Christoffel symbols, we obtain ł[g11]{} \_i Ţ\^[i0]{}(y,t)= \_4 Ţ\^[40]{} +(3 +3 ) Ţ\^[40]{}+ \_0 Ţ\^[00]{}+ (2+3) Ţ\^[00]{}+\_[ab]{}Ţ\^[ab]{}=0 ,where ł[g11a]{} Ţ\^[04]{}e\^[-y]{} =Ţ\_, Ţ\^[00]{}e\^[-2y]{} =Ţ\_,Ţ\^[44]{} =Ţ\_,Ţ\^[ab]{}e\^[-2y]{}a\^[-2]{} (0,t) =Ţ\_. In the region $0<y<\b{y}$, using the approximate metric (\[24\]), we have $$n'/n \simeq -\m, \qquad  a'/a \simeq -\m,\qquad
\dot{a}/a\simeq H(t),\qquad \dot{n}/{n}\sim H(t)\r/\m
.$$ Taking into account that in the radiation-dominated period $T\sim a(0,t)^{-1}$, we obtain that $\pa_0 \c{T}_{\b{0}\b{0}}(0,t)\simeq 
-8H(t)\c{T}_{\b{0}\b{0}}(0,t)$. [^10] Thus, Eq.(\[g11\]) can be written as ł[g12a]{} (\_y -6)Ţ\_(0,t) +He\^[y]{}(-5 Ţ\_(0,t)+\_[ab]{}Ţ\_(0,t))=0 ,where $\psi \equiv e^{\m y}(e^{\m y}N(y,t)/N(0,t) )^5$. The trace of the energy-momentum tensor of massless particles is zero, $\c{T}^i_i =0$. Using (\[i7\]), we obtain a relation $$\c{T}_{\b{a}\b{b}}\eta_{ab}=\c{T}_{\b{0}\b{0}} - 
\c{T}_{\b{4}\b{4}}.$$ Eq.(\[g12a\]) can be rewritten as ł[g13]{} \_y -6+He\^[y]{}B=0 ,where $
B=(4\c{T}_{\b{0}\b{0}}+\c{T}_{\b{4}\b{4}})/(-\c{T}_{\b{0}\b{4}})\simeq 6.44. 
$ Solving (\[g13\]), we obtain ł[g14]{} ()\^5 = {- (e\^[y]{} -1)} .

Next, let us consider the $j=4$ component of the conservation equation of the energy-momentum tensor ł[g15]{}\_i Ţ\^[i4]{}(y,t)= \_4 Ţ\^[44 ]{} +(\^0\_[04]{}+\^a\_[a4]{}) Ţ\^[44]{}+ \_0 Ţ\^[04 ]{}+(\^0\_[00]{}+\^a\_[a0]{}) Ţ\^[04]{}+ \^4\_[00]{}Ţ\^[00]{} +\^4\_[ab]{}Ţ\^[ab]{}\
=\_4 Ţ\^[44 ]{} +(+3)Ţ\^[44]{} + \_0 Ţ\^[04 ]{}+( +3)Ţ\^[04]{}+ n’nŢ\^[00]{}-a’a \_[ab]{}Ţ\^[ab]{}=0 Repeating the same steps as for the component $j=0$, we obtain ł[g16]{} \_y-5-He\^[y]{}=0 ,where $\varphi \equiv (N (y,t) e^{\m y}/N(0,t) )^5$ and $B'= -5\c{T}_{\b{0}\b{4}}/\c{T}_{\b{4}\b{4}}\simeq 6.21$. We have used that at $y=0$ 