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Relative to a timelike unit normal $n$, the electric part is $E_{ab}=C_{acbd}{n}^c{n}^d$. It describes the tidal component of the Weyl tensor measured by that observer. Antisymmetry in each curvature pair makes it spatial in both slots; pair symmetry and vanishing traces make it symmetric and trace-free.

For a proper Euclidean motion of Euclidean three-space $E(x)=Ax+b$, transport the unit normal by $\tilde{N}(E(p))=AN(p)$. The corresponding shape operators are orthogonally conjugate:Their determinant and trace, hence Gaussian curvature and mean curvature, are unchanged.

With signature $(-,+,+,+)$, future unit normal $n$ and $K_{ab}=-P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$, the Gauss–Codazzi equations become $(\perp R{)}_{abcd}={\mathcal{R}}_{abcd}+K_{ac}{K}_{db}-K_{ad}{K}_{cb}$ and $(\perp R{)}_{abcd}{n}^d=-D_a{K}_{bc}+D_b{K}_{ac}$, where in the second expression only the first three indices are projected. The intrinsic curvature and spatial covariant derivative use the spatial induced metric. The normal is timelike, so the normal-sign convention differs from a spacelike normal in a positive-definite ambient metric.

A nowhere-zero differential one-form $n$ is hypersurface orthogonal if locally $n=Fdt$ with $F\ne 0$. Its kernel consists of tangent vectors to the level hypersurfaces of $t$. Frobenius theorem makes this equivalent to $n\wedge dn=0$, where $d$ is the exterior derivative. With a torsion-free Levi-Civita connection, this is $n_{[\alpha }{\nabla }_{\beta }{n}_{\gamma ]}=0$. For a timelike unit normal, projecting the derivative on both indices gives zero antisymmetric part, which explains why hypersurface orthogonality implies symmetric extrinsic curvature.

In four spacetime dimensions, relative to a timelike unit normal $n$, define $B_{ab}=\frac{1}{2}{ϵ}_{acde}{C}^{de}{}_{bf}{n}^c{n}^f$, with the chosen spacetime metric volume tensor. This is the dual tidal part of the Weyl tensor. Normal contractions vanish by curvature and volume-tensor antisymmetry. Its sign depends on the specified orientation and dualization convention.

For a positive lapse function $\alpha$, shift vector ${\beta }^i$ and spatial induced metric ${\gamma }_{ij}$, the block metric has $g_{00}=-{\alpha }^2+{\gamma }_{ij}{\beta }^i{\beta }^j$ and $g_{0i}={\gamma }_{ij}{\beta }^j$. Its block determinant is $g=-{\alpha }^2\det \gamma$. Therefore the metric volume tensor contracted with the future unit normal restricts to the spatial volume tensor: $n^{\mu }{ϵ}_{\mu ijk}=\sqrt{\det \gamma }[ijk]$.

Relative to a unit normal $\mathbf{n}$, a vector $\mathbf{v}$ decomposes orthogonally asThe first term is its normal component and the second its tangential component.

The normal derivative of a differentiable function $u$ along a hypersurface with unit normal $N$ is the directional derivativeHigher normal derivatives at a point $x$ may be defined by differentiating the one-variable function $s\mapsto u(x+sN(x))$ repeatedly at $s=0$.

A normal-field wall imposes zero tangential magnetic field: $\mathbf{B}\times \mathbf{n}=0$ for the wall unit normal $\mathbf{n}$. In a channel with imposed normal field $B_0\hat{\mathbf{z}}$, an induced tangential component $b(z)\hat{\mathbf{x}}$ therefore satisfies $b=0$ on each wall. This condition is distinct from the no-slip boundary condition on fluid velocity.

For a minimally coupled scalar field with spatial induced metric $h_{ij}$, the field's derivative along the future unit normal is $\Pi$. If the shift vector enters as $d{x}^i-N^{i}dt$, then $\Pi =(\dot{\varphi }+N^i{\partial }_i\varphi )/N$. The canonical momentum of the density is ${\pi }_{\varphi }=\sqrt{h}\Pi$. Confusing this density with the normal derivative loses a volume factor.

An orientation of a regular surface is a continuous choice of unit normal. Reversing orientation replaces $N$ by $-N$.

Replacing the chosen unit normal $N$ by $-N$ replaces the shape operator $S$ by $-S$. On a surface, Gaussian curvature $K=\det S$ is unchanged, while mean curvature $H=\mathrm{tr}(S)/2$ changes sign.