New General Investigations of Curved Surfaces

by Carl Friedrich Gauss

NEW GENERAL INVESTIGATIONS

OF

CURVED SURFACES

[1825]

---

Although the real purpose of this work is the deduction of new theorems concerning its subject, nevertheless we shall first develop what is already known, partly for the sake of consistency and completeness, and partly because our method of treatment is different from that which has been used heretofore. We shall even begin by advancing certain properties concerning plane curves from the same principles.

1.

In order to compare in a convenient manner the different directions of straight lines in a plane with each other, we imagine a circle with unit radius described in the plane about an arbitrary centre. The position of the radius of this circle, drawn parallel to a straight line given in advance, represents then the position of that line. And the angle which two straight lines make with each other is measured by the angle between the two radii representing them, or by the arc included between their extremities. Of course, where precise definition is necessary, it is specified at the outset, for every straight line, in what sense it is regarded as drawn. Without such a distinction the direction of a straight line would always correspond to two opposite radii.

2.

In the auxiliary circle we take an arbitrary radius as the first, or its terminal point in the circumference as the origin, and determine the positive sense of measuring the arcs from this point (whether from left to right or the contrary); in the opposite direction the arcs are regarded then as negative. Thus every direction of a straight line is expressed in degrees, etc., or also by a number which expresses them in parts of the radius.

Such lines as differ in direction by ${\textstyle 360^{\circ },}$ or by a multiple of ${\textstyle 360^{\circ },}$ have, therefore, precisely the same direction, and may, generally speaking, be regarded as the same. However, in such cases where the manner of describing a variable angle is taken into consideration, it may be necessary to distinguish carefully angles differing by ${\textstyle 360^{\circ }.}$

If, for example, we have decided to measure the arcs from left to right, and if to two straight lines ${\textstyle l,}$ ${\textstyle l'}$ correspond the two directions ${\textstyle L,}$ ${\textstyle L',}$ then ${\textstyle L'-L}$ is the angle between those two straight lines. And it is easily seen that, since ${\textstyle L'-L}$ falls between ${\textstyle -180^{\circ }}$ and ${\textstyle +180^{\circ },}$ the positive or negative value indicates at once that ${\textstyle l'}$ lies on the right or the left of ${\textstyle l,}$ as seen from the point of intersection. This will be determined generally by the sign of ${\textstyle \sin(L'-L).}$

If ${\textstyle aa'}$ is a part of a curved line, and if to the tangents at ${\textstyle a,}$ ${\textstyle a'}$ correspond respectively the directions ${\textstyle \alpha ,}$ ${\textstyle \alpha ',}$ by which letters shall be denoted also the corresponding points on the auxiliary circles, and if ${\textstyle A,}$ ${\textstyle A'}$ be their distances along the arc from the origin, then the magnitude of the arc ${\textstyle \alpha \alpha '}$ or ${\textstyle A'-A}$ is called the amplitude of ${\textstyle aa'.}$

The comparison of the amplitude of the arc ${\textstyle aa'}$ with its length gives us the notion of curvature. Let ${\textstyle l}$ be any point on the arc ${\textstyle aa',}$ and let ${\textstyle \lambda ,}$ ${\textstyle \Lambda }$ be the same with reference to it that ${\textstyle \alpha ,}$ ${\textstyle A}$ and ${\textstyle \alpha ',}$ ${\textstyle A'}$ are with reference to ${\textstyle a}$ and ${\textstyle a'.}$ If now ${\textstyle \alpha \lambda }$ or ${\textstyle \Lambda -A}$ be proportional to the part ${\textstyle al}$ of the arc, then we shall say that ${\textstyle aa'}$ is uniformly curved throughout its whole length, and we shall call

$${\frac {\Lambda -A}{al}}$$

the measure of curvature, or simply the curvature. We easily see that this happens only when ${\textstyle aa'}$ is actually the arc of a circle, and that then, according to our definition, its curvature will be ${\textstyle \pm {\frac {1}{r}},}$ if ${\textstyle r}$ denotes the radius. Since we always regard ${\textstyle r}$ as positive, the upper or the lower sign will hold according as the centre lies to the right or to the left of the arc ${\textstyle aa'}$ (${\textstyle a}$ being regarded as the initial point, ${\textstyle a'}$ as the end point, and the directions on the auxiliary circle being measured from left to right). Changing one of these conditions changes the sign, changing two restores it again.

On the contrary, if ${\textstyle \Lambda -A}$ be not proportional to ${\textstyle al,}$ then we call the arc non-uniformly curved and the quotient

$${\frac {\Lambda -A}{al}}$$

may then be called its mean curvature. Curvature, on the contrary, always presupposes that the point is determined, and is defined as the mean curvature of an element at this point; it is therefore equal to

$${\frac {d\Lambda }{d\,al}}.$$

We see, therefore, that arc, amplitude, and curvature sustain a similar relation to each other as time, motion, and velocity, or as volume, mass, and density. The reciprocal of the curvature, namely,

$${\frac {d\,al}{d\Lambda }},$$

is called the radius of curvature at the point ${\textstyle l.}$ And, in keeping with the above conventions, the curve at this point is called concave toward the right and convex toward the left, if the value of the curvature or of the radius of curvature happens to be positive; but, if it happens to be negative, the contrary is true.

If we refer the position of a point in the plane to two perpendicular axes of coordinates to which correspond the directions ${\textstyle 0}$ and ${\textstyle 90^{\circ },}$ in such a manner that the first coordinate represents the distance of the point from the second axis, measured in the direction of the first axis; whereas the second coordinate represents the distance from the first axis, measured in the direction of the second axis; if, further, the indeterminates ${\textstyle x,}$ ${\textstyle y}$ represent the coordinates of a point on the curved line, ${\textstyle s}$ the length of the line measured from an arbitrary origin to this point, ${\textstyle \phi }$ the direction of the tangent at this point, and ${\textstyle r}$ the radius of curvature; then we shall have

$${\begin{aligned}dx&=\cos \phi .ds,\\dy&=\sin \phi .ds,\\r&={\frac {ds}{d\phi }}.\end{aligned}}$$

If the nature of the curved line is defined by the equation ${\textstyle V=0,}$ where ${\textstyle V}$ is a function of ${\textstyle x,}$ ${\textstyle y,}$ and if we set

$$dV=p\,dx+q\,dy,$$

then on the curved line

$$p\,dx+q\,dy=0.$$

Hence

$$p\cos \phi +q\sin \phi =0,$$

and therefore

$$\tan \phi =-{\frac {p}{q}}.$$

We have also

$$\cos \phi .dp+\sin \phi .dq-(p\sin \phi -q\cos \phi )\,d\phi =0.$$

If, therefore, we set, according to a well known theorem,

$${\begin{aligned}dp&=P\,dx+Q\,dy,\\dq&=Q\,dx+R\,dy,\end{aligned}}$$

then we have

$$(P\cos ^{2}\phi +2Q\cos \phi \sin \phi +R\sin ^{2}\phi )\,ds=(p\sin \phi -q\cos \phi )\,d\phi ,$$

therefore

$${\frac {1}{r}}={\frac {P\cos ^{2}\phi +2Q\cos \phi \sin \phi +R\sin ^{2}\phi }{p\sin \phi -q\cos \phi }},$$

or, since

$$\cos \phi ={\frac {\mp q}{\sqrt {p^{2}+q^{2}}}},\qquad \sin \phi ={\frac {\pm p}{\sqrt {p^{2}+q^{2}}}};$$

$$\pm {\frac {1}{r}}={\frac {Pq^{2}-2Qpq+Rp^{2}}{(p^{2}+q^{2})^{3/2}}}.$$

4.

The ambiguous sign in the last formula might at first seem out of place, but upon closer consideration it is found to be quite in order. In fact, since this expression depends simply upon the partial differentials of ${\textstyle V,}$ and since the function ${\textstyle V}$ itself merely defines the nature of the curve without at the same time fixing the sense in which it is supposed to be described, the question, whether the curve is convex toward the right or left, must remain undetermined until the sense is determined by some other means. The case is similar in the determination of ${\textstyle \phi }$ by means of the tangent, to single values of which correspond two angles differing by ${\textstyle 180^{\circ }.}$ The sense in which the curve is described can be specified in the following different ways.

I. By means of the sign of the change in ${\textstyle x.}$ If ${\textstyle x}$ increases, then ${\textstyle \cos \phi }$ must be positive. Hence the upper signs will hold if ${\textstyle q}$ has a negative value, and the lower signs if ${\textstyle q}$ has a positive value. When ${\textstyle x}$ decreases, the contrary is true.

II. By means of the sign of the change in ${\textstyle y.}$ If ${\textstyle y}$ increases, the upper signs must be taken when ${\textstyle p}$ is positive, the lower when ${\textstyle p}$ is negative. The contrary is true when ${\textstyle y}$ decreases.

III. By means of the sign of the value which the function ${\textstyle V}$ takes for points not on the curve. Let ${\textstyle \delta x,}$ ${\textstyle \delta y}$ be the variations of ${\textstyle x,}$ ${\textstyle y}$ when we go out from the curve toward the right, at right angles to the tangent, that is, in the direction ${\textstyle \phi +90^{\circ };}$ and let the length of this normal be ${\textstyle \delta \rho .}$ Then, evidently, we have

$${\begin{aligned}\delta x&=\delta \rho .\cos(\phi +90^{\circ }),\\\delta y&=\delta \rho .\sin(\phi +90^{\circ }),\end{aligned}}$$

or

$${\begin{aligned}\delta x&=-\delta \rho .\sin \phi ,\\\delta y&=+\delta \rho .\cos \phi .\end{aligned}}$$

Since now, when ${\textstyle \delta \rho }$ is infinitely small,

$${\begin{aligned}\delta V&=p\,\delta x+q\,\delta y\\&=(-p\sin \phi +q\cos \phi )\,\delta \rho \\&=\mp \delta \rho {\sqrt {p^{2}+q^{2}}}\end{aligned}}$$

and since on the curve itself ${\textstyle V}$ vanishes, the upper signs will hold if ${\textstyle V,}$ on passing through the curve from left to right, changes from positive to negative, and the contrary. If we combine this with what is said at the end of Art. 2, it follows that the curve is always convex toward that side on which ${\textstyle V}$ receives the same sign as

$$Pq^{2}-2Qpq+Rp^{2}.$$

For example, if the curve is a circle, and if we set

$$V=x^{2}+y^{2}-a^{2}$$

then we have

$${\begin{array}{c}p=2x,\qquad q=2y,\\P=2,\qquad Q=0,\qquad R=2,\\Pq^{2}-2Qpq+Rp^{2}=8y^{2}+8x^{2}=8a^{2},\\(p^{2}+q^{2})^{3/2}=8a^{3},\\r=\pm a\end{array}}$$

and the curve will be convex toward that side for which

$$x^{2}+y^{2}>a^{2},$$

as it should be.

The side toward which the curve is convex, or, what is the same thing, the signs in the above formulæ, will remain unchanged by moving along the curve, so long as

$${\frac {\delta V}{\delta \rho }}$$

does not change its sign. Since ${\textstyle V}$ is a continuous function, such a change can take place only when this ratio passes through the value zero. But this necessarily presupposes that ${\textstyle p}$ and ${\textstyle q}$ become zero at the same time. At such a point the radius of curvature becomes infinite or the curvature vanishes. Then, generally speaking, since here

$$-p\sin \phi +q\cos \phi$$

will change its sign, we have here a point of inflexion.

5.

The case where the nature of the curve is expressed by setting ${\textstyle y}$ equal to a given function of ${\textstyle x,}$ namely, ${\textstyle y=X,}$ is included in the foregoing, if we set

$$V=X-y.$$

If we put

$$dX=X'\,dx,\qquad dX'=X''\,dx,$$

then we have

$${\begin{array}{c}p=X',\qquad q=-1,\\P=X'',\qquad Q=0,\qquad R=0,\end{array}}$$

therefore

$$\pm {\frac {1}{r}}={\frac {X''}{(1+X'^{2})^{3/2}}}.$$

Since ${\textstyle q}$ is negative here, the upper sign holds for increasing values of ${\textstyle x.}$ We can therefore say, briefly, that for a positive ${\textstyle X''}$ the curve is concave toward the same side toward which the ${\textstyle y}$-axis lies with reference to the ${\textstyle x}$-axis; while for a negative ${\textstyle X''}$ the curve is convex toward this side.

6.

If we regard ${\textstyle x,}$ ${\textstyle y}$ as functions of ${\textstyle s,}$ these formulæ become still more elegant. Let us set

$${\begin{alignedat}{2}{\frac {dx}{ds}}&=x',\qquad &{\frac {dx'}{ds}}&=x'',\\{\frac {dy}{ds}}&=y',\qquad &{\frac {dy'}{ds}}&=y''.\end{alignedat}}$$

Then we shall have

$${\begin{alignedat}{2}x'&=\cos \phi ,\qquad &y'&=\sin \phi ,\\x''&=-{\frac {\sin \phi }{r}},\qquad &y''&={\frac {\cos \phi }{r}};\end{alignedat}}$$

or

$${\begin{alignedat}{2}y'&=-rx'',\qquad &x'&=ry'',\end{alignedat}}$$

or also

$$1=r(x'y''-y'x''),$$

so that

$$x'y''-y'x''$$

represents the curvature, and

$${\frac {1}{x'y''-y'x''}}$$

the radius of curvature.

7.

We shall now proceed to the consideration of curved surfaces. In order to represent the directions of straight lines in space considered in its three dimensions, we imagine a sphere of unit radius described about an arbitrary centre. Accordingly, a point on this sphere will represent the direction of all straight lines parallel to the radius whose extremity is at this point. As the positions of all points in space are determined by the perpendicular distances ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z}$ from three mutually perpendicular planes, the directions of the three principal axes, which are normal to these principal planes, shall be represented on the auxiliary sphere by the three points ${\textstyle (1),}$ ${\textstyle (2),}$ ${\textstyle (3).}$ These points are, therefore, always ${\textstyle 90^{\circ }}$ apart, and at once indicate the sense in which the coordinates are supposed to increase. We shall here state several well known theorems, of which constant use will be made.

1) The angle between two intersecting straight lines is measured by the arc [of the great circle] between the points on the sphere which represent their directions.

2) The orientation of every plane can be represented on the sphere by means of the great circle in which the sphere is cut by the plane through the centre parallel to the first plane.

3) The angle between two planes is equal to the angle between the great circles which represent their orientations, and is therefore also measured by the angle between the poles of the great circles.

4) If ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z;}$ ${\textstyle x',}$ ${\textstyle y',}$ ${\textstyle z'}$ are the coordinates of two points, ${\textstyle r}$ the distance between them, and ${\textstyle L}$ the point on the sphere which represents the direction of the straight line drawn from the first point to the second, then

$${\begin{alignedat}{2}x'&=x&&+r\cos(1)L,\\y'&=y&&+r\cos(2)L,\\z'&=z&&+r\cos(3)L.\end{alignedat}}$$

5) It follows immediately from this that we always have

$$\cos ^{2}(1)L+\cos ^{2}(2)L+\cos ^{2}(3)L=1$$

[and] also, if ${\textstyle L'}$ is any other point on the sphere,

$$\cos(1)L.\cos(1)L'+\cos(2)L.\cos(2)L'+\cos(3)L.\cos(3)L'=\cos LL'.$$

We shall add here another theorem, which has appeared nowhere else, as far as we know, and which can often be used with advantage.

Let ${\textstyle L,}$ ${\textstyle L',}$ ${\textstyle L'',}$ ${\textstyle L'''}$ be four points on the sphere, and ${\textstyle A}$ the angle which ${\textstyle LL'''}$ and ${\textstyle L'L''}$ make at their point of intersection. [Then we have]

$$\cos LL'.\cos L''L'''-\cos LL''.\cos L'L'''=\sin LL'''.\sin L'L''.\cos A.$$

The proof is easily obtained in the following way. Let

$$AL=t,\qquad AL'=t',\qquad AL''=t'',\qquad AL'''=t''';$$

we have then

$${\begin{alignedat}{6}&\cos LL'&&=\cos t&&\cos t'&&+\sin t&&\sin t'&&\cos A,\\&\cos L''L'''&&=\cos t''&&\cos t'''&&+\sin t''&&\sin t'''&&\cos A,\\&\cos LL''&&=\cos t&&\cos t''&&+\sin t&&\sin t''&&\cos A,\\&\cos L'L'''&&=\cos t'&&\cos t'''&&+\sin t'&&\sin t'''&&\cos A.\end{alignedat}}$$

Therefore

$${\begin{alignedat}{1}\cos LL'&\cos L''L'''-\cos LL''\cos L'L''\\&=\cos A\{\cos t\cos t'\sin t''\sin t'''+\cos t''\cos t'''\sin t\sin t'\\&\qquad \qquad -\cos t\cos t''\sin t'\sin t'''-\cos t'\cos t'''\sin t\sin t''\}\\&=\cos A(\cos t\sin t'''-\cos t'''\sin t)(\cos t'\sin t''-\cos t''\sin t')\\&=\cos A\sin(t'''-t)\sin(t''-t')\\&=\cos A\sin LL'''\sin L'L''.\end{alignedat}}$$

Since each of the two great circles goes out from ${\textstyle A}$ in two opposite directions, two supplementary angles are formed at this point. But it is seen from our analysis that those branches must be chosen, which go in the same sense from ${\textstyle L}$ toward ${\textstyle L'''}$ and from ${\textstyle L'}$ toward ${\textstyle L''.}$

Instead of the angle ${\textstyle A,}$ we can take also the distance of the pole of the great circle ${\textstyle LL'''}$ from the pole of the great circle ${\textstyle L'L''.}$ However, since every great circle has two poles, we see that we must join those about which the great circles run in the same sense from ${\textstyle L}$ toward ${\textstyle L'''}$ and from ${\textstyle L'}$ toward ${\textstyle L'',}$ respectively.

The development of the special case, where one or both of the arcs ${\textstyle LL'''}$ and ${\textstyle L'L''}$ are ${\textstyle 90^{\circ },}$ we leave to the reader.

6) Another useful theorem is obtained from the following analysis. Let ${\textstyle L,}$ ${\textstyle L',}$ ${\textstyle L''}$ be three points upon the sphere and put

$${\begin{alignedat}{6}&\cos L&&(1)=x,&&\cos L&&(2)=y,&&\cos L&&(3)=z,\\&\cos L'&&(1)=x',&&\cos L'&&(2)=y',&&\cos L'&&(3)=z',\\&\cos L''&&(1)=x'',\quad &&\cos L''&&(2)=y'',\quad &&\cos L''&&(3)=z''.\end{alignedat}}$$

We assume that the points are so arranged that they run around the triangle included by them in the same sense as the points ${\textstyle (1),}$ ${\textstyle (2),}$ ${\textstyle (3).}$ Further, let ${\textstyle \lambda }$ be that pole of the great circle ${\textstyle L'L''}$ which lies on the same side as ${\textstyle L.}$ We then have, from the above lemma,

$${\begin{alignedat}{3}&y'z''&&-z'y''&&=\sin L'L''.\cos \lambda (1),\\&z'x''&&-x'z''&&=\sin L'L''.\cos \lambda (2),\\&x'y''&&-y'x''&&=\sin L'L''.\cos \lambda (3).\end{alignedat}}$$

Therefore, if we multiply these equations by ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z}$ respectively, and add the products, we obtain

$$xy'z''+x'y''z+x''yz'-xy''z'-x'yz''-x''y'z=\sin L'L''.\cos \lambda L,$$

wherefore, we can write also, according to well known principles of spherical trigonometry,

$${\begin{alignedat}{2}\sin L'L''.&\sin LL''&&.\sin L'\\=\sin L'L''.&\sin LL'&&.\sin L''\\=\sin L'L''.&\sin L'L''&&.\sin L,\end{alignedat}}$$

if ${\textstyle L,}$ ${\textstyle L',}$ ${\textstyle L''}$ denote the three angles of the spherical triangle. At the same time we easily see that this value is one-sixth of the pyramid whose angular points are the centre of the sphere and the three points ${\textstyle L,}$ ${\textstyle L',}$ ${\textstyle L''}$ (and indeed positive, if etc.).

8.

The nature of a curved surface is defined by an equation between the coordinates of its points, which we represent by

$$f(x,y,z)=0.$$

Let the total differential of ${\textstyle f(x,y,z)}$ be

$$P\,dx+Q\,dy+R\,dz,$$

where ${\textstyle P,}$ ${\textstyle Q,}$ ${\textstyle R}$ are functions of ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z.}$ We shall always distinguish two sides of the surface, one of which we shall call the upper, and the other the lower. Generally speaking, on passing through the surface the value of ${\textstyle f}$ changes its sign, so that, as long as the continuity is not interrupted, the values are positive on one side and negative on the other.

The direction of the normal to the surface toward that side which we regard as the upper side is represented upon the auxiliary sphere by the point ${\textstyle L.}$ Let

$$\cos L(1)=X,\qquad \cos L(2)=Y,\qquad \cos L(3)=Z.$$

Also let ${\textstyle ds}$ denote an infinitely small line upon the surface; and, as its direction is denoted by the point ${\textstyle \lambda }$ on the sphere, let

$$\cos \lambda (1)=\xi ,\qquad \cos \lambda (2)=\eta ,\qquad \cos \lambda (3)=\zeta .$$

We then have

$$dx=\xi \,ds,\qquad dy=\eta \,ds,\qquad dz=\zeta \,ds,$$

therefore

$$P\xi +Q\eta +R\zeta =0,$$

and, since ${\textstyle \lambda L}$ must be equal to ${\textstyle 90^{\circ },}$ we have also

$$X\xi +Y\eta +Z\zeta =0.$$

Since ${\textstyle P,}$ ${\textstyle Q,}$ ${\textstyle R,}$ ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z}$ depend only on the position of the surface on which we take the element, and since these equations hold for every direction of the element on the surface, it is easily seen that ${\textstyle P,}$ ${\textstyle Q,}$ ${\textstyle R}$ must be proportional to ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z.}$ Therefore

$$P=X\mu ,\qquad Q=Y\mu ,\qquad R=Z\mu .$$

Therefore, since

$${\begin{array}{c}X^{2}+Y^{2}+Z^{2}=1;\\\mu =PX+QY+RZ\end{array}}$$

and

$$\mu ^{2}=P^{2}+Q^{2}+R^{2},$$

or

$$\mu =\pm {\sqrt {P^{2}+Q^{2}+R^{2}}}.$$

If we go out from the surface, in the direction of the normal, a distance equal to the element ${\textstyle \delta \rho ,}$ then we shall have

$$\delta x=X\,\delta \rho ,\qquad \delta y=Y\,\delta \rho ,\qquad \delta z=Z\,\delta \rho$$

and

$$\delta f=P\,\delta x+Q\,\delta y+R\,\delta z=\mu \,\delta \rho .$$

We see, therefore, how the sign of ${\textstyle \mu }$ depends on the change of sign of the value of ${\textstyle f}$ in passing from the lower to the upper side.

9.

Let us cut the curved surface by a plane through the point to which our notation refers; then we obtain a plane curve of which ${\textstyle ds}$ is an element, in connection with which we shall retain the above notation. We shall regard as the upper side of the plane that one on which the normal to the curved surface lies. Upon this plane we erect a normal whose direction is expressed by the point ${\textstyle {\mathfrak {L}}}$ of the auxiliary sphere. By moving along the curved line, ${\textstyle \lambda }$ and ${\textstyle L}$ will therefore change their positions, while ${\textstyle {\mathfrak {L}}}$ remains constant, and ${\textstyle \lambda L}$ and ${\textstyle \lambda {\mathfrak {L}}}$ are always equal to ${\textstyle 90^{\circ }.}$ Therefore ${\textstyle \lambda }$ describes the great circle one of whose poles is ${\textstyle {\mathfrak {L}}.}$ The element of this great circle will be equal to ${\textstyle {\frac {ds}{r}},}$ if ${\textstyle r}$ denotes the radius of curvature of the curve. And again, if we denote the direction of this element upon the sphere by ${\textstyle \lambda ',}$ then ${\textstyle \lambda '}$ will evidently lie in the same great circle and be ${\textstyle 90^{\circ }}$ from ${\textstyle \lambda }$ as well as from ${\textstyle {\mathfrak {L}}.}$ If we now set

$$\cos \lambda '(1)=\xi ',\qquad \cos \lambda '(2)=\eta ',\qquad \cos \lambda '(3)=\zeta ',$$

then we shall have

$$d\xi =\xi '\,{\frac {ds}{r}},\qquad d\eta =\eta '\,{\frac {ds}{r}},\qquad d\zeta =\zeta '\,{\frac {ds}{r}},$$

since, in fact, ${\textstyle \xi ,}$ ${\textstyle \eta ,}$ ${\textstyle \zeta }$ are merely the coordinates of the point ${\textstyle \lambda }$ referred to the centre of the sphere.

Since by the solution of the equation ${\textstyle f(x,y,z)=0}$ the coordinate ${\textstyle z}$ may be expressed in the form of a function of ${\textstyle x,}$ ${\textstyle y,}$ we shall, for greater simplicity, assume that this has been done and that we have found

$$z=F(x,y).$$

We can then write as the equation of the surface

$$z-F(x,y)=0,$$

or

$$f(x,y,z)=z-F(x,y).$$

From this follows, if we set

$${\begin{array}{c}dF(x,y)=t\,dx+u\,dy,\\P=-t,\qquad Q=-u,\qquad R=1,\end{array}}$$

where ${\textstyle t,}$ ${\textstyle u}$ are merely functions of ${\textstyle x}$ and ${\textstyle y.}$ We set also

$$dt=T\,dx+U\,dy,\qquad du=U\,dx+V\,dy.$$

Therefore upon the whole surface we have

$$dz=t\,dx+u\,dy$$

and therefore, on the curve,

$$\zeta =t\xi +u\eta .$$

Hence differentiation gives, on substituting the above values for ${\textstyle d\xi ,}$ ${\textstyle d\eta ,}$ ${\textstyle d\zeta ,}$

$${\begin{aligned}(\zeta '-t\xi '-u\eta '){\frac {ds}{r}}&=\xi \,dt+\eta \,du\\&=(\xi ^{2}T+2\xi \eta U+\eta ^{2}V)\,ds,\end{aligned}}$$

or

$${\begin{aligned}{\frac {1}{r}}&={\frac {\xi ^{2}T+2\xi \eta U+\eta ^{2}V}{-\xi 't-\eta 'SIC{\mu }{u}+\zeta '}}\\&={\frac {Z(\xi ^{2}T+2\xi \eta U+\eta ^{2}V)}{X\xi '-Y\eta '+Z\zeta '}}\\&={\frac {Z(\xi ^{2}T+2\xi \eta U+\eta ^{2}V)}{\cos L\lambda '}}.\end{aligned}}$$

10.

Before we further transform the expression just found, we will make a few remarks about it.

A normal to a curve in its plane corresponds to two directions upon the sphere, according as we draw it on the one or the other side of the curve. The one direction, toward which the curve is concave, is denoted by ${\textstyle \lambda ',}$ the other by the opposite point on the sphere. Both these points, like ${\textstyle L}$ and ${\textstyle {\mathfrak {L}},}$ are ${\textstyle 90^{\circ }}$ from ${\textstyle \lambda ,}$ and therefore lie in a great circle. And since ${\textstyle {\mathfrak {L}}}$ is also ${\textstyle 90^{\circ }}$ from ${\textstyle \lambda ,}$ ${\textstyle {\mathfrak {L}}L=90^{\circ }-L\lambda ',}$ or ${\textstyle =L\lambda '-90^{\circ }.}$ Therefore

$$\cos L\lambda '=\pm \sin {\mathfrak {L}}L,$$

where ${\textstyle \sin {\mathfrak {L}}L}$ is necessarily positive. Since ${\textstyle r}$ is regarded as positive in our analysis, the sign of ${\textstyle \cos L\lambda '}$ will be the same as that of

$$Z(\xi ^{2}T+2\xi \eta U+\eta ^{2}V).$$

And therefore a positive value of this last expression means that ${\textstyle L\lambda '}$ is less than ${\textstyle 90^{\circ },}$ or that the curve is concave toward the side on which lies the projection of the normal to the surface upon the plane. A negative value, on the contrary, shows that the curve is convex toward this side. Therefore, in general, we may set also

$${\frac {1}{r}}={\frac {Z(\xi ^{2}T+2\xi \eta U+\eta ^{2}V)}{\sin {\mathfrak {L}}L}},$$

if we regard the radius of curvature as positive in the first case, and negative in the second. ${\textstyle {\mathfrak {L}}L}$ is here the angle which our cutting plane makes with the plane tangent to the curved surface, and we see that in the different cutting planes passed through the same point and the same tangent the radii of curvature are proportional to the sine of the inclination. Because of this simple relation, we shall limit ourselves hereafter to the case where this angle is a right angle, and where the cutting plane, therefore, is passed through the normal of the curved surface. Hence we have for the radius of curvature the simple formula

$${\frac {1}{r}}=Z(\xi ^{2}T+2\xi \eta U+\eta ^{2}V).$$

11.

Since an infinite number of planes may be passed through this normal, it follows that there may be infinitely many different values of the radius of curvature. In this case ${\textstyle T,}$ ${\textstyle U,}$ ${\textstyle V,}$ ${\textstyle Z}$ are regarded as constant, ${\textstyle \xi ,}$ ${\textstyle \eta ,}$ ${\textstyle \zeta }$ as variable. In order to make the latter depend upon a single variable, we take two fixed points ${\textstyle M,}$ ${\textstyle M'}$ ${\textstyle 90^{\circ }}$ apart on the great circle whose pole is ${\textstyle L.}$ Let their coordinates referred to the centre of the sphere be ${\textstyle \alpha ,}$ ${\textstyle \beta ,}$ ${\textstyle \gamma ;}$ ${\textstyle \alpha ',}$ ${\textstyle \beta ',}$ ${\textstyle \gamma '.}$ We have then

$$\cos \lambda (1)=\cos \lambda M.\cos M(1)+\cos \lambda M'.\cos M'(1)+\cos \lambda L.\cos L(1).$$

If we set

$$\lambda M=\phi ,$$

then we have

$$\cos \lambda M'=\sin \phi ,$$

and the formula becomes

$${\begin{aligned}\xi &=\alpha \cos \phi +\alpha '\sin \phi ,\end{aligned}}$$

and likewise

$${\begin{aligned}\eta &=\beta \cos \phi +\beta '\sin \phi ,\\\zeta &=\gamma \cos \phi +\gamma '\sin \phi .\end{aligned}}$$

Therefore, if we set

$${\begin{aligned}A&=(\alpha ^{2}T+2\alpha \beta U+\beta ^{2}V)Z,\\B&=(\alpha \alpha 'T+(\alpha '\beta +\alpha \beta ')U+\beta \beta 'V)Z,\\C&=(\alpha '^{2}T+2\alpha '\beta 'U+\beta '^{2}V)Z,\end{aligned}}$$

we shall have

$${\begin{aligned}{\frac {1}{r}}&=A\cos ^{2}\phi +2B\cos \phi \sin \phi +C\sin ^{2}\phi \\&={\frac {A+C}{2}}+{\frac {A-C}{2}}\cos 2\phi +B\sin 2\phi .\end{aligned}}$$

If we put

$${\begin{aligned}{\frac {A-C}{2}}&=E\cos 2\theta ,\\B&=E\sin 2\theta ,\end{aligned}}$$

where we may assume that ${\textstyle E}$ has the same sign as ${\textstyle {\frac {A-C}{2}},}$ then we have

$${\frac {1}{r}}={\tfrac {1}{2}}(A+C)+E\cos 2(\phi -\theta ).$$

It is evident that ${\textstyle \phi }$ denotes the angle between the cutting plane and another plane through this normal and that tangent which corresponds to the direction ${\textstyle M.}$ Evidently, therefore, ${\textstyle {\frac {1}{r}}}$ takes its greatest (absolute) value, or ${\textstyle r}$ its smallest, when ${\textstyle \phi =\theta ;}$ and ${\textstyle {\frac {1}{r}}}$ its smallest absolute value, when ${\textstyle \phi =\theta +90^{\circ }.}$ Therefore the greatest and the least curvatures occur in two planes perpendicular to each other. Hence these extreme values for ${\textstyle {\frac {1}{r}}}$ are

$${\tfrac {1}{2}}(A+C)\pm {\sqrt {\left({\frac {A-C}{2}}\right)^{2}+B^{2}}}.$$

Their sum is ${\textstyle A+C}$ and their product ${\textstyle AC-B^{2},}$ or the product of the two extreme radii of curvature is

$$={\frac {1}{AC-B^{2}}}.$$

This product, which is of great importance, merits a more rigorous development. In fact, from formulæ above we find

$$AC-B^{2}=(\alpha \beta '-\beta \alpha ')^{2}(TV-U^{2})Z^{2}.$$

But from the third formula in [Theorem] 6, Art. 7, we easily infer that

$$\alpha \beta '-\beta \alpha '=\pm Z,$$

therefore

$$AC-B^{2}=Z^{4}(TV-U^{2}).$$

Besides, from Art. 8,

$${\begin{aligned}Z&=\pm {\frac {R}{\sqrt {P^{2}+Q^{2}+R^{2}}}}\\&=\pm {\frac {1}{\sqrt {1+t^{2}+u^{2}}}},\end{aligned}}$$

therefore

$$AC-B^{2}={\frac {TV-U^{2}}{(1+t^{2}+u^{2})^{2}}}.$$

Just as to each point on the curved surface corresponds a particular point ${\textstyle L}$ on the auxiliary sphere, by means of the normal erected at this point and the radius of the auxiliary sphere parallel to the normal, so the aggregate of the points on the auxiliary sphere, which correspond to all the points of a line on the curved surface, forms a line which will correspond to the line on the curved surface. And, likewise, to every finite figure on the curved surface will correspond a finite figure on the auxiliary sphere, the area of which upon the latter shall be regarded as the measure of the amplitude of the former. We shall either regard this area as a number, in which case the square of the radius of the auxiliary sphere is the unit, or else express it in degrees, etc., setting the area of the hemisphere equal to ${\textstyle 360^{\circ }.}$

The comparison of the area upon the curved surface with the corresponding amplitude leads to the idea of what we call the measure of curvature of the surface. If the former is proportional to the latter, the curvature is called uniform; and the quotient, when we divide the amplitude by the surface, is called the measure of curvature. This is the case when the curved surface is a sphere, and the measure of curvature is then a fraction whose numerator is unity and whose denominator is the square of the radius.

We shall regard the measure of curvature as positive, if the boundaries of the figures upon the curved surface and upon the auxiliary sphere run in the same sense; as negative, if the boundaries enclose the figures in contrary senses. If they are not proportional, the surface is non-uniformily curved. And at each point there exists a particular measure of curvature, which is obtained from the comparison of corresponding infinitesimal parts upon the curved surface and the auxiliary sphere. Let ${\textstyle d\sigma }$ be a surface element on the former, and ${\textstyle d\Sigma }$ the corresponding element upon the auxiliary sphere, then

$${\frac {d\Sigma }{d\sigma }}$$

will be the measure of curvature at this point.

In order to determine their boundaries, we first project both upon the ${\textstyle xy}$-plane. The magnitudes of these projections are ${\textstyle Z\,d\sigma ,}$ ${\textstyle Z\,d\Sigma .}$ The sign of ${\textstyle Z}$ will show whether the boundaries run in the same sense or in contrary senses around the surfaces and their projections. We will suppose that the figure is a triangle; the projection upon the ${\textstyle xy}$-plane has the coordinates

$$x,\ y;\qquad x+dx,\ y+dy;\qquad x+\delta x,\ y+\delta y.$$

Hence its double area will be

$$2Z\,d\sigma =dx.\delta y-dy.\delta x.$$

To the projection of the corresponding element upon the sphere will correspond the coordinates:

$${\begin{array}{c}X,\\X+{\frac {\partial X}{\partial x}}.dx+{\frac {\partial X}{\partial y}}.dy,\\X+{\frac {\partial X}{\partial x}}.\delta x+{\frac {\partial X}{\partial y}}.\delta y,\end{array}}\qquad {\begin{array}{c}Y,\\Y+{\frac {\partial Y}{\partial x}}.dx+{\frac {\partial Y}{\partial y}}.dy,\\Y+{\frac {\partial Y}{\partial x}}.\delta x+{\frac {\partial Y}{\partial y}}.\delta y,\end{array}}$$

From this the double area of the element is found to be

$${\begin{aligned}2Z\,d\Sigma &={\phantom {-}}\left({\frac {\partial X}{\partial x}}.dx+{\frac {\partial X}{\partial y}}.dy\right)\left({\frac {\partial Y}{\partial x}}.\delta x+{\frac {\partial Y}{\partial y}}.\delta y\right)\\&{\phantom {={}}}-\left({\frac {\partial X}{\partial x}}.\delta x+{\frac {\partial X}{\partial y}}.\delta y\right)\left({\frac {\partial Y}{\partial x}}.dx+{\frac {\partial Y}{\partial y}}.dy\right)\\&={\phantom {-}}\left({\frac {\partial X}{\partial x}}.{\frac {\partial Y}{\partial y}}-{\frac {\partial X}{\partial y}}.{\frac {\partial Y}{\partial x}}\right)(dx.\delta y-dy.\delta x).\end{aligned}}$$

The measure of curvature is, therefore,

$$={\frac {\partial X}{\partial x}}.{\frac {\partial Y}{\partial y}}-{\frac {\partial X}{\partial y}}.{\frac {\partial Y}{\partial x}}=\omega .$$

Since

$${\begin{array}{c}X=-tZ,\qquad Y=-uZ,\\(1+t^{2}+u^{2})Z^{2}=1,\end{array}}$$

we have

$${\begin{aligned}dX&=-Z^{3}(1+u^{2})\,dt+Z^{3}tu.du,\\dY&=+Z^{3}tu.dt-Z^{3}(1+t^{2})\,du,\end{aligned}}$$

therefore

$${\begin{alignedat}{2}{\frac {\partial X}{\partial x}}&=Z^{3}{\bigl \{}-(1+u^{2})T+tuU{\bigr \}},\qquad &{\frac {\partial Y}{\partial x}}&=Z^{3}{\bigl \{}tuT-(1+t^{2})U{\bigr \}},\\{\frac {\partial X}{\partial y}}&=Z^{3}{\bigl \{}-(1+u^{2})U+tuV{\bigr \}},\qquad &{\frac {\partial Y}{\partial y}}&=Z^{3}{\bigl \{}tuU-(1+t^{2})V{\bigr \}},\end{alignedat}}$$

and

$${\begin{aligned}\omega &=Z^{6}(TV-U^{2}){\bigl (}(1+t^{2})(1+u^{2})-t^{2}u^{2}{\bigr )}\\&=Z^{6}(TV-U^{2})(1+t^{2}+u^{2})\\&=Z^{4}(TV-U^{2})\\&={\frac {TV-U^{2}}{(1+t^{2}+u^{2})^{2}}},\end{aligned}}$$

the very same expression which we have found at the end of the preceding article. Therefore we see that

"The measure of curvature is always expressed by means of a fraction whose numerator is unity and whose denominator is the product of the maximum and minimum radii of curvature in the planes passing through the normal."

12.

We will now investigate the nature of shortest lines upon curved surfaces. The nature of a curved line in space is determined, in general, in such a way that the coordinates ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z}$ of each point are regarded as functions of a single variable, which we shall call ${\textstyle w.}$ The length of the curve, measured from an arbitrary origin to this point, is then equal to

$$\int {\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}.dw.$$

If we allow the curve to change its position by an infinitely small variation, the variation of the whole length will then be

$${\begin{aligned}&\qquad \qquad =\int {\frac {{\frac {dx}{dw}}.d\,\delta x+{\frac {dy}{dw}}.d\,\delta y+{\frac {dz}{dw}}.d\,\delta z}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}}={\frac {{\frac {dx}{dw}}.\delta x+{\frac {dy}{dw}}.\delta y+{\frac {dz}{dw}}.\delta z}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}}\\&-\int \left\{\delta x.d{\frac {\frac {dx}{dw}}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}}\right.+\delta y.d{\frac {\frac {dy}{dw}}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}}\\&\qquad \qquad +\left.\delta z.d{\frac {\frac {dz}{dw}}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}}\right\}.\end{aligned}}$$

The expression under the integral sign must vanish in the case of a minimum, as we know. Since the curved line lies upon a given curved surface whose equation is

$$P\,dx+Q\,dy+R\,dz=0,$$

the equation between the variations ${\textstyle \delta x,}$ ${\textstyle \delta y,}$ ${\textstyle \delta z}$

$$P\,\delta z+Q\,\delta y+R\,\delta z=0$$

must also hold. From this, by means of well known principles, we easily conclude that the differentials

$$d.{\frac {\frac {dx}{dw}}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}},\qquad d.{\frac {\frac {dy}{dw}}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}},$$

$$d.{\frac {\frac {dz}{dw}}{\sqrt {\left({\frac {dx}{dw}}\right)^{2}+\left({\frac {dy}{dw}}\right)^{2}+\left({\frac {dz}{dw}}\right)^{2}}}}$$

must be proportional to the quantities ${\textstyle P,}$ ${\textstyle Q,}$ ${\textstyle R}$ respectively. If ${\textstyle ds}$ is an element of the curve; ${\textstyle \lambda }$ the point upon the auxiliary sphere, which represents the direction of this element; ${\textstyle L}$ the point giving the direction of the normal as above; and ${\textstyle \xi ,}$ ${\textstyle \eta ,}$ ${\textstyle \zeta ;}$ ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z}$ the coordinates of the points ${\textstyle \lambda ,}$ ${\textstyle L}$ referred to the centre of the auxiliary sphere, then we have

$${\begin{array}{c}dx=\xi \,ds,\qquad dy=\eta \,ds,\qquad dz=\zeta \,ds,\\\xi ^{2}+\eta ^{2}+\zeta ^{2}=1.\end{array}}$$

Therefore we see that the above differentials will be equal to ${\textstyle d\xi ,}$ ${\textstyle d\eta ,}$ ${\textstyle d\zeta .}$ And since ${\textstyle P,}$ ${\textstyle Q,}$ ${\textstyle R}$ are proportional to the quantities ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z,}$ the character of the shortest line is such that

$${\frac {d\xi }{X}}={\frac {d\eta }{Y}}={\frac {d\zeta }{Z}}.$$

13.

To every point of a curved line upon a curved surface there correspond two points on the sphere, according to our point of view; namely, the point ${\textstyle \lambda ,}$ which represents the direction of the linear element, and the point ${\textstyle L,}$ which represents the direction of the normal to the surface. The two are evidently ${\textstyle 90^{\circ }}$ apart. In our former investigation (Art. 9), where [we] supposed the curved line to lie in a plane, we had two other points upon the sphere; namely, ${\textstyle {\mathfrak {L}},}$ which represents the direction of the normal to the plane, and ${\textstyle \lambda ',}$ which represents the direction of the normal to the element of the curve in the plane. In this case, therefore, ${\textstyle {\mathfrak {L}}}$ was a fixed point and ${\textstyle \lambda ,}$ ${\textstyle \lambda '}$ were always in a great circle whose pole was ${\textstyle {\mathfrak {L}}.}$ In generalizing these considerations, we shall retain the notation ${\textstyle {\mathfrak {L}},}$ ${\textstyle \lambda ',}$ but we must define the meaning of these symbols from a more general point of view. When the curve ${\textstyle s}$ is described, the points ${\textstyle L,}$ ${\textstyle \lambda }$ also describe curved lines upon the auxiliary sphere, which, generally speaking, are no longer great circles. Parallel to the element of the second line, we draw a radius of the auxiliary sphere to the point ${\textstyle \lambda ',}$ but instead of this point we take the point opposite when ${\textstyle \lambda '}$ is more than ${\textstyle 90^{\circ }}$ from ${\textstyle L.}$ In the first case, we regard the element at ${\textstyle \lambda }$ as positive, and in the other as negative. Finally, let ${\textstyle {\mathfrak {L}}}$ be the point on the auxiliary sphere, which is ${\textstyle 90^{\circ }}$ from both ${\textstyle \lambda }$ and ${\textstyle \lambda ',}$ and which is so taken that ${\textstyle \lambda ,}$ ${\textstyle \lambda ',}$ ${\textstyle {\mathfrak {L}}}$ lie in the same order as ${\textstyle (1),}$ ${\textstyle (2),}$ ${\textstyle (3).}$

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 ${\textstyle 4}$ points describes a line upon the auxiliary sphere, whose elements we shall express by ${\textstyle dL,}$ ${\textstyle d\lambda ,}$ ${\textstyle d\lambda ',}$ ${\textstyle d{\mathfrak {L}}.}$ 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 ${\textstyle \rho ,}$ 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, ${\textstyle \xi ',}$ ${\textstyle \eta ',}$ ${\textstyle \zeta '}$ must be proportional to the quantities ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z.}$ 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,

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

and since we always choose the point ${\textstyle \lambda '}$ so that

$$\lambda 'L<90^{\circ },$$

then for the shortest line

$$\lambda 'L=0,$$

or ${\textstyle \lambda '}$ and ${\textstyle L}$ must coincide. Therefore

$${\begin{aligned}\rho \,d\xi &=X\,ds,\\\rho \,d\eta &=Y\,ds,\\\rho \,d\zeta &=Z\,ds,\end{aligned}}$$

and we have here, instead of ${\textstyle 4}$ curved lines upon the auxiliary sphere, only ${\textstyle 3}$ to consider. Every element of the second line is therefore to be regarded as lying in the great circle ${\textstyle L\lambda .}$ And the positive or negative value of ${\textstyle \rho }$ refers to the concavity or the convexity of the curve in the direction of the normal.

14.

We shall now investigate the spherical angle upon the auxiliary sphere, which the great circle going from ${\textstyle L}$ toward ${\textstyle \lambda }$ makes with that one going from ${\textstyle L}$ toward one of the fixed points ${\textstyle (1),}$ ${\textstyle (2),}$ ${\textstyle (3);}$ e.g., toward ${\textstyle (3).}$ In order to have something definite here, we shall consider the sense from ${\textstyle L(3)}$ to ${\textstyle L\lambda }$ the same as that in which ${\textstyle (1),}$ ${\textstyle (2),}$ and ${\textstyle (3)}$ lie. If we call this angle ${\textstyle \phi ,}$ then it follows from the theorem of Art. 7 that

$$\sin L(3).\sin L\lambda .\sin \phi =Y\xi -X\eta ,$$

or, since ${\textstyle L\lambda =90^{\circ }}$ and

$$\sin L(3)={\sqrt {X^{2}+Y^{2}}}={\sqrt {1-Z^{2}}},$$

we have

$$\sin \phi ={\frac {Y\xi -X\eta }{\sqrt {X^{2}+Y^{2}}}}.$$

Furthermore,

$$\sin L(3).\sin L\lambda .\cos \phi =\zeta ,$$

or

$$\cos \phi ={\frac {\zeta }{\sqrt {X^{2}+Y^{2}}}}$$

and

$$\tan \phi ={\frac {Y\xi -X\eta }{\zeta }}={\frac {\zeta '}{\zeta }}.$$

Hence we have

$$d\phi ={\frac {\zeta Y\,d\xi -\zeta X\,d\eta -(Y\xi -X\eta )\,d\zeta +\xi \zeta \,dY-\eta \zeta \,dX}{(Y\xi -X\eta )^{2}+\zeta ^{2}}}.$$

The denominator of this expression is

$${\begin{aligned}&=Y^{2}\xi ^{2}-2XY\xi \eta -X^{2}\eta ^{2}+\zeta ^{2}\\&=-(X\xi +Y\eta )^{2}+(X^{2}+Y^{2})(\xi ^{2}+\eta ^{2})+\zeta ^{2}\\&=-Z^{2}\zeta ^{2}+(1-Z^{2})(1-\zeta ^{2})+\zeta ^{2}\\&=1-Z^{2},\end{aligned}}$$

or

$$d\phi ={\frac {\zeta Y\,d\xi -\zeta X\,d\eta +(X\eta -Y\xi )\,d\zeta -\eta \zeta \,dX+\xi \zeta \,dY}{1-Z^{2}}}.$$

We verify readily by expansion the identical equation

$${\begin{array}{c}\eta \zeta (X^{2}+Y^{2}+Z^{2})+YZ(\xi ^{2}+\eta ^{2}+\zeta ^{2})\\=(X\xi +Y\eta +Z\zeta )(Z\eta +Y\zeta )+(X\zeta -Z\xi )(X\eta -Y\xi )\end{array}}$$

and likewise

$${\begin{array}{c}\xi \zeta (X^{2}+Y^{2}+Z^{2})+XZ(\xi ^{2}+\eta ^{2}+\zeta ^{2})\\=(X\xi +Y\eta +Z\zeta )(X\zeta +Z\xi )+(Y\xi -X\eta )(Y\zeta -Z\eta ).\end{array}}$$

We have, therefore,

$${\begin{aligned}\eta \zeta &=-YZ+(X\zeta -Z\xi )(X\eta -Y\xi ),\\\xi \zeta &=-XZ+(Y\xi -X\eta )(Y\zeta -Z\eta ).\end{aligned}}$$

Substituting these values, we obtain

$$d\phi ={\frac {Z}{1-Z^{2}}}(Y\,dX-X\,dY)+{\frac {\zeta Y\,d\xi -\zeta X\,d\eta }{1-Z^{2}}}$$

$$+{\frac {X\eta -Y\xi }{1-Z^{2}}}{\bigl \{}d\zeta -(X\zeta -Z\xi )\,dX-(Y\zeta -Z\eta )\,dY{\bigr \}}.$$

Now

$${\begin{alignedat}{4}&X\,dX&&+Y\,dY&&+Z\,dZ&&=0,\\&\xi \,dX&&+\eta \,dY&&+\zeta \,dZ&&=-X\,d\xi -Y\,d\eta -Z\,d\zeta .\end{alignedat}}$$

On substituting we obtain, instead of what stands in the parenthesis,

$$d\zeta -Z(X\,d\xi +Y\,d\eta +Z\,d\zeta ).$$

Hence

$$d\phi ={\frac {Z}{1-Z^{2}}}(Y\,dX-X\,dY)+{\frac {d\xi }{1-Z^{2}}}\{\zeta Y-\eta X^{2}Z+\xi XYZ\}$$

$${\begin{aligned}&-{\frac {d\eta }{1-Z^{2}}}\{\zeta X+\eta XYZ-\xi Y^{2}Z\}\\&+d\zeta (\eta X-\xi Y).\end{aligned}}$$

Since, further,

$${\begin{aligned}\eta X^{2}Z-\xi XYZ&=\eta X^{2}Z+\eta Y^{2}Z+\zeta ZYZ\\&=\eta Z(1-Z^{2})+\zeta YZ^{2},\\\eta XYZ-\xi Y^{2}Z&=-\xi X^{2}Z-\zeta XZ^{2}-\xi Y^{2}Z\\&=-\xi Z(1-Z^{2})-\zeta XZ^{2},\end{aligned}}$$

our whole expression becomes

$${\begin{aligned}d\phi &={\frac {Z}{1-Z^{2}}}(Y\,dX-X\,dY)\\&\quad +(\zeta Y-\eta Z)\,d\xi +(\xi Z-\zeta X)\,d\eta +(\eta X-\xi Y)\,d\zeta .\end{aligned}}$$

15.

The formula just found is true in general, whatever be the nature of the curve. But if this be a shortest line, then it is clear that the last three terms destroy each other, and consequently

$$d\phi =-{\frac {Z}{1-Z^{2}}}(X\,dY-Y\,dX).$$

But we see at once that

$${\frac {Z}{1-Z^{2}}}(X\,dY-Y\,dX)$$

is nothing but the area of the part of the auxiliary sphere, which is formed between the element of the line ${\textstyle L,}$ the two great circles drawn through its extremities and 

${\textstyle (3),}$ and the element thus intercepted on the great circle through ${\textstyle (1)}$ and ${\textstyle (2).}$ This surface is considered positive, if ${\textstyle L}$ and ${\textstyle (3)}$ lie on the same side of ${\textstyle (1)\ (2),}$ and if the direction from ${\textstyle P}$ to ${\textstyle P'}$ is the same as that from ${\textstyle (2)}$ to ${\textstyle (1);}$ negative, if the contrary of one of these conditions hold; positive again, if the contrary of both conditions be true. In other words, the surface is considered positive if we go around the circumference of the figure ${\textstyle LL'P'P}$ in the same sense as ${\textstyle (1)\ (2)\ (3);}$ negative, if we go in the contrary sense.

If we consider now a finite part of the line from ${\textstyle L}$ to ${\textstyle L'}$ and denote by ${\textstyle \phi ,}$ ${\textstyle \phi '}$ the values of the angles at the two extremities, then we have

$$\phi '=\phi +\operatorname {Area} LL'P'P,$$

the sign of the area being taken as explained.

Now let us assume further that, from the origin upon the curved surface, infinitely many other shortest lines go out, and denote by ${\textstyle A}$ that indefinite angle which the first element, moving counter-clockwise, makes with the first element of the first line; and through the other extremities of the different curved lines let a curved line be drawn, concerning which, first of all, we leave it undecided whether it be a shortest line or not. If we suppose also that those indefinite values, which for the first line were ${\textstyle \phi ,}$ ${\textstyle \phi ',}$ be denoted by ${\textstyle \psi ,}$ ${\textstyle \psi '}$ for each of these lines, then ${\textstyle \psi '-\psi }$ is capable of being represented in the same manner on the auxiliary sphere by the space ${\textstyle LL'_{1}P'_{1}P.}$ Since evidently ${\textstyle \psi =\phi -A,}$ the space

$${\begin{aligned}LL'_{1}P'_{1}P'L'L&=\psi '-\psi -\phi '+\phi \\&=\psi '-\phi '+A\\&=LL'_{1}L'L+L'L'_{1}P'_{1}P'.\end{aligned}}$$

If the bounding line is also a shortest line, and, when prolonged, makes with ${\textstyle LL',}$ ${\textstyle LL'_{1}}$ the angles ${\textstyle B,}$ ${\textstyle B_{1};}$ if, further, ${\textstyle \chi ,}$ ${\textstyle \chi _{1}}$ denote the same at the points ${\textstyle L',}$ ${\textstyle L'_{1},}$ that ${\textstyle \phi }$ did at ${\textstyle L}$ in the line ${\textstyle LL',}$ then we have

$${\begin{aligned}\chi _{1}&=\chi +\operatorname {Area} L'L'_{1}P'_{1}P',\\\psi '-\phi '+A&=LL'_{1}L'L+\chi _{1}-\chi ;\end{aligned}}$$

but

$${\begin{aligned}\phi '&=\chi +B,\\\psi '&=\chi _{1}+B_{1},\end{aligned}}$$

therefore

$$B_{1}-B+A=LL'_{1}L'L.$$

The angles of the triangle ${\textstyle LL'L'_{1}}$ evidently are

$$A,\qquad 180^{\circ }-B,\qquad B_{1},$$

therefore their sum is

$$180^{\circ }+LL'_{1}L'L.$$

The form of the proof will require some modification and explanation, if the point ${\textstyle (3)}$ falls within the triangle. But, in general, we conclude

"The sum of the three angles of a triangle, which is formed of shortest lines upon an arbitrary curved surface, is equal to the sum of ${\textstyle 180^{\circ }}$ and the area of the triangle upon the auxiliary sphere, the boundary of which is formed by the points ${\textstyle L,}$ corresponding to the points in the boundary of the original triangle, and in such a manner that the area of the triangle may be regarded as positive or negative according as it is inclosed by its boundary in the same sense as the original figure or the contrary."

Wherefore we easily conclude also that the sum of all the angles of a polygon of ${\textstyle n}$ sides, which are shortest lines upon the curved surface, is [equal to] the sum of ${\textstyle (n-2)180^{\circ }+}$ the area of the polygon upon the sphere etc.

16.

If one curved surface can be completely developed upon another surface, then all lines upon the first surface will evidently retain their magnitudes after the development upon the other surface; likewise the angles which are formed by the intersection of two lines. Evidently, therefore, such lines also as are shortest lines upon one surface remain shortest lines after the development. Whence, if to any arbitrary polygon formed of shortest lines, while it is upon the first surface, there corresponds the figure of the zeniths upon the auxiliary sphere, the area of which is ${\textstyle A,}$ and if, on the other hand, there corresponds to the same polygon, after its development upon another surface, a figure of the zeniths upon the auxiliary sphere, the area of which is ${\textstyle A',}$ it follows at once that in every case

$$A=A'.$$

Although this proof originally presupposes the boundaries of the figures to be shortest lines, still it is easily seen that it holds generally, whatever the boundary may be. For, in fact, if the theorem is independent of the number of sides, nothing will prevent us from imagining for every polygon, of which some or all of its sides are not shortest lines, another of infinitely many sides all of which are shortest lines.

Further, it is clear that every figure retains also its area after the transformation by development.

We shall here consider 4 figures:

1. an arbitrary figure upon the first surface,
3. the figure on the auxiliary sphere, which corresponds to the zeniths of the previous figure,
5. the figure upon the second surface, which No. 1 forms by the development,
7. the figure upon the auxiliary sphere, which corresponds to the zeniths of No. 3.

Therefore, according to what we have proved, 2 and 4 have equal areas, as also 1 and 3. Since we assume these figures infinitely small, the quotient obtained by dividing 2 by 1 is the measure of curvature of the first curved surface at this point, and likewise the quotient obtained by dividing 4 by 3, that of the second surface. From this follows the important theorem:

"In the transformation of surfaces by development the measure of curvature at every point remains unchanged."

This is true, therefore, of the product of the greatest and smallest radii of curvature.

In the case of the plane, the measure of curvature is evidently everywhere zero. Whence follows therefore the important theorem:

"For all surfaces developable upon a plane the measure of curvature everywhere vanishes,"

or

$$\left({\frac {\partial ^{2}z}{\partial x\,\partial y}}\right)^{2}-\left({\frac {\partial ^{2}z}{\partial x^{2}}}\right)\left({\frac {\partial ^{2}z}{\partial x^{2}}}\right)=0,$$

which criterion is elsewhere derived from other principles, though, as it seems to us, not with the desired rigor. It is clear that in all such surfaces the zeniths of all points can not fill out any space, and therefore they must all lie in a line.

17.

From a given point on a curved surface we shall let an infinite number of shortest lines go out, which shall be distinguished from one another by the angle which their first elements make with the first element of a definite shortest line. This angle we shall call ${\textstyle \theta .}$ Further, let ${\textstyle s}$ be the length [measured from the given point] of a part of such a shortest line, and let its extremity have the coordinates ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z.}$ Since ${\textstyle \theta }$ and ${\textstyle s,}$ therefore, belong to a perfectly definite point on the curved surface, we can regard ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z}$ as functions of ${\textstyle \theta }$ and ${\textstyle s.}$ The direction of the element of ${\textstyle s}$ corresponds to the point ${\textstyle \lambda }$ on the sphere, whose coordinates are ${\textstyle \xi ,}$ ${\textstyle \eta ,}$ ${\textstyle \zeta .}$ Thus we shall have

$$\xi ={\frac {\partial x}{\partial s}},\qquad \eta ={\frac {\partial y}{\partial s}},\qquad \zeta ={\frac {\partial z}{\partial s}}.$$

The extremities of all shortest lines of equal lengths ${\textstyle s}$ correspond to a curved line whose length we may call ${\textstyle t.}$ We can evidently consider ${\textstyle t}$ as a function of ${\textstyle s}$ and ${\textstyle \theta ,}$ and if the direction of the element of ${\textstyle t}$ corresponds upon the sphere to the point ${\textstyle \lambda '}$ whose coordinates are ${\textstyle \xi ',}$ ${\textstyle \eta ',}$ ${\textstyle \zeta ',}$ we shall have

$$\xi '.{\frac {\partial t}{\partial \theta }}={\frac {\partial x}{\partial \theta }},\qquad \eta '.{\frac {\partial t}{\partial \theta }}={\frac {\partial y}{\partial \theta }},\qquad \zeta '.{\frac {\partial t}{\partial \theta }}={\frac {\partial z}{\partial \theta }}.$$

Consequently

$$(\xi \xi '+\eta \eta '+\zeta \zeta ')\,{\frac {\partial t}{\partial \theta }}={\frac {\partial x}{\partial s}}.{\frac {\partial x}{\partial \theta }}+{\frac {\partial y}{\partial s}}.{\frac {\partial y}{\partial \theta }}+{\frac {\partial z}{\partial s}}.{\frac {\partial z}{\partial \theta }}.$$ This magnitude we shall denote by ${\textstyle u,}$ which itself, therefore, will be a function of ${\textstyle \theta }$ and ${\textstyle s.}$

We find, then, if we differentiate with respect to ${\textstyle s,}$

$${\begin{aligned}{\frac {\partial u}{\partial s}}&={\frac {\partial ^{2}x}{\partial s^{2}}}.{\frac {\partial x}{\partial \theta }}+{\frac {\partial ^{2}y}{\partial s^{2}}}.{\frac {\partial y}{\partial \theta }}+{\frac {\partial ^{2}z}{\partial s^{2}}}.{\frac {\partial z}{\partial \theta }}+{\tfrac {1}{2}}\,{\frac {\partial \left\{({\tfrac {\partial x}{\partial s}})^{2}+({\tfrac {\partial y}{\partial s}})^{2}+({\tfrac {\partial z}{\partial s}})^{2}\right\}}{\partial \theta }}\\&={\frac {\partial ^{2}x}{\partial s^{2}}}.{\frac {\partial x}{\partial \theta }}+{\frac {\partial ^{2}y}{\partial s^{2}}}.{\frac {\partial y}{\partial \theta }}+{\frac {\partial ^{2}z}{\partial s^{2}}}.{\frac {\partial z}{\partial \theta }},\end{aligned}}$$

because

$$\left({\frac {\partial x}{\partial s}}\right)^{2}+\left({\frac {\partial y}{\partial s}}\right)^{2}+\left({\frac {\partial z}{\partial s}}\right)^{2}=1,$$

and therefore its differential is equal to zero.

But since all points [belonging] to one constant value of ${\textstyle \theta }$ lie on a shortest line, if we denote by ${\textstyle L}$ the zenith of the point to which ${\textstyle s,}$ ${\textstyle \theta }$ correspond and by ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z}$ the coordinates of ${\textstyle L,}$ [from the last formulæ of Art. 13],

$${\frac {\partial ^{2}x}{\partial s^{2}}}={\frac {X}{p}},\qquad {\frac {\partial ^{2}y}{\partial s^{2}}}={\frac {Y}{p}},\qquad {\frac {\partial ^{2}z}{\partial s^{2}}}={\frac {Z}{p}},$$

if ${\textstyle p}$ is the radius of curvature. We have, therefore,

$$p.{\frac {\partial u}{\partial s}}=X\,{\frac {\partial x}{\partial \theta }}+Y\,{\frac {\partial y}{\partial \theta }}+Z\,{\frac {\partial z}{\partial \theta }}={\frac {\partial t}{\partial \theta }}(X\xi '+Y\eta '+Z\zeta ').$$

But

$$X\xi '+Y\eta '+Z\zeta '=\cos L\lambda '=0,$$

because, evidently, ${\textstyle \lambda '}$ lies on the great circle whose pole is ${\textstyle L.}$ Therefore we have

$${\frac {\partial u}{\partial s}}=0,$$

or ${\textstyle u}$ independent of ${\textstyle s,}$ and therefore a function of ${\textstyle \theta }$ alone. But for ${\textstyle s=0,}$ it is evident that ${\textstyle t=0,}$ ${\textstyle {\frac {\partial t}{\partial \theta }}=0,}$ and therefore ${\textstyle u=0.}$ Whence we conclude that, in general, ${\textstyle u=0,}$ or

$$\cos \lambda \lambda '=0.$$

From this follows the beautiful theorem:

"If all lines drawn from a point on the curved surface are shortest lines of equal lengths, they meet the line which joins their extremities everywhere at right angles."

We can show in a similar manner that, if upon the curved surface any curved line whatever is given, and if we suppose drawn from every point of this line toward the same side of it and at right angles to it only shortest lines of equal lengths, the extremities of which are joined by a line, this line will be cut at right angles by those lines in all its points. We need only let ${\textstyle \theta }$ in the above development represent the length of the given curved line from an arbitrary point, and then the above calculations retain their validity, except that ${\textstyle u=0}$ for ${\textstyle s=0}$ is now contained in the hypothesis.

18.

The relations arising from these constructions deserve to be developed still more fully. We have, in the first place, if, for brevity, we write ${\textstyle m}$ for ${\textstyle {\frac {\partial t}{\partial \theta }},}$

(1)

${\textstyle \displaystyle {\begin{alignedat}{3}{\frac {\partial x}{\partial s}}&=\xi ,\qquad &{\frac {\partial y}{\partial s}}&=\eta ,\qquad &{\frac {\partial z}{\partial s}}&=\zeta ,\quad \end{alignedat}}}$

(2)

${\textstyle \displaystyle {\begin{alignedat}{3}{\frac {\partial x}{\partial \theta }}&=m\xi ',\quad &{\frac {\partial y}{\partial \theta }}&=m\eta ',\quad &{\frac {\partial z}{\partial \theta }}&=m\zeta ',\end{alignedat}}}$

(3)

${\textstyle {\begin{alignedat}{4}&\xi ^{2}&&+\eta ^{2}&&+\zeta ^{2}&&=1,\end{alignedat}}}$

(4)

${\textstyle {\begin{alignedat}{4}&\xi '^{2}&&+\eta '^{2}&&+\zeta '^{2}&&=1,\end{alignedat}}}$

(5)

${\textstyle {\begin{alignedat}{4}&\xi \xi '&&+\eta \eta '&&+\zeta \zeta '&&=0.\end{alignedat}}}$

Furthermore,

(6)

${\textstyle \displaystyle {\begin{alignedat}{4}&X^{2}&&+Y^{2}&&+Z^{2}&&=1,\end{alignedat}}}$

(7)

${\textstyle \displaystyle {\begin{alignedat}{4}&X\xi &&+Y\eta &&+Z\zeta &&=0,\end{alignedat}}}$

(8)

${\textstyle \displaystyle {\begin{alignedat}{4}&X\xi '&&+Y\eta '&&+Z\zeta '&&=0,\end{alignedat}}}$

and

[9]

${\textstyle {\begin{aligned}&\left\{{\begin{alignedat}{2}X&=\zeta \eta '&&-\eta \zeta ',\\Y&=\xi \zeta '&&-\zeta \xi ',\\Z&=\eta \xi '&&-\xi \eta ';\end{alignedat}}\right.\end{aligned}}}$

[10]

${\textstyle \displaystyle \left\{{\begin{alignedat}{2}\xi '&=\eta Z&&-\zeta Y,\\\eta '&=\zeta X&&-\xi Z,\\\zeta '&=\xi Y&&-\eta X;\end{alignedat}}\right.}$

[11]

${\textstyle \displaystyle \left\{{\begin{alignedat}{2}\xi &=Y\zeta '&&-Z\eta ',\\\eta &=Z\xi '&&-X\zeta ',\\\zeta &=X\eta '&&-Y\xi '.\end{alignedat}}\right.}$

Likewise, ${\textstyle {\frac {\partial \xi }{\partial s}},}$ ${\textstyle {\frac {\partial \eta }{\partial s}},}$ ${\textstyle {\frac {\partial \zeta }{\partial s}}}$ are proportional to ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z,}$ and if we set

$${\frac {\partial \xi }{\partial s}}=pX,\qquad {\frac {\partial \eta }{\partial s}}=pY,\qquad {\frac {\partial \zeta }{\partial s}}=pZ,$$

where ${\textstyle {\frac {1}{p}}}$ denotes the radius of curvature of the line ${\textstyle s,}$ then

$$p=X\,{\frac {\partial \xi }{\partial s}}+Y\,{\frac {\partial \eta }{\partial s}}+Z\,{\frac {\partial \zeta }{\partial s}}.$$

By differentiating (7) with respect to ${\textstyle s,}$ we obtain

$$-p=\xi \,{\frac {\partial X}{\partial s}}+\eta \,{\frac {\partial Y}{\partial s}}+\zeta \,{\frac {\partial Z}{\partial s}}.$$

We can easily show that ${\textstyle {\frac {\partial \xi '}{\partial s}},}$ ${\textstyle {\frac {\partial \eta '}{\partial s}},}$ ${\textstyle {\frac {\partial \zeta '}{\partial s}}}$ also are proportional to ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z.}$ In fact, [from 10] the values of these quantities are also [equal to]

$$\eta \,{\frac {\partial Z}{\partial s}}-\zeta \,{\frac {\partial Y}{\partial s}},\qquad \zeta \,{\frac {\partial X}{\partial s}}-\xi \,{\frac {\partial Z}{\partial s}},\qquad \xi \,{\frac {\partial Y}{\partial s}}-\eta \,{\frac {\partial X}{\partial s}},$$

therefore

$${\begin{aligned}Y\,{\frac {\partial \xi '}{\partial s}}-X\,{\frac {\partial \eta '}{\partial s}}&=-\zeta \left({\frac {Y\,\partial Y}{\partial s}}+{\frac {X\,\partial X}{\partial s}}\right)+{\frac {\partial Z}{\partial s}}(Y\eta +X\xi )\\&=-\zeta \left({\frac {X\,\partial X+Y\,\partial Y+Z\,\partial Z}{\partial s}}\right)+{\frac {\partial Z}{\partial s}}(X\xi +Y\eta +Z\zeta )\\&=0,\end{aligned}}$$

and likewise the others. We set, therefore,

$${\frac {\partial \xi '}{\partial s}}=p'X,\qquad {\frac {\partial \eta '}{\partial s}}=p'Y,\qquad {\frac {\partial \zeta '}{\partial s}}=p'Z,$$

whence

$$p'=\pm {\sqrt {\left({\frac {\partial \xi '}{\partial s}}\right)^{2}+\left({\frac {\partial \eta '}{\partial s}}\right)^{2}+\left({\frac {\partial \zeta '}{\partial s}}\right)^{2}}}$$

and also

$$p'=X\,{\frac {\partial \xi '}{\partial s}}+Y\,{\frac {\partial \eta '}{\partial s}}+Z\,{\frac {\partial \zeta '}{\partial s}}.$$

Further [we obtain], from the result obtained by differentiating (8),

$$-p'=\xi '\,{\frac {\partial X}{\partial s}}+\eta '\,{\frac {\partial Y}{\partial s}}+\zeta '\,{\frac {\partial Z}{\partial s}}.$$

But we can derive two other expressions for this. We have

$${\frac {\partial m\xi '}{\partial s}}={\frac {\partial \xi }{\partial \theta }},\qquad \left[{\frac {\partial m\eta '}{\partial s}}={\frac {\partial \eta }{\partial \theta }},\qquad {\frac {\partial m\zeta '}{\partial s}}={\frac {\partial \zeta }{\partial \theta }},\right]$$

therefore [because of (8)]

$$mp'=X\,{\frac {\partial \xi }{\partial \theta }}+Y\,{\frac {\partial \eta }{\partial \theta }}+Z\,{\frac {\partial \zeta }{\partial \theta }}.$$

[and therefore, from (7),]

$$-mp'=\xi \,{\frac {\partial X}{\partial \theta }}+\eta \,{\frac {\partial Y}{\partial \theta }}+\zeta \,{\frac {\partial Z}{\partial \theta }}.$$

After these preliminaries [using (2) and (4)] we shall now first put ${\textstyle m}$ in the form

$$m=\xi '\,{\frac {\partial x}{\partial \theta }}+\eta '\,{\frac {\partial y}{\partial \theta }}+\zeta '\,{\frac {\partial z}{\partial \theta }},$$

and differentiating with respect to ${\textstyle s,}$ we have^[1]

$${\begin{aligned}{\frac {\partial m}{\partial s}}&={\frac {\partial x}{\partial \theta }}.{\frac {\partial \xi '}{\partial s}}+{\frac {\partial y}{\partial \theta }}.{\frac {\partial \eta '}{\partial s}}+{\frac {\partial z}{\partial \theta }}.{\frac {\partial \zeta '}{\partial s}}+\xi '\,{\frac {\partial ^{2}x}{\partial s\,\partial \theta }}+\eta '\,{\frac {\partial ^{2}y}{\partial s\,\partial \theta }}+\zeta '\,{\frac {\partial ^{2}z}{\partial s\,\partial \theta }}\\&=mp'(\xi 'X+\eta 'Y+\zeta 'Z)+\xi '\,{\frac {\partial \xi }{\partial \theta }}+\eta '\,{\frac {\partial \eta }{\partial \theta }}+\zeta '\,{\frac {\partial \zeta }{\partial \theta }}\\&=\xi '\,{\frac {\partial \xi }{\partial \theta }}+\eta '\,{\frac {\partial \eta }{\partial \theta }}+\zeta '\,{\frac {\partial \zeta }{\partial \theta }}.\end{aligned}}$$

If we differentiate again with respect to ${\textstyle s,}$ and notice that

$${\frac {\partial ^{2}\xi }{\partial s\,\partial \theta }}={\frac {\partial (pX)}{\partial \theta }},\quad {\text{etc.}},$$

and that

$$X\xi '+Y\eta '+Z\zeta '=0,$$

we have

$${\begin{aligned}{\frac {\partial ^{2}m}{\partial s^{2}}}&=p\left(\xi '\,{\frac {\partial X}{\partial \theta }}+\eta '\,{\frac {\partial Y}{\partial \theta }}+\zeta '\,{\frac {\partial Z}{\partial \theta }}\right)+p'\left(X{\frac {\partial \xi }{\partial \theta }}+Y{\frac {\partial \eta }{\partial \theta }}+Z{\frac {\partial \zeta }{\partial \theta }}\right)\\&=p\left(\xi '\,{\frac {\partial X}{\partial \theta }}+\eta '\,{\frac {\partial Y}{\partial \theta }}+\zeta '\,{\frac {\partial Z}{\partial \theta }}\right)+mp'^{2}\\&=-\left(\xi \,{\frac {\partial X}{\partial s}}+\eta \,{\frac {\partial Y}{\partial s}}+\zeta \,{\frac {\partial Z}{\partial s}}\right)\left(\xi '\,{\frac {\partial X}{\partial \theta }}+\eta '\,{\frac {\partial Y}{\partial \theta }}+\zeta '\,{\frac {\partial Z}{\partial \theta }}\right)\\&{\phantom {={}}}+\left(\xi '\,{\frac {\partial X}{\partial s}}+\eta '\,{\frac {\partial Y}{\partial s}}+\zeta '\,{\frac {\partial Z}{\partial s}}\right)\left(\xi \,{\frac {\partial X}{\partial \theta }}+\eta \,{\frac {\partial Y}{\partial \theta }}+\zeta \,{\frac {\partial Z}{\partial \theta }}\right)\\&=\left({\frac {\partial Y}{\partial \theta }}\,{\frac {\partial Z}{\partial s}}-{\frac {\partial Y}{\partial s}}\,{\frac {\partial Z}{\partial \theta }}\right)X+\left({\frac {\partial Z}{\partial \theta }}\,{\frac {\partial X}{\partial s}}-{\frac {\partial Z}{\partial s}}\,{\frac {\partial X}{\partial \theta }}\right)Y+\left({\frac {\partial X}{\partial \theta }}\,{\frac {\partial Y}{\partial s}}-{\frac {\partial X}{\partial s}}\,{\frac {\partial Y}{\partial \theta }}\right)Z.\end{aligned}}$$

[But if the surface element

$$m\,ds\,d\theta$$

belonging to the point ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z}$ be represented upon the auxiliary sphere of unit radius by means of parallel normals, then there corresponds to it an area whose magnitude is

$$\left\{X\left({\frac {\partial Y}{\partial s}}\,{\frac {\partial Z}{\partial \theta }}-{\frac {\partial Y}{\partial \theta }}\,{\frac {\partial Z}{\partial s}}\right)+Y\left({\frac {\partial Z}{\partial s}}\,{\frac {\partial X}{\partial \theta }}-{\frac {\partial Z}{\partial \theta }}\,{\frac {\partial X}{\partial s}}\right)+Z\left({\frac {\partial X}{\partial s}}\,{\frac {\partial Y}{\partial \theta }}-{\frac {\partial X}{\partial \theta }}\,{\frac {\partial Y}{\partial s}}\right)\right\}ds\,d\theta .$$

Consequently, the measure of curvature at the point under consideration is equal to

$$\left.-{\frac {1}{m}}\,{\frac {\partial ^{2}m}{\partial s^{2}}}.\right]$$

1. It is better to differentiate ${\textstyle m^{2}.}$ [In fact from (2) and (4) $$m^{2}=\left({\frac {\partial x}{\partial \theta }}\right)^{2}+\left({\frac {\partial y}{\partial \theta }}\right)^{2}+\left({\frac {\partial z}{\partial \theta }}\right)^{2},$$ therefore $${\begin{aligned}m\,{\frac {\partial m}{\partial s}}&={\frac {\partial x}{\partial \theta }}.{\frac {\partial ^{2}x}{\partial \theta \,\partial s}}+{\frac {\partial y}{\partial \theta }}.{\frac {\partial ^{2}y}{\partial \theta \,\partial s}}+{\frac {\partial z}{\partial \theta }}.{\frac {\partial ^{2}z}{\partial \theta \,\partial s}}\\&=\left.m\xi '\,{\frac {\partial \xi }{\partial \theta }}+m\eta '\,{\frac {\partial \eta }{\partial \theta }}+m\zeta '\,{\frac {\partial \zeta }{\partial \theta }}.\right]\end{aligned}}$$