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Purpose

The pioneer, Bernstein, studied a single black hole which is non-rotating and distorted in azimuthal line symmetry of 2 dimensional case [37]. In this non-rotating case, one chooses the condition, $K_{ij} = 0$, and

\begin{displaymath}
\gamma_{ab} = \psi^4 \hat \gamma_{ab},
\end{displaymath} (part331)

where $\gamma_{ab}$ is the physical three metric and $\hat{\gamma}_{ab}$ is some chosen conformal three metric.

The Hamiltonian constraint reduces to

\begin{displaymath}
\hat \Delta \psi = \frac{1}{8}\psi \hat R,
\end{displaymath} (part332)

where $\hat \Delta$ is the covariant Laplacian and $\hat R$ is the Ricci tensor for the conformal three metric. This form allows us to choose an arbitrary conformal three metric, and then solve an elliptic equation for the conformal factor, therefore satisfying the constraint equations ($K_{ij} = 0$ trivially satisfies the momentum constraints in vacuum). This approach was used to create ``Brill waves'' in a spacetime without black holes [38]. Bernstein extended this to the black hole spacetime. Using spherical-polar coordinates, one can write the 3-metric,
\begin{displaymath}
ds^2 = \psi^4 (e^{2q} (dr^2 + r^2 d \theta^2) + r^2 \sin \theta d
\phi^2),
\end{displaymath} (part333)

where $q$ is the Brill ``packet'' which takes some functional form. Using this ansatz with (D13.2) leads to an elliptic equation for $\psi$ which must be solved numerically. Applying the isometry condition on $\psi$ at a finite radius, and applying $M/2r$ falloff conditions on $\psi$ at the outer boundary (the ``Robin'' condition), along with a packet which obeys the appropriate symmetries (including being invariant under the isometry operator), will make this solution describe a black hole with an incident gravitational wave. The choice of $q=0$ produces the Schwarzschild solution. The typical $q$ function used in axisymmetry, and considered here in the non-rotating case, is
\begin{displaymath}
q = Q_0 \sin^n \theta \left [ \exp\left(\frac{\eta -
\eta...
... \exp\left(\frac{\eta +
\eta_0^2}{\sigma^2}\right ) \right ].
\end{displaymath} (part334)

Note regularity along the axis requires that the exponent $n$ must be even. Choosing a logarithmic radial coordinate
\begin{displaymath}
\eta = \ln{\frac{2r}{m}}.
\end{displaymath} (part335)

(where $m$ is a scale parameter), one can rewrite (D13.3) as
\begin{displaymath}
ds^2 = \psi(\eta)^4 [ e^{2 q} (d \eta^2 + d\theta^2) + \sin^2
\theta d\phi^2].
\end{displaymath} (part336)

The scale parameter $m$ is equal to the mass of the Schwarzschild black hole, if $q=0$. In this coordinate, the 3-metric is

\begin{displaymath}
ds^2 = \tilde{\psi}^4 (e^{2q} (d\eta^2+d\theta^2)+\sin^2 \theta
d\phi^2),
\end{displaymath} (part337)

and the Schwarzschild solution is
\begin{displaymath}
\tilde{\psi} = \sqrt{2M} \cosh (\frac{\eta}{2}).
\end{displaymath} (part338)

We also change the notation of $\psi$ for the conformal factor is same as $\tilde{\psi}$ [39], for the $\eta$ coordinate has the factor $r^{1/2}$ in the conformal factor. Clearly $\psi(\eta)$ and $\psi$ differ by a factor of $\sqrt{r}$. The Hamiltonian constraint is
\begin{displaymath}
\frac{\partial^2 \tilde{\psi}}{\partial \eta^2} + \frac{\p...
...partial \eta^2} + \frac{\partial^2 q}{\partial
\theta^2} -1).
\end{displaymath} (part339)

For solving this Hamiltonian constraint numerically. At first we substitute

$\displaystyle \delta \tilde{\psi}$ $\textstyle =$ $\displaystyle \tilde{\psi}+\tilde{\psi}_0$ (part3310)
  $\textstyle =$ $\displaystyle \tilde{\psi}-\sqrt{2m} \cosh(\frac{\eta}{2}).$ (part3311)

to the equation (D13.9), then we can linearize it as
\begin{displaymath}
\frac{\partial^2 \delta\tilde{\psi}}{\partial \eta^2} + \fra...
...rtial
\eta^2} + \frac{\partial^2 q}{\partial \theta^2} -1).
\end{displaymath} (part3312)

For the boundary conditions, we use for the inner boundary condition an isometry condition:
\begin{displaymath}
\frac{\partial \tilde{\psi}}{\partial \eta}\vert _{\eta = 0} = 0,
\end{displaymath} (part3313)

and outer boundary condition, a Robin condition:
\begin{displaymath}
(\frac{\partial \tilde{\psi}}{\partial \eta} + \frac{1}{2}
\tilde{\psi})\vert _{\eta=\eta_{max}} = 0.
\end{displaymath} (part3314)


next up previous contents
Next: The Resulting Slice Up: IDAxiBrillBH Previous: Abstract   Contents