a}j_{b}+j_{a}^2)]\\ =\;& (j_{b}-j_{a})\frac{1}{2}(\omega_{a}+\omega_{b}), \end{aligned} $
(33) dividing both sides of eq.(33) by
$ (j_{b}-j_{a}) $ yields$ \begin{aligned}[b]& c_{10}\overline{t}+c_{20}\overline{t}^2+\frac{1}{2}c_{11}\overline{t}(j_{b}+j_{a})+c_{30}\overline{t}^3\\&+\frac{1}{4}c_{03}(j_{b}+j_{a})(j_{b}^2+j_{a}^2)+\frac{1}{2}c_{21}\overline{t}^2(j_{b}+j_{a})\\& +\frac{1}{3}c_{12}\overline{t}(j_{b}^2+j_{a}j_{b}+j_{a}^2)\\ =\;& c_{10}\overline{t}+c_{20}\overline{t}^2+\frac{1}{2}c_{11}\overline{t}(j_{b}+j_{a})+c_{30}\overline{t}^3\\ &+\frac{1}{2}c_{03}(j_{b}^3+j_{a}^3) +\frac{1}{2}c_{21}\overline{t}^2(j_{b}+j_{a})+\frac{1}{2}c_{12}\overline{t}(j_{b}^2+j_{a}^2). \end{aligned} $
(34) Eliminating similar terms on both sides of eq.(34) gives
$ \begin{aligned}[b]& \frac{1}{4}c_{03}(j_{b}+j_{a})(j_{b}^2+j_{a}^2)+\frac{1}{3}c_{12}\overline{t}(j_{b}^2+j_{a}j_{b}+j_{a}^2)\\ =\;& \frac{1}{2}c_{03}(j_{b}^3+j_{a}^3)+\frac{1}{2}c_{12}\overline{t}(j_{b}^2+j_{a}^2). \end{aligned}$
(35) Then we can get
$ \begin{aligned}[b]& \frac{1}{4}c_{03}(j_{b}^3+j_{b}^2j_{a}+j_{b}j_{a}^2+j_{a}^3)+\frac{1}{3}c_{12}\overline{t}(j_{b}^2+j_{a}j_{b}+j_{a}^2)\\ =\;& \frac{1}{2}c_{03}(j_{b}^3+j_{a}^3)+\frac{1}{2}c_{12}\overline{t}(j_{b}^2+j_{a}^2), \end{aligned} $
(36) which implies
$ \begin{aligned}[b] 0 =\;& \frac{1}{4}c_{03}(j_{a}^3-j_{a}j_{b}^2-j_{a}^2j_{b}+j_{b}^3)+\frac{1}{6}c_{12}\overline{t}(j_{b}^2-2j_{a}j_{b}+j_{a}^2)\\ =\;& \frac{1}{4}c_{03}(j_{b}+j_{a})(j_{b}-j_{a})^2+\frac{1}{6}c_{12}\overline{t}(j_{b}-j_{a})^2. \end{aligned}$
(37) Simplify the above equation, and we find
$ \begin{split} (j_{b}-j_{a})^2\big[\frac{1}{4}c_{03}(j_{a}+j_{b})+\frac{1}{6}c_{12}\overline{t}\big] = 0. \end{split} $
(38) Denote
$ j_{-}\equiv j_{b}-j_{a} = J_{b}-J_{a}, j_{+}\equiv j_{b}+j_{a} $ , from eq.(38), one can solve$ j_{+} $ as$ j_+ = -\frac{2c_{12}\overline{t}}{3c_{03}}, $
(39) substituting eq.(39) into eq.(29) yields
$ j_- = \sqrt{\frac{-4c_{11}\overline{t}+\left(\dfrac{4c_{12}^2}{3c_{03}}-4c_{21}\right)\overline{t}^2}{c_{03}}}. $
(40) Near the critical point, the temperature dependence of the order parameter is
$ \frac{J_{b}-J_{a}}{2}\approx A_{\beta}(\hat{T}-\hat{T}_{c})^{1/2}. $
(41) Thus we read
$ \beta = \dfrac{1}{2} $ . -
By definition (21), the critical exponent of heat capacity can be obtained as follows. Recall that the energy
$ \hat{M} $ is given by$ \hat{M} = \frac{\left(a^2+r_+^2\right) \left(\dfrac{r_+^2}{l^2}+1\right)}{2 r_+ \left(1-\dfrac{a^2}{l^2}\right)^{3/2}}, $
(42) the heat capacity
$ C_{J} $ can be calculated as$ C_{J} = \big(\frac{\partial\hat{M}}{\partial\hat{T}}\big)_{J}\rvert_{c} = \frac{3}{8P}\neq0. $
(43) Due to the heat capacity neither diverges nor vanishes, it follows that α and
$ \alpha' $ are both zero, i.e.,$ \alpha = \alpha' = 0 $ . -
The isothermal comprssibility
$ \kappa{_{T}} $ is defined as$ \kappa{_{T}} = -\frac{1}{J}\left(\frac{\partial J }{\partial \hat{\Omega}_{H}}\right)_{\hat T}, $
(44) which diverges at the critical point. Introducing
$ \widetilde{\omega},\widetilde{t} $ and$ \widetilde{j} $ as$ \widetilde{\omega} = \frac{\hat{\Omega}_{H}-\hat{\Omega}_{Hc}}{\hat{\Omega}_{Hc}},\quad \widetilde{t} = \frac{\hat{T}-\hat{T}_{c}}{\hat{T}_{c}},\quad \widetilde{j} = \frac{J-J_{c}}{J_{c}}, $
(45) then
$ \hat{\Omega}_{H} = \hat{\Omega}_{Hc}\widetilde{\omega}+\hat{\Omega}_{Hc},\quad \hat{T} = \hat{T}_{c}\widetilde{t}+\hat{T}_{c},\quad J = J_{c}\widetilde{j}+J_{c}, $
(46) and
$ \omega = \hat{\Omega}_{Hc}\widetilde{\omega},\quad \overline{t} = \hat{T}_{c}\widetilde{t},\quad j = J_{c}\widetilde{j}. $
(47) Eq.(27) can be rewritten as
$ \widetilde{\omega} = \widetilde{c_{10}}\widetilde{t}+\widetilde{c_{20}}\widetilde{t}^2+\widetilde{c_{11}}\widetilde{t}\widetilde{j}+\widetilde{c_{30}}\widetilde{t}^3+\widetilde{c_{03}}\widetilde{j}^3+\widetilde{c_{21}}\widetilde{t}^2\widetilde{j}+\widetilde{c_{12}}\widetilde{t}\widetilde{j}^2. $
(48) Using eq.(44), we have
$ \frac{-1}{J\kappa{_{T}}} = \frac{1}{\left(\dfrac{\partial J }{\partial \hat{\Omega}_{H}}\right)_{\hat T}} = \left(\frac{\partial \hat{\Omega}_{H} }{\partial J}\right)_{\hat T}. $
(49) Combining eqs.(41), (46), (47) and (48) yields
$ \frac{-1}{J\kappa{_{T}}} = \left(\frac{\partial \hat{\Omega}_{H} }{\partial J}\right)_{\hat T} = \left(\frac{\partial\hat{\Omega}_{H}}{\partial \widetilde{j}}\frac{\partial \widetilde{j}}{\partial J}\right)_{\hat{T}} = \frac{\hat{\Omega}_{Hc}}{J_{c}}\left(\frac{\partial \widetilde{\omega}}{\partial \widetilde{j}}\right)_{\widetilde{t}}\propto\left(\frac{\partial \omega}{\partial j }\right)_{\overline{t}}, $
(50) namely
$ \kappa{_{T}}^{-1}\propto\big(\frac{\partial \omega}{\partial j }\big)_{\overline{t}} = c_{11}\overline{t}+3c_{03}j^2+o(\overline t) = C_{1}\overline{t}+o(\overline t), $
(51) where
$ C_{1} $ is a real constant and eq.(41) is used in the last step. By the definition of γ and$ \gamma' $ in eq.(22), we get$ \gamma = \gamma' = 1 $ . -
According to the differential expression
$ {\rm{d}}G = -S{\rm{d}}\hat{T}+J{\rm{d}}\hat{\Omega}_{H} $ , at constant angular velocity$ \hat{\Omega}_{H} $ , we can make use of the Eqs.(3), (10) and (11) to plot$ G-\hat{T} $ diagram.From Fig. 4, the swallowtail structure is analogous to the van der Waals system. This further confirms the similarity between the modified thermodynamics and the PVT system.
Figure 4. (color online) This diagram describes the qualitative behavior of the free energy as a function of temperature for various angular velocity with
$ l = 1 $ . The angular velocity decreases from bottom to top. The green solid line corresponds to$ \hat{\Omega}_{H}>\hat{\Omega}_{Hc} $ , the yellow solid line to$ \hat{\Omega}_{H} = \hat{\Omega}_{Hc} $ , and the remaining red solid lines display$ \hat{\Omega}_{H}<\hat{\Omega}_{Hc} $ . For$ \hat{\Omega}_{H}<\hat{\Omega}_{Hc} $ , the existence of swallowtail structure implies there is a first-order phase transition in the system. -
A well-known fact is that in the vicinity of the critical point, the free energy can be expressed as a homogenous function, with corresponding homogenous indices p and q [28]. Based on the previous discussions, we see that the critical exponents satisfy the following expected relations:
$ \alpha+2\beta+\gamma = 2, \quad \alpha+\beta(\gamma+1) = 2, $
(52) $ \gamma(\delta+1) = (2-\alpha)(\delta-1), \quad \gamma = \beta(\delta-1). $
(53) The free energy has the following scaling symmetry
$ g_{s}(\Lambda^p \epsilon,\Lambda^q j) = \Lambda g_{s}(\epsilon,j),\quad p = \frac{1}{2},\quad q = \frac{3}{4}. $
(54) In terms of p and q [28], the critical exponets read
$ \alpha = \frac{2p-1}{q}, $
(55) $ \beta = \frac{1-q}{p}, $
(56) $ \gamma = \frac{2q-1}{p}, $
(57) $ \delta = \frac{q}{1-q}. $
(58) The above scaling relations could be regarded as the consistency check for the critical exponents we got in this section.
-
In standard black hole thermodynamics, the phase structure on
$ (P,V) $ section has been investigated [17−19, 23, 30]. In this section, we study the phase structures on (P,$ \hat{V} $ ) section in the modified black hole thermodynamics. In Fig. 5, we plot the$ P-\hat{V} $ diagram corresponding to the Kerr-AdS black hole.Figure 5. (color online) P-
$ \hat{V} $ diagram of Kerr-AdS black hole with$ J = 1 $ . The brown solid line is the critical isotherm and the red dot is the critical point.Following the method used in [30], after neglecting all higher order terms of J, the equation of state can be written as
$ P = \frac{\sqrt[3]{\dfrac{\pi }{6}}\hat{T} }{\sqrt[3]{\hat{V}}}-\frac{1}{2\ 6^{2/3} \sqrt[3]{\pi }\hat{V}^{2/3}}+\frac{2 \pi J^2 \left(4 \sqrt[3]{6} \pi ^{2/3} \hat{T} \sqrt[3]{\hat{V}}+3\right)}{3 \left(\sqrt[3]{6} \pi ^{2/3} \hat{T} \hat{V}^{4/3}+\hat{V} \right)^2}, $
(59) where
$ \hat{V} $ satisfies$ \hat{V} = \frac{4 \pi r_+^3}{3}+\frac{48 \pi \left(4 \pi P r_+^2+1\right)J^2}{r_+ \left(8 \pi P r_+^2+3\right)^2}. $
(60) Following the critical point condition
$ \dfrac{\partial P}{\partial \hat V}\big|_{T_C} = 0, \dfrac{\partial^2 P}{\partial \hat V^2}\big|_{T_C} = 0 $ , we can get$ P_{c} = \frac{0.003}{ J},\quad\hat{V}_{c} = 129.603 J^{3/2},\quad\hat{T}_{c} = \frac{0.041}{ \sqrt{J}}. $
(61) Now we consider the phase transition which happens below the critical temperature
$ \hat T_{c} $ . In Fig. 5, this van der Waals-like phase structure is clearly visualized. The behavior of the physical variables near its critical point is quantitatively described by the critical exponents. Following a similar method in Section 4, we have$ \alpha = \alpha' = 0, \quad \beta = \frac{1}{2},\quad \gamma = 1,\quad \delta = 3. $
(62) The above results show that the phase structure for the modified black hole thermodynamics in
$ (P,\hat{V}) $ section is almost the same as the standard one [31, 32], the only difference is that all thermal quantities have a deformation factor$ \dfrac{1}{\sqrt{\Xi}} $ , which is shown in eq.(7).
Phase transition of modified thermodynamics of the Kerr-AdS black hole
- Received Date: 2024-09-09
- Available Online: 2025-01-01
Abstract: We investigate the critical phenomena of Kerr-AdS black holes in the modified first law of thermodynamics. The modified black hole thermodynamics exhibits the van der Waals-like phase structure. All the critical exponents are calculated, and the swallowtail diagram of free energy is plotted. Comparing with existing results, the main difference is the correspondence between thermal quantities of the Kerr-AdS black holes and the van der Waals system. In previous work[