Expressions for curvature of parametric surfaces

If we define a curve $\alpha(t) = (u(t), v(t))$ in parameter space, (and thus $(x(u(t)), x(v(t))$ on the surface) and restrict the normal field to the curve, then we get a function of normals along the curve $N(t)$ . The derivative $N'(t) = N_u u' + N_v v'$ is precisely $dN(v)$ , where $v = \alpha'(t)$ . Futhermore, given a parameterization of the surface $x$ , we have a basis for vectors in the tangent plane $\{x_u, x_v\}$ and $N_u$ , $N_v$ can be represented in this basis:

$$N_u = a_{11} x_u + a_{21} x_v$$ (2.41)

$$N_v = a_{12} x_u + a_{22} x_v$$ (2.42)

Labouring on, we have

$$N'(t) = (a_{11} x_u + a_{21} x_v) u' + (a_{12} x_u + a_{22} x_v) v'$$ (2.43)

$$= (a_{11} u' + a_{12} v') x_u + (a_{21} u' + a_{22} x_v) v',$$ (2.44)

or in vector form,

$$dN \begin{pmatrix}u' \\ v'\end{pmatrix} = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix}u' \\ v' \end{pmatrix}.$$ (2.45)

So in the basis $\{x_u, x_v\}$ , $dN$ is the linear map given by the $a_{ij}$ . This is convienient, and we now want to find expressions for the $a_{ij}$ in terms of the parameterization $x$ and its derivatives.

The shape operator can be expressed in terms of the parameterization by invoking it on the tangent to a curve $\alpha(t)$ :

$$-\ensuremath{\langle dN_p(\alpha'(t)), \alpha'(t) \rangle} = \ensuremath{\langle N'(t), x_u u' + x_v v' \rangle}$$ (2.46)

$$= -\ensuremath{\langle N_u u' + N_v v', x_u u' + x_v v' \rangle}$$ (2.47)

$$= -\ensuremath{\langle N_u, x_u \rangle} u'\,^2 - \ensuremath{\la... ...{\langle N_v, x_u \rangle} u' v' - \ensuremath{\langle N_v, x_v \rangle} v'\,^2$$ (2.48)

$$= e u'\,^2 + 2 f u' v' + g v'\,^2,$$ (2.49)

where we have defined

$$e = -\ensuremath{\langle N_u, x_u \rangle} = \ensuremath{\langle N, x_{uu} \rangle}$$ (2.50)

$$f = -\ensuremath{\langle N_u, x_v \rangle} = \ensuremath{\langle N,... ...\ensuremath{\langle N, x_{vu} \rangle} = -\ensuremath{\langle N_v, x_u \rangle}$$ (2.51)

$$g = -\ensuremath{\langle N_v, x_v \rangle} = \ensuremath{\langle N, x_{vv} \rangle}.$$ (2.52)

The transformations producing the second derivatives of $x$ are generated by noticing that since, say, $x_u$ is in the tangent plane, it must be true that $\ensuremath{\langle N, x_u \rangle} = 0$ , hence $\ensuremath{\langle N_u, x_u \rangle} + \ensuremath{\langle N, x_{uu} \rangle} = 0$ .

Substituting (2.41)-(2.42) into (2.50)-(2.51) generates a set of relations between $a_{ij}$ , $\{e,f,g\}$ , and $\{E,F,G\}$ , for example, $-e = \ensuremath{\langle N_u, x_u \rangle} = a_{11} E + a_{21} F$ . When solved, these give

$$\begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} =... ...e & f \\ f & g \end{pmatrix} \begin{pmatrix}E & F \\ F & G \end{pmatrix} ^{-1}.$$ (2.53)

Hence, given a parameterization $x$ of a surface, we can write down the differential or shape operator in the basis $\{x_u, x_v\}$ .