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For $t>0$ the nonzero Christoffel symbols, apart from lower-index symmetry, arePut $S=\sum p_i$ and $Q=\sum p_i^2$. Their substitution in the Ricci tensor gives $R_{00}=(S-Q)/t^2$ and $R_{ii}=p_i(S-1)t^{2p_i-2}$, with all off-diagonal entries zero. Thus either every $p_i=0$, or some exponent is nonzero andThe latter are the Kasner metric conditions; the all-zero static Minkowski solution is a separate vacuum possibility.

In an orthonormal frame the independent curvature components are proportional to $p_i(p_i-1)/t^2$ and $p_i{p}_j/t^2$. They all vanish precisely when the exponents are all zero or a permutation of $(1,0,0)$. The latter flat metric is Minkowski space in a Milne-type coordinate chart.

All three exponents cannot be strictly positive: in the nonstatic branch, $S^2-Q=2{\sum }_{i<j}{p}_i{p}_j=0$, contradicting positive pairwise products. If all three are nonzero, there is exactly one negative exponent: two negative ones would force the positive one above one, violating $Q=1$. Therefore one scale factor contracts and two expand as $t$ increases in that branch.

PutContracting the curvature components givesin the orthonormal frame, with no sum on $i$. Vacuum requires $S_1=S_2$ and $p_i(S_1-1)=0$ for every $i$. Since the $p_i$ are not all zero, some $p_i\ne 0$, forcing $S_1=1$ and then $S_2=1$. Conversely these conditions make every Ricci component vanish. Thus the Kasner metric is vacuum exactly whenso $A=B=1$.