1 Introduction
There are many solutions of Einstein's equation. The de Sitter space-time has fantastic features, which is the solution of Einstein gravity with a positive cosmological constant. It is considered that the de Sitter space-time is a best description of our universe at very early stage after the fiducial Big Bang singularity and compatible with the current observations of the accelerating universe.[1-4]
The de Sitter space time normally contains two real horizons. The inner one is the event horizon and the outer one is the cosmological horizon. The thermodynamic properties on the two horizons are usually studied independently.[5] And the total entropy of the black hole is the sum of the entropies on the two horizons.
Based on the following considerations, we notice that the event horizon and the cosmological horizon might not be independent. Moreover, we guess that the thermodynamic properties might not independent either.
Firstly, the first laws for the de Sitter black hole on event horizon $r_h$ and cosmological horizon $r_c$ are respectively:[6]
$ dM=T_hdS_h+\Phi_h dQ+V_h d\Lambda , \\ dM=-T_cdS_h+\Phi_c dQ+V_c d\Lambda ,$
where $M$ is the conserved mass in de Sitter space-time, $\Phi$ stands for the electric potential, $Q$ is the charge of the black hole, and $\Lambda$ is the cosmological constant. One could see that the two equations in Eq. (1) are not independent for they share the same $M$, $Q$ and $\Lambda$. Thus, the event horizon and cosmological horizon should be related.
Secondly, particular attention is paid on studying the thermodynamics of black holes in the extended phase space recently, in which the cosmological constant is identified as a thermodynamic pressure.[7-10] Then, in first law of thermodynamics, the cosmological constant could be regarded as the thermodynamics pressure and its conjugate ``thermodynamics volume'' appears, thus the mass of the black hole should be explained as enthalpy instead of internal energy of the system. However, there are still many open questions about the thermodynamics of de Sitter space-time remaining to be settled down. The definition of the black hole temperature associated with the surface gravity fails in the absence of the asymptotically-flat limit. So that the authors[11-12] proposed that the normalized temperature should be used. Moreover, multi-horizons appear in the situation of de Sitter space-time and the cosmological horizon plays a crucial role in the studying of the black hole thermodynamics. It was shown that the mass independent entropy relation of the multi-horizons may be helpful for us to understand the microscopic details of the gravity system,[13-14] which might imply that the cosmological horizon might effect the thermodynamics system, and the thermodynamic properties might not be independent anymore. The thermodynamic properties might be related to each other. This suggests us the total entropy of the de Sitter black hole should not be considered easily as the sum of the entropies on the two horizons. We should also consider the entanglement part to the total entropy.
Because of the different first laws of different horizons in Eq. (1), the temperatures of the two horizons are not equal generally, which means the whole Schwarzschild-de Sitter (SdS) system cannot be in equilibrium thermodynamically.
If we consider an observer who is located between the event horizon and cosmological horizon, she/he must be interacted by these two hyper-surfaces since there is a difference of the bare temperatures of two horizons. Normally, the effect is negligible when the event horizon and cosmological horizon are separated far away. But it is worth remarking that the problem is inevitable when the two horizons merge in the Nariai limit, which the two horizons are very close to each other.
In this work, we simply assume an extra term from the contribution of the correlations of the two horizons should be included in the total entropy of the black hole like $S=S_h+S_c+S_{\rm extra}$. Based on the works of Refs. [11] and [12]. Now we investigate the Schwarzschild-de Sitter (SdS) black hole to find the correlation $S_{\rm extra}$.
2 Entropy with Extra Term Correction
The line element of the SdS black hole is given by
$ ds^2=-f(r)dt^2+\frac{1}{f(r)}dr^2+r^2d\theta^2+r^2\sin^2\theta d\varphi^2 ,$
where $f(r)$ is
$ f(r)=1-\frac{2M}{r}-\frac{\Lambda r^2 }{3} .$
The horizons of the SdS space-time is followed from the equation $f(r)=0$, which call for $0<\Lambda M^2 / 9 <1$.[17] For a de Sitter space-time, there are three roots of this function normally. One is negative and two are positive. We just investigate the two positive roots. The smaller positive one $r_h$ is the event horizon, and the bigger positive one $r_c$ is the cosmological horizon.
By taking the cosmological constant $\Lambda$ as the pressure in the extended phase space and considering the connection between the two roots, one could derive the effective volume and the effective thermodynamic quantities. Then the corresponding first law of black hole thermodynamics as
$ dM = T_{\rm eff}dS + \Phi_{\rm eff}dQ + V_{\rm eff}d\Lambda ,$
where $V_{\rm eff}$ is the thermodynamic volume between the black hole horizon and the cosmological horizon, i.e.
$ V_{\rm eff}=\frac{4\pi}{3}(r_c^3-r_h^3)=\frac{4\pi}{3}r_c^3(1-x^3) ,$
where $x \equiv r_h / r_c$. We assume the total entropy with extra term as
$ S=S_h+S_c+S_{\rm extra}=\pi r_c^2[1+x^2+g(x)] . $
The undetermined function $g(x)$ in Eq. (6) represents the extra contribution from the correlations of the two horizons. If we find the exact formulation of $g(x)$, we could get the modified total entropy.
The mass $M$ now could be written with $r_c$ and $x$ as
$ M=\frac{x r_c (1+x)}{2(1+x+x^2)} . $
As mentioned above, the temperatures on both horizons are not equal normally, which means that the system cannot be in equilibrium thermodynamically. However, in Nariai space-time, the temperatures on both horizons are regarded as the same when the two horizons coincide[17] with each other. This gives us a way to find the expression of $g(x)$.
According to Eqs. (4) and (7), and notice that $Q=0$ in SdS black hole and $\Lambda$ is a constant, one could get the effective temperature $T_{\rm eff}$ in Eq. (4) as
$ T_{\rm eff} =(\frac{\partial M}{\partial S})_{Q,\Lambda} =\frac{({1}/{r_c})({\partial M}/{\partial x})_{r_c}(1-x^3)+({\partial M}/{\partial r_c})_x x^2}{2\pi r_c [(2x+g'(x))(1-x^3)/2+(1+x^2+g(x))x^2]} \\ =-\frac{1+x-2x^2+x^3+x^4}{2\pi r_c {1+x+x^2[-2x(1+x)-2x^2g(x)+(-1+x^3)g'(x)]}} . $
This effective temperature stands for the region of the effective volume, not the horizons. If the effective temperature of region are different with the temperatures of the horizons, there will exist heat exchanges with both horizons.
The temperature of event horizon $T_h$ and temperature of cosmological horizon $T_c$ are[17]
$ T_h=\frac{1-\Lambda r_h^2}{4 \pi r_h}=\frac{1-\Lambda x^2 r_c^2}{4 \pi x r_c} , $
$ T_c=-\frac{1-\Lambda r_c^2}{4 \pi r_c} .$
In the Nariai space-time,[17] the two temperatures become approximately equal, i.e. $T_h \approx T_c$. Then the cosmological horizon under this condition is
$ r_c=\pm\frac{\sqrt{1+x}}{\sqrt{x \Lambda + x^2 \Lambda}} . $
We take the positive one. Then the temperatures $T_h$ and $T_c$ are now
$ T_h=T_c=\frac{(1-x)\sqrt{1+x}\Lambda}{4 \pi \sqrt{x(1+x)\Lambda}} . $
Under the Nariai limit, all the three temperatures, i.e. $T_{\rm eff}$, $T_h$, and $T_c$, should be equal with each other. One could also notice that when $r_h=r_c$, which means that $x=1$, the temperature would be zero. This corresponds to the extremal black hole.[18-19] From Eqs. (8) and (12), we have
$ -\frac{1+x-2x^2+x^3+x^4}{2\pi r_c \{1+x+x^2[-2x(1+x)-2x^2g(x)+(-1+x^3)g'(x)]\}} =\frac{(1-x)\sqrt{1+x}\Lambda}{4 \pi \sqrt{x(1+x)\Lambda}} .$
Solving Eq. (3), we could get
$ g(x)=(-1+x^3)^{2/3} C_1+\frac{1}{15 r_c \Lambda} \bigl \{-15 r_c(1+x^2)\Lambda+ \frac{3(3-6x-4x^2+x^3+2x^4)\sqrt{x(1+x)\Lambda}}{\sqrt{1+x}(-1+x^3)} \\ +\frac{1}{\sqrt{1+x}}(1-x^3)^{2/3}\sqrt{x(1+x)\Lambda}\bigl[9 F(1/6,2/3,7/6,x^3)+ 2 x F(1/2,2/3,3/2,x^3)\bigr]\bigr\} ,$
where $C_1$ is an integral constant and $F$ is confluent hypergeometric function. From the constrain $g(0)=0$, one could get
$ C_1=-(-1)^{1/3} . $
Finally, we get $g(x)$ as
$ g(x) =(-1+x^3)^{2/3} (-(-1)^{1/3})+\frac{1}{15 r_c \Lambda} \Bigl\{-15 r_c(1+x^2)\Lambda+ \frac{3(3-6x-4x^2+x^3+2x^4)\sqrt{x(1+x)\Lambda}}{\sqrt{1+x}(-1+x^3)} \\ + \frac{1}{\sqrt{1+x}}(1-x^3)^{2/3}\sqrt{x(1+x)\Lambda} [9 F(1/6,2/3,7/6,x^3)+ 2 x F(1/2,2/3,3/2,x^3)]\Bigr\} .$
Substituting Eq. (16) into Eq. (8) and taking $r_c=1$, $\Lambda=0.5$, one could get the plot of $T_{\rm eff}\sim x$ in Fig. 1. We could see in Fig. 1 that the effective temperature is always positive. The temperature is descending and goes to zero when $x \rightarrow 1$. This is different with Reissner-Nordstr\"{o}m-de Sitter in Ref. [11] and Kerr-de Sitter in Ref. [12].
Fig. 1 Plot of $T_{\rm eff}\sim x$ with $r_c=1$ and $\Lambda=0.5$. |
The plot of total entropy with $x$ is shown in Fig. 2. In Fig. 2, the blue dashed line is the traditional sum rule of total entropy $S=S_h+S_c$ with $x$. When $x \rightarrow 1$, the entropy goes to a finite value. While the orange line is the total entropy with extra term of the SdS black hole in Eq. (6), which shows that the entropy will diverge as $x \rightarrow 1$. Besides, one could see the entropy is monotonically increasing with the increase of $x$.
When $x$ is very small, the temperature will be very high. The black hole might have heat exchange with the environment. Figure 3 shows the relationship between mass $M$ and $x$.
Fig. 2 (Color online) Plot of $S \sim x$ with $r_c=1$ and $\Lambda=0.5$. |
Fig. 3 Plot of $M \sim x$ with $r_c=1$ and $\Lambda=0.5$. |
From Figs. 1 and 3 one could see that when the mass is descending, the two horizons will be separated far away and the effective temperature is increasing. This means that if we regard the black hole as pure energy, when the black hole radiates some energy, the temperature will be getting higher, which indicate that the specific heat of the black hole should always be negative, and the black hole could not be thermodynamically stable. From the general definition of heat capacity $C={\partial M}/{\partial T}=T({\partial S}/{\partial T})$,[12] we have
$ C_{\rm eff}=\frac{\partial M}{\partial T_{\rm eff}}=T\frac{\partial S}{\partial T_{\rm eff}} . (17) $
Substituting Eqs. (6), (8), (16) into Eq. (17), we get the plot of $C_{\rm eff} \sim x$ in Fig. 4.
Fig. 4 (Color online) Plot of $C_{\rm eff} \sim x$ with $r_c=1$ and $\Lambda=0.5$. The blue dashed line stands for the $S=S_h+S_c$ case, the orange line stands for the $S=S_h+S_c+S_{\rm extra}$ case. |
One could see that the heat capacity for the $S=S_h+S_c$ case and the $S=S_h+S_c+S_{\rm extra}$ case are always negative. This result could also be seen in Refs. [20], [21], and [22]. However, when $x \rightarrow 1$, The $C_{\rm eff}$ in $S=S_h+S_c+S_{\rm extra}$ case will be divergent rather than to zero, which is different with Ref. [22]. This means that there is no thermal equilibrium.
3 Summary
In this note, we reconsidered the thermodynamics of Schwarzschild-de Sitter black hole. The regime between the event horizon and cosmological horizon are regarded as the total thermodynamical system, in which these two horizons are entanglement with each other. The observer located at the point between the two horizons will inevitably be affected by the thermal flux because of the existence of difference in temperatures of the horizons. So that it is reasonable for us to redefine the effective thermodynamics quantities.
We obtained the effective temperature of the system, and showed that there is an extra entropy term from the correlation between the event horizon and the cosmological horizon rather than the sum of the entropies. As we can clearly see from the plot of $S(x)\sim x$ that the extra entropy becomes larger and the specific heat would be divergent rather than close to zero as the two horizons get closer and can not be ignored when the Nariai limit is taken. The similar result could also be find in Refs. [11] and [12].
The extra term of entropy might offer an explanation of effective temperatures on the two horizons that have the same value in the Nariai case. And it might bring a physical explanation or motivation for effective first law of thermodynamics.