Definition of ‘‘static spacetimes’’ from Killing vectors

Question

In Harvey Reall's Black Holes lecture notes, he defines static spacetimes as follows

A spacetime is said to be static if it admits a hypersurface-orthogonal timelike Killing vector field.

I am not sure why this would not be true for a stationary rotating spacetime. Assume for instance that we had a 2+1 dimensions spacetime such that each spatial hypersurface at a particular value for the time coordinate looks like a disk. So at each moment in time, the space looks like a disk that has shifted — for instance — in the $\phi$ direction. If a timelike Killing vector field $\xi^a = (1,0,0)$ was orthogonal to the disk at $t$ it will then remain so at $t'$.

What am I missing?

Answer 1

Let us consider the following metric in $2+1$ dimensions: $$ \mathrm{d}s^2 = \tilde{g}_{tt} \, \mathrm{d}t^2 + g_{rr} \, \mathrm{d} r^2 + g_{\phi \phi} \,( \mathrm{d} \phi - \omega \, \mathrm{d} t)^2, $$ where the metric components are functions of $r$ and $\phi$ only. (Note that the metric component $g_{tt} = \tilde{g}_{tt} + \omega^2 g_{\phi \phi}<0$.)

While it is true that two-dimensional hypersurface defined by a constant $t$ will have the same line-element for all $t$, it is not true that the Killing vector $\xi^\mu = (1, 0, 0)$ will be orthogonal to this hypersurface.

Note that when we say some vector is orthogonal to a surface, it means the vector will be orthogonal to all the vectors that are tangent to the surface. The hypersurface of constant $t=t_0$ can be described by the vector equation, $$ x^\mu = ( t_0, \, r , \, \phi), $$ where $r$ and $\phi$ parametrise the surface. Let $y_a$ be coordinates on the hypersurface. The natural coordinates on the hypersurface are, of course, $r$ and $\phi$. The set of tangent vectors on the hypersurface are given by, $$ e^\mu_{(a)} = \frac{\partial x^\mu}{\partial y^a}. $$ Explicitly, the components of the two tangent vectors are given by, $$ e^\mu_{(r)} = (0, 1, 0) , $$ and $$ e^\mu_{(\phi)} = (0, 0, 1) . $$ We will say the Killing vector is orthogonal to the hypersurface if for each $a$, $$ g_{\mu \nu} \xi^\mu e^\nu_{(a)} = 0. $$ Note that this condition is satisfied when $a=r$. However, because of the presence of the non-zero off-diagonal component of the metric, $$ g_{t\phi} = - \omega g_{\phi \phi}, $$ we would have, $$ g_{\mu \nu} \xi^\mu e^\nu_{(\phi)} = g_{t\phi} = - \omega g_{\phi \phi}. $$ Thus, when $\omega \neq 0$, the Killing vector is never orthogonal to the hypersurface of constant $t$. If $\omega = 0$, the Killing vector would be hypersurface orthogonal and the spacetime would be static.

Intuitively, a spacetime is static when the line element is invariant under time reversal $t \to - t$, in the usual coordinate system. To make a more precise statement, if a timelike Killing vector field $\xi$ satisfies $$ \xi_{[\mu} \nabla_\nu \xi_{\rho]} =0, $$ then it is hypersurface orthogonal and the spacetime is static. See the discussion in section 1.3 of the notes you are referring to. You can also have a look at Wald's GR textbook.

Answer 2

As an alternative, more geometrical point of view, consider the Killing vector $\partial_t$ at each point, and imagine the (hyper-)plane orthogonal to each vector. The vector field will be hypersurface orthogonal if it's possible to make all these planes fit as tangent planes of a family of hypersurfaces.

As an example, consider a radial vector field in Euclidean 3D space, and again imagine all the planes orthogonal to the vector at each point in space. Is it possible to fill space with surfaces such that each plane is tangent to one of the surfaces? Yes, of course, by using spheres. The planes corresponding to all the vectors at a fixed radius fit together to form a sphere.

But now take Kerr spacetime and look just at the equatorial plane, so that we have something three-dimensional that we can picture. The "horizontal" planes, those spanned by the vectors $\{\partial_r, \partial_\phi\}$ at each point, are not orthogonal to the Killing vector! That is, of course, because of the $g_{t\phi}$ element in the metric. Instead, the plane orthogonal to each vector $\partial_t$ is tilted; it is spanned by the vectors $\partial_t$ and $- g_{t\phi} \partial_t + g_{tt}\partial_\phi$, and so it points a bit into the direction of rotation.

And this is what makes the timelike Killing vector not hypersurface orthogonal, and you can see why that is closely related to the rotation of the spacetime and the frame dragging. If there was no rotation, each $\partial_t$ would be orthogonal to a "horizontal" plane, and these planes could fit together into a big $t = \text{const}$ surface. But since the planes are tilted, you can't make them tangent to a surface