content_image/423016.jpg content_image/423025.jpg content_image/423029.jpg content_image/423026.jpg content_image/423027.jpg content_image/423028.jpg content_image/423015.jpg content_image/423020.jpg content_image/423032.jpg
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Substituting this relation in the screened Poisson equa- tion (64), we obtain the screened Emden equation 

<img src='content_image/423025.jpg'>

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 

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 

<img src='content_image/423029.jpg'>

<img src='content_image/423026.jpg'>

<img src='content_image/423027.jpg'>

<img src='content_image/423028.jpg'>

## C. The temperature 

<img src='content_image/423015.jpg'>

<img src='content_image/423020.jpg'>

<img src='content_image/423032.jpg'>

