Singularity Properties of Spacelike Circular Surfaces

Abstract

The aim of the paper is on spacelike circular surfaces and singularities in Minkowski 3-space ${\mathbb{E}}_1^3$. A spacelike circular surface with a stationary radius can be swept out by movable a Lorentzian circle following a non-null curve, which acts as the spine curve. In the study, we have represented spacelike circular surface and have furnished its geometric properties such as singularities and striction curves contrasting with those of ruled surfaces. Subsequently, a new type of spacelike circular surface was distinguished and named as the spacelike roller coaster surface. Meanwhile, we support the outcomes of the work by some examples.

1. Introduction

Singularity properties of curves and surfaces is an energetic subject of experimentation in several sections of mathematics and physics. In the light of differential geometry, theory of curves and surfaces are defined by functions with one parameter and two parameters, respectively. In the latest years, singularity properties of curves and surfaces has become a fundamental tool for distinct interesting domains such as computer vision, and medical imaging [1,2,3,4].

In spatial kinematics, the movement of a one-parameter set of circles with a stationary radius creates a circular surface; however, the movement of a one-parameter set of directed lines creates a ruled surface. A circular surface has a spine curve and ruled surface has a striction curve. The wrapper of a set of circle planes is a tangential ruled surface. The characteristic of a tangential surface is straight lines, which are at a tangent to the edge of regression. The edge of regression includes the singular points of the tangential develop-able surface [1,2,3,4,5,6]. With an analogous notion, geometers have examined circular surfaces in Euclidean and Minkowski 3-spaces. For example, Izumiya et al. [7] addressed considerable geometric advantage and singularity properties of circular surfaces by contrasting them with convenient aspects of ruled surfaces. In [8], the authors located great circular surfaces as one-parameter set of great circles in the these spheres. They also expanded a mutual categorization for singularities of such surfaces and discussed the geometric versions in view of spherical geometry. In [9], a novel set of circular surfaces in Euclidean 3-space is displayed by a curve and a congruence of circles. In particular, the authors inspected almost geometrical descriptions of circular surfaces when the directrix (base) curve is an algebraic curve. In [10] Alluhaibi inspected the singularity properties of circular surfaces in Euclidean 3-space. Abdel-Baky and Unluturk defined and studied spacelike circular surfaces in Minkowski 3-space employing a differentiable one-parameter set of spacelike circles and found some geometric possessions [10]. Further, the authors realized spacelike roller coaster surfaces as spacelike circular surfaces whose creating circles are curvature lines, and specified a parametric representation of a spacelike roller coaster surface over a spacelike base curve with a spacelike principal normal. Tuncer et al. [11] recreated equations of spacelike circular surfaces and spacelike roller coaster surfaces by version employing unit split quaternions and homothetic movements. R. Abdel-Baky et al. [12] inspected many geometric assets of timelike circular surfaces by using suitable aspects of ruled surfaces. However, they did not consider the properties of the singularities.

In this work, we check the properties of the singularities of spacelike circular surfaces with stationary radius that can be swept out by moving a Lorentzian circle with its center following a non-null curve, which acts as the spine curve. Then, we find the condition for a non-null curve to act as a striction curve on a spacelike circular surface, similar to those in [7]. Moreover, we inspect the local singularities of spacelike circular surfaces. By taking into account the case in which all creating Lorentzian circles are curvature lines, except at umbilical points or singular points, we classified such spacelike circular surfaces into a spacelike canal surface, a hyperbolic sphere, a special kind of spacelike surface that is smoothly attached to these other three spacelike surfaces. Finally, we found spacelike roller coaster surfaces and support our result by giving some examples with figures. This work aims to address the singularities and symmetry properties of spacelike circular surfaces with those of ruled surfaces in 3-Minkowski 3-space ${\mathbb{E}}_1^3$.

2. Basic Concepts

In this section, we give some definitions and basic notions that we use in this paper [13,14,15]. Let ${\mathbb{R}}^3=\{\left({\beta }_1,{\beta }_2,{\beta }_3\right)\mid$, ${\beta }_i\in \mathbb{R}$ (i = 1, 2, 3)} be a 3-dimensional Cartesian space. For any $\mathit{\alpha }$$=\left({\alpha }_1,{\alpha }_2,{\alpha }_3\right)$, and $\mathit{\beta }$$=\left({\beta }_1,{\beta }_2,{\beta }_3\right)\in {\mathbb{R}}^3$, the Lorentzian scalar product of $\alpha$, and $\beta$ is realized by:

$$<\mathbf{\alpha },\mathbf{\beta }>=-{\alpha }_1{\beta }_1+{\alpha }_2{\beta }_2+{\alpha }_3{\beta }_3.$$

We call $({\mathbb{R}}^3,<,>)$ Minkowski 3-space. We use ${\mathbb{E}}_1^3$ instead of (${\mathbb{R}}^3,<,>)$. We say that a non-zero vector $\mathbf{\alpha }$$\in {\mathbb{E}}_1^3$ is spacelike, lightlike, or timelike if <$\mathbf{\alpha }$,$\mathbf{\alpha }$$>>0$, <$\mathbf{\alpha }$,$\mathbf{\alpha }$$>=0$ or <$\mathbf{\alpha }$,$\mathbf{\alpha }$>$><0$ respectively. The norm of the vector $\mathbf{\alpha }$$\in {\mathbb{E}}_1^3$ is defined to be $∥\mathbf{\alpha }∥=\sqrt{\left|<\mathbf{\alpha },\mathbf{\alpha }>\right|}$. Furthermore, for any two vectors $\mathbf{\alpha }$, and $\mathbf{\beta }$ the cross product $\mathbf{\alpha }$×$\mathbf{\beta }$ is:

$$\mathbf{\alpha }\times \mathbf{\beta }=\left|\begin{array}{ccc}-\mathbf{i}& \mathbf{j}& \mathbf{k}\\ {\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\end{array}\right|=\left(-({\alpha }_2{\beta }_3-{\alpha }_3{\beta }_2),({\alpha }_3{\beta }_1-{\alpha }_1{\beta }_3),({\alpha }_1{\beta }_2-{\alpha }_2{\beta }_1)\right),$$

where $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ is the canonical basis of ${\mathbb{E}}_1^3$. The hyperbolic and Lorentzian unit spheres, respectively, are:

$${\mathbb{H}}_{+}^2=\left\{\mathbf{\beta }\in {\mathbb{E}}_1^3\mid {∥\mathbf{\beta }∥}^2=-{\beta }_1^2+{\beta }_2^2+{\beta }_3^2=-1,{\beta }_1>0\right\},$$

and

$${\mathbb{S}}_1^2=\left\{\mathbf{\beta }\in {\mathbb{E}}_1^3\mid {∥\mathbf{\beta }∥}^2=-{\beta }_1^2+{\beta }_2^2+{\beta }_3^2=1\right\}.$$

Definition 1.

The angles among vectors in ${\mathbb{E}}_1^3$ are realized by the following [13,14,15]:

(A). For two spacelike vectors $\mathbf{a}$ and $\mathbf{b}$ in ${\mathbb{E}}_1^3$,

(i). If $\mathbf{a}$ and $\mathbf{b}$ lie in a timelike plane, then there is a unique real number $\vartheta \ge 0$ such that $<\mathbf{a},\mathbf{b}>=∥\mathbf{a}∥∥\mathbf{b}∥cosh\vartheta$,

(ii). If $\mathbf{a}$ and $\mathbf{b}$ lie in a spacelike plane, then there is a unique real number $\vartheta \ge 0$ such that $<\mathbf{a},\mathbf{b}>=∥\mathbf{a}∥∥\mathbf{b}∥cos\vartheta$.

(B). For two vectors $\mathbf{a}$ and $\mathbf{b}$ in the same timelike cone, there is a unique real number $\vartheta \ge 0$ such that $<\mathbf{a},\mathbf{b}>=-∥\mathbf{a}∥∥\mathbf{b}∥cosh\vartheta$. This number is named the hyperbolic angle among the vectors $\mathbf{a}$ and $\mathbf{b}$.

(C). For a spacelike vector $\mathbf{a}$ and a timelike vector $\mathbf{b}$, there is a unique real number $\vartheta \ge 0$ such that $<\mathbf{a},\mathbf{b}>=∥\mathbf{a}∥∥\mathbf{b}∥sinh\vartheta$.

We reference a surface M in ${\mathbb{E}}_1^3$ by

$$M:\mathbf{y}(u,\theta )=\left(y_1(u,\theta ),y_2\left(u,\theta \right),(y_3\left(u,\theta \right)\right),(u,\theta )\in \mathbb{D}\subseteq {\mathbb{R}}^2.$$

(1)

The unit normal vector is realized by $\mathbf{\xi }(u,\theta )={\mathbf{y}}_u\times {\mathbf{y}}_{\theta }{∥{\mathbf{y}}_u\times {\mathbf{y}}_{\theta }∥}^{-1}$, where ${\mathbf{y}}_i=\frac{\partial \mathbf{y}}{\partial i}$. Then, the metric (first fundamental form) I of a surface M is defined by

$$I=l_{11}d{s}^2+2l_{12}dsd\theta +l_{22}d{\theta }^2,$$

(2)

where $l_{11}=$<$y_u,$$y_{\theta }$>, $l_{12}=$<$y_u,$$y_{\theta }$>, $l_{22}=$<$y_{\theta },$$y_{\theta }$>. We define the second fundamental form $II$ of M by

$$II=m_{11}d{u}^2+2m_{12}dud\theta +m_{22}d{\theta }^2,$$

(3)

where $m_{11}$ = <$y_{uu},\mathbf{\xi }$>, $m_{12}=$<$y_{u\theta },\mathbf{\xi }$>, $m_{22}=$<$y_{\theta \theta },\mathbf{\xi }$>. The Gaussian curvature K and the mean curvature H are:

$$K(u,\theta )=ϵ\frac{m_{11}{m}_{22}-m_{12}^2}{l_{11}{l}_{22}-l_{12}^2},H(u,\theta )=\frac{m_{11}{l}_{11}-2m_{12}{l}_{12}+m_{22}{l}_{22}}{2(l_{11}{l}_{22}-l_{12}^2)},$$

(4)

where $<\mathbf{\xi },\mathbf{\xi }>=ϵ(\pm 1)$.

Definition 2.

M in ${\mathbb{E}}_1^3$ is named a spacelike (timelike) surface if the induced metric on the surface is a positive (negative) definite Riemannian metric. This is equivalent to say that the normal vector on the spacelike (timelike) surface is a timelike (spacelike) vector [13,14,15].

3. Spacelike Circular Surfaces

In this section, we address the idea of spacelike circular surfaces in ${\mathbb{E}}_1^3$. Consider a non-null smooth curve $\mathbf{\alpha }\left(u\right)$, that is, $∥{\mathbf{\alpha }}^{\prime }∥\ne 0$ for all $u\in \mathbb{R}$, and a positive real number $r>0$, a spacelike circular surface is considered as the surface which is swept out by a set of Lorentzian circles with its center following the curve $\mathbf{\alpha }\left(u\right)$. All circles are lying in Lorentzian circular plane

Given a spacelike hyperbolic spherical curve ${\mathbf{e}}_1\left(u\right)\in {\mathbb{H}}_{+}^2$, then the unit tangent vector of ${\mathbf{e}}_1\left(u\right)$ is ${\mathbf{e}}_2={\mathbf{e}}_1^{\prime }\left(u\right)$. We define a vector ${\mathbf{e}}_3={\mathbf{e}}_1\times {\mathbf{e}}_2$, then we have

$$\begin{array}{c}-<{\mathbf{e}}_1,{\mathbf{e}}_1>=<{\mathbf{e}}_2,{\mathbf{e}}_2>=<{\mathbf{e}}_3,{\mathbf{e}}_3>=1,\\ <{\mathbf{e}}_1,{\mathbf{e}}_2>=<{\mathbf{e}}_2,{\mathbf{e}}_3>=<{\mathbf{e}}_3,{\mathbf{e}}_1>=0,\\ {\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3,{\mathbf{e}}_3\times {\mathbf{e}}_1={\mathbf{e}}_2,{\mathbf{e}}_2\times {\mathbf{e}}_3=-{\mathbf{e}}_1.\end{array}$$

(5)

Excluding ${\mathbf{e}}_1\left(u\right)$ is stationary or null or ${\mathbf{e}}_1^{\prime }\left(u\right)$ null. This representation gives rise to exploring the kinematic-geometry and relevant geometric characteristics. Then, we have the following Blaschke formulae [16,17,18]:

$$\left(\begin{array}{c}{\mathbf{e}}_1^{\prime }\left(u\right)\hfill \\ {\mathbf{e}}_2^{\prime }\left(u\right)\hfill \\ {\mathbf{e}}_3^{\prime }\left(u\right)\hfill \end{array}\right)=\left(\begin{array}{ccc}0\hfill & 1\hfill & 0\hfill \\ 1\hfill & 0\hfill & \gamma \left(u\right)\hfill \\ 0\hfill & -\gamma \left(u\right)\hfill & 0\hfill \end{array}\right)\left(\begin{array}{c}{\mathbf{e}}_1\left(u\right)\hfill \\ {\mathbf{e}}_2\left(u\right)\hfill \\ {\mathbf{e}}_3\left(u\right)\hfill \end{array}\right);{(}^{\prime }=\frac{d}{du}),$$

(6)

where $\gamma \left(u\right)$ is named the spherical (geodesic) curvature of the curve ${\mathbf{e}}_1\left(u\right)$ on ${\mathbb{H}}_{+}^2$. The tangent vector ${\mathbf{\alpha }}^{\prime }$ can be demonstrated by

$${\mathbf{\alpha }}^{\prime }\left(u\right)=\delta {\mathbf{e}}_1+\sigma {\mathbf{e}}_2+\eta {\mathbf{e}}_3,$$

(7)

where $\delta \left(u\right)$, $\sigma \left(u\right)$ and $\eta \left(u\right)$ are its coordinate functions. Thus, for a positive number $r>0$, and by the solutions of the differential system (6) a spacelike circular surface M can be given by:

$$M:\mathbf{y}(u,\theta )=\mathbf{\alpha }\left(u\right)+r(cos\theta {\mathbf{e}}_2\left(u\right)+sin\theta {\mathbf{e}}_3\left(u\right)),0\le \theta \le 2\pi .$$

(8)

The four functions $\gamma \left(u\right),\alpha \left(u\right)$, $\sigma \left(u\right)$ and $\eta \left(u\right)$ constitute a complete system of Lorentzian invariants of circular surfaces. This can be clarified likely from the the primary theorem of ruled surfaces [19]. The Lorentzian circlers $\theta \to \mathbf{\alpha }\left(u\right)+r(cos\theta {\mathbf{e}}_2\left(u\right)+sin\theta {\mathbf{e}}_3\left(u\right))$ are named tracing circlers. Equation (7) offers a way for creating spacelike circular surfaces with a given radius $r>0$ through the equation

$$\mathbf{\alpha }\left(s\right)={\mathbf{\alpha }}_0+\stackrel{u}{\underset{0}{\int }}(\delta {\mathbf{e}}_1+\sigma {\mathbf{e}}_2+\eta {\mathbf{e}}_3)du.$$

(9)

Excluding circular surfaces with stationary vector ${\mathbf{e}}_1$, whose geometric ownerships are of little interest. By simple calculations, one can easily obtain:

$$\begin{array}{c}{\mathbf{y}}_{\theta }=r(-sin\theta {\mathbf{e}}_2+cos\theta {\mathbf{e}}_3),\\ {\mathbf{y}}_u=(rcos\theta +\mathbf{\alpha }){\mathbf{e}}_1+\delta {\mathbf{y}}_{\theta }+\sigma {\mathbf{e}}_2+\eta {\mathbf{e}}_3.\end{array}$$

(10)

Then,

$$\begin{array}{c}{l}_{11}=-{(rcos\theta +\delta )}^2+r^2{\gamma }^2+{\sigma }^2+{\eta }^2+2r\gamma (\eta cos\theta -\sigma sin\theta ),\\ l_{12}=r(r\gamma -\sigma sin\theta +\eta cos\theta ),l_{22}=r^2,\end{array}$$

(11)

Then the timelike unit normal is:

$$\mathbf{\xi }(u,\theta )=\frac{-r(\sigma cos\theta +\eta sin\theta ){\mathbf{e}}_1-r(rcos\theta +\delta )\left(cos\theta {\mathbf{e}}_2+sin\theta {\mathbf{e}}_3\right)}{∥{\mathbf{y}}_u\times {\mathbf{y}}_{\theta }∥},$$

(12)

where

$$∥{\mathbf{y}}_u\times {\mathbf{y}}_{\theta }∥=r\sqrt{{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\delta )}^2}.$$

(13)

Furthermore, we have:

$$\begin{array}{c}{\mathbf{y}}_{\theta \theta }=-r(cos\theta {\mathbf{e}}_2+sin\theta {\mathbf{e}}_3),\\ {\mathbf{y}}_{u\theta }=-r(sin\theta {\mathbf{e}}_1+\gamma cos\theta {\mathbf{e}}_2+\gamma sin\theta {\mathbf{e}}_3),\\ {\mathbf{y}}_{uu}=({\delta }^{\prime }-\sigma ){\mathbf{e}}_1+({\sigma }^{\prime }+rcos\theta +\alpha ){\mathbf{e}}_2+({\eta }^{\prime }+\sigma \gamma ){\mathbf{e}}_3+\gamma {\mathbf{y}}_{\theta u}.\end{array}$$

(14)

This leads to

$$\begin{array}{ccc}\hfill m_{11}& =& \frac{1}{\sqrt{{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\delta )}^2}}\hfill \\ & & \times r(rcos\theta +\delta )[r\gamma -({\eta }^{\prime }+\sigma \gamma )sin\theta -(\alpha +rcos\theta +{\sigma }^{\prime })cos\theta ]\hfill \\ & & +({\delta }^{\prime }-\sigma -r\gamma sin\theta )(\eta sin\theta +\sigma cos\theta ),\hfill \\ \hfill m_{12}& =& \frac{(rcos\theta +\alpha )[r\gamma -({\eta }^{\prime }+\sigma \gamma )sin\theta -(\alpha +rcos\theta +{\sigma }^{\prime })cos\theta ]}{\sqrt{{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\delta )}^2}},\hfill \\ \hfill m_{22}& =& \frac{r(rcos\theta +\alpha )}{\sqrt{{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\delta )}^2}}.\hfill \end{array}$$

(15)

The Gaussian and mean curvatures at a regular point can be obtained, respectively, as

$$\begin{array}{ccc}\hfill K(u,\theta )& =& \frac{1}{\sqrt{{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\delta )}^2}}\{[rsin\theta (\sigma cos\theta +\eta sin\theta )]\hfill \\ & & \times [sin\theta (\sigma cos\theta +\eta sin\theta )+\gamma (rsin\theta +\delta \left)\right]\hfill \\ & & -(rsin\theta +\delta )\times [({\delta }^{\prime }+\sigma )(\sigma cos\theta +\eta sin\theta )\hfill \\ & & +(rsin\theta +\delta )(\delta sin\theta +r{sin}^2\theta \hfill \\ & & +{\sigma }^{\prime }sin\theta +{\eta }^{\prime }cos\theta +\gamma (\sigma cos\theta +\eta sin\theta )\left)\right]\},\hfill \end{array}$$

(16)

and

$$\begin{array}{ccc}\hfill H(u,\theta )& =& \frac{1}{2r{\left[{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\alpha )}^2\right]}^{3/2}}\{r(\sigma cos\theta +\eta sin\theta )\hfill \\ & & \times [{\sigma }^{\prime }+\sigma +\gamma r(cos\theta -sin\theta )+2(\sigma cos\theta +\eta sin\theta )+\gamma \sigma ]\hfill \\ & & -(rsin\theta +\delta )\left[\right(rsin\theta +\delta \left)\right(2rsin\theta +\sigma )\hfill \\ & & +r({\sigma }^{\prime }sin\theta -{\eta }^{\prime }cos\theta )+r({\sigma }^2-{\eta }^2)\}.\hfill \end{array}$$

(17)

Definition 3.

Let M be a spacelike circular surface Equation (8). Then, at $u\in I\subseteq \mathbb{R}$, the following holds:

(1) M is named a spacelike canal (tubular) surface if the spine curve is orthogonal to the Lorentzian circular plane, that is, ${\mathbf{\alpha }}^{\prime }\left(u\right)$, ${\mathbf{e}}_1\left(u\right),{\mathbf{e}}_2\left(u\right)$, and ${\mathbf{e}}_3\left(u\right)$ satisfy

$$\delta \left(u\right)=<{\mathbf{e}}_1,{\mathbf{\alpha }}^{\prime }>\ne 0,\mathrm{and}<{\mathbf{e}}_2,{\mathbf{\alpha }}^{\prime }>=<{\mathbf{e}}_3,{\mathbf{\alpha }}^{\prime }>=0\iff \sigma \left(u\right)=\eta \left(u\right)=0.$$

(18)

(2) M is named a non-canal (roller coaster) spacelike circular surface if the spine curve is tangent to the Lorentzian circular plane, that is, ${\mathbf{\alpha }}^{\prime }$, ${\mathbf{e}}_1\left(u\right),{\mathbf{e}}_2\left(u\right),$ and ${\mathbf{e}}_3\left(u\right)$ satisfy

$$\delta \left(u\right)=<{\mathbf{e}}_1,{\mathbf{\alpha }}^{\prime }>=0,\mathrm{and}\sigma \left(u\right)=<{\mathbf{e}}_2,{\mathbf{\alpha }}^{\prime }>=0\mathrm{or}\eta \left(u\right)=<{\mathbf{e}}_3,{\mathbf{\alpha }}^{\prime }>=0.$$

(19)

There are three types of develop-able ruled surfaces: cylinders, cones, and tangent surfaces. Then, the hyperbolic sphere, the spacelike canal surface, the roller coaster surface, respectively, are with symmetrical properties to the cone, the cylinder surface, and the tangent develop-able surface. Thus, it is normal to investigate circular surfaces as an identification with the ruled surfaces.

3.1. Striction Curves

In different manner than Izumiya et al. [7], we now realize the notion of striction curves of spacelike circular surfaces. Then, for the spacelike circular surface M, the curve

$$\mathbf{c}\left(u\right)=\mathbf{\alpha }\left(u\right)+r\left(cos\theta \left(u\right){\mathbf{e}}_2\left(u\right)+sin\theta \left(u\right){\mathbf{e}}_3\left(u\right)\right),$$

(20)

is the striction curve on M if $\mathbf{c}\left(u\right)$ satisfies

$$<{\mathbf{c}}^{\prime },cos\theta \left(u\right){\mathbf{e}}_2\left(u\right)+sin\theta \left(u\right){\mathbf{e}}_3\left(u\right)>=0.$$

This is equivalent to

$$\sigma \left(u\right)cos\theta \left(u\right)+\eta \left(u\right)sin\theta \left(u\right)=0.$$

(21)

From Equation (21) it follows that striction points only happen when

$$sin\theta \left(u\right)=\pm \frac{\sigma \left(u\right)}{\sqrt{{\sigma }^2+{\eta }^2}},\mathrm{and}cos\theta \left(u\right)=\mp \frac{\eta \left(u\right)}{\sqrt{{\sigma }^2+{\eta }^2}}.$$

(22)

From Equations (8) and (22) it follows that the striction curves are:

$$\left.\begin{array}{c}{\mathbf{c}}_1\left(u\right)=\mathbf{\alpha }\left(u\right)+\frac{r}{\sqrt{{\sigma }^2+{\eta }^2}}\left(\eta \left(u\right){\mathbf{e}}_2\left(u\right)-\sigma \left(u\right){\mathbf{e}}_3\left(u\right)\right),\\ \\ {\mathbf{c}}_2\left(u\right)=\mathbf{\alpha }\left(u\right)-\frac{r}{\sqrt{{\sigma }^2+{\eta }^2}}\left(\eta \left(u\right){\mathbf{e}}_2\left(u\right)-\sigma \left(u\right){\mathbf{e}}_3\left(u\right)\right).\end{array}\right\}$$

(23)

From the above analysis, any non-canal (roller coaster) spacelike circular surface has two striction curves. Subsequently, via Equations (18) and (23), all curves on the spacelike canal surface transverse to the tracing circles guarantee the status of the striction curves; that is, $\mathbf{c}\left(u\right)=\mathbf{\alpha }\left(u\right)$. Thus, the spacelike canal surfaces are a comparable sub-set of hyperbolic cylindrical surfaces.

3.2. Curvature Lines and Local Singularities

Curvature lines and singularities are main possessions of circular surfaces and are analyzed in the following.

3.2.1. Curvature Lines

Since the curvature lines are geometric properties of surfaces, it is motivating to know in what conditions the tracing circles are curvature lines. A sufficient and necessary condition for a regular curve $\mathbf{\beta }\left(t\right)$ on a surface to be a curvature line is that ${\mathbf{\zeta }}_t\left(t\right)\Vert {\mathbf{\beta }}_t\left(t\right)$ [19]. From Equations (11) and (15) it can be seen that the $\theta$, and u-curves are not curvature lines. Therefore, after several algebraic calculations, it can be gained that the tracing circles are curvature lines if and only if

$${\mathbf{\zeta }}_{\theta }\Vert {\mathbf{y}}_{\theta }\iff \left(rcos\theta +\delta \right)(r\eta +\delta \eta cos\theta -\alpha \sigma sin\theta )=0,\forall \theta \in \mathbb{R}.$$

(24)

Thus, we consider the following cases:

Case (a)- If $\eta \left(u\right)=\sigma \left(u\right)=\delta \left(u\right)=0$, the tangent vector ${\mathbf{\alpha }}^{\prime }=\mathbf{0}$, that is, the spine curve is a stationary point. This means that M is a hyperbolic sphere with radius r, that is,

$$M=\{\mathbf{y}\in {\mathbb{E}}_1^3\mid {∥\mathbf{y}-{\mathbf{\alpha }}_0∥}^2=r^2\}.$$

Case (b)- If $\sigma \left(u\right)=\eta \left(u\right)=0$, the spine curve is orthogonal to the Lorentzian circular plane, that is, ${\mathbf{\alpha }}^{\prime }\perp {\mathbf{e}}_1$. This means that ${\mathbf{e}}_1$ orthogonal to the normal plane at each point of the timelike spine curve $\mathbf{\alpha }$. In this case, M is a spacelike canal surface with a timelike spine curve.

Case (c)- If $\delta \left(u\right)=\eta \left(u\right)=0$, the tangent vector ${\mathbf{\alpha }}^{\prime }\left(u\right)=\sigma \left(u\right){\mathbf{e}}_2\left(u\right)$, that is, the spacelike tangent vector of the spine curve lies in the the normal plane at each point of the spine curve. In this case, M is a spacelike tangent circular (roller coaster) surface with a spacelike spine curve. In the special case, if $\sigma$ is stationary, it follows that:

$$\mathbf{\alpha }\left(u\right)={\mathbf{\alpha }}_0+{\sigma }_0{\mathbf{e}}_1\left(u\right),$$

(25)

where ${\mathbf{\alpha }}_0$ is a stationary vector. From Equations (8) and (25) it can be found that

$${∥\mathbf{y}-{\mathbf{\alpha }}_0∥}^2=-{\sigma }_0^2+r^2.$$

(26)

This means that all the Lorentzian circles lie on a hyperbolic sphere of radius $\sqrt{-{\sigma }_0^2+r^2}$ < r, with ${\mathbf{\alpha }}_0$ being its center point in ${\mathbb{E}}_1^3$. Hence, the following categorizations can be given:

Theorem 1.

In addition to the spacelike circular surfaces, there are two sets of non-spacelike circular surfaces whose tracing Lorentzian circles are curvature lines. These two sets are the spacelike roller coaster surfaces and the hyperbolic spheres with the radius being less than that of the tracing Lorentzian circles.

A thorough treatment on spacelike tangent circular (roller coaster) surfaces will be given in the last.

3.2.2. Local Singularities

Now, we search singularities of spacelike circular surfaces. It is interesting to note that singular points happen at the striction curve. From Equation (13), we can show that M has a singular point at ($u_0$, ${\theta }_0$) if and only if

$$∥y_u\times y_{\theta }∥=r\sqrt{{(\sigma cos\theta +\eta sin\theta )}^2-{(rcos\theta +\delta )}^2}=0$$

which yields two (linearly dependent) equations:

$$rcos{\theta }_0+\delta \left(u_0\right)=0,\mathrm{and}\sigma \left(u_0\right)cos{\theta }_0+\eta \left(u_0\right)sin{\theta }_0=0.$$

(27)

Thus, we gain the following:

Case (A). When $rcos{\theta }_0+\delta \left(u_0\right)=0$. For M to have singular point it is necessary that $\sigma \left(u_0\right)cos{\theta }_0+\eta \left(u_0\right)sin{\theta }_0=0$. Thus, we consider the following sub-cases:

(a) If $\delta \left(u_0\right)\ne 0$, and $\eta \left(u_0\right)\ne 0$, then the singular points are located at $sin{\theta }_0=-\left(\frac{\delta \left(u_0\right)\sigma \left(u_0\right)}{r\eta \left(u_0\right)}\right)$. Since ${cos}^2{\theta }_0+{sin}^2{\theta }_0=1$, we find ${\delta }^2\left(u_0\right)({\sigma }^2\left(u_0\right)+{\eta }^2\left(u_0\right))=r^2{\eta }^2\left(u_0\right)$. Further, one can easily see that

$${\theta }_0=-{tan}^{-1}\left(\frac{\sigma \left(u_0\right)}{\eta \left(u_0\right)}\right).$$

Then, the singular points occur at

$$r=\frac{\delta \left(u_0\right)}{\eta \left(u_0\right)}\sqrt{{\sigma }^2\left(u_0\right)+{\eta }^2\left(u_0\right)}>0,\mathrm{and}{\theta }_0=-{tan}^{-1}\left(\frac{\sigma \left(u_0\right)}{\eta \left(u_0\right)}\right).$$

(28)

(b) If $\eta \left(u_0\right)=0$, and $\sigma \left(u_0\right)=0$, the singular points occur at ${\theta }_0={cos}^{-1}\left(\frac{\delta \left(u_0\right)}{r}\right)$.

Case (B). When $\sigma \left(u_0\right)cos{\theta }_0+\eta \left(u_0\right)sin{\theta }_0=0$. For M to have singular points it is necessary that $rcos{\theta }_0+\delta \left(u_0\right)=0$. If $\delta \left(u_0\right)=0$, there are two singular points on every tracing circle existing at ${\theta }_0=\frac{\pi }{2}$, and ${\theta }_0=\frac{3\pi }{2}$. Gathering these two sets of singular points gives two striction curves that contains all the singular points. From Equation (23) it follows that the two striction curves are

$${\mathbf{c}}_1\left(u\right)=\mathbf{\alpha }\left(u\right)-r{\mathbf{e}}_3\left(u\right),\mathrm{and}{\mathbf{c}}_2\left(u\right)=\mathbf{\alpha }\left(u\right)+r{\mathbf{e}}_3\left(u\right).$$

If $\left|\delta \right|>r$, there are no singular points. If $\delta =\mp r$, the singular point existing at ${\theta }_0=\frac{\pi }{2}$ or ${\theta }_0=\frac{3\pi }{2}$. If $\left|\delta \right|<r$, there are singular points on the tracing circle, existing at ${\theta }_0=\pm {cos}^{-1}\delta \left(u_0\right)/r$.

The above discussions are demonstrated by the following example:

Example 1.

Let ${\mathbf{e}}_1\left(u\right)=(sinhu,0,coshu)$ which is a timelike vector. It is easy to show that ${\mathbf{e}}_2\left(u\right)=(coshu,0,sinhu)$, and ${\mathbf{e}}_3\left(u\right)=(0,1,0)$.

(i) Let $\sigma \left(u\right)=\eta \left(u\right)=1$, and $\delta \left(u\right)=1/\sqrt{2}$. Then,

$${\mathbf{\alpha }}^{\prime }\left(u\right)=(\frac{1}{\sqrt{2}}sinhu+coshu,1,\frac{1}{\sqrt{2}}coshu+sinhu).$$

Taking integral with zero integration constants yields

$$\mathbf{\alpha }\left(u\right)=(\frac{1}{\sqrt{2}}coshu+sinhu,u,\frac{1}{\sqrt{2}}sinhu+coshu).$$

One can easily see that $\mathbf{\alpha }\left(u\right)$ has no singular points (Figure 1). Then, according to the conditions of Equation (28), we have

$$r=1,\mathrm{and}{\theta }_0=\frac{3\pi }{4}.$$

Consequently, in view of Equations (23), the striction curves are:

$$\left.\begin{array}{c}{\mathbf{c}}_1\left(u\right)=(\sqrt{2}coshu+sinhu,u-\frac{1}{\sqrt{2}},\sqrt{2}sinhu+coshu),\\ \\ {\mathbf{c}}_2\left(u\right)=(sinhu,u+1,coshu).\end{array}\right\}$$

The spacelike circular surface M with its spine curve $\mathbf{\alpha }\left(u\right)$ is

$$M:\mathbf{y}(u,\theta )=(\frac{1}{\sqrt{2}}(cos\theta +1)coshu+sinhu,u+sin\theta ,\frac{1}{\sqrt{2}}(cos\theta +1)sinu+coshu),$$

where $-0.5\le u\le 0.5$, and $0\le \theta \le 2\pi$ (Figure 2).

(ii) Let $\eta \left(u\right)=u$, $\sigma \left(u\right)=1$, and $\delta \left(u\right)=0$. Then ${\mathbf{\alpha }}^{\prime }\left(u\right)={\mathbf{e}}_2\left(u\right)+u{\mathbf{e}}_3\left(u\right)$. Similarly, we derive

$$\mathbf{\alpha }\left(u\right)=(sinhu,\frac{u^2}{2},coshu),$$

which probably has no singular points (Figure 3). Then, the striction curves are:

$$\left.\begin{array}{c}{\mathbf{c}}_1\left(u\right)=(sinhu,\frac{u^2}{2}-r,coshu),\\ \\ {\mathbf{c}}_1\left(u\right)=(sinhu,\frac{u^2}{2}+r,coshu).\end{array}\right\}$$

The spacelike circular surface with its spine curve $\mathbf{\alpha }\left(u\right)$ is:

$$M:\mathbf{y}(u,\theta )=(sinhu+rcos\theta coshu,\frac{u^2}{2}+rsin\theta ,coshu+rcos\theta sinhu),$$

Singularities appears on the striction curves (blue); where $r=1$, −0.7 $\le u\le 0.7$, and $0\le \theta \le 2\pi$ (Figure 4).

(iii) If $\delta \left(u\right)=u$, and $\sigma \left(u\right)=\eta \left(u\right)=0$. Then, we have ${\mathbf{\alpha }}^{\prime }\left(u_0\right)=u{\mathbf{e}}_1\left(u\right)$ which is a timelike. Similarly, we have

$$\mathbf{\alpha }\left(u\right)=(ucoshu-sinhu,0,usinhu-coshu),$$

which has has a cusp point at $u_0=0$ (Figure 5). The spacelike canal circular surface with the spine curve $\mathbf{\alpha }\left(u\right)$ is

$$M:\mathbf{y}(u,\theta )=(ucoshu-(1-rcos\theta )sinhu,rsin\theta ,usinhu-(1-rcos\theta )coshu).$$

For $r=1$, $0\le \theta \le 2\pi$, and $-0.7\le u\le 0.7$ the surface is illustrated in (Figure 6).

3.3. Spacelike Roller Coaster Surfaces

As stated earlier, a spacelike roller coaster surface is a surface traced by a one-parameter set of curvature lines described as ${\mathbf{\alpha }}^{\prime }\left(u\right)$ lies in the tracing circle plane at each point. This means that $\delta \left(u\right)=0,$ and $\sigma \left(u\right)=0$ or $\eta \left(u\right)=0$. Through this work, we will only occupy that $\delta \left(u\right)=\eta \left(u\right)=0$, and $\sigma \left(u\right)\ne 0$. If so, such a surface is named a spacelike roller coaster surface with a spacelike spine curve, that is, ${\mathbf{\alpha }}^{\prime }\left(u\right)=\sigma \left(u\right){\mathbf{e}}_2\left(u\right)$. By putting $\delta \left(u\right)=\eta \left(u\right)=0$ into Equation (13), we gain

$$∥y_v\times y_{\theta }∥=rcos\theta \sqrt{{\sigma }^2-r^2}.$$

(29)

From Equation (29) it follows that singularities only existing when $\theta =\pm \frac{\pi }{2}$, and $\left|\sigma \right|>r$. Then, from Equation (23), there are two striction curves given by

$${\mathbf{c}}_1\left(u\right)=\mathbf{\alpha }\left(u\right)+r{\mathbf{e}}_3\left(u\right),{\mathbf{c}}_2\left(u\right)=\mathbf{\alpha }\left(u\right)-r{\mathbf{e}}_3\left(u\right).$$

(30)

Proposition 1.

Any spacelike roller coaster surface has precisely two striction curves and intersections with any tracing circle are antipodal points each other.

By setting $\delta \left(u\right)=\eta \left(u\right)=0$ into Equations (16) and (17), we find

$$K(u,\theta )=\frac{\left({\sigma }^2-r^2\right)cos\theta +r{\sigma }^{\prime }}{{\left({\sigma }^2-r^2\right)}^{2}cos\theta },H(u,\theta )=\frac{\left({\sigma }^2-r^2\right)(1+r^2)cos\theta +r^2{\sigma }^{\prime }}{2r{\left({\sigma }^2-r^2\right)}^{3/2}cos\theta }.$$

(31)

Remarkable fact that both the Gaussian and mean curvatures does not based on the spherical curvature of ${\mathbf{e}}_1\left(u\right)\in {\mathbb{H}}_{+}^2$ but only on $\sigma$ and $\theta$. As a result, the following theorem can be given:

Theorem 2.

If a family of spacelike roller coaster surfaces has the same radius r, scalar function $\sigma \left(u\right)$ and its derivative ${\sigma }^{\prime }\left(u\right)$, the Gaussian and the mean curvatures at matching points are the same. In addition, these values are independent of the spherical curvature of ${\mathbf{e}}_1\left(u\right)\in {\mathbb{H}}_{+}^2$.

Since every tracing circle is a curvature line on the spacelike roller coaster surface, the value of one principal curvature is

$${\chi }_1(u,\theta ):=∥{\mathbf{y}}_{\theta }\times {\mathbf{y}}_{\theta \theta }∥{∥{\mathbf{y}}_{\theta }∥}^{-3}=\frac{1}{r}.$$

The other principal curvature is

$${\chi }_2(u,\theta )=\frac{K(u,\theta )}{{\chi }_1}=\frac{\left({\sigma }^2-r^2\right)cos\theta +r{\sigma }^{\prime }}{{\left({\sigma }^2-r^2\right)}^{2}cos\theta }.$$

Hence, we state the following:

Corollary 1.

The principal curvature of a spacelike roller coaster surface is stationary along each tracing circle.

Furthermore, to give parametric equation of the spacelike roller coaster surface, it is necessary to build the Serret–Frenet of the spine curve $\mathbf{\alpha }\left(v\right)$. Then, let v be the arc-length of the spacelike spine curve $\mathbf{\alpha }\left(u\right)$ and $\sigma \left(u\right)>0$, ∀$u\in \mathbb{R}$; where the Serret–Frenet frame of the spine curve can be written as

$$\begin{array}{ccc}\hfill \mathbf{t}\left(v\right)& =& {\mathbf{\alpha }}^{\prime }{∥{\mathbf{\alpha }}^{\prime }∥}^{-1}={\mathbf{e}}_2,\mathbf{n}\left(v\right)=\frac{d\mathbf{t}}{dv}{∥\frac{d\mathbf{t}}{dv}∥}^{-1}=\frac{{\mathbf{e}}_1+\gamma {\mathbf{e}}_3}{\sqrt{{\gamma }^2-1}},\hfill \\ \hfill \mathbf{b}\left(v\right)& =& \frac{\gamma {\mathbf{e}}_1+{\mathbf{e}}_3}{\sqrt{{\gamma }^2-1}},\mathrm{with}\left|\gamma \right|>1.\hfill \end{array}$$

If $cosh\phi \left(v\right)=\frac{\gamma }{\sqrt{{\gamma }^2-1}},sinh\phi \left(v\right)=\frac{1}{\sqrt{{\gamma }^2-1}}$, it follows that:

$$\left(\begin{array}{c}\mathbf{t}\hfill \\ \mathbf{n}\hfill \\ \mathbf{b}\hfill \end{array}\right)=\left(\begin{array}{ccc}0\hfill & 1\hfill & 0\hfill \\ -cosh\phi \hfill & 0\hfill & sinh\phi \hfill \\ -sinh\phi \hfill & 0\hfill & cosh\phi \hfill \end{array}\right)\left(\begin{array}{c}{\mathbf{e}}_1\hfill \\ {\mathbf{e}}_2\hfill \\ {\mathbf{e}}_3\hfill \end{array}\right),$$

(32)

where

$$\mathbf{t}\left(v\right)\times \mathbf{n}\left(v\right)=-\mathbf{b}\left(v\right),\mathbf{b}\left(v\right)\times \mathbf{t}\left(v\right)=\mathbf{n}\left(v\right),\mathbf{n}\left(v\right)\times \mathbf{b}\left(v\right)=\mathbf{t}\left(v\right).$$

(33)

Then, the Serret–Frenet equations are:

$$\frac{d}{dv}\left(\begin{array}{c}\mathbf{t}\hfill \\ \mathbf{n}\hfill \\ \mathbf{b}\hfill \end{array}\right)=\left(\begin{array}{ccc}0\hfill & \kappa \left(v\right)\hfill & 0\hfill \\ -\kappa \left(v\right)\hfill & 0\hfill & \tau \left(v\right)\hfill \\ 0\hfill & \tau \left(v\right)\hfill & 0\hfill \end{array}\right)\left(\begin{array}{c}\mathbf{t}\hfill \\ \mathbf{n}\hfill \\ \mathbf{b}\hfill \end{array}\right),$$

(34)

where

$$\kappa \left(v\right)=\frac{\sqrt{{\gamma }^2-1}}{\sigma },\tau \left(v\right)+\frac{d\phi }{dv}=0,\frac{d\phi }{dv}=\frac{d\gamma /du}{\sigma ({\gamma }^2-1)}.$$

(35)

It follows from Equation (35) that if $\gamma$ = const., the torsion $\tau \left(v\right)=0$, that is, $\mathbf{\alpha }\left(u\right)$ is a spacelike planar curvature line. In terms of $\left\{\mathbf{t}\right(v),$$\mathbf{n}\left(v\right)$, $\mathbf{b}\left(v\right)\},$ the striction curves are

$$\left.\begin{array}{c}{\mathbf{c}}_1\left(v\right)=\mathbf{\alpha }\left(v\right)+r(-sinh\phi \mathbf{n}+cosh\phi \mathbf{b}),\\ \\ {\mathbf{c}}_2\left(v\right)=\mathbf{\alpha }\left(v\right)-r(-sinh\phi \mathbf{n}+cosh\phi \mathbf{b}).\end{array}\right\}$$

(36)

The curvatures ${\kappa }_i$ and torsions ${\tau }_i$ (i=1, 2) of the striction curves can be gained as

$$\left.\begin{array}{c}{\kappa }_1\left(v\right)=\frac{\kappa }{\sqrt{\left|1-r\kappa cosh\phi \right|}},{\tau }_1\left(v\right)=\frac{\tau }{\sqrt{\left|1-r\kappa cosh\phi \right|}},\\ {\kappa }_2\left(v\right)=\frac{\kappa }{\sqrt{1+r\kappa cosh\phi }},{\tau }_2\left(v\right)=\frac{\tau }{\sqrt{1+r\kappa cosh\phi }}.\end{array}\right\}$$

(37)

Then, if the spine curve $\mathbf{\alpha }\left(v\right)$ is a spacelike plane curve (a spacelike general helix), then the striction curves ${\mathbf{c}}_{1,2}\left(v\right)$ are spacelike plane curve (spacelike general helix). In addition, from Equations (32) and (8) we have:

$$M:\mathbf{y}(v,\theta )=\mathbf{\alpha }\left(v\right)+rcos\theta \mathbf{t}\left(s\right)+rsin\theta (-sinh\phi \mathbf{n}+cosh\phi \mathbf{b}).$$

(38)

Equation (38) not only give the existence of the spacelike roller coaster, but also gives the specified expression of the surface. This is very significant in workable application.

Flat and Minimal Roller Spacelike Coaster Surfaces

A surface with vanishing Gaussian curvature is named as a flat surface. It is readily seen that for M to be a flat surface is

$$K(u,\theta )=0\iff \left({\sigma }^2-r^2\right)cos\theta +r{\sigma }^{\prime }=0.$$

Hence, for all $\theta \in I\subseteq \mathbb{R}$, we have

$$\frac{{\partial }^{2}K(u,\theta )}{{\partial }^2\theta }+K(u,\theta )=0\iff {\sigma }^{\prime }\left(u\right)=0.$$

(39)

From Equations (35) and (39) the innuendo of ${\sigma }^{\prime }\left(u\right)$ in terms of the Serret–Frenet’s invariants is

$${\sigma }^{\prime }\left(u\right)=0\iff \kappa \left(v\right)\tau \left(v\right)cos\phi \left(v\right)+\frac{d\kappa \left(v\right)}{dv}sin\phi \left(v\right)=0.$$

Thus in a neighborhood of all point on M with $\kappa \left(v\right)\ne 0$, we find that $\frac{d\kappa \left(v\right)}{dv}=\tau \left(v\right)=0$. Therefore, a spacelike roller coaster surface with vanishing Gaussian curvature is a part of a spacelike plane. Likwise, we find that M is a minimal flat spacelike surface. Hence, we state the following:

Corollary 2.

All the minimal flat spacelike roller coaster surfaces are sub-sets of spacelike planes.

Example 2.

Given the unit-speed spacelike helix

$$\mathbf{\alpha }\left(v\right)=(bv,acosv,asinv),$$

where $a>0$, $b\ne 0$, and $b^2-a^2=1$. With ordinary calculation, we have:

$$\left.\begin{array}{c}\mathbf{t}\left(v\right)=(b,-asinv,acosv),\\ \mathbf{n}\left(v\right)=(0,-cosv,-sinv),\\ \mathbf{b}\left(v\right)=(-a,bsinv,-bcosv),\\ \kappa \left(v\right)=a,\mathrm{and}\tau \left(v\right)=-b.\end{array}\right\}$$

Taking ${\phi }_0=0$ we have $\phi \left(v\right)=bv$. Thus, the spacelike roller coaster surface family is:

$$\begin{array}{ccc}\hfill \mathbf{y}(v,\theta )& =& (bv,acosv,asinv)+r(cos\theta ,-sin\theta sinhbv,sin\theta coshbv)\times \hfill \\ & & \left(\begin{array}{ccc}b& -asinv& acosv\\ 0& -cosv& -sinv\\ -a& bsinv& -bcosv\end{array}\right).\hfill \end{array}$$

For $a=1$ and $r=1$, $-0.5\pi \le v\le 0.5\pi$, and $0\le \theta \le 2\pi ,$ the spacelike roller coaster surface with the spine curve $\mathbf{\alpha }\left(v\right)$ (green) is shown in (Figure 7). Singularities appear on the striction curves (blue). It is clear that $\mathbf{\alpha }\left(v\right)$ has no singular point, see (Figure 8).

4. Conclusions

According to the unit normal vector of the Lorentzian circular plane and the arc length of its hyperbolic spherical image curve, the parametrization of spacelike circular surface is attained. Then, significant matching properties of spacelike circular surfaces with ruled surfaces are researched. In addition, singularities of spacelike circular surfaces are examined and some properties are implied to determine whether a spacelike surface is a spacelike canal or non-canal surface. Then, we offered spacelike roller coaster surfaces as a special class of spacelike circular surfaces. Subsequently, the conditions desired for spacelike roller coaster surfaces to be flat or minimal surfaces are gained. Meanwhile, we support the results of this approach with some examples.