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Solutions with an equation of state p = p(µ)16.2 Non-static solutions 259Solutions which obey an equation of state have been discussed byWyman (1946), and special cases by Taub (1968). Because of (15.46),(15.47) and ν =ln˙λ − f(t), vanishing shear impliesp ′ = −(µ + p) ˙λ ′ / ˙λ, ˙µ = −3(µ + p) ˙λ. (16.52)If µ + p vanishes, then µ and p are constant, and the solution in questionis the static vacuum solution of Kottler, see Theorem 15.5 and Table 15.1.If µ + p is different from zero, but µ ′ vanishes (µ = µ(t),p = p(t)), thenwe have µ ′ = p ′ = ν ′ = ˙λ ′ =0. We choose the time coordinate t so that νis zero, and infer from (16.24) thatλ = λ 1 (r)+λ 2 (t), ˙λ = ˙λ2 =e f (16.53)holds. This special time dependence of λ is compatible with (16.24) onlyif ϕ =0=F. So the solutions with µ = µ(t),p= p(t) are the subcasee −λ = A(t)[1 + εr 2 /4], ε =0, ±1, e ν = 1 (16.54)of (16.50). These are exactly the Friedmann-like universes (§14.2).If neither µ + p nor µ ′ vanishes, then ˙λ ′ is not zero and can be eliminatedfrom (16.52), which leads to ˙µµ ′ =(µ + p)˙µ ′ . This is integrated by∫ln ˙µ =lnM(µ) + ln ˙α(t), ln M(µ) ≡ [µ + p(µ)] −1 dµ, (16.55)and in a further step byH(µ) =α(t)+β(t),∫H(µ) ≡M −1 (µ)dµ. (16.56)Choosing the time coordinate so that α = t (˙α = 0 is prohibited byµ + p ̸= 0, ˙λ ̸= 0⇒ ˙µ ̸= 0, cp. (16.52)), we see that because of (16.56)and (16.52) the functions µ and λ must have the special t dependenceµ(r, t) =µ(v), λ(r, t) =λ 1 (v)+λ 2 (r 2 ), v ≡ t + G(r 2 ). (16.57)To determine the functions λ 1 ,λ 2 ,G and the function F occurring in thefirst integral (16.27), we insert (16.57) into (16.27). Writinge −λ = L = u(v)l(x), x = r 2 , (16.58)we obtain˙u(G ,xx l +2G ,x l, x )+ul ,xx +ülG 2 ,x = u 2 l 2 F. (16.59)