Translation:On the spacetime lines of a Minkowski world/Paragraph 7

We are now going to investigate the nature of these curves of constant curvatures more closely, for which we require results from the differential geometry of curves of three-fold curvature in $$S_{4}$$, which should be developed in advance. As it is in the nature of things, many applications to the theory of the world lines in $$S_{4}$$ can then be made.

The comoving tetrad.
Let there be a curve in parameter representation

$x^{(1)}=x^{(1)}(t),\quad x^{(2)}=x^{(2)}(t),\quad x^{(3)}=x^{(3)}(t),\quad x^{(4)}=x^{(4)}(t)$

then the directions of

$\frac{dx}{dt}$

determine a line through $$x$$ (the tangent) and a neighboring point of the curve,

$\left[\frac{dx}{dt}\frac{d^{2}x}{dt^{2}}\right]$|undefined

determine a plane through $$x$$ (osculating plane) and two neighboring points of the curve,

$\left[\frac{dx}{dt}\frac{d^{2}x}{dt^{2}}\frac{d^{3}x}{dt^{3}}\right]$|undefined

determine a $$S_{3}$$ through $$x$$ (osculating space) and three neighboring points of the curve.

Now, the direction cosines for the following four lines through $$x$$ are determined:


 * $$c_{1}^{(\alpha)}$$ $$\alpha=1,2,3,4$$ direction cosines of the tangent;


 * $$c_{2}^{(\alpha)}$$ $$\alpha=1,2,3,4$$ direction cosines of the principal normal, i.e. that normal which is perpendicular to the tangent in the osculating plane;


 * $$c_{3}^{(\alpha)}$$ $$\alpha=1,2,3,4$$ direction cosines of the binormal; it is perpendicular to the osculating plane in the osculating space;


 * $$c_{4}^{(\alpha)}$$ $$\alpha=1,2,3,4$$ direction cosines of the trinormal; is the normal of the osculating space.

These four directions form an orthogonal tetrad [Vierkant]; the determinant is

$\left

with the other ten orthogonality conditions

$\sum_{\alpha=1}^{4}c_{i}^{(\alpha)}c_{k}^{(\alpha)}=[ik]$

One forms the symmetric matrix:

$D\equiv\left\Vert \begin{matrix}\sum\frac{dx}{dt}\frac{dx}{dt} & \sum\frac{d^{2}x}{dt^{2}}\frac{dx}{dt} & \sum\frac{dx}{dt}\frac{d^{3}x}{dt^{3}} & \sum\frac{dx}{dt}\frac{d^{4}x}{dt^{4}}\\ \sum\frac{d^{2}x}{dt^{2}}\frac{dx}{dt} & \sum\frac{d^{2}x}{dt^{2}}\frac{d^{2}x}{dt^{2}} & \sum\frac{d^{2}x}{dt^{2}}\frac{d^{3}x}{dt^{3}} & \sum\frac{d^{2}x}{dt^{2}}\frac{d^{4}x}{dt^{4}}\\ \sum\frac{d^{3}x}{dt^{3}}\frac{dx}{dt} & \sum\frac{d^{3}x}{dt^{3}}\frac{d^{2}x}{dt^{2}} & \sum\frac{d^{3}x}{dt^{3}}\frac{d^{3}x}{dt^{3}} & \sum\frac{d^{3}x}{dt^{3}}\frac{d^{4}x}{dt^{4}}\\ \sum\frac{d^{4}x}{dt^{4}}\frac{dx}{dt} & \sum\frac{d^{4}x}{dt^{4}}\frac{d^{2}x}{dt^{2}} & \sum\frac{d^{4}x}{dt^{4}}\frac{d^{3}x}{dt^{3}} & \sum\frac{d^{4}x}{dt^{4}}\frac{d^{4}x}{dt^{4}} \end{matrix}\right\Vert =\left\Vert d_{ik}\right\Vert \ i,k=1,2,3,4$|undefined

and denotes its principal subdeterminants as follows:

$\begin{aligned}D^{(1)} & =d_{11} & D^{(2)} & =\left

If $$D_{ik}^{(r)}$$ denotes the subdeterminants of $$r-1^{\mathrm{th}}$$ order of $$D^{(r)}$$, it follows

$c_{r}^{(\alpha)}=\frac{\sum_{k=1}^{r}D_{rk}^{(r)}\frac{d^{k}x^{(\alpha)}}{dt^{k}}}{\sqrt{D^{(r)}}\sqrt{D^{(r-1)}}}\quad\begin{matrix}\alpha=\\ r= \end{matrix}1,2,3,4$|undefined

The three radii of curvature $$\mathrm{R}_{1}\mathrm{R}_{2}\mathrm{R}_{3}$$.
Two consecutive tangents form the angle $$d\omega_{1}$$; when $$s$$ is the arc defined by $$\left(\frac{ds}{dt}\right)^{2}=d_{11}=D^{(1)}$$, it follows:

$\left(\frac{d\omega_{1}}{ds}\right)^{2}=\left(\frac{1}{\mathrm{R}_{1}}\right)^{2}=\frac{D^{(2)}D^{(0)}}{\left(D^{(1)}\right)^{2}}\cdot\frac{1}{D^{(1)}}=\frac{D^{(2)}\cdot D^{(0)}}{\left(D^{(1)}\right)^{3}}$|undefined

in which $$D^{(0)}\equiv1$$.

For the angle $$d\omega_{2}$$ of two consecutive osculating planes it is furthermore

$\left(\frac{d\omega_{2}}{ds}\right)^{2}=\left(\frac{1}{\mathrm{R}_{2}}\right)^{2}=\frac{D^{(3)}D^{(1)}}{\left(D^{(2)}\right)^{2}}\cdot\frac{1}{D^{(1)}}=\frac{D^{(3)}}{\left(D^{(2)}\right)^{2}}$|undefined

Eventually it follows for the angle $$d\omega_{3}$$ of two consecutive osculating spaces:

$\left(\frac{d\omega_{3}}{ds}\right)^{2}=\left(\frac{1}{\mathrm{R}_{3}}\right)^{2}=\frac{D^{(4)}D^{(2)}}{\left(D^{(3)}\right)^{2}}\cdot\frac{1}{D^{(1)}}$|undefined

For curves located entirely in $$S_{3}$$ or $$S_{2}$$, it is

$\frac{1}{\mathrm{R}_{3}}\equiv0$|undefined

or

$\frac{1}{R_{3}}\equiv\frac{1}{R_{2}}\equiv0$|undefined

or

$\frac{1}{\mathrm{R}_{3}}\equiv\frac{1}{\mathrm{R}_{2}}\equiv\frac{1}{\mathrm{R}_{1}}\equiv0$ (straight line)|undefined

If $$\frac{1}{\mathrm{R}_{3}},\frac{1}{\mathrm{R}_{2}},\frac{1}{\mathrm{R}_{1}}$$ are all constant along the curve, it can be displaced into itself, as when can read off from the “natural equations” of the curve.

The formulas of and  provide the change of the direction cosines of the axes of the comoving tetrad with $$s$$:

{{MathForm2|(19)|$$\left.\begin{matrix}\frac{dc_{1}^{(\alpha)}}{ds}= & \ast & \frac{c_{2}^{(\alpha)}}{\mathrm{R}_{1}} & \ast & \ast\\ \frac{dc_{2}^{(\alpha)}}{ds}= & -\frac{c_{1}^{(\alpha)}}{\mathrm{R}_{1}} & \ast & \frac{c_{3}^{(\alpha)}}{\mathrm{R}_{1}} & \ast\\ \frac{dc_{3}^{(\alpha)}}{ds}= & \ast & -\frac{c_{2}^{(\alpha)}}{\mathrm{R}_{1}} & \ast & \frac{c_{4}^{(\alpha)}}{\mathrm{R}_{1}}\\ \frac{dc_{4}^{(\alpha)}}{ds}= & \ast & \ast & -\frac{c_{3}^{(\alpha)}}{\mathrm{R}_{1}} & \ast \end{matrix}\ \alpha=1,2,3,4\right\} $$}}

Kinematic interpretation of the formulas.
Let us introduce a curve $$x$$ being congruent to $$y$$. The corresponding point-pairs $$y$$ and $$z$$ shall be marked by the same $$s$$. Then it follows

$\frac{d}{ds}(y-x)=0$

If one writes

$\sum_{\alpha=1}^{4}\left(y^{(\alpha)}-x^{(\alpha)}\right)c_{1}^{(\alpha)}=\left(y-x,\ c_{i}\right)$

then it follows from the formulas:

{{MathForm2|(20)|$$\left.\begin{matrix}d\left(y-x,\ c_{1}\right)= & \ast & \left(y-x,\ c_{2}\right)d\omega_{1} & \ast & \ast\\ d\left(y-x,\ c_{2}\right)= & -\left(y-x,\ c_{1}\right)d\omega_{1} & \ast & +\left(y-x,\ c_{3}\right)d\omega_{2} & \ast\\ d\left(y-x,\ c_{3}\right)= & \ast & -\left(y-x,\ c_{2}\right)d\omega_{2} & \ast & \left(y-x,\ c_{4}\right)d\omega_{2}\\ d\left(y-x,\ c_{4}\right)= & \ast & \ast & -\left(y-x,\ c_{3}\right)d\omega_{3} & \ast \end{matrix}\right\} $$}}

This means though, that the radius vector $$y-x$$ is constant with respect to magnitude and direction, it is an infinitesimal orthogonal transformation of the axis cross. The passage from the comoving tetrad of point $$x$$ to that of point $$x+dz$$ thus happens, neglecting the translation along the tangent around $$ds$$, by an infinitesimal orthogonal transformation, namely

a rotation in the osculating planes $$\left[c_{1}c_{2}\right]$$ around $$d\omega_{1}$$ from $$c_{1}$$ to $$c_{2}$$;

a rotation in the plane $$\left[c_{2}c_{3}\right]$$ around $$d\omega_{2}$$ from $$c_{2}$$ to $$c_{3}$$;

a rotation in the plane $$\left[c_{3}c_{4}\right]$$ around $$d\omega_{3}$$ from $$c_{3}$$ to $$c_{4}$$;

as it can be easily seen from the previously written form (20) – because $$\left(y-x,\ c_{i}\right)$$ is indeed the component with respect to $$c_{i}$$ of the radius vector $$y-z$$ –, if one additionally uses the law of the addition of infinitesimal rotation. This could also have been derived from the definition of the angles $$d\omega_{1}$$$$d\omega_{2}$$$$d\omega_{3}$$.

Radius vectors fixed in normal-space $$\left[c_{2}c_{3}c_{4}\right]$$.
These are evidently the ones, which are located in the normal space and which do not participate in the rotations $$d\omega_{2}d\omega_{3}$$; if we, for a moment $$d\omega_{1}$$, set the change in location of the normal-space or its normal equal to zero, then such vectors must remain parallel to themselves, thus they must be given by the previous formulas (20) for $$d\omega_{1}=0$$:

{{MathForm2|(21)|$$\left.\begin{matrix}\frac{d}{ds}\left(y-x,\ c_{1}\right)= & 0\\ \frac{d}{ds}\left(y-x,\ c_{2}\right)= & & \ast & \frac{\left(y-x,\ c_{3}\right)}{\mathrm{R}_{2}} & \ast\\ \frac{d}{ds}\left(y-x,\ c_{3}\right)= & & -\frac{\left(y-x,\ c_{2}\right)}{\mathrm{R}_{2}} & \ast & \frac{\left(y-x,\ c_{4}\right)}{\mathrm{R}_{3}}\\ \frac{d}{ds}\left(y-x,\ c_{4}\right)= & & \ast & -\frac{\left(y-x,\ c_{3}\right)}{\mathrm{R}_{3}} & \ast \end{matrix}\right\} $$}}

If we put

$\left(y-x,\ c_{i}\right)=\eta^{(i)},\ i=1,2,3,4$

then a vector fixed in normal-space is given by $$\eta^{(1)}=0$$, while generally a vector rigidly connected with the tangent can be split into a vector $$\eta^{(2)}\eta^{(3)}\eta^{(4)}$$ fixed in normal-space and a constant component along the tangent $$\eta^{(1)}$$.

For a vector fixed in normal-space we have $$\eta^{(1)}\equiv0$$

which is the form of the equations in $$S_{3}$$, leading to two identical  equations for

$\frac{\eta^{(2)}+i\eta^{(3)}}{\eta-\eta^{(2)}}=u$ and $-\frac{\eta^{(2)}+i\eta^{(3)}}{\eta-\eta^{(4)}}=v$. |undefined

Because we have by (22)

$\eta^{(2)}\frac{d\eta^{(2)}}{ds}+\eta^{(3)}\frac{d\eta^{(3)}}{ds}+\eta^{(4)}\frac{d\eta^{(4)}}{ds}=\frac{1}{2}\frac{d}{ds}\left\{ \left(\eta^{(2)}\right)^{2}+\left(\eta^{(3)}\right)^{2}+\left(\eta^{(4)}\right)^{2}\right\} =\frac{1}{2}\frac{d}{ds}(\eta)^{2}=0$|undefined

thus, which is also clear in terms of geometry:

$\left(\eta^{(2)}\right)^{2}+\left(\eta^{(3)}\right)^{2}+\left(\eta^{(4)}\right)^{2}=(\eta)^{2}=\text{const.}$

For the general integral system of (22) it follows by employment of the initial values $$\eta_{0}^{(2)}\eta_{0}^{(3)}\eta_{0}^{(4)}$$ for $$s=0$$:

Due to the constancy of $$\eta$$ it follows that the matrix

$\left

is orthogonal. Introducing $$\vartheta_{k}^{(i)}$$ in the expression for

if follows

by which we introduce the vectors

These are of course special cases of vectors fixed in normal-space, which emerge from (25) by $$\eta_{0}^{(2)}=1$$, $$\eta_{0}^{(3)}=\eta_{0}^{(4)}=0$$ or $$\eta_{0}^{(2)}=\eta_{0}^{(4)}=0$$, $$\eta_{0}^{(3)}=1$$. They will serve us to form a framework which is rigidly connected with the tangent (of course, they are all perpendicular to the tangent, as they are perpendicular among themselves), i.e. if the tangent is not changing its direction, they remain parallel with themselves in all directions. As a reference system in the sense of relativity theory, comoving with the moving point in the most general case, we have to call a system varying from location to location in such a manner, that the direction cosines of its spacelike axes are given as functions of $$s$$ (the arc of the world line of the moving point) by $$b_{2}^{(\alpha)}$$ or $$b_{3}^{(\alpha)}$$ or $$b_{4}^{(\alpha)}$$ with $$\alpha=1,2,3,4$$, while those of its timelike axes are given by $$c_{1}^{(\alpha)}$$ (direction cosine of the tangent). Because an arbitrary point $$y$$ will steadily be at rest in the “comoving system” $$\left[b_{2}b_{3}b_{4}c_{1}\right]$$, as long as this point only satisfies the condition that its world line can be related to world line $$x$$ by equal values of $$s$$ in such a way, that $$y-x$$ becomes a vector fixed in normal-space, so that within

$y^{(\alpha)}(s)=x^{(\alpha)}(s)+\eta^{(2)}(s)c_{2}^{(\alpha)}(s)+\eta^{(3)}(s)c_{3}^{(\alpha)}(s)+\eta^{(4)}(s)c_{4}^{(\alpha)}(s)\quad\alpha=1,2,3,4$

the $$\eta$$ satisfy the differential equations (22). In order to prove this – note that $$s$$ acts as a timelike parameter (even though not as arc $$s_{y}$$!) also on the world line of $$y$$ – we consider the differential quotient

$\frac{dy^{(\alpha)}}{ds}=c_{1}^{(\alpha)}+\frac{d\eta^{(2)}}{ds}c_{2}+\frac{d\eta^{(3)}}{ds}c_{3}+\frac{d\eta^{(3)}}{ds}c_{4}+\eta^{(2)}\frac{dc_{2}}{ds}+\eta^{(3)}\frac{dc_{3}}{ds}+\eta^{(4)}\frac{dc_{4}}{ds}$|undefined

It follows by (19) and (22)

i.e. the tangent of the wordline of $$y$$ is steadily directed parallel to the tangent in the related point of world line $$x$$; the former therefore is called a parallel curve of the latter, and it is the orthogonal trajectory of all $$\infty^{1}$$ normal-spaces of the latter. We can also introduce the vectors $$b_{k}$$ instead of $$c_{k}$$, and then we have

1736

$y_{(s)}^{(\alpha)}(s)=x^{(\alpha)}(s)+\eta_{0}^{(2)}b_{2}^{(\alpha)}(s)+\eta_{0}^{(3)}b_{3}^{(\alpha)}(s)+\eta_{0}^{(4)}b_{4}^{(\alpha)}(s)$

with

$\frac{dy^{(\alpha)}}{ds}=\left(1-\frac{\vartheta_{2}^{(2)}\eta_{0}^{(2)}+\vartheta_{3}^{(2)}\eta_{0}^{(3)}+\vartheta_{4}^{(2)}\eta_{0}^{(4)}}{\mathrm{R}_{1}}\right)c_{1}^{(\alpha)}$|undefined

by (26)

In this representation one recognizes the principal type of 's rigid body.

Vectors rigidly connected with the tangent.
For those, we of course have

$y_{(s)}^{(\alpha)}(s)=x^{(\alpha)}(s)+\eta^{(1)}c_{1}^{(\alpha)}(s)+\eta^{(2)}(s)c_{2}^{(\alpha)}(s)+\eta^{(3)}(s)c_{3}^{(\alpha)}(s)+\eta^{(4)}(s)c_{4}^{(\alpha)}(s)\quad\alpha=1,2,3,4$

and $$\eta^{(1)}=\text{const}$$, while the $$\eta_{0}^{(2)}\eta_{0}^{(3)}\eta_{0}^{(4)}$$ satisfy the differential equations (22). Thus by (19) and (22)

Characterization of the world lines.
1. Using the real parameter $$t$$:

$D^{(1)}=\sum\left(\frac{dx}{dt}\right)^{2}<0$

(tangent is timelike); furthermore

$D^{(2)}\leqq0,\ D^{(3)}\leqq0,\ D^{(4)}\leqq0,$

therefore the angle $$d\omega_{1}$$ is purely imaginary, $$d\omega_{2}$$ and $$d\omega_{3}$$ are real; regarding the radii of the three curvatures: $$\mathrm{R}_{1}$$ is real, $$\mathrm{R}_{2}$$ and $$\mathrm{R}_{3}$$ are imaginary.

2. Using the imaginary arc $$s$$ gives:

$D^{(1)}=1,\ D^{(2)}\geqq0,\ D^{(3)}\leqq0,\ D^{(4)}\leqq0,$

3. From $$\frac{1}{\mathrm{R}_{1}}\equiv0$$ it follows

$\frac{d^{2}x}{ds^{2}}\equiv\frac{d^{3}x}{ds^{3}}\equiv\frac{d^{4}x}{ds^{4}}\equiv0$|undefined

and therefore

$\frac{1}{\mathrm{R}_{2}}\equiv\frac{1}{\mathrm{R}_{3}}\equiv0$|undefined

From $$\frac{1}{\mathrm{R}_{2}}\equiv0$$ it follows

$D^{(3)}=\left(\left\Vert \frac{dx}{ds}\frac{d^{2}x}{ds^{2}}\frac{d^{3}x}{ds^{3}}\right\Vert \right)^{2}\equiv0$|undefined

But from the four subdeterminants of this matrix $$\left\Vert \frac{dx}{ds}\frac{d^{2}x}{ds^{2}}\frac{d^{3}x}{ds^{3}}\right\Vert $$ we can compute the trinormal, which is spacelike and never minimal. Therefore, for $$D^{(3)}=0$$ these four subdeterminants must vanish individually, and therefore also $$D^{(4)}=0$$ or $$\frac{1}{\mathrm{R}_{3}}\equiv0$$.

4. Infinities or discontinuities in the arising quantities are excluded.

Curves, whose three curvatures are constant.
We have the linear system (19) in four dependent variables $$z_{1}z_{2}z_{3}z_{4}$$ with constant coefficients:

{{center|$$\left.\begin{matrix}\frac{dz_{1}}{ds}= & \ast & \frac{z_{2}}{\mathrm{R}_{1}} & \ast & \ast\\ \frac{dz_{2}}{ds}= & -\frac{z_{1}}{\mathrm{R}_{1}} & \ast & \frac{z_{3}}{\mathrm{R}_{2}} & \ast\\ \frac{dz_{3}}{ds}= & \ast & -\frac{z_{2}}{\mathrm{R}_{2}} & \ast & \frac{z_{4}}{\mathrm{R}_{3}}\\ \frac{dz_{4}}{ds}= & \ast & \ast & -\frac{z_{3}}{\mathrm{R}_{3}} & \ast \end{matrix}\right\} $$}}

which provides us four linear independent systems of integrals. However, when $$z$$ or $$z'$$ are two such systems, it follows from the things previously stated:

1738

$\sum_{i=1}^{4}\left(z_{i}\right)^{2}=\text{const,}\ \sum_{i=1}^{4}z_{i}z_{i}^{\prime}=\text{const,}\ \sum_{i=1}^{4}\left(z_{i}^{\prime}\right)^{2}=\text{const}$

there are quadratic relations, which we can be arranged by suitable choice of the integration constants, so that we obtain an orthogonal matrix of 16 direction cosines. From the four quantities $$z_{1}^{(\alpha)}$$ it then follows

$x^{(\alpha)}=\int z_{1}^{(\alpha)}ds\quad\alpha=1,2,3,4$

In respect to that, we have to consider the characteristic determinant:

$\Delta(\lambda)\equiv\left

There are the following cases in the denotation of the elementary divisors:

(A) $$[1111]$$, where there arise two opposite equal roots;

(B) multiple roots:


 * 1. $$[(11)11]$$ one root-pair is zero: It follows

$\frac{1}{\mathrm{R}_{3}}\equiv0$ and$\frac{1}{\mathrm{R}_{1}}\ne0$|undefined


 * while $$\frac{1}{\mathrm{R}_{2}}$$ can be $$=0$$ or $$\ne0$$


 * 2. $$[(1111)]$$ both root-pairs are zero:

$\frac{1}{\mathrm{R}_{3}}=0,\ \frac{1}{\mathrm{R}_{2}}=0,\ \frac{1}{\mathrm{R}_{1}}=0.$|undefined

3. [(31)] both-root pairs are zero:

$\frac{1}{\mathrm{R}_{3}}=0,\ \frac{1}{\mathrm{R}_{1}^{2}}+\frac{1}{\mathrm{R}_{2}^{2}}=0,\ \frac{1}{\mathrm{R}_{1}}=\pm i\frac{1}{\mathrm{R}_{2}}$|undefined

Other cases are not possible; thus with respect to the world lines it is impossible:

$$[(11)11]$$ with $$\frac{1}{\mathrm{R}_{1}}=0,\ \frac{1}{\mathrm{R}_{3}}\ne0$$ or $$\frac{1}{\mathrm{R}_{1}}=\frac{1}{\mathrm{R}_{3}}=0$$ with $$\frac{1}{\mathrm{R}_{2}}\ne0$$.

Furthermore, the double root

$\lambda^{2}=-\frac{1}{2}\sum\frac{1}{\mathrm{R}_{2}}$|undefined

is impossible; because it leads to

$\sum\frac{1}{\mathrm{R}_{2}}=\pm\frac{2}{\mathrm{R}_{1}\mathrm{R}_{3}}$ or $\left(\frac{1}{\mathrm{R}_{1}}\mp\frac{1}{\mathrm{R}_{3}}\right)^{2}+\frac{1}{\mathrm{R}_{2}^{2}}=0$,|undefined

thus

$\frac{1}{\mathrm{R}_{1}}=\pm\frac{1}{\mathrm{R}_{3}}\pm i\frac{1}{\mathrm{R}_{2}}$|undefined

However, by the things previously said, $$\mathrm{R}_{1}$$ is real, $$\mathrm{R}_{3}$$ and $$\mathrm{R}_{2}$$ are purely imaginary; therefore it should be $$\frac{1}{\mathrm{R}_{3}}=0$$ (case 3) or $$\frac{1}{\mathrm{R}_{3}}=\frac{1}{\mathrm{R}_{2}}=\frac{1}{\mathrm{R}_{1}}=0$$ (case 2).

Integration of the system; case (A).
Let us set:

$M=\frac{1}{\mathrm{R}_{1}^{2}}+\frac{1}{\mathrm{R}_{2}^{2}}+\frac{1}{\mathrm{R}_{3}^{2}}+\frac{2}{\mathrm{R}_{1}\mathrm{R}_{3}},$|undefined

$N=\frac{1}{\mathrm{R}_{1}^{2}}+\frac{1}{\mathrm{R}_{2}^{2}}+\frac{1}{\mathrm{R}_{3}^{2}}-\frac{2}{\mathrm{R}_{1}\mathrm{R}_{3}},$|undefined

$\phi=s\frac{\sqrt{M}+\sqrt{N}}{2},\ \chi=s\frac{\sqrt{M}-\sqrt{N}}{2},$|undefined

$\begin{aligned}g^{(12)} & =\frac{d\omega_{1}d\phi-d\omega_{3}d\chi}{(d\phi)^{2}-(d\chi)^{2}}, & g^{(12)} & =0, & g^{(14)} & =-\frac{d\omega_{2}d\chi}{(d\phi)^{2}-(d\chi)^{2}},\\ g^{(34)} & =\frac{d\omega_{3}d\phi-d\omega_{1}d\chi}{(d\phi)^{2}-(d\chi)^{2}}, & g^{(24)} & =0, & g^{(23)} & =\frac{d\omega_{2}d\phi}{(d\phi)^{2}-(d\chi)^{2}}, \end{aligned} $|undefined

Furthermore, the four pair-wise orthogonal unit vectors $$e_{0}f_{0}g_{0}h_{0}$$ shall be defined for the value $$s=0$$ as follows:

{{center|$$\left.\begin{aligned}e_{0}^{(\alpha)} & =c_{1,0}^{(\alpha)}\frac{g^{(12)}}{\sqrt{\left(g^{(12)}\right)^{2}+\left(g^{(23)}\right)^{2}}}-c_{3,0}^{(\alpha)}\frac{g^{(23)}}{\sqrt{\left(g^{(12)}\right)^{2}+\left(g^{(23)}\right)^{2}}}\\ f_{0}^{(\alpha)} & =c_{2,0}^{(\alpha)}\frac{g^{(12)}}{\sqrt{\left(g^{(12)}\right)^{2}+\left(g^{(14)}\right)^{2}}}+c_{4,0}^{(\alpha)}\frac{g^{(14)}}{\sqrt{\left(g^{(12)}\right)^{2}+\left(g^{(14)}\right)^{2}}}\\ g_{0}^{(\alpha)} & =c_{1,0}^{(\alpha)}\frac{g^{(34)}}{\sqrt{\left(g^{(34)}\right)^{2}+\left(g^{(14)}\right)^{2}}}-c_{3,0}^{(\alpha)}\frac{g^{(14)}}{\sqrt{\left(g^{(34)}\right)^{2}+\left(g^{(14)}\right)^{2}}}\\ h_{0}^{(\alpha)} & =c_{2,0}^{(\alpha)}\frac{g^{(34)}}{\sqrt{\left(g^{(34)}\right)^{2}+\left(g^{(23)}\right)^{2}}}+c_{4,0}^{(\alpha)}\frac{g^{(23)}}{\sqrt{\left(g^{(34)}\right)^{2}+\left(g^{(23)}\right)^{2}}} \end{aligned} \right\} \ \alpha=1,2,3,4$$}}

Then the integral reads:

$\begin{aligned}x^{(\alpha)}=x_{0}^{(\alpha)}+\mathrm{R}_{1}c_{2,0}^{(\alpha)}+\frac{\mathrm{R}_{1}\mathrm{R}_{3}}{\mathrm{R}_{2}}c_{4,0}^{(\alpha)} & +\rho_{\phi}\sin\varphi\cdot e_{0}^{(\alpha)}-\rho_{\varphi}\cos\phi f_{0}^{(\alpha)}\\ & +\rho_{\chi}\sin\chi\cdot g_{0}^{(\alpha)}-\rho_{\chi}\cos\chi h_{0}^{(\alpha)},\\ & \quad\alpha=1,2,3,4 \end{aligned} $|undefined

where

$\rho_{\phi}=\sqrt{\frac{\left(d\omega_{1}\right)^{2}-(d\chi)^{2}}{\left(d\phi\right)^{2}-(d\chi)^{2}}}\frac{ds}{d\phi}$ and $\rho_{\chi}=\sqrt{\frac{\left(d\phi\right)^{2}-d\omega_{1}^{2}}{\left(d\phi\right)^{2}-(d\chi)^{2}}}\frac{ds}{d\chi}$|undefined

Here, the $$g^{\left(\alpha_{1}\alpha_{2}\right)}$$ are the location parameter of a plane $$E_{\phi}=[ef]$$, and the dual system $$g^{\ast\left(\alpha_{1}\alpha_{2}\right)}=g^{\left(\alpha_{3}\alpha_{4}\right)}$$ gives the plane $$E_{\chi}=[gh]$$ which is completely perpendicular to it. In these planes $$E_{\phi}$$ or $$E_{\chi}$$, around the point

$x_{0}+\mathrm{R}_{1}c_{2,0}+\frac{\mathrm{R}_{1}\mathrm{R}_{3}}{\mathrm{R}_{2}}c_{4,0}$|undefined

there is a rotation around the angles $$\phi$$ or $$\chi$$ on the circle with radius $$\rho_{\phi}$$ or $$\rho_{\chi}$$. The curve $$A$$ lies therefore on a hypersphere, which is its osculating hypersphere at point $$x_{0}$$, which is at the same time the osculating hypersphere for all points of the curve, as one can see after some computations using the formula by l.c. for the center of the osculating hypersphere

$x+\mathrm{R}_{1}c_{2}+\mathrm{R}_{2}\frac{d\mathrm{R}_{1}}{ds}c_{3}+\mathrm{R}_{3}\left(\frac{d}{ds}\left[\mathrm{R}_{2}\frac{d\mathrm{R}_{1}}{ds}\right]+\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)c_{4}$.|undefined

The meaning of the mentioned rotations $$\phi$$ and $$\chi$$ is as follows: Every orthogonal infinitesimal transformation in $$S_{4}$$ is (similar as in $$S_{3}$$) given by a skew-symmetric $$4\times4$$-row matrix or also by the related vector of second kind, which in the general case is not represented by a surface-piece, but by two mutually perpendicular surface-pieces, i.e. its components $$\varepsilon^{(\alpha\beta)}$$ do not define a straight line in the infinitely distant $$S_{3}^{\infty}$$ of $$S_{4}$$, but a linear complex, because of

$\varepsilon^{(12)}\varepsilon^{(34)}+\varepsilon^{(13)}\varepsilon^{(32)}+\varepsilon^{(14)}\varepsilon^{(23)}\ne0$

It is known, that in the general case this complex shares two conjugated polars with every surface of second kind; for the absolute measure-surface, they become the infinitely distant lines of two mutually perpendicular planes. In the case of the infinitesimal orthogonal transformations underlying the formulas, it is just

$\varepsilon^{(12)}=d\omega_{1},\ \varepsilon^{(23)}=d\omega_{2},\ \varepsilon^{(34)}=d\omega_{3};\ \varepsilon^{(13)}=\varepsilon^{(14)}=\varepsilon^{(24)}=0,$

thus

$\varepsilon^{(12)}\varepsilon^{(34)}+\varepsilon^{(13)}\varepsilon^{(42)}+\varepsilon^{(14)}\varepsilon^{(23)}=d\omega_{1}d\omega_{3}\ne0$

becomes a complex in case A, and the mentioned two perpendicular planes representing it, $$E_{\phi}$$ or $$E_{\chi}$$.

This is at first true for every curve of three-fold curvature. If now $$\frac{1}{\mathrm{R}_{1}}=\text{const}$$, $$\frac{1}{\mathrm{R}_{2}}=\text{const}$$, $$\frac{1}{\mathrm{R}_{3}}=\text{const}$$, then the elements of the rotations remain the same from point to point, then the individual infinitesimal orthogonal transformations are added together to a finite orthogonal transformation having the same rotation planes; we write

$$x+\mathrm{R}_{1}c_{2}+\frac{\mathrm{R}_{1}\mathrm{R}_{3}}{\mathrm{R}_{2}}c_{4}\equiv x_{0}+\mathrm{R}_{1}c_{2,0}+\frac{\mathrm{R}_{1}\mathrm{R}_{3}}{\mathrm{R}_{2}}c_{4,0}=(0,0,0,0)$$

$f_{0}\Vert-x^{(1)},\ e_{0}\Vert x^{(2)},\ h_{0}\Vert-x^{(3)},\ g_{0}\Vert x^{(4)}$

and we obtain for curve A (simultaneously the trajectory of the most general orthogonal transformation of determinant +1 and fixed origin)

$x^{(1)}=\rho_{\phi}\cos\phi,\ x^{(2)}=\rho_{\phi}\sin\phi,\ x^{(3)}=\rho_{\chi}\cos\chi,\ x^{(4)}=\rho_{\chi}\sin\chi$

The related family.
The other trajectories of the family are indeed curves (A) as well. They are only distinguished by one another by the mutual relation of their points. Thus it will be

$y^{(1)}=\rho_{\phi}^{\prime}\cos\phi',\ x^{(2)}=\rho_{\phi}^{\prime}\sin\phi',\ y^{(3)}=\rho_{\chi}^{\prime}\cos\chi',\ y^{(4)}=\rho_{\chi}^{\prime}\sin\chi'$

$$\phi'=\phi_{0}^{\prime}$$ or $$\chi'=\chi_{0}^{\prime}$$ shall belong to the value $$s=\phi=\chi=0$$; we write

$\phi'=\lambda\bar{\phi}+\phi_{0}^{\prime}\quad\chi'=\mu\bar{\phi}+\chi_{0}^{\prime}$

and in this way we obtain $$\phi=0$$ at $$s=0$$ or also

$\chi'=\mu\bar{\phi}+\chi_{0}^{\prime}=\mu\phi$

$\phi'=\lambda\bar{\phi}+\phi_{0}^{\prime}=\lambda\bar{\phi}+\frac{\lambda}{\mu}\chi_{0}^{\prime}+-\frac{\lambda}{\mu}\chi_{0}^{\prime}+\phi_{0}^{\prime}=\lambda\phi-\frac{\lambda}{\mu}\chi_{0}^{\prime}+\phi_{0}^{\prime}=\lambda\left(\phi-\phi_{y}^{0}\right)$

where $$\phi_{y}^{0}$$ is the angle, which corresponds to the value $$\lambda\bar{\phi}=-\phi_{0}^{\prime}$$, thus $$\phi'=0$$, by which we find (if we also set $$\rho_{\phi}^{\prime}=a_{y}$$, $$\rho_{\chi}^{\prime}=b_{y}$$):

$\begin{aligned}y^{(1)} & =a_{y}\cos\lambda\left(\phi-\phi_{y}^{0}\right), & y^{(2)} & =a_{y}\text{sin }\lambda\left(\phi-\phi_{y}^{0}\right),\\ y^{(3)} & =b_{y}\cos\mu\varphi, & y^{(4)} & =b_{y}\sin\mu\varphi. \end{aligned}$

with $$a_{y},b_{y},\phi_{0}$$ as the parameter of the family, $$\lambda$$, $$\mu$$, are constant for one and the same family.

Representation of the family as the totality of curves, which “participate” in the “windings” of the principal curve.
Every point, which is fixed in the comoving tetrad, thus

$y=x+\Lambda^{(1)}c_{1}+\Lambda^{(2)}c_{2}+\Lambda^{(3)}c_{3}+\Lambda^{(4)}c_{4}$

where the $$\Lambda$$ are constants, will evidently participate in its rotation, i.e. it describes a trajectory of the family, which is definitely determined by curve $$x$$. Thus the previous curves $$y$$ must also be representable in this shape, of which one can easily convince oneself when one computes the axes $$c_{k}^{(\alpha)}$$ of the comoving tetrad for

$\begin{aligned}x^{(1)} & =a_{x}\cos\lambda\left(\phi-\phi_{x}^{0}\right), & x^{(2)} & =a_{x}\text{sin }\lambda\left(\phi-\phi_{x}^{0}\right),\\ x^{(3)} & =b_{x}\cos\mu\varphi, & x^{(4)} & =b_{x}\sin\mu\varphi. \end{aligned} $

One will find, that the quantities

$a_{y}b_{y}\cos\lambda\left(\phi_{x}^{0}-\phi_{y}^{0}\right)\text{sin }\lambda\left(\phi_{x}^{0}-\phi_{y}^{0}\right)$

within

$y^{(1)}=a_{y}\cos\lambda\left(\phi-\phi_{y}^{0}\right)=a_{y}\cos\lambda\left(\phi-\phi_{x}^{0}\right)\cos\lambda\left(\phi_{x}^{0}-\phi_{y}^{0}\right)+a_{y}\text{sin }\lambda\left(\phi-\phi_{x}^{0}\right)\text{sin }\lambda\left(\phi_{x}^{0}-\phi_{y}^{0}\right)$ etc.

can be computed as linear functions of the $$\Lambda$$, where only those quantities arise as coefficients, which depend on $$\mathrm{R}_{1}\mathrm{R}_{2}\mathrm{R}_{3}$$, i.e. constants. The directions

$\frac{\partial y}{\partial a_{y}}\quad\frac{\partial y}{\partial b_{y}}\quad\frac{\partial y}{\partial\phi_{y}^{0}}$ and $\frac{\partial y}{\partial\phi}$|undefined

therefore allow for the representation

$\sum_{k=1}^{4}A^{(k)}c_{k},$,

in which the $$A^{(k)}$$ are constants with respect to $$\phi$$ or $$s$$. This system varying with the light-point or reference-point, is therefore evidently nothing other than the comoving tetrad of the light-point or of the reference-point, because the direction for the latter can again be computed in the form

$\sum_{k=1}^{4}A^{(k)}c_{k}$

Generalization.
Let there be a curve of three-fold curvature of $$S_{4}$$ and a position $$x_{0}$$, where we have the values

$\frac{1}{\mathrm{R}_{1,0}}\ne0,\ \frac{1}{\mathrm{R}_{2,0}}\ne0,\ \frac{1}{\mathrm{R}_{3,0}}\ne0$.|undefined

If we determine (by using these values) a curve (A) that goes through $$x_{0}$$:

$\begin{aligned}X^{(\alpha)}=x_{0}^{(\alpha)}+\mathrm{R}_{1,0}c_{2,0}^{(\alpha)}+\frac{\mathrm{R}_{1,0}\mathrm{R}_{3,0}}{\mathrm{R}_{2,0}}c_{4,0}^{(\alpha)} & +\rho_{\phi}^{0}\sin\phi\cdot e_{0}^{(\alpha)}-\rho_{\phi}^{0}\cos\phi\cdot f_{0}^{(\alpha)}\\ & +\rho_{\chi}^{0}\sin\chi\cdot g_{0}^{(\alpha)}-\rho_{\chi}^{0}\cos\chi\cdot h_{0}^{(\alpha)},\\ & \quad\alpha=1,2,3,4 \end{aligned} $|undefined

where

$\phi=\left(\frac{d\phi}{ds}\right)_{0}s,\ \chi=\left(\frac{d\chi}{ds}\right)_{0}s$

and $$\left(\frac{d\phi}{ds}\right)_{0}$$, $$\left(\frac{d\chi}{ds}\right)_{0}$$ are composed of the values $$\frac{1}{\mathrm{R}_{1,0}},\ \frac{1}{\mathrm{R}_{2,0}},\ \frac{1}{\mathrm{R}_{3,0}}$$ in accordance with the things previously said, then this curve has a contact of second order with the given one in $$x_{0}$$. It will become a contact of third order, if

$\left(\frac{d\mathrm{R}_{1}}{ds}\right)_{s=0}=0$|undefined

Thus a curve will go through the point $$y$$, which participates in the windings of curve $$x$$, if again

$y=x+\Lambda^{(1)}c_{1}+\Lambda^{(2)}c_{2}+\Lambda^{(3)}c_{3}+\Lambda^{(4)}c_{4}$

where the $$\Lambda$$ are constants. The emergence of this curve can be imagined by drawing the trajectory of family (A), which in $$x$$ is determined by the local values $$\mathrm{R}_{1}\mathrm{R}_{2}\mathrm{R}_{3}$$, through the respective location $$y$$, and by displacing $$y$$ always by an infinitesimal piece along the thus determined trajectories (which are varying from place to place); but such infinitesimal piece has only a contact of first order with the eventual curve

$y=x+\Lambda^{(1)}c_{1}+\Lambda^{(2)}c_{2}+\Lambda^{(3)}c_{3}+\Lambda^{(4)}c_{4}$,

and a contact of second order for

$\left(\frac{d\mathrm{R}_{1}}{ds}\right)_{s=0}=0$.|undefined

===Case (B) 1. Common helix, or circular line of $$S_{3}$$.===

If $$\frac{1}{\mathrm{R}_{3}}\equiv0,\ \frac{1}{\mathrm{R}_{1}}=\text{const}\ne0,\ \frac{1}{\mathrm{R}_{2}}=\text{const}$$ and at first $$\ne0$$, then one has to set

$\begin{aligned}\vartheta & =s\cdot\sqrt{\frac{1}{\mathrm{R}_{1}^{2}}+\frac{1}{\mathrm{R}_{2}^{2}}},\\ k^{(\alpha)} & =c_{1}^{(\alpha)}\frac{d\omega_{1}}{d\vartheta}-c_{3}^{(\alpha)}\frac{d\omega_{2}}{d\vartheta},\\ j^{(\alpha)} & =c_{1}^{(\alpha)}\frac{d\omega_{2}}{d\vartheta}+c_{3}^{(\alpha)}\frac{d\omega_{1}}{d\vartheta},\quad\alpha=1,2,3,4 \end{aligned} $|undefined

where $$kc_{2}j$$ forms an orthogonal triad, and one obtains

$x=x_{0}+c_{2,0}\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}+k_{0}\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}\sin\vartheta-c_{2,0}\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}\cos\vartheta+h_{0}\frac{d\omega_{2}}{d\vartheta}\frac{ds}{d\vartheta}\vartheta$|undefined

which is a helix having its axis parallel to $$h_{0}$$ through the center of the circular cylinder $$x_{0}+c_{2,0}\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}$$, with slope parameter $$\frac{d\omega_{2}}{d\vartheta}\frac{ds}{d\vartheta}$$. If one takes

$-c_{2,0}\Vert x^{(1)},\ \varkappa_{0}\Vert x^{(2)},\ h_{0}\Vert x^{(4)}$

$x+c_{2,0}\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}=(0,0,0,0)$|undefined

then one obtains:

$\begin{aligned}x^{(1)} & =\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}\cos\vartheta=a_{x}\cos\vartheta,\\ x^{(2)} & =\frac{d\omega_{1}}{d\vartheta}\frac{ds}{d\vartheta}\sin\vartheta=a_{x}\sin\vartheta,\\ x^{(3)} & =x_{0}^{(3)},\\ x^{(4)} & =\frac{d\omega_{2}}{d\vartheta}\frac{ds}{d\vartheta}\vartheta=\mu\vartheta \end{aligned} $|undefined

which at the same time is an orthogonal transformation at which a rotation in the $$x^{(1)}x^{(2)}$$ plane arises, while the plane $$x^{(3)}x^{(4)}$$ experiences a displacement along $$x^{(4)}$$, from which it follows for the family:

$\begin{aligned}y^{(1)} & =a_{y}\cos\lambda\left(\phi-\phi_{y}^{0}\right), & y^{(2)} & =a_{y}\text{sin }\lambda\left(\phi-\phi_{y}^{0}\right),\\ y^{(3)} & =y_{0}^{(3)}, & y^{(4)} & =\mu\varphi_{y}, \end{aligned} $

which again can be represented by

$y^{(\alpha)}=x^{(\alpha)}+\Lambda^{(1)}c_{1}^{(\alpha)}+\Lambda^{(2)}c_{2}^{(\alpha)}+\Lambda^{(3)}c_{3}^{(\alpha)}+\Lambda^{(4)}c_{4}^{(\alpha)},\quad\alpha=1,2,3,4$

when $$\Lambda$$ are constant; this representation can be extended to curves that are participating in the windings of an arbitrary given one for which $$\frac{1}{\mathrm{R}_{3}}=0$$. To that end, the given curve only needs to osculate through the curve B, where the latter is of second order (for $$\left(\frac{d\mathrm{R}_{1}}{ds}\right)_{s=0}=0$$ of the third order) and is computed in terms of $$\frac{1}{\mathrm{R}_{1}}=\frac{1}{\mathrm{R}_{1,0}}$$, $$\frac{1}{\mathrm{R}_{2}}=\frac{1}{\mathrm{R}_{2,0}}$$ (these shall be the values in $$x_{0}$$ at the given curve). Then one finds curve $$y$$ by compilation of the infinitesimal trajectory pieces, which leave behind the orthogonal transformations belonging to the respective $$x_{0}$$.

In case B, the directions

$\frac{\partial y}{\partial a_{y}},\ \frac{\partial y}{\partial\phi_{y}^{0}},\ \frac{\partial y}{\partial y_{0}^{(3)}},\ \frac{\partial y}{\partial\phi}$|undefined

are again representable in the form

$\sum_{k=1}^{4}A^{(k)}c_{k}$

with constant $$A^{(k)}$$.

The special case $$\frac{1}{\mathrm{R}_{2}}=0$$ does not require a particular discussion (circle).

Case (B) 3. Lyon curve (Helix on a cylinder of minimal lines).
Let us introduce within the integral (B) 1:

$\frac{d\omega_{2}}{ds}=\pm\sqrt{\left(\frac{d\vartheta}{ds}\right)^{2}-\left(\frac{d\omega_{1}}{ds}\right)^{2}}$|undefined

and then we let $$\frac{d\vartheta}{ds}$$ converge against zero. Then one obtains the case

$\frac{1}{\mathrm{R}_{1}^{2}}+\frac{1}{\mathrm{R}_{2}^{2}}=0$;|undefined

it follows

$x=x_{0}+c_{1,0}\left(s-\frac{s^{3}}{6\mathrm{R}_{1}^{2}}\right)+c_{2,0}\frac{s^{2}}{2\mathrm{R}_{1}}\pm c_{3,0}\frac{is^{3}}{6\mathrm{R}_{1}^{2}}$|undefined

($$s$$ is the arc, $$\mathrm{R}_{1}$$ thre radius of the first curvature, $$\mathrm{R}_{2}=\pm i\mathrm{R}_{1}$$). In this case, the direction $$h_{0}$$ coincides with one of the two minimal lines of plane $$\left[c_{1,0}c_{3,0}\right]$$. The curve therefore lies on a cylinder, whose generators are directed parallel to this direction. The tangent

$c_{1}=c_{1,0}\left(1-\frac{s^{2}}{2\mathrm{R}_{1}^{2}}\right)+c_{2,0}\frac{s}{\mathrm{R}_{1}}\pm c_{3,0}\frac{is^{2}}{2\mathrm{R}_{1}^{2}}$|undefined

forms a constant product $$\pm i$$ with the minimal vector in this direction

$\pm c_{1,0}+c_{3,0}$,

which is a known property of all helices; therefore it has to be denoted as helix.

If

$x_{0}=(0,0,0,0),\ s=i\vartheta\alpha,\ \frac{s}{\mathrm{R}_{1}}=i\vartheta,\ \frac{s}{\mathrm{R}_{1}}=\frac{1}{\alpha}$|undefined

and

$c_{1,0}\Vert x^{(4)},\ c_{2,0}\Vert x^{(1)},\ c_{3,0}\Vert x^{(3)}$

are taken, it follows

$x^{(1)}=-\frac{1}{2}\alpha\vartheta^{2},\ x^{(2)}=0,\ x^{(3)}=\pm\frac{1}{6}\alpha\vartheta^{3},\ x^{(4)}=i\frac{\alpha\vartheta^{3}}{6}+i\alpha\vartheta$,|undefined

which is simultaneously the trajectory of the respective orthogonal transformation. For $$\vartheta=\vartheta_{0}$$ we shall have $$x_{0}^{(1)}x_{0}^{(2)}x_{0}^{(3)}x_{0}^{(4)}$$, thus if we now replace $$\vartheta$$ by $$\vartheta+\vartheta_{0}$$:

$\begin{aligned}x^{(1)} & =-\frac{1}{2}\alpha\vartheta^{2}-\alpha\vartheta_{0}-\frac{1}{2}\alpha\vartheta_{0}^{2}=-\frac{1}{2}\alpha\vartheta^{2}+\vartheta\left(\pm x_{0}^{(3)}+ix_{0}^{(4)}+x_{0}^{(1)}\right),\\ x^{(2)} & =x_{0}^{(2)},\\ x^{(3)} & =\pm\left\{ \frac{1}{6}\alpha\vartheta^{3}-\frac{1}{2}\vartheta^{2}\left(\pm x_{0}^{(3)}+ix_{0}^{(4)}\right)-\vartheta x_{0}^{(1)}\pm x_{0}^{(3)}\right\} \\ x^{(4)} & =i\left\{ \frac{1}{6}\alpha\vartheta^{3}-\frac{1}{2}\vartheta^{2}\left(\pm x_{0}^{(3)}+ix_{0}^{(4)}\right)-\vartheta x_{0}^{(1)}\pm x_{0}^{(3)}\right\} +i\alpha\vartheta-i\left(\pm x_{0}^{(3)}+ix_{0}^{(4)}\right) \end{aligned} $

One can additionally assume

$\pm x_{0}^{(3)}+ix_{0}^{(4)}=0$

and obtains the representation of the family:

$x^{(1)}=x_{0}^{(1)}-\frac{1}{2}\alpha\vartheta^{2},\ x^{(2)}=x_{0}^{(2)},\ x^{(3)}=x_{0}^{(3)}\pm\left\{ -\vartheta x_{0}^{(1)}+\frac{1}{6}\alpha\vartheta^{3}\right\} ,\ x^{(4)}=\pm x^{(3)}+i\alpha\vartheta$.

The remarks made earlier are valid again, concerning the form

$y=x+\Lambda^{(1)}c_{1}+\Lambda^{(2)}c_{2}+\Lambda^{(3)}c_{3}+\Lambda^{(4)}c_{4}$

and the representation of $$\frac{\partial y}{\partial y_{0}^{(1)}},\ \frac{\partial y}{\partial y_{0}^{(2)}},\ \frac{\partial y}{\partial y_{0}^{(3)}},\ \frac{\partial y}{\partial\vartheta}$$ as

$\sum_{k=1}^{4}A^{(k)}c_{k}$

Reality questions.
Regarding A it follows, that when for instance $$E_{\phi}$$ only contains spacelike directions, then $$E_{\chi}$$ as the completely perpendicular plane must also contain timelike directions (its infinitely distant line cuts the absolute measure-surface in a real way). In our representation, $$\chi$$ must be an imaginary angle. Regarding B (1) it follows, that the rotation angle $$\vartheta$$ can be real or imaginary and correspondingly the displacement can be imaginary or real: cases (B) 1 and 2 of § 6. The types of curves of constant curvatures are thus obtained the same way as there.