NUPHB14505S0550-3213(18)30334-110.1016/j.nuclphysb.2018.11.018The Author(s)High Energy Physics – TheoryHolographic entanglement thermodynamics for higher dimensional charged black holeSouravKararabsourav.karar91@gmail.comDebabrataGhoraibdebanuphy123@gmail.comSunandanGangopadhyayb⁎sunandan.gangopadhyay@bose.res.insunandan.gangopadhyay@gmail.comaDepartment of Physics, Government General Degree College, Muragachha 741154, Nadia, IndiaDepartment of PhysicsGovernment General Degree CollegeMuragachhaNadia741154IndiabS.N. Bose National Centre for Basic Sciences, Block-JD, Sector III, Kolkata 700106, IndiaS.N. Bose National Centre for Basic SciencesBlock-JDSector IIIKolkata700106India⁎Corresponding author.Editor: Stephan StiebergerAbstractIn this paper, we have investigated the entanglement thermodynamics for d-dimensional charged AdS black hole by studying the holographic entanglement entropy in different cases. We have first computed the holographic entanglement entropy in extremal and non-extremal cases in two different regimes, namely, the low temperature and high temperature limits. We then obtain the first law of entanglement thermodynamics for boundary field theory in the low temperature regime in d-dimensions.1IntroductionThere has been an immense amount of interest in information theory recently. This is primarily due to the fact that quantum correlations, which is an important ingredient in the theory of information, play a crucial role in various branches of physics, namely, condensed matter physics, statistical mechanics as well as in quantum theories of gravity. It has been realized that the fundamental laws of physics can be given an information theoretic interpretation [1,2]. In classical information theory, information is quantified by a measure called Shannon entropy. The counterpart of this concept in quantum information theory is entanglement entropy (EE). EE is a fundamental quantity in quantum information theory as it provides a measure for quantum correlation in a bipartite quantum system. In the last few years EE has been successfully used as a probe in quantum phases of matter [3–8]. Other areas where information theory has provided important insights are the thermodynamic derivation of Einstein's equations of general relativity [9] and in resolving the black hole information loss paradox [10–12]. So it is realized that in understanding the geometry of spacetime, information theory and specially EE would play a vibrant role.Obtaining the EE of 1+1-dimensional conformal field theories has been an important problem in theoretical physics. The computation was first done in [3–5] by a method known as the replica trick. Interestingly, the behavior of the EE for a 1+1-dimensional conformal field theory (CFT) exhibited an universal logarithmic behavior [3]. Recently, the gauge/gravity correspondence has played a key role in computing the EE of a boundary CFT holographically from its bulk gravitational dual. The insight comes from the fact that the holographic principle [13–15] states that the number of degrees of freedom in a region of space is equal in number to the degrees of freedom on the boundary that surrounds the space. This principle first proposed in the context of black hole entropy, became one of the most cherished ideas in modern theoretical physics with the advent of the AdS/CFT correspondence [16,17]. This correspondence is the most successful realization of the holographic principle in theoretical physics. It relates the gravitational theory in AdS space to the CFT that lives on the boundary of AdS space. It has evolved as a powerful theoretical input in condensed matter physics, nuclear physics and QCD [18–20].The prescription of computing the holographic entanglement entropy (HEE) was first proposed in [21], [22]. According to the prescription, the HEE for a subsystem A in a (d−1)-dimensional boundary field theory is given bySA=Area(ΓA)4G(d) where ΓA is the minimal surface area of the bulk extension (on a fixed time slice) whose boundary coincides with the edges of the subsystem living at the boundary and G(d) is the d-dimensional Newton's gravitational constant. This formula is very similar to the black hole entropy formula suggested by Bekenstein and Hawking [23–25]SBH=Area(σ)4Gd where σ is the area of the horizon of the black hole. This striking similarity between HEE and black hole entropy inspired many to suggest that EE is the origin of black hole entropy [26–28].Studies of the HEE of AdS black holes which are dual to a field theory at finite temperature have also been carried out [29–31]. In [32], the HEE has been computed for charged AdS black hole to observe the effect of temperature and charge on the EE of a strip like subregion in the boundary field theory dual to the charged AdS black hole in the bulk. An important question in this regard is whether the HEE satisfies a relation analogous to the first law of thermodynamics, which is indeed satisfied by thermal entropy. In [33], the difference in HEE of a thermally excited AdS spacetime and pure AdS spacetime has been computed. This led them to conclude that the change in HEE (ΔSE) is proportional to the change in energy (ΔE). The proportionality constant was identified as the inverse of entanglement temperature (Tent). The same question was addressed in [34] and a similar relation between ΔSE and ΔE was obtained. Entanglement thermodynamics has been explicitly studied in different backgrounds including non-conformal and non-relativistic backgrounds [35–38]. Entanglement thermodynamics for charged black hole in AdS4 background has also been studied in [32]. In this investigation, a relation like the first law of entanglement thermodynamics in the low temperature limit was obtained.In this paper, we extend the analysis for a charged black hole in d-dimensions. We look at different temperature and charge limits. We have explicitly calculated the EE for charged black hole (AdS-RN black hole) in d-dimensions for different temperature and charge limits. A law like the first law of entanglement thermodynamics has also been obtained in the small temperature limit. The expressions for the EE in different limits have explicit dependence on the dimension of spacetime d. The analysis helps us to understand the implications of the dimension of spacetime on information theoretic quantities.The paper is organized as follows. We have first reviewed the AdS-RN black hole in arbitrary spacetime dimension in section 2. From the definition of black hole temperature, we then find the extremality condition for the AdS charged black hole. In section 3, we have investigated the HEE in small charge and large charge limit for extremal black hole. Then we have computed the HEE in the low temperature and high temperature regime for small charged non-extremal black hole as well as for large charged non-extremal black hole. The form of the HEE expression for small charged extremal black hole is the same as the HEE in the low temperature regime for small charged non-extremal black hole. It is to be noted that the bulk dual extremal black hole is considered to be the ground state. Similarly the bulk dual non-extremal black hole is considered as the excited state. We then discuss the first law of entanglement thermodynamics in the small charge limit in section 4. We conclude in section 5.2Charged AdS black holeIn this section, we start by writing down the charged AdS black hole metric with planar horizon in d-dimensions. This reads(1)ds2=−r2R2f(r)dt2+R2dr2r2f(r)+r2R2(dx12+dx22+......+dxd−22)f(r)=1−Mrd−1+Q2r2(d−2) where R is the radius of AdS spacetime, M and Q are the mass and charge of the black hole respectively. We shall set R=1 for the rest of this paper.The horizon of the black hole is given by f(r)|r=rh=0. This yields(2)M=rhd−1(1+Q2rh2(d−2)) which relates the mass M of the AdS-RN black hole and its charge Q. The Hawking temperature for this black hole is given by(3)TH=r2f′(r)4π|rh=(d−1)rh4π[1−(d−3d−1)Q2rh2(d−2)]. The lapse function f(r) can now be expressed in terms of only the charge Q and the radius of the horizon rh as(4)f(r)=1−(rhr)d−1+Q2(1r2(d−2)−1rd−1rhd−3).3Computation of HEEWith the basic setup in hand we are now ready to calculate the HEE of the AdS-RN black hole. We consider an entangling region at the boundary in the form of a straight belt of width l given by(5)−l2≤x1≤l2;0≤x2,x3,⋯,xd−2≤L. According to the proposal in [21,22] we have to find the minimal codimension two hypersurface in the bulk whose boundary coincides with the two ends of the interval −l2≤x1≤l2. Then the entanglement entropy is given by the minimal area divided by 4G(d), where G(d) is the Newton's gravitational constant in d dimensions.The area of the hypersurface for the system (1) is given by(6)A=Ld−3∫−l2l2dx1r2(d−2)+(r′)2f(r)r2(d−4) where the surface is parametrized by r=r(x1). Using the standard procedure of minimization we get(7)A=2Ld−3∫rt∞rd−4drf(r){1−(rtr)2d−4} with the minimal surface characterized by(8)drdx1=f(r)r4{(r2rt2)(d−2)−1}, where rt is the turning point of the extremal surface satisfying r′|r=rt=0. Integration of the above equation gives the length of the system to be(9)l2=∫rt∞rtd−2drrdf(r){1−(rtr)2d−4}. It is not difficult to see that the integral (7) is divergent as we reach the boundary. Therefore we have to introduce an infrared (IR) cutoff at r=rb, where rb is very large. This IR cutoff is related holographically to the field theory counterpart which is the ultraviolet (UV) cutoff a by the relation rb=1/a. In field theory this UV cutoff is nothing but the lattice spacing. The finite part of the entangling entropy can be used to study the high and low charge (or temperature) behavior of the field theory which is dual to the AdS-RN black hole.To compute the integrals (7) and (9), we change the integration variable r by u=rtr, which makes the lapse function(10)f(u)=1−(rhrt)d−1ud−1−Q2rhd−3(urt)d−1+Q2(urt)2(d−2). The length of the subsystem along x1 and the area of subsystem now read(11)l=2rt∫01duud−21−u2(d−2)(1−(rhrt)d−1ud−1−Q2rhd−3(urt)d−1+Q2(urt)2(d−2))−1/2(12)A=2(Lrt)d−3∫01duu−(d−2)1−u2(d−2)(1−(rhrt)d−1ud−1−Q2rhd−3(urt)d−1+Q2(urt)2(d−2))−1/2. In the following sections we compute the HEE for both the extremal black hole and the black hole at finite temperature. The extremal black hole is characterized by its zero Hawking temperature (TH=0). An important relation that arises in this context is the relation between the charge Q of the black hole and its horizon radius rh(13)rh2(d−2)≥(d−3d−1)Q2. In the above expression, the equality sign holds for the extremal black hole and the inequality holds for the non-extremal black hole. It is interesting to note that the AdS/CFT correspondence tells that the field theory counterpart of the AdS-RN black hole is in its ground state for the extremal black hole and in its excited state for the non-extremal black hole.3.1Extremal black holeLet us first calculate the HEE for the case of the extremal black hole. The extremality condition says that the charge of the black hole has to be(14)Q2=(d−1d−3)rh2(d−2). Using the above condition in eq. (10), we can rewrite the lapse function in terms of u as(15)f(u)=1−2(d−2)d−3(rhurt)d−1+d−1d−3(rhurt)2(d−2). We can also express the integrals (11), (12) as(16)l=2rt∫01duud−21−u2(d−2)(1−2(d−2)d−3(rhurt)d−1+d−1d−3(rhurt)2(d−2))−1/2(17)A=2(Lrt)d−3∫01duu−(d−2)1−u2(d−2)(1−2(d−2)d−3(rhurt)d−1+d−1d−3(rhurt)2(d−2))−1/2. It is not hard to see that these integrals cannot be evaluated analytically. In order to evaluate the integrals analytically we have to take certain limits. In the subsequent discussion we have taken two extreme limits: small charge limit and large charge limit. In the following subsections we have considered these two limits to calculate the HEE.3.1.1Small charge limitWe can see from eq. (14) that if Q is small, then rh will be small. To be specific we have l(d−3d−1)12(d−2)Q1d−2≤lrh≪1 in the small charge limit. As the horizon radius rh is very small, the turning point rt is far away from it. Therefore (rhrt) is a very small quantity. So we can neglect higher order terms of (rhrt). Using this approximation we can now Taylor expand to write(18)1f(u)≈1+d−2d−3(rhurt)d−1. Using the above equation, we can now simplify eq. (16) as(19)l≈2rc∫01ud−2du1−u2(d−2)(1+d−2d−3(rhurt)d−1)=2rt[∫01ud−2du1−u2(d−2)+d−2d−3(rhrt)d−1∫01u2d−3du1−u2(d−2)]. Therefore the turning point rt reads(20)rt=2l[πΓ(d−12(d−2))Γ(12(d−2))+π2(d−3)(rhrt)d−1Γ(d−1d−2)Γ(3d−42(d−2))]. The form of the above expression suggests that we cannot solve rt exactly. Hence using the perturbative approach, we obtain(21)rt=2l[πΓ(d−12(d−2))Γ(12(d−2))+π2(d−3)(lrh2)d−1(Γ(12(d−2))πΓ(d−12(d−2)))d−1Γ(d−1d−2)Γ(3d−42(d−2))]. Now using same approximation (18), the area of the extremal surface reads(22)A=2(Lrt)d−3[∫01u−(d−2)du1−u2(d−2)+d−2d−3(rhrt)d−1∫01udu1−u2(d−2)]. It is observed that the first integral in A is divergent as u→0. To regularize the integral we introduce the UV cut-off rtrb and add a counter term (−2(Lrb)d−3d−3) in order to get a finite value of the extremal area which reads(23)Afinite=2(Lrt)d−3[∫rtrb1u−(d−2)du1−u2(d−2)+d−2d−3(rhrt)d−1∫01udu1−u2(d−2)]−2(Lrb)d−3d−3=2(Lrt)d−3[π(d−2)Γ(3−d2(d−2))Γ(12(d−2))+π(d−3)(rhrt)d−1Γ(1d−2)Γ(d2(d−2))]. Substituting the turning point rt from eq. (21) into eq. (23) and keeping terms upto O((lrh)d−1) and then simplifying, we obtain(24)Afinite=(Ll)d−3[−(2π)d−2d−3(Γ(d−1d−2)Γ(12(d−2)))d−2+d−2d(d−3)(lrh)d−14π×(Γ(12(d−2))Γ(d−12(d−2)))2Γ(1d−2)Γ(d2(d−2))]. The finite holographic entanglement entropy reads (Afinite4G(d)) from the above equation(25)SAfinite=SAAdS+SAext where SAAdS is the entanglement entropy for pure AdS spacetime and the extra piece comes from the extremality of black holes. The expressions for SAAdS and SAext read(26)SAAdS=−(2π)d−24GNd(d−3)(Ll)d−3(Γ(d−12(d−2))Γ(12(d−2)))d−2(27)SAext=Ld−3l2rhd−116πG(d)d−2d(d−3)(Γ(12(d−2))Γ(d−12(d−2)))2Γ(1d−2)Γ(d2(d−2)). But we know that the relation between the mass of the extremal black hole with its horizon is given by(28)rhd−1=d−32(d−2)Mext. Substituting this in eq. (27), we obtain(29)SAext=kLd−3l2Mext where(30)k=132dG(d)π(Γ(12(d−2))Γ(d−12(d−2)))2Γ(1d−2)Γ(d2(d−2)). It is reassuring to note that the above expression reduces to the result in [32] in the d=4 limit.3.1.2Large charge limitIn this subsection we are going to compute the HEE of an extremal AdS-RN black hole whose charge Q is large. By the extremality condition (13), this means that the horizon radius rh is also large. This in turn implies rhl≫1. As the horizon radius is very large, so we can assume that the horizon is very close to the turning point of the extremal surface (rh∼rt). Now looking at the area integral (17), we find that the dominant contribution to the finite part of the integral comes from u→1 limit. On the other hand, defining u0=rtrh, we see that u0∼1. Hence most of the contribution to the finite part of the area integral comes from the near horizon limit. We should then Taylor expand the lapse function (15) around u0 to evaluate the area integral. For this Taylor expansion to be valid one must show that u−u0 is small enough. As rt and rb are very large, u is very close to u0 throughout the integral. Hence we can now Taylor expand eq. (15) and neglect higher order terms to obtain(31)f(u)=f(u0)+f′(u0)(u−u0)+f″(u0)2!(u−u0)2+O((u−u0)3)=(d−1)(d−2)(1−uu0)2+O((u−u0)3)≃(d−1)(d−2)(1−rhurt)2. Using this approximated value of f(u), the length of the entangling region becomes(32)l=2rt(d−1)(d−2)∫01ud−2du1−u2(d−2)1(1−rhrtu). To simplify this integral we make a binomial expansion and the length integral now takes the form(33)lrt2=1(d−1)(d−2)∑n=0∞(rhrt)n∫01un+d−2du1−u2(d−2)=12(d−2)3/2πd−1∑n=0∞Γ(n+d−12(d−2))Γ(n+2d−32(d−2))(rhrt)n. For large value of n, this expression is divergent. Using gamma function properties and Stirling formula, one can check that for large value of n, the above summation goes as 2(d−2)n(rhrt)n. To get a finite value we isolate the divergent terms to obtain(34)lrt=1(d−2)3/2πd−1Γ(d−12(d−2))Γ(2d−32(d−2))+π(d−1)(d−2)∑n=1∞(Γ(n+d−12(d−2))(d−2)Γ(n+2d−32(d−2))−2(d−2)n)(rhrt)n+1(d−2)2πd−1Li12(rhrt) where(35)Li12(rhrt)=∑n=1∞1n(rhrt)n is polylogarithmic function. As the horizon radius rh is very close to the turning point rt of the extremal surface, we can assume that rt=rh(1+ϵ) where ϵ is a very small positive number [39]. Substituting it in eq. (34), we get(36)lrh=k1+2d−1(πd−2)1ϵ+O(ϵ) where(37)k1=πd−11(d−2)3/2Γ(d−12(d−2))Γ(2d−32(d−2))+2πd−11(d−2)ζ(12)+π(d−1)(d−2)∑n=1∞(Γ(n+d−12(d−2))(d−2)Γ(n+2d−32(d−2))−2(d−2)n). We can also invert eq. (36) to obtain(38)ϵ≈2π2(d−1)(d−2)21(lrh−k1)2. The rh appearing in the above equation can be replaced by eq. (14) to get ϵ in terms of the black hole charge Q.Now the extremal surface area (17) reads(39)A=2(Lrt)d−3(d−1)(d−2)∫01duu−(d−2)1−u2(d−2)1(1−rhrtu). Again by using the binomial expansion we get(40)A=2(Lrt)d−3(d−1)(d−2)∑n=0∞(rhrt)n∫01un−d+2du1−u2(d−2). The above equation is divergent for the terms corresponding to n<(d−2). Let us regularize it from n=0 to n=d−3. To regularize the divergent terms we have to introduce IR cut-off rb in the integrals. Let us start with the integral corresponding to n=0:(41)A0finite=2(Lrt)d−3(d−1)(d−2)∫rtrb1du1ud−21−u2(d−2)−2(Lrb)d−3(d−1)(d−2)=−2π(Lrt)d−3(d−3)(d−1)(d−2)Γ(d−12(d−2))Γ(12(d−2)). The term corresponding to n=1 is given by(42)A1finite=2(Lrt)d−3(d−1)(d−2)(rhrt)∫01duu−(d−3)1−u2(d−2)=2rhLd−3rtd−4(d−1)(d−2)[∫rtrb1du1ud−3+∑k=1∞Γ(k+12)πΓ(k+1)∫01duu3−d+2k(d−2)]. In the above equation, we have separated the first term as it is divergent and we have regularized it. Since rt and rb are both large (rt∼rb) hence the first term do not contribute. The finite value of the integral is(43)A1finite=2rhLd−3rtd−4(d−1)(d−2)[1d−4+π2(d−2)Γ(4−d2(d−2))Γ(22(d−2))]. In general the expression for the regularized terms are(44)Amfinite=2(Lrt)d−3(d−1)(d−2)(rhrt)m[1d−m−3+π2(d−2)Γ(m−d+32(d−2))Γ(m+12(d−2))] for m=1,2,…,(d−4). Let us now look at the n=d−3 term:(45)Ad−3=2(Lrh)d−3(d−1)(d−2)∫01du1u1−u2(d−2)=2(Lrh)d−3(d−1)(d−2)[∫rtrb1du1u+∑k=1∞Γ(k+12)πΓ(k+1)∫01duu2k(d−2)−1]=2(Lrh)d−3(d−1)(d−2)[−log(rtrb)+log42(d−2)]≈2(Lrh)d−3(d−1)(d−2)log42(d−2).The remaining terms in eq. (40) corresponding to n≥(d−2) are given as(46)An≥(d−2)=2(Lrt)d−3(d−1)(d−2)∑n=(d−2)∞(rhrt)n∫01un−d+2du1−u2(d−2)=2(Lrt)d−3(d−1)(d−2)∑n=(d−2)∞π2(d−2)Γ(n−d+32(d−2))Γ(n+12(d−2))(rhrt)n. Therefore the above contribution diverges as rt approaches to rh, as for large n the factor inside the summation goes as ∼1n(rhrt)n. To remove the divergence we use the identity Γ(n+1)=nΓ(n) to rewrite it as(47)An≥(d−2)=(Lrt)d−3π(d−1)(d−2)∑n=(d−2)∞{1(d−2)+1(n−d+3)}Γ(n+d−12(d−2))Γ(n+2d−32(d−2))(rhrt)n. Now for large value of n, the second term goes as 2(d−2)nn(rhrt)n. Therefore it is convergent. Using eq. (33) in the above equation we obtain(48)An≥(d−2)finite=(Lrt)d−3π(d−1)(d−2)[(d−2)(d−1)πlrt−∑m=0d−3Γ(m+d−12(d−2))(d−2)Γ(m+2d−32(d−2))(rhrt)m+∑n=d−2∞1(n−d+3)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))(rhrt)n]. The leading contribution in An≥(d−2)finite comes from the limit rt=rh. The second series in eq. (48) is convergent at leading order. If we want to find the subleading order contributions, we have to put rt=rh(1+ϵ) and expand it binomially. Now the second series in eq. (48) may not be convergent at the subleading order. Therefore we isolate the divergent terms of the series up to O(ϵ) and rewrite eq. (48) as(49)An≥(d−2)finite=(Lrt)d−3π(d−1)(d−2)[(d−2)(d−1)πlrt−∑m=0d−3Γ(m+d−12(d−2))(d−2)Γ(m+2d−32(d−2))(rhrt)m+∑n=d−2∞(1n−d+3Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2(d−2)nn)(rhrt)n+∑n=d−2∞2(d−2)nn(rhrt)n]. Now we can write(50)∑n=d−2∞2(d−2)nn(rhrt)n=2(d−2)[Li32[rhrt]−∑m=1d−31mm(rhrt)m]. We can now put all the results together in eq. (40) to write the total area of the extremal surface as(51)Afinite=(Lrt)d−3(d−1)(d−2)[(d−2)(d−1)lrt−(d−2)πd−3Γ(d−12(d−2))Γ(12(d−2))+∑n=1d−4(1d−n−3+π2(d−2)Γ(n−d+32(d−2))Γ(n+12(d−2))−π2(d−2)Γ(n+d−12(d−2))Γ(n+2d−32(d−2)))(rhrt)n+(log4(d−2)−π(d−2)Γ(3/2))(rhrt)d−3+π∑n=d−2∞(1n−d+3Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2(d−2)nn)(rhrt)n2π(d−2)(Li32[rhrt]−∑n=1d−31nn(rhrt)n)]. Now we substitute rt=rh(1+ϵ) in the above expression. After simplification we finally obtain(52)Afinite=Ld−3lrhd−2+(Lrh)d−3(K1+K2ϵ+K3ϵ)+O(ϵ3/2) where(53)K1=1(d−1)(d−2)[−2πd−2d−3Γ(d−12(d−2))Γ(12(d−2))+log4(d−2)−π(d−2)Γ(3/2)+2π(d−2)[ζ(32)−∑n=1d−31nn]+π∑n=d−2∞[1n−d+3Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2(d−2)nn]+∑n=1d−3(1d−n−3+d−2n+1Γ(n−d+32(d−2))Γ(n+12(d−2)))]K2=−22πd−1K3=2(d−1)(d−2)[1+dd−1Γ(−12(d−2))Γ(d−32(d−2))+π(d−2)2((d−1)ξ(32)−ξ(12))]. Therefore the renormalized holographic entanglement entropy in the large charge regime for the extremal black hole is given by(54)SAfinite=Ld−3lSBHext+(Lrh)d−34GNd(K1+K2ϵ+K3ϵ)+O(ϵ3/2) where SBHext=rhd−24GNd.3.2Non-extremal black hole3.2.1Small charge limitLow temperature limitIn this subsection we shall compute the HEE for a subsystem in the boundary theory whose bulk dual is the non-extremal black hole which has small charge Q. We further assume that the Hawking temperature TH of the black hole is also small. The non-extremality condition implies Q2≠d−1d−3rh2(d−2). As the temperature of the black hole TH is small, eq. (3) implies that rh has to be small. Again eq. (13) in small charge limit also suggest the same criterion. Thus in small charge and temperature limit, the horizon radius rh of the black hole is very small, so we can make the assumption rhrt≪1. Now we replace the charge Q by the quantity α=Qrhd−2 in eq. (10) which makes the lapse function to be(55)f(u)=1−{1+α2−(rhurt)d−3}(rhurt)d−1. To find the values of the expressions (11) and (12), we have to use an approximated expression for 1f(u). As we have already made the assumption of low temperature, eq. (3) implies that Qrhd−2=α∼1. We can now Taylor expand 1f(u) around rhrt∼0 and neglect higher order terms to obtain(56)1f(u)=1+1+α22(rhurt)d−1+O((rhurt)2(d−2)). Using this in eq. (11) we obtain the length of subsystem to be(57)l=2rt[∫01ud−2du1−u2(d−2)+1+α22(rhrt)d−1∫01u2d−3du1−u2(d−2)]⇒rt=2l[πΓ(d−12(d−2))Γ(12(d−2))+1+α22(rhrt)d−1π2(d−2)2Γ(1d−2)Γ(3d−42(d−2))]. To get the solution of the turning point rt of the extremal surface in terms of the length (l) of the subsystem, we use perturbative technique to get(58)rt=2πl[Γ(d−12(d−2))Γ(12(d−2))+1+α2(d−2)2(lrh)d−12d+1(Γ(12(d−2))πΓ(d−12(d−2)))d−1Γ(1d−2)Γ(3d−42(d−2))]. Now the extremal surface area reads(59)A=2(Lrt)d−3[∫01u−d+2du1−u2(d−2)+1+α22(rhrt)d−1∫01udu1−u2(d−2)]. The first integral of the above expression is divergent. This can be regularized by introducing UV cut-off 1rb. So, the finite part of the extremal surface area is(60)Afinite=2(Lrt)d−3[∫rtrb1u−d+2du1−u2(d−2)+1+α22(rhrt)d−1π2(d−2)Γ(1d−2)Γ(d2(d−2))]−2(lrb)d−3d−3=(Lrt)d−3πd−2[Γ(3−d2(d−2))Γ(12(d−2))+1+α22(rhrt)d−1Γ(1d−2)Γ(d2(d−2))]. Now we shall substitute eq. (58) in the above equation to express the extremal surface area in terms of the subsystem length l. This yields(61)Afinite=(Ll)d−3[−(2π)d−2d−3(Γ(d−1d−2)Γ(12(d−2)))d−2+1+α28πd(lrh)d−1×(Γ(12(d−2))Γ(d−12(d−2)))2Γ(1d−2)Γ(d2(d−2))]. From the definition of the entanglement entropy we therefore obtain(62)SAfinite=Afinite4GNd=SAAdS+SAnon-ext where(63)SAAdS=−(2π)d−24GNd(d−3)(Ll)d−3(Γ(d−12(d−2))Γ(12(d−2)))d−2(64)SAnon-ext=(1+α2)Ld−3l2rhd−132dπG(d)(Γ(12(d−2))Γ(d−12(d−2)))2Γ(1d−2)Γ(d2(d−2)). These expressions reproduce the results in [32] in d=4 limit. We can now use the mass of non-extremal black hole which is M=rhd−1(1+α2) to express SAnon-ext in a more convenient way as(65)SAnon-ext=kLd−3l2Mnon-ext where(66)k=132dG(d)π(Γ(12(d−2))Γ(d−12(d−2)))2Γ(1d−2)Γ(d2(d−2)) which is same as eq. (30).High temperature limitIn this subsection we will investigate the behavior of HEE for a subsystem in the boundary whose bulk dual is a non-extremal black hole with small charge but a high Hawking temperature. In the high temperature limit the expression for temperature of black hole (3) and the non-extremality condition (13) together suggest that the horizon radius rh has to be large. Therefore one may easily see that Q2rh2(d−2)≪1. We now define a new the quantity δ2=(d−3)Q2(d−1)rh2(d−2)≪1 to express lapse function as(67)f(u)=1−(rhurt)d−1−d−1d−3(rhurt)d−1δ2(1−(rhurt)d−3)(68)⇒1f(u)≈11−(rhurt)d−1[1+(d−1)δ22(d−3)(rhurt)d−11−(rhurt)d−31−(rhurt)d−1]. In the last line we have made a Taylor expansion of the function around δ=0 and neglected the higher order terms. We can use this expression of lapse function in eq. (11) to find the subsystem length to be(69)l=2rt[∫01ud−2du1−u2(d−2)11−(rhurt)d−1+(d−1)δ22(d−3)(rhrt)d−1×∫01u2d−3du1−u2(d−2)(1−(rhurt)d−3)(1−(rhurt)d−1)3/2]. It is not possible to evaluate both the above integrals analytically in this existing form. We use the following identities(70)11−y=∑n=0∞Γ(n+1/2)πΓ(n+1)yn;1(1−y)32=∑n=0∞2Γ(n+3/2)πΓ(n+1)yn to express the above integrals in a convenient form which makes analytical solution possible. We rewrite eq. (69) as(71)lrt2=∑n=0∞Γ(n+12)πΓ(n+1)(rhrt)(d−1)n∫01u(d−1)n+d−2du1−u2(d−2)+(d−1)δ22(d−3)(rhrt)d−1∑n=0∞2Γ(n+32)πΓ(n+1)(rhrt)(d−1)n×∫01u(d−1)n+2d−3du1−u2(d−2)(1−(rhurt)d−3)⇒lrt=∑n=0∞Γ(n+12)(d−2)Γ(n+1)Γ((d−1)n+d−12(d−2))Γ((d−1)n+2d−32(d−2))(rhrt)(d−1)n+(d−1)δ2d−3∑n=0∞Γ(n+32)(d−2)Γ(n+1)×(rhrt)(d−1)n+d−1[Γ((d−1)n+2d−22(d−2))Γ((d−1)n+3d−42(d−2))−(rhrt)d−3Γ((d−1)n+3d−52(d−2))Γ((d−1)n+4d−72(d−2))]. Let us now look at the form of divergence for different terms of the above expression. For large value of n, the first term behaves as ∼1n(rhrt)n and the second and third terms behave in same way as ∼(rhrt)n. Therefore the divergences of the second and third term cancels out each other. We isolate the divergent terms to get(72)lrt=πd−2Γ(d−12(d−2))Γ(2d−32(d−2))+∑n=1∞[Γ(n+12)(d−2)Γ(n+1)Γ((d−1)n+d−12(d−2))Γ((d−1)n+2d−32(d−2))−2(d−2)(d−1)1n](rhrt)(d−1)n+(d−1)δ2(d−3)(d−2)∑n=0∞[Γ(n+32)Γ(n+1)Γ((d−1)n+2d−22(d−2))Γ((d−1)n+3d−42(d−2))−2(d−2)(d−1)](rhrt)(d−1)n+d−1−(d−1)δ2(d−3)(d−2)∑n=0∞[Γ(n+32)Γ(n+1)Γ((d−1)n+3d−52(d−2))Γ((d−1)n+4d−72(d−2))−2(d−2)(d−1)](rhrt)(d−1)n+2d−4+δ2d−32(d−1)d−2(1−(rhrt)d−3)(1−(rhrt)d−1)(rhrt)d−1−2(d−1)(d−2)log(1−(rhrt)d−1). Now in the high temperature limit, rh takes large value. We can say that rh∼rt. Hence we can write rt=(1+ϵ)rh where ϵ is a very small positive number. With this the above equation reads(73)lrh=−2(d−1)(d−2)log((d−1)ϵ)+C1+δ2C2+O(ϵ) where(74)C1=πd−2Γ(d−12(d−2))Γ(2d−32(d−2))+∑n=1∞[Γ(n+12)(d−2)Γ(n+1)Γ((d−1)n+d−12(d−2))Γ((d−1)n+2d−32(d−2))−2(d−2)(d−1)1n]C2=2(d−2)(d−1)+(d−1)(d−3)(d−2)∑n=0∞[Γ(n+32)Γ(n+1)Γ((d−1)n+2d−22(d−2))Γ((d−1)n+3d−42(d−2))−2(d−2)(d−1)]−(d−1)δ2(d−3)(d−2)∑n=0∞[Γ(n+32)Γ(n+1)Γ((d−1)n+3d−52(d−2))Γ((d−1)n+4d−72(d−2))−2(d−2)(d−1)]. From the definition of δ, we had earlier obtained(75)T=(d−1)rh4π(1−δ2). Substituting the value of rh from the above equation in eq. (73) and simplifying, we obtain(76)ϵ≈ϵente−d−22(d−1)4πTl(1+δ2) where ϵent=1d−1eC1+C2δ2. Now the surface area reads(77)A=2(Lrt)d−3[∫01duu−d+21−u2(d−2)11−(rhurt)d−1+(d−1)δ22(d−3)(rhrt)d−1∫01duu(1−(rhurt)d−3)1−u2(d−2)(1−(rhurt)d−1)3/2]=2(Lrt)d−3[∫01duu−d+21−u2(d−2)(1+∑n=1∞Γ(n+12)πΓ(n+1)(rhurt)(d−1)n)+(d−1)δ22(d−3)(rhrt)d−1×∫01duu(1−(rhurt)d−3)1−u2(d−2)(∑n=0∞2Γ(n+32)πΓ(n+1)(rhurt)(d−1)n)]. In the last line of the above equation we have used the relation (70) to evaluate the integrals analytically. The first integral corresponds to the area integral for pure AdS spacetime. Only this integral is divergent near u→0. To remove this divergence, we introduce UV cut-off 1rb and add a counter term to obtain a finite value of surface area(78)Afinite=2(Lrt)d−3[∫rtrb1duu−d+21−u2(d−2)+∑n=1∞Γ(n+12)πΓ(n+1)(rhrt)(d−1)n∫01duu(d−1)n−d+21−u2(d−2)+(d−1)δ22(d−3)∑n=0∞2Γ(n+32)πΓ(n+1)(rhurt)(d−1)n+d−1×∫01duu(d−1)n+11−u2(d−2)(1−(rhrt)d−3)]−2(Lrb)d−3d−3=(Lrt)d−3[πd−2Γ(3−d2(d−2))Γ(12(d−2))+∑n=1∞1d−2Γ(n+1/2)Γ(n+1+d(n−1)2(d−2))Γ(n+1)Γ(n+1+nd2(d−2))(rhrt)(d−1)n+δ2(d−1)(d−2)(d−3)∑n=0∞Γ(n+3/2)Γ(2+n(d−1)2(d−2))Γ(n+1)Γ(d+n(d−1)2(d−2))(rhrt)(n+1)(d−1)−δ2(d−1)(d−2)(d−3)∑n=0∞Γ(n+3/2)Γ((d−1)(n+1)2(d−2))Γ(n+1)Γ(n(d+1)+2d−32(d−2))(rhrt)n(d−1)+2(d−2)](79)=(Lrt)d−3[πd−2Γ(3−d2(d−2))Γ(12(d−2))+∑n=1∞1d−2(1+d−2n+1+(n−1)(d−2))×Γ(n+1/2)Γ((n+1)(d−1)2(d−2))Γ(n+1)Γ(n+1+(n+2)(d−2)2(d−2))(rhrt)(d−1)n+δ2(d−1)(d−2)(d−3)∑n=0∞(1+d−22+n(d−1))×Γ(n+3/2)Γ((n+2)(d−1)2(d−2))Γ(n+1)Γ(n+2+(n+3)(d−2)2(d−2))(rhrt)(n+1)(d−1)−δ2(d−1)(d−2)(d−3)∑n=0∞(1+d−2n(d−1)+d−1)×Γ(n+3/2)Γ(n+1+(n+3)(d−2)2(d−2))Γ(n+1)Γ(n+1+(n+4)(d−2)2(d−2))(rhrt)n(d−1)+2(d−2)]. Now we can use eq. (71) to recast the surface area as(80)Afinite=(Lrt)d−3[π3−dΓ(d−12(d−2))Γ(2d−32(d−2))+lrt+∑n=1∞Γ(n+12)Γ((d−1)n+d−12(d−2))(rhrt)(d−1)n((d−1)n−(d−3))Γ(n+1)Γ((d−1)n+2d−32(d−2))+(d−1)δ2d−3×∑n=0∞Γ(n+32)Γ(n+1){Γ((d−1)n+2(d−1)2(d−2))(rhrt)(d−1)n+d−1((d−1)n+2)Γ((d−1)n+3d−42(d−2))−Γ((d−1)n+3d−52(d−2))(rhrt)(d−1)n+2d−4((d−1)n+d−1)Γ((d−1)n+4d−72(d−2))}]. The first summation term goes as ∼1n2(rhrt)n, where as the second and third summation terms goes as ∼1n(rhrt)n for large n. The leading contribution to the area Afinite comes from rh=rt. Those summation terms are not divergent in this limit. But as we know from [39] that the turning point of the extremal surface cannot penetrate the horizon radius rh, we have to substitute rt=rh(1+ϵ) in eq. (80) and expand binomially to get terms that are higher order in ϵ. Then it is easy to see that the summation terms are not convergent at the order O(ϵ) and also at higher order. As in this case rh is very large, ϵ is likely to be a tiny quantity. So we consider contribution to surface area only upto order O(ϵ). We now use rt=rh(1+ϵ) and separate the divergent terms from the summations to express eq. (80) as(81)Afinite=Ld−3lrhd−2+Ld−3rhd−3(K1+δ2K2)+Ld−3rhd−3(K3ϵ+δ2(K4ϵ+K5ϵlogϵ)) where(82)K1=π3−dΓ(d−12(d−2))Γ(2d−32(d−2))+2(d−2)(d−1)32ξ(2)+∑n=1∞(1((d−1)n+3−d)Γ(n+1/2)Γ((n+1)(d−1)2(d−2))Γ(n+1)Γ((d−1)n+2d−32(d−2))−1n22(d−2)(d−1)32)K2=(d−1d−3)[(Γ(d−1d−2)2Γ(3d−42(d−2))−Γ(3d−52(d−2))(d−1)Γ(4d−72(d−2)))Γ(3/2)+∑n=1∞(12+n(d−1)Γ(n+32)Γ((d−1)(n+2)2(d−2))Γ(n+1)Γ((d−1)n+3d−42(d−2))−1n2(d−2)(d−1)32)+∑n=1∞(1(n+1)(d−1)Γ(n+32)Γ((d−1)n++3d−52(d−2))Γ(n+1)Γ((d−1)n+3d−72(d−2))−1n2(d−2)(d−1)32)K3=2(d−2)d−1(log(d−1)−1)K4=(d−1d−3)[2(d−2)Γ(3d−52(d−2))(d−1)Γ(4d−72(d−2))−(d−1)Γ(d−1d−2)2Γ(3d−42(d−2))]Γ(32)−2(d−2)d−1log(d−1)K5=−2(d−2)d−1. Therefore the renormalized HEE in the small charge regime is given by(83)SAfinite=Ld−3lSBH+(Lrh)d−34GNd(K1+δ2K2)+Ld−3rhd−34GNd(K3ϵ+δ2(K4ϵ+K5ϵlogϵ)) where SBH=rhd−24GNd.3.2.2Large charge limitThe HEE of a non-extremal AdS-RN black hole with high charge limit has been computed in this section. The non-extremality condition (13) sets the horizon radius (rh) at large value and we have rhl≫1. Hence all the assumptions made in section 3.1.2 are also applicable in this case. Hence we can now Taylor expand eq. (15) around u0=rtrh and neglect higher order terms to obtain the lapse function as(84)f(u)=f(u0)+f′(u0)(u−u0)+O((u−u0)2)≈[(d−1)−(d−3)Q2rh2(d−2)](1−uu0). Let us denote the prefactor infront of the above equation by σ, so that σ=(d−1)−(d−3)Q2rh2(d−2). This factor also had appeared in the expression for black hole temperature in eq. (3). In the low temperature limit we have σ→0 and in high temperature limit we have σ→(d−1). Now the length of the subsystem reads(85)l=2rtσ∫01ud−2du1−u2(d−2)11−rhurt⇒lrt=2σ∑n=0∞Γ(n+12)πΓ(n+1)(rhrt)n∫01un+d−2du1−u2(d−2)=1(d−2)σ∑n=0∞Γ(n+12)πΓ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))(rhrt)n where we have used eq. (70). For large value of n, this expression is divergent. Using gamma function properties and Stirling formula, one can see that for large value of n, the above summation goes as 2(d−2)n(rhrt)n. Therefore it is divergent as rt→rh. To get a finite value we isolate the divergent terms to get(86)lrt=π(d−2)σΓ(d−12(d−2))Γ(2d−32(d−2))+1σ∑n=1∞(Γ(n+12)(d−2)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2d−21n)(rhrt)n−2(d−2)σlog(1−rhrt). Now we substitute rt=rh(1+ϵ) and expand around ϵ to finally obtain(87)σlrh=−2d−2log(ϵ)+D1+O(ϵ) where(88)D1=π(d−2)Γ(d−12(d−2))Γ(2d−32(d−2))+∑n=1∞{Γ(n+12)(d−2)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2d−21n}. From the above equation, one can find(89)ϵ=ϵente−(d−2)σ2lrh where ϵent=ed−22D1. Now we shall compute the extremal surface area which reads(90)A=2(Lrt)d−3σ∫01du1ud−21−u2(d−2)1σ(1−rhrtu). Since we want an analytical expression of area we use eq. (70) to rewrite the area integral as(91)A=2(Lrt)d−3σ∑n=0∞Γ(n+12)πΓ(n+1)(rhrt)n∫01un−d+2du1−u2(d−2). This integral seems to be divergent for n<(d−2). The divergence occurs when u→0. So we can put an UV cutoff rt/rb in the integrals to regularize it from n=0 to n=d−3. We use the same procedure as used in section 3.1.2 to obtain the finite part of area integrals as(92)A0finite=−2π(Lrt)d−3(d−3)σΓ(d−12(d−2))Γ(12(d−2))Amfinite=2(Lrt)d−3σΓ(n+12)πΓ(n+1)(rhrt)m×[1d−m−3+π2(d−2)Γ(m−d+32(d−2))Γ(m+12(d−2))];(m=1,2,...,d−4)Ad−3finite=2(Lrt)d−3Γ(2d−52)πσΓ(d−2)(rhrt)d−3log42(d−2). For n≥(d−2), we get(93)An≥(d−2)=2(Lrt)d−3σ∑n=(d−2)∞Γ(n+12)πΓ(n+1)(rhrt)n∫01un−d+2du1−u2(d−2)=2(Lrt)d−3σ∑n=(d−2)∞12(d−2)Γ(n+12)Γ(n+1)Γ(n−d+32(d−2))Γ(n+12(d−2))(rhrt)n=(Lrt)d−3σ∑n=(d−2)∞(1d−2+1n−d+3)Γ(n+12)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))(rhrt)n. Using eq. (85) we can recast the above expression in terms of the subsystem length as below(94)An≥(d−2)finite=(Lrt)d−3σ[σlrt−∑m=0d−31d−2Γ(m+12)Γ(m+1)Γ(m+d−12(d−2))Γ(m+2d−32(d−2))(rhrt)m+∑n=d−2∞1n−d+3Γ(n+12)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))(rhrt)n]. For large value of n, the third term goes as 2(d−2)n2(rhrt)n. Hence this term is finite as rt→rh. Hence we can get a finite leading order contribution to this term. But as rt=rh(1+ϵ), one can see that the third term in eq. (94) is divergent at first order in ϵ. So we separate the divergent terms to rewrite eq. (94) as(95)An≥(d−2)finite=(Lrt)d−3σ[σlrt−∑m=0d−31d−2Γ(m+12)Γ(m+1)Γ(m+d−12(d−2))Γ(m+2d−32(d−2))(rhrt)m+∑n=d−2∞{1n−d+3Γ(n+12)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2(d−2)n2}(rhrt)n+∑n=d−2∞2(d−2)n2(rhrt)n]. Now we can write(96)∑n=d−2∞2(d−2)n2(rhrt)n=2(d−2)[Li2[rhrt]−∑m=1d−31m2(rhrt)m]. Using this, the total surface area reads(97)Afinite=(Lrt)d−3σ[σlrt−2π(d−2)d−3Γ(d−12(d−2))Γ(12(d−2))+∑n=1d−4Γ(n+12)πΓ(n+1)(πn+1Γ(n−d+32(d−2))Γ(n+12(d−2))+2[d−(n+3)])(rhrt)n+Γ(2d−52)(d−2)πΓ(d−2)(log4−Γ(3/2))(rhrt)d−3+2(d−2)(Li2[rhrt]−∑n=1d−31n2(rhrt)n)+∑n=d−2∞(1n−d+3Γ(n+12)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2(d−2)n2)(rhrt)n]. Now we substitute rt=rh(1+ϵ) in eq. (97) to know the sub-leading term upto order ϵ. After simplification we finally obtain the finite part of the area of extremal surface to be(98)Afinite=Ld−3lrhd−2+(Lrh)d−3σ{K1′+K2′ϵ+O(ϵ2)} where(99)K1′=−2πd−2d−3Γ(d−12(d−2))Γ(12(d−2))+d−22ξ(2)−d2∑n=1d−31n2+∑n=1d−4Γ(n+12)πΓ(n+1)[πn+1Γ(n−d+32(d−2))Γ(n+12(d−2))+2d−(n+3)]+Γ(2d−52)π(d−2)Γ(d−2)(log4−πΓ(3/2))+∑n=d−2∞[1n−d+3Γ(n+12)Γ(n+1)Γ(n+d−12(d−2))Γ(n+2d−32(d−2))−2(d−2)n2]K2′=Γ(2d−72)πΓ(d−3)(2+πΓ(−12(d−2))(d−3)Γ(d−32(d−2)))−2(d−2). Therefore the renormalized holographic entanglement entropy is given by (in the large charge regime for non-extremal black hole)(100)SAfinite=Ld−3lSBHext+(Lrh)d−34GNσ{K1′+K2′ϵ+O(ϵ2)} where SBHext=rhd−24GNd.4Entanglement thermodynamicsIn this section, we investigate the entanglement thermodynamics in the small charge limit. To carry out this study, one has to note from the AdS/CFT duality that in the presence of a RN black hole, an extremal black hole has to be considered as the dual to the ground state (zero temperature state) of the boundary field theory. Further, a non-extremal black hole has to be considered as the dual to the excited state (finite temperature state) of the boundary field theory. The difference in the entanglement entropies of these two states would lead to the first law of entanglement thermodynamics. Hence we obtain(101)ΔSA=ΔEATent where(102)ΔSA=SA−SAext=kLd−3l2(M−Mext)(103)ΔEA=∫Adx1dx2...dxd−3Ttttemp≠0−∫Adx1dx2...dxd−3Ttttemp=0=d−216πGNdLd−3l(M−Mext). Substituting these two relation in eq. (101), the entanglement temperature reads(104)Tent=2(d−2)2π(Γ(d−12(d−2))Γ(12(d−2)))2[1Γ(1d−2)Γ(d2(d−2))−Γ(d−12(d−2))Γ(12(d−2))].5ConclusionWe have explicitly investigated the entanglement thermodynamics for d-dimensional charged black hole by studying the holographic entanglement entropy in different cases. We computed the holographic entanglement entropy in extremal and non-extremal cases in two different regimes, namely, the small charge limit and the large charge limit. For non-extremal black hole, there are two limiting cases, namely, the low temperature limit and the high temperature limit. We have calculated the holographic entanglement entropy for all these cases. It is observed that the holographic entanglement entropy for small charged extremal black hole is the same as the holographic entanglement entropy in the low temperature regime for small charged non-extremal black hole. 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