Understanding Kerr Black Holes: Metric, Killing Vectors & Event Horizons

[JD_PM]

TL;DR Summary

I would like to gain more insight and discuss rotating black holes.

I am particularly interested in understanding the isometries related to the metric (i.e. Killing vectors), the conserved quantities associated to them and in what cases we have event horizons.

I will summarize what I know so far about it and then make some questions to get started. Please feel free to correct me if any of my statements is wrong.

I study from Carroll's book and the beautiful Tong's lecture notes on GR.

Before explicitly stating the Kerr metric let us discuss a bit what to expect, comparing it to the easiest solution to (in-vacuum) Einstein's equations that I know: the Schwarzschild metric.

I studied that the Schwarzschild metric is derived under the following assumptions: the metric must be spherically symmetric and static (the latter word means that the metric has a Killing vector associated to the time coordinate $K=\partial_t$ and that the metric is invariant under time reversal, $t \rightarrow -t$).

For the Kerr metric I see we do not make such assumptions. It makes sense to me to not assume spherical symmetry, as a black hole (I think) cannot be always approximated to be a sphere. Besides, (I think) its shape can change over time so indeed it makes sense to not assume that the metric is time-independent.

Let's now state the Kerr metric

\begin{align*}
(ds)^2 &= -\left(1-\frac{2GMr}{\rho^2} \right)(dt)^2 - \frac{2GMar \sin^2 \theta}{\rho^2}(dt d\phi + d\phi dt) \\
&+ \frac{\rho^2}{\Delta}(dr)^2 + \rho^2 (d \theta)^2 + \frac{\sin^2 \theta}{\rho^2} \left[ (r^2+a^2)^2-a^2 \Delta \sin^2 \theta \right] (d \phi)^2
\end{align*}

Where

\begin{equation*}
\Delta = r^2 -2GMr+a^2
\end{equation*}

And

\begin{equation*}
\rho^2 = r^2+a^2 \cos^2 \theta, \ \ \ \ a=\mathcal{J}/M
\end{equation*}

Where $\mathcal{J}$ is the (Komar) angular momentum.

Let us now study the Killing vectors associated to such metric. We notice that no coefficient of the coordinates depends neither on time nor $\phi$ so we indeed have $K=\partial_t$ and $K=\partial_{\phi}$. Incidentally we notice that indeed the metric is static, as $t \rightarrow -t$ does not hold due to $(dt d\phi + d\phi dt)$.

We have $K=\partial_t$ and $K=\partial_{\phi}$ as Killing vectors so we expect to have 2 conserved quantities associated to this metric: energy and angular momentum. Regarding the latter: my guess is that what is conserved is the magnitude of the angular momentum, am I right?

Let me make the Schwarzschild metric comparison again: that one has 4 Killing vectors: one associated to energy conservation and the other three to conservation of (3D) angular momentum, which indeed makes sense as the Schwarzschild metric is spherically symmetric. However, in Kerr's metric, I did not expect the presence of a Killing vector associated to angular momentum as we made no assumption regarding shape/motion.

To find at what values of $r$ we have an event horizon, we search for values which blow up our metric i.e. $\Delta = 0$.

Thus we obtain two physical event horizons out of solving

\begin{equation*}
\Delta = r^2 -2GMr+a^2=0
\end{equation*}

Those are

\begin{equation*}
r_{\pm} = GM \pm \sqrt{G^2 M^2 - a^2}
\end{equation*}

My questions are

1) Is $K=\partial_{\phi}$ associated to angular momentum's magnitude?

2) Why are not $\rho = 0 \Rightarrow r=\pm a \cos \theta$ valid event horizon locations?

Please feel free to add extra insight, I really enjoy learning in PF!

Thank you!

 

[PeterDonis]

I studied that the Schwarzschild metric is derived under the following assumptions: the metric must be spherically symmetric and static

No, the assumptions are that the metric is spherically symmetric and vacuum (zero stress-energy tensor). The presence of the extra Killing vector field $\partial_t$ is not assumed; it is derived in the course of solving the Einstein Field Equation for a spherically symmetric, vacuum spacetime. This result is known as Birkhoff's Theorem.

For the Kerr metric I see we do not make such assumptions.

We don't make the same assumptions, but we do make assumptions. The assumptions are that the metric is axially symmetric and vacuum (instead of spherically symmetric and vacuum), and (unlike in the Schwarzschild case) that the spacetime is stationary (since there is no analogue of Birkhoff's Theorem for this case).

we notice that indeed the metric is static, as $t \rightarrow - t$ does not hold

No, the Kerr metric is not static; it's only stationary. That has nothing to do with the metric not being invariant under the transformation $t \rightarrow - t$. It has to do with the timelike Killing vector field $\partial_t$ not being hypersurface orthogonal, which is shown by the presence of the $dt d \phi$ cross term in the metric, not the fact that it switches sign under $t \rightarrow - t$.

in Kerr's metric, I did not expect the presence of a Killing vector associated to angular momentum as we made no assumption regarding shape/motion.

Yes, we did; we assumed axial symmetry (see above). That is where the Killing vector field $\partial_\phi$ comes from.

1) Is $K=\partial_{\phi}$ associated to angular momentum's magnitude?

It's associated with the magnitude of the angular momentum of a test object in free-fall motion in the spacetime of the black hole. It is not associated with the angular momentum of the hole itself.

2) Why are not $\rho = 0 \Rightarrow r=\pm a \cos \theta$ valid event horizon locations?

First, $\rho = 0$ is the wrong condition to ask about here; the condition you mean, I think, is that $g_{tt} = 0$, which means $\rho = 2 G M r$.

That said, the reason this surface is not an event horizon is that the event horizon is the boundary of the region that cannot send light signals to future null infinity. Objects inside $\rho = 2 G M r$ can still do that. What they can't do is remain static, i.e., have zero angular velocity about the hole. They have to orbit the hole in the same sense that the hole is rotating.

In terms of Killing vector fields, the boundary $\rho = 2 G M r$ is where the Killing vector field $\partial_t$ becomes null; but because the hole is rotating, that is not the right Killing vector field to look at to locate the event horizon. The right Killing vector field for that is the KVF $\partial_t + \Omega \partial_\phi$, where $\Omega$ is called the "angular velocity of the horizon". The surface where this KVF becomes null marks the event horizon.

 

[JD_PM]

@PeterDonis thanks for your reply.

I will re-read section 6.6 in Carroll's book and also learn about what is meant with a Killing vector not being hypersurface orthogonal (i.e. I will read appendix D on hypersurfaces in Carroll's book) and then come back

 

[JD_PM]

Hi @PeterDonis my apologies for taking too much time.

I have been thinking how would conserved quantities in Kerr's spacetime look like, by comparing it with Schwarzschild's metric. What I know related to the later is what follows

In Schwarzschild's metric, we let a particle obey the homogeneous geodesic equation i.e.

\begin{equation*}
U^{\lambda} \nabla_{\lambda} U^{\mu} = \frac{d^2 x^{\mu}}{d \tau^2} + \Gamma_{\rho \sigma}^{\mu}\frac{d x^{\rho}}{d \tau}\frac{d x^{\sigma}}{d \tau} = 0
\end{equation*}

Where we parameterized a geodesic by the affine parameter $\tau$ (i.e. proper time).

Under such conditions, there is a four-velocity vector $U^{\mu}=dx^{\mu}/d \tau$ and a Killing vector $K_{\mu}$ whose product yields the subsequent conserved quantity (along the given geodesic)

\begin{equation*}
K_{\mu}U^{\mu} = \text{constant}
\end{equation*}

Spoiler: Proof

\begin{align*}
\frac{d}{d \tau} \left( K_{\mu} U^{\mu} \right) &= \frac{d x^{\nu}}{d \tau} \nabla_{\nu} \left( K_{\mu} U^{\mu} \right)\\
&= U^{\nu} \nabla_{\nu} \left( K_{\mu} U^{\mu} \right) \nonumber \\
&= U^{\nu} U^{\mu} \nabla_{\nu} K_{\mu} + K_{\mu} U^{\nu} \nabla_{\nu} U^{\mu} \nonumber \\
&= U^{\mu} U^{\nu} \nabla_{(\nu} K_{\mu)} + K_{\mu} U^{\nu} \nabla_{\nu} U^{\mu} \\
&= 0
\end{align*}

Where to start off we used the chain rule then the product rule, noticed that $U^{\mu} U^{\nu}$ is symmetric (so only the symmetric part of $\nabla_{\nu} K_{\mu}$ contributes to the product), used the Killing equation and the geodesic equation.

Schwarzschild's metric has 4 Killing vectors: one associated to the $t-$coordinate independence of the metric and the other 3 associated to the spherical symmetry of the metric: they span $SO(3)$. Actually such symmetry allows us to choose $\theta= \pi /2$ to be the plane in which the particle moves. Thus the angular momentum of the particle points in the z-direction only.

It is customary to choose two of these to find the conserved quantities:

\begin{equation*}
K_{\mu}=(\partial_{t})_{\mu}=\left( -\left(1 - \frac{R_S}{r} \right),0,0,0 \right)
\end{equation*}

\begin{equation*}
R_{\mu}=(\partial_{\phi})_{\mu}=\left( 0,0,0, r^2 \right)
\end{equation*}

So we get as conserved quantities (where we introduce a -ive sign for E to give physical sense to it)

\begin{equation*}
E := -K_{t} U^{t} =\left(1 - \frac{R_S}{r} \right) \frac{dt}{d \tau}
\end{equation*}

\begin{equation*}
L := K_{\phi} U^{\phi} = r^2 \left( \frac{d \phi}{d \tau} \right)
\end{equation*}

The idea is to get conserved quantities out of Kerr's metric

The first difference is that now we only have 2 Killing vectors. I would say they look as follows

\begin{equation*}
K_{\mu}=(\partial_{t})_{\mu}=\left( -\left(1 - \frac{2 G M r}{\rho^2} \right),0,0,0 \right)
\end{equation*}

\begin{equation*}
R_{\mu}=(\partial_{\phi})_{\mu}=\left( 0,0,0, \frac{\sin^2 \theta}{\rho^2} \left[ (r^2+a^2)^2-a^2 \Delta \sin^2 \theta \right] \right)
\end{equation*}

However I am not convinced, as I am not taking the mixed terms $dt d\phi$ into account...

 

[PeterDonis]

now we only have 2 Killing vectors. I would say they look as follows

Those are the two that are standardly used to give conserved quantities in the Kerr metric, yes. For a test object with 4-momentum $p^\mu$, the quantity $K_\mu p^\mu$ gives the energy at infinity, and the quantity $R_\mu p^\mu$ gives the orbital angular momentum. (If you use the 4-velocity $U^\mu$ of the object, you get the same quantities per unit rest mass.)

Note that, with the usual convention in Schwarzschild spacetime where we look at orbits in the equatorial plane, $\theta = \pi / 2$, we only look at the corresponding two Killing vectors and get the same conserved quantities. The only difference is the exact mathematical form of the Killing vector components, which reflects the difference in the spacetime geometries.

 

[JD_PM]

Hi @PeterDonis

No, the Kerr metric is not static; it's only stationary. That has nothing to do with the metric not being invariant under the transformation $t \rightarrow - t$. It has to do with the timelike Killing vector field $\partial_t$ not being hypersurface orthogonal, which is shown by the presence of the $dt d \phi$ cross term in the metric, not the fact that it switches sign under $t \rightarrow - t$.

I now see that the Kerr metric is not static. But I thought it was because it is not invariant under the transformation $t \rightarrow - t$.

I took this assertion from Tong's notes in GR, section 'a closer look at time independence'

 

[PeterDonis]

I now see that the Kerr metric is not static. But I thought it was because it is not invariant under the transformation $t \rightarrow - t$.

The most general way of stating the condition for a spacetime being static is that it has a timelike Killing vector field $K$ (stationary) which is hypersurface orthogonal (static). That way of stating the condition is independent of any choice of coordinates.

Tong and I both are choosing coordinates in which $t$ is a coordinate along integral curves of $K$. But his way of stating the condition actually gets the implication backwards. Once we have chosen the $t$ coordinate this way, if $K$ is hypersurface orthogonal, we can choose the space coordinates such that there are no $dt \ dX$ cross terms in the metric. But if $K$ is not hypersurface orthogonal (i.e., if the spacetime is stationary but not static), it is impossible to choose such coordinates; no matter what space coordinates you choose, you will have at least one $dt \ dX$ cross term in the metric.

If there is an unavoidable $dt \ dX$ cross term in the metric because the metric is not static, then it will be true that the metric is not invariant under the transformation $t \rightarrow - t$. However, the reverse implication is not true; you cannot deduce from fact that the metric, in a coordinate chart where $t$ is the coordinate along integral curves of $K$, is not invariant under the transformation $t \rightarrow - t$ (whether because of a $dt \ dX$ cross term in the metric or anything else), that the spacetime is not static. It is perfectly possible to find coordinates on a static spacetime in which the metric is not invariant under the transformation $t \rightarrow - t$; a well known example is Painleve coordinates on Schwarzschild spacetime. The only thing you can say about a static spacetime is that you are not forced to choose coordinates such that the metric is not invariant under $t \rightarrow - t$; you will always be able to find coordinates in which the metric is invariant under that transformation. But you cannot say that all coordinates on a static spacetime that have $t$ as the coordinate along integral curves of $K$ will have that property.

 

[JD_PM]

Thank you.

Let me make a naïve question: what is the definition of a hypersurface orthogonal? I think I really need to understand such a concept first, in order to follow your argument (I've been looking in Tong's notes but did not find it).

I have only learned about spacelike and timelike hypersurfaces and the normal vectors associated to them.

More precisely, I know that we can define a $(d-1)$ -dimensional hypersurface $\Sigma$ by fixing $f(x^{\mu}) = \text{const.}$ and define $\xi_{\mu}:= \nabla_{\mu} f$ to be normal to $\Sigma$

 

[PeterDonis]

what is the definition of a hypersurface orthogonal?

It means that there exists a family of spacelike hypersurfaces that fill the entire spacetime and that are everywhere orthogonal to the integral curves of $K$. With appropriately chosen coordinates, this orthogonality manifests itself as the absence of any $dt \ dX$ cross term in the metric; the absence of that term means that the surfaces of constant $t$ are the orthogonal family of spacelike hypersurfaces in question (because any spacelike vector lying in a surface of constant $t$ will be orthogonal to any timelike vector that points along an integral curve of $K$). But, as noted, not all coordinates on a static spacetime will have this property.