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A 3+1 decomposition of spacetime describes spacetime as spatial hypersurfaces labelled by a time coordinate $t$, with a positive-definite spatial induced metric ${\gamma }_{ij}$. In units $c=1$ and metric signature $(-,+,+,+)$,The positive lapse function $\alpha$ measures proper time between nearby slices along the future unit normal; the shift vector ${\beta }^i$ describes the spatial displacement of the coordinate grid relative to that normal. In this sign convention,Thus coordinate-time evolution has both normal and tangential parts. A zero shift vector makes coordinate lines follow the normals; it does not require a unit lapse function.

Flat Minkowski spacetime illustrates the distinction. In Cartesian coordinates $\alpha =1$, $\beta =0$ and ${\gamma }_{ij}={\delta }_{ij}$. Changing spatial labels to ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$ gives the same geometry with $\beta =\mathbf{v}$ and the same spatial induced metric. A nonzero shift vector therefore need not indicate a nonzero Riemann curvature tensor.

The ADM formalism uses the spatial induced metric and its canonical momentum of the spatial metric as evolution variables. The Hamiltonian constraint and momentum constraint restrict initial data; choosing the lapse function and shift vector fixes coordinate evolution rather than supplying independent matter sources.

The Arnowitt-Deser-Misner energy is the Hamiltonian boundary charge generating an asymptotic time translation in the ADM formalism. For a spatial metric tensor $h_{ij}$ in asymptotically Cartesian coordinates, use geometrized units $G=c=1$:Here $n^i$ is the outward Euclidean unit normal vector. Standard differentiable asymptotic flatness gives $h_{ij}={\delta }_{ij}+O(r^{-1})$ and ${\partial }_k{h}_{ij}=O(r^{-2})$; suitable additional decay and integrability conditions ensure that the limit exists. Together with ADM momentum, this charge forms the asymptotic energy-momentum four-vector.^[ref]

For a time-symmetric Schwarzschild spacetime slice in isotropic coordinates, $h_{ij}={\psi }^4{\delta }_{ij}$ with $\psi =1+M/(2r)$. Its surface integrand is $-2{\partial }_r{\psi }^4=4M/r^2+O(r^{-3})$. Integrating over $S_r$ therefore gives $E_{ADM}=M$. The corresponding slice of Minkowski spacetime has constant $h_{ij}={\delta }_{ij}$ and zero ADM energy. In a moving asymptotic frame, the energy component need not equal the invariant ADM mass.