It is known that to each unit speed spacelike curve *x* = *x*(*s*): *I* → *Q*^{3} ⊂
$\begin{array}{}{E}_{1}^{4}\end{array}$, one can associate a pseudo orthonormal frame {*x*,*α*,*y*,*β*}. In this condition, the Frenet Frame for unit speed spacelike curve *x* = *x*(*s*): *I* → *Q*^{3} ⊂
$\begin{array}{}{E}_{1}^{4}\end{array}$
are given as
$$\begin{array}{}{\mathbf{x}}^{\mathrm{\prime}}=\alpha \\ {\alpha}^{\mathrm{\prime}}=\kappa \mathbf{x}-\mathbf{y}\\ {\beta}^{\mathrm{\prime}}=\tau \mathbf{x}\\ {\mathbf{y}}^{\mathrm{\prime}}=-\kappa \alpha -\tau \beta .\end{array}$$(4.1)

Let *α* = *x*′ and choose *β* such that
$$\begin{array}{}det\phantom{\rule{thinmathspace}{0ex}}(\mathbf{x},\alpha ,\beta ,\phantom{\rule{thinmathspace}{0ex}}\mathbf{y}\phantom{\rule{thinmathspace}{0ex}})=1.\end{array}$$(4.2)

From [6], for any asymptotic orthonormal frame {**x**,*α*, **y**,*β*} of the unit speed spacelike curve *x* = *x*(*s*): *I* → *Q*^{3} ⊂
$\begin{array}{}{E}_{1}^{4}\end{array}$
with
$$\begin{array}{}<\mathbf{x}\mathbf{,}\phantom{\rule{thinmathspace}{0ex}}\alpha >=<\mathbf{x}\mathbf{,}\phantom{\rule{thinmathspace}{0ex}}\beta >=<\mathbf{y}\mathbf{,}\alpha >=<\mathbf{y}\mathbf{,}\phantom{\rule{thinmathspace}{0ex}}\beta >=<\alpha \mathbf{,}\phantom{\rule{thinmathspace}{0ex}}\beta >=\phantom{\rule{thinmathspace}{0ex}}0,\\ \\ \phantom{\rule{thinmathspace}{0ex}}<\alpha \mathbf{,}\alpha >=<\beta \mathbf{,}\phantom{\rule{thinmathspace}{0ex}}\beta >=\u3008\mathbf{x}\mathbf{,}\mathbf{y}\u3009\mathbf{=}\mathbf{1}.\end{array}$$(4.3)

The frame field {**x**,*α*,**y**,*β*}} is the cone frenet frame of the curve *x*(*s*).

#### Definition 4.1

Let the functions *κ* and *τ* in (4.1) be the (first) cone curvature and cone torsion of the spacelike curve *x* in *Q*^{3} ⊂
$\begin{array}{}{E}_{1}^{4}\end{array}$.

By definition, for a spacelike normal curve in the 3-dimensional lightlike cone, the position vector *x* satisfies
$$\begin{array}{}x(s)=\lambda (s)x(s)+\mu (s)\alpha (s)+\gamma (s)\beta (s)\end{array}$$(4.4)
for some differentiable functions *λ*, *μ* and *β*.

#### Theorem 4.1

Denote *x* be a unit speed spacelike normal curve with curvatures *κ*,*τ* ≠ 0 in *Q*^{3}. Then the coefficients of curve *x* is as follows:
$$\begin{array}{}\gamma =c\\ \mu =0\\ \lambda =1,\end{array}$$
or
$$\begin{array}{}\lambda =1-{c}_{1}\sqrt{K}{e}^{\sqrt{K}s}+{c}_{2}\sqrt{K}{e}^{-\sqrt{K}s}\\ \mu ={c}_{1}{e}^{\sqrt{K}s}+{c}_{2}{e}^{-\sqrt{K}s}-\frac{c\tau}{\kappa}\\ \gamma =c\end{array}$$(4.5)
or
$$\begin{array}{}\lambda =1-a\sqrt{K}\mathrm{sinh}(\sqrt{K}s)+b\sqrt{K}\mathrm{cosh}(\sqrt{K}s)\\ \mu =a\mathrm{cosh}(\sqrt{K}s)+b\mathrm{sinh}(\sqrt{K}s)-\frac{c\tau}{\kappa}\\ \gamma =c\end{array}$$(4.6)
where
$\begin{array}{}{c}_{1}=\frac{a+b}{2},{c}_{2}=\frac{a-b}{2},\end{array}$
*a*,*b*,*c*,*c*_{1},*c*_{2} *ϵ* *IR*.

#### Proof

Let *x* be a unit speed spacelike normal curve in *Q*^{3} and *κ*,*τ* be cone curvature functions of *x*. Differentiating (4.4) with respect to *s* and using (4.1), we have
$$\begin{array}{}\alpha =\left({\lambda}^{\mathrm{\prime}}+\mu \kappa +\gamma \tau \right)x-\mu y+\left(\lambda +{\mu}^{\mathrm{\prime}}\right)\alpha +{\gamma}^{\mathrm{\prime}}\beta .\end{array}$$(4.7)

Exposing the inner product *y*,*x*,*α*,*β* to the both sides of (4.7), respectively, we find
$$\begin{array}{}{\lambda}^{\mathrm{\prime}}+\mu \kappa +\gamma \tau =0\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\mu =0\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\lambda +{\mu}^{\mathrm{\prime}}=1\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}{\gamma}^{\mathrm{\prime}}=0.\end{array}$$(4.8)

Making the necessary mathematical operations, we get
$$\begin{array}{}\gamma =c\\ \mu =0\\ \lambda =1,\end{array}$$(4.9)
where *c* ∈ *IR*.

Again making necessary calculations in (4.8), we can write the following differential equation:
$$\begin{array}{}{\mu}^{\mathrm{\prime}\mathrm{\prime}}-\mu \kappa =c\tau .\end{array}$$(4.10)

Solving (4.10), we obtain (4.5) and considering the definition of hyperbolic functions sinh *s* and cosh *s*, we have (4.6). Thus the proof is completed. □

#### Theorem 4.2

Denote *x* be a unit speed spacelike normal curve in *Q*^{3}. If the following statements are provided, *x* is a normal curve so that the curvature of curve *x* is *κ*, *τ* ≠ 0.

the distance function *d* = *constant*.

the components of vector fields *x*, *α* and *β* of curve are respectively given by
$$\begin{array}{}\u3008x(s),y(s)\u3009=\lambda (s)=1-{c}_{{}_{1}}\sqrt{K}{e}^{\sqrt{K}s}+{c}_{2}\sqrt{K}{e}^{-\sqrt{K}s}\\ \u3008x(s),\alpha (s)\u3009=\mu (s)={c}_{1}{e}^{\sqrt{K}s}+{c}_{2}{e}^{-\sqrt{K}s}-\frac{c\tau}{\kappa}\\ \u3008x(s),\beta (s)\u3009=\gamma (s)=c.\end{array}$$(4.11)

the normal component *x*^{N} of *x* is constant.

#### Proof

Suppose that *x* is a unit speed spacelike normal curve in *Q*^{3} and *x* is a normal curve such that its curvature is *κ*,
*τ* ≠ 0. The position vector, *x*, satisfies (4.4). If *λ* and *μ* hold (4.5), then we have
$$\begin{array}{}{d}^{2}(s)={\u2225x\left(s\right)\u2225}^{2}={\mu}^{2}+{\gamma}^{2}.\end{array}$$

From (4.8), since *μ* = 0, *γ* = *c*, *c* ∈ ℝ_{0} then *d* = *c* = *cons*.

Furthermore, multiplying with *μ*′ and *γ* both sides of the second and fourth equations of (4.8) respectively, we get
$$\begin{array}{}{\mu}^{2}={m}^{2}\\ {\gamma}^{2}={n}^{2},\end{array}$$(4.12)
where *m* ∈ ℝ and *n* ∈ ℝ_{0}. From (4.12), we find
$$\begin{array}{}{d}^{2}(x(s))={\u2225x\left(s\right)\u2225}^{2}={\mu}^{2}+{\gamma}^{2}={m}^{2}+{n}^{2},\end{array}$$(4.13)
which means that *d*^{2} = *m*^{2} + *n*^{2} = *cons*. Thus (1) is proved.

Conversely, suppose that (1) holds. Then be *d* is constant. Differentiating the second time *d*^{2}(*x*(*s*)) = ∥*x*(*s*)∥^{2} = *constant* with respect to *s*, we obtain *κ*〈*x*(*s*),*x*(*s*)〉 = 0. Since *κ* ≠ 0, 〈*x*(*s*),*x*(*s*)〉 = 0. This means that *x* is a normal curve.

For the proof of (2), from (4.6) and (4.4), respectively, we have (4.11).

Conversely, differentiating the second time 〈*x*(*s*),*y*(*s*)〉 = *λ*(*s*) and using (4.1), we find (*κ*^{2}+*τ*^{2})〈*x*(*s*),*x*(*s*)〉 = 0. For *κ*,*τ* ≠ 0, since *κ*^{2} + *τ*^{2} ≠ 0, 〈*x*(*s*),*x*(*s*)
〉 = 0. This means that *x* is a normal curve.

Again differentiating 〈*x*(*s*),*β*(*s*)〉 = *γ*(*s*), we have 〈*x*(*s*),*x*(*s*)〉 = 0. Hence *x* is a normal curve. Thus (2) is proved.

For the proof of (3), the normal component *x*^{N} of curve *x* is *μ**α* + *γ**β*. Hence
$\begin{array}{}\parallel {x}^{N}{\parallel}^{2}\end{array}$
= *μ*^{2} + *γ*^{2}. From (4.13), we have *μ*^{2} + *γ*^{2} = *m*^{2} + *n*^{2}. Since *m* ∈ ℝ and *n* ∈ ℝ _{0},
$\begin{array}{}\parallel {x}^{N}{\parallel}^{2}\end{array}$
= *m*^{2}+*n*^{2} = *constant*. Thus (3) is proved.

Conversely, suppose that (3) holds. From (4.4), we can give
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\left(s\right)=\lambda \left(s\right)x\left(s\right)+{x}^{N}\left(s\right)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\left(s\right)=\lambda \left(s\right)x\left(s\right)+\mu \left(s\right)\alpha \left(s\right)+\gamma \left(s\right)\beta \left(s\right)\\ \u3008x\left(s\right),x\left(s\right)\u3009\left(1-\lambda \left(s\right)\right)=\mu \left(s\right)\u3008x\left(s\right),\alpha \left(s\right)\u3009+\gamma \left(s\right)\u3008x\left(s\right),\beta \left(s\right)\u3009\\ \u3008x\left(s\right),x\left(s\right)\u3009\left(1-\lambda \left(s\right)\right)=\u3008{x}^{N}\left(s\right),x\left(s\right)\u3009.\end{array}$$

Using (4.9), we get
$$\begin{array}{}0=0+c\u3008x\left(s\right),\beta \left(s\right)\u3009.\end{array}$$(4.14)

Differentiating (4.14), we have 0 = *c**τ*〈*x*(*s*),*x*(*s*)〉. Since *c**τ* ≠ 0, 〈*x*(*s*),*x*(*s*)〉 = 0. This means that *x* is a normal curve. Thus the theorem is proved. □

#### Theorem 4.3

Denote *x* be a unit speed spacelike normal curve in *Q*^{3}. Then *x* lies on lightlike cone *Q*^{3}(*m*) with vertex at *m* if and only if *x* is congruous to a normal curve by
$$\begin{array}{}\left(\lambda \left(s\right)-1\right)x\left(s\right)+\mu \left(s\right)\alpha \left(s\right)+\gamma \left(s\right)\beta \left(s\right)=0.\end{array}$$(4.15)

#### Proof

From (4.4), let’s write
$$\begin{array}{}m=x\left(s\right)-\lambda \left(s\right)x\left(s\right)-\mu \left(s\right)\alpha \left(s\right)-\gamma \left(s\right)\beta \left(s\right),\end{array}$$(4.16)
differentiating (4.16), we obtain
$$\begin{array}{}{m}^{\mathrm{\prime}}=\left(1-\lambda \left(s\right)-{\mu}^{\mathrm{\prime}}\left(s\right)\right)\alpha \left(s\right)-{\gamma}^{\mathrm{\prime}}\left(s\right)\beta \left(s\right)\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{1em}{0ex}}-\left({\lambda}^{\mathrm{\prime}}\left(s\right)+\kappa \left(s\right)\mu \left(s\right)+\tau \left(s\right)\gamma \left(s\right)\right)x\left(s\right)-\mu \left(s\right)y\left(s\right),\end{array}$$
from (4.8), *m*′ = 0. Thus *m* = *constant*, which means that *x* is congruous to a normal curve.

Further
$$\begin{array}{}{\u3008x\left(s\right)-m,x\left(s\right)-m\u3009}^{\mathrm{\prime}}=0\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u3008x\left(s\right)-m,\alpha \left(s\right)\u3009=0.\end{array}$$(4.17)

Hence *x* lies precisely on a lightlike cone *Q*^{3} and since 〈*x*(*s*) − *m*,*α*(*s*)〉 = 0, *x*(*s*) − *m* ∈ *span*{*α*} ^{⊥}, which means that *x* is a normal curve.

Furthermore, from 〈*x*(*s*) − *m*,*x*(*s*) − *m*〉′ = 0, we have ∥*x*(*s*) − *m*∥ = *cons*., which means that *x* lies on lightlike cone *Q*^{3} with a vertex at *m*.

Let’s consider a pseudo orthonormal frame {*x*,*α*,*β*,*y*} in *Q*^{3}. Then we can write
$$\begin{array}{}x-m=ax+b\alpha +{c}^{\ast}y+d\beta ,\phantom{\rule{1em}{0ex}}a,b,{c}^{\ast},d\in I{R}_{0}^{+}.\end{array}$$(4.18)

Exposing the inner product with *y*,*α*,*x*,*β* to (4.18), we have
$$\begin{array}{}\u3008x-m,y\u3009=a\\ \u3008x-m,\alpha \u3009=b\\ \u3008x-m,\beta \u3009=d\\ \u3008x-m,x\u3009={c}^{\ast}.\end{array}$$

Differentiating 〈*x* − *m*,*α*〉 = *b*, we obtain
$$\begin{array}{}1+\kappa \u3008x-m,x\u3009-\u3008x-m,y\u3009=0.\end{array}$$(4.19)

Exposing the inner product with *y*,*α*,*β* to both sides of (4.16), we get
$$\begin{array}{}\u3008x-m,y\u3009=\lambda =1\\ \u3008x-m,\alpha \u3009=\mu =0\\ \u3008x-m,\beta \u3009=\gamma =c.\end{array}$$(4.20)

Substituting (4.19) in (4.16), we find
$$\begin{array}{}x-m=x+c\beta .\end{array}$$(4.21)

Therefore, considering the first equation of (4.20) in (4.19), we have *κ*〈*x* − *m*,*x*〉 = 0. Since *κ* ≠ 0, 〈*x* − *m*,*x*〉 = 0 = *c*^{∗}. Since *x* normal curve lies on a lightlike cone *Q*^{3}(*m*), we can write 〈*x*(*s*) − *m*,*x*(*s*) − *m*〉 = 0 and since 〈*x*(*s*) − *m*,*x*(*s*) − *m*〉 = 〈*c**β*,*c**β*〉 = *c*^{2}, we get *c* = 0. Hence we have *x* − *m* = *x*, therefore we get *m* = 0. Thus the theorem is proved. □

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