16 0 25 50 75 100 75 100 125 150 175 n=1 n=2 d=1 c 0 10 20 30 40 50 20 40 60 80 100 n=1 n=2 d=1 c FIG. 23: Inverse temperature η as a function of α for the first two branches in d = 1. We have taken µ = 10 > µ c . FIG. 22: Inverse temperature η as a function of α for the first two branches in d = 1. We have taken µ = 1 < µ c . FIG. 24: Inverse temperature η as a function of α for the first two branches in d = 2. For α → $^{+}$∞, the inverse temperature of the first branch tends to η c = 4. We have taken µ = 1. 0 50 100 150 200 5 10 15 20 25 30 35 n=2 n=1 d=2 $_{c}$=4 Substituting this relation in the screened Poisson equa- tion (64), we obtain the screened Emden equation where κ = k$_{0}$/ ( S$_{d}$Gβmρ$_{0}$ ) 1 / 2 and λ = − k 2 $_{0}$Φ$_{0}$ /S$_{d}$Gρ$_{0}$ . The boundary conditions at the origin are The normalized box radius is α = ( S$_{d}$Gβmρ$_{0}$ ) 1 / $^{2}$R and the boundary condition Φ $^{′}$( R ) = 0 becomes Introducing the normalized screening length 1 ξ d − 1 d dξ ( ξ d − $_{1}$dψ dξ ) − κ $^{2}$ψ = e − ψ − λ, (68) ψ (0) = ψ $^{′}$(0) = 0 . (69) ψ $^{′}$( α ) = 0 . (70) µ = k$_{0}$R, (71) M = ρ$_{0}$S$_{d}$ ( R α ) $^{d}$∫ α 0 e − $^{ψ}$ξd − 1 dξ. (72) the parameter κ can be rewritten κ = µ/α . For given µ , we solve the problem as follows: (i) We fix α . (ii) κ = µ/α is then given. (iii) We determine λ by an iterative method such that ψ $^{′}$( α ) = 0. (iv) We obtain different solutions λ$_{n}$ ( α ) determining different branches n = 1, n = 2 etc. This procedure determines for each value of α , and for each branch, the normalized density profile e − ψ ( ξ $^{)}$. The homogeneous solution corresponds to ψ = 0 and λ = 1. This solution is degenerate because the boundary condition (70) is satisfied for any value of α . We must now relate the parameter α to the tempera- ture T . Introducing the dimensionless variables defined previously and recalling that r = Rξ/α , the mass can be written 0.01 1 100 10000 5 25 n=1 n=2 d=3 FIG. 25: Inverse temperature η as a function of α for the first two branches in d = 3. For α → $^{+}$∞, the inverse temperature of the first branch undergoes damped oscillations around the value η s ≃ 3 . 25. We have taken µ = 1. C. The temperature