# What is a static spacetime?

1. Apr 5, 2012

### andrewkirk

I am trying to understand the derivation of a solution of Einstein's field equations for a static, spherically symmetric spacetime (chapter 10 of Schutz's 'A first course in general relativity').

The derivation defines a static spacetime as one satisfying two conditions, one of which is that the metric tensor g is independent of t (time). I am confused by this definition as whether something is independent of a given coordinate depends on the coordinate system. g may be independent of t in one coordinate system but dependent on it in another.

This seems to imply that a spacetime may be static in some coordinate systems but non-static in others. Is that correct? If so, what do we mean by a static spacetime?

2. Apr 5, 2012

### Staff: Mentor

A static spacetime is static regardless of the coordinate chart used, but Schutz's language does make it a bit hard to see how that works.

The key point about the Schwarzschild time coordinate, t, is that it is not just any "time" coordinate; it happens to match up exactly with the time translation symmetry of the spacetime. In other words, saying the metric is independent of t, in this particular case, is saying something which has a coordinate-independent, invariant meaning; observers who are "static" with respect to t (i.e., their worldlines have all Schwarzschild coordinates constant *except* for t) see the actual, physical geometry of the spacetime as unchanging. This means that, if such an observer does a physical experiment to measure the geometry at one time, t1, the results will be exactly the same as if he does the same experiment at any other time, t2--i.e., the experiment can be "translated in time" along the observer's worldline without changing anything.

There is a way to express all this in more explicitly coordinate-independent language, by describe the time translation symmetry in terms of a Killing vector field; see the Wikipedia page here:

http://en.wikipedia.org/wiki/Killing_vector_field

The vector field $\partial / \partial t$ in Schwarzschild coordinates is a Killing vector field on Schwarzschild spacetime, the "time translation" Killing vector field. So what Schutz really should have said is that a static spacetime admits a coordinate chart in which the "time" coordinate is associated with a time translation Killing vector field in this way, *and* in which the metric is independent of that time coordinate.

Actually, the definition I just gave, as it stands, does not quite guarantee that the spacetime is static; it only guarantees that it is stationary. You mention that there are two conditions given by Schutz; if the second one has something to do with static observers' worldlines being "hypersurface orthogonal", or with the absence of off-diagonal or "cross terms" in the metric, that would be the necessary additional condition to guarantee that the metric is static, instead of just stationary.

3. Apr 5, 2012

### andrewkirk

Thank you Peter. That helps a lot.

The second condition Schutz listed for a spacetime to be static is that the geometry must be unchanged by time-reversal. He gives a rotating star as an example of a spacetime that would satisfy the first but not the second condition, and hence be stationary but not static. Is Schutz's second condition equivalent to yours?

Do you (or anybody else) know of where a nicely laid out derivation of the geometry of a static, spherically symmetrical spacetime would be available? I find Schutz is often insufficiently precise, which causes confusion, as it has on this point.

4. Apr 6, 2012

### Matterwave

I believe the definitions are equivalent. If your system is time-reversal invariant, the metric can't have terms linear in dt (else the sign in these terms would change). This is the same as precluding the possibility of cross terms of the form dxdt, etc., in the metric which implies orthogonality of the time-like vector field with your space-like hypersurfaces.

If you want mathematical rigor, you can look in Wald, chapter 6.

5. Apr 6, 2012

### julian

There is also Ray D'Inverno's book, start of chap 14...maths with pictures. In fact full derivation given, maths fairly straightforward...although you have to learn about isometries at the end of chap 7.

Last edited: Apr 6, 2012
6. Apr 6, 2012

### julian

Like D'Inverno a lot, even though he confuses active diffeomorphisms with coordinate transformations...that's an issue of mine. But dont worry about that too much on a first reading as most books on GR seem to get them confused...he does describe an active transformation, he just refers to it as a coordinate invariance in the first paragraph (end of chap 7).

Last edited: Apr 6, 2012
7. Apr 6, 2012

### julian

A symmetry of spacetime, an isometry, is a physical property and NOTHING physical can be attributed to a mere coordinate transformation.

8. Apr 6, 2012

### andrewkirk

Thank you all for the comments and suggestions.

9. Apr 6, 2012

### dini andriani

how to imagine coordinate of time?

10. Apr 6, 2012

### pervect

Staff Emeritus
In coordinate independent terms, a stationary space-time has both a time-like Killing vector, and closed spatial hypersurfaces which are orthogonal to the timelike Killing vector.

If a space-time has a time-like Killing vector, but doesn't satisfy the second condition, then it is only stationary, and not static. This happens when for instance in the Kerr metric.

If any coordinate system exists in which none of the metric coefficients are functions of time, you're guaranteed to have time-like Killing vectors (well, as long as your time coordinate generates a timelike vector field, i.e. isn't a spatial coordinate). So it's really a coordinate independent property of the metric, but the rigorous definition involves getting into Lie derivatives and such.

11. Apr 6, 2012

### Staff: Mentor

It looks that way, yes. As Matterwave said, a metric with "cross terms" linear in dt would not be unchanged by time reversal, because the sign of the dt term would change. The spacetime of a rotating object, such as a star, or a rotating black hole (the Kerr metric) is a good example of a spacetime which is stationary but not static, yes.

12. Apr 6, 2012

### Staff: Mentor

This qualification brings up a key point: Schwarzschild spacetime is actually *not* static inside the horizon, because the Killing vector field $\partial / \partial t$ is not timelike there, it is spacelike. So the term "static" may apply only to a portion of a spacetime. Similar remarks apply to the term "stationary"--the Kerr metric, for example, is not stationary inside the horizon, because the Killing vector field $\partial / \partial t$ (where "t" is now the Boyer-Lindquist time coordinate) is not timelike there.

I've seen some references that relax the definitions of static and stationary to say that the Killing vector field only needs to be "asymptotically" timelike, apparently to be able to say that spacetimes like the above are static or stationary everywhere instead of just outside the horizon. However, I don't really buy this qualification, because one of the key reasons for testing to see if a spacetime is stationary or static is to be able to define a family of stationary or static observers, whose worldlines we can use as convenient references since they are orbits of the "time translation" Killing vector. But you can only do this in the portion of the spacetime where the Killing vector is actually timelike, i.e., outside the horizon. For example, there are no observers inside the horizon of a black hole who can "hover" at a constant radial coordinate r, since to do so such an observer would have to move on a spacelike worldline (i.e., "faster than light"). So it makes sense to me to say that these spacetimes are only static or stationary outside the horizon, since that's the only region where the appropriate kinds of observers can exist.

13. Oct 1, 2013

### YoungPhysicist

I have a problem with symmetries, a static metric must be spherically symmetric?
İf not, ls there any symmetry that must be in a static metric.

14. Oct 1, 2013

### WannabeNewton

No a static space-time is not necessarily spherically symmetric. A static space-time is one which has a time-like killing field $\xi$ that is hypersurface orthogonal: $\xi \wedge d\xi = 0$. This is the only defining property of a static space-time.