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 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.

The formulas derived for Schwarzschild (\ref{BH-energy}) and (\ref{BH-lambda}) still work, just now $f(\rho)$ is not fixed.
(a)As for almost all particles $d\tau /d\rho$ stays bounded, the proper time of reaching the horizon candidate at $\rho=\rho^\star$ is finite if and only if $\rho^\star$ itself is finite. Otherwise we would have the "horizon" at infinite proper distance. It can be called a remote horizon, but extension across such a surface is not possible and not needed, as spacetime in that direction is already geodesically complete.
(b) The Killing vector that forms the horizon is $\xi_t =\partial_t$, its norm $|\xi_t |^2 =g_{tt}=f$. Thus $\rho=\rho^\star$ is indeed a Killing horizon by definition.

(a) Integrating $dx=f^{-1}d\rho$, one obtains \begin{equation} \rho\sim \left\{\begin{array}{l} e^{xF(0)},\quad \text{for}\quad q=1,\\ |x|^{-1/(q-1)}\quad\text{for}\quad q>1 . \end{array}\right. \end{equation} We see that $x\to -\infty$ at the horizon. It converges for $q<1$, but in this case one can show that the Kretschmann scalar $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$ diverges, so this is not a horizon but a singularity.
(b) $ds^2 =f \; dV\;dW$; the horizon is at $x\to -\infty$ while $t$ is fixed, so in terms of $V$ and $W$ it is split into two parts: the future horizon at $V\to -\infty$ while $W$ is fixed, and the past horizon at $W\to +\infty$ while $V$ is fixed.
(c) Near the past horizon in terms of $(v,W)$ the metric takes form \begin{equation} ds^2 =\Big(f \frac{dW}{dw}\Big)\; dV\; dw, \end{equation} while $W\approx -x$, so for it to be regular there we need \begin{equation} \frac{dW}{dw}\sim f^{-1}\sim \rho^{q} \sim \left\{ \begin{array}{l} |x|^{(1-q)/q}\sim W ^{(1-q)/q}\quad\text{for}\quad q>1 ,\\ e^{xF(0)}\sim e^{-WF(0)}\quad\text{for}\quad q= 1, \end{array}\right. \end{equation} and after integrating \begin{equation} W\sim \left\{\begin{array}{l} w^{1-q}\quad\text{for}\quad q>1 ,\\ \ln w\quad\text{for}\quad q= 1. \end{array}\right. \end{equation} The same way the asymptotic relation $V(v)$ is found near the future horizon. Thus a horizon candidate $\rho=\rho^\star$ is defined by condition $f(\rho^\star)=0$. If $\rho^\star =\pm\infty$, then the proper time of reaching it diverges, so it is not a horizon, but an infinitely remote boundary of spacetime. It can be a remote horizon if $x^\star =\pm \infty$. In case $x^\star =O(1)$ the candidate is not a horizon, but a singularity, so again there is no continuation. An asymptotically flat spacetime by definition has the same structure of infinity as that of Minkowski spacetime; in general this is not always the case---recall the de Sitter and other cosmological spacetimes.

(a) Horizontal line, approached either from above or from below; it acts as the boundary of spacetime -- there is no continuation across (see Fig.);

Space-like singularity.

(b) Vertical line, approached either from the left or from the right; as it is singular, there is again no continuation (see Fig.);

Time-like singularity.

(c) This is the infinitely remote part of Minkowski spacetime, from future infinity, to future null infinity, to spacelike infinity, to past null infinity, and finally to past infinity; it can constitute either the left or the right boundary of a conformal block and has the shape of "$>$" or "$<$" (see Fig. \ref{Conf-BH-InfMink});

Minkowski infinity.

(d) The same as the Schwarzschild horizon as approached from the exterior region: the two sides of a triangle pointed left "$<$" or right "$<$" and approached from the inside; thus the shape is the same as above, but now the horizon allows continuation across (see Fig.);

Horizon as boundary of an R-region.

(e) The same as Schwarzschild horizon as approached from the interior region: two sides of a triangle pointed up "$\wedge$" or down "$\vee$" and approached from the inside (see Fig.).

Horizon as boundary of a T-region

(f) Two diagonal lines -- a corner of a triangle, -- pointed up or down, and approached from the inside. Thus the shape is the same as in the previous case, "$\wedge$" or "$\vee$"; the difference is that now there is no continuation across (see Fig.).

Remote horizon as a boundary of a T-region

For arbitrary $f(\rho)$ we can
1) split the full spacetime into regions between the zeros $\rho_i$ of $f(\rho)$;
2) draw for each region the conformal diagram, which is a square $\Diamond$ or half of it $\triangle$, $\nabla$, $\triangleleft$, $\triangleright$;
3) glue the pieces together along the horizons $\rho=\rho_i$, while leaving singularities and infinities as the boundary.

1) In the generic case there are two positive roots $+\infty>r_+ >r_- >0$, which give horizons, while $r=0$ is a singularity. This gives us three regions between the roots: (a) $r\in(r_+ ,+\infty)$ between the horizon and asymptotically flat infinity, thus having the same structure as the Schwarzschild exterior region and the shape of a "diamond" $\Diamond$; (b) $r\in (r_- ,r_+)$ is a T-region between the horizons (the shape is the same $\Diamond$) and (c) $r\in (0,r_-)$ is again an R-region between the timelike singularity and the horizon. The timelike singularity turns the latter conformal diagram into a triangle, $\triangleright$ or $\triangleleft$. Gluing all the blocks together gives us the left diagram of Figs.

2) Due to degeneracy the T-region is absent, instead the two kinds of R-regions are separated by double horizons; the III-blocks on the right vanish, while on the left the singularity merges into a solid vertical line.

3) There can be up to three positive roots of $f$ in case $\Lambda>0$: $r_{-}<r_+<r_\Lambda$. Then there are four different conformal blocks:
a) block IV for the R-region between the timelike singularity and the inner horizon $r\in (0,r_-)$: $\triangleleft,\triangleright$;
b) block III for the T-region between the two horizons $r\in(r_- ,r_+)$; it is the same as for Reissner-Nordstr\"{o}m: $\Diamond$;
c) block II for the R-region between the horizons $r\in (r_+ ,r_\Lambda)$; it differs from the exterior region of Reissner -Nordstrom or Schwarzschild by replacement of asymptotic infinity with another pair of horizons. The shape is the same, $\Diamond$, but the boundary now allows continuation across it in all directions;
d) block I for the T-region between the external (cosmological) horizon and the de-Sitter-like infinity. The structure of infinity is determined by $f(r)$ for large $r$, where the $\sim 1/r$ and $1/r^2$ terms can be neglected, so effectively we have the de Sitter spacetime. Thus the infinity is spacelike and represented by one horizontal line. The block is the triangle: $\triangle,\nabla$ (the same as the upper and lower sectors of the exact full de Sitter spacetime).

The full diagram is shown on Fig.

Conformal diagram for the Reissner-Nordstrom-de Sitter black hole solution. T-regions are shaded with darker grey. The diagram is infinitely continued up and down.