Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/42

orthogonally the lines of the second system; i.e., in such a way that we have generally ${\textstyle \omega =90^{\circ },}$ or ${\textstyle F=0.}$ Then the formula for the measure of curvature becomes

$$4E^{2}G^{2}k=E.{\frac {\partial E}{\partial q}}.{\frac {\partial G}{\partial q}}+E\left({\frac {\partial G}{\partial p}}\right)^{2}+G.{\frac {\partial E}{\partial p}}.{\frac {\partial G}{\partial p}}+G\left({\frac {\partial E}{\partial q}}\right)^{2}-2EG\left({\frac {\partial ^{2}E}{\partial q^{2}}}+{\frac {\partial ^{2}G}{\partial p^{2}}}\right),$$

and for the variation of the angle ${\textstyle \theta }$

$${\sqrt {EG}}.d\theta ={\frac {1}{2}}.{\frac {\partial E}{\partial q}}.dp-{\frac {1}{2}}.{\frac {\partial G}{\partial p}}.dq$$

Among the various cases in which we have this condition of orthogonality, the most important is that in which all the lines of one of the two systems, e.g., the first, are shortest lines. Here for a constant value of ${\textstyle q}$ the angle ${\textstyle \theta }$ becomes equal to zero, and therefore the equation for the variation of ${\textstyle \theta }$ just given shows that we must have ${\textstyle {\frac {\partial E}{\partial q}}=0,}$ or that the coefficient ${\textstyle E}$ must be independent of ${\textstyle q;}$ i.e., ${\textstyle E}$ must be either a constant or a function of ${\textstyle p}$ alone. It will be simplest to take for ${\textstyle p}$ the length of each line of the first system, which length, when all the lines of the first system meet in a point, is to be measured from this point, or, if there is no common intersection, from any line whatever of the second system. Having made these conventions, it is evident that ${\textstyle p}$ and ${\textstyle q}$ denote now the same quantities that were expressed in Arts. 15, 16 by ${\textstyle r}$ and ${\textstyle \phi ,}$ and that ${\textstyle E=1.}$ Thus the two preceding formulæ become:

$${\begin{aligned}4G^{2}k&=\left({\frac {\partial G}{\partial p}}\right)^{2}-2G\,{\frac {\partial ^{2}G}{\partial p^{2}}}\\{\sqrt {G}}.d\theta &=-{\frac {1}{2}}.{\frac {\partial G}{\partial p}}.dq\end{aligned}}$$

or, setting ${\textstyle {\sqrt {G}}=m,}$

$$k=-{\frac {1}{m}}.{\frac {\partial ^{2}m}{\partial p^{2}}},\quad d\theta =-{\frac {\partial m}{\partial p}}.dq$$

Generally speaking, ${\textstyle m}$ will be a function of ${\textstyle p,}$ ${\textstyle q,}$ and ${\textstyle m\,dq}$ the expression for the element of any line whatever of the second system. But in the particular case where all the lines ${\textstyle p}$ go out from the same point, evidently we must have ${\textstyle m=0}$ for ${\textstyle p=0.}$ Furthermore, in the case under discussion we will take for ${\textstyle q}$ the angle itself which the first element of any line whatever of the first system makes with the element of any one of the lines chosen arbitrarily. Then, since for an infinitely small value of ${\textstyle p}$ the element of a line of the second system (which can be regarded as a circle described with radius ${\textstyle p}$) is equal to ${\textstyle p\,dq,}$ we shall have for an infinitely small value of ${\textstyle p,}$ ${\textstyle m=p,}$ and consequently, for ${\textstyle p=0,}$ ${\textstyle m=0}$ at the same time, and ${\textstyle {\frac {\partial m}{\partial p}}=1.}$