text:
\section{Introduction}

In this paper we study the motion of an axisymmetric, current carrying relativistic string
loop in the background of a Kerr black hole. This interesting dynamical system
has been studied before in its own right~\cite{Larsen:1993nt,Frolov:1999pj}),
and our goal here is to better understand its behavior.

Our investigation is loosely motivated
by the problem of the production of collimated
astrophysical jets
of matter. Systems that exhibit collimated jets range from accreting young stars, neutron stars and black holes to supermassive black holes
in quasars and
active galactic nuclei~\cite{livio}.
Magnetized plasmas interacting with an accretion disk are believed to play
a central role in
jet production, but despite much work and many prosposals for the mechanism,
the process is still not yet well understood.
Such plasmas are governed by magnetohydrodynamics
(MHD), a complicated nonlinear field theory. Under some circumstances, plasmas
exhibit associated string-like behavior, either via the dynamics of magnetic
field lines embedded in the plasma~\cite{Christensson:1999tp,semenov,Semenov:2004ib}, or in the dynamics of relatively thin isolated
flux tubes of plasma which can be approximately described by an effective
one-dimensional string~\cite{spruit}.
In either of these cases, the string would be described by an
energy density
and a tension, and might be carrying currents of mass and/or charge.
The hope is that some essential aspects of the physics could be
captured by string dynamics, which is tremendously simpler than MHD.

The dynamical
formalism for describing relativistic current carrying strings is
well-developed~\cite{Hindmarsh:1994re,vilshell,Larsen:1993ha,Carter:1996rn},
and a generalization to a wider variety of equations of state
is given in Ref.~\cite{Carter:1997pb}.
Here we restrict attention to a relatively simple class of strings, namely
axisymmetric loops characterized by a constant tension and a conserved
mass current on the string worldsheet,
in order to begin developing some insight into the factors influencing the string dynamics.
This could be generalized in future work to allow for different equations of state,
coupling of a charge current to an external
electromagnetic field~\cite{Larsen:1992xz,Larsen:1992ya}, and deviations from
axisymmetry~\cite{Dyadechkin:2008zz, Dubath:2006vs, Snajdr:2002aa, Snajdr:2002rw}.

The string tension prevents a string loop from expanding beyond some radius,
and the worldsheet current can produce an angular momentum barrier preventing
the  loop from collapsing into the black hole. Thus, depending on its energy and current,
a string may be trapped in a toroidal region surrounding the black hole, or its motion may be
confined to a cylindrical shell that extends to infinity. In the latter case, one
can envisage processes wherein internal energy of the string is converted into
translational kinetic energy via the propagation on the black hole background.
This is one of the
phenomena we aim to understand. The other key question is what role may be
played by the spin of the black hole, as a result of the associated dragging
of inertial frames.

\section{Current carrying string in curved spacetime}
\label{covfeq}

The string worldsheet is described by giving its spacetime coordinates
$X^\lambda(\sigma^a)$ ($\lambda=0,1,2,3$)
as functions of the two worldsheet coordinates $\sigma^a$ ($a=0,1$).
The induced metric on the worldsheet is
\begin{equation}
\label{induced}
h_{ab}= g_{\mu\nu}X^\mu_{,a}X^\nu_{,b},
\end{equation}
where $g_{\mu\nu}$ is the ambient spacetime metric,
for which we adopt the signature ($-$$+$$+$$+$).
To describe the current we introduce a scalar field $\varphi(\sigma^a)$
living on the worldsheet.
We consider dynamics generated by the action
\begin{equation}
\label{scs}
S=-\int d^2\sigma\, \sqrt{-h}\left(\mu/c+h^{ab}\varphi_{,a} \varphi_{,b}\right),
\end{equation}
where the constant $\mu$ denotes the string tension,
and $c$ is the speed of light which is
hereafter set to unity.
We assign the line element $g_{\mu\nu} dx^\mu dx^\nu$
the dimensions of length squared, and take the
worldsheet coordinates to be dimensionless,
so that the dimensions of $h_{ab}$ are length squared,
those of $\mu$ are energy over length,
and those of $\varphi$ are square root of action.\footnote{Strings described
by action (\ref{scs}) were originally introduced~\cite{Witten:1984eb}
as an effective description of
``superconducting strings", a type of topological defect that
might occur in a theory with multiple scalar fields undergoing spontaneous
symmetry breaking.}

The action is the integral of a worldsheet scalar density,
hence it is independent of the choice of worldsheet
coordinates. Note that the part of the action involving the scalar field $\varphi$ is
invariant under conformal rescalings of the metric.
One could of course consider
other scalar functions of the invariant $h^{ab}\varphi_{,a} \varphi_{,b}$ in
the Lagrangian, but for the exploratory purposes of the present
investigation the simple choice in (\ref{scs}) is sufficient.
In any case, one ultimately wants to generalize the system to
allow for a wider class of equations of state that are perhaps more
suitable for MHD applications.

Varying the action with respect to $h_{ab}$ yields the worldsheet
stress-energy tensor density $\tilde{\Sigma}^{ab}$,
\begin{equation}
\delta_h S = \textstyle{\frac{1}{2}}\int d^2\sigma\, \tilde{\Sigma}^{ab}\delta h_{ab},
\end{equation}
where
\begin{equation}\label{tS}
\tilde{\Sigma}^{ab}=\sqrt{-h}\Bigl(2j^a j^b - (\mu+j^2)h^{ab}\Bigr),
\end{equation}
and
\begin{equation}
j_a = \varphi_{,a}, \qquad j^a = h^{ab}j_b, \qquad j^2 = h^{ab}j_aj_b.
\end{equation}
The tilde on $\tilde{\Sigma}^{ab}$ is a reminder that this quantity has density weight one
with respect to worldsheet coordinate transformations.
Because of the conformal invariance, the $j$-dependent part of $\tilde{\Sigma}^{ab}$ is traceless.
The contribution from the string tension
is proportional to the metric, and with $\mu>0$ has
a positive energy density and
an opposite, negative pressure, i.e.\ a tension.
The contribution from the
current is traceless, due to the
conformal invariance of that part of the action.
It can be viewed as a 1+1 dimensional massless radiation fluid,
with positive energy density and equal pressure.

Varying the action with respect to $X^\mu$ yields the equation of
motion\footnote{One could rewrite eq.~(\ref{eom1}) in terms of the
Christoffel symbols of $g_{\mu\nu}$, however
it is more convenient to work with the explicit form (\ref{eom1}).}
\begin{equation}
\label{eom1}
(\tilde{\Sigma}^{ab}g_{\mu\lambda}X^\mu_{,a})_{,b}-\textstyle{\frac{1}{2}} \tilde{\Sigma}^{ab}g_{\mu\nu,\lambda} X^\mu_{,a} X^\nu_{,b}=0.
\end{equation}
Varying the action with respect to $\varphi$ yields the 1+1 dimensional
wave equation,
\begin{equation}
\label{eomj}
(\sqrt{-h} h^{ab}\varphi_{,a})_{,b}=0,
\end{equation}
which also implies that the current is divergenceless, $(\sqrt{-h}j^a)_{,a}=0$.
The worldsheet stress-energy tensor is also divergenceless,
$\tilde{\Sigma}^{ab}{}_{;b}=0$, with respect to the $h_{ab}$-compatible
covariant derivative. (To verify this directly  one must use $j_{a,b}=j_{b,a}$,
which follows from $j_a=\varphi_{,a}$. Conversely, $(\sqrt{-h}j^a)_{,a}=0$
and $\tilde{\Sigma}^{ab}{}_{;b}=0$, together imply that $j_a=\varphi_{,a}$ for some scalar field
$\varphi$.)

\section{Axisymmetric dynamics in a stationary, axisymmetric spacetime}
\label{dynamics}

\subsection{Conformal gauge}
Every two dimensional metric is conformal to a locally flat metric,
hence in particular
\begin{equation}\label{O}
h_{ab}=\O^2 \eta_{ab},
\end{equation}
where $\eta_{ab}$ is locally flat and
$\O$ is a worldsheet scalar function.
We can always adopt coordinates
$\sigma^a=(\tau,\sigma)$
such that the coordinate components
of $\eta_{ab}$ are $\eta_{\tau\sigma}=0$ and $\eta_{\tau\t}=-\eta_{\sigma\s}=-1$,
i.e.\ the constant Minkowski metric.
This ``conformal gauge" choice
is equivalent to requiring $h_{\tau\sigma}=0$ and $h_{\tau\t}+h_{\sigma\s}=0$,
which as conditions on $X^\mu$ read
\begin{equation}\label{confgauge}
g_{\mu\nu}X^\mu_{,\tau}X^\nu_{,\sigma}=0
=g_{\mu\nu}(X^\mu_{,\tau}X^\nu_{,\tau}+X^\mu_{,\sigma}X^\nu_{,\sigma}).
\end{equation}
In this gauge
$h=-\O^4$, and the metric components
satisfy $\sqrt{-h}\, h^{ab}=\eta^{ab}$,
where $\eta^{ab}$ is the inverse of the Minkowski
metric. The conformal gauge conditions (\ref{confgauge})
do not completely fix the coordinates: the lightlike
combinations $\tau\pm\sigma$ may be replaced by any smooth
monotonic functions of themselves.

We aim to study axisymmetric string motion in
an axisymmetric, stationary spacetime. Such a
spacetime is
described by a metric, written in coordinates $(t,r,\theta,\phi)$,
of the general form
\begin{equation} \label{ds2}
ds^2 = g_{tt}dt^2 + 2g_{t\phi}dt d\phi
+g_{\phi\phi}d\phi^2 +g_{rr}dr^2
+g_{\theta\theta}d\theta^2
\end{equation}
where all the metric components are independent of $t$ and $\phi$.

It is tempting to choose the worldsheet coordinate $\sigma$ to be
equal to the spacetime coordinate $\phi$, in which case
a general axisymmetric worldsheet would take the form
$X^\mu(\sigma,\tau)=(t(\tau),r(\tau),\theta(\tau),\sigma)$. However, this
would imply $h_{\tau\sigma}=g_{t\phi}t_{,\tau}$, which
does not satisfy the conformal gauge condition
if $g_{t\phi}\ne0$. We therefore allow for a $\tau$ dependent relative shift
$\phi=\sigma+f(\tau)$, so that
\begin{equation}\label{X}
X^\mu(\sigma,\tau)=(t(\tau),r(\tau),\theta(\tau),\sigma+f(\tau)).
\end{equation}
The gauge condition $h_{\tau\sigma}=0$ then becomes
\begin{equation}\label{Xd-phi}
g_{\mu\phi}\dot{X}^\mu=g_{t\phi} \dot{t} + g_{\phi\phi} \dot{f}=0,
\end{equation}
which determines $f$ via
\begin{equation}
 \dot{f}=-(g_{t\phi}/g_{tt}) \dot{t},\label{fdot}
\end{equation}
where the dot stands for $d/d\tau$.
Equation (\ref{Xd-phi}) means that
$\dot{X}^\mu$, and thus the worldsheet vector
$\partial_\tau$, are orthogonal to the
axial rotation Killing vector $\partial_\phi$.
That is, the curves of constant $\sigma$ are
zero angular momentum (ZAMO) worldlines.
For later reference we note that this
same gauge condition yields the useful relation
\begin{equation}\label{Xd-t}
g_{\mu t}\dot{X}^\mu=(g_{tt}-g_{t\phi}^2/g_{\phi\phi})\dot{t}.
\end{equation}

The second gauge condition $h_{\tau\t}+h_{\sigma\s}=0$
becomes
\begin{equation}
(g_{tt}-g_{t\phi}^2/g_{\phi\phi})\dot{t}^2+g_{rr}\dot{r}^2+g_{\theta\theta}\dot{\theta}^2+g_{\phi\phi}=0.
\label{gauge2}
\end{equation}
The conformal factor $\O^2$ is determined in this gauge
by the equation $h_{\sigma\s}=\O^2\eta_{\sigma\s}=\O^2$, i.e.
\begin{equation}\label{Omega}
\O^2=g_{\phi\phi}.
\end{equation}

\subsection{Equations of motion}

In the conformal gauge (\ref{O})
the equation of motion (\ref{eomj}) for $\varphi$ takes the simple form
\begin{equation}\label{waveeqn}
\varphi_{,\tau\t}-\varphi_{,\sigma\s}=0.
\end{equation}
The general solution is
$\varphi=\varphi_+(\tau+\sigma)+\varphi_-(\tau-\sigma)$, where $\varphi_\pm$ are arbitrary functions.
On the other hand,
the assumption of
axisymmetry implies that the current is
independent of $\sigma$. Thus $j_{a,\sigma}=0$, which implies
$ \varphi_{,\tau\sigma}=\varphi_{,\sigma\s}=0$. Together with (\ref{waveeqn}) this
implies that $\varphi$ must be linear in both $\sigma$ and $\tau$,
\begin{equation}\label{alpha}
\varphi=j_\sigma\s+ j_\tau\t,
\end{equation}
where $j_\sigma$ and $j_\tau$ are constants.\footnote{Note that while
$j_s$ is single-valued, the scalar $\varphi$ is multi-valued. When the
string action is taken to be an effective description of a field theory
defect~\cite{Witten:1984eb}, $\varphi$ is a phase defined modulo
$2\pi$, which implies that the possible values of the current are
quantized.}
Then $j^2=\Omega^{-1}(j_\sigma^2-j_\tau^2)$,
and the components of $\tilde{\Sigma}^{ab}$ (\ref{tS}) become
\begin{eqnarray}\label{Sigmatilde}
\tilde{\Sigma}^{\tau\t}&=& g_{\phi\phi}^{-1}(j_\sigma^2+ j_\tau^2)+\mu\label{tStt}\\
\tilde{\Sigma}^{\sigma\s}&=&g_{\phi\phi}^{-1}(j_\sigma^2+ j_\tau^2)-\mu\label{tSss}\\
\tilde{\Sigma}^{\sigma\tau}&=&-2g_{\phi\phi}^{-1}j_\sigma j_\tau.\label{tSst}
\end{eqnarray}

The quantity $\tilde{\Sigma}^{\tau\t}$ in the conformal gauge
is equal to the string energy density
as measured by observers co-moving with the string at
constant $\sigma$.
These are the co-moving ZAMO observers
mentioned in the discussion following eq.~(\ref{fdot}).
To verify
the claim about the relation between $\tilde{\Sigma}^{\tau\t}$ and the string energy density, note that
since the 2-velocity $u^a$ of those observers is
orthogonal to the constant $\tau$ surfaces, we have
$u_a=\O \partial_a \tau$, the prefactor $\O$ being
determined by the unit normalization of $u^a$.
Also $h=-\O^4$. The energy density is thus
given by $(-h)^{-1/2}\tilde{\Sigma}^{ab}u_a u_b=\tilde{\Sigma}^{\tau\t}$.
In particular, $\mu$ is the contribution form the
string tension, and $(j_\sigma^2+ j_\tau^2)/g_{\phi\phi}$ is the
energy density of the current.

The equations of motion (\ref{eom1}) can now be written out
more explicitly in our chosen gauge. Using the gauge conditions
and the axisymmetry, they take the form
\begin{align}
\label{eom1a}
&(\tilde{\Sigma}^{\tau\t}g_{\mu\lambda}\dot{X}^\mu + \tilde{\Sigma}^{\sigma\tau}g_{\phi\lambda})_{,\tau}\\
&\quad-\textstyle{\frac{1}{2}}(\tilde{\Sigma}^{\tau\t}g_{\mu\nu,\lambda}\dot{X}^\mu\dot{X}^\nu+\tilde{\Sigma}^{\sigma\s}g_{\phi\phi,\lambda}+\tilde{\Sigma}^{\tau\sigma}g_{t\phi,\lambda}\,\dot{t})=0.\nonumber
\end{align}
For $\lambda$ equal to $\phi$
this equation is identically satisfied. For $\lambda=t$,  it yields
(using (\ref{Xd-t})) the energy conservation law,
\begin{equation}\label{eomt}
\tilde{\Sigma}^{\tau\t}(g_{tt}-g_{t\phi}^2/g_{\phi\phi})\dot{t} + \tilde{\Sigma}^{\sigma\tau}g_{t\phi}=-E,
\end{equation}
where $E$ is a constant to be identified below with the total Killing energy
of the string (divided by $2\pi$).
For $\lambda$ equal to $r$ or $\theta$ it yields
\begin{align}
 \left(\tilde{\Sigma}^{\tau\t}g_{rr}\dot{r}\right)_{,\tau} -&\Big(\tilde{\Sigma}^{\tau\t}g_{\mu\nu,r}\dot{X}^\mu\dot{X}^\nu+\tilde{\Sigma}^{\sigma\s}g_{\phi\phi,r}\Big)=0\label{eomr}\\
\left(\tilde{\Sigma}^{\tau\t}g_{\theta\th}\dot{\theta}\right)_{,\tau} -&\Big(\tilde{\Sigma}^{\tau\t}g_{\mu\nu,\theta}\dot{X}^\mu\dot{X}^\nu +\tilde{\Sigma}^{\sigma\s}g_{\phi\phi,\theta}\Big)=0.\label{eomth}
\end{align}

The dynamics depends on the current only through the worldsheet
stress tensor (\ref{tStt}-\ref{tSst}), in which the current enters only
in the
two quadratic combinations $j_\sigma^2+ j_\tau^2$ and $j_\sigma j_\tau$,
both of which are symmetric under interchange of
$j_\sigma$ and $j_\tau$.
Note that $j_\sigma j_\tau $ appears only via $\dot{t}$, through the
energy conservation law
(\ref{eomt}), and then only when $g_{t\phi}\ne0$.

To parametrize the solutions, we will later use the first of
these combinations, together with the ratio of the current components,
\begin{eqnarray}
J^2&\equiv &j_\sigma^2+ j_\tau^2\\
\omega}\def\O{\Omega &\equiv &-j_\sigma/j_\tau.
\end{eqnarray}
(The minus
sign is included so that positive $\omega}\def\O{\Omega$ will
correspond to positive angular momentum.
Note that $-j_\sigma/j_\tau=j^\sigma/j^\tau$.)
The product $j_\sigma j_\tau $ can be expressed
in terms of $J$ and $\omega}\def\O{\Omega$
as $j_\sigma j_\tau = -\omega}\def\O{\Omega J^2/(1+\omega}\def\O{\Omega^2)$.
Since the dynamics is
symmetric under interchange of $j_\sigma$ and $j_\tau$,
the range $-1\le\omega}\def\O{\Omega\le1$ covers all distinct
cases. The extreme values $\omega}\def\O{\Omega=\pm1$ corresond
to null currents.

\subsection{Conserved quantities}

Suppose $\xi^\lambda$ is a spacetime Killing vector. If the
spacetime coordinates are chosen so the components of $\xi^\lambda$
are constant everywhere, then Killing's
equation implies $g_{\mu\nu,\lambda} \xi^\lambda = 0$.
Contracting (\ref{eom1}) with $\xi^\lambda$ then
yields a conserved Killing current. That is,
the worldsheet vector density
\begin{equation}
\label{killcur}
{\cal J}_\xi^b = \tilde{\Sigma}^{ab} X^{\mu}_{,a} g_{\mu\lambda} \xi^\lambda
\end{equation}
satisfies
\begin{equation}
\label{killcur2}
{\cal J}_\xi^b{}_{,b}=0.
\end{equation}
Although we argued for its conservation using an adapted coordinate system,
${\cal J}_\xi^b$ is manifestly spacetime coordinate independent.
Note that it is just the worldsheet energy-momentum tensor contracted
with the pullback of the Killing one-form to the worldsheet. Thus
integrating it over any closed cross section of the worldsheet gives
the conserved quantity associated with the Killing vector,
\begin{equation}
 Q_\xi=\int {\cal J}_\xi^b dS_b.
\end{equation}
If we take the integral over a surface of constant $\tau$
and use the axisymmetry, the integral becomes simply
\begin{equation}
Q_\xi=2\pi{\cal J}_\xi^\tau=2\pi \tilde{\Sigma}^{a\tau}X^\mu_{,a} g_{\mu\lambda}\xi^\lambda.
\end{equation}

When $\xi^\lambda$ is the Killing vector $\partial_t$ corresponding to $t$ translation symmetry, then
(\ref{X}) and (\ref{Xd-t}) yield
\begin{equation}
-Q_{\partial_t}={\cal E}=2\pi E,
\end{equation}
where $E$ is the constant introduced in eq.~(\ref{eomt}).
Thus $E$ is in
fact the Killing energy of the string divided by $2\pi$ .
When $\xi^\lambda$ is the Killing vector $\partial_\phi$
corresponding to $\phi$ translation symmetry (rotation), then
using (\ref{X}), (\ref{Xd-phi}) and (\ref{tSst})
we find the angular momentum $L$ of the string,
\begin{equation}
Q_{\partial_\phi}=L=-4\pi  \, j_\sigma j_\tau.
\end{equation}
This is manifestly constant for the solutions (\ref{alpha}) under consideration.
Without the current, the string carries no angular momentum, since the
stress tensor is then proportional to $h_{ab}$, which is
boost invariant along the worldsheet.

\subsection{Effective potential}

We can solve (\ref{eomt}) for $\dot{t}$
and substitute into the gauge condition (\ref{gauge2}), yielding
\begin{equation}\label{gaugeKerr}
(g_{t\phi}^2/g_{\phi\phi}-g_{tt})(\tilde{\Sigma}^{\tau\t})^2(g_{rr}\dot{r}^2+g_{\theta\th}\dot{\theta}^2) + V(r,\theta)=0,
\end{equation}
where
\begin{equation}\label{VKerr}
V(r,\theta)=-(E+g_{t\phi}\tilde{\Sigma}^{\sigma\tau})^2+(g_{t\phi}^2-g_{tt}g_{\phi\phi})(\tilde{\Sigma}^{\tau\t})^2
\end{equation}
is what we call here the ``effective potential".
The motion outside the horizon
is confined to the region where $V(r,\theta)\le0$, since
the first term in (\ref{gaugeKerr}) is non-negative
outside the horizon.\footnote{The determinant
of the metric (\ref{ds2}) is
$(g_{tt}g_{\phi\phi}-g_{t\phi}^2)g_{rr}g_{\theta\th}$, which must
be negative for a Lorentzian metric. For the metrics we consider,
$g_{\theta\th}>0$ everywhere and $g_{rr}, g_{\phi\phi}>0$
outside the horizon,  so evidently $g_{t\phi}^2/g_{\phi\phi}-g_{tt}$ is positive
outside the horizon.}
More explicitly, the motion is bounded by the contour\footnote{To obtain
this we have taken a square root, but the root
is unique since, according to (\ref{eomt}) the combination
$E+g_{t\phi}\tilde{\Sigma}^{\sigma\tau}$ is positive for trajectories outside
the horizon.}
\begin{eqnarray}\label{VKerr0}
E=E_b&\equiv&(g_{t\phi}^2-g_{tt}g_{\phi\phi})^{1/2}\,
\tilde{\Sigma}^{\tau\t}-g_{t\phi}\tilde{\Sigma}^{\sigma\tau}\\
&=&A(r,\theta)\mu + B(r,\theta,\omega}\def\O{\Omega) J^2,
\end{eqnarray}
where $A$ and $B$ are functions determined by the metric components and, in the case of $B$, the value of $\omega}\def\O{\Omega$.
The initial conditions and the value of the current
determine $E$ through (\ref{gauge2}) and (\ref{eomt}), and the
subsequent motion is confined to the region contained within
the contour (\ref{VKerr0}).  We shall determine that motion
in several representative cases by numerical
integration.

\section{Newtonian elastic ring}
\label{newtanal}

Before presenting the results for the relativistic string, we
develop a Newtonian analogue of the system, to help interpret the
dynamics. Indeed, it turns out that most of the salient features
are independent of relativistic effects.

Consider a Newtonian elastic ring of total mass $m$, moving axisymmetrically in the
gravitational field of a point mass $M$. Using cylindrical coordinates $(\rho,z,\phi)$,
the conserved Newtonian energy of the system is given by
\begin{equation}
E_N= \textstyle{\frac{1}{2}} m (\dot{\rho}^2+\dot{z}^2 + \rho^2 \omega}\def\O{\Omega^2)+\textstyle{\frac{1}{2}} k \rho^2-GMm/(\rho^2+z^2)^{1/2}\, ,
\end{equation}
where $\omega}\def\O{\Omega$ is the angular velocity and $k$ the elastic constant.
The angular momentum $l=m\rho^2\omega}\def\O{\Omega$ about the axis is conserved.
Using $l$,
energy conservation can be expressed as
\begin{equation}
\textstyle{\frac{1}{2}} m (\dot{\rho}^2+\dot{z}^2)+V_N(\rho,z) =0,
\end{equation}
with
\begin{equation}\label{VN}
V_N(\rho,z)=-E_N + \frac{l^2}{2m\rho^2} + {\textstyle{\frac{1}{2}} k \rho^2}-\frac{GMm}{\sqrt{\rho^2+z^2}}.
\end{equation}
The motion of the ring is confined to the region
$V_N(\rho,z)\le0$, and thus bounded by the contour
$V_N(\rho,z)=0$, i.e. the curve in the $\rho$-$z$ plane given by
\begin{equation}\label{Ncontour}
E_N = \frac{l^2}{2m\rho^2} + {\textstyle{\frac{1}{2}} k \rho^2}-\frac{GMm}{\sqrt{\rho^2+z^2}}.
\end{equation}

\begin{figure}[t ]
\subfigure[ ~$l=0$ \label{fig4a}]{\includegraphics[width=4cm]{newtplot0.pdf}}
\subfigure[~$l=0.12$ \label{fig4b}]{\includegraphics[width=4cm]{newtplot0p12.pdf}}\\
\subfigure[~$l=1.5$ \label{fig4c}]{\includegraphics[width=4cm]{newtplot1p5.pdf}}
\subfigure[~$l=2$ \label{fig4d}]{\includegraphics[width=4cm]{newtplot2.pdf}}
\caption{\label{fig4}  Newtonian elastic spinning ring of total mass $m$, moving axisymmetrically in the gravitational field of a point mass $M$. Contour plots of the boundaries on the motion for different energies and for angular momentum
$l=0,\, 0.12,\, 1.5,\, 2$ (in units with $m=k=GM=1$). Both bound
trajectories and escape trajectories can be present,
depending on the energy.}
\end{figure}

The bounding contour is given in Fig.~\ref{fig4} for
different values of the angular momentum and energy, in units with $m=k=GM=1$.
The angular momentum term $\propto \rho^{-2}$ dominates at small $\rho$ and makes an
inner $\rho$ wall,
while the elastic energy term $\propto\rho^2$ dominates at large $\rho$
and makes an outer $\rho$.
The gravitational potential
well deforms the inner wall inward toward the center,
and the outer wall outward.
The figure illustrates that,
depending on
the energy and angular momentum, the ring can either oscillate in a confined toroidal
region around the equatorial plane,  or escape to infinity confined to a channel
along the azimuthal direction. In the exceptional case of
zero angular momentum, the ring can be confined to a spherical region
including the origin.

\section{String dynamics}
\label{motion}

\subsection{String in flat background}
\label{flatbackground}

In order to demonstrate the effect of the current on the motion of the string
in a simplified setting,
and to obtain a useful formula for the string energy far from the black hole,
let us first consider the case in which the background is just flat spacetime.
We use cylindrical coordinates $(t,\rho,z,\phi)$, so $g_{\mu\nu}={\rm diag}(-1,1,1,\rho^2)$, and we consider
axisymmetric motion.
The effective potential (\ref{VKerr}) then takes the form
\begin{equation}\label{Vflat}
V(\rho)=-E^2 + \rho^2(\tilde{\Sigma}^{\tau\t})^2.
\end{equation}
The motion is confined to the region where $V(\rho)\le0$.

The bounds of the motion are defined by
$V(\rho)=0$, i.e.\
\begin{equation}\label{V0flat}
E=E_b=\rho \tilde{\Sigma}^{\tau\t}=\mu\rho +J^2/\rho.
\end{equation}
The right hand side is $1/2\pi$ times
the Killing energy
of the string when $\dot{r}=\dot{\theta}=0$.
Since the string is then at rest, and there is no
redshift factor in flat spacetime, this is the same as
the energy measured in the rest frame of the string.
The first term represents the
elastic energy of the string, while the
second term is
the angular momentum barrier arising
from the current circulating in both
directions around the string. Alternatively, it
can be viewed as a
variable ``rest mass"
of the string, arising from the
energy density $J^2/\rho^2$ of the
current.
In the presence of the current, the string
oscillates between two
$\rho$ values, the roots of (\ref{V0flat}),
never collapsing to a point.

The elastic energy and angular momentum
play similar roles in the relativistic case
(\ref{V0flat}) and the Newtonian
case (\ref{Ncontour}), but with different powers
of $\rho$. The relativistic elastic energy is proportional
to the string length rather than its square,
and the angular momentum barrier comes
from a kinetic energy that is linear rather than
quadratic in the momentum.
A more important difference between the Newtonian
elastic ring and the relativistic string is that
in the latter the current can
circulate in either direction. Both directions can
simultaneously contribute to the angular momentum barrier, so the
barrier is not determined by the net angular momentum.
Indeed it is present even when the net angular momentum
vanishes, which is the case when either $j_\sigma$ or $j_\tau$ vanishes.

\subsection{String in Kerr background}

\subsubsection{General setting}

Next we consider the full problem,
{\em i.e.}\ the motion of the string in Kerr spacetime, with line element
(in Boyer-Lindquist coordinates)
\begin{align}\label{Kerr}
ds^2=&-\left(1-\frac{2Mr}{R^2}\right) dt^2
-\frac{4M r a\sin^2\theta}{R^2} dt d\phi\nonumber\\
&\quad+\left(r^2+a^2+\frac{2M r a^2}{R^2} \sin^2\theta\right)\sin^2\theta d \phi^2\nonumber\\
&\qquad +\frac{R^2}{\Delta^2} dr^2
+R^2 d\theta^2.
\end{align}
Here $M$ and $a=J/M$ are the mass and the specific
angular momentum of the black hole respectively,
\begin{equation}
R^2=r^2+a^2\cos^2\theta, \quad \Delta^2=r^2-2M r+a^2,
\end{equation}
and we are using units in which $G=c=1$.
(Below we shall instead choose units with $GM=1$.)
The event horizon is located where $\Delta=0$, at
\begin{equation}
r_h=M+\sqrt{M^2-a^2},
\end{equation}
and the boundary of the ergoregion lies where the coefficient of $dt^2$ vanishes, at
\begin{equation}
r_e=M+\sqrt{M^2-a^2 \cos^2\theta}.
\end{equation}
We consider only motion outside the horizon, so
this coordinate system is adequate.

\begin{figure}[t]
\includegraphics[width=8cm]{initialconditions.pdf}
\caption{\label{figinit} Graphical representation of the initial conditions, in units with $GM/c^2=1$.
The area inside the horizon is covered in grey, the dotted red line marks the equatorial plane and the dashed
black curve is the boundary of the ergoregion. $P_1$ corresponds to $r=r_h+1.25$ and $P_2$ to $r=r_{\rm ISCO}$.
At both points, $\theta=\pi/2+0.02$, and we choose $\dot{r}=\dot{\theta}=0$. The specific angular momentum
is $a=0.99$ and, therefore,  $P_2$ lies within the ergoregion ($r_{\rm ISCO} < r_e$).}
\end{figure}

The explicit form of the effective potential (\ref{VKerr})
is too complicated to be illuminating, unlike the flat background case.
A given orbit is determined once  $j_s$, $j_t$, and
initial conditions are specified.
The initial conditions determine the values of the energy
$E$ and angular momentum $L$,
and the
subsequent motion is confined to the region contained in
the contour (\ref{VKerr0}).
The equations of motion can be integrated numerically
to explore the detailed behavior of the string motion.

We will present here only results for the string motion around a rapidly spinning
black hole, with $a=0.99$.
This is complementary to the $a=0$ case presented by Larsen~\cite{Larsen:1993nt}.
The role of the black hole spin is addressed below in subsection \ref{spin}.
We consider
 just two sets of initial conditions for several different values of $J^2$ and $\omega}\def\O{\Omega$,
 which appear to
 be adequate for understanding the features of the motion. For both sets
 the string starts out at rest, $\dot{r}_0=\dot{\theta}_0=0$.
 The initial polar angle is  fixed in both cases at
 $\theta_0=\pi/2+0.02$, just below the equatorial plane.
 The two cases differ only in the initial value of $r$.
 The two values we will explore are (somewhat arbitrarily)
 $r_0=r_h+1.25 M\approx 2.39\, M$ and $r_0=r_{\rm ISCO}\approx 1.45\, M $,
where
$r_{\rm ISCO}$ is the radius of the innermost stable co-rotating circular orbit (ISCO)
which is given in terms of $M$ and $a$ in, e.g., Ref.~\cite{Bardeen:1972fi}.

In Fig.~\ref{figinit} the two sets initial conditions correspond to points $P_1$ and $P_2$ respectively.
Each point in this figure
other than those on the $y$ axis
corresponds to a circular ring in space.
The coordinates $x$ and $y$, used as labels in this and later
graphs of the $(r,\theta)$ plane, are defined by
 \begin{align}
 & x=r \sin \theta,\\
 & y=r\cos\theta.
 \end{align}

The grey area indicates the region inside the horizon, the dotted red line marks the equatorial plane and the dashed
black curve is the boundary of the ergoregion. For $\omega}\def\O{\Omega$ we consider only the three values $\omega}\def\O{\Omega=0, \pm1$,  since these
span the full dynamical range and can be expected to frame the full picture.
Note that changing the sign of $\omega$ is equivalent to changing the sign of $a$, since
$\omega}\def\O{\Omega$ appears always multiplied by $a$ (via $g_{t\phi}$)
in the equations of motion, and all other occurrences of $a$ are via $a^2$.
Positive $\omega}\def\O{\Omega$ corresponds to co-rotation, while negative
$\omega}\def\O{\Omega$ corresponds to counter-rotation.

\subsubsection{Boundaries of motion}
\label{boundaries}

By studying the boundaries of the motion we can classify the types of dynamics and determine how
they depend on the current, energy, and initial conditions. These boundaries are determined
by eq.~(\ref{VKerr0}). For a given initial condition and current, a particular boundary is determined.
These boundaries
are the relativistic version of the contours plotted in Fig.~\ref{fig4} for a
Newtonian elastic spinning ring and, as we will see shortly, the share many qualitative characteristics
with their Newtonian counterparts.
At the end of this subsection we give examples of such boundaries, but first we present some graphs,
deduced from the potential, that allow one to see which ranges of current or energy will correspond
to different qualitative types of boundaries.

\begin{figure}[t]
\subfigure[  $~\omega=1$ \label{Ebwm1}]{\includegraphics[width=6.5cm]{Ebw1.pdf}}\\
\subfigure[$~\omega=0$ \label{Ebw0}]{\includegraphics[width=6.5cm]{Ebw0.pdf}}\\
\subfigure[$~\omega=-1$ \label{Ebw1}]{\includegraphics[width=6.5cm]{Ebwm1.pdf}}
\caption{\label{Ebfig}  Plots of the energy of a string at rest (or energy at the boundary of motion) as a function of $r$ (plots starting at $r=r_h$) and
$\sqrt{j_\sigma^2+j_\tau^2}$,
for three values of the ratio between the components of the current $\omega=-j_\sigma/j_\tau$ and at $\theta=\pi/2+0.02$ ($GM=c=\mu=1$).
All graphs possess
minima, which implies that bound motion is possible for energies
greater than $E_b^{\min}$, the energy
associated with the minimum. However, only for $\omega=0,-1$
is there a maximum located outside the horizon.
}
\end{figure}

\begin{figure*}[t]
\subfigure[ ~$r=r_h+1.25$, $\omega=1$ \label{Ejqswm1}]{\includegraphics[width=5.5cm]{Ejsqwm1.pdf}}
\subfigure[~$r=r_h+1.25$, $\omega=0$ \label{Ejqsw0}]{\includegraphics[width=5.5cm]{Ejsqw0.pdf}}
\subfigure[~$r=r_h+1.25$, $\omega=-1$ \label{Ejqsw1}]{\includegraphics[width=5.5cm]{Ejsqw1.pdf}}\\
\subfigure[ ~$r=r_{\rm ISCO}$, $\omega=1$ \label{Ejqswm1in}]{\includegraphics[width=5.5cm]{Ejsqwm1in.pdf}}
\subfigure[~$r=r_{\rm ISCO}$, $\omega=0$ \label{Ejqsw0in}]{\includegraphics[width=5.5cm]{Ejsqw0in.pdf}}
\subfigure[~$r=r_{\rm ISCO}$, $\omega=-1$ \label{Ejqsw1in}]{\includegraphics[width=5.5cm]{Ejsqw1in.pdf}}
\caption{\label{Ejsq}  Plots of the energy of a string at rest (or at the boundary of motion) as a function of
$J=\sqrt{j_\sigma^2+j_t^2}$
for specific values of  $\omega$ and $r$ as indicated, and for $\theta=\pi/2+0.02$ ($GM=c=\mu=1$).
The solid black curve,
which can be thought of as a constant $r$ slice of the graphs in Fig.~\ref{Ebfig},
is the energy of the string at rest at the given radius $r$, and is the only
curve on the graph that depends on the value of $r$.
The dashed red line corresponds to the energy of a string at rest at infinity (the escape energy).
The shaded region is the area between the curves corresponding to the minimum and the
maximum of $E_b$ for each $J$
(cf.~Fig.~\ref{Ebfig}), with the exception of the $\omega=1$
cases where the dotted-dashed black curve corresponds to the value of $E_b$ at the horizon.
These graphs
allow one to infer
the properties of the motion.
For a given energy or value of $J$, there is a corresponding point on the solid black curve.
If this point lies
outside the shaded area and above the dashed red line then the motion has just an outer boundary
in the $x$ direction,  but
the string has enough energy to potentially escape
(there is no boundary in the $y$ direction).
If the point lies
outside the shaded area and
below the dashed red line then the motion
is bounded in all directions.
If instead the point lies
within the shaded region, then the motion has boundaries both at low and high values of
$x$.
Additionally, if the point lies above the dashed red line then there is no boundary on the
$y$ motion
and the string can potentially escape. If the point
lies below the dashed red line then
the motion
can be trapped in some toroidal volume.
Notice that for $\omega=-1$ and inside the ergosphere,
Fig.~\ref{Ejqsw1in}, there can never be an inner boundary.}
\end{figure*}

To begin with,
in Fig.~\ref{Ebfig} we have plotted $E_b$ as a function of
$J=\sqrt{j_\sigma^2+j_\tau^2}$ and $r$ for fixed $\theta=\pi/2+0.02$,
which corresponds to our initial conditions, and for three different values of $\omega$.
In this and all subsequent plots we adopt
units with $GM=c=\mu=1$. That is, in terms of the gravitational radius
$r_g=GM/c^2$, what is plotted
is $E_b/\mu r_g$ and $J\sqrt{c/\mu}/r_g$ vs. $r/r_g$.

In the cases $\omega}\def\O{\Omega=0,-1$, where there is
a maximum outside the horizon for any
given value of $J$,
the motion has an outer boundary in $r$ but no inner boundary
if either $E<E_b^{\rm min}$ or $E>E_b^{\rm max}$,
where $E_b^{\rm min}$ and $E_b^{\rm max}$ correspond to the minimum and the maximum of $E_b$ respectively. If $E_b^{\rm max}>E>E_b^{\rm min}$ then there will be three solutions to eq.~(\ref{VKerr0}), corresponding to an inner and an outer boundary and yet another outer boundary at low values of $r$.
In the co-rotating case $\omega}\def\O{\Omega=1$, where there is no maximum,
the role of $E_b^{\rm max}$ is played by $E_b^h$,
the value of $E_b$ at $r=r_h$. For both an inner and an outer boundary to exist, the energy must
fall in the range $E_b^{h}>E>E_b^{\rm min}$. Notice that
in this case no string can exist outside the horizon with energy $E<E_b^{\rm min}$.

So far we have been discussing bounds in the radial direction near the equatorial plane.
There is also a boundary in the motion along the axis if the total energy of the string
is less than the energy of a string at rest at infinity, since the string must have at least
this much energy in order to be able to escape. We call this energy $E_{\rm esc}$.
To determine how it depends on $J$, we can use the
flat background example of section \ref{flatbackground}.
Eqn.~(\ref{V0flat})  gives $E_b=\mu\rho+J^2/\rho$, whose minimum lies at
$E_{\rm esc}(J)=2J\sqrt{\mu}$.

 \begin{figure}[t ]
\subfigure[~$E=50$, $\omega=1$ \label{E50wm1}]{\includegraphics[width=4cm]{E50wm1.pdf}}
\subfigure[~$E=17$, $\omega=1$ \label{E17wm1}]{\includegraphics[width=4cm]{E17wm1.pdf}}\\
\subfigure[~$E=50$, $\omega=0$ \label{E50w0}]{\includegraphics[width=4cm]{E50w0.pdf}}
\subfigure[~$E=17$, $\omega=0$ \label{E17w0}]{\includegraphics[width=4cm]{E17w0.pdf}}\\
\subfigure[~$E=50$, $\omega=-1$ \label{E50w1}]{\includegraphics[width=4cm]{E50w1.pdf}}
\subfigure[~$E=17$, $\omega=-1$ \label{E17w1}]{\includegraphics[width=4cm]{E17w1.pdf}}
\caption{\label{Eoutfig}  Boundaries of motion for different values of the energy and
the $\omega$ parameter, for  string starting at rest at $r=r_h+1.25$ and $\theta=\pi/2+0.02$.
The grey area represent the region inside the event horizon, the dashed black curve the
boundary of the ergoregion and the dashed red line the boundaries of motion.
The form of the boundaries is exactly what one could have predicted by using
Figs.~\ref{Ejqswm1}-\ref{Ejqsw1}.}
\end{figure}

Fig.~\ref{Ejsq} presents plots that allow one to see how the qualitative properties of the
motion depend on $J$,
for specific values of  $\omega$ and the initial $r$ as indicated,
and for $\theta=\pi/2+0.02$.
The solid black curve,
which can be thought of as constant $r$ slice of the graphs in Fig.~\ref{Ebfig},
 is the energy (\ref{VKerr0}) of the string at rest at the given radius $r$, and is the only
curve on the graph which depends on the value of the initial $r$.
The dependence on $\omega}\def\O{\Omega$ comes in only through the $g_{t\phi}$ term
in (\ref{VKerr0}). Increasing $\omega}\def\O{\Omega$ acts to raise the
energy for a given $J$.
The dashed red line corresponds to the energy of a string at rest at infinity (the escape energy).
The shaded region is the area between the curves corresponding to the minimum and the
maximum of $E_b$ for each $J$
(cf.~Fig.~\ref{Ebfig}), with the exception of the $\omega=1$ cases where the
dotted-dashed black curve corresponds to the value of $E_b$ at the
horizon.\footnote{It is rather surprising to have a string (instantaneously)
at rest at the horizon
with non-zero Killing energy.
This comes about
as a singular limit whose nature is hidden by
a singular gauge.
 The first term in eq.~(\ref{VKerr0}) vanishes on the horizon, so we have there
$E_b=-g_{t\phi}\tilde{\Sigma}^{\sigma\tau}=2(g_{t\phi}/g_{\phi\phi})j_\sigma j_\tau$, using eq.~(\ref{tSst}) in the last step.
The metric factors are finite and non-zero as long as the black hole is spinning,
in which case
the energy is non-zero as long as both $j_\sigma$ and $j_\tau$
are non-zero. But if the string is instantaneously static
at the horizon, then the worldsheet is null on that slice,
rather than being time-like, so the conformal gauge
is not accessible. In the limit as such a configuration
is approached, the conformal gauge tangent
vector $X^\mu_{,\tau}$ approaches the null horizon
generator, becoming infinitely stretched since the
gauge condition requires that it has
a nonzero timelike norm.
Thus the current 4-vector
$j^\mu = X^\mu_{,\tau}j^\tau + X^\mu_{,\sigma}j^\sigma$ itself diverges
as long as $j^\tau=h^{\tau\t}j_\tau=g_{\phi\phi}^{-1} j_\tau$ is non-zero.
A non-zero Killing energy configuration of an
instantaneously static string at
the horizon arises from this
singular limit of the current, but only, as seen above,
provided the black hole has spin.
The underlying reason for this
last requirement is that only in the presence of spin
is the Killing vector $\partial_t$,
with respect to which the energy is defined,
distinct from the null Killing vector that is normal to the
horizon.}

\begin{figure}[t ]
\subfigure[~$E=50$, $\omega=0$ \label{E50w0zoom}]{\includegraphics[width=4cm]{E50w0zoom.pdf}}
\subfigure[~$E=50$, $\omega=-1$ \label{E50w1zoom}]{\includegraphics[width=4cm]{E50w1zoom.pdf}}
\caption{\label{Eoutfig2}  Magnified segments of Fig.~\ref{E50w0} and \ref{E50w1} which demonstrate the existence of yet a third boundary of motion for $\omega=0,-1$. This acts as a second outer boundary, therefore forbidding any string starting from inside the region bounded by the innermost red dotted curve to ever leave this region.}
\end{figure}

We now demonstrate how one can infer all the qualitative information about the boundaries of the motion from Fig.~\ref{Ejsq}.
For a given energy or value of $J$, there is a corresponding point on the solid black curve.
This point can lie in several distinct regions of the graph, which correspond to qualitatively different types of
boundary:
\begin{enumerate}[(i)]
\item\label{item1} A region outside the shaded area and above the dashed red line. In this case there is
just an outer boundary, {\em i.e.}~there is only one value of $r$ outside the horizon that yields a
solution to eq.~(\ref{VKerr0}) for each $\theta$ for given $E$ and $J$.  The string  can potentially escape in the $y$ direction, as its energy is higher than the $E_{\rm esc}$.
\item A region outside the shaded area and below the dashed red line. Again, there is
just an outer boundary.
However, now the string  cannot escape in the $y$ direction as its energy is lower than the $E_{\rm esc}$.
\item A region inside the shaded area and above the dashed red line. In this case
there are always outer and inner boundaries,
and for the examples with $\omega}\def\O{\Omega=0,-1$ there is
yet another outer boundary at a lower value of $x$.
That is,
eq.~(\ref{VKerr0}) has up to three solutions
for $r$ outside the horizon
 each value of $\theta$ and for given $E$ and $J$. However, there is no boundary on the
$y$ motion
and the string can potentially escape.
\item  A region inside the shaded area and below the dashed red line.
The boundaries of the motion are the same in number as
in the previous case, but
with a qualitative difference: the inner and the outer boundaries at higher
values of $x$ meet at a certain value of $y$ and together form a closed curve.
Therefore, there is also effectively a boundary in the $y$ motion and the string
can be trapped  in some toroidal volume.
\end{enumerate}
There is a subtle point that should not pass unnoticed:
depending on the initial conditions,
a string with given $E$ and $J$
which corresponds to a point in the shaded region in Fig.~\ref{Ejsq} can either be moving
in between the inner and outer boundary at higher $x$'s, or it can be
inside
the second outer boundary at smaller $x$'s.
Finally it should be stressed that a second outer boundary at small $x$'s can
only exist when there $E_b$ has a maximum outside the horizon.

\begin{figure}[t ]
\subfigure[~$E=38$, $\omega=1$ \label{E38wm1}]{\includegraphics[width=4cm]{E38wm1in.pdf}}
\subfigure[~$E=8$, $\omega=1$ \label{E8wm1}]{\includegraphics[width=4cm]{E8wm1in.pdf}}\\
\subfigure[~$E=38$, $\omega=1$ \label{E38wm1z}]{\includegraphics[width=4cm]{E38wm1inzoom.pdf}}
\subfigure[~$E=8$, $\omega=1$ \label{E8wm1z}]{\includegraphics[width=4cm]{E8wm1inzoom.pdf}}\\
\subfigure[~$E=38$, $\omega=0$ \label{E38w0}]{\includegraphics[width=4cm]{E38w0in.pdf}}
\subfigure[~$E=8$, $\omega=0$ \label{E8w0}]{\includegraphics[width=4cm]{E8w0in.pdf}}
\caption{\label{Einfig}   Boundaries of motion for different values of the energy and the $\omega$ parameter, for  string starting at rest at $r=r_{\rm ISCO}$ and $\theta=\pi/2+0.02$.  The form of the boundaries is exactly what one could have predicted by using Figs.~\ref{Ejqswm1in}-\ref{Ejqsw1in}. Figs.~\ref{E38wm1z} and \ref{E8wm1z} focus on the region close to the black hole of Figs.~\ref{E38wm1} and \ref{E8wm1} respectively. Clearly, the inner boundary of motion lies within the ergoregion.}
\end{figure}

Next we give in Fig.~\ref{Eoutfig} a few examples of plots of the boundaries of motion corresponding to
given initial conditions and current (or energy), for $r=r_h+1.25$.
The qualitative nature of these boundary plots can
be read off from Fig.~\ref{Ejsq}.
For example, with $E=50$ there exist both inner and outer boundaries in the $x$ direction
for all values of $\omega}\def\O{\Omega$, corresponding to the fact that the solid black curve
in Figs.~\ref{Ejqswm1}, \ref{Ejqsw0}, \ref{Ejqsw1} falls inside the shaded region when $E=50$.
Additionally, for $\omega=1,0$ there is no boundary along the $y$ direction,
whereas for $\omega=-1$ there is such a boundary, corresponding to the fact that
in the latter case the $E=50$ point on the
solid black curve lies below the dashed red line in Fig.~\ref{Ejqsw1}.
In Fig.~\ref{Eoutfig2}, one can see the region closer to the black hole: for the $\omega=0,-1$
cases, there exists a third boundary, which is an outer boundary. This is in accordance with
the form of $E_b$ in Figs. \ref{Ebw0} and \ref{Ebw1}
and highlights the subtlety discussed above right after outlining the various regions of Fig.~\ref{Ejsq}.
 For $E=17$ on the other hand, only the $\omega=1$ choice
 allows motion which is unbounded in the $y$ direction. For $\omega=0$
 the string is trapped in a toroidal volume, and for $\omega=-1$ there is
 only an outer boundary.
Fig.~\ref{Einfig} is similar to Fig.~\ref{Eoutfig}, but for $r=r_{\rm ISCO}$. (We do not plot the $\omega}\def\O{\Omega=-1$
case since
there is no energy for which there will be an internal boundary for the motion.)
In this case,
the inner boundary of motion lies within the ergoregion.

Before closing this section it is worth stressing once more the similarity between
the contour plots of the boundaries of motion for a Newtonian elastic spinning ring
presented in Fig.~\ref{fig4} and the boundaries of motion for the relativistic strings presented in this section.

\subsubsection{Trajectories}

In this subsection we look at examples of individual trajectories, and in particular
we exhibit processes whereby a string is ejected along the axis, with some of the
initial internal energy of the string converted to translational kinetic energy.
To this end, we employed numerical integration with Mathematica 6.0
to solve the equations of motion.
As already mentioned we consider here two different initial conditions:
one for which the string starts outside the ergoregion, and one
for which the string starts at $r=r_{\rm ISCO}$, which lies well within the
ergoregion (points $P_1$ and $P_2$ in Fig.~\ref{figinit} respectively).
Starting at rest from these initial positions,
there are
three types of trajectories, as
indicted by the types of
boundaries of motion:
those that
are trapped inside an outer boundary with no
inner boundary and simply
fall in across the horizon,
those that are trapped in some toroidal volume,
and those that can reach spatial infinity
along the direction of the axis. The trajectories of the first type
display no interesting features so we do not present any
explicit examples of them
here.\footnote{Note that one can have trajectories of this type
even if there exists an inner boundary, provided that an outer boundary
at smaller $x$'s exists as well, as discussed earlier. This is the case,
for instance, when the boundaries of motion are those of
Fig.~\ref{E50w0zoom} or Fig.~\ref{E17w0} and the string start at
rest at $r=r_h+1.25$.}

\begin{figure}[t]
\subfigure[~$E=9$, $r=r_h+1.25$ \label{E9wm1mot}]{\includegraphics[width=4cm]{E9wm1mot.pdf}}
\subfigure[~$E=9$, $r=r_h+1.25$  \label{E9wm1motz}]{\includegraphics[width=4cm]{E9wm1motzoom.pdf}}\\
\subfigure[~$E=8$, $r=r_{\rm ISCO}$  \label{E8wm1inmot}]{\includegraphics[width=4cm]{E8wm1inmot.pdf}}
\subfigure[~$E=8$, $r=r_{\rm ISCO}$  \label{E8wm1inmotz}]{\includegraphics[width=4cm]{E8wm1inmotzoom.pdf}}
\caption{\label{boundmotionfig} Confined trajectories of the string for different initial positions and different energies.
The blue continuous line corresponds to the trajectory of a point on the string ($\phi$ dimension suppressed).
No qualitative difference seems to exist between the motion of a string starting inside the ergoregion and
one starting outside. In both case the motion seems to follow no clear pattern.}
\end{figure}

We begin by discussing
bound trajectories.
The energies we considered previously
do not
lead to confined motion for a string starting at $P_1$.
This can be seen in Fig.~\ref{Eoutfig}: For the cases where there can be confined motion,
namely Figs.~\ref{E17w0} and \ref{E50w1}, our initial position lies on the innermost
outer boundary. Thus the string would fall straight into the black hole.
However, for other energies there can be
confined motion for a string starting at $P_1$. Also, according to Fig.~\ref{E8wm1}
we have already found an example where the string's motion is confined when it starts
from $P_2$, {\em i.e.}~inside the ergoregion. In Fig.~\ref{boundmotionfig} we present
the motion of a string starting from $P_1$ with $E=9$ and that of a string starting from
$P_2$ with $E=8$ for qualitative comparison. Both cases refer to $\omega=1$ (co-rotation).
In both cases the strings move somewhat randomly, without following
any clear pattern.
Fig.~\ref{E9wm1motz} and \ref{E8wm1inmotz} focus on the area inside and around the
ergoregion.
The string is sometimes pulled by gravity rather
sharply back toward the equatorial plane after reflecting from the inner boundary.
We have monitored the accuracy by running the code backward in time from the point the
forward run ended, and checking how closely this backward run
approaches
the initial conditions.
In the plots presented we stop the simulation before the accuracy becomes lower than
$10^{-4}$.\footnote{Using only the default settings of Mathematica, we
were able to run the simulation for $\tau\sim 150-200$ for trajectories starting from
$P_1$ and for $\tau\sim 50-100$ for trajectories starting from $P_2$ (provided that
the string did not end up in the inside the horizon before that time) before dropping
below the required accuracy. Remarkably, for the ``focused" trajectories, such as
those in Fig.~\ref{focusfig}, the accuracy was still two orders of magnitude higher
than our minimum requirement at $\tau=500$. Note that $\tau$ is the coordinate
time on the worldsheet.}

\begin{figure}[t ]
\subfigure[~$E=8$, $r=r_h+1.25$ \label{E8wm1mot}]{\includegraphics[width=4cm]{E8wm1mot.pdf}}
\subfigure[~$E=8$, $r=r_h+1.25$  \label{E8wm1motz}]{\includegraphics[width=4cm]{E8wm1motzoom.pdf}}\\
\subfigure[~$E=5.5$, $r=r_{\rm ISCO}$  \label{E5.5wm1inmot}]{\includegraphics[width=4cm]{E5p5wm1inmot.pdf}}
\subfigure[~$E=5.5$, $r=r_{\rm ISCO}$  \label{E5.5wm1inmotz}]{\includegraphics[width=4cm]{E5p5wm1inmotzoom.pdf}}
\caption{\label{focusfig} ``Focusing'' of the trajectories. In this case a pattern seems to be arising. In Figs.~\ref{E8wm1motz} and \ref{E5.5wm1inmotz} we have distorted the aspect ratio in order to make this manifest.}
\end{figure}

Before proceeding further to trajectories that can reach spatial infinity,
there is another interesting feature of confined motion which should not be missed.
In Fig.~\ref{focusfig} we present confined trajectories starting from $P_1$ with
$E=8$ and from $P_2$ with $E=5.5$, always for $\omega=1$. There
is evidently
some remarkable ``focusing" of the motion close to the equatorial plane in both cases,
and a very regular pattern is generated. In
Figs.~\ref{E8wm1motz}
and \ref{E5.5wm1inmotz} we have distorted the aspect ratio
in order to make manifest the pattern that arises in the motion.
We do not understand why these cases are so different from those shown in Fig.~\ref{boundmotionfig}. The focussing cannot be explained by considering only the boundary of allowed motion (dotted curve) as inferred using the effective potential.  The appearance of a Lissajous-like pattern indicates that when the string stays  close to the equatorial plane the motion becomes similar to complex periodic oscillation.

Finally,
we discuss
trajectories that can reach spatial
infinity.\footnote{These trajectories are similar to the
time-reversed version of Larsen's adiabatic capture trajectories found in Ref.~\cite{Larsen:1993nt}.}
In Fig.~\ref{E17wm1mot} we have plotted the trajectory of a string starting from $P_1$ with $E=17$
and $\omega=1$. After a couple of oscillations around the equatorial plane, which are exhibited
more clearly in Fig.~\ref{E17wm1motz}, the string escapes along the $z$ direction.
The behavior of a string starting from $P_2$ with $E=38$ and $\omega=1$ is quite similar.
In both cases, there is some conversion of internal energy of the string into kinetic energy along the $y$ axis.

\begin{figure}[t ]
\subfigure[~$E=17$, $r=r_h+1.25$ \label{E17wm1mot}]{\includegraphics[width=4cm]{E17wm1mot.pdf}}
\subfigure[~$E=17$, $r=r_h+1.25$  \label{E17wm1motz}]{\includegraphics[width=4cm]{E17wm1motzoom.pdf}}\\
\subfigure[~$E=38$, $r=r_{\rm ISCO}$  \label{E38wm1inmot}]{\includegraphics[width=4cm]{E38wm1inmot.pdf}}
\subfigure[~$E=38$, $r=r_{\rm ISCO}$  \label{E38wm1inmotz}]{\includegraphics[width=4cm]{E38wm1inmotz.pdf}}
\caption{\label{freefig} Escape trajectories for strings of different energy and angular momentum,
start at $P_1$ and $P_2$ respectively. After some oscilations around the equatorial plane the
strings escape to infinity along the axis of symmetry. Clearly there is some conversion of
internal to kinetic energy.}
\end{figure}

Another interesting class of non-confined trajectories are
the scattering trajectories
found by Larsen in Ref.~\cite{Larsen:1993nt}.
An example of such a trajectory for a string with $E= 11.2929$, $\omega=1$, and initial conditions  $r_0=200$, $\theta_0=3.11$, $\dot{r}_0= -3.4$ and $\dot{\theta}_0=-0.007$ is shown in Fig.~\ref{scattering}.
The string comes in along the axis and after a couple of bounces
around the black hole,
which can be better appreciated in Fig.~\ref{scattzoom}, escapes
to infinity out the other side.
The scattering
in this case reduces the mean radius in the
$x$ direction, signaling
a conversion of
internal
energy to kinetic energy along the $y$ direction,
as explained below in the next subsection, \ref{ejection}.

In Figs.~\ref{collapse} and \ref{collapsezoom} we show a case where the string
actually falls into the black hole instead if getting scattered. For this case,
$E= 11.8417 $, $\omega=1$, and the initial conditions  are $r_0=200$, $\theta_0=3.11$,
$\dot{r}_0= -2.4$ and $\dot{\theta}_0=-0.007$. In this case
the string has more energy than a string at rest on the horizon
and, hence,
its motion has no inner boundary close to the equatorial plane
({\em cf.}~the discussion in section \ref{boundaries} about Figs.~\ref{Ebwm1} and \ref{Ejsq}).

\begin{figure}[t ]
\subfigure[~$E=11.3$, scattering \label{scattering}]{\includegraphics[width=4cm]{scattering.pdf}}
\subfigure[~$E=11.3$, scattering  \label{scattzoom}]{\includegraphics[width=4cm]{scatteringzoom.pdf}}\\
\subfigure[~$E=11.8$, collapse  \label{collapse}]{\includegraphics[width=4cm]{capturefall.pdf}}
\subfigure[~$E=11.8$, collapse  \label{collapsezoom}]{\includegraphics[width=4cm]{capturefallzoom.pdf}}
\caption{\label{scattcaptfig} Scattering and collapse on the black hole of strings coming from spatial infinity.
The scattering can lead to energy conversion, as can
be seen from Fig.~\ref{scattering} by the fact that the
average radius of oscillation along the $x$ direction has been
slightly reduced after the scattering.
 On the other hand, if the inner boundary of motion lies within the horizon close to
 the equatorial plane, then instead of a scattering there can be a collapse, see
 Figs.~\ref{collapse} and \ref{collapsezoom}. }
\end{figure}

\subsubsection{Ejection energy}
\label{ejection}
As noted, on some trajectories internal energy of the string can be converted
to translational kinetic energy along the axis. To characterize this energy
we may associate a relativistic boost factor $\G=E/E_0$ with the asymptotic
string, where $E_0$ is the energy of the string in its center of mass frame
at infinity. In the center
of mass frame, all velocities vanish at the inner and outer boundaries
$\rho=\rho_{i,o}$ of oscillation, so we may use (\ref{V0flat}) to obtain
$E_0=\rho  + J^2/\rho$. Eliminating $J$ we find $E_0=\rho_i+\rho_o$.
Thus, using the present notation $x$ for the cylindrical radius, we have
\begin{equation}
\G=\frac{E}{x_i+x_o}.
\end{equation}
With this relation the $\G$ factors corresponding to the processes shown
in the previous examples can be computed. For example, for the $E=17$ case
in Fig.~\ref{freefig} we find $\G\approx  1.01$, or $v/c\approx 0.15$, and
for the $E=38$ case we find $\G\approx 1.09$, or $v/c\approx 0.39$.
We expect that these mildly relativistic boost factors
are typical of what one can expect in generic cases.

\subsubsection{The role of black hole spin}\label{spin}

The black hole spin enters the string dynamics
via the $a$ dependence of the various metric
coefficients. While the value of $a$, certainly affects the boundaries of motion and dynamics when other quantities
are held fixed, we do not see any systematic qualitative effect
that can be attributed uniquely to the spin. That is,
by varying the current, energy, and location of the string similar
effects can be produced. This statement is based both
on examining cases (not explicitly reported here)
for which $a=0$, as well as by comparing the behaviors
for different values of $\omega}\def\O{\Omega$. The dynamics only depends
on $\omega}\def\O{\Omega$ when $a$ is non-zero, but the effect of
changing $\omega}\def\O{\Omega$ can also
generally be reproduced by changing other quantities.

The one exception to this statement concerns the
issue of negative Killing energy states. In the presence
of black hole spin, there is an ergoregion, inside of which
such states exist for point particles.
The same is true for the string, though the mechanism
is a bit different.

For a particle,
negative angular momentum orbits  in the ergoregion have negative
Killing energy.
Longitudinal motion of a string is meaningless, so it
is not the ``orbital" motion, but rather the current
on the worldsheet that can access negative energy
states. In particular,
the second term on the right hand side of
the energy (\ref{VKerr0}), i.e.\
$-g_{t\phi}\tilde{\Sigma}^{\sigma\tau}=|g_{t\phi}/g_{\phi\phi}| L/2\pi$,
is negative there when the angular momentum
$L=4\pi\omega}\def\O{\Omega J^2/(1+\omega}\def\O{\Omega^2)$ is negative.
The first term in eq.~(\ref{VKerr0}) vanishes
at the horizon, so the energy is certainly negative there
when $L<0$.
The negative
energy states extend out
a certain distance (which depends on the values
of $\omega}\def\O{\Omega$ and $J$)
from the horizon into the ergoregion.
For example, one can see in Fig.~\ref{Ejqsw1in}
the negative energy
states at $r_{\rm ISCO}$ in the case $\omega}\def\O{\Omega=-1$, except for the
smallest values of $J$.
(For the limiting case of zero tension and $\omega}\def\O{\Omega=-1$, there
are negative energy states everywhere in the ergoregion.)

\section{Relevance to Astrophysics?}

As stated in the Introduction,
although our primary aim
in this paper
was to study the motion of a current carrying strings in a Kerr background
as an interesting dynamical system,
our investigation was also loosely motivated
by the problem of the production of collimated
jets in astrophysics.
We should
therefore like to comment on what
might be the relevance of our findings to
astrophysical systems.

First of all,
the current carried by the string
in the model need not represent
an electric current, but it can
just serve to model the
role of
angular momentum. In this sense, a current carrying string
can serve as an idealization of some MHD matter configuration
with tension and angular momentum, as mentioned in the Introduction.
Simplistic as it may be,
this allows
for the identification of a
mechanism
that might conceivably be relevant:
the tension sets a barrier for the motion that keeps the string
within a certain radius from the axis of symmetry,
while
the angular momentum sets a barrier that keeps the string
outside a certain radius.  Gravity deforms these
barriers creating
either a closed region in which the string is trapped,
or a ``restricted path"  that leads to motion along the symmetry
axis.
If some physical process were to
``feed" such a
system
with strings close to the equatorial plane, then
under the right circumstances one might end up with a very well
collimated stream of strings moving out along the axis of symmetry.
Of course, the axisymmetry we are imposing might be too restrictive,
and in particular, even if a string started out approximately axisymmetric,
it might wobble or twist and reconnect. Perhaps the tension
could act to suppress such effects, but this requires further investigation.

Still, one may speculate that the general flavor of the mechanism
exhibited by the strings might generically be present in a
class of astrophysical systems.
For instance, given a system with an accretion disk, plasma
leaving the inner edge of the disk would carry some angular momentum,
and plasma flux tubes do possess a significant tension (which however
is not constant but grows in proportion to the length).
It might even be that differential rotation can stretch such a tube,
storing in it a large amount of electromagnetic energy before launching
it towards the axis. External magnetic fields, which have been left
out of consideration in our initial study, might also play an important
role in the subsequent dynamics. It remains to be seen
whether aspects of this picture could actually be astrophysically
relevant.

Regarding the energetics, it should be noted that the mechanism
described here does not lead to significant (ultra-relativistic) acceleration
of the strings. While internal energy of the string can be converted to kinetic
energy of motion along the axis of symmetry, the resulting $\Gamma$ factors
in the examples we studied are rather close to unity, the largest velocity
we reported being $0.39 c$.
Therefore, even if the mechanism we have found is somehow relevant to collimation,
it  does not appear able to explain the acceleration
of relativistic jets.
The acceleration would have to occur due to
the presence of electromagnetic fields, or to some other mechanism.
In particular, the process discussed here does not tap into the
rotational energy of the black hole, which is a prime candidate
for the source of ultra-relativistic jet energy.

On the contrary, the collimation effect
we find persists even at the Newtonian level, which may be
advantageous: collimated jets seem to be present not only around
black holes, but also around objects with significantly
weaker gravitational fields, such as accreting young stars
for instance~\cite{livio}.
Therefore, if there is a common mechanism for the collimation,
it should not depend on relativistic effects. Hence,
the effects described here could perhaps be part of a more
involved process that collimates and drives astrophysical jets.

\section{Conclusions}

We have studied the motion of relativistic, current carrying strings moving axisymmetrically
on the background of a Kerr black hole. First the equations of motion and
conserved quantities in the conformal gauge were found.
Next,  we determined how the energy of a string
that is instantaneously at rest depends on the current and the position of the
string. Contours of this quantity determine boundaries of the possible motion of a string with
that energy and current. We compared this analysis with that for a Newtonian elastic ring
moving around a point mass, and found that system to be qualitatively similar.

By considering a number of examples and various plots of the relevant functions,
the possible types of motion were mapped out.
Regions of parameters for which the string falls into the black hole, or is trapped
in a toroidal volume,  or can escape to infinity, were identified. After this general analysis,
we examined representative trajectories found by numerical integration, illustrating
various interesting behaviors. In particular, we found that a string can start out at
rest near the equatorial plane and, after bouncing around, be ejected out along the axis,
some of its internal (elastic or rotational kinetic) energy having been transformed into
translational kinetic energy. The resulting velocity can be an order unity fraction of the
speed of light.

Finally, we addressed the question of possible astrophysical significance of this
system as a simple model of MHD plasma flux tubes, which might conceivably
 play a role in jet formation and collimation.

\begin{acknowledgments}
We thank Vijay Kaul for collaboration in the early stages of this research.
This work was supported by the National Science Foundation under grant PHYS-0601800.
\end{acknowledgments}

redpajama_set_name: RedPajamaArXiv


This walkthrough gives explicit instructions on playing SIMON THE SORCERER to a successful conclusion. It does, however, assume you have read the Game Manual and are familiar with the various methods of controlling your actions.
In directing you from place to another, I either supply instructions relative to the screen, right, left, up or down, or assume you will use the on-line map to zap you to the nearest location to your next destination. After you have visited a place once, I assume you will know how to find that place again. However, if the directions are not clear, I have provided maps as described in the following General Hints.
Be pleasant, cooperative and always offer to help. When conversing with potential or definite enemies never back off unless you have completely run out of options. If the conversation doesn't turn out the way I say, just try another response until it does.
Don't be afraid to take risks. Do so. You can't die and you may need to experience the downside before you can solve the puzzle.
Visit the Wise Owl in the Woods for random hints. He will always tell you something useful, but it may be for past or future use. Hopefully he will occasionally tell you something of immediate use.
Quite honestly, you don't need to save at all, except before quitting. You can't die and the game never crashed on me. (The inventory did become invisible from time to time though!) Anyway, you may want to replay some sequences again in a different order. Save as often as feels comfortable. For instance, I saved frequently during a thunderstorm!
Watch the credits! Simon is lounging around in his bedroom bored with his homework. Meanwhile Simon's dog is exploring the attic. Hearing barking from upstairs Simon climbs up to the attic and opens the chest. The dog jumps out and Simon discovers a spellbook. Handling the spellbook invokes a spell which causes a small local rift in the fabric of space and time. The dog walks through the rift and Simon follows . . .
Simon finds himself in the middle of a Goblin food invocation ceremony. The dog hides, but Simon loses his spellbook and is put into the Goblin's cooking pot! Doggie to the rescue! They both escape and arrive at Calypso's cottage. Calypso, a Grand High Wizard, has left a letter for Simon explaining that he has been captured by the evil wizard Sordid and has brought Simon here to find and rescue him.
But first Simon must become a wizard himself so that he can use Calypso's spellbook . . . where ever it is now!
Take the magnet from the refrigerator door. Open the desk drawer and take the scissors.
Go outside and go right into the village and at the Blacksmith's, up to the tavern. Open the door and go into the tavern. Look at the fruit machine and take the matches. Talk to the Nubile Valkyries about Sordid. Use the scissors to cut off the sleeping dwarf's beard. Go into the room on the right and converse with the wizards until they agree to tell you how to become a wizard. They tell you must find and return with a magical staff. The staff is last known to be in the possession of, the now deceased, Nafflin the Necromancer.
Leave the tavern. Head left past the Shoppe, up to the Dodgy Geezer. Get away from this guy as quickly as you can. You'll find him useful later on. Go left to the Apothecary's. Pick up the ladder outside the apothecary's cottage. Go inside and take the cold remedy and specimen jar.
Return to outside the Shoppe and go left into the woods. When you reach the four way junction, take the first path on the left. Continue along this path until you see the wise owl. Talk to the owl. Pick up a hint or two. Take the feather from the left center of the screen.
Return to the four way junction and take the first path on the right. Continue on this path until you see the tree stump. Talk to the woodworm. They will ask you to find some mahogany for them.
Take the path down and continue left to the Witch's Cottage. Turn the handle of the well and haul up a bucket of water. Take the bucket of water.
Return along the path and then go left to the four way junction. Take the second path on the right and continue on this path until you meet the Barbarian. Help the Barbarian and he will give you a whistle.
Return left to the four way junction. Take the lower right path and walk past the Woodworm tree stump to the Troll Bridge. As soon as the action stops talk to the troll so that he takes and uses your whistle. The Barbarian comes and clears the bridge.
Walk across the Troll Bridge and continue along this path, taking the upper right path at the junction, until you meet the Oaf. Talk to the Oaf about his beans. Suggest they need watering. Pour the bucket of water over the beans. You will leave. Return and pick the beans up out of the puddle.
Return to the village and go to Calypso's cottage. Plant the beans in the compost behind Calypso's cottage. Pick the water melon.
Return to the Troll Bridge and go to the upper left to the sousaphone player. You will try to escape the noise but you also want that sousaphone! Visit the sousaphone player again and throw the melon into the sousaphone. The sousaphone player will hand the instrument over to you for repair.
Return left to the five way junction, then take the upper right path to the cordoned off hole. Talk to the paleontologist who is inside the hole. He is looking for a fossil.
Continue right along the path to the Center of the Forest. Take the higher path, that starts on the left, to the Goblin Fortress. Look for the shopping list under the stone. Take it.
Return to the lower path and continue right to the Woodcutter's Cottage. Talk to the woodcutter. He is miserable because he can't cut down the magic trees. He needs a magic axe made from Milrith. He will give you a metal detector.
Return to the Center of the Forest and take the upper right path. Continue through the cave entrance and on into the swamp. Attempt to open the door of the Swampling's residence and the Swampling will invite you in. Talk to the Swampling. Accept the stew. Eat as much Swamp stew as you can. Ugh! Fill the specimen jar with the stew when you can. Keep eating the stew until the Swampling has to leave. Leave and continue right to the mountains.
Keep walking right through the mountains until you reach the Sleeping Giant. (Why have all those people turned to stone?) Disturb the sleeping giant by playing the sousaphone. This will make him push the tree over so you can walk across the chasm.
Continue right to the Dragon's Cave. Visit the dragon. Blackened! Click on Simon to recover. Go into the cave again. Give the cold remedy to the Dragon. Go into the cave and take the fire extinguisher. Leave and walk behind the cave. Pick up the rock with the fossil.
Go right, and right again, at the climbing pins, continue right and talk to the Talking Tree.
Return to the Sleeping Giant then go left. Use the metal detector.
Return to the village. Open the Shoppe door. Hand over the shopping list to the shopkeeper.
Visit the Blacksmith and give him the rock. You will leave with a fossil.
Return to the Sleeping Giant and go left. Search the dirt and rocks and pick up the milrith ore.
Return to the village and visit the Blacksmith. Give the milrith to the Blacksmith and he will make you an axe head. Pick up the bell clapper from the bench. Also take the rope and attach it to the magnet.
Return to the Center of the Forest and go right to the Woodcutter's cottage. Hand the axe head over to the woodcutter and he will leave. Go into his cottage. Pick up the climbing pin on the left-hand side. Put out the fire in the fireplace with the fire extinguisher. Move the hook over the fireplace and you will go down to the basement. Take the mahogany.
Go to the Troll Bridge and take the lower left path to the tree stump where the woodworm reside. Talk to the woodworm. The woodworm become part of your inventory.
Go to the Cross Roads. Take the lower right path to the tower. Put the bell clapper in the bell and sound the bell. Climb the long braid of hair. Talk to the girl behind the screen and kiss her. Well, she did say "when she was a piglet"! The pig 'Repulser' becomes part of your inventory. Use the woodworm on the floorboards. Go down through the holes to the basement, using the ladder on the second hole.
Open the tomb. Could this be Nafflin the Necromancer with the magic staff? You will be forced to leave, but you can now return here through the ground floor doorway. Go in and go back down to the tomb, open it again. Do not look at the mummy but pick up the stringy bandage at his back. The mummy will unwind and be destroyed. Take the staff.
Return to the Cross Roads and take the path up, on the right, to the waterfall. Climb down the vines at the lower left of the screen. Talk to the Golum. He will tell you he needs food for a party. Give the Swamp stew to the Golum fisherman. You will sit down to fish and eventually find the ring of invisibility.
Return to the village. Go to the path outside the Apothecary and go left to the Beehive Cottage. Have Repulser, the pig, eat the Chocolate truffle door. Go into the cottage. Take the smokebox and beekeeper's hat. Go outside. Use the smokebox on the beehive. Pick up the wax.
Go to the tavern and ask the barman to fix you a drink. While he is preoccupied put the wax on the nozzle of the barrel behind the bar. The barman will think this barrel is empty and leave it for you outside. He will also give you a free beer voucher.
Pick up the barrel of beer when you leave.
Return to the Center of the Forest. Look at the rock at the mine entrance. Pick it up and look at the rock again. It says "beer". This is the password to get into the mine. Put on the beard. Go into the mine. Try to go to the left. Bribe the guard with the barrel of beer. Return up the steps and go to the left. Talk to the dwarves. Find out how to get into the treasury (through the golden door on the upper left). Take the hook. Return to the steps to the right and go down to the beer room. Use the feather on the sleeping guard to make him turn over. Take the key.
Return to the treasury door and open it with the key. Talk to the dwarf. Bribe him with the beer voucher. He will give you a gem.
Return to the village and visit the Dodgy Geezer. Drive a hard bargain and sell the gem to him for 20 gold coins.
Return to the Shoppe. Buy the hammer and white spirit. Leave. Outside the shop, open the box and get in.
Open the box you are hiding in. Look at the boxes. Pick up the spellbook. Look at the spellbook and you will find a loose piece of paper. Pick up the rat bone. Put the piece of paper under the door. Use the rat bone on the lock. Pick up the piece of paper. Use the key to unlock the door. Open the door and leave.
Pick up the bucket and go down the stairs. Talk to the Druid. Remove the ring. Talk to the Druid again until he tells you he needs to see a full moon. Put the bucket over the druids head. Pick up the mints. Pick up the flaming brand and hold it over the Druid's head. The Druid escapes as a frog.
When you hear Goblin guards approaching, open the iron maiden and hide in it. The frog will return with a hacksaw. Open the iron maiden. Take the hacksaw and use it on the bars. Leave.
Return to the Dragon's Cave and use the hook to climb up to the boulder above the cave entrance. Use the rope and magnet on the hole (this is the skylight in the Dragon's Cave) and you will collect some gold coins. Repeat this three times until the magnet comes up empty.
Climb down and return to the Talking Tree. Clean up the pink splotch with the white spirit. Talk to the tree. The tree will tell you four magic words.
Return to the village. Visit the wizards in the tavern. Talk to them. Hand over the staff and pay the fee to become a wizard. Ask the wizards about Sordid.
Visit the Apothecary. The Druid is at home and he wants you to find him some Frogsbane.
Go to the Center of the Forest. Return to the Swampling's hut. (Thank goodness the Swampling's not back yet!) Move the box and open the trap door. Go through the trap door and walk along the planks. Fix the loose plank with the hammer and nail. Continue right to Skull Island. Pick the Frogsbane.
Return to the village and visit the Apothecary. Hand over the Frogsbane and you will receive a mysterious potion labeled "Drink Me".
Go to the witch's cottage. Open the door and go in. Try to take her broom. You will get chucked out but enter the cottage again. The witch will offer to play a game with you. Accept.
Use the magic word "Sausages" repeatedly until you win the game. You may not win the first game, but keep returning to the cottage and playing the game until you do. The witch will then cheat and become a dragon. Escape by saying "Abracadabra" and becoming a mouse. Run away through the mouse-hole near the rocking chair. You now have the broom.
Return to the Dragon's Cave and go to the climbing pins. There is one missing but you have one in your inventory. Put the spare climbing pin in the hole. Walk across the climbing pins.
Eat the mints and the snowman will melt.
Continue right to Sordid's Tower. Use the witch's broom to get across the chasm. Consume the "Drink Me" potion. You will walk through a crack in the door. Your dog will grab you and put you in the garden.
Pick up the leaf. Pick up the stone. Look in the bucket and take the matchstick. Go left to the puddle. Pick up the lily-pad. Put the matchstick on the lily-pad and the leaf on the match to make a sailing boat. Sail over to the seeds. Take a seed. Return to shore. Crush the seed with the stone. Attach the dog hair to the tap. Use the seed oil on the tap and turn on the water by pulling the dog hair.
Use your makeshift sailing boat to get to the other side of the puddle. On leaving the boat you are nearly swallowed by a frog! Look at the water and pick up the tadpole. Threaten the frog and that you will harm the tadpole unless the frog gets out of your way. When the frog leaves look at the mushrooms. Pick the mushroom labeled "Eat Me" and eat it.
Open the door and go inside. Terrified, you run outside. Pick up the branch. Go inside again and put the branch in the snapping chest's mouth. Pick up the shield and the spear.
Go downstairs to the torture chamber. Pick up the chest. Move the lever to open the press. Put the chest on the block and move the lever so the chest is crushed. Move the lever again to raise the press. Pick up the candles. Use the spear to fetch down the skull hanging from the ceiling on the left. Pick up the skull.
Go upstairs, then upstairs again, to the bedroom. Pick up the sock, the pouch and the magic wand. Pick up and read the book. Now you know what to do with Sordid's wand. Talk to the mirror. Put the sock in the pouch. Put the pouch in the mouse-hole at the foot of the stairs. You will catch the mouse.
Go upstairs and talk to the demons. Keep talking until they tell you they want to return to the pit. Pick up the book and read it. You now know how to return these demons to their home. Talk to the demons some more and find out more information if you can. Pick up the chemical, the magical metal polish, and use it to polish the shield. Place the shield on the hook.
Go downstairs to the bedroom. Talk to the mirror and ask to see the lab. You will hear the demon's true names.
Go upstairs and keep talking to the demons until they give you the chalk so you can start the ritual to send them back. When you have completed the ritual, get into the teleporter, select the Fiery Pits of Rondor as your destination.
Pick up the sapling and the pebble. Talk to the attendant. He will give you some brochures. Look at the brochures and then use the elastic band on the sapling. Use the catapult on the fire alarm. Pick up the souvenir matches from the counter.
Go right, across the bridge and pick up the floor wax. Go right to Sordid. Use the wand on Sordid to turn him to stone. Now light the pits with the matches. Throw the wand into the lava. Uh-oh! Sordid tries to kill you. Return to the lava pits and deal with Sordid. How? Use the floor wax so that he loses his footing . . . and falls into the lava!
Congratulations! Enjoy the closing sequences!
This walkthrough is copyright (c) 1995 by Diana Griffiths. All rights reserved. Not to be distributed without permission.


text:
The Scarlet Scroll • April 16, 2018 • https://thescarletscroll.com/7190/opinion/the-post-promotes-a-journalists-right-to-freedom-of-the-press/
"The Post" promotes a journalist's right to freedom of the press
Courtesy FOX Movies
Ben Bradlee (Tom Hanks) conversing with Katharine Graham (Meryl Streep) about opportunity to expose government secrets to the public.
Nicole Kresich, Staff Reporter
Steven Spielberg's drama The Post highlights an important subject to all journalists: their First Amendment freedoms. Dating back to The Washington Post's 1971 brave decision to expose and publish the truths behind the Vietnam War's Pentagon Papers, The Post tells a true story from a journalistic perspective to emphasize the public's right to the truth, and the journalist's constitutional freedom to express it.
With the knowledge that the federal government had already censured The New York Times' publication of the papers, along with the threat against The Washington Post to do the same, the publisher of The Washington Post, Katharine Graham, expertly portrayed by Meryl Streep, anguished over the decision to publish these government secrets and risk losing her paper altogether. Electing not to publish, after having already obtained the confidential documents, Graham would now be guilty of "saving face" to preserve long term relationships, just as the government had done to the American people with so many years of knowing that a Vietnam victory was truthfully unattainable. Going against the advice of her council and inner circle, Graham, along with longtime editor Ben Bradlee, played by Tom Hanks, who shared her passion for the First Amendment freedoms, courageously published portions of the Pentagon Papers, regardless of the consequences.
Courtesy FOX Movies Ben Bradlee (Tom Hanks) story planning with reporters from The Washington Post.
In a review of The Post by The New York Times, the credibility and trust, of journalists specifically, is brought to the forefront with Spielberg's movie. Spielberg's film illustrates the public's present day mistrust of the media and their characterization of journalists as either good or evil, heroes or villains, truth tellers or liars.
"There's more than a little corn and wishful thinking in the high-minded moments in The Post; movies like either to glorify or demonize journalists, relying on heroes and villains. Yet given the recent assaults on journalism and the truth, this heroizing is also irresistible," wrote The New York Times reviewer, Manohla Dargis.
The production of The Post reinforces the promise that under the United States Constitution, no political hierarchy has the ability to suppress the truth from the American people. As American citizens we look to the media to report the truths that affect us as people of this democracy, regardless of its favor. Nixon's attempt, as portrayed in The Post, to cease the publication of the government secrets was not effective under a court of law due to protections under the First Amendment freedoms of speech, expression and the press.
Courtesy FOX Movies The Washington Post staff surround the radio as they listen to the Pentagon Papers leak.
According to a survey by the American Press Institute, "Americans are skeptical of 'the news media' in the abstract, but generally trust the news they themselves rely on."
While there are numerous outlets and sources to obtain newsworthy information, the truth remains the same despite the bias, origin, or platform. With something as significant as the Pentagon Papers leak, the information will surface, regardless of perceived biases.
Journalists reporting on such significant topics have a professional responsibility to be truth tellers, supported by the freedoms permitted under the first amendment. The Post masterfully addresses the present day threats against press freedom and affirms the battle between real and fake news. Spielberg's movie effectively informs the public of the hurdles and the constantly treaded boundaries journalists face just to assert their First Amendment freedoms.
Ben Bradlee, The Washington Post editor, battled the government's oppressions of the First Amendment, in the name of journalistic freedom, to secure his duty to report the truth to the people.
"The only way to protect the right to publish is to publish," Bradlee said.

redpajama_set_name: RedPajamaCommonCrawl


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Amongst one of the most widespread electrical wiring queries is on how to wire a switch. While applying switches at your house is very uncomplicated, wiring one particular might not be that uncomplicated for everyone. An ON-OFF switch is really quite simple to wire. You'll find several types of switches, but for this example, let's say you are putting in a single-pole toggle switch, an extremely typical swap (and also the most basic).
You'll find three shades of wires in a usual single-pole toggle switch: black, white, and inexperienced. Splice the black wire in two and connect them around the terminal screws - a single on top and the other within the bottom screw of your change. The white wire serves like a source of uninterrupted electric power which is generally related to the light-weight colored terminal screw (e.g., silver). Hook up the inexperienced wire for the ground screw within your change.
These ways would generally be ample to produce an average change get the job done with no issue. Nevertheless, when you will not be assured which you can complete the job properly and safely you superior permit the pros get it done as an alternative. Soon after all, you can find a motive why this process is one of the commonest electrical wiring queries questioned by a lot of people.
For many rationale, ways to wire a ceiling fan can also be certainly one of the most common electrical wiring queries. To simplify this job, you may use an individual change for a single ceiling fan. To wire the supporter, it is simply a matter of connecting the black wire from the ceiling fan into the black wire of your switch. When there is a lightweight, the blue wire need to be linked into the black wire on the swap too.
You will discover explanations why these are typically quite possibly the most commonly questioned electrical wiring questions. A single, a lot of think it can be basic to complete, and two, they are the prevalent electrical duties at your home. But then you definitely shouldn't put your basic safety in danger as part of your aim to save cash. The stakes could even be substantially higher for those who seek to spend less and do an electrical wiring work with out ample understanding or expertise.