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In four spacetime dimensions, the deformation tensor decomposes aswhere $\theta =B^{ab}{h}_{ab}$ is the expansion scalar, ${\sigma }_{ab}=B_{(ab)}-\theta h_{ab}/3$ is the symmetric trace-free shear tensor of a timelike congruence, and ${\omega }_{ab}=B_{[ab]}$ is the vorticity tensor of a timelike congruence.

For the normal flow with zero shift vector, $a_{\mathrm{loc}}=(\det h{)}^{1/6}$ obeys $H_{\mathrm{loc}}=N^{-1}{\partial }_t\log a_{\mathrm{loc}}=-K/3$ in the negative extrinsic curvature of a spatial hypersurface convention. The expansion scalar is $\theta =-K=3H_{\mathrm{loc}}$.

Set $h=\det h_{ij}$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. The Jacobi determinant derivative formula and metric compatibility giveConsequentlyThe local volume element is $\sqrt{h}{d}^{3}x$, so for zero shift vector the proper-time derivative along the normal is $d/ds=N^{-1}{\partial }_t$ andHere $\theta$ is the expansion scalar. Writing $a_{\mathrm{loc}}=h^{1/6}$ defines a local linear scale from this volume, whence the local volume Hubble parameter isThe factor of three converts volume expansion into linear expansion, and $N$ converts coordinate time into proper time. This agrees with the usual Hubble parameter in a homogeneous FLRW metric.

For a volume-shape decomposition the unimodular spatial metric is ${\gamma }_{ij}=a_{\mathrm{loc}}^{-2}{h}_{ij}$, with determinant one. If instead the printed positive power $a_{\mathrm{loc}}^2{h}_{ij}$ is taken literally, its determinant is $a_{\mathrm{loc}}^{12}$; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.

A timelike geodesic vector field is a smooth unit timelike field $u^a$ whose integral curves obey $u^b{\nabla }_b{u}^a=0$. Its expansion scalar, shear tensor of a timelike congruence, and vorticity tensor of a timelike congruence evolve according to the timelike Raychaudhuri equation.