text:
\section{Introduction and motivation}\label{s1}

 A non-degenerate surface in $3$-dimensional de Sitter space $\mathbb S_1^3$ is called a {\it minimal surface} if the mean curvature $H$ vanishes on the surface.  Although the equation $H=0$ is equivalent that the trace of the second fundamental form vanishes on the surface,  there is no a    variational interpretation of these surfaces such as it occurs in  Euclidean space. For spacelike surfaces of $\mathbb S_1^3$, the induced metric is Riemannian and it makes sense to define the area of a surface. In this case, a   surface of $\mathbb S_1^3$ with zero constant mean curvature is a   local maximizer of the area functional. If the surface is timelike, the induced metric has index one and there is not a notion of the area of a timelike surface.  Despite these differences with the Euclidean case, and in order to simplify the statements our results,  the terminology of minimal surfaces  to name a non-degenerate surface of $\mathbb S_1^3$ with zero constant mean curvature   is kept in this article.

An interesting class of surfaces of $\mathbb S_1^3$ are the surfaces of revolution, also called, rotational surfaces. The family of surfaces of revolution in $\mathbb S_1^3$  is richer than the Euclidean setting because the rotational axis can be of three different causal character.  If, in addition, we investigate which of these surfaces are minimal, we need to distinguish  if the surface is spacelike and timelike. The  property that the surface is rotational implies that the equation $H=0$  is an ordinary differential equation given  in terms of the generating curve of the surface. This makes that the classification of rotational minimal surfaces of $\mathbb S_1^3$ can be obtained:  \cite{ak,li1,ll1,mo}; see also  \cite{br,li2,ya}. Recently,   rotational surfaces in $\mathbb S_1^3$ with a Weingarten relation between its curvatures have been investigated  because the Weingarten relation is described again  in terms of   ordinary differential equations \cite{de,ll2}.

If the classification of rotational minimal surfaces of $\mathbb S_1^3$ is deduced by solving ordinary differential equations, there is not known a special property of the generating curves. This contrasts to the Euclidean situation which it is  recalled giving a brief historical account.  Euler proved in 1744 that the catenoid is the only non-planar rotational minimal surface in  $\mathbb R^3$ being the catenary its generating curve.  This curve is also  the solution of the question that asks by the shape of a chain hanging by its own weight. This problem  attracted the interest of many scientistics some centuries above, beginning with Galileo. The answer, the catenary, was obtained some decades before to the Euler's result joining efforts of Hooke, Leibniz, Huygens and Bernouilli, among others.

Coming back to the de Sitter space, and motivated by the property  that  the generating curve of the catenoid is variationally  characterized as the shape of a hanging chain, the purpose of this paper is to answer the following

\begin{quote}
{\it Question:} Can the generating curves of rotational minimal surfaces of  $\mathbb S_1^3$ be characterized from a variational viewpoint?
\end{quote}

   In $\mathbb R^2$, the problem of a hanging chain is equivalent to find a curve that minimizes the center of mass where the weight of the curve  is measured with respect to a straight-line of $\mathbb R^2$. The variational problem is posed among all curves of $\mathbb R^2$ with the same endpoints and the same length. Here the curve is viewed as an ideal incompressible chain of constant density.  In the $2$-dimensional  de Sitter space $\mathbb S_1^2$,  the   {\it catenary problem} asks what is the shape of a hanging chain in $\mathbb S_1^2$. It is then necessary to precise what it is the meaning of the `weight' of a curve of $\mathbb S_1^2$ indicating what is the reference to measure the center of mass.  If  we are able to solve this problem, the solution curve will be the analog catenary in $\mathbb S_1^2$. Now the natural way to proceed is to immerse $\mathbb S_1^2$ in the $3$-dimensional de Sitter space $\mathbb S_1^3$, and construct the surface of revolution whose generating curve is the above catenary. It is natural to ask if the mean curvature of this surface vanishes constantly on the surface.

 A first problem that  appears is that we need to distinguish if   curves in $\mathbb S_1^2$ and surfaces in $\mathbb S_1^3$ are  spacelike or timelike.  A second  problem is how to define the center of mass  of a curve in $\mathbb S_1^2$. Here we need to precise the model of the de Sitter space. Let $\mathbb L^3$ be the Lorentz-Minkowski space, that is, the vector space $\mathbb R^3$ equipped with the Lorentzian metric $\langle,\rangle=(dx)^2+(dy)^2-(dz)^2$, where $(x,y,z)$ stand for the canonical coordinates of $\mathbb R^3$.  The $2$-dimensional  de Sitter space is defined as the set  $\mathbb S_1^2=\{(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1\}$ endowed with the induced metric from $\mathbb L^3$. The de Sitter space $\mathbb S_1^2$ is a Lorentzian surface of constant curvature equal to $1$.

 In order to measure the center of mass of a curve of $\mathbb S_1^2$, we fix a reference plane $\Pi$ of $\mathbb L^3$ and the center of mass is measured using the distance to $\Pi$. Since in $\mathbb L^3$ there are three types of planes  depending on the causal character, the plane $\Pi$ can be  spacelike, timelike and degenerate.  Each one of these planes separates $\mathbb S_1^2$ in two domains and let us fix one of these domains, which it is denoted by $(\mathbb S_1^2)^+$. Let   $\gamma\colon [a,b]\to (\mathbb S_1^2)^+$, $\gamma=\gamma(t)$,  be a non-degenerate curve.  Assuming that the density is constantly $1$, the weight of $\gamma$ is defined by
$\int_a^b d(t)\, ds,$
where $d(t)$ is the distance between $\gamma(t)$ and $\Pi$ and $ds$ is the arc-length of $\gamma$. Thus the center of mass of $\gamma$ is
$$\frac{\int_a^b d(t)\, ds}{\int_a^b \, ds}=\frac{\int_a^b d(t)|\gamma'(t)|\, dt}{\int_a^b |\gamma'(t)|\, dt},$$
where $|\gamma'(t)|=\sqrt{|\langle \gamma'(t),\gamma'(t)\rangle|}$. The variational problem posed in $\mathbb S_1^2$ consists to find the curve that minimizes the center of mass in the class of  all curves of $\mathbb S_1^2$ with prescribed endpoints and fixed length. Introducing a Lagrange multiplier due to the length constraint, the energy to minimize is
\begin{equation}\label{ee}
\mathcal{E}_\Pi[\gamma]=\int_a^b (d(t)+\lambda)|\gamma'(t)|\, dt.
\end{equation}

 \begin{definition} \label{ds1}
A critical point of $\mathcal{E}_\Pi$  is called a catenary with respect to the plane $\Pi$.
\end{definition}

 The following  step in our investigation  consists in   constructing a surface of revolution in $\mathbb S_1^3$ having a catenary as generating curve. For this, we immersed $\mathbb S_1^2$ into the $3$-dimensional de Sitter space $\mathbb S_1^3$ as follows. Let   $\mathbb L^4$ be the $4$-dimensional Lorentz-Minkowski space where the metric is $(dx_1)^2+(dx_2)^2-(dx_3)^2+(dx_4)^2$ and $(x_1,x_2,x_3,x_4)$ are Cartesian coordinates of $\mathbb R^4$. The $3$-dimensional de Sitter space $\mathbb S_1^3$ is defined by
 $$\mathbb S_1^3=\{p=(x_1,x_2,x_3,x_4)\in\mathbb L^4:\langle p,p\rangle=1\}$$
 endowed with the induced metric from $\mathbb L^4$.  The space $\mathbb L^3$ is immersed in $\mathbb L^4$ by $(x,y,z)\mapsto (x,y,z,0)$. In particular, the $2$-dimensional de Sitter space $\mathbb S_1^2$ is immersed in $\mathbb S_1^3$ as $\mathbb S_1^2\times\{0\}$.   The steps to follow are now clear:
  \begin{enumerate}
  \item Let $\gamma$ be a catenary of $\mathbb S_1^2$ with respect to a plane $\Pi$.
  \item Let $L$ be the geodesic obtained by intersecting $\mathbb S_1^2$ with $\Pi$, $L=\mathbb S_1^2\cap\Pi$.
  \item Let $L\equiv L\times\{0\}\subset\mathbb S_1^3$.  Consider $\mathcal{G}$ the one-parameter group of rotations of $\mathbb S_1^3$ that leave pointwise fixed $L$.
  \item Rotate $\gamma\equiv\gamma\times\{0\}$ about $\mathcal{G}$, obtaining a surface of revolution $S_\gamma$ of $\mathbb S_1^3$.
  \item Compute the mean curvature $H$ of $S_\gamma$ and investigate if $H$ is $0$.
  \end{enumerate}

 As a consequence of all this work, the main result can be stated as follows.

\begin{theorem}\label{t-main} Let $M$ be a non-degenerate surface of revolution in $\mathbb S_1^3$ whose rotation axis is $L=\mathbb S_1^2\cap \Pi$.  Then $M$ is a minimal surface if and only if its generating curve is a catenary with respect to $\Pi$.
\end{theorem}

This theorem provides a variational characterization of the generating curves of rotational minimal surfaces and it   shows that our approach to catenaries  in the de Sitter space  extends the classical notions of Euclidean catenary and catenoid.

 The proof of Theorem \ref{t-main} is separated in three cases depending in   the three types of surfaces of revolution of $\mathbb S_1^3$. Within a more general context, these surfaces are  described in \cite{cd}. To be precise, fix a $2$-plane $P^2\subset \mathbb L^4$ and consider the one-parameter group $\mathcal{G}$ of isometries of $\mathbb L^4$ that leaves $P^2$  pointwise fixed. Let $P^3$ be a $3$-dimensional subspace of $\mathbb L^4$ such that $P^2\subset P^3$ and consider $\gamma$  a non-degenerate curve in $\mathbb S_1^3\cap (P^3-P^2)$. The surface  $S_\gamma$  obtained by acting $\mathcal{G}$ on $\gamma$   is called a surface of revolution of $\mathbb S_1^3$ and $\gamma$ is called its generating curve. This surface is said to be spherical (resp. hyperbolic, parabolic) if the induced metric on $P$ from $\mathbb L^4$ is Riemannian (resp. Lorentzian, degenerate).  Explicit expressions of the groups $\mathcal{G}$ and of the parametrizations of the rotational surfaces will be described in the following sections: spherical case (Section \ref{sec2}), hyperbolic case (Section \ref{sec3}) and parabolic case (Section \ref{sec4}).

Following the above scheme, in each one of the three sections, it is defined the energy $\mathcal{E}_\Pi$ which depends on the distance to the reference plane $\Pi\subset\mathbb S_1^2$. Catenaries will be obtained as solutions of the Euler-Lagrange equation of these energies (Theorems \ref{t11}, \ref{t21} and \ref{t41}). Next, catenaries will be also characterized   as solutions of a prescribing curvature equation involving the distance to the plane $\Pi$ and the angle that make the principal normal vector of the curve with the unit vector field orthogonal to $\Pi$ (Theorems \ref{t12}, \ref{t22}  and \ref{t42}). Finally and answering to the question proposed in this paper, we prove Theorem \ref{t-main} in each one of these cases (Theorems \ref{t13}, \ref{t23}  and \ref{t43}).  A last Section \ref{sec5} will investigate the concept of intrinsic catenary where the center of mass is calculated using the (intrinsic) distance of $\mathbb S_1^2$.

\section{The catenary problem: spherical case}\label{sec2}

The first case to investigate are the catenaries of $\mathbb S_1^2$ where the center of mass is measured with respect to a spacelike plane. Without loss of generality,  we can assume that this plane is the plane $\Pi_{xy}$  of equation $z=0$. The catenary problem  consists into find the shape of a curve contained in $(\mathbb S_1^2)_{xy}^+=\{(x,y,z)\in \mathbb S_1^2:z>0\}$  whose shape is obtained as a critical point of the center of mass with respect to $\Pi_{xy}$.  The distance of a point $(x,y,z)\in(\mathbb S_1^2)_{xy}^+$ to $\Pi$ is $z$.

In order to obtain a manageable expression of this energy, let  us parametrize the de Sitter space  $\mathbb S_1^2$   by $\Psi\colon\mathbb R^2\to\mathbb S_1^2$, where
$$\Psi(u,v)=(\cosh(u)\cos{v},\cosh(u)\sin{v},\sinh(u)).$$
Let $\gamma\colon [a,b]\to(\mathbb S_1^2)_{xy}^+$ be a non-degenerate curve  and let  $\gamma(t)=\Psi(u(t),v(t))$. Then $\langle\gamma'(t),\gamma'(t)\rangle=v'^2\cosh(u)^2-u'^2$. If $\gamma$ is spacelike (resp. timelike) this number is positive (resp. negative). Along this paper, we will write   $\epsilon(v'^2\cosh(u)^2-u'^2)>0$ where $\epsilon=1$ (resp.  $\epsilon=-1$) if $\gamma$ is spacelike (resp. timelike). Then $|\gamma'(t)|=\sqrt{\epsilon(v'^2\cosh(u)^2-u'^2)}$ and $|\gamma'(t)|^2=\epsilon\langle\gamma'(t),\gamma'(t)\rangle$.

The distance of $\gamma(t)$ to $\Pi_{xy}$ is $\sinh u(t)$. According to \eqref{ee},  the energy of the  catenary problem with respect to the plane $\Pi_{xy}$ is
\begin{equation*}
\mathcal{E}_{xy}[\gamma] =\int_a^b (\sinh(u)+\lambda) \sqrt{\epsilon(v'^2\cosh(u)^2-u'^2)}\, dt,
\end{equation*}
where $\lambda$ is a Lagrange multiplier due to the length of the curve being constant.

We will see that the critical points of $\mathcal{E}_{xy}$ will be expressed in terms of the curvature $\kappa$ of $\gamma$. Here $\kappa$ is the curvature of $\gamma$  viewed as a curve in the space $\mathbb S_1^2$, or in other words, $\kappa$ is the geodesic curvature of $\gamma$. For the rest of  computations of this paper, it is necessary to have an expression of $\kappa$ when $\gamma$ is written as $\gamma(t)=\Psi(u(t),v(t))$.

\begin{proposition} Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $\mathbb S_1^2$. Then its (geodesic) curvature $\kappa$ is
\begin{equation}\label{eqk}
\kappa=\epsilon
\frac{v'(v'^2\sinh(u)\cosh(u)^2-2u'^2\sinh(u))-\cosh(u)(u'v''-v'u'')}{|\gamma'(t)|^3}.
\end{equation}
\end{proposition}

\begin{proof} Let $\xi(p)=-p$, $p\in \mathbb S_1^2$, be the unit normal vector of $\mathbb S_1^2$. Then
$$\kappa=\epsilon\frac{\langle\gamma'',\gamma'\times \xi(\gamma)\rangle}{|\gamma'(t)|^3}=\epsilon\frac{\mbox{det}(\gamma,\gamma',\gamma'')}{|\gamma'(t)|^3},$$
and \eqref{eqk} follows by a straightforward calculation.
\end{proof}

For further purposes, we need the following computations. First,
\begin{equation}\label{k1}
\begin{split}\frac{d}{dt}\left(\frac{u'}{|\gamma'|}\right)&=\frac{\epsilon v'\cosh(u)}{|\gamma'|^3}(-u'^2v'\sinh(u)+\cosh(u)(v'u''-u'v''))\\
&=\frac{\epsilon v'\cosh(u)}{|\gamma'|^3}(u'^2v'\sinh(u)-v'^3\sinh(u)\cosh(u)^2+\epsilon \kappa|\gamma'|^3)\\
&=v'\cosh(u)\left(\kappa-\frac{v'\sinh(u)}{|\gamma'|}\right),
\end{split}
\end{equation}
where in the second identity we have replaced the expression $v'u''-u'v''$ in terms of $\kappa$ thanks to \eqref{eqk}. A similar computation yields
\begin{equation}\label{k2}
\begin{split}\frac{d}{dt}\left(\frac{v'\cosh(u)}{|\gamma'|}\right)&=u'\left(\kappa-\frac{v'\sinh(u)}{|\gamma'|}\right)\end{split}
\end{equation}

\begin{theorem} \label{t11}
Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{xy}^+$. Then $\gamma$ is a  catenary with respect to $\Pi_{xy}$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c11}
\kappa=-\frac{v'\cosh(u)^2}{(\sinh(u)+\lambda)|\gamma'|}.
\end{equation}
\end{theorem}

\begin{proof} The Euler-Lagrange equations of $\mathcal{E}_{xy}$ are calculated using
 \begin{equation}\label{sel}
\frac{\partial J}{\partial u}-\frac{d}{dt} \left(\frac{\partial J}{\partial u'}\right)=0,\quad
 \frac{\partial J}{\partial v}-\frac{d}{dt} \left(\frac{\partial J}{\partial v'}\right)=0,
\end{equation}
where $J=J[u,v]$ is the integrand of $\mathcal{E}_{xy}$,
$$J [u,v]=(\sinh(u)+\lambda) \sqrt{\epsilon(v'^2\cosh(u)^2-u'^2)}.$$
Equations \eqref{sel} are, respectively,
$$\frac{ v'^2\cosh(u)\left(\cosh(u)^2+\sinh(u)(\sinh(u)+\lambda)\right)}{|\gamma'|}+(\sinh(u)+\lambda)\frac{d}{dt}\left(\frac{u'}{|\gamma'|}\right)=0.$$
$$\left(\cosh(u)^2+\sinh(u)(\sinh(u)+\lambda)\right)\cosh(u)\frac{u'v'}{|\gamma'|}+\sinh(u)\cosh(u) \frac{d}{dt}\left(\frac{v'\cosh(u)}{|\gamma'|}\right)=0.$$
 Using \eqref{k1} and \eqref{k2},  we obtain
$$v'\cosh(u)\left(\frac{v'\cosh(u)^2}{|\gamma'|}+\kappa(\sinh(u)+\lambda)\right)=0.$$
$$u'\cosh(u)\left(\frac{v'\cosh(u)^2}{|\gamma'|}+\kappa(\sinh(u)+\lambda)\right)=0.$$
This yields the result because $u'$ and $v'$ cannot simultaneously vanish.
\end{proof}

\begin{remark} The case $v'=0$ is equivalent to $\kappa=0$ and $\gamma$ is a meridian $\gamma(t)=\Psi(t,v_0)$ for some constant $v_0\in\mathbb R$. This situation also occurs in the Euclidean case: see Remark \ref{rem2} below.
\end{remark}

Notice that the Lagrangian $J[u,v]$ does not depend on $v$. Thus the second equation of \eqref{sel} gives a first integration of $\gamma$, namely,
$$\frac{\partial J}{\partial v}=\frac{\epsilon v'\cosh(u)^2(\sinh(u)+\lambda)}{|\gamma'|}=c,$$
for some non-zero constant $c\in\mathbb R$. Without loss of generality, we can assume that $\gamma(t)=\Psi(u(t),t)$. From the above equation, we deduce
\begin{equation}\label{first}
u(t)=m\pm\frac{1}{c}\int^t\cosh(u)\sqrt{c^2-\epsilon\cosh(u)^2(\sinh(u)+\lambda)}\, dt,\quad c,m\in\mathbb R.
\end{equation}

We characterize the   catenaries with respect to $\Pi_{xy}$  in terms of the angle that makes the principal normal vector ${\bf n}$ of $\gamma$ with a vector field of $\mathbb L^3$.  The vector $\mathbf{n}$ is understood as a unitary tangent  of  $\mathbb S_1^2$ at $\gamma$ which is orthogonal to $\gamma'$.  On the other hand, the vector field of $\mathbb L^3$ is the unit vector field orthogonal to the plane $\Pi_{xy}$. Define  $Z=\partial_z\in\mathfrak{X}(\mathbb L^3)$.

\begin{theorem}\label{t12} Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{xy}^+$. Then $\gamma$ is a catenary with respect to $\Pi_{xy}$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c12}
\kappa(t)=\frac{\langle{\bf n}(t), Z\rangle}{d_{xy}(t)+\lambda},
\end{equation}
where $d_{xy}(t)$ is the distance of $\gamma(t)$ to the plane $\Pi_{xy}$.
\end{theorem}
\begin{proof}
Since the unit normal vector $\xi$ to $(\mathbb S_1^2)_{xy}^+$ along $\gamma$ is $\xi(\gamma)=-\gamma$,   the principal normal vector ${\mathbf n}$ is
$${\bf n}=\frac{ \xi(\gamma)\times \gamma'}{|\gamma'|}=-\frac{\gamma\times\gamma'}{|\gamma'|}.$$
Thus
$$\langle{\bf n}(t),Z\rangle=-\frac{\mbox{det}(\gamma,\gamma',\partial_z)}{ |\gamma'|}= -\frac{v'\cosh(u)^2}{|\gamma'|}.$$
Since $d_{xy}=\sinh(u)$, it follows \eqref{c12} from \eqref{c11}.
\end{proof}

\begin{remark}\label{rem2}
Equation \eqref{c12} is analogous to the Euclidean case. In the catenary problem in $\mathbb R^2$,    the reference line is assumed to be the $x$-axis. The energy to minimize is   $[y]\mapsto \int_a^b (y+\lambda)\sqrt{1+y'^2}\, dx$ for curves $y=y(x)$, $x\in [a,b]$, $y(x)>0$. A critical point $y$ is characterized by the equation
\begin{equation}\label{euclideo}
\frac{y''}{1+y'^2}=\frac{1}{y+\lambda},
\end{equation}
 whose solution   is the catenary
\begin{equation}\label{cat}
y(x)=\frac{1}{c}\cosh(cx+a)-\lambda,\quad a,c\in\mathbb R, c>0.
\end{equation}
 The curvature of $y(x)$ is $y''/(1+y'^2)^{3/2}$ and the principal normal vector $\mathbf{n}$ is $(-y',1)/\sqrt{1+y'^2}$. If $W(x,y)=\partial_y$ is the unit vector field of $\mathbb R^2$ in the direction of the gravity,  then equation \eqref{euclideo} is equivalent to
$$\kappa(x)=\frac{\langle \mathbf{n}(x),W\rangle}{d(x)+\lambda},$$
where $d(x)=y$ is the distance of $(x,y(x))$ to the $x$-axis. Notice that vertical straight-lines of $\mathbb R^2$ are solutions of this equation because $\kappa=0$ and $\langle{\bf n},W\rangle=0$.
\end{remark}

Once the catenaries of $\mathbb S_1^2$ with respect to the plane $\Pi_{xy}$ have been established, we  answer to the Question posed in the Introduction. We will rotate a catenary curve with respect to the plane $\Pi_{xy}$ about the geodesic $L_{xy}=\Pi_{xy}\cap\mathbb S_1^2$.  For this, let us immerse  $\mathbb S_1^2$ into $\mathbb S_1^3$ considering a Riemannian factor in the fourth coordinate such as it was explained in Section \ref{s1}.    Consider the geodesic $L_{xy}\subset\mathbb S_1^2\times\{0\}\subset\mathbb S_1^3$, which will be the rotation axis. The one-parametric group of rotations of $\mathbb S_1^3$ that leave pointwise the geodesic $L_{xy}$ is $\mathcal{G}_{xy}=\{\mathcal{R}_s^{xy}:s\in\mathbb R\}$, where
 $$\mathcal{R}_s^{xy}= \left(\begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&\cosh(s)&\sinh(s)\\ 0&0& \sinh(s) &\cosh(s)\end{array}\right).$$
 Let $\gamma=\gamma(t)$ be a curve in $(\mathbb S_1^2)_{xy}^+\subset\mathbb S_1^3$. Let $S_\gamma^{xy}=\{\mathcal{R}^{xy}_s\cdot\gamma(t)\colon s\in\mathbb R, t\in [a,b]\}$ be the surface of revolution obtained by the orbit of $\gamma$ under the group $\mathcal{G}_{xy}$. Since the plane $\Pi_{xy}$ containing the rotation axis is spacelike, the surface of revolution is of spherical type according to the terminology of \cite{cd}.

 \begin{theorem}\label{t13}
  The surface of revolution $S_\gamma^{xy}$ is minimal if and only if $\gamma$ is a catenary  with respect to $\Pi_{xy}$  for the Lagrange multiplier $\lambda=0$.
 \end{theorem}

 \begin{proof}
 In order to simplify the arguments, we can assume without loss of generality that  $\gamma(t)=\Psi(u(t),t)$. Let us see $\gamma(t)\equiv  (\Psi(u(t),t),0)$ as a curve in $\mathbb S_1^3$.   A parametrization of $S_\gamma^{xy}$ is
 $$\mathbf{r}(t,s)=(\cosh(u)\cos(t),\cosh(u)\sin(t),\cosh(s)\sinh(u),\sinh(s)\sinh(u)).$$
 It is necessary the expression of  the mean curvature $H$ of $S_\gamma^{xy}$ in terms of the parametrization $\mathbf{r}$. As usually, let $\{E,F,G\}$ and $\{h_{11},h_{12},h_{22}\}$ be the coefficients of the first and second fundamental form of $S_\gamma^{xy}$ for the parametrization $\mathbf{r}$:
 $$E=\langle \mathbf{r}_{t},\mathbf{r}_{t}\rangle,\quad F=\langle \mathbf{r}_{t},\mathbf{r}_{s}\rangle,\quad G=\langle \mathbf{r}_{s},\mathbf{r}_{s}\rangle,$$
 $$h_{11}=\langle N,\mathbf{r}_{tt}\rangle,\quad h_{12}=\langle N,\mathbf{r}_{ts}\rangle,\quad h_{22}=\langle N,\mathbf{r}_{ss}\rangle.$$
 Then
 $$H=\frac{\delta}{2}\frac{Eh_{22}-2Fh_{12}+G h_{11}}{EG-F^2},$$
 where $\delta=1$ if $S_\gamma^{xy}$ is spacelike and $\delta=-1$ if $S_\gamma^{xy}$ is timelike. Notice that if $\epsilon=-1$, then necessarily $\delta=-1$, but if $\epsilon=1$, then $\delta$ may be $1$ or $-1$. In the proof of this theorem, and in the subsequent sections, the coefficient $F$ vanishes, so
 \begin{equation}\label{hh}
 H=\frac{\delta}{2}\frac{Eh_{22} +G h_{11}}{EG}.
 \end{equation}
 In fact, we will investigate when $S_{\gamma}^{xy}$ is minimal, so our interest focuses under what hypotheses on $\gamma$  we have $Eh_{22}+Gh_{11}$ is $0$ on the surface.

 The calculation of $N$ is obtained knowing that $N$ is not only orthogonal to $\mathbf{r}_{t}$ and $\mathbf{r}_{s}$, but also to $\mathbf{r}$ since $N$ is a tangent vector of $\mathbb S_1^3$.   A straightforward computations leads to,
  $$N=\frac{1}{|\gamma'|}\left(\begin{array}{l}\cos (t) \sinh(u)\cosh(u)-u'\sin (t)   \\  u' \cos (t)  +\sin(t)\sinh(u)\cosh(u) \\
  \cosh (s) \cosh(u)^2\\ \sinh (s) \cosh(u)^2\end{array}   \right).$$
  In particular, $\langle N,N\rangle=-\epsilon$. We also have
 $$E=  \cosh (u)^2-u'^2=\epsilon|\gamma'|^2,\quad G=\sinh(u)^2.$$
 On the other hand, the coefficients $h_{11}$ and $h_{22}$ of the second fundamental form are
 \begin{equation*}
 \begin{split}
 h_{11}&=\frac{-\cosh(u)u''+2 u'^2 \sinh(u)-\sinh(u) \cosh(u)^2}{|\gamma'|},\\
 h_{22}&=-\frac{\sinh(u) \cosh(u)^2}{|\gamma'|}.
 \end{split}
 \end{equation*}
 Then
$$
H=-\frac{\delta\sinh(u)}{4|\gamma'|EG}\left((1-3\cosh(2u))u'^2+\cosh(u)(\cosh(u)+\cosh(3u)+2\sinh(u)u'')\right).
$$
In this expression of $H$, the term $u''$ is replaced in function of the curvature $\kappa$. So, from  \eqref{eqk} we have
$$u''=\frac{\epsilon\kappa|\gamma'|^3-\sinh(u)\cosh(u)^2+2u'^2\sinh(u)}{\cosh(u)}.$$
Then
 \begin{equation}\label{h1}
 H=-\frac{\delta\epsilon\sinh(u)|\gamma'|}{2EG}\left(\cosh(u)^2+\kappa\sinh(u)|\gamma'|\right).
 \end{equation}
 Then $H=0$ if and only if $\cosh(u)^2+\kappa\sinh(u)|\gamma'|$, proving the result thanks to \eqref{c11}.
 \end{proof}

 \begin{remark} The fact that $\lambda$ must be $0$ in Theorem \ref{t13} in order to ensure that $S_\gamma^{xy}$ is   minimal is expectable because the same situation occurs in Euclidean space $\mathbb R^3$.
 If we immerse $\mathbb R^2$ into $\mathbb R^3$ by $(x,y)\mapsto (x,0,y)$ and if   the catenary $y(x)$ given in \eqref{cat} is rotated about the $x$-axis of $\mathbb R^3$, the mean curvature $H$ of the surface is
 $$H=\frac12\left(\frac{1}{y\sqrt{1+y'^2}}-\frac{y''}{(1+y'^2)^{3/2}}\right)= \frac{\lambda c^2}{2\cosh(cx+a)^2(\cosh(cx+a)-\lambda c)}.$$
 Thus $H=0$ if and only if $\lambda=0$.
 \end{remark}

 \section{The catenary problem: hyperbolic case}\label{sec3}

In the case of the catenary problem with respect to a Lorentzian plane, we have two possible choices of Lorentzian coordinate planes. After a rigid motion of $\mathbb S_1^2$, we can assume that the plane is the plane $\Pi_{xz}$ of equation $y=0$.  In this section it will be studied the curves of $\mathbb S_1^2$ that are critical points of the center of mass when it  is measured with respect to the  plane $\Pi_{xz}$. The distance of $(x,y,z)\in\mathbb S_1^2$ to $\Pi_{xz}$ is $|y|$. Let $(\mathbb S_1^2)_{xz}^+=\{(x,y,z)\in\mathbb S_1^2:y>0\}$. Let $\gamma=\gamma(t)$, $t\in [a,b]$, be a curve in $(\mathbb S_1^2)_{xz}^+$ and let $\gamma(t)=\Psi(u(t),v(t))$. The expression of the energy \eqref{ee} is  now
$$\mathcal{E}_{xz}[\gamma]=\int_a^b (\cosh(u)\sin(v)+\lambda) \sqrt{\epsilon(v'^2\cosh(u)^2-u'^2)}\, dt.$$

\begin{theorem}  \label{t21}
Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{xz}^+$. Then $\gamma$ is a  catenary with respect to $\Pi_{xz}$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c21}
\kappa=-\frac{u'\cos(v)+v'\sinh(u)\cosh(u) \sin(v)}{(\cosh(u)\sin(v)+\lambda)|\gamma'|}.
\end{equation}
\end{theorem}

\begin{proof}
We compute the Euler-Lagrange equations \eqref{sel} for the energy $\mathcal{E}_{xz}$. The first equation
$\frac{\partial J}{\partial u}-\frac{d}{dt} \left(\frac{\partial J}{\partial u'}\right)=0$ writes as
\begin{equation*}
\begin{split}
&\frac{v'\cosh(u)}{|\gamma'|}\left(v'\cosh(u)\sinh(u)\sin(v)+v'\sinh(u)(\cosh(u)\sin(v)+\lambda)+u'\cos(v)\right)\\
&=-(\cosh(u)\sin(v)+\lambda)\frac{d}{dt}\left(\frac{u'}{|\gamma'|}\right).
\end{split}
\end{equation*}
Using \eqref{k1}, this equation is simplified into
  $$\frac{v'\sinh(u)\cosh(u) \sin(v)+u'\cos(v)}{|\gamma'|}=-(\cosh(u)\sin(v)+\lambda) \kappa.$$
The second equation of \eqref{sel} coincides with the above one, obtaining \eqref{c21}.
\end{proof}

As in Theorem \ref{t12}, we characterize the catenaries  with respect to $\Pi_{xz}$ in terms of the   angle that makes the principal normal vector ${\bf n}$ of $\gamma$ with a vector field of $\mathbb L^3$. Consider the unit vector field $Y=\partial_y\in\mathfrak{X}(\mathbb L^3)$. Notice that $Y$  is   orthogonal to the plane $\Pi_{xz}$.

\begin{theorem}\label{t22} Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{xz}^+$. Then $\gamma$ is a catenary with respect to $\Pi_{xz}$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c22}
\kappa(t)=-\frac{\langle{\bf n}(t), Y\rangle}{d_{xz}(t)+\lambda},
\end{equation}
where $d_{xz}(t)$ is the distance of $\gamma(t)$ to the plane $\Pi_{xz}$.
\end{theorem}

\begin{proof} Following the same arguments as in Theorem \ref{t12}, we have
$$\langle{\bf n}(t),Y\rangle=-\frac{ \mbox{det}(\gamma,\gamma',\partial_y)}{|\gamma'|}= \frac{v'\sinh(u)\cosh(u) \sin(v)+ u'\cos(v)}{|\gamma'|}.$$
Since $d_{xz}=\cosh(u)\sin(v)$, it follows the result from \eqref{c21}.
\end{proof}

Finally we construct the surfaces of revolution in $\mathbb S_1^3$ of hyperbolic type using the catenaries with respect to $\Pi_{xz}$. Consider the geodesic $L_{xz}=\mathbb S_1^2\cap\Pi_{xz}$ viewed as a curve in $\mathbb S_1^3$. This geodesic $L_{xz}$ will be the rotation axis of a rotational surface of $\mathbb S_1^3$ of hyperbolic type because the plane $\Pi_{xz}$ is Lorentzian.  The one-parameter family of rotations fixing $L_{xz}$  is the group $\mathcal{G}_{xz}=\{R_s^{xz}:s\in\mathbb R\}$, where
 $$ R_s^{xz}: =\left(\begin{array}{cccc}1&0&0&0\\ 0&\cos s&0&-\sin s\\ 0&0&1&0\\ 0&\sin s&0&\cos s\end{array}\right).$$
 If  $\gamma=\gamma(t)$ is a curve in $(\mathbb S_1^2)_{xz}^+$, let $S_\gamma^{xz}$ denote the surface of revolution of $\mathbb S_1^3$ obtained by rotating $\gamma$  with respect to $L_{xz}$, $S_\gamma^{xz}=\{\mathcal{R}_s^{xz}\cdot\gamma(t):s\in\mathbb R, t\in [a,b]\}$.

 \begin{theorem} \label{t23}
The surface of revolution $S_\gamma^{xz}$ is minimal if and only if $\gamma$ is a catenary with respect to $\Pi_{xz}$ for the Lagrange multiplier $\lambda=0$.
\end{theorem}
\begin{proof}
Without loss of generality, we assume that $\gamma$ is parametrized by $\gamma(t)=\Psi(u(t),t)$, $t\in [a,b]$.
 A parametrization of $S_\gamma^{xz}$ is
 $$\mathbf{r}(t,s) =( \cos(t) \cosh(u),\sin(t) \cosh(u)\cos(s),\sinh(u),\sin(s) \sin(t) \cosh(u)).$$
 The unit normal is
$$N=\frac{1}{|\gamma'|}\left(\begin{array}{c}
\cos (t) \sinh (u) \cosh (u)-\sin (t) u'\\
\cos (s) \left(\cos (t) u'+\sin (t) \sinh (u) \cosh (u)\right)\\
\cosh ^2(u)\\
\sin (s) \left(\cos (t) u'+\sin (t) \sinh (u) \cosh (u)\right)
\end{array}\right),$$
and $\langle N,N\rangle=-\epsilon$.
The coefficients of the first fundamental form are $F=0$ and
$$E=\cosh(u)^2-u'^2=\epsilon|\gamma'|^2,\quad G= \sin(t)^2\cosh(u)^2.$$
We also have
\begin{equation*}
\begin{split}
h_{11}&=\frac{2u'^2\sinh(u)-\cosh(u)(u''+\sinh(u)\cosh(u))}{|\gamma'|},\\
 h_{22}&=-\frac{\cosh(t)\sin(t)(u'\cos(t)+\sinh(u)\cosh(u)\sin(t))}{|\gamma'|}.
 \end{split}
 \end{equation*}
Using \eqref{hh}, we have
\begin{equation*}
\begin{split}
 H&=\frac{\delta\cosh(u)\sin(t)}{4EG|\gamma'|}\\
 & \Big(-2 \sin (t) \cosh(u)^2 \left(u''+\sinh (2 u)\right)+2 \cos (t) u'^3\\
 &+3 \sin (t) u'^2 \sinh (2 u)-2 \cos (t) u' \cosh ^2(u)\Big).
 \end{split}
 \end{equation*}
Using \eqref{eqk}, we replace $u''$ in terms of the curvature $\kappa$, obtaining
$$
H=-\frac{\delta\epsilon|\gamma'|}{2EG}\left(u'\cos(t)+\sinh(u)\cosh(u)\sin(t)+\cosh(u)\sin(t)|\gamma'|\kappa\right).
$$
The result is now immediate from \eqref{c21}.
\end{proof}

\section{The catenary problem: parabolic case} \label{sec4}

This section is devoted to consider catenaries of $\mathbb S_1^2$ when the center of mass is calculated with the distance with respect to a degenerate plane. Without loss of generality, let $\Pi_{y-z}$ be the plane of equation $y-z=0$. The distance of    a point  $(x,y,z)$ to $\Pi_{y-z}$ is   $|y-z|$ up to the factor $\sqrt{2}$. Let $(\mathbb S_1^2)_{y-z}^+$ be half-space $\{(x,y,z)\in\mathbb S_1^2:y-z>0\}$. For a curve $\gamma(t)=\Psi(u(t),v(t))$, $t\in  [a,b]$, contained in $(\mathbb S_1^2)_{y-z}^+$, define   the energy
$$\mathcal{E}_{y-z}[\gamma]=\int_a^b (\cosh(u)\sin(v)-\sinh(u) +\lambda)\sqrt{\epsilon(v'^2\cosh(u)^2-u'^2)}\, dt.$$
where $\lambda$ is a Lagrange multiplier.

\begin{theorem}  \label{t41}
Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{y-z}^+$. Then $\gamma$ is a  catenary with respect to $\Pi_{y-z}$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c41}
\kappa= -\frac{v'\cosh(u)(\sinh(u)\sin(v)-\cosh(u))+u' \cos(v)}{(\cosh(u)\sin(v)-\sinh(u)+\lambda)|\gamma'|}.
\end{equation}
\end{theorem}

\begin{proof}
The computations are similar as in Theorem \ref{t11}. For this, it suffices to compute $\frac{\partial J}{\partial u}-\frac{d}{dt} \left(\frac{\partial J}{\partial u'}\right)=0$. Then we have
\begin{equation*}
\begin{split}
&\frac{v'^2\left(\cosh(u)^2(\sinh(u)\sin(v)-\cosh(u))+\sinh(u)\cosh(u)(\cosh(u)\sin(v)-\sinh(u)+\lambda)\right)}{|\gamma'|}\\
&=
-\frac{\cosh(u)\cos(v)u'v'}{|\gamma'|}-(\cosh(u)\sin(v)-\sinh(u)+\lambda)\frac{d}{dt}\left(\frac{u'}{|\gamma'|}\right).\end{split}
\end{equation*}
Using \eqref{k1}, and simplifying,
\begin{equation*}
\begin{split}
&\frac{v'^2\cosh(u)^2(\sinh(u)\sin(v)-\cosh(u))+u'v'\cosh(u)\cos(v)}{|\gamma'|}\\
&=-(\cosh(u)\sin(v)-\sinh(u)+\lambda)v'\cosh(u)\kappa.
\end{split}
\end{equation*}

 \end{proof}

We characterize the catenaries of Theorem \ref{t41}  in terms of the   angle that makes the  principal normal vector ${\bf n}$ of $\gamma$ with the vector field $T=\partial_y+\partial_z\in\mathfrak{X}(\mathbb L^3)$. This vector field is      orthogonal to the plane $\Pi_{y-z}$.

\begin{theorem}\label{t42} Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{y-z}^+$. Then $\gamma$ is a catenary with respect to $\Pi_{y-z}$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c42}
\kappa(t)=-\frac{\langle{\bf n}(t), T\rangle}{d_{y-z}(t)+\lambda},
\end{equation}
where $d_{y-z}(t)$ is the distance of $\gamma(t)$ to the plane $\Pi_{y-z}$.
\end{theorem}

\begin{proof} It suffices to compute $\langle{\bf n}(t), T\rangle$,
$$\langle{\bf n}(t),T\rangle=\frac{v' \cosh(u)\left(\sinh(u)\sin(v)-\cosh(u)\right)-u' \cos(v)}{|\gamma'|}$$
and use   \eqref{c41}.
\end{proof}

We now consider rotational surfaces of $\mathbb S_1^3$ about the rotation axis $L_{y-z}=\Pi_{y-z}\cap\mathbb S_1^2\subset\mathbb S_1^3$. Since the plane $\Pi_{y-z}$ is degenerate, the surface of revolution is of parabolic type.     The one-parameter group $\mathcal{G}_{y-z}$ of rotations of $\mathbb S_1^3$ fixing $L_{y-z}$  is the group $\mathcal{G}_{y-z}=\{R_s^{y-z}:s\in\mathbb R\}$, where
$$R_s^{y-z}=\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1-\frac{s^2}{2} & \frac{s^2}{2} & s \\
 0 & -\frac{s^2}{2} & \frac{s^2}{2}+1 & s \\
 0 & -s & s & 1 \\
\end{array}
\right).$$
Let $\gamma=\gamma(t)$ be a curve in $\mathbb S_1^2$ and let $S_\gamma^{y-z}=\{\mathcal{R}_s^{y-z}\cdot\gamma(t):s\in\mathbb R,t\in [a,b]\}$ the surface of revolution defined by $\gamma$.

\begin{theorem}\label{t43}
The surface of revolution $S_\gamma^{y-z}$ is  minimal if and only if $\gamma$ is a catenary with respect to $\Pi_{y-z}$ for the Lagrange multiplier $\lambda=0$.
\end{theorem}

\begin{proof}
The computations are simplified again if the generating curve is assumed to be   $\gamma(t)=(\Psi(u(t),t),0)$. The parametrization of $S_\gamma^{y-z}$  is
 $$\mathbf{r}(t,s) =\left(\begin{array}{l}
 \cos (t) \cosh (u)\\
  \frac{s^2}{2} \sinh (u)-\frac{s^2-2}{2} \sin (t) \cosh (u) \\
\frac{s^2+2}{2} \sinh (u)-\frac{s^2}{2} \sin (t) \cosh (u)\\
 s (\sinh (u)-\sin (t) \cosh (u))
 \end{array}\right).$$
We compute the mean curvature $H$ of $S_{\gamma}^{y-z}$. The unit normal $N$ of $S_{\gamma}^{y-z}$ is
$$N=\frac{1}{|\gamma'|}\left(\begin{array}{c}
  \cos (t) \sinh (2 u)-2 \sin (t) u'\\
 - \left(s^2-2\right) \cos (t) u'+ s^2 \cosh ^2(u)-\frac{s^2-2}{2} \sin (t) \sinh (2 u)\\
 -s^2 \cos (t) u'+  \left(s^2+2\right) \cosh ^2(u)-\frac{s^2}{2}\sin (t)) \sinh (2 u)\\
 - s \left(2 \cos (t) u'-2 \cosh ^2(u)+\sin (t) \sinh (2 u)\right)
\end{array}\right)$$
and $\langle N,N\rangle=-\epsilon$.
The coefficients of the first fundamental form are $F=0$ and
$$E=\cosh(u)^2-u'^2=\epsilon|\gamma'|^2,\quad G=(\sinh(u)-\cosh(u)\sin(t))^2.$$
The coefficients of the second fundamental form are $h_{12}=0$ and
\begin{equation*}
\begin{split}
h_{11}&=\frac{2\sinh(u)u'^2-\sin(u)\cosh(u)^2-\cosh(u)u''}{|\gamma'|},\\
h_{22}&=\frac{(\cosh(u)\sin(t)-\sinh(u))(\cosh(u)^2-\sinh(u)\cosh(u)\sin(t)-u'\cos(t))}{|\gamma'|}.
\end{split}
\end{equation*}
As in the above cases, the mean curvature $H$ is calculated using the formula \eqref{hh}. After substituting $u''$ in function of $\kappa$ thanks to \eqref{eqk}, we obtain
\begin{equation*}
\begin{split}
H&=\frac{\delta\epsilon(\cosh(u)\sin(t)-\sinh(u))}{2EG}\\
&\Big( \kappa(\cosh(u)\sin(t)-\sinh(u))|\gamma'|+\cosh(u)(\sinh(t)\sin(t)\\
&-\cosh(u))+u'\cos(t)\Big).
\end{split}
\end{equation*}
Then \eqref{c41} completes the proof.
\end{proof}
InFigure \ref{fig1}  shows pictures of catenaries of $\mathbb S_1^2$ with respect to the three types of causal character of the plane reference.

\begin{figure}[hbtp]
\begin{center}
\includegraphics[width=.3\textwidth]{fig11.pdf} \includegraphics[width=.26\textwidth]{fig12.pdf} \includegraphics[width=.32\textwidth]{fig13.pdf}
\end{center}
\caption{Catenaries of $\mathbb S_1^2$: with respect to spacelike plane (left), timelike plane (middle) and degenerate plane (right). }\label{fig1}
\end{figure}

\section{The intrinsic catenary problem}\label{sec5}

The catenaries studied in the previous sections are critical points of energies  that measure the center of mass with respect to  planes of $\mathbb L^3$.  However, if one wants to pose an `intrinsic catenary problem' involving only data in $\mathbb S_1^2$,  it would be desirable that the distance were measured in $\mathbb S_1^2$   with respect to a geodesic of $\mathbb S_1^2$. In this section we will investigate this problem in the particular case that this geodesic is spacelike. Without loss of generality, consider the spacelike geodesic $L=\mathbb S_1^2\cap\Pi_{xy}$, which will serve as reference line to measure the center of mass of a curve.

 The distance of a point $(x,y,z)\in\mathbb S_1^2$ from $L$ is computed as follows. Let $(x,y,z)=\Psi(u,v)$. Then the geodesic on $\mathbb S_1^2$ passing through to $(x,y,z)$ and orthogonal to $L$ is $\alpha(t)=\Psi(u+t,v)$, $t\in [-u,0]$. The intrinsic distance of $(x,y,z)$ to $L$ is
$$\int_{-u}^{0}|\alpha'(t)|\, dt=\int_0^{-u}dt=u=\mbox{arcsinh}(z).$$
In particular, it is necessary that  $z\not=0$. Consider the half-space   $(\mathbb S_1^2)_{xy}^+=\{(x,y,z)\in \mathbb S_1^2:z>0\}$. If $\gamma$ is a curve in $(\mathbb S_1^2)_{xy}^+$ parametrized by $\gamma(t)=\Psi(u(t),v(t))$,  the functional energy is defined by
$$\mathcal{E}_{in}[\gamma]=\int_a^b (u+\lambda)\sqrt{\epsilon(v'^2\cosh(u)^2-u'^2)}\, dt,$$
where $\lambda$ is a Lagrange multiplier that indicates that in the variational problem it is assumed that all curves have fixed length.
A critical point of $\mathcal{E}_{in}$ is called an {\it intrinsic catenary} of $\mathbb S_1^2$ with respect to the geodesic $L$.

\begin{theorem}\label{t1} Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{xy}^+$. Then $\gamma$ is an intrinsic catenary with respect to $L$ if and only if its curvature $\kappa$ satisfies
\begin{equation}\label{c51}
\kappa=-\frac{v'\cosh(u)}{(u+\lambda)|\gamma'|}.
\end{equation}
\end{theorem}

\begin{proof} We compute the Euler-Lagrange equations of $\mathcal{E}_{in}$ using \eqref{sel}. Both equations are
$$\frac{v'^2\cosh(u)(\cosh(u)+(u+\lambda)\sinh(u))}{|\gamma'|}=-(u+\lambda)\frac{d}{dt}\left(\frac{u'}{|\gamma'|}\right),$$
$$ \frac{u'v'\cosh(u)(\cosh(u)+(u+\lambda)\sinh(u))}{|\gamma'|}+(u+\lambda)\cosh(u)\frac{d}{dt}\left(\frac{v'\cosh(u)}{|\gamma'|}\right)=0.$$

Using the expressions \eqref{k1} and \eqref{k2}, we have
$$v'\cosh(u)\left((u+\lambda)\kappa+\frac{v'\cosh(u)}{|\gamma'|}\right)=0,$$
$$u'\cosh(u)\left((u+\lambda)\kappa+\frac{v'\cosh(u)}{|\gamma'|}\right)=0,$$
proving the result.
\end{proof}

It is possible to  characterize intrinsic catenaries of $\mathbb S_1^2$ as solutions a prescribing curvature equation involving the principal normal vector of the curve and vector fields. In contrast to Theorems \ref{t12}, \ref{t22} and \ref{t42}, it is natural that now the vector field is a vector field on $\mathbb S_1^2$. Furthermore, this vector field will indicate the direction of how is measuring the distance to $L$. Since this distance is done using the geodesics orthogonal to $L$, that is, the meridians of $\mathbb S_1^2$, the vector field will be tangent to the meridians. Hence, define   the vector field $V\in\mathfrak{X}(\mathbb S_1^2)$
\begin{equation*}
V(\Psi(u,v))=\Psi_u=\sinh(u)\cos(v)\partial_x+\sinh(u)\sin(v)\partial_y+\cosh(u)\partial_z.
\end{equation*}

\begin{theorem} \label{t2}Let $\gamma(t)=\Psi(u(t),v(t))$ be a non-degenerate curve in $(\mathbb S_1^2)_{xy}^+$. Then $\gamma$ is an intrinsic catenary with respect to $L$ if and only if its curvature $\kappa$ satisfies
\begin{equation*}
\kappa(t)=\frac{\langle{\bf n}(t), V\rangle}{d(t)+\lambda},
\end{equation*}
where $d(t)$ is the intrinsic distance of $\gamma(t)$ to $L$.
\end{theorem}

\begin{proof}
Using that  ${\bf n}= - \gamma\times\gamma'/|\gamma'|$, from $\gamma(t)=\Psi(u(t),v(t))$ we have $\gamma'=u'\Psi_u+v'\Psi_v$. Thus
$$\langle{\bf n}(t),V\rangle=-\frac{\mbox{det}(\gamma,\gamma',V)}{|\gamma'|}=-\frac{v'}{|\gamma'|}\mbox{det}(\Psi,\Psi_v,\Psi_u)=-\frac{v'\cosh(u)}{|\gamma'|}.$$
This expression together \eqref{c51} gives the result.
\end{proof}

Finally, notice that the surface of revolution $S_\gamma^{xy}$ of spherical type obtained by rotating an intrinsic catenary $\gamma$ about $L$ is not minimal. Its mean curvature is given from \eqref{h1}, which together \eqref{c51} gives for $\lambda=0$,
\begin{equation*}
\begin{split}
H&=-\frac{\delta}{2|\gamma'|\sinh(u)}\left(\cosh(u)^2+\kappa\sinh(u)|\gamma'|\right)\\
&=-\frac{\delta\cosh(u)}{2|\gamma'|\sinh(u)}\left(\cosh(u)-\frac{v'\sinh(u)}{u}\right).
\end{split}
\end{equation*}
\section*{Acknowledgements}
Rafael  L\'opez  is a member of the Institute of Mathematics  of the University of Granada. This work  has been partially supported by  the Projects  I+D+i PID2020-117868GB-I00, supported by MCIN/ AEI/10.13039/501100011033/,  A-FQM-139-UGR18 and P18-FR-4049.

\section*{Data Availability Statement}

My manuscript has no associate data. No funding was received for conducting this study. The authors have no competing interests to declare that are relevant to the content of this article.

redpajama_set_name: RedPajamaArXiv


For fans of scrupulously researched historical fiction, Tricia Goyer releases Dawn of a Thousand Nights. Following on the heels of Goyer's From Dust and Ashes and Night Song, readers journey with Dan Fletcher and Libby Conner, both U.S. pilots fighting in WWII's Pacific Theater. Fletcher is captured by the Japanese and is forced to endure the horror of the Bataan Death March in the Philippines. It is here that he encounters Natsu Hidiki, a guard whose nagging conscience won't let him ignore the human degradation in which he is participating. Libby Conner is a WASP (Women's Air Force Service Pilot), who ferries military planes between Hawaii and the South Pacific. By the time of the attack on Pearl Harbor, she has turned the head of Fletcher in no small way. Readers will be captivated by this tale of dignity and honor triumphing in the midst of a very dark time in history.
The latest transfer orders published by the War Department show that some 118 second lieutenants who have recently won their wings have been ordered to embark for the Pacific outposts in the next three weeks.
That necessity has in turn raised the problems of how to give young pilots actual army flying experience in high-speed, complex combat ships once they have completed their work with the slower trainers.
Until a pilot has had about a year of such experience he is not considered ready to engage in combat operations.
Dan Lukens's leather-gloved hand rammed the throttle. The force of the screaming Allison V-12 engine shoved his body into the back of his seat. As the rumble from a half dozen fighters grew louder, Dan blinked sweat from his eyes. The metallic black "birds" came into view, distorting the air behind them into a fuzzy haze.
Grimacing, Dan shoved the stick full forward and rudder pedals full right and cranked his Curtiss P-40 into the Chennault maneuver. His nimble fighter spun into an outside barrel roll, jerking his body against his harness. He pulled out of the roll and converted his diving momentum to a high-speed climb, mentally preparing for the dogfight ahead.
"Come on, boys. Show me what you've got," Dan hissed through clenched teeth, his neck tensed and senses peaking.
But the attackers held back. They circled Dan's group cautiously. Finally, when minutes passed and they didn't strike, Dan took the initiative.
He broke left through a wispy cloud and targeted the lead plane. "You're mine now!" he yelled over the roaring engine. He pinpointed the enemy plane through his sights-a clear shot. He squeezed the trigger. "Got ya!" he yelled with excitement into his radio. "You're out of here, pal. Yes, I'm talking to you." When the plane fell away, he turned his attention to Gabriel Lincoln's plane-now being trailed by one of the attackers. Gabe attempted to shake him, to no avail.
"Keep it up, Gabe. Don't let him get you. I'm on my way."
Dan dropped into the sights of Gabe's attacker, hoping to draw his attention. The attacker pulled off Gabe, making a yard-wide turn and leaving a trail of exhaust behind him.
In their pilots' training, Dan's instructor had told him never to follow a Japanese Zero into a tight turn. Yet with adrenaline pumping, he turned after the plane in tigerish pursuit.
Dan pulled against his harness as if trying to urge every bit of power from the P-40. "That's the way, doll face. You've got it." He dove straight into the attacker's flight path, cutting him off. The attacker had no choice but to accelerate and pop up into Dan's line of vision, or slam into his plane. Dan was betting his life on the first option.
"Pull up. Come on." A loud roaring sound from beneath him drowned out his words.
Seconds seemed like hours as the long nose of Dan's P-40 blocked his view of the other fighter, but he held his position, worrying for the briefest instant whether he'd overdone it, cut the guy off too short.
Suddenly, the attacker increased his speed and altitude, popping into view not more than fifty yards in front of Dan. The pilot's sweaty, glaring face turned Dan's direction.
Dan grinned, spotting the plane in his sights. "Got ya!" he called into his radio, then waved to the pilot.
Dan noted his kill. The attacker dropped toward the ground, and though the temptation was strong to visually follow the plane's path to the harbor, Dan had better sense. Especially with other attackers around.
After picking off the rest of the "enemy" fighters, Dan headed back to once again claim the day's most kills. His heart still thumped as the harbor came into view, sparkling in the morning sunlight.
As long as I've got this plane, I can tackle anything thrown at me. He sucked in a calming breath as the Curtiss P-40 broke through thin clouds into the blue. A surge of confidence as strong as the plane's engine welled in his chest. Confidence in his flying and in his team. While others fretted that war was inevitable, Dan trusted that the U.S. Army Air Corps would hold its own ... trusted that his leaders knew what they were doing.
As the P-40 continued its descent, Pearl Harbor greeted him, its blue waters spreading out in the shape of a cloverleaf-the stalk being the entrance to the channel and the leaves unfolding to the west, middle, east, and southeast lochs.
A four-leaf clover. Dan felt like the luckiest guy alive to be playing war games with fast planes in such a beautiful place. It reminded him of the toy battles he and his friends used to play in the sandlot at the end of the street. Only these weren't model planes, but rather the army's best, that he and his buddies maneuvered around the sky.
Between the east and southeast lochs, Ford Island bustled with activity. It was a compact airfield complete with tower and hangars, administrative buildings, and living quarters. The movement of people and machines on the island stirred in full gear.
Alongside the island, docked in the harbor, sat a fleet of battleships moored in pairs. Dan took note of their bows pointed toward the southwest channel. Intermingled among them, civilian ships dropped off tourist groups to be entertained on Waikiki Beach, lovingly referred to as the "Coney Island of the Pacific."
Dan had taken off from Ford Island numerous times, but today his maneuvers originated from another nearby air base, Hickam Field-due south of Ford Island and the Navy Yard-on the island of Oahu itself.
Within a matter of minutes, he had executed his landing at Hickam and parked his plane wingtip to wingtip next to the others. He joined the other pilots in leather flight jackets, who now formed a dozen-man circle on the tarmac. Dan slid off his flight helmet as he approached the group and ran his hand through his sweat-drenched blond hair. Perspiration had made a wet stain on his flight suit, ringing his underarms and making a dark spot on his chest.
A cheer rose from two or three of the guys as he approached. Others called out "Just lucky!" and "I'll get you next time!"
"I'm just glad you're on our side." His buddy Zeke Olson gave him a strong shoulder squeeze. "I guess that means you win the bet. I owe you a drink at the Black Cat."
"Nah. How about a football game on the beach? I'm getting tired of all the commotion and noise in the honky-tonk district."
"You're on. UCLA paired up against Oregon State once again-Hawaiian style."
The aroma of tropical flowers wafted through the cool morning air as Libby Conners, dressed in a flight suit, strode across the tarmac of John Rodgers Airport to do a quick check of the Piper Cub. The plane wasn't as fancy as those she'd flown in the States, but it was simple and economical. And this newer Cub benefited from a wider front seat, allowing teacher and student to sit side by side rather than tandem. She considered the Piper safe and sturdy-until the trade winds blew in.
But to Libby, the best type of plane was any one she could fly. That's why she'd come to Hawaii and why she had stayed. Distanced from mainland prejudices against women, she had found a place where she could soar.
Satisfied the plane was in top shape, she patted its sleek metal belly.
"Mr. Conners?" said a youthful voice from behind her.
Libby turned, her tall frame causing her to meet the sailor eye-to-eye. They're looking younger all the time. She stretched out her hand. "Miss Conners, actually. Good to have you. This your first flying lesson?"
The boy's eyes widened as he studied her face and the medium brown locks she'd tucked into her flight helmet.
"Better close that jaw before you swallow a fly, kid." Libby took two long strides and pointed to the open door of the Cub. "The clock is ticking."
The sailor didn't budge. Libby glanced back.
"Uh, I didn't know I was gettin' lessons from a girl." His hands and shoulders moved as he spoke, as if he shrugged off his words as well as spoke them.
Libby crossed her arms, straightened her shoulders, and waited.
His eyes darted back toward the office, as if hoping someone would emerge and tell him there'd been a misunderstanding. When no one came to his rescue, he dug his hands into his pockets and offered a slight shrug. "Uh, okay, I guess. You're not gonna get me killed, are you?"
Been flying since you were still in elementary school learning how to read, Libby wanted to respond. Instead she reached over and patted the kid's shoulder. "Got my adequate teaching licenses all in order. Besides, during this first lesson I doubt you'll learn much anyway. Most guys are so excited by the sight of the island and ocean from the air that nothing sticks. So we'll sign you up for a lesson next weekend too. Saturday, if possible. Sunday's nearly always full."
The kid nodded and climbed into the small cockpit. His face reddened as Libby slid in next to him, their shoulders and hips pressed together in the tight space.
He'll get over that too. Libby quickly scanned the runway, checking for other planes before starting the engine. Before long, he'll just think of me as one of the guys. A buddy, a pal. And a darn good instructor.
Sometimes, in the first waking heartbeat when dawn stirred Natsuo Hidki from his slumber, he believed he was there again. In the California town of modern living. In the country that could not comprehend the traditions of his homeland. At the college where friendships were made and soon abandoned.
It had been two years since he'd returned to Japan. His time at UCLA had been short. Why, then, did his mind keep taking him back? Perhaps as a curse. An aching reminder of his turning away from the Land of the Rising Sun.
Father had asked him to go. Believed that Western thinking and technology were important for the future of Japan. And though his mother dared not disagree, in the split second before she looked away Natsuo had seen fear in her gaze. She worried that her son would not come back. Just as his sister had not.
But Natsuo had returned. Smarter in the understanding of the world, and witnessing his homeland in a new light. Though still bound by his country's codes of honor, Natsuo's California teachers had succeeded in planting the sturdy oaks of Western thinking within the ancient soil of his soul.
America. Why couldn't it leave him alone?
As Natsuo lay on his mat and breathed in the scent of cherry blossoms through the open window, he wondered why he'd been transported back. It would be better to forget. Especially now, with his upcoming position as a soldier.
Through the thin walls of his childhood home, his mother's humming caused the corners of Natsuo's lips to flicker a slight smile. She was preparing his favorite breakfast, rice and miso soup. His call to service weighed on her heart too.
Natsuo had memorized the words long ago at UCLA, in a class his father had urged him not to take. How, after all, could American poetry ever make a difference to the future of Japan?
He only wished it would stop haunting his thoughts.
Excerpted from DAWN OF A Thousand Nights by TRICIA GOYER Copyright © 2005 by Tricia Goyer.
Compelling, breathtaking, and sometimes brutal, Dawn of a Thousand Nights pulled me into its WWII setting and kept me reading late into the night again and again. Ms. Goyer is a masterful storyteller who knows how to bring history to life.
A great story! A great writer! Historically accurate. I know. I was there.
Heartbreaking and triumphant, Dawn of a Thousand Nights makes real the struggles and sacrifices of those men and women at home and abroad who fought to give us the freedoms we often take for granted today.
It is rare to see a novel pay such attention to historical detail.
Tricia Goyer's Dawn of a Thousand Nights is a deeply emotional read, rich with historical detail and true to the heart of God.
Tricia Goyer has written one of the best-researched and most accurate accounts of World War II I have ever read. She has woven her characters into the events in Pearl Harbor, the Philippines, and Japan so realistically that it is hard to remember that this is actually a book of fiction.
Having lived through WWII and actively participated in the military for the Air Force, I found Dawn of a Thousand Nights not only an easy read but an accurate portrayal. It brought to mind the suffering of those who had been captured and the realization of how man can be so cruel to man. Tricia Goyer is a very talented storyteller.
Dawn of a Thousand Nights gives the reader an amazing peek into history and into the human heart. This story is gripping and inspiring, written with grace and intelligence. The reader is in for a soaring flight of the mind, heart, and soul.
Tricia Goyer's masterful storytelling techniques will grip the reader with magnificence of human emotions set in the midst of a tragic part of US history. Share in the victory and defeat, good and evil, and humility and pride as heroes are made within the pages of this unforgettable drama.
Tricia Goyer is a seamstress of fiction. She dares to weave designs full of suspense, history, humor, romance, and spiritual truth. In the same manner that Bodie Thoene's early novels brought new understanding, Tricia Goyer's stories never sugarcoat the horror, but always highlight God's workings in the midst of suffering.
Dawn of a Thousand Nights 4.9 out of 5 based on 0 ratings. 7 reviews.
In June 1941, World War II rages in Europe and a romance begins in Hawaii. Dan Lukens, former UCLA football star, is now a top fighter pilot for the US Army Air Corps and is stationed in Hawaii. Libby Conners fled the mainland US for the friendlier skies of Hawaii. She is a female pilot and flight instructor, occupations frowned upon in California. Dan falls for Libby after she nearly drowns him in the ocean. They become engaged the day before Dan is shipped off to Clark Field in the Philippines. Libby is giving flight lessons on December 7, 1941, when the Japanese attack Pearl Harbor. On December 8th, the Japanese attack the Philippines, destroying the United States' air power. When the islands are finally overrun, Dan becomes a prisoner of war. The one thing that keeps him going amid the horrors of the POW camps is Libby and the life he plans with her after the war, if he survives. Libby joins the WAFS-Women's Air Ferrying Squadron-delivering plans all over the country for the military. She struggles through the loss of close friends and not knowing if Dan is dead or alive. Tricia Goyer does an excellent job bringing to life an often overlooked aspect of World War II-the war with Japan. Historical details woven throughout the book bring the years 1941-1945 draw the reader into this rich story.
I want to buy more of her books now. Such a sweet book and easy read.
Dawn of a Thousand Nights is a fictional novel with the historical backdrop of World War II. Tricia Goyer gives us the subtext of the story with the subtitle, ¿A Story of Honor.¿ The novel revolves around a trio of main characters: two American pilots (one man and one women) and a Japanese interpreter. All three are fully focused at the beginning of the book with the desire to help and serve their respective countries. As the story unfolds, the characters individually advance through a wide-ranging emotional spectrum ¿ from bitter hatred against the enemy, to fear, love, despair, the works. But it all comes back around to honor ¿ how each handles the circumstances of their lives in the midst of a devastating war. These are not vague one-dimensional characters dropped into a cheap story line. Quite the opposite - Dawn of a Thousand Nights is multi-dimensional and richly detailed with accurate period references. The novel focuses on specific events of WWII in the Pacific Theater: the bombing of Pearl Harbor by the Japanese, the battle for supremacy in the Philippines, the Bataan Death March. And in the same time frame on the west coast of America, a separate but equally important battle is fought in the arena of ideas: specifically, in regards to the use of women as pilots in the war. The expanded role of women during WWII emphasizes their vast and valuable contributions to the war effort in their own right. Dawn of a Thousand Nights is an amazing read ¿ bittersweet at times, yet complex and satisfying through and through. Tricia Goyer is an incredibly talented writer. She is able to deliver past history as if it were fresh news, all the while weaving a diverse mix of romance, patriotism, friendship, courage and drama into a resonant cast of characters. This novel will definitely stand the test of time as a stand-out work from a real stand-up author.
This is the first of Tricia's books that I read, and I must say, I was very impressed. Her attention to detail was wonderful, and I could tell that the book was well researched and documented. Set as a historical romance with the emphasis on historical, this book is definitely not fluff that you'll forget in a week or two. But, before you start to think that Dawn of a Thousand nights is nothing more than a WWII textbook, let me assure you that Tricia Goyer is a gifted storyteller. I found myself drawn into the text, staying up late reading until it was finished. The characters are deep, well written, complex creations, as is the setting. Tricia is not one to gloss over the difficult days of war. She draws you into her story and you are at Pearl Harbor. You are with Dan Connors during the Bataan Death March. But, within such a difficult time, she shows the hope that was and is always available to those who trust God. Definitely a 'couldn't put it down' kind of book. There was more than one point where I had tears in my eyes. This may have been my first experience reading a Tricia Goyer novel, but it won't be my last.


text:
Dans la mythologie grecque, Déimaque (en grec ancien ) est un fils d'Éléon roi de la ville homonyme en Béotie qui accompagna Héraclès dans son expédition contre Troie. Il y rencontre Glaucia, la fille du dieu-fleuve Scamandre, qui tomba amoureuse du héros. Elle se retrouva enceinte, mais Déimaque périt en combattant les Troyens. Craignant que l'on ne découvre son état, Glaucia se réfugia auprès d'Héraclès, qui la prit sous sa protection. Elle donna naissance à un fils, nommé Scamandre, et Héraclès les ramena tous deux à Eléon où Scamandre devenu adulte succéda à son grand-père.

Notes et références

Mortel de la mythologie grecque
Mythologie grecque de Béotie
Mythologie grecque de Troade
Geste d'Héraclès

redpajama_set_name: RedPajamaWikipedia


Wondering what level 250,000 XP is in Fortnite Season 6 so you can get that slick looking, fully-upgraded Calamity skin? Well, here's what you need to know to get there.
Fortnite Season 6 is right around the corner, bringing along a slew of new surprises for eager fans. Another season means a new battle pass, so players should prepare for yet another grind if they want all of the latest skins in the game. If you want the stage 6 variant of the Calamity skin, you'll need to earn at least 250,000 XP. That said, here's what level 250,000 XP is in Fortnite Season 6.
So far, we still haven't gotten an official XP chart for Fortnite Season 6. While we don't know the exact level equivalent for 250,000 XP, we can try and estimate it through Season 5's XP chart. Of course, once we have an updated Season 6 XP chart, we'll give you a better estimation on what level 250,000 XP is in Fortnite Season 6.
Assuming you actively play Fortnite, you should be around level 60 or 61 when you have 250,000 XP. However, do take this with a grain of salt, as Season 6 might mix things up a bit with their computation. Nonetheless, rest assured that we will be updating this post as soon as we learn the exact computation and level for 250,000 XP.
If you're looking for a quick way to boost your XP in Season 6, you have a few options. You can either earn it by playing the game or choose to purchase all of the tiers instead. There's no going wrong with either method, depending on how badly you really want that Calamity skin. That said, the latest season has plenty of other cosmetics just waiting for you to unlock. If you want to collect them all, try your best to grind beyond level 60.
Alternatively, by playing, you can just grind out the XP you need. Completing daily challenges, as well as the week 1 challenges in Fortnite Season 6 will help you to get the 250,000 XP nice and quickly.
That's all you need to know about what level 250,000 XP is in Fortnite Season 6.
For more tips and tricks on Season 6, make sure to take a good look through our nifty Fortnite Season 6 wiki. The page is packed with all sorts of helpful guides that'll help you complete your weekly challenges and score some sweet victories. For anything else you need help on, be sure to drop it in the comments below.


text:
layout: post
title:  "Gravedad: ¿Cuál te atrae más, el lejano Sol ó una muy cercana araña?"
author: rafa_aguilar
date:   2015-04-22 11:00:00
categories: entradas
comments: true
---

Parece algo bastante incoherente, pero este artículo intenta es potenciar la curiosidad sobre la física de manera lúdica, simplemente basándose en el dicho *"Donde quieras que te encuentras, siempre estarás a pocos metros de una araña"*... Conozco a algunos personajes que con solo leer esto ya se les erizará la piel, así que si tu grado de arácnofobia es tal que tienes pesadillas con estos insectos de solo verlos en el dibujo de un niño, te recomiendo que no lo leas.  Si por el contrario, puedes soportarlo un poco y te llama la atención la física, quédate, es interesante ;) .

Advertencia

_Este artículo es una burda pseudo-traducción del artículo original de  **xkcd** que podrán encontrar [aquí][SpiderVsSun], así que si quieres ver el original y no la copia, ya estás advertido._
{: .notice}

Todo comenzó con una pregunta de la usuario *Marina Fleming* del portal de **[xkcd][WhatIf]**, del cual estaré <s>plagiando</s> importando otras cosas, la pregunta, en resumen, decía mas o menos esto:

>¿Quien aplica más fuerza gravitacional sobre mí? ¿Una araña o el Sol?, la duda surge porque aprendí en mi escuela de física que la fuerza gravitacional es inversamente proporcional al cuadrado de la distancia. Y como el Sol está tan lejos...  --Marina Fleming

Esta pregunta parece totalmente razonable, sin embargo es muy fácil confundirla y con pequeños giros convertirla en una totalmente irracional.

![][irracional]

De hecho, la fuerza gravitacional que pueda ejercer una *sola* araña jamás será mas fuerte que la que ejerce el Sol sobre nosotros, ni siquiera si la tarántula *[come-pájaros][bird-eating]* tiene la voluntad, <s>y tu la valentía de no salir corriendo</s>, de acercarse lo más que se pueda a ti, nunca podrá vencer al Sol, aún así en esas *macabras* condiciones, la fuerza del Sol será 50 millones de veces más fuerte.

![][cobarde]

*¿Y que tal si fueran **Todas las arañas del mundo juntas**?*
{: .notice}

Si, como si no fuese suficiente con la *come-pájaros*, ahora te toca imaginarte todas tus pesadillas juntas y cerquitas de ti para el próximo ejercicio.  Basémonos en un estudio hecho en el [2009 en Brasil][estudio] para lanzar acá unos números gruesos, en dicho estudio estimaron una cantidad de *miligramos* por metro cuadrado en algunas áreas de muestra de ese país, si establecemos que esa densidad de arañas se encuentran igualmente distribuidas en un 10% de la superficie terrestre y en más ningún otro lado, totalizará la <s>espantosa</s> cifra de 200.000.000 Kilogramos (ó 200 millones KG) de arañas en todo el mundo.

![][humanos]

Aún cuando estos números disten drásticamente de la realidad, es suficiente para dar respuesta a la pregunta original, y es que si calculamos la gravedad que generaría dicha masa de arañas(suena igual de horrible que cuando lo escribí), la gravedad generada por el Sol aún la supera en 13 órdenes de magnitud.

No hay cálculo que de como resultado una mayor fuerza de gravedad para cualquier cantidad de arañas que la provista por el Sol. y es que enfrentémoslo, vivimos en un mundo donde hay arañas por todas parte pero es dominado por una estrella gigantesca... Al menos no es al revés!

![][SistemaAracnido]

[SpiderVsSun]:http://what-if.xkcd.com/136/
[WhatIf]:http://what-if.xkcd.com
[irracional]:http://what-if.xkcd.com/imgs/a/136/question.png
[bird-eating]:https://en.wikipedia.org/wiki/Goliath_birdeater
[cobarde]:http://what-if.xkcd.com/imgs/a/136/force.png
[estudio]:http://www.bioone.org/doi/pdf/10.1636/T08-21.1
[humanos]:http://what-if.xkcd.com/imgs/a/136/feet.png
[SistemaAracnido]:http://what-if.xkcd.com/imgs/a/136/planet.png

redpajama_set_name: RedPajamaGithub


(Early Show) The Saint Presents From Nashville, TN, Ben Sparaco and The New Effect and special guests. On Sale Friday 2/15 at 10am. Ben Sparaco and The New Effect is a genre-defying trio based out of Nashville, Tennessee. Led by young guitarist/singer/songwriter Ben Sparaco, the band features Adam Discipio and Anthony Quirk on bass and drums respectively. Between the three you an hear influences from modern indie rock, classic jam bands, Blue Note era jazz, and everything in between. Ben Sparaco and The New Effect formed in 2017 when Florida-born singer, songwriter, and guitarist began playing with drummer Anthony Quirk and bassist Adam Discipio. Now based in the flourishing musical oasis of Nashville, TN, Ben and the band have maintained a massive touring schedule, cutting their teeth playing coast to coast in clubs and festivals. Sparaco spent his teen years making a name for himself in the jam scene of the southeast, playing with members of the Allman Brothers Band, Dead and Company, Tedeschi Trucks Band, and more before he had even turned 21. This included a touring stint in late Allman Brothers drummer Butch Trucks' band. After recording his 2016 release "Bring the Jubilee EP" with a Florida-based group of musicians, he moved to Nashville. There, he amassed a group of some of Nashville's finest to record his debut solo record "Wooden" with producer Ross Holmes (Mumford and Sons, Bruce Hornsby, Warren Haynes), which also featured a guest appearance from Luther Dickinson. After finalizing The New Effect's powerful lineup with Berklee College of Music graduates Quirk and Discipio, Sparaco's songwriting started drifting from his southern rock roots, showing indie rock, new wave, and pop influences, evident on the EP "Greetings, from Ben Sparaco and The New Effect." Following this with the single "Scared of The Dark" in the fall of 2018, Sparaco and his band have earned glowing press reviews, further establishing themselves as up and comers to watch in the indie rock world.
The Jersey Shore Festival is a FREE ADMISSION, rain-or-shine music and arts festival that kicks off the summer festival season and proves that summer really does start sooner at the Jersey Shore. With performances by over 150 artists from around the world, the Jersey Shore Festival is the premier emerging artist festival in New Jersey! Aside from the performances, many other fun activities will be available for the whole family!


text:
The MONEY Book and Review of top Canadian Finance and Personal Finance Books. MONEY Books – Mutual Fund Guide – Mutual Fund Book – The Mutual Fund Book – incudes Mutual Fund Review. Canadian Finance Books – Personal Finance and Canadian Investing books, guides and manuals. The Book Review – Canadian book review and financial author library.
The Ascent of Money
Niall Ferguson (born April 18, 1964, in Glasgow, Scotland) is a British historian who specialises in financial and economic history as well as the history of empire. He is the Laurence A. Tisch Professor of History at Harvard University and the William Ziegler Professor of Business Administration at Harvard Business School. He was educated at the private Glasgow Academy in Scotland, and at Magdalen College, Oxford where he was a Demy and graduated with first class honours in 1985, where he did completed both his undergraduate degree and PhD. After two years as a Hanseatic Scholar in Hamburg and Berlin, he took up a Research Fellowship at Christ's College, Cambridge, in 1989, subsequently moving to a Lectureship at Peterhouse. He returned to Oxford in 1992 to become Fellow and Tutor in Modern History at Jesus College, a post he held until 2000, when he was appointed Professor of Political and Financial History at Oxford. Mr. Ferguson taught at Jesus College Oxford for nearly a decade before he decided to cross the pond because the U.S. higher education system was doing things at his speed; a quicker more efficient way of doing things impressed him a lot. Niall was soon a professor at New York University in 2001-2002 and then 2 years later he accepted positions with Harvard University and Harvard Business School in 2004 he remarks Harvard took it's time, but it's one institution that attains the highest standards in higher education regularly and worth waiting for them do their homework. Dr. Ferguson is also a Senior Fellow of the Hoover Institution, Stanford University. Niall Ferguson is a regular contributing editor of the Financial Times and his latest book "The Ascent of Money" is on the New York Times Best Seller List.
He is now best known for telling the story of money and the history of money in all its lustre, lore and affect on humans since the beginning of commerce and trade. Outside of the academic world Ferguson is known for his revisionist type views that try to rehabilitate imperialism and colonialism. Within academia circles he is the champion of counterfactual history and is often times the subject of some considerable controversy.
I had the chance to sit down with Niall Ferguson for sometime and I let him know that I am personally a great fan of his work but at the same time I pointed out the fact that he has some critics out there that doubt some or a lot of your work as not more than sensationalism.
Ferguson goes on to respond ' I am only interested in one thing and that is historical truth only to understand the world better. Nobody gains if historians take liberties with the past so I use my best efforts to research widely as I possible can to make sure that I've read as much of the existing specialist literature as I can to try to use the archives but research in my own right. I am trying to draw conclusions that are as accurate as I am able to make them. Explaining the complexities of the past financial history is not easy because this is a technical subject and there is an enormous complexity to the international financial system. So for television particularly we try to make it simpler. I think simplification is essential for teaching, as long as you don't simplify to the point of distortion and misrepresentation, then it seems to me that it is indeed perfectly legitimate and most desirable.
The Intellectual Process; Ferguson goes on to sum up how it all works: I think every serious historian wants to stimulate debate on a subject because the historical process is not like science, we don't just sit around in laboratories re-running human history. We offer our interpretations on the basis of the facts that we've been able to establish and then we discuss them and debate them intelligently.
Ferguson on the critics, fools and the nay Sayers: Throughout the time I have been publishing I have encountered critics of my work and feel it is an integral part of the historical scholarship that one's work should be subject to criticism. I think it's often through criticism that we progress closer to a better understanding of the past. There is no doubt in my mind the next book I write will find it's own set of critics. I look forward to any errors or mistakes in accuracy to clarify for future editions or paperback to be better. The other form of criticism is to say there is nothing wrong with the evidence presented here but you or others interpret it differently than me. So I have my critics, and I don't doubt that each new book I publish will find it's own set of new critics. I don't mind that as long as the criticism is fair. The kind of criticism you can point out like a mistake or an inaccuracy and if somebody does that, I hasten to correct the error so that the information in the paperback version or in future editions is better. The other form of criticism is to say, "Well there's nothing wrong with the evidence presented here but I interpret it differently from Ferguson on the difference of opinions." Well that's fine but if somebody chooses to offer a different interpretation from me, there's no particular reason why I should accept it. It would simply be a difference of emphasis. So those are the ways that historical criticism works and I think I've had my share of fair, and unfair, criticism over the years. It comes with the territory. If you said something that everybody agreed with, it probably wouldn't be worth saying. You'd simply be restating the conventional wisdom. My aim is always to challenge that received view of thinking.
Ferguson the passionate historian and author not a businessman: If one was interested in making money, one would have become an investment banker in 1985 when I graduated from Oxford and I never considered that for a moment. I dedicated my working life to understanding the way the world works, the modern world since around 1700. And the research I've done is a quest on my part to understand that better and then to explain it. And so if I run out of questions that I want to ask about the past then I'll retire from history. But there are always new questions and so there are always new projects. Right now I'm just putting the very last touches to a biography of an eminent banker in many ways the most important figure in London after the Second World War—Siegmund Warburg.
I asked Mr. Ferguson if he knew much about the war of 1812, The Plains of Abraham and Canada in general.
Ferguson: Well of course! A historian of the British Empire would surely know and understand a lot of about what pivotal things happened in upper and lower Canada to shape the entire world.
I asked Ferguson what he thinks of Canada as a nation and are we on the global radar?
Certainly, I mean the work I did on the British Empire that produced the book, Empire, was in part homage to Canada. I have longstanding family ties to Toronto and, indeed, to Saskatchewan. I have traveled widely in Canada and for me the history of North America is just not the history of the United States, as it often seems to be.
For me the history of North America is the history of two, if not three, major experiments: the experiments, of course of the republican form of government, after 1776 and 13—and subsequently more—colonies. But the other experiment was the experiment with constitutional monarchy within the British Empire that endured right the way through into the 20th century. And still retains I think a real importance in Canadian life and then within Canada, of course, you have at least two different cultures co-existing.
Ferguson goes on about Canada and it's Language Culture: Anglophone and the Francophone. It seems to be very, very interesting subjects for a historian of the modern period. I constantly try to remind my US-born students that if you compare the performance of Canada and the United States in economic terms, or in social terms, that it doesn't seem as if the, the choice of new political institutions in 1776 in those 13 colonies had a huge amount of difference. It's not like Canada's a desperately impoverished backwater. On the contrary, by some measures today, Canada is quite a way ahead of the United States. So this is a tremendously interesting field for historical study. Comparative history of North America is something we need more of. Frankly too few North American historians do that. There are historians of Canada, and there are historians of the United States. They tend to talk past one another and this seems to me to miss the whole point. The really interesting thing is that, despite a massive divergence in political institutions, these two societies have evolved in many similar ways. They've certainly remained materially, economically, on a par. But they've also developed some really profound institutional differences. And we see that very clearly in the current crisis, whether you look at banking regulation or healthcare. Canada looks to be in a stronger position than the United States.
It's a global economy now and not just domestic. The G8 will meet in Huntsville, Ontario next year; where does Canada rank and how do you think we will fair amongst other world economies?
Ferguson: Well I think Canada is in the strongest position it's been in, perhaps, internationally and in all of its history because I think Canadians can legitimately say that they've conducted financial regulation in a better way than the U.S. Given Canada's wealth in terms of resources, it doesn't have the kind of fiscal problems that are going to attack nearly all the other members of the G8. Canada's is the only public debt that isn't soaring up towards 100% of GDP in the next ten years. So it's a time when Canadians can perhaps legitimately put aside their historic inferiority complex and walk tall, because Canada's institutions look like they've done quite well in the last ten years, better certainly than those of the United States. The problem of course is that the world is changing so rapidly that the G8 signifies less as an institution than the G20. And it's striking that all the discussion in the American press is of the G20 in Pittsburgh rather than the G8.
Throughout history, time and the geography of the world we have been plagued by many wars. What is your opinion on war and in particular Afghanistan and Iraq?
Well I've written about American Empire and explored the question of the legitimacy of military action in a book called Colossus, which was published in 2004. I made it fairly clear that I don't regard war as always an illegitimate evil. I'm not a pacifist in that sense. Sometimes war is necessary—necessary of course in the case of an act of aggression which must be resisted, but it can also be necessary where preemption or prevention is preferable to waiting to be attacked. The United States certainly had no alternative but to take military action in Afghanistan after 9/11. Where it had a choice, by contrast, in the case of Iraq. I think, with the benefit of hindsight, it might have made more sense to focus on Afghanistan and not to invade Iraq. The benefits of that invasion of Iraq seem at this point to be exceeded by the costs not only in terms of money but also in terms of human life.
On the other hand, it's quite hard to see a very good future for Iraq or, indeed, that region if, if the status quo with Saddam in power had simply been left in place. So there wasn't an easy, happy alternative. It was a choice between two evils. But I must say that I think at this point, knowing what we know, this is not, on balance, a good decision, not least because overthrowing Saddam Hussein greatly strengthened the relative power of Iran. And I think the Bush administration underestimated that consequence of their action. Afghanistan is the simpler case as I think I've already said. It would be of course completely insane to abandon the effort that's currently being made in Afghanistan to create a stable government there. To allow the Taliban to come back into power in Afghanistan would be a complete disaster and anybody who thinks that that's an option must be suffering from some kind of amnesia.
You often speak of "Chimerica" referring to America's relationship with China please explain:
Well China's doing better than almost any economy in the world in terms of output. Its economy is one of the few that's growing strongly this year. China's growth of course is still way down compared with the pre-crisis period. It's almost been cut in half, so one shouldn't exaggerate the achievement. This still a very dramatic slowdown in relative terms and China faces at least three major problems. Right here and now the first problem is that by powering an economic recovery with a very large-scale state infrastructure program and very loose credit conditions, the Chinese have created something of a bubble in their own stock market which is now deflating rapidly. The second problem is that they've, by pursuing a strategy of export-led growth, accumulated 2 trillion dollars' worth of international reserves, a very large proportion of them denominated in US dollars, which doesn't look like the greatest investment in history. And the third problem that they have is that their rapid growth over the past 20 years has created a certain mismatch between their political system (still a one-party state) and their social system, which is changing very rapidly with the emergence of the middle class and indeed a super-rich elite. Historically that's a pretty difficult combination: political stagnation and rapid social change. So I think China's economic miracle is something that conceals at least three serious structural problems. Over the long run, they need to develop their own consumer society. They need to become less reliant on exports. They need to make their own income distribution more equal. But these things can't be done overnight. In the long run they also have major demographic problems because of the one-child policy. And so over a 20-year timeframe, China's prospects aren't actually quite so rosy.
Using some of his own verbiage I asked Ferguson to tell me about "Chindia":
He laughs on goes on to say: Well in some ways India resembles the tortoise in the story of "The Tortoise and the Hare" because China would represent the hare that zooms ahead in the early part of the race, but the tortoise, which moves more slowly, ultimately wins that race. And I think because India has a democracy, and it has the rule of law, and in many ways it relies less on government and more on the market. Its long-run prospects look more appealing. If India could address its infrastructural problem the way the Chinese have addressed theirs, they would be I think, a huge payoff. But it's hard to build a whole new system of highways, or high-speed rail links, if you have meaningful property rights. In China, you just tell the peasants, "Piss off, we're building a new, new highway here." You can't really do that in India because the peasants say, "No, you piss off because this is our land." So there's a big difference there. I have a relatively optimistic long-run view of India's trajectory. I think over a 50-year time horizon, things are looking good. Short-run, it's not going to grow, as fast as China and from that point of view, we shouldn't expect the kind of returns of investment that may be possible in China now.
In 2008, Allen Lane published his most recent book, The Ascent of Money: A Financial History of the World that he also presented as a PBS Special in the United States and on Channel 4 in Great Britain. Google has also chose to broadcast the video throughout the Internet making the author and the well-authored book 'The Ascent of Money' all that much more popular.
Niall believes that television and Internet will have a more vast reach, but books are important; after all it's just one way I can make a living by selling some books but getting the reach is as important; after all I am still just a Scottish professor who is only asking questions and learning more to teach better and sharing the wealth of that knowledge is what I do best.
"When I stop asking questions to satisfy even my self than that will be the day I stop learning and also seize being an historian and human, I suppose".
The Ascent of Money is a new comprehensive four-part series presents like thee epic film about money and the history of the global economic system. Niall Ferguson says, "I want to reveal financial history as the essential back-story behind all history. From ancient Mesopotamia, right down to the present the day, the ascent of money has been an indispensable part of the ascent of man."
How do you eat an elephant? Book Review
April 25, 2013 May 14, 2013 James Dean
Book Review: how to eat an elephant
MONEY.CA Book Review
HOW TO EAT AN ELEPHANT: THE MONEY BOOK REVIEW
Written by Ian R. Whiting, Senior Editor
Not exactly what you had in mind when you think elephants but this analogy turned idiom is a great way to describe and explain the complexities of personal finance in an easy and manageable way.
Frank Wiginton has developed a reputation as an easy-to-understand financial educator and has written an interesting book entitled "HOW TO EAT AN ELEPHANT – ACHIEVING FINANCIAL SUCCESS ONE BITE AT A TIME" that includes access to a self-help website (http://www.howtoeatanelephant.ca) that takes people through his entire process. His chosen title is very appropriate as developing a sound financial plan is perceived as a daunting exercise for most people.
Frank uses a variety of stories and scenarios to share his thoughts and perspectives on each part of your plan. Frank takes people through a logical process beginning with setting goals followed by that most feared word – budget! To his credit, his process helps relieve people of the much of the stress that often accompanies this challenging exercise.
Frank takes his readers on a journey through all of the important parts of a sound plan including debt and cash-flow management, life and disability insurance along with critical illness and long term care needs. He discusses pensions and other retirement resources including OAS and CPP/QPP, your RRSPs and TFSAs. He provides sound guidance on savings for education requirements and the legal niceties required to ensure our wishes are carried out in the event we are unable to act for ourselves and after we pass away.
After reading the book, I came to a series of conclusions about his process and style:
a) he emphasises that achieving your own definition of financial success is a journey, not a destination;
b) each person is accountable for their own successes and failures;
c) working with a professional team of advisors will reduce your overall level of risk of failure and enhance your probability of success, but you can't abdicate your own responsibility to your team;
d) common-sense is your greatest ally;
e) you need to have patience – starting with the patience to devote time each month to your own financial affairs; and
f) you need to monitor your progress regularly against your goals and make needed adjustments as soon as the need becomes apparent.
Like every effective author, Frank has strong personal beliefs and has done a very good job at sharing them with his readers. Even if you choose not to agree with his perspective or views, the process he uses to explain each part of the plan ensures that you go through a logical evaluation of the various options before making a decision rather than shooting from the hip.
Throughout the book, there are "One Frank Thought" boxes in which his personality shines through while sharing some unique perspectives on the topic in question – they add greatly to both the material and the concepts he presents.
A very worthwhile investment of $21.95 plus taxes is available at fine retail establishments including Chapters and Indigo through Wiley and Sons Limited http://ca.wiley.com/WileyCDA/ the publisher or www.howtoeatanelephant.ca.
To view the photo associated with this press release, please visit the following link: http://www.marketwire.com/library/20130425-How_To_Eat_An_Elephant_ORIG.jpg
To view the video associated with this press release, please visit the following link: http://www.youtube.com/watch?v=VySIxLCtQsk

redpajama_set_name: RedPajamaCommonCrawl


The Park Foundation was formed in 1966. Its original focus was on education and grant-making in communities where Park Communications had interests. When he died in 1993, Mr. Park bequeathed more than 70% of his holdings to the Foundation. The Foundation is dedicated to the aid and support of education, public broadcasting, environment, and other selected areas of interest to the Park family. Scholarship programs have been established in Mr. Park's name at the two institutions with which he was so close — Ithaca College and North Carolina State University. The two scholarship programs emphasize academic excellence, leadership, and community service — in keeping with Mr. Park's values. Public broadcasting is a particularly meaningful recipient of funding because the Foundation had its origin in the world of communications. More recently, the Foundation's interest in environmental causes has been refined to focus on issues of freshwater, particularly in the eastern United States.


and slightly exhausted at the end.
This week, Winning Wednesday is a book I read last year that I want to share with someone – "A Thousand Splendid Suns" by Khaled Hosseini (he also wrote The Kite Runner) It is a good book and I experienced another culture in a way I hadn't before and I was exhausted at the end because I wanted to help every woman in Afghanistan. I wanted to fix their situation! Not that I think their world should look like mine, but I was sad at the centuries of traditions those women are burdened with and the degrading treatment and/or abuse that is a part of too many of their lives.
Reading "A Thousand Splendid Suns" opened my eyes to other world, taking my view beyond the pictures on the evening news. And if I allow it, I'll be more compassionate because I will better understand individuals with a different life than my own. Individuals created by the same creator as me – and in his image, just like I am.
Do you want to read it? Leave your name below for a chance to win it.
Also, I would love to hear your thoughts about how far one 'drop' can ripple. Have you seen that in your own life or someone elses?
Sounds very interesting. Women of different cultures have been on my heart letely.
Sounds like something I would like to read. I always read the same stuff and I feel that I should read about life and get to know what it going on.
How far one drop can ripple…ponder this. I have a heart for victims of human trafficking. Over 27 million victims worldwide, with many experts saying that is grossly understimated. Mind boggling, isn't it? Almost too much to even tackle. BUT….if I help free 2, who are then empowered to help 2 more, who are then empowered to help 2 more….in less than 2 months trafficking would be eliminated. Of course, the process actually takes years, but it proves how a little really goes SO far. Bless you Janet!!!!
Gayle – agree with you about the human trafficking. I find myself not even able to think about it at times, because its so overwhelming. Interesting that 2 by 2 by 2 could equal such a major effect in so short a time. That does make a big problem look more manageable.
Don't pick me…have it and already read it. I wholeheartedly agree. Life-changing and thought-provoking. Just like his first, but very different too. As far as the ripple…this summer, I met a man who was affected 20 years ago by someone's writing, and to this day it affects his daily life and ministry. Profound. Best part of my summer!
This one has been on my "to read" list for a while…..I'm feeling lucky!!
.there are so many ways a drop of water has effected my life….the biggest one was when I was hospitalized for 23 days. I developed a massive blockage in intestines that was beginning to poison my body. I was scheduled for emergency surgery which my doctor told me would be very very risky and that I should call my family. I called Jon first…that was the drop of water. He was on his way to church when I called. He told Pastor John what was happening. Then the ripple began….Pastor John asked everyone to pray for me. (I didn't know the whole church was praying for me long after Jon arrived) When Jon got to the hospital, they took me for one last ultrasound to check where the huge blockage was and it was gone! Nothing there! The surgeon was astounded but I just smiled at him and said "Believe…..just believe in miracles…..God knew my needs and He worked a miracle" The surgeon just shook his head and said "I've never seen this happen before….maybe it's time to start believing". Yeah!! To this day I pray for the doctor that he has learned to believe in and trust God, that miracles DO happen.


I'm fairly new new here, and it seems some friends might be helpful.
Name above stable door is "Jurgen fischer", please add me.
How do you add people? My name is Jane Marie Nash would really like more friends but not sure how to add you all if I search for names it's too vague for FB.
Scott Parks. Only through the HoF stand alone site. I won't accept friend invites on facebook.
Don't let your mind wander. It's too small and fragile to be out alone.
add me plss.. nid more friens. Alp Aquino..
ps... where is the turn on chat button?
the instructions to turn on/off chat are in the Wiki, under the heading "chat rooms". You can get to the Wiki by clicking on "game help" in the racing screen.
Could you please add me as friend also..