Γ \to \infty

requires

v_{b} = - 3 Θ_{1} + O (Γ^{- 1}) .

Let

S = 3 Θ_{1} + v_{b}

. Rearranging the baryon Euler equation and substituting the leading relation only on its right-hand side gives

S = - \frac{R}{Γ} (v_{b}^{'} + H v_{b} + k Ψ) = \frac{R}{Γ} (3 Θ_{1}^{'} + 3 H Θ_{1} - k Ψ) + O (Γ^{- 2}) .

Insert this slip into the photon Euler equation:

3 (1 + R) Θ_{1}^{'} + 3 H R Θ_{1} = k Θ_{0} + k (1 + R) Ψ.

Θ_{1} = - (Θ_{0}^{'} - Φ^{'}) / k

. Eliminating the dipole yields

Θ_{0}^{''} + \frac{H R}{1 + R} Θ_{0}^{'} + c_{s}^{2} k^{2} Θ_{0} = - \frac{k ^{2}}{3} Ψ + Φ^{''} + \frac{H R}{1 + R} Φ^{'},

c_{s}^{2} = \frac{1}{3 ( 1 + R )} .

Solved by gpt-5.6-sol high.

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