Transformations of normals under parametric scalings

Introduction

Consider a surface in \(\R^3\) defined parametrically by

\[\begin{align*} \bv{f} : U \subset \R^2 &\to \R^3 \\\ (u, v) &\mapsto \bv{f}(u, v) \end{align*}\]

Define the two tangent vectors

\[\begin{align*} \hat{\bv{u}} &= \frac{\pderiv{\bv{f}}{u}}{\norm*{\pderiv{\bv{f}}{u}}} & \hat{\bv{v}} &= \frac{\pderiv{\bv{f}}{v}}{\norm*{\pderiv{\bv{f}}{v}}} \end{align*}\]

The normal to the surface at a point \((u, v)\) is given by

\[\begin{align*} \hat{\bv{n}} &= \frac{\hat{\bv{u}} \times \hat{\bv{v}}}{\norm*{\hat{\bv{u}} \times \hat{\bv{v}}}} \propto \pderiv{\bv{f}}{u} \times \pderiv{\bv{f}}{v} \end{align*}\]

Applying a Scale

Consider the new surface

\[\bv{g}(u, v) = \lambda(u, v) \bv{f}(u, v)\]

for some function \(\lambda : U \to \R\) which we assume to be strictly positive. We can compute the (unnormalised) normal to \(\bv{g}\) as

\[\begin{align*} \bv{n}_g &= \pderiv{\bv{g}}{u} \times \pderiv{\bv{g}}{v} \\ &= \left[ \pderiv{\lambda}{u} \bv{f} + \lambda \pderiv{\bv{f}}{u} \right] \times \left[ \pderiv{\lambda}{v} \bv{f} + \lambda \pderiv{\bv{f}}{v} \right] \\ &= \lambda \pderiv{\lambda}{v} \cdot \pderiv{\bv{f}}{u} \times \bv{f} \,+\, \lambda \pderiv{\lambda}{u} \cdot \bv{f} \times \pderiv{\bv{f}}{v} \,+\, \lambda^2 \cdot \pderiv{\bv{f}}{u} \times \pderiv{\bv{f}}{v} \end{align*}\]

Introduce the normalisation operator \(\mathcal{N}(\bv{x}) = \frac{\bv{x}}{\norm{\bv{x}}}\) and note that for $k > 0$, for all \(\bv{x} \neq 0\), we have \(\mathcal{N}(k \bv{x}) = \mathcal{N}(\bv{x})\). Then

\[\begin{align*} \hat{\bv{n}}_g &= \mathcal{N}(\bv{n}_g) \\ &= \mathcal{N}\left( \pderiv{\lambda}{v} \abs*{\pderiv{\bv{f}}{u}} \cdot \hat{\bv{u}}_f \times \bv{f} \,+\, \pderiv{\lambda}{u} \abs*{\pderiv{\bv{f}}{v}} \cdot \bv{f} \times \hat{\bv{v}}_f \,+\, \lambda \abs*{\pderiv{\bv{f}}{u}} \cdot \abs*{\pderiv{\bv{f}}{v}} \cdot \hat{\bv{u}}_f \times \hat{\bv{v}}_f \right) \end{align*}\]

Example

Consider a torus with radii \(R > r > 0\),

\[\bv{f}(\theta, \phi) = \begin{pmatrix} (R + r \cos \theta) \cos \phi \\ (R + r \cos \theta) \sin \phi \\ r \sin \theta \end{pmatrix}\]

Then we have

\[\begin{align*} \pderiv{\bv{f}}{\theta} &= \begin{pmatrix} -r \sin \theta \cos \phi \\ -r \sin \theta \sin \phi \\ r \cos \theta \end{pmatrix} & \pderiv{\bv{f}}{\phi} &= \begin{pmatrix} -(R + r \cos \theta) \sin \phi \\ (R + r \cos \theta) \cos \phi \\ 0 \end{pmatrix} \\ \abs*{\pderiv{\bv{f}}{\theta}} &= r & \abs*{\pderiv{\bv{f}}{\phi}} &= R + r \cos \theta \\ \hat{\bv{\theta}} &= \begin{pmatrix} -\sin \theta \cos \phi \\ -\sin \theta \sin \phi \\ \cos \theta \end{pmatrix} & \hat{\bv{\phi}} &= \begin{pmatrix} -\sin \phi \\ \cos \phi \\ 0 \end{pmatrix} \end{align*}\]