0 50 100 150 200 250 0.005 0.01 0.015 0.02 0.025 0.03 10 15 20 25 5e-05 0.0001 0.00015 0.0002 d = 1 m 0 25 50 0.45 0.5 0.55 0.6 0.65 n=1 n=2 d=1 c 17 FIG. 28: Energy Λ as a function of α for the first two branches in d = 1. We have taken µ = 15 > µ m . FIG. 26: Energy Λ as a function of α for the first two branches in d = 1. We have taken µ = 1 < µ c . FIG. 29: Energy Λ as a function of α for the first two branches in d = 2. We have taken µ = 1. FIG. 27: Energy Λ as a function of α for the first two branches in d = 1. We have taken µ = 10 > µ c . 1 100 1 2 3 n=1 n=2 d = 2 0 50 100 150 200 0.01 0.02 0.03 0.04 0.05 n=1 n=2 d = 1 c D. The energy We must also relate α to the energy E . The total energy is given by Using α = ( S$_{d}$Gβmρ$_{0}$ ) 1 / $^{2}$R and introducing the dimen- sionless temperature (51), we obtain Using the Maxwell-Boltzmann distribution (10), the ki- netic energy is simply Using the screened Poisson equation (64) and integrating by parts, the potential energy can be written The total energy E = K + W is therefore given by This equation gives the relation between the inverse tem- perature η and α for the n -th branch. In Figs. 22, 23, 24 and 25, we plot the inverse temperature η as a func- tion of α for the first two branches n = 1 , 2 in different dimensions of space d = 1 , 2 , 3. The discussion is similar to the one given in Sec. III C. We have also represented the branch corresponding to the homogeneous solution. Its equation is given by η = α $^{2}$/d . The branch n = 1 of inhomogeneous solutions connects the branch of homo- geneous solutions at α 2 c = dη ∗ c (see Appendix F). E = ∫ f v 2 2 d r d v + 1 2 ∫ ρ Φ d r . (74) K = d 2 Nk$_{B}$T. (75) W = − 1 2 S$_{d}$G ∫ [ ( ∇ Φ) 2 + k 2 $_{0}$Φ $^{2}$] d r . (76) E = d 2 Nk$_{B}$T − 1 2 S$_{d}$G ∫ [ ( ∇ Φ) 2 + k 2 $_{0}$Φ $^{2}$] d r . (77) η = 1 α d − 2 ∫ α 0 e − $^{ψ}$ξd − $^{1}$dξ. (73)