# The Nonequilibrium Back Reaction of Hawking Radiation to a Schwarzschild Black Hole

Authors: He Wang; Jin Wang
Journal: Advances in High Energy Physics
 (2020)

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## Abstract

We investigate the nonequilibrium back reaction on the Schwarzschild black hole from the radiation field. The back reactions are characterized by the membrane close to the black hole. When the membrane is thin, we found that larger temperature difference can lead to more significant negative surface tension, larger thermodynamic dissipation cost, and back reaction in energy and entropy as well as larger black hole area. This may be relevant to the primordial black holes in early universe. Moreover, our nonequilibrium model can resolve the inconsistency issue of the black hole back reaction under zero mass limit in the equilibrium case. In the thick membrane case, the nonequilibrium back reaction is found to be more significant than that in the thin membrane case. The nonequilibrium temperature difference can increase the energy and entropy loss as well as the thermodynamic dissipation of the black hole and the membrane back reactions. The nonequilibrium dissipation cost characterized by the entropy production rate appears to be significant compared to the entropy rate radiated by the black hole under finite temperature difference. This may shed light on the black hole information paradox due to the information loss from the entropy production rate in the nonequilibrium cases. The nonequilibrium thermodynamic fluctuations can also reflect the effects of the back reactions of the Hawking radiation on the evolution of a black hole.


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## Body

## 1. Introduction

The Hawking radiation has been suggested to emerge near the black hole event horizon with a thermal spectrum [1]. The radiated particles carry energy momentum so that the radiation can have a back reaction on the black hole. One way of quantifying this back reaction is to find a rational renormalized energy-momentum tensor 0∣Tμv∣0 and regard it as a source. In this way, one can study the back reaction by solving the semiclassical Einstein equation approximatively as Gμv=−k0∣Tμv∣0 [2]. The back reaction and/or quantum corrections will modify the Hawking temperature and entropy (quantum corrections and back reaction are often intertwined) [3–6]. However, we often have poor knowledge about the source 0∣Tμv∣0. Therefore, this dynamical approach is often limited. On the other hand, a thermodynamic approach can be used to study the black hole back reaction problem without the direct use of 0∣Tμv∣0. The thermodynamic approach has been applied to quantify the back reaction to the Schwarzschild black hole [7, 8], Kerr black hole, and Kerr-Newmann black hole [9, 10].Both dynamical and thermodynamic approaches are under a precondition: black hole is in equilibrium with the radiation. However, due to a negative thermal capacity, it is difficult for black hole to maintain the equilibrium unless the black hole is in the box. A negative thermal capacity can magnify the small temperature fluctuations, leading to thermodynamic instability.In this study, we investigate the nonequilibrium back reaction of the radiation field on the Schwarzschild black hole by considering both the thin membrane and thick membrane cases. The nonequilibriumness is characterized by the temperature difference between the black hole and the radiation field,∆T. We found that in general the nonequilibrium effects characterized by the temperature difference can increase the energy and entropy loss as well as the entropy production rate of the black hole and the membrane back reactions. We uncover that the nonequilibrium back reaction can lead to significant changes in the energy, entropy of some primordial black holes created in the early universe for the thin membrane. We also found that larger temperature difference can lead to more significant negative surface tension and back reaction as well as larger black hole area. Moreover, our nonequilibrium model can resolve the inconsistency issue of the black hole back reaction under zero mass limit in the equilibrium case. As the mass of the black hole MBH⟶0, the equilibrium dressed energy and entropy of the black hole do not appear to vanish. By considering the nonequilibrium scenario in the present case, we found that as MBH⟶0, both Mdressed and Sdressed will vanish. This provides another way of resolving the issue of inconsistency of the black hole back reaction under zero mass limit without using the quantum correction. This leads to the conclusion that our nonequilibrium model is more natural and physically reasonable than the equilibrium model for the black hole back reaction.On the other hand, the associated Hawking temperatureT=1/8πMBH is divergent at zero black hole mass or size. Both the equilibrium and the nonequilibrium scenarios have a temperature divergence with or without the back reaction. Due to the quantum uncertainty principle, the size of the black hole cannot exactly approach to zero. This is bounded by the large energy/energy-momentum fluctuations at the small black hole size. Therefore, one can avoid such singularity point where the radius of the black hole approaches zero r⟶0 and the associated temperature approaches infinity T⟶0 when considering quantum effects. The quantum corrections can give rise to well-behaved temperature and entropy in the MBH⟶0 limit [3]. For the thick membrane case, we consider the case where the temperature varies uniformly in the membrane. The back reaction from the thick membrane is found to be more significant compared to that in the thin membrane case. Importantly, we quantified the nonequilibrium thermodynamic dissipation represented by the entropy production rate for both the thin membrane and thick membrane which increases significantly as the temperature difference increases. We show that once there is a finite temperature difference, the nonequilibrium entropy production rate appears to be significant compared to the entropy rate radiated by the black hole. The nonequilibrium entropy production rate can thus have an impact on the black hole evaporation rate. This may be significant for black hole information paradox due to the information loss by the entropy production rate originated from the nonequilibriumness characterized by the temperature difference between the black hole and the radiation environment. We explored the nonequilibrium thermodynamic fluctuations of the black hole, which can reflect the effects of the back reactions of the Hawking radiation on the evolution of a black hole evaporation.In Section 2, we firstly developed a thermodynamic model for studying back reaction, and then, we mainly focus on the thin membrane case. In Section 3, we consider the thick membrane case when the temperature varies uniformly within the membrane. We found the back reaction to the black hole mass, the entropy, and the nonequilibrium thermodynamic cost characterized by the entropy production rate can be significant compared to that of the black hole and its evaporation rate, respectively, under the finite temperature difference. In Section 4, nonequilibrium thermodynamic fluctuations of the black hole are studied, and we found the more massive the black hole is, the weaker its back reaction is. In Section 5, we give a conclusion for our study.


## 2. Back Reactions to the Black Holes: Thin Membrane Case

### 2.1. Treat the Back Reaction as a Membrane: Equilibrium Scenario

In this subsection, we briefly review the thermodynamic approach to back reaction on the Schwarzschild black hole at an equilibrium state [7, 8]. Let us consider a Schwarzschild black hole in a large adiabatic bulk volume filled with thermal radiation field, as shown in Figure 1. One can approximately assume that the radiation field is filled in the whole universe, and its temperature is unchanged during the dynamical processes. The bulk is connected to a piston in the asymptotically flat regime by a narrow tube. The piston can do work on the radiation field.Figure 1
The illustration of a model for back reaction with a Schwarzschild black hole in a bulk volume surrounded by the radiation field connected to a piston to the outside on the edge.The total volume change of the radiation field can be described asδV=δVBH+δVex.δVBH represents the volume change of the black hole, and δVex represents the volume change of the radiation field. Then, the total energy and entropy of the system including the back reaction are E=MBH+MR+MHandS=SBH+SR+SH, respectively. MBH represents the mass of black hole, which is also the energy of black hole in the natural units. MR=aT4V represents the energy of radiation field where a represents a positive definite coefficient and T=1/8πMBH represents the temperature of both the black hole and the radiation field. SBH=4πMBH2 represents the entropy of black hole, and SR=4/3aT3V represents the entropy of the radiation field where V is the volume of the radiation field. MH and SH represent the energy and entropy of back reaction, respectively.From the above definitions and analysis, one can easily obtain that the change of the energy to be:(1)dE=dMH−TdSH−13aT4dV=dMH−TdSH−13aT4dVBH−13aT4dVex.The above expression is obtained since the system is assumed to be in an adiabatic bulk. The first law of thermodynamics for such a system readsdE=−1/3aT4dVex, so dMH=TdSH+1/3aT4dVBH.The back reaction of the radiation field to the black hole can be described by a 2-dimensional membrane with energyMH and temperature T. The rational is as follows. The first law of thermodynamics of the membrane reads
(2)dMH=TdSH+σdA,where A represents the area of the membrane and σ represents the surface tension of the membrane. If σdA=1/3aT4dVBH, a natural inference is that the back reaction can be viewed as a membrane with temperature T. Therefore, to investigate the back reaction to the Schwarzschild black hole from the thermal radiation under thermal equilibrium in the curved spacetime, one can consider a thermodynamic system composed of a Schwarzschild black hole, a thermal radiation, and an effective two-dimensional thermodynamic membrane near the event horizon in flat spacetime. The membrane is formed by and physically equivalent to the back reaction.Let us assume that the membrane is near to the event horizon, thereforeA also represents the area of the horizon, thus σ=εa/24πT3 and ε is a dimensionless constant. The two sides of the membrane are connected to the black hole and radiation field, respectively. Moreover, this viewpoint will greatly simplify the thermodynamics of curved spacetime. Now the system contains three subsystems: the black hole, the membrane attached to horizon, and the radiation field, so that one can easily derive the energy and entropy of the membrane (or the back reaction) by using thermodynamic relations. The free energy of the membrane is given as FH=MH−TSH, whose differential form reads as
(3)dFH=−SHdT+σdA.From the differential expressions of the energy and free energy from the above expression with respect to the membrane, one gets(4)MH=4ε3a8π3MBH−1+CE,SH=2εa8π2+CS.CE and CS are two integral constants. The energy and entropy of black hole with back reaction become
(5)Mdressed=MBH1+4ε3a8π3MBH−2+CE,Sdressed=4πMBH21+4εa8π3MBH−2+CS,respectively. At certain condition (choose the proper integration constant, i.e., CE=0 and ε=3C0, the above expressions for the energy and entropy considering back reactions exactly correspond to the result of York’s [2]. There are two masses now: a naked black hole mass (without back reaction) MBH and a dressed black hole mass (with back reaction) Mdressed=MBH+MH.Now the temperature ambiguity may appear. Which one of the above masses is related to the equilibrium temperature of the system or further, the temperature of the black hole? Reference [7] shows that only the naked mass of the black hole determines the equilibrium temperature of the system. According to T=1/8πMBH, one can rewrite it as
(6)T≈18πMdressed1+4ε3a8π3Mdressed−2.Ifε=k0+12π2/10a−1, it is exactly what York obtained [2]
(7)T=18πMdressed1+k0+123840πMdressed−2.We have reviewed the discussions on the effectiveness of thermodynamics approach to back reaction. The main idea is to consider back reaction as a membrane attached to the black hole and study the thermodynamics of the system, rather than directly calculating through the semiclassical Einstein equationGμv=−k0∣Tμv∣0. In other words, one can just study the back reaction of the black hole through an effective membrane in flat Minkowski spacetime without considering the gravity or curved spacetime explicitly, giving equivalent thermodynamics. However, if we take a limit MBH⟶0, it appears that Mdressed and Sdressed do not vanish. This creates an obvious inconsistency issue here with the nonzero back reaction when the black hole is not present any more. In what follows, we show that this inconsistency issue of the black hole back reaction under zero mass limit in the equilibrium scenario can be resolved by considering nonequilibrium effect. Therefore, a natural question becomes then: what is the effect of the membrane (back reaction) in nonequilibrium case between the black hole and the radiation field?


### 2.2. Nonequilibrium Surface Tension

To consider the more general nonequilibrium cases, let us assume that the radiation field has the temperatureTR, and the black hole (horizon) has the temperature TBH, while the temperature of the membrane is TH. If we consider that the membrane is then and very close to the black hole surface, we can set TH≅TBH. In this study, we consider near but away from the equilibrium case. We assume that the black hole, the membrane, and the radiation field have the same temperature at the beginning TBH=TH=TR. As the time passes by, TBH and TR can become different (caused by the thermodynamic fluctuations and by black hole evaporation from the Hawking radiation), but TH and TBH approximately keep the same, TH≅TBH. We also assume that three subsystems (black hole and the membrane and the radiation field) are all at local equilibrium although the whole system is not at equilibrium. The irreversible processes caused by the heat or radiation transfer energy can emerge between the membrane and the radiation field. Therefore, one obtains:
(8)dMBH=TBHdSBH,(9)dMH=THdSH+σdA,(10)dMR=TRdSR−PdV.Here,MBH represents the energy of the black hole, MH represents the energy of the back reaction, and MR represents the energy of the radiation field. P=1/3aTR4, P represents the pressure of the black body radiation, TR represents the temperature of the radiation field, a represents a positive definite coefficient, and σ represents the equivalent tension coefficient on the membrane. Equation (8) is derived from the first law of thermodynamics of black hole [11].In a large adiabatic bulk, the total changes of the energy (internal energy) of the whole system are given as:(11)dMBH+MH+MR=−PdVex.From (8), (9), and (10), by using TH≅TBH(12)PdVBH=TBHdSBH+dSH+dSR−∆TdSR+σdA.Here,∆T=TBH−TR. Assuming that the underlying process is adiabatic, then, the total change of the reversible entropy is dSBH+dSH+dSR=0. Therefore, equation (12) turns into
(13)σ=PdVBH+∆TdSRdA=1/3aTR4dVBH+4/3aTR3∆TdVdA.Here,dV=dVBH+dVex. Consider the situation that dVex is far less than dVBH (it is reasonable as long as box is very large). In this way, we can omit dVex. We can see that in ordinary units
(14)σ=−aћcTR424πkBTBH−aћcTR3∆T6πkBTBH.Figure2 shows the variation of the tension coefficient with respect to the temperature difference. The tension coefficient σ has a negative value. The tension coefficient of the membrane of the black hole decreases (the absolute value increases) as the temperature difference increases as shown in Figure 2. When ∆T=0, we recover the result of equilibrium state in [7].Figure 2
Shows the relationship between the tension coefficient of the membrane of the black hole and temperature difference with an initial temperature5.6×1030k (for the rational, we choose such a high temperature, see below).We found that the larger the mass of the black hole, the smaller the absolute value of the surface tension coefficient, namely, the weaker the back reaction is. For various masses of black holes, the surface tension coefficients have the same trend as the temperature difference increases,σ always decreases as shown in Figure 2. The absolute value becomes larger, indicating that the larger the temperature difference is, the more significant the surface tension is and therefore more significant the effect of the membrane is, i.e., back reaction is more prominent. The surface tension coefficient is negative. This indicates that if the area of the membrane becomes larger, the free energy of the membrane is reduced. Due to the existence of the temperature difference, the absolute value of the tension coefficient in the nonequilibrium scenario is significantly larger than that in the equilibrium state. While the Hawking radiation has a tendency to shrink the area of the black hole horizon, the back reaction has a tendency to enlarge the area. This might be why the surface tension coefficient is negative. The negative tension coefficient may also be expected due to the intrinsic gravitational interactions with negative heat capacity.


### 2.3. The Reversible Entropy Changes of the Membrane and the Black Hole against Temperature Difference

The differential expression of the free energy on the membrane is given as(15)dFH=−SHdTBH+σdA.SHis the entropy caused by the exchange of matter and energy between the membrane and the two sides. From (15), according to the Maxwell relation, we can get
(16)SH=−∂σ∂TBHA=aTR4ћ3c332π2TBH4kB3SH is the reversible entropy differentiating from TR (temperature of the black hole in equilibrium state) to TBH (temperature of the black hole at nonequilibrium state). Black holes of various masses have the same trend as the temperature difference becomes larger, and the reversible entropy gradually decreases. Because of the temperature difference, the back reaction is distinctly different from the equilibrium case. The power emitted causes the mass to decrease at the rate dMBH/dt=−ћc4/G2α/MBH2, where α is a numerical coefficient. For MBH≫1017g, α=2.001×10−4, for 5×1014g≪MBH≪1017g, α=3.6×10−4 [12]. We can obtain ṠBH=−8παkB2/ћTBH≫ṠH=−64παGaTR4ћ/TBH. That means the reversible entropy change of the back reaction is not significant compared to the black hole’s, for a black hole with such mass. From (16), the total entropy of the black hole with the membrane (only consider reversible part) is Sdressed=ћc5/16πGkBTBH21+aTR4ћ2G/2πc2TBH2kB2. If the back reaction has a significant effect, it requires the mass of black hole to be around the magnitude of ћ4c4a/128Gπ3kB4≈1.4×10−10kg. When the mass of black hole is much larger than 1.4×10−10kg, the entropy of the black hole is much larger than the reversible entropy on the membrane, so the effect of back reaction is weak. For a primordial black hole (PBH, suggested by Hawking [13]) with the minimum mass whose mass is Mp=ћc/G=2.17×10−8kg, which is formed in the early universe and evaporated rapidly over a very short life, its back reaction is more prominent.Figure3(a) shows that the reversible entropy of the membrane decreases as the temperature difference increases while Figure 3(b) shows that the entropy of the black hole decreases as the temperature difference increases. Thus, the nonequilibrium effects characterized by the temperature difference can reduce the reversible entropy of the black hole. The back reaction of the membrane to the black hole on the entropy change is relatively small.Figure 3
(a) shows the relationship between the reversible entropy of membrane and temperature difference with an initial temperature5.6×1030k (it was only in the very early universe that such high temperatures were possible). Black holes of various masses have the same trend as the temperature difference becomes larger, and the reversible entropy monotonically decreases. (b) shows change of entropy of the black hole with an initial temperature 5.6×1030k.
(a)(b)


### 2.4. The Energy Changes of the Membrane and the Black Hole against Temperature Difference

From equation (9), we can derive that
(17)MH=−∂σ∂TBHTBHA+σA=aTR4ћ3c348π2TBH3kB3−aTR3ћ3c3∆T24π2TBH3kB3.The energy of the black hole with the membrane is:(18)MH+MBH=aTR4ћ3c348π2TBH3kB3−aTR3ћ3c3∆T24π2TBH3kB3+c5ћ8πGkBTBH=c5ћ8πGkBTBH1+aTR4ћ2G6πc2TBH2kB2−aTR3ћ2G∆T3πc2TBH2kB2.Figure4(a) shows that the energy of the membrane monotonically decreases as the temperature difference increases. Figure 4(b) shows the energy of the black hole also decreases as the temperature difference increases. The nonequilibrium effect characterized by the temperature difference can reduce the reversible entropy and the energy of the black hole. The back reaction is relatively small compared to the black hole energy change.Figure 4
(a) shows the energy of membrane monotonically decreases as the temperature difference increases. (b) shows the relationship between the energy of the black hole and the temperature difference with an initial temperature5.6×1030k.
(a)(b)It is easy to check if the back reaction has a significant effect; it also requires mass of the black hole to be around the magnitude ofћ4c4a/128Gπ3kB4≈1.4×10−10kg. We can conclude that only small mass primordial black hole can have a more significant back reaction. The energy and reversible entropy of the back reaction relative to the energy and entropy of the black holes can be measured by ∝1/MBH2; the more massive the black hole is, the weaker its back reaction is. Moreover, in our nonequilibrium scenario, we can resolve the inconsistency issue of the black hole back reaction under zero mass limit appeared in the equilibrium scenario. For example, if we take a limit MBH⟶0 (i.e., TBH⟶∞), Mdressed and Sdressed will naturally vanish in our nonequilibrium model. We see that the nonequilibriumness can influence the evolution and back reaction of the black hole. Our nonequilibrium model is more natural and physically reasonable than the equilibrium model for the black hole back reaction and evolution.


### 2.5. The Entropy Production Rate of the Membrane and the Rate of Entropy Change for the Black Hole against Temperature Difference

For a black hole with an ideal black body temperatureTBH at the horizon immersed in a radiation field with a temperature TR, the entropy production rate per unit surface area per unit time on the black body surface caused by temperature difference and the thermal radiation is given as Θ=−4/3ωTR3+4/3ωTBH3+ωTR4−ωTBH4/TBH, ω is the Stefan-Boltzmann constant [14, 15]. The entropy production rate formula above can be understood as the total entropy production from the system (the black hole) and environment (the radiation field). The last two terms represent the difference between the input radiation energy flux and the output black hole energy flux giving rise to the entropy production of the black hole. The first two terms represent the difference between the input black hole entropy flux and the output radiation field entropy flux giving rise to the entropy production of the radiation field. Let us assume that the thermal radiation process relevant to the black hole only occurs near the horizon, and the entropy production of the surface of the black hole per unit time is dSi/dt=ΘA, where A is the area of the horizon. We can approximately estimate the increase of irreversible entropy. According to TBH=ћc3/8πGkBMBH and A=16πG2MBH2/c4, then, we can obtain the nonequilibrium entropy production rate to be:
(19)dSidt=ћ2c2ω4πkB2TBH2−43TR3+43TBH3+TR4−TBH4TBH.Black holes under various masses have the same trend of the changes as the temperature difference becomes larger. As the temperature difference becomes larger, the nonequilibrium thermodynamic dissipation quantified by the entropy production rate as well as the entropy change rate of the black hole increases significantly.Importantly, we found that once there is a finite temperature difference, the additional entropy change caused by the radiation energy transfer of the total system in a unit time is significant as shown in Figure5. Furthermore, as seen clearly in Figure 5, the nonequilibrium effect characterized by the temperature difference can enhance both the dissipation cost rate of the membrane and the entropy change rate of the black hole.Figure 5
Shows the variation of the irreversible entropy production rate (red line) and the rate of entropy change of black hole (green line) with temperature difference with an initial temperature 2.73 k. We find entropy production rate is significant compared to the rate of entropy change of a black hole at a finite temperature; it also holds for other black holes with different mass.For the entropy change of the whole system, the effect of the irreversible nonequilibrium entropy production rate from radiation energy transfer appears to be significant compared to the rate of the entropy change of a black hole at a finite temperature difference as shown in Figure5. The nonequilibrium entropy production rate can thus have an impact on the black hole evaporation rate. This may provide some insights towards the black hole information paradox since there can be significant information loss due to the entropy production rate in the nonequilibrium cases from the temperature difference between the black hole and the radiation environment.


## 2.1. Treat the Back Reaction as a Membrane: Equilibrium Scenario

In this subsection, we briefly review the thermodynamic approach to back reaction on the Schwarzschild black hole at an equilibrium state [7, 8]. Let us consider a Schwarzschild black hole in a large adiabatic bulk volume filled with thermal radiation field, as shown in Figure 1. One can approximately assume that the radiation field is filled in the whole universe, and its temperature is unchanged during the dynamical processes. The bulk is connected to a piston in the asymptotically flat regime by a narrow tube. The piston can do work on the radiation field.Figure 1
The illustration of a model for back reaction with a Schwarzschild black hole in a bulk volume surrounded by the radiation field connected to a piston to the outside on the edge.The total volume change of the radiation field can be described asδV=δVBH+δVex.δVBH represents the volume change of the black hole, and δVex represents the volume change of the radiation field. Then, the total energy and entropy of the system including the back reaction are E=MBH+MR+MHandS=SBH+SR+SH, respectively. MBH represents the mass of black hole, which is also the energy of black hole in the natural units. MR=aT4V represents the energy of radiation field where a represents a positive definite coefficient and T=1/8πMBH represents the temperature of both the black hole and the radiation field. SBH=4πMBH2 represents the entropy of black hole, and SR=4/3aT3V represents the entropy of the radiation field where V is the volume of the radiation field. MH and SH represent the energy and entropy of back reaction, respectively.From the above definitions and analysis, one can easily obtain that the change of the energy to be:(1)dE=dMH−TdSH−13aT4dV=dMH−TdSH−13aT4dVBH−13aT4dVex.The above expression is obtained since the system is assumed to be in an adiabatic bulk. The first law of thermodynamics for such a system readsdE=−1/3aT4dVex, so dMH=TdSH+1/3aT4dVBH.The back reaction of the radiation field to the black hole can be described by a 2-dimensional membrane with energyMH and temperature T. The rational is as follows. The first law of thermodynamics of the membrane reads
(2)dMH=TdSH+σdA,where A represents the area of the membrane and σ represents the surface tension of the membrane. If σdA=1/3aT4dVBH, a natural inference is that the back reaction can be viewed as a membrane with temperature T. Therefore, to investigate the back reaction to the Schwarzschild black hole from the thermal radiation under thermal equilibrium in the curved spacetime, one can consider a thermodynamic system composed of a Schwarzschild black hole, a thermal radiation, and an effective two-dimensional thermodynamic membrane near the event horizon in flat spacetime. The membrane is formed by and physically equivalent to the back reaction.Let us assume that the membrane is near to the event horizon, thereforeA also represents the area of the horizon, thus σ=εa/24πT3 and ε is a dimensionless constant. The two sides of the membrane are connected to the black hole and radiation field, respectively. Moreover, this viewpoint will greatly simplify the thermodynamics of curved spacetime. Now the system contains three subsystems: the black hole, the membrane attached to horizon, and the radiation field, so that one can easily derive the energy and entropy of the membrane (or the back reaction) by using thermodynamic relations. The free energy of the membrane is given as FH=MH−TSH, whose differential form reads as
(3)dFH=−SHdT+σdA.From the differential expressions of the energy and free energy from the above expression with respect to the membrane, one gets(4)MH=4ε3a8π3MBH−1+CE,SH=2εa8π2+CS.CE and CS are two integral constants. The energy and entropy of black hole with back reaction become
(5)Mdressed=MBH1+4ε3a8π3MBH−2+CE,Sdressed=4πMBH21+4εa8π3MBH−2+CS,respectively. At certain condition (choose the proper integration constant, i.e., CE=0 and ε=3C0, the above expressions for the energy and entropy considering back reactions exactly correspond to the result of York’s [2]. There are two masses now: a naked black hole mass (without back reaction) MBH and a dressed black hole mass (with back reaction) Mdressed=MBH+MH.Now the temperature ambiguity may appear. Which one of the above masses is related to the equilibrium temperature of the system or further, the temperature of the black hole? Reference [7] shows that only the naked mass of the black hole determines the equilibrium temperature of the system. According to T=1/8πMBH, one can rewrite it as
(6)T≈18πMdressed1+4ε3a8π3Mdressed−2.Ifε=k0+12π2/10a−1, it is exactly what York obtained [2]
(7)T=18πMdressed1+k0+123840πMdressed−2.We have reviewed the discussions on the effectiveness of thermodynamics approach to back reaction. The main idea is to consider back reaction as a membrane attached to the black hole and study the thermodynamics of the system, rather than directly calculating through the semiclassical Einstein equationGμv=−k0∣Tμv∣0. In other words, one can just study the back reaction of the black hole through an effective membrane in flat Minkowski spacetime without considering the gravity or curved spacetime explicitly, giving equivalent thermodynamics. However, if we take a limit MBH⟶0, it appears that Mdressed and Sdressed do not vanish. This creates an obvious inconsistency issue here with the nonzero back reaction when the black hole is not present any more. In what follows, we show that this inconsistency issue of the black hole back reaction under zero mass limit in the equilibrium scenario can be resolved by considering nonequilibrium effect. Therefore, a natural question becomes then: what is the effect of the membrane (back reaction) in nonequilibrium case between the black hole and the radiation field?


## 2.2. Nonequilibrium Surface Tension

To consider the more general nonequilibrium cases, let us assume that the radiation field has the temperatureTR, and the black hole (horizon) has the temperature TBH, while the temperature of the membrane is TH. If we consider that the membrane is then and very close to the black hole surface, we can set TH≅TBH. In this study, we consider near but away from the equilibrium case. We assume that the black hole, the membrane, and the radiation field have the same temperature at the beginning TBH=TH=TR. As the time passes by, TBH and TR can become different (caused by the thermodynamic fluctuations and by black hole evaporation from the Hawking radiation), but TH and TBH approximately keep the same, TH≅TBH. We also assume that three subsystems (black hole and the membrane and the radiation field) are all at local equilibrium although the whole system is not at equilibrium. The irreversible processes caused by the heat or radiation transfer energy can emerge between the membrane and the radiation field. Therefore, one obtains:
(8)dMBH=TBHdSBH,(9)dMH=THdSH+σdA,(10)dMR=TRdSR−PdV.Here,MBH represents the energy of the black hole, MH represents the energy of the back reaction, and MR represents the energy of the radiation field. P=1/3aTR4, P represents the pressure of the black body radiation, TR represents the temperature of the radiation field, a represents a positive definite coefficient, and σ represents the equivalent tension coefficient on the membrane. Equation (8) is derived from the first law of thermodynamics of black hole [11].In a large adiabatic bulk, the total changes of the energy (internal energy) of the whole system are given as:(11)dMBH+MH+MR=−PdVex.From (8), (9), and (10), by using TH≅TBH(12)PdVBH=TBHdSBH+dSH+dSR−∆TdSR+σdA.Here,∆T=TBH−TR. Assuming that the underlying process is adiabatic, then, the total change of the reversible entropy is dSBH+dSH+dSR=0. Therefore, equation (12) turns into
(13)σ=PdVBH+∆TdSRdA=1/3aTR4dVBH+4/3aTR3∆TdVdA.Here,dV=dVBH+dVex. Consider the situation that dVex is far less than dVBH (it is reasonable as long as box is very large). In this way, we can omit dVex. We can see that in ordinary units
(14)σ=−aћcTR424πkBTBH−aћcTR3∆T6πkBTBH.Figure2 shows the variation of the tension coefficient with respect to the temperature difference. The tension coefficient σ has a negative value. The tension coefficient of the membrane of the black hole decreases (the absolute value increases) as the temperature difference increases as shown in Figure 2. When ∆T=0, we recover the result of equilibrium state in [7].Figure 2
Shows the relationship between the tension coefficient of the membrane of the black hole and temperature difference with an initial temperature5.6×1030k (for the rational, we choose such a high temperature, see below).We found that the larger the mass of the black hole, the smaller the absolute value of the surface tension coefficient, namely, the weaker the back reaction is. For various masses of black holes, the surface tension coefficients have the same trend as the temperature difference increases,σ always decreases as shown in Figure 2. The absolute value becomes larger, indicating that the larger the temperature difference is, the more significant the surface tension is and therefore more significant the effect of the membrane is, i.e., back reaction is more prominent. The surface tension coefficient is negative. This indicates that if the area of the membrane becomes larger, the free energy of the membrane is reduced. Due to the existence of the temperature difference, the absolute value of the tension coefficient in the nonequilibrium scenario is significantly larger than that in the equilibrium state. While the Hawking radiation has a tendency to shrink the area of the black hole horizon, the back reaction has a tendency to enlarge the area. This might be why the surface tension coefficient is negative. The negative tension coefficient may also be expected due to the intrinsic gravitational interactions with negative heat capacity.


## 2.3. The Reversible Entropy Changes of the Membrane and the Black Hole against Temperature Difference

The differential expression of the free energy on the membrane is given as(15)dFH=−SHdTBH+σdA.SHis the entropy caused by the exchange of matter and energy between the membrane and the two sides. From (15), according to the Maxwell relation, we can get
(16)SH=−∂σ∂TBHA=aTR4ћ3c332π2TBH4kB3SH is the reversible entropy differentiating from TR (temperature of the black hole in equilibrium state) to TBH (temperature of the black hole at nonequilibrium state). Black holes of various masses have the same trend as the temperature difference becomes larger, and the reversible entropy gradually decreases. Because of the temperature difference, the back reaction is distinctly different from the equilibrium case. The power emitted causes the mass to decrease at the rate dMBH/dt=−ћc4/G2α/MBH2, where α is a numerical coefficient. For MBH≫1017g, α=2.001×10−4, for 5×1014g≪MBH≪1017g, α=3.6×10−4 [12]. We can obtain ṠBH=−8παkB2/ћTBH≫ṠH=−64παGaTR4ћ/TBH. That means the reversible entropy change of the back reaction is not significant compared to the black hole’s, for a black hole with such mass. From (16), the total entropy of the black hole with the membrane (only consider reversible part) is Sdressed=ћc5/16πGkBTBH21+aTR4ћ2G/2πc2TBH2kB2. If the back reaction has a significant effect, it requires the mass of black hole to be around the magnitude of ћ4c4a/128Gπ3kB4≈1.4×10−10kg. When the mass of black hole is much larger than 1.4×10−10kg, the entropy of the black hole is much larger than the reversible entropy on the membrane, so the effect of back reaction is weak. For a primordial black hole (PBH, suggested by Hawking [13]) with the minimum mass whose mass is Mp=ћc/G=2.17×10−8kg, which is formed in the early universe and evaporated rapidly over a very short life, its back reaction is more prominent.Figure3(a) shows that the reversible entropy of the membrane decreases as the temperature difference increases while Figure 3(b) shows that the entropy of the black hole decreases as the temperature difference increases. Thus, the nonequilibrium effects characterized by the temperature difference can reduce the reversible entropy of the black hole. The back reaction of the membrane to the black hole on the entropy change is relatively small.Figure 3
(a) shows the relationship between the reversible entropy of membrane and temperature difference with an initial temperature5.6×1030k (it was only in the very early universe that such high temperatures were possible). Black holes of various masses have the same trend as the temperature difference becomes larger, and the reversible entropy monotonically decreases. (b) shows change of entropy of the black hole with an initial temperature 5.6×1030k.
(a)(b)


## 2.4. The Energy Changes of the Membrane and the Black Hole against Temperature Difference

From equation (9), we can derive that
(17)MH=−∂σ∂TBHTBHA+σA=aTR4ћ3c348π2TBH3kB3−aTR3ћ3c3∆T24π2TBH3kB3.The energy of the black hole with the membrane is:(18)MH+MBH=aTR4ћ3c348π2TBH3kB3−aTR3ћ3c3∆T24π2TBH3kB3+c5ћ8πGkBTBH=c5ћ8πGkBTBH1+aTR4ћ2G6πc2TBH2kB2−aTR3ћ2G∆T3πc2TBH2kB2.Figure4(a) shows that the energy of the membrane monotonically decreases as the temperature difference increases. Figure 4(b) shows the energy of the black hole also decreases as the temperature difference increases. The nonequilibrium effect characterized by the temperature difference can reduce the reversible entropy and the energy of the black hole. The back reaction is relatively small compared to the black hole energy change.Figure 4
(a) shows the energy of membrane monotonically decreases as the temperature difference increases. (b) shows the relationship between the energy of the black hole and the temperature difference with an initial temperature5.6×1030k.
(a)(b)It is easy to check if the back reaction has a significant effect; it also requires mass of the black hole to be around the magnitude ofћ4c4a/128Gπ3kB4≈1.4×10−10kg. We can conclude that only small mass primordial black hole can have a more significant back reaction. The energy and reversible entropy of the back reaction relative to the energy and entropy of the black holes can be measured by ∝1/MBH2; the more massive the black hole is, the weaker its back reaction is. Moreover, in our nonequilibrium scenario, we can resolve the inconsistency issue of the black hole back reaction under zero mass limit appeared in the equilibrium scenario. For example, if we take a limit MBH⟶0 (i.e., TBH⟶∞), Mdressed and Sdressed will naturally vanish in our nonequilibrium model. We see that the nonequilibriumness can influence the evolution and back reaction of the black hole. Our nonequilibrium model is more natural and physically reasonable than the equilibrium model for the black hole back reaction and evolution.


## 2.5. The Entropy Production Rate of the Membrane and the Rate of Entropy Change for the Black Hole against Temperature Difference

For a black hole with an ideal black body temperatureTBH at the horizon immersed in a radiation field with a temperature TR, the entropy production rate per unit surface area per unit time on the black body surface caused by temperature difference and the thermal radiation is given as Θ=−4/3ωTR3+4/3ωTBH3+ωTR4−ωTBH4/TBH, ω is the Stefan-Boltzmann constant [14, 15]. The entropy production rate formula above can be understood as the total entropy production from the system (the black hole) and environment (the radiation field). The last two terms represent the difference between the input radiation energy flux and the output black hole energy flux giving rise to the entropy production of the black hole. The first two terms represent the difference between the input black hole entropy flux and the output radiation field entropy flux giving rise to the entropy production of the radiation field. Let us assume that the thermal radiation process relevant to the black hole only occurs near the horizon, and the entropy production of the surface of the black hole per unit time is dSi/dt=ΘA, where A is the area of the horizon. We can approximately estimate the increase of irreversible entropy. According to TBH=ћc3/8πGkBMBH and A=16πG2MBH2/c4, then, we can obtain the nonequilibrium entropy production rate to be:
(19)dSidt=ћ2c2ω4πkB2TBH2−43TR3+43TBH3+TR4−TBH4TBH.Black holes under various masses have the same trend of the changes as the temperature difference becomes larger. As the temperature difference becomes larger, the nonequilibrium thermodynamic dissipation quantified by the entropy production rate as well as the entropy change rate of the black hole increases significantly.Importantly, we found that once there is a finite temperature difference, the additional entropy change caused by the radiation energy transfer of the total system in a unit time is significant as shown in Figure5. Furthermore, as seen clearly in Figure 5, the nonequilibrium effect characterized by the temperature difference can enhance both the dissipation cost rate of the membrane and the entropy change rate of the black hole.Figure 5
Shows the variation of the irreversible entropy production rate (red line) and the rate of entropy change of black hole (green line) with temperature difference with an initial temperature 2.73 k. We find entropy production rate is significant compared to the rate of entropy change of a black hole at a finite temperature; it also holds for other black holes with different mass.For the entropy change of the whole system, the effect of the irreversible nonequilibrium entropy production rate from radiation energy transfer appears to be significant compared to the rate of the entropy change of a black hole at a finite temperature difference as shown in Figure5. The nonequilibrium entropy production rate can thus have an impact on the black hole evaporation rate. This may provide some insights towards the black hole information paradox since there can be significant information loss due to the entropy production rate in the nonequilibrium cases from the temperature difference between the black hole and the radiation environment.


## 3. Nonequilibrium Black Hole Thermodynamics and Back Reaction of the Thick Membrane

Now let us consider the case that the membrane is thick. When the membrane has a thickness, one has to take into account the changes in the temperature on the membrane, from the near black hole side atTBH where TBH is the black hole temperature to near radiation field side TR where TR is the environmental radiation field temperature surrounding the black hole and the back reaction membrane. If the membrane is not very thick, then, it is reasonable to consider the situation where the temperature is linear in the length. The change in the temperature within the membrane is related only to the distance r to the black hole and not to the angle assuming isotropic for simplicity. For the system not too far from equilibrium and we can assume that the temperature within the membrane is linear in distance and can be expressed as T=TBH+r−rBHTR−TBH/L, where L is the thickness of the membrane, rBH is the black hole radius. We assume local equilibrium to be maintained as each layer of a sphere with radius r so that local equilibrium thermodynamics can apply, although the overall system is in nonequilibrium state.


### 3.1. Nonequilibrium Thermodynamics under Local Equilibrium Assumption

For a general thermodynamic system, the second law applies:dS≥dQ/T (dS is the entropy change of the system, dQ is the heat exchange between the system and the environment while T is the temperature), which can be generalized to the open system as dS=deS+diS, where deS represents the change in entropy of the exchange of matter and energy between the system and the environment while diS represents the change in entropy caused by the irreversible processes within the system. Let us consider the following situation. The overall system is in nonequilibrium state while its small subsystem is still in local equilibrium and the equilibrium thermodynamic laws still hold for the subsystems, Tds=du, where s, T, and u are entropy, temperature, and energy of the subsystem, respectively. Therefore, for the whole system, S=∫sdV,U=∫udV.In the simple process of heat transport, the local change of the internal energy is equal to the thermal heat flux in and out of the subsystem:∂u/∂t=−∇∙Jq, where Jq is the heat flow per unit area per unit time. Therefore, ∂s/∂t=1/T∂u/∂t=−1/T∇∙Jq=−∇∙Jq/T+Jq∙∇1/T. In the general cases, under local equilibrium for the subsystems, the entropy change rate is determined by ∂s/∂t=−∇∙Js+⊙, where Js is the entropy flow per unit area per unit time and ⊙ is entropy production rate per unit volume [16]. The entropy change rate of the whole system is given as dS/dt=d/dt∫sdV=∫∂s/∂tdV=∫−∇∙Js+⊙dV=−∮JsdA+∫⊙dV. The first term is the entropy flow from the outside through the surface of the system per unit time. The second term is the sum of the entropy production of all the subsystems per unit time. Comparing the heat transport case with the general entropy change formula under local equilibrium assumption with ∂s/∂t=−∇∙Js+⊙, we can see that Js=Jq/T and ⊙=Jq∙∇1/T.


### 3.2. Nonequilibrium Black Hole Thermodynamics under Local Equilibrium Assumption

In our model, for both ends of the membrane, the input energy flows per unit time and per unit area are given asJBH=ωTBH4 from the black hole and JR=ωTR4 from the radiation field, ω is the Stefan-Boltzmann constant. The outward radiation along the radius is in the positive direction. The radiation energy flow in the membrane is Jq=ωTBH4−ωTR4. Although we assume the local equilibrium at each layer of the radius r, the system overall is in nonequilibrium state with temperature distribution varying in space. The energy flows are assumed mostly through the radiation rather than thermal conduction in our study.The rate of the exchange in entropy between the back reaction membrane system and the environment per unit area in the membrane is given asṡe=−∇∙Jq/T=Jq/T2∇T=Jq∆T/T2L, where ∇T=dT/dr=∆T/Lisfrom our setting T=TBH+r−rBHTR−TBH/L and ∆T=TR−TBH here. The radiation energy flow in the membrane Jq=ωTBH4−ωTR4 is constant analogous to a case of an electric current Jq moving through a resistance of length L under a voltage is ∇1/T. The entropy production rate per unit area is given as ṡi=⊙=Jq∙∇1/T=−Jq∆T/T2L>0. Since the energy flux Jq is a constant in the membrane, we can check that the system entropy of the membrane does not change and is therefore in steady state. Since the total entropy production rate per unit area of the membrane system and the environment is the sum of the system entropy is equal to the entropy change rate of the system and its exchange with the environments (same as the environmental entropy change) per unit area according to the nonequilibrium thermodynamics, the entropy production rate is equal to the entropy exchange rate with the environment or the environmental entropy change rate. For the whole membrane system, we can obtain the total entropy exchange between the membrane system and the environments as Ṡe=∫ṡedV=∫ṡe4πr2dr. Suppose the thickness of the membrane is much less than the radius of the black hole, L≪rBH. Then, we can obtain approximately Ṡe≈ABH∫ṡedr=JqABH1/TBH−1/TR=Jqћ2c2/4πTBH2kB21/TBH−1/TR, where the area of the black hole is given as ABH=ћ2c2/4πTBH2kB2. For a black hole at temperature 2.7 k, its radius is given as rBH=ћc/4πTBHkB=6.68×10−5m, so the thickness of the membrane should be less than 10−7m at least such that 4πrBH+L2=4πrBH2+4πL2+8πrBHL≈4πrBH2. Also, the total entropy production rate of the membrane system is approximately given as Ṡi=∫ṡidV≈Jqћ2c2/4πTBH2kB21/TR−1/TBH>0.


### 3.3. The Rate of Energy and Entropy Change of the Back Reaction on the Black Hole against Temperature Difference

We now explore how the entropy change of the black hole, the energy change, the mass change of the black hole, and the mass or energy of the membrane change with respect to time,ṠBH=−ωTBH3ABH=−ωћ2c2TBH/4πkB2, where ωTBH3 is the entropy flow from black hole and ṀBH=−JBHABH=−ωћ2c2TBH2/4πkB2, ṀH=−ωTBH4ABH−ωTR4×4πrBH+L2, where the energy flux JBH=ωTBH4 is from the black hole and the energy flux JR=ωTR4 is from the radiation field. Using the assumption that the thickness of the membrane is much less than the black hole radius, we obtain ṀH≈−ωTBH2/4π−ωTR4/4πTBH2ћ2c2/kB2.Let us assume that the expression ofṠBH and ṀBH above are approximately valid for the “naked” black hole, and the back reactions of the entropy rate and energy rate from the membrane are given as Ṡe and ṀH, respectively. After considering the back reaction, the effective or dressed entropy rate and energy rate of the black hole are given as Ṡdressed=ṠH+ṠBH, Ṁdressed=ṀH+ṀBH, respectively.When∆T=0, the energy and entropy loss per unit time of “dressed” black hole is equal to that of the “naked” black hole’s as shown in Figures 6(a) and 6(b). The system is in equilibrium. When the temperature difference is larger, the system becomes more nonequilibrium. Compared to the thin membrane case, we find that the thick membrane can lead to more significant back reactions on the black holes. In both Figures 6(a) and 6(b), we see that the energy and entropy change rate of a dressed black hole after taking into account of the back reaction are larger than that of the “naked” black hole, and the distances between the green line and red line increase as ∆T increases. This shows that the back reaction on the energy rate and entropy rate of the black hole is more significant as the temperature difference increases. The evaporation rate of the dressed black hole when taking into account of the thickness of the membrane back reaction is larger than that of the “naked black hole.” On the other hand, from both Figures 6(a) and 6(b), the nonequilibrium effects characterized by the temperature difference can enhance the energy and entropy loss rate of the black hole and the membrane back reaction.Figure 6
(a) and (b) The energy and entropy loss per unit time of “naked” black hole and “dressed” black hole varying with the temperature difference with an initial temperature 2.73 k. Here,∆T=TBH−TR.
(a)(b)


### 3.4. Energy and Entropy Change of the Black Hole due to the Back Reaction against Temperature Difference

We would not only like to see how significant the rate of the back reaction but also the back reaction of the membrane itself is on the black hole. To do so, we need to compare the contribution of the entropy and energy from the membrane towards the black hole. Let us estimate the time scale given the entropy and energy rate of the membrane above. The response of a black hole to a perturbation will be dominated by a set of damped oscillations called quasi-normal modes. The quasi-normal mode frequency corresponds to a natural frequency of a black hole. It is divided into two parts: the real part represents the oscillation, and the imaginary part represents damping [17]. For large damping, i.e., for a large imaginary part of ω, the real part of the frequency approaches a nonzero value, and the imaginary part becomes equally spaced. It is given as Mω=0.04371235+in+1/2 [18]. The constant real part of the quasi-normal frequency can be approximated as wQNM=ln3c3/8πGM. Thus, the natural time scale can be estimated as τ=2π/wQNM=16π2GM/ln3c3=2πћ/ln3kBTBH.In the time intervalτ, ∆Se=Ṡeτ=Jqћ3c2/2ln3kB3TBH31/TBH−1/TR, ∆Si=Jqћ3c2/2ln3kB3TBH31/TR−1/TBH, ∆SBH=−ωћ3c2/2ln3kB3, ∆MBH=−ωћ3c2TBH/2ln3kB3, ∆MH=−ωTBH/2−ωTR4/2TBH3ћ3c2/ln3kB3. We measure the back reaction as ∆Se/∆SBH=TBH4−TR41/TBHTR3−1/TBH4 and ∆MH/∆MBH=1−TR4/TBH4. In the case of thick membrane, the energy and entropy change of back reaction during certain periods are dependent on the temperature difference.When∆T=0, the system is in equilibrium, ∆MH=∆Se=0 in both Figures 7(a) and 7(b). According to Figures 7(a) and 7(b), both ∆MH/∆MBH and ∆Se/∆SBH increase as the temperature difference increases. The back reaction becomes more significant as temperature difference increases. Also, for a black hole with a mass MBH≫5×1014g, the effect of thick membrane Ṡe=Jqћ2c2/4πTBH2kB21/TBH−1/TR is much more significant than that of thin membrane ṠH=−ω2ћ4GTR4/4π2kB4TBHc under finite temperature difference. Under an infinitesimally small temperature difference, the effect of a thick membrane is similar to that of a thin membrane; they are both weak. Therefore, we can conclude that the back reaction from the membrane to the black hole can be significant in terms of the energy (mass) and entropy change. The nonequilibrium effect characterized by the temperature difference can enhance the back reactions in terms of the energy (mass) and entropy change to the black hole.Figure 7
(a) and (b) show∆MH/∆MBH, ∆Se/∆SBH varying with the temperature difference with an initial temperature 2.73 k, respectively.
(a)(b)


### 3.5. Nonequilibrium Entropy Production and Entropy Change of the Black Hole against Temperature Difference

For irreversible nonequilibrium entropy production, once there is a finite temperature difference, the additional entropy change caused by the radiation energy transfer of the total system in a unit time can be significant as shown in Figure8(a). It can be quantified by ∆Si/∆SBH=TBH4−TR41/TBHTR3−1/TBH4.Figure 8
(a) shows the variation of the irreversible entropy production rate (red line) and the rate of entropy change of the black hole (green line) with respect to the temperature difference with an initial temperature 2.73 k. (b) shows the entropy change caused by irreversible entropy production (red line) and the entropy change of the black hole caused by radiation (green line) against the temperature difference with an initial temperature 2.73 k in time intervalτ.
(a)(b)Figure8(a) shows the variation of the irreversible entropy production rate (red line) and the rate of entropy change of black hole (green line) with respect to the temperature difference at initial temperature 2.73 k. We found as the temperature difference increases, both the entropy production rate and rate of entropy change of the black hole increase. This indicates that the nonequilibrium effects characterized by the temperature difference can increase the dissipation cost rate and entropy change rate of the black hole. We also find that the entropy production rate is significant compared to the rate of entropy change of the black hole at a finite temperature difference. This also holds for black holes with different masses. Figure 8(b) shows the entropy change caused by irreversible entropy production rate (red line) and the entropy change of the black hole caused by radiation (green line) with the temperature difference at initial temperature 2.73 k in time interval τ. Notice that the green line in Figure 8(b) is a constant. This is due to the underlying assumption that the entropy (area) of a black hole radiating out in time interval τ is quantized from the quasi-normal mode frequency expression [19]. This shows τ as a natural time of the black hole. We found as the temperature difference increases, both the entropy production and entropy change of the black hole increase. This shows that the nonequilibrium effects characterized by the temperature difference can increase the dissipation cost and entropy change of the black hole. The entropy change caused by irreversible entropy production is significant compared to the entropy change of a black hole caused by radiation at a finite temperature difference. This also holds for black holes with different mass. The nonequilibrium entropy production rate can thus have an impact on the black hole evaporation rate.This indicates that there can be significant entropy or information loss under the nonequilibrium conditions characterized by the temperature difference between the black hole and the radiation environment. This extra loss of the entropy or information may give some insights towards the understanding the source of the black hole information paradox.


## 3.1. Nonequilibrium Thermodynamics under Local Equilibrium Assumption

For a general thermodynamic system, the second law applies:dS≥dQ/T (dS is the entropy change of the system, dQ is the heat exchange between the system and the environment while T is the temperature), which can be generalized to the open system as dS=deS+diS, where deS represents the change in entropy of the exchange of matter and energy between the system and the environment while diS represents the change in entropy caused by the irreversible processes within the system. Let us consider the following situation. The overall system is in nonequilibrium state while its small subsystem is still in local equilibrium and the equilibrium thermodynamic laws still hold for the subsystems, Tds=du, where s, T, and u are entropy, temperature, and energy of the subsystem, respectively. Therefore, for the whole system, S=∫sdV,U=∫udV.In the simple process of heat transport, the local change of the internal energy is equal to the thermal heat flux in and out of the subsystem:∂u/∂t=−∇∙Jq, where Jq is the heat flow per unit area per unit time. Therefore, ∂s/∂t=1/T∂u/∂t=−1/T∇∙Jq=−∇∙Jq/T+Jq∙∇1/T. In the general cases, under local equilibrium for the subsystems, the entropy change rate is determined by ∂s/∂t=−∇∙Js+⊙, where Js is the entropy flow per unit area per unit time and ⊙ is entropy production rate per unit volume [16]. The entropy change rate of the whole system is given as dS/dt=d/dt∫sdV=∫∂s/∂tdV=∫−∇∙Js+⊙dV=−∮JsdA+∫⊙dV. The first term is the entropy flow from the outside through the surface of the system per unit time. The second term is the sum of the entropy production of all the subsystems per unit time. Comparing the heat transport case with the general entropy change formula under local equilibrium assumption with ∂s/∂t=−∇∙Js+⊙, we can see that Js=Jq/T and ⊙=Jq∙∇1/T.


## 3.2. Nonequilibrium Black Hole Thermodynamics under Local Equilibrium Assumption

In our model, for both ends of the membrane, the input energy flows per unit time and per unit area are given asJBH=ωTBH4 from the black hole and JR=ωTR4 from the radiation field, ω is the Stefan-Boltzmann constant. The outward radiation along the radius is in the positive direction. The radiation energy flow in the membrane is Jq=ωTBH4−ωTR4. Although we assume the local equilibrium at each layer of the radius r, the system overall is in nonequilibrium state with temperature distribution varying in space. The energy flows are assumed mostly through the radiation rather than thermal conduction in our study.The rate of the exchange in entropy between the back reaction membrane system and the environment per unit area in the membrane is given asṡe=−∇∙Jq/T=Jq/T2∇T=Jq∆T/T2L, where ∇T=dT/dr=∆T/Lisfrom our setting T=TBH+r−rBHTR−TBH/L and ∆T=TR−TBH here. The radiation energy flow in the membrane Jq=ωTBH4−ωTR4 is constant analogous to a case of an electric current Jq moving through a resistance of length L under a voltage is ∇1/T. The entropy production rate per unit area is given as ṡi=⊙=Jq∙∇1/T=−Jq∆T/T2L>0. Since the energy flux Jq is a constant in the membrane, we can check that the system entropy of the membrane does not change and is therefore in steady state. Since the total entropy production rate per unit area of the membrane system and the environment is the sum of the system entropy is equal to the entropy change rate of the system and its exchange with the environments (same as the environmental entropy change) per unit area according to the nonequilibrium thermodynamics, the entropy production rate is equal to the entropy exchange rate with the environment or the environmental entropy change rate. For the whole membrane system, we can obtain the total entropy exchange between the membrane system and the environments as Ṡe=∫ṡedV=∫ṡe4πr2dr. Suppose the thickness of the membrane is much less than the radius of the black hole, L≪rBH. Then, we can obtain approximately Ṡe≈ABH∫ṡedr=JqABH1/TBH−1/TR=Jqћ2c2/4πTBH2kB21/TBH−1/TR, where the area of the black hole is given as ABH=ћ2c2/4πTBH2kB2. For a black hole at temperature 2.7 k, its radius is given as rBH=ћc/4πTBHkB=6.68×10−5m, so the thickness of the membrane should be less than 10−7m at least such that 4πrBH+L2=4πrBH2+4πL2+8πrBHL≈4πrBH2. Also, the total entropy production rate of the membrane system is approximately given as Ṡi=∫ṡidV≈Jqћ2c2/4πTBH2kB21/TR−1/TBH>0.


## 3.3. The Rate of Energy and Entropy Change of the Back Reaction on the Black Hole against Temperature Difference

We now explore how the entropy change of the black hole, the energy change, the mass change of the black hole, and the mass or energy of the membrane change with respect to time,ṠBH=−ωTBH3ABH=−ωћ2c2TBH/4πkB2, where ωTBH3 is the entropy flow from black hole and ṀBH=−JBHABH=−ωћ2c2TBH2/4πkB2, ṀH=−ωTBH4ABH−ωTR4×4πrBH+L2, where the energy flux JBH=ωTBH4 is from the black hole and the energy flux JR=ωTR4 is from the radiation field. Using the assumption that the thickness of the membrane is much less than the black hole radius, we obtain ṀH≈−ωTBH2/4π−ωTR4/4πTBH2ћ2c2/kB2.Let us assume that the expression ofṠBH and ṀBH above are approximately valid for the “naked” black hole, and the back reactions of the entropy rate and energy rate from the membrane are given as Ṡe and ṀH, respectively. After considering the back reaction, the effective or dressed entropy rate and energy rate of the black hole are given as Ṡdressed=ṠH+ṠBH, Ṁdressed=ṀH+ṀBH, respectively.When∆T=0, the energy and entropy loss per unit time of “dressed” black hole is equal to that of the “naked” black hole’s as shown in Figures 6(a) and 6(b). The system is in equilibrium. When the temperature difference is larger, the system becomes more nonequilibrium. Compared to the thin membrane case, we find that the thick membrane can lead to more significant back reactions on the black holes. In both Figures 6(a) and 6(b), we see that the energy and entropy change rate of a dressed black hole after taking into account of the back reaction are larger than that of the “naked” black hole, and the distances between the green line and red line increase as ∆T increases. This shows that the back reaction on the energy rate and entropy rate of the black hole is more significant as the temperature difference increases. The evaporation rate of the dressed black hole when taking into account of the thickness of the membrane back reaction is larger than that of the “naked black hole.” On the other hand, from both Figures 6(a) and 6(b), the nonequilibrium effects characterized by the temperature difference can enhance the energy and entropy loss rate of the black hole and the membrane back reaction.Figure 6
(a) and (b) The energy and entropy loss per unit time of “naked” black hole and “dressed” black hole varying with the temperature difference with an initial temperature 2.73 k. Here,∆T=TBH−TR.
(a)(b)


## 3.4. Energy and Entropy Change of the Black Hole due to the Back Reaction against Temperature Difference

We would not only like to see how significant the rate of the back reaction but also the back reaction of the membrane itself is on the black hole. To do so, we need to compare the contribution of the entropy and energy from the membrane towards the black hole. Let us estimate the time scale given the entropy and energy rate of the membrane above. The response of a black hole to a perturbation will be dominated by a set of damped oscillations called quasi-normal modes. The quasi-normal mode frequency corresponds to a natural frequency of a black hole. It is divided into two parts: the real part represents the oscillation, and the imaginary part represents damping [17]. For large damping, i.e., for a large imaginary part of ω, the real part of the frequency approaches a nonzero value, and the imaginary part becomes equally spaced. It is given as Mω=0.04371235+in+1/2 [18]. The constant real part of the quasi-normal frequency can be approximated as wQNM=ln3c3/8πGM. Thus, the natural time scale can be estimated as τ=2π/wQNM=16π2GM/ln3c3=2πћ/ln3kBTBH.In the time intervalτ, ∆Se=Ṡeτ=Jqћ3c2/2ln3kB3TBH31/TBH−1/TR, ∆Si=Jqћ3c2/2ln3kB3TBH31/TR−1/TBH, ∆SBH=−ωћ3c2/2ln3kB3, ∆MBH=−ωћ3c2TBH/2ln3kB3, ∆MH=−ωTBH/2−ωTR4/2TBH3ћ3c2/ln3kB3. We measure the back reaction as ∆Se/∆SBH=TBH4−TR41/TBHTR3−1/TBH4 and ∆MH/∆MBH=1−TR4/TBH4. In the case of thick membrane, the energy and entropy change of back reaction during certain periods are dependent on the temperature difference.When∆T=0, the system is in equilibrium, ∆MH=∆Se=0 in both Figures 7(a) and 7(b). According to Figures 7(a) and 7(b), both ∆MH/∆MBH and ∆Se/∆SBH increase as the temperature difference increases. The back reaction becomes more significant as temperature difference increases. Also, for a black hole with a mass MBH≫5×1014g, the effect of thick membrane Ṡe=Jqћ2c2/4πTBH2kB21/TBH−1/TR is much more significant than that of thin membrane ṠH=−ω2ћ4GTR4/4π2kB4TBHc under finite temperature difference. Under an infinitesimally small temperature difference, the effect of a thick membrane is similar to that of a thin membrane; they are both weak. Therefore, we can conclude that the back reaction from the membrane to the black hole can be significant in terms of the energy (mass) and entropy change. The nonequilibrium effect characterized by the temperature difference can enhance the back reactions in terms of the energy (mass) and entropy change to the black hole.Figure 7
(a) and (b) show∆MH/∆MBH, ∆Se/∆SBH varying with the temperature difference with an initial temperature 2.73 k, respectively.
(a)(b)


## 3.5. Nonequilibrium Entropy Production and Entropy Change of the Black Hole against Temperature Difference

For irreversible nonequilibrium entropy production, once there is a finite temperature difference, the additional entropy change caused by the radiation energy transfer of the total system in a unit time can be significant as shown in Figure8(a). It can be quantified by ∆Si/∆SBH=TBH4−TR41/TBHTR3−1/TBH4.Figure 8
(a) shows the variation of the irreversible entropy production rate (red line) and the rate of entropy change of the black hole (green line) with respect to the temperature difference with an initial temperature 2.73 k. (b) shows the entropy change caused by irreversible entropy production (red line) and the entropy change of the black hole caused by radiation (green line) against the temperature difference with an initial temperature 2.73 k in time intervalτ.
(a)(b)Figure8(a) shows the variation of the irreversible entropy production rate (red line) and the rate of entropy change of black hole (green line) with respect to the temperature difference at initial temperature 2.73 k. We found as the temperature difference increases, both the entropy production rate and rate of entropy change of the black hole increase. This indicates that the nonequilibrium effects characterized by the temperature difference can increase the dissipation cost rate and entropy change rate of the black hole. We also find that the entropy production rate is significant compared to the rate of entropy change of the black hole at a finite temperature difference. This also holds for black holes with different masses. Figure 8(b) shows the entropy change caused by irreversible entropy production rate (red line) and the entropy change of the black hole caused by radiation (green line) with the temperature difference at initial temperature 2.73 k in time interval τ. Notice that the green line in Figure 8(b) is a constant. This is due to the underlying assumption that the entropy (area) of a black hole radiating out in time interval τ is quantized from the quasi-normal mode frequency expression [19]. This shows τ as a natural time of the black hole. We found as the temperature difference increases, both the entropy production and entropy change of the black hole increase. This shows that the nonequilibrium effects characterized by the temperature difference can increase the dissipation cost and entropy change of the black hole. The entropy change caused by irreversible entropy production is significant compared to the entropy change of a black hole caused by radiation at a finite temperature difference. This also holds for black holes with different mass. The nonequilibrium entropy production rate can thus have an impact on the black hole evaporation rate.This indicates that there can be significant entropy or information loss under the nonequilibrium conditions characterized by the temperature difference between the black hole and the radiation environment. This extra loss of the entropy or information may give some insights towards the understanding the source of the black hole information paradox.


## 4. Nonequilibrium Thermodynamic Fluctuations of the Black Hole and Back Reaction

The power emitted causes the mass of the black hole to decrease at the ratedMBH/dt=−ћc4/G2α/MBH2. So, if the black hole is massive enough, i.e., its mass exceeds 5×1014g,the loss rate of these quantities is almost constant, and the evaporation process may be approximated by a nonequilibrium quasi steady-state, provided that the time scale of any realistic measurement is much shorter than the time scale of the process ∝MBH3 [20]. It is easy to check that for MBH>1014g, the characteristic decay time is of the order of the age of the universe. The electromagnetic radiation and its back reaction on the metric were considered in a stochastic Langevin approach [21]. The results yielded a lower limit on the statistical variations of the metric itself, which could be thought as the “seismology” of the spacetime’s geometry, as a consequence of the back reaction. In this spirit, one can consider the stochastic equation ṀBH=−α/MBH2+δṀ,where δṀ represents a spontaneous thermal fluctuation of ṀBH, such a thermal fluctuation in the black hole should be reflected in the fluctuations of the background metrics and thereby may be regarded as reflecting the effects of back reaction of the Hawking radiation to the evolution of a black hole evaporation within a phenomenological level [21, 22].By using the Landau-Lifshitz theory [23], for a general fluctuating dissipative system, ẋi=−∑jΓijXj+δẋ, ẋi is a general flux and Xi is the thermodynamic force conjugate to the flux ẋi, Γij is phenomenological transport coefficients which is defined as Γij=∂ẋi/∂Xj, and δẋ stands for fluctuations. The entropy production rate can be given as Ṡ=−∑iẋiXi from the nonequilibrium thermodynamics. Because the entropy of the Schwarzschild black hole is not concave, the above expression must be modified as Ṡ=∑i±ẋiXi (for more details see [20]). Therefore, for the Schwarzschild black hole, Ṡ=ṀBH/TBH=−8πα/MBH, ṀBH can be interpreted as the flux or current and 1/TBH can be interpreted as the thermodynamic driving force conjugate to the flux ṀBH. Then, the second order moments in the fluctuations of the fluxes can be expressed as δẋiδẋj=Γij+Γjiδij.Now one can derive thatδṀδṀ=α4πMBH3−1 from ΓMM=∂ṀBH/∂1/TBH and δṠδṠ=∂S/∂MBH2δṀδṀ=16παMBH−1, where S=4πMBH2. The second order moments reflect the fluctuation intensity of the black hole, namely, the intensity of the back reaction. Therefore, we can also conclude that the more massive the black hole is, the weaker its back reaction is. This argument holds for a black hole with mass exceeding 5×1014g, since we can consider the evaporation of the black hole as a nonequilibrium quasi steady-state process only in this situation.


## 5. Conclusion

In this study, we developed a nonequilibrium thermodynamics model for investigating the back reaction of the Hawking radiation on the Schwarzschild black hole. In our study, two scenarios are considered: the thin membrane and the thick membrane. The thin membrane formed has a negative tension coefficient. We found that larger temperature difference can lead to more significant negative surface tension and back reaction as well as larger black hole area. The thin membrane has the tendency to expand to counteract the tendency of the Hawking radiation for reducing the black hole area. We showed that the more massive the black hole is, the weaker its back reaction is based on a nonequilibrium thermodynamic approach of the thin membrane. We also confirm this using a phenomenological fluctuation analysis. We conclude that at the nonequilibrium state, the back reaction may contribute to the change of the entropy and energy for the primordial black hole. The larger temperature difference between the black hole and the radiation environment gives rise to more significant back reaction. Moreover, our nonequilibrium model can resolve the inconsistency issue of the black hole back reaction under zero mass limit in the equilibrium case. Our nonequilibrium model appears to be more natural and physically reasonable than the equilibrium model for the black hole back reaction and evolution. We also study the thick membrane assuming the thickness of the membrane is much less than the radius of the black hole such that the temperature can be viewed to vary uniformly in the membrane. We find that the thick membrane shows more significant effects on back reactions compared to the thin membrane.Importantly, we also quantified the nonequilibrium thermodynamic dissipation represented by the entropy production rate. We showed that the entropy production rate increases significantly as the temperature difference increases for both the thin membrane and the thick membrane. We found that in general the nonequilibrium effects characterized by the temperature difference can increase the energy and entropy loss as well as the entropy production rate of the black hole and the membrane back reactions. We show that once there is a finite temperature difference, the nonequilibrium entropy production rate appears to be significant compared to the entropy rate radiated by the black hole. The nonequilibrium entropy production rate can thus have an impact on the black hole evaporation rate. This may have impacts on the black hole information paradox due to the significant information loss from the entropy production rate under nonequilibrium scenarios (temperature difference between the black hole and the surrounding radiation environment).



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