The geometry around a rotating uncharged axially-symmetric black hole is described by the Kerr metric. Being stationary and axisymmetric, the Kerr metric admits two Killing vector fields: $$\partial_t=\frac{\partial}{\partial t},\quad \partial_\phi=\frac{\partial}{\partial\phi}$$

The Kerr metric permits only the above two Killing vectors, and thus any Killing vector field is a linear combination of them. The notion of static and stationary observers are associated with the Killing vectors as follows: $$\partial_t\rightarrow\textrm{static observer}$$ $$\partial_t+\omega\partial_\phi\rightarrow\textrm{stationary observer}$$ Here, $\omega=-(g_{t\phi}/g_{\phi\phi})$ is the angular velocity of the Zero Angular Momentum Observers (ZAMOs) that corotates with the black hole.

Question: What is the physical interpretation and differences between the static and stationary observers as associated with the Killing vectors in Kerr geometry?


3 Answers 3


In Boyer-Lindquist coordinates, the coordinate time observers, or the field lines of the coordinate time vector field $\partial_{t}$, are by definition maintaining constant spatial coordinates.

One can easily convince oneself of this fact by recalling that the vector field is related to its one-parameter family of local diffeomorphisms $\{\mathcal{G}_{\lambda}\}_{\lambda \in I}$, where $\mathcal{G}_{\bullet}(\bullet):I \times \mathcal{M} \rightarrow \mathcal{M} $ through:

$$ X(\mathcal{G}_{\lambda_{0}}(p)) = \frac{d}{d\lambda}_{\vert \lambda=\lambda_{0}} \mathcal{G}_{\lambda}(p). $$

In an oftentimes more familiar language of vector field integral curves, one picks a specific point $p \in \mathcal{M}$ and creates a curve $\gamma(\bullet):= \mathcal{G}_{\bullet}(p) : I \rightarrow \mathcal{M}$.

$$ X(\gamma(\lambda)) = \dot{\gamma}(\lambda) $$

Representing a curve in Boyer-Lindquist coordinates:

$$ x^{\mu} \circ \gamma (\lambda) = (t(\lambda),r(\lambda), \theta(\lambda), \varphi(\lambda)) $$

and differentiating with respect to the $\lambda$ parameter: $$ \frac{d}{d\lambda} x^{\mu} \circ \gamma (\lambda) = \dot{t}\partial_{t}+ \dot{r}\partial_{r}+\dot{\theta}\partial_{\theta}+\dot{\varphi}\partial_{\varphi}$$ we can see that the $\partial_{t}$ corresponds to a curve with just $\frac{dt}{d\lambda} = 1$, so that $t(\lambda) = \lambda$, and all the other derivatives of the curve components vanish, so that $x^{i}(\lambda) = x^{i}_{0} = \rm const$. For these coordinates and for this vector field, the vector field lines represent no movement in spatial coordinates.

Meanwhile, the static observers $\partial_{t} + \omega \partial_{\varphi}$ have a nonvanishing $\dot{\varphi}$ and thus will change their angular position. You're welcome to plug in the $\omega$ and try to integrate the coordinate representations of the curves, fixing the angle $\theta$ and the $r$ through the two trivial spatial equations ($\dot{r}=0, \dot{\theta}=0$).

Another characterisation that can be made about these different observers is through the 3+1 split, so the type of the hypersurfaces that are created by $t(\lambda)=\lambda=\rm const$ or the stationary observers.

The property of a static spacetime is that there exists a timelike Killing vector field $K$ which is hypersurface orthogonal.

We say a vector field $K$ is hypersurface orthogonal, if the whole spacetime can be foliated by the parameter of the one-parameter family of diffeomorphisms ($K=\frac{\partial }{\partial \lambda}$) associated with the vector field $K$, such that the hypersurfaces of constant $\lambda$ are everywhere orthogonal to the vector field $K$.

See the definition, for example here.

It turns out, most easily through a special case of (dual) Frobenius theorem out that the coordinate vector field $\partial_{t}$ is not hypersuface orthogonal. Formally, that involves calculating $K^{\flat}= g(\partial_{t}, \bullet)$ and checking that $K^{\flat}\wedge dK^{\flat} \neq 0$.

In Wald's book, where this characterisation of the Frobenius theorem can be found (along with proofs, p. 434-438) the condition of being hypersurface orthogonal is equivalently phrased as $K^{\flat}_{[\mu}\nabla_{\nu}K^{\flat}_{\rho]} = 0$.

Of course, the spacetime $\mathcal{M}$ can be foliated into $\mathbb{R}\times \Sigma$, but the foliation parameter that corresponds to $\partial_{t}$ does not to lead to hypersurfaces whose tangent vector fields $\Gamma(T\Sigma)$ are orthogonal to said vector field. This is clearly visible from $g(\partial_{t}, \partial_{\varphi}) \neq 0$.

Note: This part I haven't revised in some time, but I have done my best.

How can we see in another way that the distribution spanned by the vector fields orthogonal to $\partial_{t}$ is not integrable?

The idea coming from the proof and claim of the Frobenius theorem is that when the one-parameter diffeomorphism groups of tangent fields in the distribution combine to a single diffeomorphism, this provides for the identification of a subset of the original manifold as a submanifold with a chart.

Now, the vector field distribution orthogonal to $\partial_{t}$ is

$$ \Gamma(T\mathcal{M}) \supset \Gamma(\Delta) = \langle \partial_{r}, \partial_{\theta}, \partial_{t}+ \omega\partial_{\varphi}\rangle_{\mathcal{C}^{\infty}(\mathcal{M})} $$

Here, the $\langle \rangle_{\mathcal{C}^{\infty}(\mathcal{M})}$ means the span of the basis vector fields spanned by their $\mathcal{C}^{\infty}(\mathcal{M})$-combinations - this is a module of vector fields.

The above distribution is not involutive, because it is not true that $[X,Y] \in \Gamma(\Delta)$ for all $X,Y \in \Gamma(\Delta)$. For example, $[\partial_{r}, \partial_{t}+ \omega\partial_{\varphi}] \notin \Gamma(\Delta)$, because $\omega$ contains $r$-dependent terms, which upon differentiating you can no longer put as a $\mathcal{C}^{\infty}(\mathcal{M})$-combination of the distribution's basis vector fields.

Meanwhile, the diffeomorphism groups of the tangent fields of the distribution
$$ \Gamma(T\mathcal{M}) \supset \Gamma(\Delta) = \langle \partial_{r}, \partial_{\theta}, \partial_{\varphi}\rangle_{\mathcal{C}^{\infty}(\mathcal{M})}, $$ combine to a well-defined diffeomorphic mapping (for any $t\in \mathbb{R}$) once you pick a point $p\in\mathcal{M}$ $$\mathcal{G}:I \times J \times K \times \mathcal{M} \rightarrow \mathcal{M}$$ $$ \mathcal{G}:= \mathcal{G}^{\partial_{r}}_{\lambda_{1}} \circ \mathcal{G}^{\partial_{\theta}}_{\lambda_{2}} \circ \mathcal{G}^{\partial_{\varphi}}_{\lambda_{3}} $$

such that $\mathcal{G}^{-1}$ exists and provides a coordinate chart for the subset of the manifold at constant $t$.

This follows because the distribution is trivially involutive ($[X,Y] \in \Gamma(\Delta)$ for all $X, Y\in \Gamma(\Delta)$) and thus integrable to a submanifold $\Sigma$, on which we have the coordinates $(r,\theta, \varphi)$.

An interesting characterisation of the static and stationary observers can also be made when you contract their vector fields with the $\partial_{\phi}$ vector field. You'll see that, contrary to static observers, the stationary ones (ZAMOs), do not report any angular momentum.

$$ L_{\partial_{t}} = g(\partial_{t}, \partial_{\varphi}) \neq 0$$ $$ L_{\text ZAMO} = g(\partial_{t} + \omega\partial_{\varphi}, \partial_{\varphi}) = 0$$ But this of course you essentially mentioned.

Last interesting thing I can think of now is that when you perform a 3+1 split for the Boyer-Lindquist coordinates and calculate the normal vector field $\mathbf{n}^{\flat}=-\alpha dt$ with suitable normalization $g(\mathbf{n}, \mathbf{n}) = -1$, you will (by manipulating the expression for $\mathbf{n}$) find that the coordinate time vector field $\partial_{t}$ is a combination of the $\mathbf{n}$ with a suitable factor in front, as well as a non-zero shift vector $\boldsymbol{\beta}$, i.e.:

$$ \partial_{t} = \alpha \mathbf{n} + \boldsymbol{\beta}.$$

In the 3+1 language, what this means is that the coordinate time lines stay at constant spatial coordinates thanks to the presence of the nonvanishing $\beta$ vector, which, as you'd be able to see, is equal to

\begin{gather} \boldsymbol{\beta} = \gamma^{\varphi\varphi}g_{t\varphi}\partial_{\varphi}, \end{gather}

As noted by the other answer, this shift, countering the co-rotation, will have to become superluminal inside the ergosphere.

$\gamma^{ij}$ above stands for the inverse three-metric $\gamma^{ij}$, coming from the matrix inversion of the spatial 3-metric $\gamma_{ij}= g_{ij} = (g_{\mu\nu})_{dt=0}$. If you expand the $\beta$ found above, you'll find that it features the negative sign wrt. to the black hole spin $a=\frac{J}{M}$. More details on the 3+1 split in the script.

For a given coordinate system you can also find the 3+1 quantities (lapse $\alpha$, shift vector field $\boldsymbol{\beta}$, spatial metric $\gamma_{ij}$) from the following representation:

$$ g = (-\alpha^{2} + \gamma_{ij}\beta^{i}\beta^{j})dt\otimes dt + \gamma_{ij}\beta^{i}(dt \otimes dx^{j} + dx^{j}\otimes dt) + \gamma_{ij }dx^{i}\otimes dx^{j}$$

  • $\begingroup$ Thank you for the detailed answer. $\endgroup$
    – Richard
    Jul 7, 2022 at 10:12

What you call the stationary observer is in other words the ZAMO at constant r, and the static one does not corotate with the frame dragging, so such an observer can only exist outside the ergosphere.


A static observer sees a static image: the black hole's shadow doesn't move against the background of fixed stars.

A stationary observer measures no Sagnac effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.