# Event-horizon from the blackhole metric

1. Sep 19, 2014

### Ravi Mohan

What is the most general method of obtaining the event-horizon from the given blackhole metric.

Let us consider Kerr blackhole in Kerr coordinates given by
$$ds^2 = -\frac{\Delta-a^2sin^2\theta}{\Sigma}dv^2+2dvdr -\frac{2asin^2\theta(r^2+a^2-\Delta)}{\Sigma}dvd\chi-2asin^2\theta d\chi dr + \frac{(r^2+a^2)^2-\Delta a^2sin^2\theta}{\Sigma}sin^2\theta d\chi^2+\Sigma d\theta^2,$$
where
$$\Sigma = r^2+a^2cos^2\theta\\ \Delta = r^2 - 2Mr+a^2.$$

How do we find the killing vector in a coordinate system?
Any hint or reference would be of great help.

2. Sep 21, 2014

### michael879

I'm not really sure what a general method for finding event horizons would be, as "event horizon" is not a well defined term (it is used to describe multiple horizons). However, if you are strictly limiting yourself to black holes, the Kerr-Newman metric is the most general metric and its event horizons can be used to find the event horizon of any black hole. The event horizons for the Kerr-Newman metric are located at the radii where $g_{rr}$ diverges, which occurs when $\Delta = 0$. So in the chargeless case (Kerr metric) this is:
$$r_{\pm} = M \pm \sqrt{M^2-a^2}$$

3. Sep 22, 2014

### bapowell

The event horizons of exact black hole solutions are also Killing horizons, surfaces in spacetime along which the timelike Killing vector is null: http://en.wikipedia.org/wiki/Killing_horizon

4. Sep 22, 2014

### pervect

Staff Emeritus
As I recall, killing horizons, absolute horizons, and apparent horizons were all different (though closely related) concepts. If I had to guess, I'd guess that the interest was in finding the "absolute" horizon. This requires the ability of light like geodesics to escape the black hole, I believe this is formally defined as "reaching future null infinity". I don't recall the exact formal definition of future null infinity, a quick lookup in Wald didn't find one either. I don't know of any method more efficient than just writing the null geodesics and seeing if they reach a suitably defined infinity or not. In a more general metric than the example, non asymptotically flat for instance, defining "infinity" formally may take a bit of effort. And if you want something other than the absolute horizon, you need to specify what sort of horizon you're looking for.

5. Sep 22, 2014

### Staff: Mentor

Absolute and apparent horizons are definitely different concepts. I think a Killing horizon is a particular kind of absolute horizon (i.e., any Killing horizon must also be an absolute horizon, but not vice versa). However, I'm not 100% sure about that (and my copy of Hawking & Ellis is buried too deep right now to check it ).

Yes.

I think Hawking & Ellis is probably the best reference for a formal definition of this as well. Informally, you either construct or (if you're lazy like me ;)) find somewhere online a Penrose diagram of the spacetime you're interested in; future null infinity is just a section of the boundary of the diagram. The diagram makes it easy to see from what regions light rays can reach future null infinity (since light rays are 45 degree lines).

I'm not sure, but I think a spacetime might be required to be asymptotically flat in order to have a future null infinity.

6. Sep 22, 2014

### Staff: Mentor

The easiest way, if you're lucky to be working in an appropriate coordinate chart, is to look at what coordinates the metric does not depend on. (For example, in Kerr spacetime there are two, as you can see by looking at the line element you wrote down.) The coordinate basis vectors in those coordinate directions are Killing vectors. Also, any linear combination of Killing vectors with constant coefficients is a Killing vector.

7. Sep 23, 2014

### michael879

Or at least open. You definitely can't define null infinity in a closed universe ;) However, it seems like a reasonable generalization (to me at least) to just require the null geodesic to never fall back in to the black hole, even if it doesn't reach "infinity". I'm pretty sure that despite their definition absolute event horizons are completely one way (at least for black holes). i.e. if a null geodesic can reach r' from r then the event horizon is < r'. *IF* this is true, then you don't need to worry about defining null infinity to calculate the horizon. I am not sure if this holds for non-black hole solutions though

8. Sep 23, 2014

### PAllen

Well, for starters, try to give a coordinate independent definition of r for a rotating BH embedded in a closed universe with dynamic curvature change.What you are really hinting at is apparent horizon, which is a local trapping surface.

9. Sep 23, 2014

### michael879

Good point lol, I was only thinking of asymptotically flat, stationary black holes. That sounds right, I think I remember reading somewhere that apparent=absolute in stationary space-times which would explain my confusion

10. Sep 23, 2014

### Staff: Mentor

I don't think this is sufficient. For example, a flat FRW spacetime is open, but it has no null infinity.

As PAllen points out, this is a definition of an apparent horizon, not an absolute horizon. There can certainly be apparent horizons in spacetimes that don't have a null infinity.