Entropy Calculation using brick wall method in Kerr-Newman AdS Black Hole

Entropy Calculation using brick wall method in Kerr-Newman AdS Black Hole Chandra Prakash chandra.prak[email protected] ABSTRACT Using the brick wall method, we will calculate the entropy of Kerr-Newman AdS black Hole and arrive at the already well established result. During the calculation, we will be using the generalized equation for FNSR and FSR mode of free energy. Introduction Ever since Bekenstein and Hawking derived the expression for black hole entropy, the question of what it means has always been our priority. For the first time, it was t’Hooft who in his paper titled “On the quantum structure of a black hole”, first tried to evaluate the entropy of free particles using quantum field theory in curved spacetime and statistical physics. Though the method has some limitations, but what we will consider here is a generalized method which will apply for any black hole. In the first section, we will consider the classic klein gordan equation and then after that calculate the free energy of excitations in quantum field. This free energy will then be used to calculate the total entropy of the thermal gas of quantum field excitations outside the event horizon in a thin film. = Klein-Gordon Equation in Kerr-Newman in AdS/CFT The line element for the Kerr-Newman in AdS/CFT is given by ds2=−−a2sin2θ∆θ+ ∆r ρ2Ξ2dt2+ρ2 ∆r dr2+ρ2 ∆θ dθ2+Σ sin2θ ρ2Ξ2dφ2−2asin2θ{∆r+ (r2+a2)∆θ} ρ2Ξ2dtdφ where, ∆r= (a2+r2)1−r2 l2+Q2−2GMr −αr1−3ω ∆θ= 1 + a2cos2θ l2 Ξ = 1 + a2 l2 Σ=∆θ(r2+a2)2−a2sin2θ∆r ρ2=r2+a2cos2θ=r2+a2−a2sin2θ l2=3 Λ Chandra Prakash 2 The inverse metric of above line integral can be given via: gµν = gφφ D0 0 gtφ −D 01 grr 0 0 0 0 1 gθθ 0 gtφ −D 0 0 gtt D  The Kerr-Newman has four coordinate singularity given by: ∆r= (r−rH)(r−r−)(r−rq)(r−rc) = 0 where, rHis the event horizon, r−is the cauchy horizon and rqand rcare the cosmological horizon. Substituting the KNAdS inverse-metric into the Klein Gordan Equation of massless scalar field . ∂ ∂xµ√−ggµν ∂ ∂xνψ= 0 Expanding it and solving we get: ∂ ∂t √−ggtt ∂ ∂tψ+∂ ∂r √−ggrr ∂ ∂r ψ+∂ ∂φ √−ggφφ ∂ ∂φψ +∂ ∂θ √−ggθθ ∂ ∂θ ψ+∂ ∂t √−ggtφ ∂ ∂φψ+∂ ∂φ √−ggφt ∂ ∂tψ= 0 Using WKB approximation i.e. ψ(t, r, θ, ψ) = eiR(r)eiS(θ)eimφe−iEt: √−ggtti2E2e−iEteiR(r)eiS(θ)eimφ +i∂ ∂r (√−ggrr)∂R(r) ∂r eiR(r)eiS(θ)eimφe−iEt +√−ggrr ∂2 ∂r2eiR(r)eiS(θ)eimφe−iEt +√−ggφφ ∂2 ∂φ2eimφeiS(θ)e−iEt +i∂ ∂θ √−ggθθ∂S(θ) ∂θ eiθeimφeiR(r)e−iEt +√−ggθθ ∂2 ∂θ2eiθeimφeiR(r)e−iEt −2i2mE√−ggtφeimφeiθeiR(r)e−iEt = 0 Comparing real and imaginary parts of the equation, we arrive at: ∂R(r) ∂r 2 =1 grr "−gttE2−m2gφφ −gθθ ∂S(θ) ∂θ 2 + 2mEgtφ# k2 r=1 grr −gttE2−m2gφφ −gθθ(kθ)2+ 2mEgtφ(using standard notation) Chandra Prakash 3 Free Energy and Entropy According to the theory of canonical ensemble and using semi-classical approximation, the free energy of scalar free particles within a shell of width L−(rH+)in Kerr-Newman AdS background is: βF =X E,m ln[1 −e−β(E−mΩH)] =Zdm Z∞ 0 dΓ(E) ln[1 −e−β(E−mΩH)] =Zdm Γ(E) ln[1 −e−β(E−mΩH)]∞ 0−Z∞ 0 Γ(E)e−β(E−mΩH) 1−e−β(E−mΩH)βdE =−βZdm Z∞ 0 Γ(E)1 eβ(E−mΩH)−1dE To proceed further we need to know the expression for Γ(E) = Rg(E)dE, which describes the total number of modes with energy less than Eand a fixed m, assuming φ(r, θ, ψ, t)6= 0 for rH+≤r≤Land φ= 0 outside this shell, is obtained by integrating over the volume of phase space: Γ(E, m) = Zdθdφ ZL rH+ dr 1 πZdkθkr =Zdθdφ ZL rH+ dr 1 πZdkθ 1 √grr −gttE2−m2gφφ −gθθ (kθ)2+ 2mEgtψ1/2 Our integral becomes, with appropriate limit for k2 r≥0: Γ(E) = π 2Zdθdφ ZL rH+ dr 1 pgrrgθθ −gttE2−m2gφφ + 2mEgtφ Substituting it back to the original equation and simplifying that further, we arrive at: F=−1 2Zdm Zdθdφ ZL rH+ dr 1 pgrrgθθ Z∞ 0 dE −gttE2−m2gφφ + 2mEgtφ eβ(E−mΩH)−1 A quick look at the integrand and we observe that this integral diverges at E=mΩH, which is why we need to split it into two parts, one where 0≤E≤mΩHand other where, ΩH< E < ∞. F=−1 2Zdm Zdθdφ ZL rH+ dr 1 pgrrgθθ ZmΩH 0 dE −gttE2−m2gφφ + 2mEgtφ eβ(E−mΩH)−1 −1 2Zdm Zdθdφ ZL rH+ dr 1 pgrrgθθ Z∞ mΩH dE −gttE2−m2gφφ + 2mEgtφ eβ(E−mΩH)−1 =FNSR +FSR Chandra Prakash 4 Evaluating FNSR, we get this final expression: FNSR =−2Γ(4)ζ(4) 3β4Zdθdφ ZL rH+ dr (grr gθθ )1/2 {(−D)}3/2(gφφ)2 (where −D =g2 tφ −gttgφφ) This particular integral is quite complicated to solve which is why we wil taylor expand it at r=rHlike this and neglect all other contributions: (grr gθθ )1/2 {(−D)}3/2(gφφ)2=1 (r−rH)2F(r, θ) = F(rH, θ) (r−rH)2+F0(rH, θ) (r−rH)+O(r−rH) where F(r, θ) = (grr gθθ )1/2 {(−D)}3/2(gφφ)2(r−rH)2 Using the metric and performing simplification to the result we arrive at: (grr gθθ )1/2 (−D)3/2(gφφ)2≈f(rH, θ)1 (r−rH)2 where f(rH, θ) = (r2 H+a2) 4 Ξ 4(r2 H+a2)2 [(rH−r−)(rH−rq)(rH−rc)]2 [∆2 θ(r2 H+a2)](r2 H+a2cos2θ)2Ξ3sin θ ∆2 θ(a2+r2 H)2+ 2a2sin2θ(a2+r2 H)3/2 Using this result our integral now becomes: FNSR ≈ −ζ(4) β4Zdθdφ f(rH, θ)δ (+δ)(using L=rH++δ) FSR ≈ −ζ(4) 2β4Zdθdφf(rH, θ)δ (+δ)(using L=rH++δ) Using the above results we can easily evaluate the total free energy in the thin film as: F=FNSR +FSR =−3ζ(4) 2β4Zdθdφ f(rH, θ)δ (+δ) The above expression for free energy can be simplified further by using surface gravity: κ=2π β =1 2(r2 H+a2)(rH−r−)(rH−rq)(rH−rc)(evaluated at r=rH) Chandra Prakash 5 and: dA=√gθθgφφdθdφ =sρ2 ∆θ Σ sin2θ ρ2Ξ2dθdψ =(r2 H+a2) Ξsin θdθdψ Using these two above results in our simplication, we arrive at: F=−3ζ(4) 2β4Zdθdφ(r2 H+a2) sin θ 4 Ξ 4(r2 H+a2)2 [(rH−r−)(rH−rq)(rH−rc)]2 [∆2 θ(r2 H+a2)](r2 H+a2cos2θ)2Ξ3 ∆2 θ(a2+r2 H)2+ 2a2sin2θ(a2+r2 H)3/2δ (+δ) =−3ζ(4) 8π2β2ZdA(r2 H+a2)(r2 H+a2cos2θ)2Ξ3 4(a2+r2 H)2+ 2a2sin2θ(a2+r2 H)3/2δ (+δ) Then finally the entropy becomes: S=β2∂F ∂β =−β2∂ ∂β "3ζ(4) 8π2β2ZdA(r2 H+a2)(r2 H+a2cos2θ)2Ξ3 4(a2+r2 H)2+ 2a2sin2θ(a2+r2 H)3/2 δ (+δ)# =3ζ(4) 4π2βZdA(r2 H+a2)(r2 H+a2cos2θ)2Ξ3 4(a2+r2 H)2+ 2a2sin2θ(a2+r2 H)3/2δ (+δ) =RdA 4(using 3ζ(4) π2β (r2 H+a2)(r2 H+a2cos2θ)2Ξ3 [(a2+r2 H)2+2a2sin2θ(a2+r2 H)]3/2hδ (+δ)i= 1) Conclusions and Discussions The result we finally got, tells us that the entropy of hawking particles in thin shell near the event horizon is actually proportional to area of event horizon! 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