Difference between revisions of "Conformal diagrams: stationary black holes"

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(Problem 4: Piecing the puzzle.)
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|[[File:Conf-BH-SchwFull.png|center|thumb|365px|{The full conformal diagram for the (maximally extended) Schwarzschild solution.}
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|[[File:Conf-BH-SchwFull.png|center|thumb|365px|The full conformal diagram for the (maximally extended) Schwarzschild solution.]]
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Revision as of 01:55, 4 February 2014


In the context of black hole spacetimes there are many subtly distinct notions of horizons, with the most useful being different from those used in cosmology. In particular, particle horizons do not play any role. The event horizon is defined not with respect to some selected observer, but with respect to all external observers: in an asymptotically flat spacetime* a future event horizon is the hypersurface which separates the events causally connected to future infinity and those that are not. Likewise the past event horizon delimits the events that are causally connected with past infinity or not. Another simple but powerful concept is the Killing horizon: in a spacetime with a Killing vector field $\xi^\mu$ it is a (hyper-)surface, on which $\xi^\mu$ becomes lightlike. We will see explicitly for the considered examples that the Killing horizons are in fact event horizons by constructing the corresponding conformal diagrams. In general, in the frame of GR a Killing horizon in a stationary spacetime is (almost) always an event horizon and also coincides with most other notions of horizons there are.

This section elaborates on the techniques of constructing conformal diagrams for stationary black hole solutions. The construction for the Schwarzschild black hole, or its variation, can be found in most textbooks on GR; for the general receipt see \cite{BrRubin}.

* Meaning the spacetime possesses the infinity with the same structure as that of Minkowski, which is important.

Schwarzschild-Kruskal black hole solution

Problem 1: Schwarzschild exterior.

The simplest black hole solution is that of Schwarzschild, given by \begin{equation} ds^{2}=f(r)dt^2 -\frac{dr^2}{f(r)}-r^2 d\Omega^2,\qquad f(r)=1-\frac{r_g}{r}, \label{SchwBH} \end{equation} where $r_g$ is the gravitational radius, and $d\Omega^2$ is the angular part of the metric, which we will not be concerned with. The surface $r=r_g$ is the horizon. Focus for now only on the external part of the solution, \[\big\{-\infty<t<+\infty,\;\; r_g<r<+\infty)\big\}.\] The general procedure of building a conformal diagram for the $(t,r)$ slice, as discussed in the cosmological context earlier, works here perfectly well, but needs one additional step in the beginning:
(a) use a new radial coordinate to bring the metric to conformally flat form; (b) pass to null coordinates; (c) shrink the ranges of coordinate values to finite intervals with the help of $\arctan$; (d) return to timelike and spacelike coordinates.

Identify the boundaries of Schwarzschild's exterior region on the conformal diagram and compare it with Minkowski spacetime's.

Problem 2: Schwarzschild interior.

The region $r\in (0,r_g)$ represents the black hole's interior, between the horizon $r=r_g$ and the singularity $r=0$. Construct the conformal diagram for this region following the same scheme as before.
(a) Which of the coordinates $(t,r)$ are timelike and which are spacelike? (b) Is the singularity spacelike or timelike? (c) Is the interior solution static?
The Schwarzschild black hole's interior is an example of the T-region, where $f(r)<0$, as opposed to the R-region, where $f(r)>0$.

Problem 3: Geodesic incompleteness and horizon regularity.

Consider radial motion of a massive particle and show that the exterior and interior parts of the Schwarzschild are not by themselves geodesically complete, i.e. particle's worldlines are terminated at the horizon at finite values of affine parameter. Show that, on the other hand, the horizon is not a singularity, by constructing the null coordinate frame, in which the metric on the horizon is explicitly regular.

Problem 4: Piecing the puzzle.

Geodesic incompleteness means the full conformal diagram must be assembled from the parts corresponding to external and internal solutions by gluing them together along same values of $r$ (remember that each point of the diagram corresponds to a sphere). Piece the puzzle.

Note that a) there are two variants of both external and internal solutions' diagrams, differing with orientation and b) the boundaries of the full diagram must go along either infinities or singularities.