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

(a) In terms of the new ("tortoise") coordinate \begin{equation} x=\int \frac{dr}{f(r)}=r+r_g \ln |r-r_g | \end{equation} we get \begin{equation} ds^2 =f(r(x))\big[dt^2 -dx^2\big]. \end{equation} The horizon is pushed to infinity: $x\to -\infty$.
(b) Introduce \begin{equation} v=t+x,\quad u=t-x \end{equation} The horizon is at $v\to-\infty$ when $u=const$ (past horizon) and at $u\to -\infty$ when $v=const$ (future horizon). The asymptotically flat infinity is in the opposite direction: at $v\to+\infty$ when $u=const$ and at $u\to +\infty$ when $v=const$

Conformal diagram for the exterior region of Schwarzschild. The dashed part of the boundary is the horizon; dash-dotted thin lines represent null geodesics; thin red lines are lines of constant $r$, thin blue lines are lines of constant $t$.

(c) Pass to $V=\arctan v$ and $U=\arctan u$; the horizon now becomes two lines at $V=-\pi/2$ and $U=-\pi/2$. The infinity is $V=+\pi/2$ and $U=+\pi/2$. The full exterior region is the square enclosed by those four lines.
(d) Transformation to $T=V+U$, $R=V-U$ rotates this square by $\pi/4$ and this is the conformal diagram block that we need. There is a past and future horizon. Note that while past should be below and future above (we expect the spacetime to be oriented, so that the direction of future is always uniquely determined), there is no rule that says that the horizons must be on the left and infinities on the right. Thus there are two mirror-reflected variants, with the horizons on the left and on the right.

The procedure is exactly the same as for the exterior region, the only thing that is different is the ranges of coordinates used and reversal of the timelike and spacelike coordinates:
(a) Now $t$ is spacelike and $r$ a timelike coordinate: \begin{equation} ds^2 =|f(r)|^{-1}dr^2 -|f(r)|dt^2 . \end{equation} (b) The singularity $r=0$ is therefore a spacelike one-dimensional object: one does not "reach" it or not, instead one lives until the specified time. (c) The interior solution in not static, as the metric essentially depends on the new time coordinate $r$ (or $-r$). The null coordinates then are (note the second sign) \begin{equation} v=t+x,\qquad u=x-t. \end{equation} The horizon is at $v\to -\infty$ when $u=const$ and at $u\to -\infty$ when $v=const$. After shrinking with $\arctan$ it is pulled to $v=-\pi/2$ and $u=-\pi/2$. The third part of the boundary, corresponding to the singularity $r=0$, is $v+u=const$. Finally, we return to timelike and spacelike coordinates by introducing (there are two ways of choosing the sign) \begin{equation} T=\pm (v+u),\qquad R=V-U. \end{equation} The resulting conformal diagram is a 45 degrees right triangle with the right angle pointing either up (then the singularity is in the past) or down (then the singularity is in future). Note that we will know which of the horizons is past and which is future only when we match the blocks with the ones that have the corresponding infinities (although one can guess already where those will be).

Conformal diagram for the interior region of Schwarzschild with future singularity. The dashed part of the boundary is the horizon(s); dash-dotted thin lines represent null geodesics; thin red lines are lines of constant $r$, thin blue lines are lines of constant $t$..

The geodesic equation is obtained from normalization condition \begin{equation} \varepsilon^2=u^\mu u_\mu =f(r)\Big(\frac{dt}{d\lambda}\Big)^{2} -f^{-1}(r) \Big(\frac{dr}{d\lambda}\Big)^{2}, \end{equation} where $\varepsilon^2=0$ or $\varepsilon^2=1$ for null and timelike geodesics respectively, and conservation equation due to the Killing vector \begin{equation} E=u^\mu \xi_\mu =u_0 =f(r)\frac{dt}{d\lambda}, \label{BH-energy} \end{equation} and reads \begin{equation} \frac{d\lambda}{dr} =\Big[E^2 -f(r)\varepsilon^2 \Big]^{-1/2}. \label{BH-lambda} \end{equation} The integral $\lambda(r)$ converges at $r=r_g$, so the value of the affine parameter at the horizon is finite. In regard to regularity let us start from the exterior region. In terms of $u,v$ the metric has the overall conformal factor, which turns to zero at the horizon: \begin{equation} f(r)\sim (r-r_g)\sim e^{\ln (r-r_g)} \sim e^{x}\sim e^{(v-u)/2}, \end{equation} so in order to eliminate that, we just need to use new null coordinates $(u',v')$, such that the conformal factor in \begin{equation} ds^{2} \sim \Big(e^{-u/2}\frac{du}{du'}\Big) \sim \Big(e^{+v/2}\frac{dv}{dv'}\Big) du' dv' \end{equation} is finite and does not turn to zero. This is achieved e.g. by coordinate transformation to \begin{equation} u'=e^{u/2},\qquad v'=e^{-v/2}. \end{equation} After this one can carry out the second part of construction of the conformal diagram.

Starting from any one of the four pieces, there is only one way to assemble the diagram, and one has to use all the parts:

• The puzzle pieces: two for the exterior region (above) and two for the interior region (below).} \end{figure
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The horizon structure is very similar to that of full de Sitter spacetime in open slicing or ``static coordinates (the notions and properties of R- and T-regions apply equally well), while the structure of infinity is the same as that of Minkowski, and additionally there are singularities.

Finally, one could note that the form of the conformal block for the exterior solution was determined from the start. First, we have asymptotic flatness, therefore the Minkowski's structure of infinity. This is already half of the block's boundary.

{The full conformal diagram for the (maximally extended) Schwarzschild solution.} \end{figure