Is the event horizon of an extremal black-hole just a coordinate singularity?

Question

For generic black holes one demonstrates that the event horizon is just a coordinate singularity by changing coordinates to a system where the metric is well defined. For example we have Kruskal-type coordinates $U_\pm\sim e^{\pm \kappa u_\pm}$ with $\kappa$ the surface gravity. In terms of these coordinates one usually finds $\mathrm ds^2\sim \frac{\exp(\text{stuff})}{r}\mathrm dU_+\mathrm dU_-+\cdots$, where it becomes manifest that $r=r_h$ is fine, and only $r=0$ is a true singularity.

For an extremal black-hole, though, the surface gravity vanishes $\kappa=0$ and the trick above does not work. Equivalently, if we approach extremality from a sub-extremal black-hole, the coordinates $U_\pm$ are easily seen to be ill-defined (e.g., then are $\sim \frac{1}{r_+-r_-}$, which is singular as $r_-\to r_+$).

I haven't been able to find a proof that the event horizon of extremal black-holes is non-singular, and I don't know how to fix the standard argument myself. Is there a simple coordinate system where, say, extremal Reissner-Nordström is seen to be non-singular at $r=r_h$? FWIW, I am looking for an argument of the type above rather than some abstract, high-level proof (e.g. by proving geodesic completion or some complicated argument like that).

Answer 1

I'll use this approach:

1. Start with a metric that is non-singular for all $r\neq 0$.

2. Transform the time coordinate to get the more familiar form of an extremal black hole.

The coordinate singularity on the horizon enters in step 2, because the coordinate transform itself is singular. The fact that we started with a non-singular metric shows that the singularity on the horizon is an artifact of the coordinate system.

Non-rotating black hole with extremal charge

Let $d\Omega^2$ denote the standard metric on the unit sphere, and use the letters $w,r$ for the other two coordinates. Start with the metric $$ dw^2-dr^2-V(r)(dw+dr)^2-r^2 d\Omega^2 \tag{1} $$ where $V(r)$ is smooth and finite for all $r>0$. Define a function $f(r)$ by $$ \frac{d}{dr}f(r)=\frac{V(r)}{V(r)-1}, \tag{2} $$ and define a new coordinate $t$ by $$ w = t + f(r). \tag{3} $$ Substitute (3) into (1) and use (2) to get this identity, after a little algebra: $$ dw^2-dr^2-V(r)(dw+dr)^2-r^2 d\Omega^2 = \big(1-V(r)\big)dt^2-\frac{dr^2}{1-V(r)}-r^2d\Omega^2. \tag{4} $$ The metric (1) was nonsingular for all $r>0$, but the coordinate transform (3) introduced a singularity at the value of $r$ for which $V(r)=1$, which is thus obviously only a coordinate singularity.

To apply this to the case of an extremal charged non-rotating black hole, define the function $V(r)$ by $$ V(r) \equiv 1-\left(1-\frac{Q}{r}\right)^2. \tag{5} $$ Then (4) is the familiar form of the metric for the extreme black hole, and the metric (1) is clearly nonsingular for all $r>0$. Mission accomplished.

Actually, we need to be a little more careful before we conclude that (1) is well-behaved when $V(r)=1$, because the $dw^2$ term in (1) cancels when $V(r)=1$. One way to see that the metric is still nondegenerate there is to use the identity $dr^2+dr\,dw = (du^2-dw^2)/4$ with $u\equiv w+2r$.

The metric (1) is an example of a Kerr-Schild metric. This whole analysis also works for non-extremal charged black holes, just by generalizing the function (5).

Uncharged black hole with extremal rotation

The extremal rotating black hole can be handled in a similar way. For a Kerr black hole (extremal or not), the Kerr-Schild form of the metric is $$ \newcommand{\bfu}{\mathbf{u}} \newcommand{\bfx}{\mathbf{x}} dw^2-d\bfx^2 - V(\bfx)\big(dw+\bfu(\bfx)\cdot d\bfx\big)^2 \tag{6} $$ where the independent coordinates are $w$ and $\bfx=(x,y,z)$, and where the functions $\bfu=(u_x,u_y,u_z)$ and $V$ are defined by $$ u_x+iu_y = \frac{x+iy}{r(\bfx)+ia} \hspace{2cm} u_z = \frac{z}{r(\bfx)} \hspace{2cm} V = M\nabla\cdot\bfu, \tag{7} $$ where $\nabla$ is the gradient with respect to $\bfx$ and where the function $r(\bfx)$ is defined implicitly by the conditions $$ \bfu^2=1 \hspace{2cm} r\geq 0. \tag{8} $$ Everything in equations (6)-(8) is nonsingular for all $r>0$, even in the extremal case $a=M$. To relate this to the Boyer-Lindquist form of the metric, define new coordinates $t,\hat x,\hat y$ by $$ t = w - f(r) \hspace{2cm} \hat x = x+ay/r \hspace{2cm} \hat y = y-ax/r \tag{9} $$ with $$ \frac{d}{dr}f(r) = \frac{2Mr}{r^2-2Mr+a^2}, \tag{10} $$ and then express $\hat x,\hat y,z$ in terms of $r$ and angles as usual. After a lot of algebra, this should reproduce the familiar Boyer-Lindquist form of the metric. The coordinate transform (9)-(10) is singular where $r^2-2Mr+a^2=0$, which is why the resulting Boyer-Lindquist form of the metric has a coordinate singularity there, even though the original metric (6) has no such singularity.