2. The extrinsic catenary problem

Minimal surfaces in a three-dimensional Riemannian manifold are critical points of the area functional. They are characterized by the vanishing of the mean curvature $H$. If $g_{ij}$ and $h_{ij}$ denote the coefficients of the first and second fundamental forms of a minimal surface, the minimality condition reads

(2.1)\begin{equation} H = \frac{1}{2} \frac{g_{22}h_{11}-2g_{12}h_{12}+g_{11}g_{22}}{g_{11}g_{22}-g_{12}^2} = 0. \end{equation}

We want to classify the minimal surfaces of revolution in the hyperbolic space $\mathbb {H}^3(r)$ of constant sectional curvature $K=-({1}/{r^2})$. In other words, surfaces generated by 1-parameter subgroups of isometries whose orbits are curves of constant curvature $\kappa$ contained in a totally geodesic surface, i.e., a copy of $\mathbb {H}^2(r)$ in $\mathbb {H}^3(r)$. A surface of revolution is then foliated by hypercycles if $0<\kappa < {1}/{r}$, horocycles if $\kappa ={1}/{r}$, and circles if $\kappa >{1}/{r}$. Thus, we have three classes of minimal surfaces of revolution, and the classification of the minimal ones will be done regarding their generating curves.

Let $\mathbb {E}_1^{n+1}=(\mathbb {R}^{n+1},\langle X,Y\rangle _1=-X_0Y_0+\sum _{i=1}^nX_iY_i)$ be the $(n+1)$-dimensional Lorentz–Minkowski space. In the hyperboloid model of the hyperbolic space, we see $\mathbb {H}^n(r)$ as a hypersurface in $\mathbb {E}_1^{n+1}$ of constant sectional curvature $K=-1/r^2$:

(2.2)\begin{equation} \mathbb{H}^n(r) = \{X=(X_0,\ldots,X_n)\in\mathbb{E}_1^{n+1}: \langle X,X\rangle_1={-}r^2,\,X_0>0\}. \end{equation}

We shall denote by $\langle \cdot,\cdot \rangle$ the induced inner product of $\mathbb {H}^n(r)$. In addition, we embed $\mathbb {H}^2(r)$ into $\mathbb {H}^3(r)$ by the natural inclusion

(2.3)\begin{equation} (x,y,z)\in \mathbb{H}^2(r)\hookrightarrow (x,y,z,0)\in \mathbb{H}^3(r). \end{equation}

The extrinsic catenary problem is formulated as follows:

The extrinsic catenary problem. Given a subspace $P^2$ of $\mathbb {E}_1^{4}$, find the shape of a chain $\gamma :[a,b]\to \mathbb {H}^2(r)\subset \mathbb {H}^3(r)$ that optimizes the potential energy

(2.4)\begin{equation} \mathcal{W}[\gamma]=\int_a^b \textrm{dist}(\gamma(t),P^2)\Vert\dot{\gamma}(t)\Vert\,\mathrm{d} t, \end{equation}

where $\textrm {dist}(\gamma (t),P^2)$ is the distance in $\mathbb {E}_1^{4}$ of the point $\gamma (t)$ to $P^2$.

In the definition of extrinsic catenaries, it is implicitly assumed that $\gamma$ lies on one side of the plane $P^2$, implying that the integrand in (2.4) is positive. In contrast to how catenaries are defined in Ref. [Reference López5], the weight is now calculated using the extrinsic distance in $\mathbb {E}_1^4$, rather than the intrinsic distance to a given geodesic in $\mathbb {H}^2(r)$.

In $\mathbb {E}_1^{n+1}$, there are three types of planes $P^2$ according to their causal character. In general, if $P^k$ denotes a $k$-dimensional subspace of $\mathbb {E}_1^{n+1}$, $P^k$ is said to be Lorentzian, Riemannian, or degenerate, if the restriction of the metric $\langle \cdot,\cdot \rangle _1$ to $P^k$ is a Lorentzian, Riemannian, or degenerate metric, respectively. Therefore, we define three types of extrinsic catenaries: $\gamma$ is an extrinsic catenary of (i) elliptic type, (ii) hyperbolic type, and (iii) parabolic type if $\gamma$ is a critical point of (2.4) when $P^2$ is (i) Lorentzian, (ii) Riemannian, and (iii) degenerate, respectively.

Let $\{\mathbf {e}_x,\mathbf {e}_y,\mathbf {e}_z\}$ be the unit velocity vectors of the canonical Cartesian coordinates $(x,y,z)$ of $\mathbb {E}_1^3$ and denote $[\mathbf {v}_1,\ldots,\mathbf {v}_k]= \mbox {span}\{\mathbf {v}_1,\ldots,\mathbf {v}_k\}$. Without loss of generality, we can suppose that the subspace $P^2$ is one of the following:

1. (i) $P^2=[\mathbf {e}_{x},\mathbf {e}_{y}]$ (Lorentzian).

2. (ii) $P^2=[\mathbf {e}_y,\mathbf {e}_z]$ (Riemannian).

3. (iii) $P^2= [\tfrac {1}{\sqrt {2}}(\mathbf {e}_{y} +\mathbf {e}_{x}),\mathbf {e}_z]$ (degenerate).

To characterize the extrinsic catenaries, we will obtain the Euler–Lagrange equations of the functional (2.4). This task can be significantly simplified by choosing a suitable system of coordinates and then applying the standard techniques of the Calculus of Variations.

Without loss of generality, let $\ell$ be the geodesic in $\mathbb {H}^2(r)$ obtained from the intersection with the plane of equation $z=0$:

(2.5)\begin{equation} \ell(v) = r\left(\cosh\frac{v}{r},\sinh\frac{v}{r},0\right), \quad v\in\mathbb{R}. \end{equation}

If $\beta _v(u)$ denotes the geodesic with unit velocity $X(v)=(0,0,1)\in T_{\ell (v)}\mathbb {H}^2(r)$ at $u=0$, $\langle X(v),\ell '(v)\rangle _1=0$, then we can parametrize $\mathbb {H}^2(r)$ by

(2.6)\begin{align} \psi(u,v) & = \beta_v(u) = \ell(v)\cosh \frac{u}{r}+r\sinh \frac{u}{r}\,X\nonumber\\ & = r\left(\cosh\frac{u}{r}\cosh\frac{v}{r},\cosh\frac{u}{r}\sinh\frac{v}{r}, \sinh\frac{u}{r}\right). \end{align}

This coordinate system is known as the semi-geodesic coordinates [Reference Struik9]. Since

(2.7)\begin{equation} \begin{aligned} \psi_u & = \left(\sinh\frac{u}{r}\cosh\frac{v}{r},\sinh\frac{u}{r}\sinh\frac{v}{r},\cosh\frac{u}{r}\right)\\ \psi_v & = \left(\cosh\frac{u}{r}\sinh\frac{v}{r},\cosh\frac{u}{r}\cosh\frac{v}{r},0\right), \end{aligned} \end{equation}

the induced metric on $\mathbb {H}^2(r)$ takes the form

(2.8)\begin{equation} \mathrm{d} s^2 = \mathrm{d} u^2+\cosh^2\frac{u}{r}\,\mathrm{d} v^2. \end{equation}

Now, we compute the distance in $\mathbb {E}_1^3$ from $\gamma (t)$ to $P^2$. If we write $\gamma (t)=\psi (u(t),v(t))\in \mathbb {H}^2(r)\subset \mathbb {E}_1^3$, then from (2.6), we have

\[ \textrm{dist}(\gamma(t),[\mathbf{e}_{x},\mathbf{e}_{y}]) = r\sinh\frac{u}{r}, \quad \textrm{dist}(\gamma(t),[\mathbf{e}_y,\mathbf{e}_z]) = r\cosh\frac{u}{r}\cosh\frac{v}{r}, \]

and

\[ \textrm{dist}\left(\gamma(t),\left[\frac{1}{\sqrt{2}}(\mathbf{e}_{y}+\mathbf{e}_{x}), \mathbf{e}_z\right]\right) = \frac{r}{\sqrt{2}}\mathrm{e}^{- {v}/{r}}\cosh\frac{u}{r}. \]

The first two distances are the length of the line segment orthogonal to $P^2$, which connects $\gamma (t)$ to $P^2$. The third distance is trickier since the reference plane is degenerate. (Thus, it contains its orthogonal line.) To circumvent this difficulty, we choose $\mathbb {E}_1^3= [\tfrac {1}{\sqrt {2}}(\mathbf {e}_{y}-\mathbf {e}_{x}), \tfrac {1}{\sqrt {2}}(\mathbf {e}_{y}+\mathbf {e}_{x}),\mathbf {e}_z]$ and then consider $\tfrac {1}{\sqrt {2}}(\mathbf {e}_{y}-\mathbf {e}_{x})$ as playing the role of the director vector of the ‘line segment orthogonal to $P^2$, which connects $\gamma (t)$ to $P^2$.’ The distance $\textrm {dist}(\gamma (t),P^2)$ is then the (Euclidean) length of this line segment. (Note that $L_1=\tfrac {1}{\sqrt {2}}(\mathbf {e}_{y}-\mathbf {e}_{x})$ and $L_2=\tfrac {1}{\sqrt {2}}(\mathbf {e}_{y}+\mathbf {e}_{x})$ are two lightlike vectors orthogonal to $\mathbf {e}_z$ and normalized by the condition $\langle L_1,L_2\rangle =1$.) The definition's suitability for the degenerate case is justified a posteriori: it is the ‘distance’ allowing us to characterize minimal surfaces of revolution of parabolic type similarly to the classification of surfaces of elliptic and hyperbolic types.

Up to multiplication by a constant, the distance to $P^2$ can be taken as $\sinh {u}/{r}$ (elliptic), $\cosh {u}/{r}\cosh {v}/{r}$ (hyperbolic), and $e^{-v/r}\cosh {u}/{r}$ (parabolic). Therefore,

1. (i) Extrinsic catenaries of elliptic type are critical points of

   (2.9)\begin{equation} \mathcal{W}_{\mathcal{E}}[\gamma(t)] = \int_a^b\sinh\frac{u}{r}\sqrt{\dot{u}^2+\dot{v}^2\cosh^2\frac{u}{r}}\,\mathrm{d} t,\quad u>0. \end{equation}

2. (ii) Extrinsic catenaries of hyperbolic type are critical points of

   (2.10)\begin{equation} \mathcal{W}_{\mathcal{H}} [\gamma(t)]= \int_a^b\cosh \frac{u}{r}\cosh\frac{v}{r}\sqrt{\dot{u}^2+\dot{v}^2\cosh^2\frac{u}{r}}\,\mathrm{d} t,\quad u>0. \end{equation}

3. (iii) Extrinsic catenaries of parabolic type are critical points of

   (2.11)\begin{equation} \mathcal{W}_{\mathcal{P}}[\gamma(t)] = \int_a^b \mathrm{e}^{-{v}/{r}}\cosh\frac{u}{r}\sqrt{\dot{u}^2+\dot{v}^2\cosh^2\frac{u}{r}}\,\mathrm{d} t,\quad u>0. \end{equation}
   The assumption $u>0$ means that $\gamma$ lies on one side of the plane $P^2$.

The functionals $\mathcal {W}_{\mathcal {E}}$, $\mathcal {W}_{\mathcal {H}}$, and $\mathcal {W}_{\mathcal {P}}$ are all of the form $\int f(u,v) \Vert \dot {\gamma }\Vert \mathrm {d} t$. To compute the corresponding Euler–Lagrange equations, we first need the expression of the geodesic curvature $\kappa$ in $\mathbb {H}^2(r)$ with respect to coordinates (2.6).

Lemma 2.3 If $\gamma (t)=\psi (u(t),v(t))$ is a smooth regular curve in $\mathbb {H}^2(r)$, its signed curvature in $\mathbb {H}^2(r)$ is given by

(2.12)\begin{equation} \kappa ={-}\frac{1}{\Vert\dot{\gamma}(t)\Vert^3} \left[ \frac{2\dot{u}^2\dot{v}}{r}\sinh \frac{u}{r} +\frac{\dot{v}^3}{r}\cosh^2\frac{u}{r}\sinh\frac{u}{r} +(\ddot{u}\dot{v}-\dot{u}\ddot{v})\cosh\frac{u}{r}\right]. \end{equation}

Proof. Let $\Gamma _{ij}^k$ be the Christoffel symbols of the metric $\mathrm {d} s^2$ associated with the coordinates (2.6). The expression for $\kappa$ is then given by

\[ \kappa = \frac{\sqrt{\det g_{ij}}\left(\Gamma_{22}^1\dot{v}^3 -\Gamma_{11}^2\dot{u}^3-(2\Gamma_{12}^2-\Gamma_{11}^1)\dot{u}^2 \dot{v}+(2\Gamma_{12}^1-\Gamma_{22}^2)\dot{u}\dot{v}^2+ \ddot{u}\dot{v}-\dot{u}\ddot{v}\right)}{(g_{11}\dot{u}^2+ 2g_{12}\dot{u}\dot{u}+g_{22}\dot{v}^2)^{3/2}}. \]

For the metric $\mathrm {d} s^2$ given in (2.8), we have $g_{11}=1$, $g_{12}=0$, and $g_{22}=\cosh ^2 {u}/{r}$. As a consequence, $\Gamma _{11}^1=\Gamma _{12}^1=\Gamma _{11}^2=\Gamma _{22}^2=0$ and

\[ \Gamma_{22}^1={-}\frac{1}{r}\sinh\frac{u}{r}\cosh\frac{u}{r},\quad \Gamma_{12}^2=\frac{1}{r}\tanh \frac{u}{r}. \]

Now, using that $\Vert \dot {\gamma }\Vert =\sqrt {g_{11}\dot {u}^2+2g_{12}\dot {u}\dot {u}+g_{22}\dot {v}^2}$ and substituting the expressions for $\Gamma _{ij}^k$, Eq. (2.12) is immediate.

Lemma 2.4 Consider the functional

(2.13)\begin{equation} \mathcal{E}_f[u(t),v(t)] = \int_a^b f(u,v) \Vert\dot{\gamma}(t)\Vert\, \mathrm{d} t, \end{equation}

where $f$ is some smooth, positive function. The Euler–Lagrange equations of $\mathcal {E}_f$ are

(2.14)\begin{equation} \begin{aligned} \dot{v}\left[f\cosh\frac{u}{r}\left(\kappa-\frac{\dot{v}f_u\cosh {u}/{r}}{f\Vert\dot{\gamma}\Vert} \right)+\frac{\dot{u}f_v}{\Vert\dot{\gamma}\Vert}\right] & = 0\\ \dot{u}\left[f\cosh\frac{u}{r}\left(\kappa-\frac{\dot{v}f_u\cosh {u}/{r}}{f\Vert\dot{\gamma}\Vert}\right)+\frac{\dot{u}f_v}{\Vert\dot{\gamma}\Vert}\right] & = 0, \end{aligned} \end{equation}

where $\kappa$ is the geodesic curvature of $\gamma$ in $\mathbb {H}^2(r)$. If, in addition, $\gamma$ is regular, then $\gamma$ is a solution of the Euler–Lagrange equations if, and only if,

(2.15)\begin{equation} \kappa=\frac{1}{f\Vert\dot{\gamma}\Vert}\left(\dot{v} f_u\cosh \frac{u}{r}-\frac{\dot{u}f_v}{\cosh {u}/{r}}\right). \end{equation}

Proof. The Euler–Lagrange equations associated with the Lagrangian $\mathcal {L}=f\Vert \dot {\gamma }\Vert$ are

(2.16)\begin{equation} \frac{\partial \mathcal{L}}{\partial u}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial \mathcal{L}}{\partial \dot{u}} = 0 \quad \text{and} \quad \frac{\partial \mathcal{L}}{\partial v}-\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial \mathcal{L}}{\partial \dot{v}} = 0. \end{equation}

These equations are

\begin{align*} & f_u\Vert\dot{\gamma}\Vert+\frac{\dot{v}^2f\cosh {u}/{r} \sinh {u}/{r}}{r\Vert\dot{\gamma}\Vert}-\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{f\dot{u}}{\Vert\dot{\gamma}\Vert}\right) =0 \quad \text{and}\\ & f_v\Vert\dot{\gamma}\Vert-\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{f\dot{v}\cosh^2 {u}/{r}}{\Vert\dot{\gamma}\Vert}\right) = 0. \end{align*}

Equivalently, we have

\begin{align*} \frac{\dot{v}^2f_u\cosh^2 {u}/{r}-\dot{u}\dot{v}f_v+ ({1}/{r})\dot{v}^2f\sinh {u}/{r}\cosh {u}/{r}}{\Vert\dot{\gamma}\Vert} - f \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\dot{u}}{\Vert\dot{\gamma}\Vert}\right)& =0\\ \frac{\dot{u}^2f_v-\dot{u}\dot{v}f_u\cosh^2 {u}/{r}-\frac{2}{r}\dot{u} \dot{v}f\sinh {u}/{r}\cosh {u}/{r}}{\Vert\dot{\gamma}\Vert}- f\cosh^2\tfrac{u}{r}\frac{\mathrm{d}}{\mathrm{d} t} \left(\frac{\dot{v}}{\Vert\dot{\gamma}\Vert}\right)& =0. \end{align*}

After some manipulations, both equations can be expressed as

\begin{align*} 0 & = \dot{v}\left[\frac{\dot{u}f_v-\dot{v}f_u\cosh^2 {u}/{r} }{\Vert\dot{\gamma}\Vert}\right. \\ & \quad - \left.\frac{f\cosh {u}/{r}}{\Vert\dot{\gamma}\Vert^3} \left(\cosh\tfrac{u}{r}(\dot{u}\ddot{v}-\dot{v}\ddot{u}) +\frac{\dot{v}\sinh {u}/{r}}{r}\left(2\dot{u}^2+\cosh^2 \frac{u}{r}\dot{v}^2\right)\right)\right] \end{align*}

and

\begin{align*} 0 & = \dot{u}\left[\frac{\dot{u}f_v-f_u\dot{v}\cosh^2 {u}/{r} }{\Vert\dot{\gamma}\Vert}\right. \\ & \quad -\left.\frac{f\cosh {u}/{r}}{\Vert\dot{\gamma}\Vert^3} \left(\cosh\frac{u}{r}(\dot{u}\ddot{v}-\dot{v}\ddot{u}) +\frac{\dot{v}\sinh {u}/{r}}{r}\left(2\dot{u}^2+\cosh^2\frac{u}{r}\dot{v}^2\right)\right)\right]. \end{align*}

The identities (2.14) are now obtained using the above two equations and the expression of $\kappa$ given in Eq. (2.12).

Finally, we are in a position to characterize the extrinsic catenaries in the hyperbolic plane.

3. Characterization of extrinsic catenaries

In this section, we provide a coordinate-free characterization of extrinsic catenaries with the help of Killing vector fields of $\mathbb {E}_1^4$. The motivation comes from the Euclidean catenary $y(x)= {1}/{a}\cosh (ax+b)$ in the $(x,y)$-plane. This curve is the solution to the hanging chain problem in the Euclidean plane. In this setting, the reference line is the horizontal line $\ell$ of equation $y=0$, and the corresponding Euler–Lagrange equation is

(3.1)\begin{equation} \frac{y''}{(1+y'^2)^{3/2}}=\frac{1}{y\sqrt{1+y'^2}}. \end{equation}

The left-hand side is just the curvature $\kappa$ of the curve $y=y(x)$. The right-hand side is the Euclidean product ${\bf n}\cdot \mathbf {e}_y$ divided by the distance to $\ell$, where $\mathbf {n}=(-y',1)/\sqrt {1+y'^2}$ is the unit normal of $y=y(x)$ and $-\mathbf {e}_y=(0,-1)$ is the direction of gravity in Euclidean plane. Thus, Eq. (3.1) can be expressed as

(3.2)\begin{equation} \kappa = \frac{{\bf n}\cdot\mathbf{e}_y}{\textrm{dist}(\gamma,\ell)}. \end{equation}

Observe that the direction of gravity $-\mathbf {e}_y$ also has a geometric interpretation. It gives the direction of the geodesics orthogonal to $\ell$ used to compute $\textrm {dist}(\gamma,\ell )$.

We can provide a similar result for extrinsic catenaries in hyperbolic space. However, since the potential energy (2.4) measures the distance in $\mathbb {E}_1^4$ from the plane $P^2$, it is natural to expect that the right-hand side of Eq. (3.2) will incorporate the extrinsic nature of (2.4). Such a characterization of hyperbolic extrinsic geodesics will be obtained as a corollary of a more general result:

Theorem 3.1 A regular curve $\gamma (t)=\psi (u(t),v(t))$ in $\mathbb {H}^2(r)$ is a critical point of the functional (2.13) if, and only if, its curvature $\kappa$ in $\mathbb {H}^2(r)$ satisfies

(3.3)\begin{equation} \kappa ={-} \frac{\langle\mathbf{n}, \nabla f\rangle}{f}, \end{equation}

where $\mathbf {n}$ is the unit normal of $\gamma$ and $\nabla f$ is the gradient vector field of $f$.

Proof. The tangent vector of $\gamma$ is $\dot {\gamma }=\dot {u}\partial _u+\dot {v}\partial _v$. Thus, the unit normal $\mathbf {n}$ is

(3.4)\begin{equation} \mathbf{n}=\frac{1}{\Vert\dot{\gamma}\Vert} \left(-\dot{v}\cosh\frac{u}{r}\partial_u +\frac{\dot{u}}{\cosh {u}/{r}}\partial_v\right), \end{equation}

where $\{\partial _u,\partial _v\}$ is the coordinate basis determined by the parametrization $\psi$. The expression of the gradient of $f$ with respect to this basis is

(3.5)\begin{equation} \nabla f= f_u\partial_u +\frac{f_v}{\cosh^2 {u}/{r}} \partial_v. \end{equation}

It follows from Eqs. (3.4) and (3.5) that

\[ \langle\mathbf{n},\nabla f\rangle= \frac{1}{\Vert\dot{\gamma}\Vert}\left(-\dot{v}\cosh \frac{u}{r}f_u+\frac{\dot{u}}{\cosh {u}/{r}}f_v\right). \]

This equation and Eq. (2.15) finally prove Eq. (3.3).

Finally, we are in a position to provide the following characterization for extrinsic catenaries.

Theorem 3.2 A regular curve $\gamma (t)=\psi (u(t),v(t))$ in $\mathbb {H}^2(r)$ is an extrinsic catenary if, and only if, its curvature $\kappa$ in $\mathbb {H}^2(r)$ satisfies

(3.6)\begin{equation} \kappa(t) ={-}\frac{\langle\mathbf{n}(t), X(t)\rangle}{\mathrm{dist}(\gamma(t),P^2)}, \end{equation}

where $X$ is the Killing field in $\mathbb {E}_1^3$ of the geodesics in $\mathbb {E}_1^3$ orthogonal to $P^2$. If $\gamma$ is a catenary of hyperbolic or elliptic type, such geodesics connect $\gamma$ to a point of $P^2$.

Proof. If $X$ is a vector field in $\mathbb {E}_1^3$, its tangent part $X^\top$ in $\mathbb {H}^2(r)$ is given by

(3.7)\begin{equation} X^\top = X - \frac{1}{r^2}\langle X,\psi\rangle\psi = \langle X,\psi_u\rangle\psi_u + \frac{\langle X,\psi_v\rangle}{\cosh^2 {u}/{r}}\,\psi_v, \end{equation}

where $\psi$ is the parametrization (2.6) of $\mathbb {H}^2(r)$. Consider now the three types of Killing vector fields that determine the distance between a point of $\mathbb {H}^2(r)$ and a subspace $P^2$ depending on the causal character of $P^2$. Denote by $\{\mathbf {e}_x,\mathbf {e}_y,\mathbf {e}_z\}$ the canonical basis of $\mathbb {E}_1^3$.

1. (i) If $P^2=[\mathbf {e}_{x},\mathbf {e}_{y}]$, the Killing vector field is $X=\mathbf {e}_z$ and, from (3.7), we have

   \begin{align*} X^\top & ={-}\frac{1}{2} \sinh \frac{2 u}{r} \cosh \frac{v}{r}\mathbf{e}_x - \frac{1}{2} \sinh \frac{2 u}{r} \sinh \frac{v}{r}\mathbf{e}_y+ \left(1-\sinh ^2\frac{u}{r}\right)\mathbf{e}_z \\ & = \cosh\frac{u}{r}\psi_u. \end{align*}

2. (ii) If $P^2=[\mathbf {e}_y,\mathbf {e}_z]$, the Killing vector field is $X=-\mathbf {e}_x$ and, from (3.7), we have

   \begin{align*} X^\top & = \left({-}1 - \cosh ^2\frac{u}{r} \cosh ^2\frac{v}{r}\right) \mathbf{e}_x - \frac{1}{2} \cosh ^2\frac{u}{r} \sinh \frac{2 v}{r}\mathbf{e}_y - \frac{1}{2} \sinh \frac{2 u}{r} \cosh \frac{v}{r}\mathbf{e}_z \\ & = \sinh\frac{u}{r}\cosh\frac{v}{r}\,\psi_u + \frac{\sinh {v}/{r}}{\cosh {u}/{r}}\,\psi_v. \end{align*}

3. (iii) If $P^2= [\tfrac {1}{\sqrt {2}}(\mathbf {e}_{y}+\mathbf {e}_{x}),\mathbf {e}_z]$, the Killing vector field $X=-\tfrac {1}{\sqrt {2}}(\mathbf {e}_x+\mathbf {e}_y)$ and, from (3.7), we have

   \begin{align*} X^\top & ={-} \frac{1}{\sqrt{2}}\left(1 + \mathrm{e}^{- {v}/{r}} \cosh ^2\frac{u}{r} \cosh \frac{v}{r}\right)\mathbf{e}_x \\ & \quad - \frac{1}{\sqrt{2}}\left[1 + \mathrm{e}^{- { v/}{r}}\cosh^2 \frac{u}{r}\sinh\frac{v}{r}\right] \mathbf{e}_y - \frac{\mathrm{e}^{- {v}/{r}}}{2\sqrt{2}} \sinh \frac{2 u}{r}\mathbf{e}_z \\ & = \frac{\mathrm{e}^{- {v}/{r}}}{\sqrt{2}} \sinh\frac{u}{r} \, \psi_u - \frac{\mathrm{e}^{- {v}/{r}}}{\sqrt{2}\cosh {u}/{r}}\,\psi_v. \end{align*}

Let us compute the gradient $\nabla f$ in $\mathbb {H}^2(r)$ of the distances to each of the above planes $P^2$.

1. (i) In the elliptic case, $f(u,v)=r \sinh {u}/{r}$. Using Eq. (3.5) and $\partial _u=\psi _u$ and $\partial _v=\psi _v$ in (2.7), we obtain that $\nabla f=X^\top$.

2. (ii) In the hyperbolic case, $f(u,v)=r\cosh {u}/{r}\cosh {v}/{r}$. Using Eq. (3.5) and $\partial _u=\psi _u$ and $\partial _v=\psi _v$ in (2.7), we obtain that $\nabla f=X^\top$.

3. (iii) In the parabolic case, $f(u,v)=\tfrac {r}{\sqrt {2}}\mathrm {e}^{- {v}/{r}} \cosh {u}/{r}$. Using Eq. (3.5) and $\partial _u=\psi _u$ and $\partial _v=\psi _v$ in (2.7), we obtain that $\nabla f= X^\top$.

Finally, applying theorem 3.1 gives the desired result. Indeed,

\[ \kappa ={-}\frac{\langle\mathbf{n},\nabla f\rangle}{f} ={-}\frac{\langle\mathbf{n},X^{\top}\rangle}{\mathrm{dist}(\gamma,P^2)} ={-}\frac{\langle\mathbf{n},X\rangle}{\mathrm{dist}(\gamma,P^2)}. \]

4. Surfaces of revolution in hyperbolic space

This section describes the three types of surfaces of revolution in $\mathbb {H}^3(r)$. To define them, we first exploit the fact that all isometries of $\mathbb {H}^{3}(r)$ are induced by orthogonal transformations of $\mathbb {E}_1^{4}$. Later, after introducing a convenient coordinate system for each surface type, we compute their mean curvature.

Let $P^k$ denote a $k$-dimensional subspace of $\mathbb {E}_1^{4}$ and $O(P^k)$ be the set of orthogonal transformations of $\mathbb {E}_1^{4}$ with positive determinant and that leave $P^k$ pointwise fixed. (In the context of the Lorentzian geometry, the orthogonal transformations of $\mathbb {E}_1^{4}$ are often referred to as Lorentzian transformations.) Following do Carmo–Dajczer [Reference do Carmo and Dajczer2], we have the following classification for surfaces of revolution in $\mathbb {H}^3(r)$.

Now, we shall explain the general recipe of how to choose the planes $P^2$ and $P^3$ and the generating curve $\gamma$ in the constructions below. First, we fix a basis $\{\mathbf {e}_1,\mathbf {e}_2,\mathbf {e}_3,\mathbf {e}_4\}$ of $\mathbb {E}_1^4$ (not necessarily the canonical one). Later, we shall take $P^2=[\mathbf {e}_1,\mathbf {e}_2]$ and $P^3=[\mathbf {e}_1,\mathbf {e}_2,\mathbf {e}_3]$. Finally, to generate a surface of revolution, we take a curve $\gamma (t)=x_1(t)\mathbf {e}_1+x_2(t)\mathbf {e}_2+x_3(t)\mathbf {e}_3$ and then rotate it around $P^2$. Without loss of generality, the parametrizations of the three types of surfaces of revolution are the following:

1. (i) Surfaces of elliptic type. Consider $P^2=[\mathbf {e}_{x},\mathbf {e}_{y}]$ and $P^3=[\mathbf {e}_{x},\mathbf {e}_{y},\mathbf {e}_{z}]$. The members $\mathcal {E}_{\theta }$ of $O(P^2)$ take the form

   \[ \mathcal{E}_{\theta} = \left( \begin{array}{@{}cccc@{}} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos\theta & -\sin\theta \\ 0 & 0 & \sin\theta & \cos\theta \end{array}\right). \]
   Given a curve $\gamma (t)=(x(t),y(t),z(t),0)\in \mathbb {H}^2(r)=P^3\cap \mathbb {H}^3(r)$, we obtain a surface of revolution $S_{\mathcal {E}}$ parametrized by
   (4.1)\begin{equation} S_{\mathcal{E}}(t,\theta)=\mathcal{E}_{\theta}(\gamma(t)) =(x(t),y(t),z(t)\cos\theta,z(t)\sin\theta). \end{equation}

2. (ii) Surfaces of hyperbolic type. Consider $P^2=[\mathbf {e}_y,\mathbf {e}_z]$ and $P^3=[\mathbf {e}_y,\mathbf {e}_{z},\mathbf {e}_{x}]$. The members $\mathcal {H}_{\theta }$ of $O(P^2)$ take the form

   \[ \mathcal{H}_{\theta} = \left( \begin{array}{@{}cccc@{}} \cosh\theta & 0 & 0 & \sinh\theta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sinh\theta & 0 & 0 & \cosh\theta \\ \end{array} \right). \]
   Given a curve $\gamma (t)=(x(t),y(t),z(t),0)\in \mathbb {H}^2(r)=P^3\cap \mathbb {H}^3(r)$, we obtain a surface of revolution $S_{\mathcal {H}}$ parametrized by
   (4.2)\begin{equation} S_{\mathcal{H}}(t,\theta) = \mathcal{H}_{\theta}(\gamma(t))=(x(t)\cosh\theta,y(t),z(t),x(t)\sinh\theta). \end{equation}

3. (iii) Surfaces of parabolic type. Consider the subspaces $P^2=[ {(\mathbf {e}_{y}+\mathbf {e}_{x})}/{\sqrt {2}},\mathbf {e}_z]$ and $P^3=[ {(\mathbf {e}_{y}+\mathbf {e}_{x})}/{\sqrt {2}},\mathbf {e}_z, {(\mathbf {e}_{y}-\mathbf {e}_{x})}/{\sqrt {2}}]$. The members $\mathcal {P}_{\theta }$ of $O(P^2)$ take the form

   \[ \mathcal{P}_{\theta} = \left( \begin{array}{@{}cccc@{}} 1+\dfrac{\theta^2}{2} & -\dfrac{\theta^2}{2} & 0 & -\theta \\ \dfrac{\theta^2}{2} & 1-\dfrac{\theta^2}{2} & 0 & -\theta \\ 0 & 0 & 1 & 0 \\ - \theta & \theta & 0 & 1 \\ \end{array} \right). \]
   Now, given a curve $\gamma (t)=(x(t),y(t),z(t),0)\in \mathbb {H}^2(r)=P^3\cap \mathbb {H}^3(r)$, we obtain a surface of revolution $S_{\mathcal {P}}(t,\theta )=\mathcal {P}_{\theta }(\gamma (t))$ parametrized by
   (4.3)\begin{equation} S_{\mathcal{P}}(t,\theta) = \left(x+\frac{\theta^2}{2}(x-y),y+\frac{\theta^2}{2}(x-y),z,-\theta (x- y)\right). \end{equation}

Let us calculate the mean curvature of the surfaces of revolution parametrized as above.

Proof. First, note we may employ the ternary product $\times _1$ of $\mathbb {E}_1^4$ to define a vector product $\times$ in $\mathbb {H}^3(r)$. Indeed, if $X,Y\in T_p\mathbb {H}^3(r)$, we have

\[ X\times Y = X \times_1 Y \times_1 \frac{p}{r} = \frac{1}{r}\det\left( \begin{array}{@{}cccc@{}} x_1 & y_1 & z_1 & w_1 \\ x_2 & y_2 & z_2 & w_2 \\ x & y & z & w \\ - \mathbf{e}_x & \mathbf{e}_y & \mathbf{e}_z & \mathbf{e}_w \\ \end{array} \right), \]

where $p=(x,y,z,w)$, $X=(x_1,y_1,z_1,w_1)$, and $Y=(x_2,y_2,z_2,w_2)$. Here, the vectors of the orthonormal basis $\{\mathbf {e}_x, \mathbf {e}_y, \mathbf {e}_z, \mathbf {e}_w\}$ are placed in the last line to guarantee that $\mathbf {e}_x \times _1 \mathbf {e}_y \times _1 \mathbf {e}_z = \mathbf {e}_w$.

Consider a parametrization $\Phi =\Phi (u^1,u^2)$ of a surface of $\mathbb {H}^3(r)$. Recall that the coefficients $(g_{ij})$ of the first fundamental form are $g_{ij}=\langle {\partial \Phi }/{\partial u^i}, {\partial \Phi }/{\partial u^j}\rangle$. Fix the unit normal

\[ \xi=\frac{1}{\sqrt{\det g_{ij}}} \frac{\partial \Phi}{\partial u^1}\times\frac{\partial \Phi}{\partial u^2}. \]

The coefficients $(h_{ij})$ of the second fundamental form are

\[ h_{ij}=\left\langle\nabla_{ {\partial \Phi}/{\partial u^i}} \frac{\partial \Phi}{\partial u^j},\xi\right\rangle = \left\langle\frac{\partial^2 \Phi}{\partial u^i\partial u^j},\xi\right\rangle. \]

Then, the mean curvature $H$ of $\Phi (u^1,u^2)$ is given by

(4.7)\begin{equation} H=\frac{g_{22}h_{11}-2g_{12}h_{12}+g_{11}h_{22}}{2(g_{11}g_{22}-g_{12}^2)}. \end{equation}

We now compute (4.7) in each of the three types of surfaces of revolution.

1. (i) Let $S_{\gamma }$ be of revolution of spherical type. Without loss of generality, we may parametrize it as in (4.1): $S_{\gamma }=S_{\mathcal {E}}$. The tangent plane is spanned by

   \[ \frac{\partial S_{\mathcal{E}}}{\partial t} = (\dot{x},\dot{y},\dot{z}\cos\theta,\dot{z}\sin\theta) \quad \mbox{and} \quad \frac{\partial S_{\mathcal{E}}}{\partial \theta} = (0,0,-z\sin\theta,z\cos\theta). \]
   Thus, the first fundamental form of $S_{\mathcal {E}}$ becomes
   \[ \mathrm{d} s^2 = (-\dot{x}^2+\dot{y}^2+\dot{z}^2)\mathrm{d} t^2+z^2\mathrm{d}\theta^2 = \Vert\dot{\gamma}\Vert^2\mathrm{d} t^2+z^2\mathrm{d}\theta^2. \]

   The unit normal $\xi$ of $S_{\mathcal {E}}$ is given by

   \[ \xi = \frac{1}{r\Vert\dot{\gamma}\Vert}\Big(y\dot{z} -\dot{y}z,x\dot{z}-\dot{x}z,(\dot{x}y-x\dot{y})\cos\theta,(\dot{x}y-x\dot{y})\sin\theta \Big). \]
   Let us compute the second fundamental form's coefficients $h_{ij}$. The second derivatives of $S_{\mathcal {E}}$ are
   \[ \frac{\partial^2 S_{\mathcal{E}}}{\partial t^2} = (\ddot{x},\ddot{y},\ddot{z}\cos\theta,\ddot{z}\sin\theta), \quad \frac{\partial^2 S_{\mathcal{E}}}{\partial t\partial\theta} = (0,0,-\dot{z}\sin\theta,\dot{z}\cos\theta), \]
   and
   \[ \frac{\partial^2 S_{\mathcal{E}}}{\partial \theta^2} = (0,0,-z\cos\theta,-z\sin\theta). \]
   Thus, we have $h_{12}=0$ and
   \[ h_{11} = \frac{1}{r\Vert\dot{\gamma}\Vert}\left[x(\ddot{y}\dot{z}- \dot{y}\ddot{z})-y(\ddot{x}\dot{z}-\dot{x}\ddot{z})+z(\ddot{x}\dot{y} -\dot{x}\ddot{y})\right],\quad h_{22} = \frac{z(x\dot{y}-\dot{x}y)}{r\Vert\dot{\gamma}\Vert}. \]
   Substitution of $h_{ij}$ and $g_{ij}$ in Eq. (4.7) gives the desired expression (4.4).

2. (ii) Let $S_{\gamma }$ be a surface of revolution of hyperbolic type. Without loss of generality, we may parametrize it as in (4.2): $S_{\gamma }=S_{\mathcal {H}}$. The tangent plane is spanned by

   (4.8)\begin{equation} \frac{\partial S_{\mathcal{H}}}{\partial t} = (\dot{x}\cosh\theta,\dot{y},\dot{z},\dot{x}\sinh\theta) \quad \mbox{and} \quad \frac{\partial S_{\mathcal{H}}}{\partial \theta} = (x\sinh\theta,0,0,x\cosh\theta). \end{equation}
   Thus, the first fundamental form of $S_{\mathcal {H}}$ becomes
   (4.9)\begin{equation} \mathrm{d} s^2 = (-\dot{x}^2+\dot{y}^2+\dot{z}^2)\mathrm{d} t^2+z^2\mathrm{d}\theta^2 = \Vert\dot{\gamma}\Vert^2\mathrm{d} t^2+x^2\mathrm{d}\theta^2. \end{equation}

   The unit normal $\xi$ of $S_{\mathcal {H}}$ is given by

   \[ \xi = \frac{1}{r\Vert\dot{\gamma}\Vert}\Big((y\dot{z}-\dot{y}z) \cosh\theta,x\dot{z}-\dot{x}z,\dot{x}y-x\dot{y},(y\dot{z}-\dot{y}z)\sinh\theta\Big). \]
   Let us compute the second fundamental form's coefficients $h_{ij}$. The second derivatives of $S_{\mathcal {H}}$ are
   \[ \frac{\partial^2 S_{\mathcal{H}}}{\partial t^2} = (\ddot{x}\cosh\theta,\ddot{y},\ddot{z},\ddot{x}\sinh\theta), \quad \frac{\partial^2 S_{\mathcal{H}}}{\partial t\partial\theta} = (\dot{x}\sinh\theta,0,0,\dot{x}\cosh\theta), \]
   and
   \[ \frac{\partial^2 S_{\mathcal{H}}}{\partial \theta^2} = (x\cosh\theta,0,0,x\sinh\theta). \]
   Thus, we have $h_{12}=0$ and
   \[ h_{11} = \frac{1}{r\Vert\dot{\gamma}\Vert}\left[x(\ddot{y}\dot{z} -\dot{y}\ddot{z})-y(\ddot{x}\dot{z}-\dot{x}\ddot{z})+z(\ddot{x}\dot{y} -\dot{x}\ddot{y})\right], \quad h_{22} ={-}\frac{x(y\dot{z}-\dot{y}z)}{r\Vert\dot{\gamma}\Vert}. \]
   Again, a substitution of $h_{ij}$ and $g_{ij}$ in Eq. (4.7) gives (4.5).

3. (iii) Let $S_{\gamma }$ be a surface of revolution of parabolic type. Without loss of generality, we may parametrize it as in (4.3): $S_{\gamma }=S_{\mathcal {P}}$. The tangent plane is now spanned by

   \[ \frac{\partial S_{\mathcal{P}}}{\partial t} = \left(\left(1+\frac{\theta^2}{2}\right)\dot{x} -\frac{\theta^2}{2}\dot{y}, \frac{\theta^2}{2}\dot{x}+\left(1-\frac{\theta^2}{2}\right) \dot{y},\dot{z},\theta\dot{y}-\theta\dot{x}\right) \]
   and
   \[ \frac{\partial S_{\mathcal{P}}}{\partial \theta} = \Big(\theta (x- {y}),\theta (x- {y}),0,y-x \Big). \]
   Thus, the first fundamental form of $S_{\mathcal {P}}$ becomes
   \[ \mathrm{d} s^2 = (-\dot{x}^2+\dot{y}^2+\dot{z}^2)\mathrm{d} t^2+(x-y)^2\mathrm{d}\theta^2 = \Vert\dot{\gamma}\Vert^2\mathrm{d} t^2+(x-y)^2\mathrm{d}\theta^2. \]

   The unit normal $\xi$ of $S_{\mathcal {P}}$ is given by

   \begin{align*} \xi & = \left[ \frac{\theta^2}{2}(x\dot{z}-\dot{x}z) -\left(1+\frac{\theta^2}{2}\right)(y\dot{z}-\dot{y}z)\right] \frac{\mathbf{e}_x}{r\Vert\dot{\gamma}\Vert} \nonumber \\ & \quad -\left[\frac{\theta^2}{2}(y\dot{z}-\dot{y}z) +\left(1-\frac{\theta^2}{2}\right)(x\dot{z}-\dot{x}z) \right]\frac{\mathbf{e}_y}{r\Vert\dot{\gamma}\Vert}\nonumber \\ & \quad + \Big(x\dot{y}-\dot{x}y\Big)\frac{\mathbf{e}_z}{r\Vert\dot{\gamma}\Vert}- \theta\left[(x-y)\dot{z}-(\dot{x}-\dot{y})z\right]\frac{\mathbf{e}_w}{r\Vert\dot{\gamma}\Vert}. \end{align*}
   Let us compute the second fundamental form's coefficients $h_{ij}$. The second derivatives of $S_{\mathcal {P}}$ are
   \begin{align*} & \frac{\partial^2 S_{\mathcal{P}}}{\partial t^2} = \left(\left(1+\frac{\theta^2}{2}\right)\ddot{x}-\frac{\theta^2}{2}\ddot{y},\quad \frac{\theta^2}{2}\ddot{x}+\left(1-\frac{\theta^2}{2}\right)\ddot{y}, \ddot{z},\theta\ddot{y}-\theta\ddot{x}\right),\\ & \frac{\partial^2 S_{\mathcal{P}}}{\partial t\partial\theta} = (\dot{x}-\dot{y})\Big(\theta ,\theta ,0,-1 \Big), \quad \mbox{and} \quad \frac{\partial^2 S_{\mathcal{P}}}{\partial \theta^2} = \Big(x- {y},x- {y},0,0 \Big). \end{align*}
   Thus, we have $h_{12}=0$ and
   \begin{align*} h_{11} & ={-}\frac{1}{r\Vert\dot{\gamma}\Vert}\left[x(\ddot{y}\dot{z}-\dot{y}\ddot{z})-y(\ddot{x}\dot{z}-\dot{x}\ddot{z})+z(\ddot{x}\dot{y}-\dot{x}\ddot{y})\right],\\ h_{22} & ={-}(x-y)\frac{(x-y)\dot{z}-(\dot{x}-\dot{y})z}{r\Vert\dot{\gamma}\Vert}. \end{align*}
   Substitution of $h_{ij}$ and $g_{ij}$ in Eq. (4.7) gives (4.6).

5. Minimal surfaces of revolution

This section proves the first of our main results, theorem 5.3 below, concerning the characterization of minimal surfaces of revolution in hyperbolic space. (See figure 1.) To do that, we use proposition 4.3 to express the mean curvature in terms of the curvature $\kappa$ of the generating curve, proposition 5.2. Then, we compare the resulting expression with the curvatures of the extrinsic catenaries that appear in theorem 2.5.

Figure 1. Examples of minimal surfaces of revolution in the Poincaré ball model of $\mathbb {H}^3(1)$ obtained from the hyperboloid model (2.2) through stereographic projection $(x_0,x_1,x_2,x_3)\mapsto ({x_1}/{(1+x_0)},{x_2}/{(1+x_0)},{x_3}/{(1+x_0)})$. (In the figures, the purple spheres indicate the points in the ideal boundary of $\mathbb {H}^3(1)$.) The generating curves of the surfaces are found by numerically solving the differential equation obtained by equating the expressions for the curvature given in lemma 2.3 and theorem 2.5. (a) Elliptic type. (b) Parabolic type. (c) Hyperbolic type.

First, we need an expression for the curvature of a generic curve in $\mathbb {H}^2(r)$.

Lemma 5.1 The curvature $\kappa$ of $\gamma (t)=(x(t),y(t),z(t))$ in $\mathbb {H}^2(r)$ is given by

(5.1)\begin{equation} \kappa ={-}\frac{x(\ddot{y}\dot{z}-\dot{y}\ddot{z})-y(\ddot{x}\dot{z}-\dot{x}\ddot{z}) +z(\ddot{x}\dot{y}-\dot{x}\ddot{y})}{r\Vert\dot{\gamma}\Vert^3}. \end{equation}

Proof. The unit normal of the hyperbolic plane $\mathbb {H}^2(r)$ seen as a surface of $\mathbb {E}_1^3$ is given by $N= \gamma /r$. Thus, it follows that we may express the curvature $\kappa$ of $\gamma (t)=(x(t),y(t),z(t))$ in $\mathbb {H}^2(r)$ as

(5.2)\begin{equation} \kappa = \frac{\langle N\times_1\dot{\gamma},\nabla_{\dot{\gamma}} \dot{\gamma}\rangle}{\Vert\dot{\gamma}\Vert^3} = \frac{\langle\gamma\times_1\dot{\gamma},\ddot{\gamma}\rangle_1}{r\Vert\dot{\gamma}\Vert^3}, \end{equation}

where the vector product of $V=(v_x,v_y,v_z)$ with $W=(w_x,w_y,w_z)$ in $\mathbb {E}_1^3$ is defined as $V\times _1 W=(-v_yw_z+v_zw_y,-v_xw_z+v_zw_x,v_xw_y-v_yw_x)$. A straightforward computation shows that $\gamma \times _1\dot {\gamma } = (\dot {y}z-y\dot {z},\dot {x}z-x\dot {z},x\dot {y}-\dot {x}y)$. Therefore,

\[ \langle\gamma\times_1\dot{\gamma},\ddot{\gamma}\rangle_1 ={-}x(\ddot{y}\dot{z}-\dot{y}\ddot{z})+y(\ddot{x}\dot{z}-\dot{x}\ddot{z}) -z(\ddot{x}\dot{y}-\dot{x}\ddot{y}). \]

Substitution of $\langle \gamma \times _1\dot {\gamma },\ddot {\gamma }\rangle _1$ in Eq. (5.2) proves (5.1).

Proof. Let $\gamma (t)=\psi (u(t),v(t))$ be the generating curve of a surface of revolution. We calculate the right-hand sides of Eqs. (5.3)–(5.5). Using the parametrization $\psi$ given in (2.6), we can write

\[ \gamma(t)=r\left(\cosh\frac{u(t)}{r}\cosh\frac{v(t)}{r}, \cosh\frac{u(t)}{r}\sinh\frac{v(t)}{r},\sinh\frac{u(t)}{r}\right). \]

Then

(5.6)\begin{align} \frac{x\dot{y}-\dot{x}y}{rz\Vert\dot{\gamma}\Vert} & = \frac{\dot{v}\cosh^2 \frac{u}{r}}{r\Vert\dot{\gamma}\Vert\sinh\frac{u}{r}}, \end{align}

(5.7)\begin{align} \frac{y\dot{z}-\dot{y}z}{rx\Vert\dot{\gamma}\Vert} & = \frac{\dot{u}\sinh \frac{v}{r}-\dot{v}\cosh\frac{v}{r}\cosh\frac{u}{u}\sinh \frac{u}{r}}{r\Vert\dot{\gamma}\Vert\cosh\frac{v}{r} \cosh\frac{u}{r}}, \end{align}

(5.8)\begin{align} \frac{(x-y)\dot{z}-(\dot{x}-\dot{y})z}{r(x-y)\Vert\dot{\gamma}\Vert} & = \frac{\dot{u}+ \dot{v}\cosh\frac{u}{r}\sinh\frac{u}{r}}{r\Vert\dot{\gamma} \Vert\cosh\frac{u}{r}}. \end{align}

Finally, we conclude the proof using the expressions for the curvature $\kappa$ of an extrinsic catenary given in theorem 2.5.

6. Hyperbolic horocatenary

In this section, we consider a variation of the hanging chain problem introduced by López [Reference López5] where one measures the distance to a reference geodesic by using the so-called horocycle distance [Reference Izumiya4]. In other words, we shall employ $d_h(\ell,p)$ defined as the distance measured along the horocycle passing through $p$ and orthogonal to the geodesic $\ell$. The solutions to this problem will be called hyperbolic horocatenary [Reference López5]. Using horocatenaries will allow us to provide an intrinsic characterization of the extrinsic catenaries of the elliptic type, theorem 6.1.

As done for hyperbolic catenaries, we first introduce a coordinate system adapted to the problem. In the hyperboloid model, horocycles correspond to the curves given as the intersection of $\mathbb {H}^2(r)$ with lightlike planes of $\mathbb {E}_1^3$ not passing through the origin [Reference Spivak8]. (Thus, horocycles are implicitly written as Euclidean parabolas.) Alternatively, we may describe horocycles as the orbits of the one-parameter subgroup of parabolic rotations [Reference do Carmo and Dajczer2], i.e., rotations induced in $\mathbb {H}^2(r)$ by the orthogonal transformations of $\mathbb {E}_1^3$ that leave a lightlike plane $P^2$ pointwise fixed. As a concrete example, consider the plane $P^2=[\mathbf {e}_x+\mathbf {e}_y,\mathbf {e}_z]$, where $\{\mathbf {e}_x,\mathbf {e}_y,\mathbf {e}_z\}$ is the canonical basis given by the unit velocity vectors associated with the Cartesian coordinates $(x,y,z)$. The orthogonal transformations of $\mathbb {E}_1^3$ that leave $P^2$ pointwise fixed correspond to a lightlike rotation $L_{\theta }$ with axis $(1,1,0)$ and whose matrix is given by [Reference Nešović7]

(6.1)\begin{equation} L_{\theta} = \left( \begin{array}{ccc} 1+\dfrac{\theta^2}{2} & -\dfrac{\theta^2}{2} & -\theta \\ \dfrac{\theta^2}{2} & 1-\dfrac{\theta^2}{2} & -\theta \\ -\theta & \theta & 1 \end{array} \right). \end{equation}

Now, let $\ell (v)$ be the geodesic

(6.2)\begin{equation} v \mapsto \ell(v) = r \left(\cosh\frac{v}{r},\sinh\frac{v}{r},0\right). \end{equation}

The horocycles orthogonal to $\ell$ are given by $u\mapsto L_{u/r}(\ell (v))$. We may then parametrize $\mathbb {H}^2(r)$ as $\phi (u,v) = L_{u/r}(\ell (v))$, which gives

(6.3)\begin{equation} \phi= r\left(\begin{array}{c} \left(1+\dfrac{u^2}{2r^2}\right)\cosh\dfrac{v}{r}-\dfrac{u^2}{2r^2}\sinh\dfrac{v}{r}\\ \dfrac{u^2}{2r^2}\cosh\dfrac{v}{r}+\left(1-\dfrac{u^2}{2r^2}\right)\sinh\dfrac{v}{r}\\ \dfrac{u}{r}\left(\sinh\dfrac{v}{r}-\cosh\dfrac{v}{r}\right) \end{array}\right). \end{equation}

Alternatively, using the identity $\cosh {v}/{r}-\sinh {v}/{r}=\mathrm {e}^{- {v}/{r}}$, we may rewrite the parametrization as

(6.4)\begin{equation} \phi(u,v) = r\left(\cosh\frac{v}{r}+\frac{u^2}{2r^2}\mathrm{e}^{- {v}/{r}}, \sinh\frac{v}{r}+\frac{u^2}{2r^2}\mathrm{e}^{- {v}/{r}}, -\frac{u}{r}\,\mathrm{e}^{- {v}/{r}} \right). \end{equation}

The tangent vectors of the parametrization $\phi$ are

(6.5)\begin{align} \phi_u = \mathrm{e}^{- {v}/{r}}\left(\frac{u}{r},\frac{u}{r}, -1\right),\quad \phi_v = \left(\sinh\frac{v}{r}-\frac{u^2}{2r^2}\mathrm{e}^{- {v}/{r}}, \cosh\frac{v}{r}-\frac{u^2}{2r^2}\mathrm{e}^{- {v}/{r}}, \frac{u}{r}\,\mathrm{e}^{- {v}/{r}}\right). \end{align}

Then, the coefficients $g_{ij}$ of the metric are

\begin{align*} g_{11} & = \mathrm{e}^{- {2v}/{r}},\\ g_{12} & = \mathrm{e}^{- {v}/{r}}\left(-\frac{u}{r}\sinh\frac{v}{r} +\frac{u^3}{2r^3}\mathrm{e}^{- {v}/{r}}+\frac{u}{r}\cosh\frac{v}{r}- \frac{u^3}{2r^3}\mathrm{e}^{- {v}/{r}}-\frac{u}{r}\mathrm{e}^{- {v}/{r}}\right)=0 \end{align*}

and

\begin{align*} g_{22}& ={-}\sinh^2\frac{v}{r}+\frac{u^2}{r^2}\mathrm{e}^{- {v}/{r}} \sinh\frac{v}{r}-\frac{u^4}{4r^4}\mathrm{e}^{- {2v}/{r}} +\cosh^2\frac{v}{r}-\frac{u^2}{r^2}\mathrm{e}^{- {v}/{r}}\cosh\frac{v}{r} \\ & \quad + \frac{u^4}{4r^4}\mathrm{e}^{- {2v}/{r}}+\frac{u^2}{r^2}\,\mathrm{e}^{- {2v}/{r}} = 1. \end{align*}

In short, the metric of $\mathbb {H}^2(r)$ in the coordinates system $\phi (u,v)$ is

(6.6)\begin{equation} g = \mathrm{e}^{- {2v}/{r}}\,\mathrm{d} u^2+\mathrm{d} v^2. \end{equation}

This coordinate system is similar to the semi-geodesic coordinate system. From now on, we shall refer to $\phi$ as the horo-geodesic parametrization. Note that the coordinate curves $v\mapsto \phi (u,v)$ are geodesics: $\phi _{vv}= ({1}/{r})\phi \Rightarrow \nabla _{\phi _v}\phi _v=0$. On the other hand, the coordinate curves $u\mapsto \phi (u,v)$ are horocycles.

By construction, the coordinate horocycles are orthogonal to the geodesic $\ell (v)=\phi (0,v)$. Thus, the horocycle distance can be computed as

(6.7)\begin{equation} d_h(\phi(u,v),\ell) = \int_0^u \mathrm{e}^{- {v}/{r}}\,\mathrm{d} t = u\,\mathrm{e}^{- {v}/{r}}. \end{equation}

Note that $d_h$ is nothing but the height of $\phi$ with respect to the $xy$-plane. Therefore, we can provide an intrinsic characterization of extrinsic catenary of spherical type

Remark 6.2 The horo-geodesic coordinates are analogous to the semi-geodesic coordinates after we exchange $u$ and $v$. Thus, the Christoffel symbols are

\[ \Gamma_{11}^1 = \frac{F_u}{F}=0,\quad \Gamma_{11}^2 ={-}FF_v=\frac{1}{r}\mathrm{e}^{- {2v}/{r}},\quad \Gamma_{12}^1=\frac{F_v}{F}={-}\frac{1}{r},\quad \Gamma_{12}^2=0,\quad \Gamma_{22}^i=0, \]

where $F=\mathrm {e}^{- {v}/{r}}$. The curvature of $\gamma (t)=\phi (u(t),v(t))$ in $\mathbb {H}^2(r)$ is given by

(6.8)\begin{equation} \kappa =\frac{\ddot{u}\dot{v}-\dot{u}(\ddot{v}+ ({\dot{u}^2}/{r})\mathrm{e}^{- {2v}/{r}}+ ({2\dot{v}^2}/{r}))}{\mathrm{e}^{{v}/{r}}(\mathrm{e}^{- {2v}/{r}}\dot{u}^2+\dot{v}^2)^{{3}/{2}}}. \end{equation}