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258 16 Spherically-symmetric perfectfluid solutions(Kustaanheimo and Qvist 1948). It contains, for x 1 =0,x 2 = −4R 2 , themetric of McVittie (1933)ds 2 =(1+f) 4 e g(t) [1 + r 2 /4R 2 ] −2 (r 2 dΩ 2 +dr 2 ) − (1 − f)dt 2 /(1 + f),2f = me −g(t)/2 [1 + r 2 /4R 2 ] 1/2 /r,R = const,(16.46)which has been interpreted as a mass in a Robertson–Walker universe(for a (global) interpretation of the shearfree perfect fluids see Sussman(1988b)).For A(t) = 0 and a = 0, one obtains the solutionds 2 = S 2 (t)(1 + h) 4 (r 2 dΩ 2 +dr 2 ) − (1 − h) 2 (1 + h 2 ) −2 dt 2 ,(16.47)h ≡ S −1 (t)(αr 2 + β) −1/2 .Solutions with a homogeneous distribution of matter µ = µ(t)If we assume a homogeneous distribution of matter, µ = µ(t), then wecan differentiate the field equation (16.28a) with respect to r and eliminateλ ′′ and λ ′′′ by means of (16.25). We obtain 3ϕ + rϕ ′ = 0, i.e. because of(16.27)F (x) =(2bx) −5/2 . (16.48)To satisfy (16.28a), which requires that e −2λ (2λ ′′ + λ ′2 +4λ ′ /r) dependsonly on t, we have to make an appropriate choice of A(t) and B(t) inthe general formula (16.42). The final result is: all shearfree, sphericallysymmetric,expanding perfect fluid solutions with the energy density µdepending only on t are given bywithords 2 =e 2λ(r,t) (r 2 dΩ 2 +dr 2 ) − ˙λ 2 e −2f(t) dt 2 (16.49)e −λ = A(t)+B(t)r 2 , 12AB =3e 2f − κ 0 µ (b = 0) (16.50)e −λ = √ 2bur, b = const ̸= 0,∫ [ ] (16.51)−1/22u 3 /3+b 2 u 2 + b(3e 2f − κ 0 µ)/6 du = (ln r)/b + B(t)(Kustaanheimo 1947). In general, the pressure p will depend on both tand r; the subcase p = p(t) is contained in the solutions considered in thefollowing paragraphs.