Unit normal ($n$)

A unit normal is a normal vector of Euclidean norm one. Given a nonzero normal vector $n_0$, normalization gives $n=n_0/\Vert n_0\Vert$. This makes a dot product $w\cdot n$ measure the signed length of the normal component, without an extra factor from the length of $n_0$. On a hypersurface, the two opposite choices $n$ and $-n$ correspond to opposite normal orientations; a continuous choice on an orientable surface fixes its orientation of a surface.

Along the boundary of an oriented surface, the outward unit conormal lies in the tangent plane of the surface and is perpendicular to the boundary unit tangent vector. If $\mathbf{n}$ is the chosen unit normal vector, the positively oriented unit tangent vector is $\mathbf{n}\times \mathbf{\nu }$.

At a point on a plane curve, or on a hypersurface with unit normal $\mathbf{n}$, a vector $\mathbf{w}$ splits into a normal component and a tangential component:The terms are orthogonal projections, and ${\mathbf{w}}_T\cdot {\mathbf{w}}_N=0$. Changing $\mathbf{n}$ to $-\mathbf{n}$ reverses the scalar coefficient $\mathbf{w}\cdot \mathbf{n}$, but leaves the normal component vector unchanged.

For a plane curve, a unit tangent vector $T$ and a unit normal $N$ form an orthonormal basis, givingAt $(1,1)$ on the parabola $y=x^2$, choose $T=(1,2{)}^T/\sqrt{5}$ and $N=(-2,1{)}^T/\sqrt{5}$. For $\mathbf{w}=(3,1{)}^T$,Their sum is $(3,1{)}^T$ and their inner product is zero. A negative normal coefficient means that the normal component points opposite the chosen $N$, rather than giving it a negative Euclidean norm.

These are local directions: being tangent at a point does not mean that a straight displacement stays on the curve. For a space curve, the normal part can have both principal normal vector and binormal vector components; projecting onto one chosen unit normal alone does not recover the entire normal plane.

For a twice continuously differentiable regular curve describing motion, let $v=\Vert r^{\prime }(t)\Vert >0$, $T=r^{\prime }/v$, and let $N$ be the principal normal vector where the curvature $\kappa$ is positive. Differentiating the velocity vector $r^{\prime }=vT$ givesThe tangential component changes the speed; the normal component changes the direction of the velocity vector. A particle can therefore have nonzero acceleration at constant speed, as in circular motion with constant speed. The formula follows from $dT/dt=(dT/ds)(ds/dt)=\kappa vN$ and the product rule. When $\kappa =0$, the normal contribution vanishes even though the principal normal vector is undefined there. ^[ref]

In dimensionless coordinates, take $r(t)=(t,t^2)$. At $t=1$,The resulting vector components areBoth increasing speed and turning contribute to the same acceleration.

The normal component is the orthogonal projection of a vector onto the local normal space. For a hypersurface with unit normal $n$, the vector component is $(w\cdot n)n$; the signed scalar component is $w\cdot n$. These two meanings must be distinguished. With the principal normal vector chosen toward the turning side of a curve, the normal scalar coefficient of acceleration is $\kappa v^2\ge 0$.

The tangential component is the orthogonal projection of a vector onto the tangent space. For a plane curve, it is $(w\cdot T)T$; for a hypersurface, it is $w-(w\cdot n)n$. The scalar tangential component of acceleration is $\dot{v}$, which can be positive, zero or negative as the speed increases, remains constant or decreases.

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 $s=0$.

For a divergence-form differential operator $-\nabla \cdot K\nabla$, its conormal derivative is the flux factor appearing in integration by parts. It generally differs from the ordinary normal derivative ${\partial }_{n}u=\mathbf{n}\cdot \nabla u$ when $K$ is anisotropic. Decomposing the gradient along a smooth boundary into its normal and tangential components gives ${\partial }_{n,K}u=(\mathbf{n}\cdot K\mathbf{n}){\partial }_{n}u+(\mathbf{n}\cdot K\mathbf{t}){\partial }_{\mathbf{t}}u$ in two dimensions. Thus zero normal derivative does not in general remove the boundary term in the usual symmetric weak formulation. A zero conormal derivative does remove it and produces the natural homogeneous Neumann boundary condition for that energy.