Stationary spacetime
# Stationary spacetime

In general relativity, specifically in the Einstein field equations, a spacetime is said to be stationary if it admits a Killing vector that is asymptotically timelike.

## Description and analysis

In a stationary spacetime, the metric tensor components, $g_{\mu \nu }$, may be chosen so that they are all independent of the time coordinate. The line element of a stationary spacetime has the form $(i,j=1,2,3)$

    $ds^{2}=\lambda (dt-\omega _{i}\,dy^{i})^{2}-\lambda ^{-1}h_{ij}\,dy^{i}\,dy^{j},$

where $t$ is the time coordinate, $y^{i}$ are the three spatial coordinates and $h_{ij}$ is the metric tensor of 3-dimensional space. In this coordinate system the Killing vector field $\xi ^{\mu }$ has the components $\xi ^{\mu }=(1,0,0,0)$. $\lambda$ is a positive scalar representing the norm of the Killing vector, i.e., $\lambda =g_{\mu \nu }\xi ^{\mu }\xi ^{\nu }$, and $\omega _{i}$ is a 3-vector, called the twist vector, which vanishes when the Killing vector is hypersurface orthogonal. The latter arises as the spatial components of the twist 4-vector $\omega _{\mu }=e_{\mu \nu \rho \sigma }\xi ^{\nu }\nabla ^{\rho }\xi ^{\sigma }$(see, for example, p. 163) which is orthogonal to the Killing vector $\xi ^{\mu }$, i.e., satisfies $\omega _{\mu }\xi ^{\mu }=0$. The twist vector measures the extent to which the Killing vector fails to be orthogonal to a family of 3-surfaces. A non-zero twist indicates the presence of rotation in the spacetime geometry.

The coordinate representation described above has an interesting geometrical interpretation. The time translation Killing vector generates a one-parameter group of motion $G$ in the spacetime $M$. By identifying the spacetime points that lie on a particular trajectory (also called orbit) one gets a 3-dimensional space (the manifold of Killing trajectories) $V=M/G$, the quotient space. Each point of $V$ represents a trajectory in the spacetime $M$. This identification, called a canonical projection, $\pi :M\rightarrow V$ is a mapping that sends each trajectory in $M$ onto a point in $V$ and induces a metric $h=-\lambda \pi *g$ on $V$ via pullback. The quantities $\lambda$, $\omega _{i}$ and $h_{ij}$ are all fields on $V$ and are consequently independent of time. Thus, the geometry of a stationary spacetime does not change in time. In the special case $\omega _{i}=0$ the spacetime is said to be static. By definition, every static spacetime is stationary, but the converse is not generally true, as the Kerr metric provides a counterexample.

## Use as starting point for vacuum field equations

In a stationary spacetime satisfying the vacuum Einstein equations $R_{\mu \nu }=0$ outside the sources, the twist 4-vector $\omega _{\mu }$ is curl-free,

    $\nabla _{\mu }\omega _{\nu }-\nabla _{\nu }\omega _{\mu }=0,\,$

and is therefore locally the gradient of a scalar $\omega$ (called the twist scalar):

    $\omega _{\mu }=\nabla _{\mu }\omega .\,$

Instead of the scalars $\lambda$ and $\omega$ it is more convenient to use the two Hansen potentials, the mass and angular momentum potentials, $\Phi _{M}$ and $\Phi _{J}$, defined as

    $\Phi _{M}={\frac {1}{4}}\lambda ^{-1}(\lambda ^{2}+\omega ^{2}-1),$
    $\Phi _{J}={\frac {1}{2}}\lambda ^{-1}\omega .$

In general relativity the mass potential $\Phi _{M}$ plays the role of the Newtonian gravitational potential. A nontrivial angular momentum potential $\Phi _{J}$ arises for rotating sources due to the rotational kinetic energy which, because of mass–energy equivalence, can also act as the source of a gravitational field. The situation is analogous to a static electromagnetic field where one has two sets of potentials, electric and magnetic. In general relativity, rotating sources produce a gravitomagnetic field that has no Newtonian analog.

A stationary vacuum metric is thus expressible in terms of the Hansen potentials $\Phi _{A}$ ($A=M$, $J$) and the 3-metric $h_{ij}$. In terms of these quantities the Einstein vacuum field equations can be put in the form

    $(h^{ij}\nabla _{i}\nabla _{j}-2R^{(3)})\Phi _{A}=0,\,$
    $R_{ij}^{(3)}=2[\nabla _{i}\Phi _{A}\nabla _{j}\Phi _{A}-(1+4\Phi ^{2})^{-1}\nabla _{i}\Phi ^{2}\nabla _{j}\Phi ^{2}],$

where $\Phi ^{2}=\Phi _{A}\Phi _{A}=(\Phi _{M}^{2}+\Phi _{J}^{2})$, and $R_{ij}^{(3)}$ is the Ricci tensor of the spatial metric and $R^{(3)}=h^{ij}R_{ij}^{(3)}$ the corresponding Ricci scalar. These equations form the starting point for investigating exact stationary vacuum metrics.

