We denote by 𝕃^{2} := (ℝ^{2}, *g* = −*dx*^{2} + *dy*^{2}) the Lorentz-Minkowski plane, where (*x*, *y*) are the rectangular coordinates on 𝕃^{2}. We say that a non-zero vector *v* ∈ 𝕃^{2} is spacelike if *g*(*v*, *v*) > 0, lightlike if *g*(*v*, *v*) = 0, and timelike if *g*(*v*, *v*) < 0.

Let *γ* = (*x*, *y*): *I* ⊆ ℝ → ℝ^{2} be a curve. We say that *γ* = *γ*(*t*) is spacelike (resp. timelike) if the tangent vector *γ′*(*t*) is spacelike (resp. timelike) for all *t* ∈ *I*. A point *γ* (*t*) is called a lightlike point if *γ′*(*t*) is a lightlike vector. We study geometric properties of curves that have no lightlike points in this section, because the curvature is not in general well defined at the lightlike points.

Hence, let *γ* = (*x*, *y*) be a spacelike (resp. timelike) curve parametrized by arc-length; that is, *g*(*γ̇*(*s*), *γ̇*(*s*)) = 1 (resp. *g*(*γ̇*(*s*), *γ̇*(*s*)) = −1) ∀ *s* ∈ *I*, where *I* is some interval in ℝ. Here ˙ means derivation with respect to *s*. We will say that *γ* = *γ*(*s*) is a unit-speed curve in both cases.

We define the Frenet dihedron in such a way that the curvature *κ* has a sign and then it is only preserved by direct rigid motions (see [14]): Let *T* = *γ̇* = (*ẋ*, *ẏ*) be the tangent vector to the curve *γ* and we choose *N* = *γ̇*^{⊥} = (*ẏ*, *ẋ*) as the corresponding normal vector. We remark that *T* and *N* have different causal character. Let *g*(*T*, *T*) = *ϵ*, with *ϵ* = 1 if *γ* is spacelike, and *ϵ* = −1 if *γ* is timelike. Then *g*(*N*, *N*) = −*ϵ*. The (signed) curvature of *γ* is the function *κ* = *κ*(*s*) such that

$$\begin{array}{}{\displaystyle \dot{T}(s)=\kappa (s)N(s),}\end{array}$$(1)

where

$$\begin{array}{}{\displaystyle \kappa (s)=-\u03f5g(\dot{T}(s),N(s))=\u03f5(\ddot{x}\dot{y}-\dot{x}\ddot{y}).}\end{array}$$(2)

The Frenet equations of *γ* are given by (1) and

$$\begin{array}{}{\displaystyle \dot{N}(s)=\kappa (s)T(s).}\end{array}$$(3)

It is possible, as it happens in the Euclidean case, to obtain a parametrization by arc-length of the curve *γ* in terms of integrals of the curvature. Specifically, any spacelike curve *α* (*s*) in 𝕃^{2} can be represented by

$$\begin{array}{}{\displaystyle \alpha (s)=\left(\int \mathrm{sinh}\phi (s)ds,\int \mathrm{cosh}\phi (s)ds\right),\phantom{\rule{thinmathspace}{0ex}}\frac{d\phi (s)}{ds}=\kappa (s),}\end{array}$$(4)

and any timelike curve *β* (*s*) in 𝕃^{2} can be represented by

$$\begin{array}{}{\displaystyle \beta (s)=\left(\int \mathrm{cosh}\varphi (s)ds,\int \mathrm{sinh}\varphi (s)ds\right),\phantom{\rule{thinmathspace}{0ex}}\frac{d\varphi (s)}{ds}=\kappa (s).}\end{array}$$(5)

For example, up to a translation, any spacelike geodesic can be written as

$$\begin{array}{}{\displaystyle {\alpha}_{{\phi}_{0}}(s)=(\mathrm{sinh}{\phi}_{0}\phantom{\rule{thinmathspace}{0ex}}s,\mathrm{cosh}{\phi}_{0}\phantom{\rule{thinmathspace}{0ex}}s),\phantom{\rule{thinmathspace}{0ex}}s\in \mathbb{R},\phantom{\rule{thinmathspace}{0ex}}{\phi}_{0}\in \mathbb{R},}\end{array}$$(6)

and any timelike geodesic can be written as

$$\begin{array}{}{\displaystyle {\beta}_{{\varphi}_{0}}(s)=(\mathrm{cosh}{\varphi}_{0}\phantom{\rule{thinmathspace}{0ex}}s,\mathrm{sinh}{\varphi}_{0}\phantom{\rule{thinmathspace}{0ex}}s),\phantom{\rule{thinmathspace}{0ex}}s\in \mathbb{R},\phantom{\rule{thinmathspace}{0ex}}{\varphi}_{0}\in \mathbb{R}.}\end{array}$$(7)

On the other hand, the transformation *R*_{ν} : 𝕃^{2} → 𝕃^{2}, *ν* ∈ ℝ, given by

$$\begin{array}{}{\displaystyle {R}_{\nu}(x,y)=(\mathrm{cosh}\nu \phantom{\rule{thinmathspace}{0ex}}x+\mathrm{sinh}\nu \phantom{\rule{thinmathspace}{0ex}}y,\mathrm{sinh}\nu \phantom{\rule{thinmathspace}{0ex}}x+\mathrm{cosh}\nu \phantom{\rule{thinmathspace}{0ex}}y)}\end{array}$$

is an isometry of 𝕃^{2} that preserves the curvature of a curve *γ* and satisfies

$$\begin{array}{}{\displaystyle {R}_{\nu}\circ {\alpha}_{{\phi}_{0}}={\alpha}_{{\phi}_{0}+\nu},\phantom{\rule{thinmathspace}{0ex}}{R}_{\nu}\circ {\beta}_{{\varphi}_{0}}={\beta}_{{\varphi}_{0}+\nu}.}\end{array}$$

In this way, any spacelike geodesic is congruent to *α*_{0}, i.e. the *y*-axis, and any timelike geodesic is congruent to *β*_{0}, i.e. the *x*-axis (see Figure 1).

Fig. 1 Spacelike (blue) and timelike (red) geodesics in 𝕃^{2}.

## 2.1 Curves in 𝕃^{2} such that *κ* = *κ*(*y*)

Given a spacelike or timelike curve *γ* = (*x*, *y*) in 𝕃^{2}, we are first interested in the analytical condition *κ* = *κ*(*y*). We look for its geometric interpretation. For this purpose, we define the *Lorentzian pseudodistance* by

$$\begin{array}{}{\displaystyle \delta :{\mathbb{L}}^{2}\times {\mathbb{L}}^{2}\to [0,+\mathrm{\infty}),\phantom{\rule{thinmathspace}{0ex}}\delta (P,Q)=\sqrt{|g(\overrightarrow{PQ},\overrightarrow{PQ})|}.}\end{array}$$

We fix the timelike geodesic *β*_{0}, i.e. the *x*-axis. Given an arbitrary point *P* = (*x*, *y*) ∈ 𝕃^{2}, *y* ≠ 0, we consider all the spacelike geodesics *α*_{m} with slope *m* = coth *φ*_{0}, |*m*| > 1, passing through *P*, and let *P′* = (*x* − *y/m*, 0) the crossing point of *α*_{m} and the *x*-axis (see Figure 2). Then:

Fig. 2 Spacelike geodesics in 𝕃^{2} passing through *P*.

$$\begin{array}{}{\displaystyle 0<\delta (P,{P}^{\prime}{)}^{2}=\left(1-\frac{1}{{m}^{2}}\right){y}^{2}=\frac{{y}^{2}}{{\mathrm{cosh}}^{2}{\phi}_{0}}\le {y}^{2}}\end{array}$$

and the equality holds if and only if *φ*_{0} = 0, that is, *α*_{m} is a vertical geodesic. Thus |*y*| is the maximum Lorentzian pseudodistance through spacelike geodesics from *P* = (*x*, *y*) ∈ 𝕃^{2}, *y* ≠ 0, to the timelike geodesic given by the *x*-axis.

At a given point *γ*(*s*) = (*x*(*s*), *y*(*s*)) on the curve, the *geometric linear momentum (with respect to the* *x*-*axis)* 𝓚 is given by

$$\begin{array}{}{\displaystyle \mathcal{K}(s)=\dot{x}(s).}\end{array}$$(8)

In physical terms, using Noether’s Theorem, 𝓚 may be interpreted as the linear momentum with respect to the *x*-axis of a particle of unit mass with unit-speed and trajectory *γ*.

Given that *γ* is unit-speed, that is, −*ẋ*^{2} + *ẏ*^{2} = *ϵ*, and (8), we easily obtain that

$$\begin{array}{}{\displaystyle ds=\frac{dy}{\sqrt{\u03f5+{\dot{x}}^{2}}}=\frac{dy}{\sqrt{{\mathcal{K}}^{2}+\u03f5}},\phantom{\rule{1em}{0ex}}dx=\mathcal{K}ds.}\end{array}$$(9)

Thus, given 𝓚 = 𝓚(*y*) as an explicit function, looking at (9) one may attempt to compute *y*(*s*) and *x*(*s*) in three steps: integrate to get *s* = *s*(*y*), invert to get *y* = *y*(*s*) and integrate to get *x* = *x*(*s*).

In addition, we have that the curvature *κ* satisfies (1) and (3), i.e. *ẍ* = *κẏ*. Taking into account (8), we deduce that
$\begin{array}{}{\displaystyle \frac{d\mathcal{K}}{ds}=\kappa \dot{y}}\end{array}$
and, since we are assuming that *κ* = *κ*(*y*), we finally arrive at

$$\begin{array}{}{\displaystyle d\mathcal{K}=\kappa (y)dy,}\end{array}$$(10)

that is, 𝓚 = 𝓚(*y*) can be interpreted as an anti-derivative of *κ*(*y*).

As a summary, we have proved the following result in the spirit of Theorem 3.1 in [1].

#### Theorem 2.1

*Let* *κ* = *κ*(*y*) *be a continuous function*. *Then the problem of determining locally a spacelike or timelike curve in* 𝕃^{2} *whose curvature is* *κ*(*y*) *with geometric linear momentum* 𝓚 (*y*) *satisfying* (10) —|*y*| *being the (non constant) maximum Lorentzian pseudodistance through spacelike geodesics to the* *x*-*axis*— *is solvable by quadratures considering the unit speed curve* (*x*(*s*), *y*(*s*)), *where* *y*(*s*) *and* *x* (*s*) *are obtained through (9) after inverting* *s* = *s*(*y*). *Such a curve is uniquely determined by* 𝓚 (*y*) *up to a translation in the* *x*-*direction (and a translation of the arc parameter* *s*).

We show two illustrative examples applying steps (i)-(iii) in Remark 2.2:

#### Example 2.3

(*κ* ≡ 0). *Then* 𝓚 ≡ *c* ∈ ℝ,
$\begin{array}{}{\displaystyle s=\int dy/\sqrt{{c}^{2}+\u03f5}=y/\sqrt{{c}^{2}+\u03f5},{c}^{2}+\u03f5>0.}\end{array}$
*So* *y*(*s*) =
$\begin{array}{}{\displaystyle \sqrt{{c}^{2}+\u03f5}\phantom{\rule{thinmathspace}{0ex}}s}\end{array}$
*and* *x*(*s*) = *c s*, *s* ∈ ℝ. *If* *ϵ* = 1, *we write K* ≡ *c* := sinh *φ*_{0} *and then we arrive at the spacelike geodesics* *α*_{φ0}. *We observe that* *c* = 0 = *φ*_{0} *corresponds to the* *y*-*axis*. *If* *ϵ* = −1, *we write K* ≡ *c* := cosh *ϕ*_{0} *and then we arrive at the timelike geodesics* *β*_{ϕ0}. *We note that* *c* = 1 ⇔ *ϕ*_{0} = 0 *corresponds to the* *x*-*axis*. *See Figure 1*.

#### Example 2.4

(*κ* ≡ *k*_{0} > 0). *Now* 𝓚 (*y*) = *k*_{0}*y* + *c*, *c* ∈ ℝ. *Thus*
$\begin{array}{}{\displaystyle s=\int dy/\sqrt{({k}_{0}y+c{)}^{2}+\u03f5}.}\end{array}$
*If* *ϵ* = 1, *then* *s* = arcsinh (*k*_{0}*y* + *c*)/*k*_{0}, *y*(*s*) = (sinh(*k*_{0}*s*) − *c*)/*k*_{0} *and* *x*(*s*) = cosh(*k*_{0} *s*)/*k*_{0}. *If* *ϵ* = −1, *then* *s* = arccosh (*k*_{0}*y* + *c*)/*k*_{0}, *y*(*s*) = (cosh(*k*_{0}*s*) − *c*)/*k*_{0} *and* *x*(*s*) = sinh(*k*_{0} *s*)/*k*_{0}. *They correspond respectively to spacelike and timelike pseudocircles in* 𝕃^{2} *of radius* 1/*k*_{0} *(see Figure 3)*. *When* *c* = 0, *we obtain the branches of* *x*^{2} − *y*^{2} = *ϵ*/
$\begin{array}{}{\displaystyle {k}_{0}^{2}}\end{array}$
*with positive curvature* *k*_{0}, *that are asymptotic to the light cone of* 𝕃^{2}.

Fig. 3 Spacelike (blue) and timelike (red) pseudocircles in 𝕃^{2} with constant positive curvature.

## 2.2 Curves in 𝕃^{2} such that *κ* = *κ*(*x*)

Given a spacelike or timelike curve *γ* = (*x*, *y*) in 𝕃^{2}, we are now interested in the analytical condition *κ* = *κ*(*x*) and we search for its geometric interpretation using again the Lorentzian pseudodistance *δ*. We fix the spacelike geodesic *α*_{0}, i.e. the *y*-axis. Given an arbitrary point *P* = (*x*, *y*) ∈ 𝕃^{2}, *x* ≠ 0, we consider all the timelike geodesics *β*_{m} with slope *m* = tanh *ϕ*_{0}, |*m*| < 1, passing through *P*, and let *P′* = (0, *y* − *mx*) the crossing point of *β*_{m} and the *y*-axis (see Figure 4). Then:

Fig. 4 Timelike geodesics in 𝕃^{2} passing through *P*.

$$\begin{array}{}{\displaystyle 0>-\delta (P,{P}^{\prime}{)}^{2}=({m}^{2}-1){x}^{2}=-\frac{{x}^{2}}{{\mathrm{cosh}}^{2}{\varphi}_{0}}\ge -{x}^{2}}\end{array}$$

and the equality holds if and only if *ϕ*_{0} = 0, that is, *β*_{m} is a horizontal geodesic. Thus |*x*| is now the maximum Lorentzian pseudodistance through timelike geodesics from *P* = (*x*, *y*) ∈ 𝕃^{2}, *x* ≠ 0, to the spacelike geodesic given by the *y*-axis.

We now make a similar study to the one made in the preceding section. At a given point *γ* (*s*) = (*x*(*s*), *y*(*s*)) on the curve, the *geometric linear momentum (respect to the* *y*-*axis)* 𝓚 is given by

$$\begin{array}{}{\displaystyle \mathcal{K}(s)=\dot{y}(s).}\end{array}$$(11)

In physical terms, using Noether’s Theorem, 𝓚 may be interpreted as the linear momentum with respect to the *y*-axis of a particle of unit mass with unit-speed and trajectory *γ*.

Using that *γ* is unit-speed, that is, −*ẋ*^{2} + *ẏ*^{2} = *ϵ*, and (11), we easily obtain that

$$\begin{array}{}{\displaystyle ds=\frac{dx}{\sqrt{{\dot{y}}^{2}-\u03f5}}=\frac{dx}{\sqrt{{\mathcal{K}}^{2}-\u03f5}},\phantom{\rule{1em}{0ex}}dy=\mathcal{K}ds.}\end{array}$$(12)

Thus, given 𝓚 = 𝓚 (*x*) as an explicit function, looking at (12) one may attempt to compute *x*(*s*) and *y*(*s*) in three steps: integrate to get *s* = *s*(*x*), invert to get *x* = *x*(*s*) and integrate to get *y* = *y*(*s*).

In addition, we have that the curvature *κ* satisfies (1) and (3), i.e. *ÿ* = *κ ẋ*. Taking into account (11), we deduce that
$\begin{array}{}{\displaystyle \frac{d\mathcal{K}}{ds}=\kappa \dot{x}}\end{array}$
and, since we are assuming that *κ* = *κ*(*x*), we finally arrive at

$$\begin{array}{}{\displaystyle d\mathcal{K}=\kappa (x)dx,}\end{array}$$(13)

that is, 𝓚 = 𝓚(*x*) can be interpreted as an anti-derivative of *κ*(*x*).

As a summary, we have proved the following result, dual in a certain sense to Theorem 2.1

#### Theorem 2.6

*Let* *κ* = *κ*(*x*) *be a continuous function*. *Then the problem of determining locally a spacelike or timelike curve in* 𝕃^{2} *whose curvature is* *κ*(*x*) *with geometric linear momentum* 𝓚 (*x*) *satisfying* (13) —|*x*| *being the (non constant) maximum Lorentzian pseudodistance through timelike geodesics to the* *y*-*axis*— *is solvable by quadratures considering the unit speed curve* (*x*(*s*), *y*(*s*)), *where* *x*(*s*) *and* *y*(*s*) *are obtained through (12) after inverting* *s* = *s*(*x*). *Such a curve is uniquely determined by* 𝓚 (*x*) *up to a translation in the* *y*-*direction (and a translation of the arc parameter* *s)*.

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