= Solution
Use $\mathcal H=a'/a$ for the conformal expansion rate and $H=a'/a^2$ for the physical <Hubble parameter>. A future unit normal to a constant-$\tau$ slice is
$$
\boxed{n^\mu=(a^{-1},0,0,0),\qquad n_\mu=(-a,0,0,0).}
$$
It has norm minus one, and the <spatial projection tensor> is $P^\mu{}_\nu=\operatorname{diag}(0,1,1,1)$. The nonzero connections needed here are $\Gamma^0{}_{00}=\mathcal H$, $\Gamma^0{}_{ij}=\mathcal H\delta_{ij}$ and $\Gamma^i{}_{0j}=\mathcal H\delta^i{}_j$. Thus $\nabla_i n_j=-\Gamma^0{}_{ij}n_0=a'\delta_{ij}$. Projection removes every normal component, yielding
$$
\boxed{K_{00}=K_{0i}=0,\qquad K_{ij}=a'\delta_{ij},\qquad K^i{}_j=H\delta^i{}_j,\qquad K=3H.}
$$
This uses the <positive-expansion convention for extrinsic curvature> in the question. With the alternative convention that inserts an overall minus sign, all these extrinsic-curvature quantities change sign.
For a comoving cell, the physical volume is proportional to $a^3$. Its logarithmic derivative in <proper time> is $3\dot a/a=3H=K$. Thus \b[the trace is the expansion rate of the normal congruence]: positive for expansion and negative for contraction. The isotropic tensor $K^i{}_j=H\delta^i{}_j$ has vanishing trace-free shear. It measures homogeneous expansion, not the intrinsic spatial curvature, which is zero in this flat model.
Back to article page
