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23.3 Solutions with a G 2 on S 2 365where a, A and B are constants. The energy density and pressure obeyκ 0 p − ρ = κ 0 µ − m +1 4m +1− n2ρ =m − 1 4T 2m F 2[T ˙ 2 ]T 2 − ɛa2 ,(23.14c)H 2 T 2m F 2 ρ ≡ nG′ P ′2GP − P ′2 (3 − 4m)G′2+4P 2 4G 2 + (4m − 3 − n2 )ɛa 2 H 2,4where F 2 = G 1−2m P n . There are two differential equations linking G,H and P , so in general one function, corresponding to a choice of thecoordinate x, can be chosen freely. Four families were examined in detailby Ruiz and Senovilla (1992) and are given below. Each contains solutionswith an H 3 mentioned in §23.2. The fluid has a γ-law equation of statewith γ =2m/(m +1) if n 2 =4m + 1 (these cases were considered byVan den Bergh and Skea (1992)) or if T =e at , ɛ = 1, which implies ahomothety. The latter case contains the solutions found by Uggla (1992),see §23.2, and also the special solutions without restrictions on n andm found by Kamaniand Mansouri(1996) by taking H 2 G 1−2m P n =1,P =e qx , G =e sx for constants q and s.The first family is given by ɛ = 0. For general m and n,P k+l = C (3−4m)/(1−2m) + r, G = P k C 2(1−m)/(2m−1) ,(23.15)√nk + l = −2 +3− 4m, k = n + √ n 2 +3− 4m, H = sC ′ /P l ,1 − m4m − 3where r and s are arbitrary constants, and C(x) is an arbitrary function.The A = 0 limit gives the static cylindrically-symmetric solutions (22.27).The case C =(1+x 2 ) q , k = l = −N =1,m =(3q +2)/2(2q + 1),n =(q +1)/(2q +1) was found by Davidson (1992) and includes a γ =4/3solution (Davidson 1991) for q = −2/5. The case P =1,T = t, G = x c ,for some constant c, admits an H 3 which may act on T 3 or S 3 and withrespect to which the fluid is tilted. The formulae above have zero or infiniteexponents for three choices of m, which thus must be treated separately:they arem =(n 2 +3)/4, P = exp(ax 2n2 /(1+n 2) ),G = P 1/n (ax) (1−n2 )/(1+n 2) , H = P 1/n ;(23.16a)m = 3 4 , P4n = ax + r, G =e −ax P 1/2n , H =e −ax P 1/2n−4n ; (23.16b)m = 1 2 , Pk+l = C + r, G = P k /C, H = sC ′ /CP l , (23.16c)