Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/113

 we draw a radius of the auxiliary sphere to the point $\lambda',$ but instead of this point we take the point opposite when $\lambda'$  is more than $90^{\circ}$  from $L.$  In the first case, we regard the element at $\lambda$  as positive, and in the other as negative. Finally, let $\mathfrak{L}$ be the point on the auxiliary sphere, which is $90^{\circ}$  from both $\lambda$  and $\lambda',$  and which is so taken that $\lambda,$  $\lambda',$  $\mathfrak{L}$  lie in the same order as $(1),$  $(2),$  $(3).$

The coordinates of the four points of the auxiliary sphere, referred to its centre, are for

$$\begin{alignedat}{4} &L\qquad &&X\quad &&Y\quad &&Z \\ &\lambda &&\xi   &&\eta   &&\zeta \\ &\lambda'&&\xi'  &&\eta'  &&\zeta' \\ &\mathfrak{L}     &&\alpha &&\beta  &&\gamma. \end{alignedat}$$

Hence each of these $4$  points describes a line upon the auxiliary sphere, whose elements we shall express by $dL,$  $d\lambda,$  $d\lambda',$  $d\mathfrak{L}.$ We have, therefore,

$$\begin{aligned} d\xi  &= \xi'\, d\lambda, \\ d\eta &= \eta'\, d\lambda, \\ d\zeta &= \zeta'\, d\lambda. \end{aligned}$$

In an analogous way we now call

$$\frac{d\lambda}{ds}$$

the measure of curvature of the curved line upon the curved surface, and its reciprocal

$$\frac{ds}{d\lambda}$$

the radius of curvature. If we denote the latter by $\rho,$ then

$$\begin{aligned} \rho\, d\xi  &= \xi'\, ds, \\ \rho\, d\eta &= \eta'\, ds, \\ \rho\, d\zeta &= \zeta'\, ds. \end{aligned}$$

If, therefore, our line be a shortest line, $\xi',$  $\eta',$  $\zeta'$ must be proportional to the quantities $X,$  $Y,$  $Z.$  But, since at the same time

$$\xi'^{2} + \eta'^{2} + \zeta'^{2} = X^{2} + Y^{2} + Z^{2} = 1,$$

we have

$$\xi'  = \pm X,\quad \eta' = \pm Y,\quad \zeta' = \pm Z,$$

and since, further,

$$\begin{aligned} \xi'X + \eta'Y + \zeta'Z &= \cos \lambda'L \\ &= \pm (X^{2} + Y^{2} + Z^{2}) \\ &= \pm 1, \end{aligned}$$