
\documentclass[aps,unsortedaddress,12pt]{revtex4}
\begin{document}
\title{Fermions scattering in a three dimensional extreme black hole background}
\preprint{USACH}

\author{Samuel Lepe}
\email {slepe@ucv.cl}
\affiliation{Instituto de F\'{\i}sica, Facultad de Ciencias B\'{ a}sicas y Matem\'{a}ticas,
Universidad Cat\'{o}lica de
Valpara\' {\i}so, Casilla 4059, Valpara\'{\i}so, Chile.}
\author{Fernando M\'endez}
\email {fmendez@lauca.usach.cl}
\affiliation{Departamento de F\'{\i}sica, Universidad de Santiago de Chile, Casilla 307, Santiago
2, Chile.}
\author{Joel Saavedra}
\email {jsaavedr@lauca.usach.cl}
\affiliation{Academia Polit\'ecnica Militar, Valenzuela Llanos 623, La Reina, Santiago, Chile.}
\author{Lautaro Vergara}
\email {lvergara@lauca.usach.cl}
\affiliation{Departamento de F\'{\i}sica, Universidad de Santiago de Chile, Casilla 307, Santiago
2, Chile.}

\begin{abstract}
The absorption cross section for scattering of fermions off an extreme BTZ black hole is
calculated. It is shown that, as in
the case of scalar particles, an extreme BTZ black hole exhibits an absorption cross section equals
to zero,  what is
consistent with the vanishing entropy of such object. We also show that the Dirac equation, in this
background, can be
solved completely for a very special case where the energy of the incident wave and the azimuthal
eigenvalue must satisfy
a {\it fine tuning condition}. Additionally,  we give a general argument to prove that the particle
flux near the horizon is
zero,  no matter how they are constructed. Finally we show that the {\it reciprocal space}
introduced  previously in \cite{
gm} gives rise to the same result and,  therefore,  it could be considered as the space where the
scattering process take
place in a AdS spacetime.
\pacs{0460, 0470, 0470D}
\end{abstract}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%
\section{Introduction}


The scattering of particles off a black hole (BH) is a very interesting problem because for $D=4$
\footnote{The problem
in higher dimensions can be found in S. R. Das and S. D. Mathur, {\it Nucl. Phys.} {\bf B478},
561 (1996),
hep-th/9606185; {\it ibid}, {\it Phys. Lett.} {\bf B375}, 103 (1996) hep-th/9601152; S. R. Das, G.
Gibbons and S. D.
Mathur, {\it Phys. Rev. Lett.} {\bf 78}, 417 (1997), hep-th/9609052.}, the classical absorption
rate  is proportional to the
area of the
black hole (BH) \cite{kaplan} and one does not know whether this result could survive at quantum
level. The standard
approach to this problem  is semiclassical, that means to  solve the Klein-Gordon or Dirac
equation in a fixed BH
background \cite{ birrel}.

In three dimensions the previous result holds, except for extreme BH. The spinless relativistic
particle case  was
discussed  in a recent paper \cite{gm}, where it  was shown   that the absorption cross section  for
such BH is zero. From
the thermodynamic point of view, this result can be interpreted as a signal of zero entropy, in
agreement with previous
results found in the literature \cite{teitel}. Besides, in our approach we showed a manner to
circumvent the problem of
defining states at infinity in an AdS spacetime, introducing the so called {\it reciprocal space}.

Particles with spin 1/2 in these background has not been considered previously \footnote{For the
non extreme case see {\it
e.g.}  S. Das and  A. Dasgupta, {\it JHEP} {\bf 9910}, 025 (1999);  A. Dasgupta, {\it Phys. Lett.}
{\bf B445}, 279
(1999).}. In fact the extreme case is rather different and can not be obtained from  previous
analysis because there is
not a smooth transformation connecting both cases. The problem come from the fact that  they
correspond to different
topologies of spacetime:  the non extreme $2+1$ dimensional black hole has the topology of the
cylinder, while the
extreme case has the topology of an annulus \cite{teitel}.

In this paper we show that the absorption cross section for massive particles of spin $1/2$ in a
$2+1$-dimensional
extreme BH is zero,  as in the scalar case. We  discuss  how to construct the states at spatial
infinity, where there is no
free particle-type solutions. The Dirac equation is solved for a special case where $\omega$ and
$n$, the energy and
azimuthal eigenvalues, respectively, satisfy a {\it fine tuning} condition.

The paper is organized as follow. In the next section we will review the Dirac equation in a
curved space time and the
$2+1$ dimensional extreme black hole. Section III is devoted to write explicitly the Dirac
equation in this curved
background and to examine their asymptotic solutions. In Section IV the flux is constructed and the
cross section is
calculated; the reciprocal space approach is also  discussed here. Finally, in Section V discussion
and conclusions are
presented.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%
\section{The Dirac equation in a 2+1-dimensional BH background}

Let us consider a three dimensional Riemann manifold with Minkowski signature $(-,+,+) $ and
line element
\begin{equation}
ds^{2}=g_{\mu \nu }dx^{\mu }dx^{\nu }.
\label{1}
\end{equation}

The non-coordinate basis one-form $e^{a}=e_{\mu }^{a}dx^{\mu }$ and the affine spin
connection
$\omega _{b}^{a}=\omega _{b\mu }^{a}dx^{\mu }$ are defined by \cite{egu}
\begin{eqnarray}
ds^{2} &=&e^{a}e^{b}\eta _{ab}, \label{2} \\
de^{a}+\omega _{b}^{a}e^{b} &=&0,
\label{3}
\end{eqnarray}
where $\eta _{ab}=\mbox{diag}\left( -,+,+\right) $; latin indexes denote tangent space
components and greek indices
stand for components of objects defined on the manifold.

The Dirac equation for a particle with mass $m$ in the curved background (\ref{1}) is given by
\cite{naka}
\begin{equation}
\gamma ^{a}E_{a}^{\mu }\left( \partial _{\mu }-\frac{1}{8}\omega
_{bc\mu }
\left[ \gamma ^{b},\gamma ^{c}\right] \right) \Psi =m\Psi ,
\label{DE}
\end{equation}
where $E_{a}^{\mu }$ is the inverse triad and satisfies
$E_{a}^{\mu }e_{\mu }^{b}=\delta _{a}^{b}$ with
$\delta _{a}^{b}$ the  identity. $\left\{ \gamma ^{a}\right\} $ are the Dirac matrices in the tangent
space defined by the
Clifford algebra
\begin{equation}
\left\{ \gamma ^{a},\gamma ^{b}\right\} =2\eta ^{ab}.
\label{cliff}
\end{equation}

\subsection{ Extreme $2+1$ black hole metric}

In order to compute the absorption cross section we must solve  (\ref{DE}) in the three
dimensional extreme BH
background.

In a $2+1$ dimensional spacetime the Einstein equations with cosmological constant
$\Lambda=-\ell^{-2}$ have a
solution
\cite{btz}
\begin{equation}
ds^{2}=-N^{2}\left( r\right) dt^{2}+N^{-2}\left( r\right) dr^{2}
+r^{2}\left[d\phi +N^{\phi }\left( r\right) dt\right] ^{2},
\label{btz}
\end{equation}
where the  lapse $N^{2}(r)$ and shift $N^{\phi }(r) $ functions are given by
\begin{eqnarray}
N^{2}\left( r\right) &=&-M+\frac{r^{2}}{\ell^{2}}+\frac{J^{2}}{4
r^{2}}, \\
N^{\phi }\left( r\right) &=&-\frac{J}{2r^{2}}.
\end{eqnarray}
Here $M$ and $J$ are the mass and angular momentum of the black hole, respectively.

The lapse function vanishes when
\begin{equation}
r_{\pm }=r_{ex}\left[1\pm \sqrt{1-\frac{J^{2}}{M^{2}\ell^{2}}}
\right]^{\frac{1}{2}},
\end{equation}
and therefore, the solution (\ref{btz}) is defined for $r_+<r<\infty$, $-\pi<\phi<\pi$ and
$-\infty<t<\infty$.


The extreme solution  correspond to $J^{2}=M^{2}\ell^{2}$, in which case
$r_{\pm}=r_{ex}=\ell\sqrt{M/2}$ and the
previous line element can be written as
\begin{equation}
ds_{ex}^{2}=-\left( \frac{r^{2}}{\ell^{2}}-2\frac{r_{ex}^{2}}{
\ell^{2}}\right)dt^{2}+\frac{\ell^{2}r^{2}}{\left( r^{2}-r_{ex}^
{2}\right)^{2}}dr^{2}-2\frac{r_{ex}^{2}}{\ell}dtd\phi +r^{2}d\phi ^{2}.
\label{btze}
\end{equation}

Instead of solving (\ref{DE}) in this metric, is convenient to define a dimensionless set of
coordinates
$\left\{ u,v,\rho \right\} $
as follow
\begin{equation}
u =\frac{t}{\ell}+\phi,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v =\frac{t}{\ell}-\phi,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\,\,\,\,\,\,\,\,\,\,\,\,e^{2\rho
} =
\frac{r^{2}-r_{ex}
^{2}}{\ell^{2}},
\end{equation}
where  $-\infty<\{u,v\}<\infty$, $-\infty<\rho<\infty$. In the space $\{u \times v\}$, two points
$(u_1,v_1)$ and
$(u_2,v_2)$ are identified if they satisfy $u_1=v_2$ and $v_1=u_2$, for any value of $\rho$.

The line element (\ref{btze}) in the new coordinates reads
\begin{equation}
ds^{2}=r_{ex}^{2}dv^{2}-\ell^{2}e^{2\rho }dudv+\ell^{2}d\rho
^{2},
\end{equation}
and according to (\ref{2}) we have the triads
\begin{equation}
e^{1}=\frac{\ell^{2}~e^{2\rho }}{2~r_{ex}}~du,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e^{2}=r_{ex}~dv-
\frac{\ell^{2}~e^{2\rho
}}{2~
r_{ex}}~du,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,e^{3}=\ell ~d\rho ,
\label{15}
\end{equation}
and from (\ref{3}) the connections
\begin{equation}
\omega ^{1}\,_{2}=d\rho,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\omega ^{1}~_{3}=\frac{\ell~e^{
2\rho}}{2~r_{ex}}~du
+
\frac{r_{
ex}}{\ell}~dv,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\omega ^{2}~_{3}=-\frac{\ell~e^{2\rho }}{2~r_{ex}}du.
\label{16}
\end{equation}
The non vanishing components of the inverse triad are
\begin{equation}
E_{1}^{u}=\frac{2~r_{ex}~e^{-2\rho
}}{\ell^{2}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E_{1}^{v}=\frac{1}{r_{ex}}=E_{2}^{v},\,\,\,\,\,
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E_{3}^{\rho}=\frac{1}{\ell}.
\label{17}
\end{equation}

In the next section we will write the Dirac equation in this coordinates and will find its solution.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%
\section{ The Dirac equation and its solution}
\subsection{The Dirac Equation}
We choose the Dirac matrices as follow
\begin{equation}
\gamma ^{1} =-i\sigma ^{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\gamma ^{2} =\sigma ^{1},\,\,
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\,\, \,\,\,\,\gamma ^{3} =\sigma ^{2},
\end{equation}
where $\sigma^i$ are the Pauli matrices. This choice satisfy (\ref{cliff}).

The Dirac equation (\ref{DE}) can be written  directly by using (\ref{15}) through (\ref{17}). If
we write the spinor as
\[\Psi(u,v,\rho)=\left(
\begin{array}{c}
{\cal{U}}(u,v,\rho) \\
{\cal{V}}(u,v,\rho)
\end{array}\right),
\]
then (\ref{DE}) becomes
\begin{eqnarray}
\left[-i\left( \frac{2 r_{ex} e^{-2\rho } }{\ell^{2}}\partial _{
u}+\frac{\partial _{v}}{r_{ex}}\right) -(\frac{1}{2\ell}+m)
\right]{\cal{U}}+\left[
\frac{\partial _{v}}{r_{ex}}-\frac{i}{\ell}\left( \partial _{
\rho }-1\right)  \right]{\cal V}&=&0,\label{DEC1}
\\
\left[\frac{\partial _{v}}{r_{ex}}+\frac{i}{\ell}\left( \partial
_{\rho }-1\right) \right]{\cal{U}}+\left[ i\left( \frac{2r_{ex}e
^{-2\rho }}{\ell^{2}}
\partial _{u}+\frac{\partial _{v}}{r_{ex}}\right) -(\frac{1}{2
\ell}+m) \right]{\cal{V}}&=&0.
\label{DEC2}
\end{eqnarray}

In order to solve this equation, let us look for solutions with the shape
\begin{equation}
\Psi \left( u,v,\rho \right) =e^{i\left( \alpha u+\beta v\right)
}
\left(\begin{array}{c}
F(\rho)\\
G(\rho)
\end{array}\right),
\label{ansatz}
\end{equation}
where $\alpha $ and $\beta $ are constants related with the angular and temporal eigenvalues of
the solution of (\ref{DE})
in the coordinates $\left\{ t,\phi ,r\right\} $;  namely if  the solution behaves like
$e^{i(n\phi +\omega t)}$, then
\begin{equation}
\alpha=\frac{1}{2}\left( \omega \ell+n\right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\beta  =
\frac{1}{2}\left(
\omega
\ell-n\right).
\end{equation}

For the $\rho$-dependent part of the equation is more convenient to define $z=e^{-2\rho }$.
Therefore, using the Ansatz
(\ref{ansatz}), the $z$-part of (\ref{DEC1}) and (\ref{DEC2}) becomes \footnote{Here we have
used the same name for
the functions $F$ and $G$, independently if they depend on $\rho$ or on $z$. The notation  will be
used through all the
text unless it becomes confusing.}

\begin{eqnarray}
\left[ \frac{2 \alpha r_{ex}}{\ell^{2}}z+\frac{\beta }{r_{ex}}-
\frac{1}{2\ell }-m\right] F\left( z\right) +i\left[ \frac{\beta
}{r_{ex}}+\frac{1}{\ell}
\left( 2z\frac{d}{dz}+1\right) \right] G\left( z\right)&=&0,
\label{ED1}
\\
i\left[ \frac{\beta }{r_{ex}}-\frac{1}{\ell}\left(
2z\frac{d}{dz}+1\right) \right] F\left( z\right)-\left[ 2\frac{
\alpha r_{ex}}{\ell^{2}}z+\frac{\beta }{
r_{ex}}+\frac{1}{2\ell}+m\right] G\left(
z\right)&=&0.\label{ED2}
\end{eqnarray}

One can find a solution of this set of coupled equations by rewriting them as a second order one.
For example, if we solve
(\ref{ED1}) for $F(z)$ and replace this result in (\ref{ED2}) we find that $G(z)$ satisfy a second
order equation. By
defining the variable  $x =\alpha( \,r_{ex}/\ell)\,z$ one finds that $G(x)$ satisfies
\begin{equation}
A(x) G^{\prime \prime }(x) +B(x)G^{\prime}(x) +C(x)G(x) =0,
\label{eqg}
\end{equation}
being
\begin{eqnarray}
A(x)  &=&(\delta-x)x^2,
\\
B(x)  &=&(2\delta-x)x ,
\\
C(x)  &=&-x^3+(\delta-\tilde{\beta})\,x^2+
\frac{1}{4}\bigg[(\tilde{\beta}+1)^2+4\delta(2\tilde{
\beta}+\delta)\bigg]x-\frac{\delta}{4}\left((2\delta +\tilde{\beta})^2-1\right),
\end{eqnarray}
where the constant $\delta$, $m_{\mbox{\scriptsize{eff}}}$ and $\tilde{\beta}$ are given by
\begin{equation}
\delta =\frac{1}{2}\left(\ell\,m_{\mbox{\scriptsize{eff}}}-
\tilde{\beta}\right),\,\,\,\,\,\,\,\,\,\,\,\,m_{\mbox{\scriptsize{eff}}}=
m + \frac{1}{2\ell}\,
,\,\,\,\,\,\,\,\,\,\,\,\,\tilde{\beta}=\frac{\ell \beta}{r_{ex}}.
\end{equation}
The same procedure yields a similar equation for $F\left(z\right)$.

Note that this equation has three singular points for $\delta\neq 0$. Two of them are regular ($0$
and $\delta$) and the
other is an irregular one, located at infinity (that corresponds to  the horizon in radial coordinates).

In order to solve the equation we must consider two cases: $\delta =0$ and $\delta\neq 0$. In the
first case, the equation
lost one regular singularity, while in the second one, it keeps  the three points. Therefore, we will
be able to solve
completely one of them, while the other, must be treated in an approximate way.

The physical meaning of such cases are the next. For s-waves, the $\delta=0$ condition
corresponds to particles with
energies satisfying the relation $\omega =2 m_{\mbox{\scriptsize{eff}}} r_{ex} \Omega$, where
$\Omega=1/\ell$ is the
angular velocity of the horizon. One  can think that the previous condition just fixes the energy  of
the particle to the energy
(classical) of a particle with mass $m_{\mbox{\scriptsize{eff}}}$ rotating in a circular orbit of
radius $r_{ex}$  with
angular velocity $\Omega$  . For waves with $n\neq 0$, the relation holds, but now the energy has
a contribution of the
angular part of the wave.

Note that this condition gives a  precise relation between the energy and the azimuthal eigenvalue
and is a {\it fine tuning}
condition that, for a generic wave, should be hard to fulfill.

In the next section we will discuss all cases in  detail.


\subsection{Solutions of Dirac equation}

As we said previously, there exist two  distinct cases. We will prove that, despite the solutions
are different for both
cases, they share a common characteristic related with their behavior near the horizon.
\subsubsection{$\delta =0$.}

The equation (\ref{eqg}) now reads
\begin{equation}
x^2 G'' + x G' +\left[x^2+\tilde{\beta}x -\frac{1}{4} (\tilde{
\beta}+1)^2\right]G=0,
\label{sammy}
\end{equation}
whose solution is
\begin{equation}
G(x)=e^{-ix} x^{\frac{\tilde{\beta}+1}{2}}\left(P\,\,F\left[1
+\frac{1}{2}(1+i)\tilde{\beta}),2+\tilde{\beta};2ix\right]+Q
\,\,U\left[1+\frac{1}{2}(1+i)\tilde{\beta},2+\tilde{\beta};2
ix\right]\right),
\end{equation}
where $F[c,d;x]$ and $U[c,d;x]$ are confluent hypergeometric functions (Kummer's solution). $P$
and $Q$ are complex
constants. These functions provide a set of linearly independent solutions of (\ref{sammy}) only if
$\tilde\beta$ is not
an integer \cite{htf}.

The confluent hypergeometric $F[c,d;x]$  has a smooth limit near $x=0$, while the confluent
hypergeometric $U[c,d;x]$ is
regular at infinity. Thus, the regular piece of the solution at infinity (that is, near the horizon in
radial coordinate) is
\begin{eqnarray}
G(x)&=&e^{-ix} x^{\frac{\tilde{\beta}+1}{2}}Q
\,\,U\left[1+\frac{1}{2}(1+i)\tilde{\beta},2+\tilde{\beta};2
ix\right], \nonumber
\\
&\sim&e^{-ix}\sum _{n=0}^{N}\alpha_n\, x^{-n-1/2}\,\,\,\,\,\,\,
\,
\,\,\,\,\,(x\rightarrow \infty),
\end{eqnarray}
where $\alpha_n$ is a constant depending on $\tilde \beta$ and $N$ is an arbitrary integer
controlling the order of the
expansion.

The regular solution for $x\rightarrow 0$ (that is,  at  spatial infinity) is $F[c,d;x]$ and has a
series expansion in positive
powers of $x$, but it is not necessary write it explicitly at this moment.

Finally let us comment that, at the  intermediate region, one can match both solutions by noticing
that
\begin{equation}
U[c,d;x]=\frac{\Gamma(1-d)}{\Gamma(c-d+1)}F[c,d;x]+\frac{\Gamma(
d-1)}{\Gamma(c)}x^{1-d}F[c-d+1,2-d;x],
\label{prop}
\end{equation}
in order to find a relation among the arbitrary constants presents in the solution. By using this
method is possible to define
a solution at spatial infinity. However this is a delicate point because one needs to consider both
solutions in order to
have an asymptotic state with non vanishing flux \cite{birming}.

\subsubsection{$\delta \neq 0$.}
As we said before, in this case, the equation has three singular points, been $x\rightarrow\infty$ of
irregular type. We can
write (\ref{eqg}) in a more transparent fashion by defining the variable $y=x/\delta$ and
\begin{equation}
G(y)=\frac{1-y}{\sqrt{y}} D(y).
\end{equation}

The equation for $D(y)$ turns out to be
\begin{eqnarray}
y^2 (1-y)^2 D'' +y(1-y)(1-2y)D'  +
\bigg[\delta^2 y^4+\delta\left( \tilde {\beta}-2\delta\right)y^3
&-& \frac{\tilde{\beta}}{4}\left(2+\tilde{ \beta}+12 \delta
\right)
y^2+\nonumber
\\
\frac{1}{4}\left(5( \tilde {\beta}+\delta)^2 +2 \tilde {\beta}(
1+\delta) -4\right)y-\frac{1}{4}(2 \tilde {\beta}+\delta)^2
\bigg]D&=&0.
\label{grosa}
\end{eqnarray}

This is the equation of spheroidal wave  function and has analytic solution just for some particular
cases \cite{murphy}.

We are interested in the asymptotic behavior of the solutions in order to define flux near horizon
and at spatial infinity. So,
let us study the asymptotic version of (\ref{grosa})

We start analyzing the  $y\rightarrow 0$ case. Maintaining terms up to first order, equation (\ref{
grosa}) can be written as
\begin{equation}
y^2 D_0 ''+(1-y)yD_0'+\left(-\frac{1}{4}(2\tilde{\beta}+\delta)^
2+
\frac{1}{4}\left(\tilde{\beta}(2+4\delta)-3(\tilde{\beta}^2-
\delta^2-4) \right)y\right)D_0=0,
\end{equation}
whose solution is
\begin{equation}
D_0(y)=P_0\,\,\,y^{-\frac{b}{2}}F\left[ \frac{1}{2}\left(a+b
\right),1-
b,y\right] +Q_0\,\,\,y^{\frac{b}{2}}F\left[
\frac{1}{2}\left(a-b\right),1
+b,y\right],
\label{sol0}
\end{equation}
where $P_0$  and $Q_0$ are  complex constants,
$a=2+\frac{3}{2}(\tilde{\beta}^2-\delta^2)-\tilde{\beta} (1+2\delta)$
and $b=2\tilde{\beta}+\delta$ and $F[c,d;x]$ is the confluent hypergeometric function.

Again, the states that define the incoming flux will be a superposition of this solutions.

In order to discuss the region $y\rightarrow \infty$, we write (\ref{grosa}) in power series about
infinity and discard all
terms at order ${\cal O}(y^{-1})$. The resulting equation is
\begin{equation}
y^2 D_\infty ''+2yD_\infty '+\left(\delta^2\,y^2+\delta\tilde{\beta} \,y-\frac{1}{4} \left(2\tilde{
\beta}+\delta^2+(\tilde{
\beta}+
\delta)(\tilde{ \beta}+3\delta)\right)\right)D_\infty=0,
\end{equation}
whose solutions turn out to be
\begin{eqnarray}
D_{\infty}(y)&=&e^{-i\delta\,y}y^{\frac{1}{2}(a-1)}\bigg(P_\infty \,F\left[\frac{1}{2}\left(1+i
\tilde{\beta}+a\right),1+
a,2i\delta\,y\right]+\nonumber
\\
&&Q_\infty\,U\left[\frac{1}{2}\left(1+i \tilde{\beta}+a\right),1+ a,2i\delta\,y\right]\bigg),
\end{eqnarray}
where $a=\sqrt{(\tilde{\beta}+1)^2+4\delta(\tilde{\beta}+\delta)}$. $F[c,d;y]$ and $U[c,d;y]$
are the  confluent
hypergeometric functions discussed previously for the case $\delta =0$.

As we said in the previous section, the regular part of the solution is given by the $U[c,d;y]$
function. By the same
arguments given there, we can write it as follow
\begin{equation}
D_\infty(y)\sim e^{-i\delta y}\sum _{n=0}^{N}\gamma_n\, y^{-n-1},
\end{equation}
where $\gamma_n$ are complex constants. Therefore, the $G(x)$ function turn out to be
\begin{equation}
G(x)\sim e^{-i x}\sum _{n=0}^{N}\tilde\gamma_n\, x^{-n-1/2}.
\end{equation}

 Finally let us point out that the previous results can be understood from a physical point of view
 as follows. Because our
 choice of radial coordinate, the near horizon region corresponds to the irregular singular point
 $x \rightarrow\infty $. But
 the horizon is not a physical singularity, therefore one should find well behaved solutions near
 this point. Therefore, they
 can be written as
\begin{equation}
G_{\infty}(x)\sim \frac{1}{x^s}\left(\tilde\alpha_0 +\frac{
\tilde\alpha_1}{x}+\frac{\tilde\alpha_2}{x ^2}+\cdots\right),
\label{general}
\end{equation}
where $s$ is a positive real number. This is just the result founded for $\delta = 0$ and
$\delta \neq 0$ with $s=1/2$.
Note that the exponential $e^{ix}$ can not be obtained by this argument, but complex functions are
irrelevant in order to
compute
the flux, as we will see below.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%
\section{Ingoing and incoming fluxes and the absorption rate}

 As was said before, it is not clear how to prepare an initial scattering state (incoming flux) in an
 AdS background
 because the solution of the equations of motion at infinity ($r \to \infty$) are not plane waves. The
 problem can be
 circumvented arguing that, no matter how one defines the incoming flux, this must be non zero
 because it represents the
 particles that we are sending to the black hole, therefore, we only must care about that we are far
 enough from the horizon.

Therefore, we define the incoming flux as the number of particles described by the Dirac
equation in a metric asymptotically AdS \cite{birming}. These solutions, obviously are not plane
waves and  the
only restriction to our solution at infinity is that they can be connected with the solutions near the
horizon. This
requirement ensure the flux measured near the horizon corresponds to the particles that were sent
from infinity (no matter
if they are plane waves or not).

This last  statement is a little bit ambiguous, because there is no reason to go to infinity
considering that there are no plane
waves-type solutions in this region. The point here is that it is possible to define such flux from the
solutions found in the
previous section and no matter where one define the incoming flux, it must be different from zero.


\subsection{The current}

In order to calculate the absorption cross section it  is necessary to prepare the incoming flux (
initial state) and to obtain
the ingoing flux (near $r_{ex}$) later. The current along $\rho $- direction is
\begin{eqnarray}
j^{\rho } &=&\kappa\bar{\psi}\gamma ^{a}E_{a}^{\rho }\psi,
\\
&=&-\frac{2\kappa}{\ell }\Re \mbox{e}({\cal U}^{\ast }{\cal V}),
\end{eqnarray}
being $\kappa$ a coupling constant.

From (\ref{ansatz})  and (\ref{ED1}) we obtain
\begin{equation}
j^{\rho }(x) ={\cal A}\left(\frac{x}{
\delta-x}\right)\Re \mbox{e}\left\{i\,G(x)\frac{d}{dx}G^{\ast }(
x)
\right\},
\label{alfincarajo}
\end{equation}
where $\cal A$ is a constant.

\subsection{Flux and cross section}

As in ordinary quantum mechanics, the black hole absorption rate $\sigma _{abs}$ is related to
the ratio of the ingoing
flux at the  horizon (total number of particles entering the horizon) and incoming flux at infinity (
initial state)
\begin{equation}
\sigma _{abs}=\frac{\mathcal{F}\left( z\rightarrow \infty \right) }{\mathcal{ F}\left( z\rightarrow
0\right) }
\label{sigma},
\end{equation}
where
\begin{equation}
\mathcal{F}=\sqrt{-g }\,j^{\rho }(x),
\end{equation}
with  $g$ the determinant of the metric.

It is straightforward to prove that
\begin{equation}
\mathcal{F}={\cal A}\,\sqrt{r_{ex}^2+\frac{\alpha\ell
r_{ex}}{x}}\left( \frac{x}{1-x}\right) \Re
\mbox{e}\left\{i\,G(x)
\frac{d}{dx}G^{*}\left(x\right)\right\},
\end{equation}

Now we must evaluate this quantity for $\delta =0$ and $\delta \neq 0$, near the horizon and at
spatial infinity. However,
as we showed in the previous section, for the near horizon region there is  a general form of the
solution given by
(\ref{general}).

It is straightforward to evaluate the flux near the  horizon
\begin{eqnarray}
\mathcal{F}&=&{\cal A}\,\sqrt{r_{ex}^2+\frac{\alpha\ell \,r_{ex}}{x}} \,\frac{x}{1-x}\, \Re
\mbox{e}\left\{\frac{i}{x
^{2s}}\left( \tilde{\alpha}_0 +\frac{ \tilde{\alpha}_1}{x}+\frac{\tilde{\alpha}_2}{x^2}+ \cdots
\right) \left(\kappa _0 +
\frac{\kappa _1}{x}+\frac{\kappa _2}{x^2}\cdots\right) \right\}\nonumber
\\
&=&{\cal A}\,\sqrt{r_{ex}^2+\frac{\ell^2}{x}}
\,\frac{1}{1-x}\, \Re \mbox{e}\left\{\frac{i}{x^{2s-1}}\left(\tilde{\alpha_0}\kappa_0+\frac{
\tilde{\alpha}_0\kappa_1+
\tilde{\alpha}_1\kappa_0}{x}+\frac{\tilde{\alpha}_0\kappa_2+\tilde{\alpha}_1\kappa_1+\tilde{
\alpha}_2\kappa_0}{x^
2}\cdots\right)\right\}.\nonumber
\end{eqnarray}
Here $\tilde{\alpha}_i$  and
$\kappa_0=i\tilde{\alpha}_0^{*}, \kappa_1=i\tilde{\alpha}_1^{*}-s \tilde{\alpha}_0^{*},
\kappa_2=
i\tilde{\alpha}_2^{*}-\tilde{\alpha} _1^{*} - s \tilde{\alpha}_1^{*},\cdots $, are numerical
constants.

The last expression shows that for $s\geq 1/2$
\begin{equation}
\lim_{x\rightarrow\infty}{\cal F}\to 0.
\end{equation}

By other hand, the flux at infinity is a constant because it represents the particles that were sent
from some point far from
the horizon of the BH, as we argued previously. That means that if we are  sending  particles from,
say,  $x_0$, then
\begin{equation}
\lim_{x\rightarrow x_0}{\cal F}\to \mbox{constant}.
\end{equation}

Finally, from (\ref{sigma}) one prove
\begin{equation}
\sigma_{abs}=0.
\end{equation}
for the extreme 2+1 dimensional black hole.

Let us point out that this result has no contradictions with the non-extreme black hole. The reason
is that our argument
about the shape of the function $G(x)$ is still valid in the regular black hole background, but the
damping factor in front of
(\ref{alfincarajo}) is different for such case. The reason is that (\ref{ED1}) and (\ref{ED2}) are
not the same for the
extreme and non-extreme cases and one cannot map one into the other in a continuous manner.

\subsection{Reciprocal space}

In \cite{gm} we argued that it is possible to circumvent the problem of defining the flux of
particles at infinity in a AdS
spacetime by working in a sort of reciprocal space. That is, to map the  original problem in other,
where the notion of free
particles at infinity holds.

In the present case this  can also be done and in this section we will show how to do that. The
arguments given before
--in \cite{gm}-- are still valid and much of the discussion in this section is based on it.

The main point here is to write the equation (\ref{eqg}) in their canonical form
\begin{equation}
u''(x) + I(x)u(x)=0,
\end{equation}
where $I(x)$ is the invariant.

In our problem we must distinguish, again,  two cases:  $\delta=0$ and $\delta\neq 0$.

For $\delta=0$, one define
\begin{equation}
G(x)=\frac{u_0 (x)}{\sqrt{x}},
\end{equation}
where $u_0 (x)$ satisfy the equation
\begin{equation}
u_0''+\left(1+\frac{\tilde{\beta}}{x}-\frac{\tilde{\beta}(\tilde{\beta}+2)}{4\,x^2}\right)u_0=0.
\label{u0}
\end{equation}

For $\delta \neq 0$, by defining $\xi=x/\delta$ and
\begin{equation}
G(\xi)=\frac{\sqrt{\xi-1}}{\xi}\,u_1 (\xi),
\end{equation}
one find that $u_1 (\xi)$ satisfies
\begin{equation}
u_1''+\left(\delta^2+\frac{\tilde{\beta}(2\delta+1)-1}{2\,\xi}+\frac{(\tilde{\beta}+2\delta)^2-1}{
4\,\xi^2}+\frac{1-\tilde{
\beta}}{1-\xi}+\frac{(3/4}{(\xi-1)^2}\right)u_1=0.
\label{u1}
\end{equation}

In (\ref{u0})  and (\ref{u1}), $x$ and $\xi$ play the role of a radial coordinate, respectively, in a
Schr\"odinger-type
equation. The reciprocal spaces are then ${\cal H}_0=\{\phi,x\}$ and ${\cal H}_1=\{\phi,\xi\}$,
where $\phi$ is an
angular coordinate.

Additionally, one must demand that
\[
u_0(x=0)=0=u_1(\xi=0),
\]
in order to have continuous solutions everywhere
\cite{elipticos}.

Finally, one note that the {\it potential} terms in (\ref{u0}) and (\ref{u1}) vanishes as
$x\rightarrow\infty$  and
$\xi \rightarrow\infty$, respectively and therefore, it is possible to define asymptotic states as in
the usual scattering
theory;
that is, solutions like $A_0(\phi)e^{ix}$ for $u_0$ and $A_1e^{i\delta\xi}$ for $u_1$, where
$A_i$  are the scattering
amplitude and the exponential term, the asymptotic states.

But, by the optic theorem, one can directly prove that this gives rise to a vanishing total cross
section (see \cite{gm} for
details) and returning to the original space, it means a zero absorption cross section as was
calculated in the previous
section.

\section{Conclusions}
In this work we have discussed the problem of fermions scattering  by a 2+1-dimensional extreme
black hole
background.

We showed that the Dirac equation can be solved completely when the {\it fine tuning condition}
is satisfied.

We also gave a general argument in order to show that the solutions of the Dirac equation near the
horizon has  to
have a precise form, what was checked by solving them.

With this solution we were able to prove that the flux of incoming particles near the horizon
vanishes and therefore,
the absorption  cross section becomes zero, what is in agreement with previous results for scalar
particles.

The previous result is not contradictory with the well known fact that for regular black hole
background, the cross
section is proportional to the area (entropy) of the horizon. In fact, our result is consistent with the
fact that both
solutions of the Einstein equations (the regular black hole and the extreme black hole) define
different topologies of
the spacetime.

We also prove that the description in terms of the {\it reciprocal space} give the same results and
then, it could be
considered as the space where the scattering in AdS should be defined.

Finally, let us remark that our main  result is  in accordance with arguments that show(s) that the
extreme black hole
entropy is zero and then they could be considered as fundamental objects.

 \acknowledgments

This work was supported by UCV-VRIEA-DI under Grant N$^{0}$ 123.763/2002 (SL),
FONDECYT \ N$^{0}$
3000005 (FM),  FONDECYT \ N$^{0}$ 1020061(LV). One of us (SL) acknowledge
Departamento de F\'{\i}sica,
Universidad de Santiago de Chile for hospitality. LV and FM   would like to thanks M. Ba\~nados
for enlightening
discussions. JS and   FM would like to thanks  J. Zanelli and R. Troncoso for several discussions.
FM acknowledge
also the useful discussions with J. Gamboa who suggested this problem.
\begin{references}

\bibitem{kaplan} S. A. Kaplan, {\it Zh. Eksp. Teo. Fiz.} {\bf 19}, 951 (1949); Ya. B. Zeldovich
and I. D. Novikov, {\it
Dokl. Akad. Nauk.} {\bf 155}, 1033.

\bibitem{birrel}{\it e.g.} see N. D. Birrel and P.C.W. Davies, {\it Quantum Fields in Curved
Space times}, Cambridge
University Press (1982).

\bibitem{gm} J. Gamboa and F. M\'endez, {\it Class.  and Quantum Grav.} {\bf 18}, 225 (2001).

\bibitem{teitel}C. Teitelboim, {\it Phys.Rev.} {\bf D51}, 315 ( 1995), {\it  Erratum-ibid.} {\bf
D52}, 6201 (1995).


\bibitem{egu}T. Eguchi, P. Gilkey and A. Hanson, {\it Gravitation,  Gauge Theory and
Differential Geometry}, Phys.
Rep. {\bf 66}, 6,  (1980).

\bibitem{naka}M. Nakahara, {\it Geometry, Topology and Physics},
Adam Hilger, 1991.

\bibitem{btz}M. Ba\~nados, C. Teitelboim and J. Zanelli, {\it
Phys. Rev.  Lett.} {\bf 69}, 1819 (1992).

\bibitem{birming}D. Birmingham, I. Sachs and S. Sen, {\it Phys.
Lett.} {\bf B413}, 281 (1997).

\bibitem{murphy} The reader interested in the complete solution
of the equation and approximate techniques can see {\it Methods of
Theoretical Physics}, P. Morse and H. Feshbach,  McGraw-Hill Book
Co., Inc. (1953), Vol I Chap. V and Vol II  Chap XI. Also could
see {\it Ordinary Differential Equations and Their Solutions}, G.
Murphy, D. Van Nostrand Co., Inc. (1960).

\bibitem{htf}Staff of Bateman Manuscript Project, {\it Higher
Transcendental Function}, McGraw-Hill Book Co., Inc. (1953).

\bibitem{elipticos}T. R. Govindarajan, V. Suneta and S. Vaidya, {\it Nuc. Phys.} {\bf B58}, 291 (
2000).

\end{references}

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