# Non singular black hole solutions

#### George Jones

Staff Emeritus
Gold Member
Ok, but a static manifold has a unique timelike KV field
Not necessarily unique.
do you mean then that a manifold can have regions that fulfill this requirement and other regions that don't?
Yes.
What is that manifold called then, static and not static at the same time?
Yes. More precisely, it is called static on (open) region U if there is a timelike, hypersurface-orthogonal Killing vector field defined throughout U.

#### TrickyDicky

Not necessarily unique.
After some research on the web I guess you are referring here to the difference between a static manifold (that can be locally or partially static or as you say static on a region like in the case we are discussing) and a standard static manifold, where there truly exist a unique global orthogonal splitting of the spacetime.

#### George Jones

Staff Emeritus
Gold Member
What I meant was that in (parts of) some static spacetimes, it is possible to have linearly independent timelike Killing vectors. The additional requirement of hypersurface-orthogonality still does not seem to pin down the Killing vector. Again, consider Minkowski spacetime in a global inertial coordinate system. Then, $\mathbf{e}_t$ and $\mathbf{e}_x + 2 \mathbf{e}_t$ are both timelike, hypersurface orthogonal Killing vector fields.

#### TrickyDicky

What I meant was that in (parts of) some static spacetimes, it is possible to have linearly independent timelike Killing vectors. The additional requirement of hypersurface-orthogonality still does not seem to pin down the Killing vector. Again, consider Minkowski spacetime in a global inertial coordinate system. Then, $\mathbf{e}_t$ and $\mathbf{e}_x + 2 \mathbf{e}_t$ are both timelike, hypersurface orthogonal Killing vector fields.
I have the impression you can only do that with cartesian coordinates in flat spacetime, because points in Minkowski space can be identified with vectors but in GR curved spacetime vectors are only defined locally and $\mathbf{e}_x + 2 \mathbf{e}_t$ may not be a vector.

#### TrickyDicky

A vector field doesn't have to have the same causal character everywhere, i.e., it can be timelike in one part of spacetime, lightlike in another part, and spacelike in another.
The WP page geodesic entry after explaining the different null, timelike and spacelike types says: "Note that a geodesic cannot be spacelike at one point and timelike at another". Is there a difference in this respect between vectors and geodesics?

#### George Jones

Staff Emeritus
Gold Member
The WP page geodesic entry after explaining the different null, timelike and spacelike types says: "Note that a geodesic cannot be spacelike at one point and timelike at another". Is there a difference in this respect between vectors and geodesics?
Yes, even though vectors and curves are related, there is, in principle, a difference in this respect. In particular, the integral curves associated with the Killing vector field $\partial_t$ (curves of contant r, theta,a and phi) are not geodesics, so your result could possible fail for them. It, however, does not fail, i.e., an individual integral curve for $\partial_t$ does not change its causal character, but different integral curves for $\partial_t$ can different characters. Outside the event horizon (where spactime is static), the integral curves of $\partial_t$ are all timelike; inside the event horizon (where spacetime is not even stationary), the integral curves of $\partial_t$ are all spacelike.

The integral curves for $\partial_t$ are the curved dashed lines on

These means that the vectors that make up the vector field $\partial_t$ are tangent vectors for these curves. Note the different causal character outside and inside the event horizon.