# A Are Killing Horizon and Stationary Limit Surface the same?

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1. May 20, 2016

### Elnur Hajiyev

I know that Killing horizon is the hypersurface on which timelike Killing vector field becomes null. Beyond that surface Killing vector field becomes spacelike. But Stationary Limit Surface has also such a property. I wonder, if they are the same thing, if so, why is there different names for this concept even inside the same context(I saw both of them in Carrol's book). If not, what does Killing Horizon means other than that "once crossed the horizon, it is not possible to remain stationary"?

Last edited: May 20, 2016
2. May 20, 2016

### Simon Bridge

i.e. a Kerr black hole: the ergosphere lies between a static limit surface and a Killing surface
-- Straumann N. (2004) General Relativity: With Applications to Astrophysics; Springer Physics

3. May 20, 2016

### Elnur Hajiyev

If Killing Surface is different notion than Stationary Limit Surface, what does it exactly mean, physically?

4. May 20, 2016

### Staff: Mentor

To rephrase what I think you are trying to say in more precise terms: a black hole's event horizon and a black hole's stationary limit surface are both Killing horizons. This is true. For the case of a Kerr black hole, where the two surfaces are not the same (for a Schwarzschild black hole they are the same), the difference is which Killing vector field becomes null on the surface. For a Kerr black hole, in Boyer-Lindquist coordinates, the KVF $\partial_t$ becomes null at the stationary limit; but the KVF $\partial_t + \Omega \partial_\phi$, where $\Omega$ is the "angular velocity of the horizon" (defined in terms of the hole's mass and angular momentum), is the one that becomes null at the event horizon.

5. May 20, 2016

### Staff: Mentor

That's not what "Killing horizon"means. It means a surface on which a KVF becomes null. If there is more than one KVF in a spacetime, there will also be more than one Killing horizon.

The term "stationary" is much more restrictive. In the case under discussion, "remaining stationary" means an observer's worldline is the orbit of a timelike KVF that asymptotically approaches a Minkowski time translation at infinity. This is true of the KVF $\partial_t$ in Kerr spacetime, but not of the KVF $\partial_t + \Omega \partial_\phi$. So the former KVF's Killing horizon is a "stationary limit" surface, but the latter's Killing horizon is not. (It is, as I noted in my previous post, the event horizon of the hole, but that is a separate issue that is not related to whether observers can be stationary or not.)

6. May 20, 2016

### Elnur Hajiyev

@PeterDonis , Thanks for the excellent explanation. I wonder, then what does it mean KVF to be null? In general, what physical consequences does it have and what is the motivation to bring a notion of Killing Horizon, if it can be represented as a Stationary Limit Surface and an event horizon and what is the intuitive way of thinking about Killing Horizon?

Last edited: May 20, 2016
7. May 20, 2016

### Staff: Mentor

The same thing it means for any other vector field to be null. A vector field is a mapping of vectors to events in spacetime; if the vectors mapped to some set of events (such as the set of events making up the Killing horizon of a KVF) are null, then the vector field is set to be null on that set of events.

It depends on the specific KVF you are talking about and what physical interpretation its vectors, or integral curves, have. See below.

You are misstating things here. The Killing horizon is the general concept. The Stationary Limit surface and the event horizon are two (different) particular types of Killing horizons, corresponding to two (different) particular KVFs. So asking what the motivation is for using the concept of a Killing horizon is getting things backwards; the general concept of Killing horizon is just part of the general concept of KVFs. What you should be asking is what is the motivation for looking at particular types of Killing horizons such as stationary limit surfaces or event horizons. See below.

Consider a region of spacetime in which a particular KVF is timelike. In this region, the integral curves of the KVF describe the worldlines of a possible family of observers, such that along each observer's worldline, the spacetime geometry is unchanged. (The "unchanged geometry" part comes from the definition of a KVF; the "worldline" part comes from the fact that the KVF is timelike.)

Now consider the Killing horizon for the same KVF--this is the surface on which the KVF is null. This surface evidently forms the boundary of the region of spacetime in which the above family of observers exists--on and "below" (i.e., on the spacelike side of) the Killing horizon, the integral curves of the KVF are n longer timelike, so they can't be the worldlines of any observers.

A simple example is the KVF $\partial_t$ in Schwarzschild spacetime (expressed in standard Schwarzschild coordinates). This KVF is timelike outside the event horizon, null on the horizon, and spacelike inside it. This means that, outside the horizon, there is a family of observers whose worldlines are the integral curves of $\partial_t$. These observers are "hovering" at a constant radius and constant angular coordinates, and they are at rest with respect to the asymptotic inertial "rest frame" of the black hole at infinity (this is a heuristic description but it can be made rigorous). No other KVF in the spacetime has this property. The Killing horizon associated with this KVF is therefore a "stationary limit" surface.

However, as I implied above in describing the regions where this KVF is timelike, null, and spacelike, the Killing horizon associated with this KVF is also an event horizon, i.e., the boundary of the region of spacetime that can send light signals to infinity. Unfortunately, although we can say that the event horizon must be a Killing horizon (more precisely, the EH of a stationary black hole must be a Killing horizon--this was proved as a theorem by Hawking, IIRC, in the early 1970s), I don't know of any intuitive way to show precisely which KVF's Killing horizon is the event horizon. You have to already know the global properties of the spacetime.

For example, consider Kerr spacetime again: here the stationary limit surface is the Killing horizon associated with $\partial_t$, but the event horizon is the Killing horizon associated with $\partial_t + \Omega \partial_\phi$. Showing that $\partial_t$'s Killing horizon is the stationary limit can be done the same way as for Schwarzschild spacetime, above. But I don't know of any intuitive way to show why the event horizon is not that Killing horizon but a different one, and which one it is. (Since any linear combination of KVFs with constant coefficients is also a KVF, there are a infinite number of possible KVFs combining $\partial_t$ and $\partial_\phi$, in both Kerr and Schwarzschild spacetime.) You just have to know the global properties of Kerr spacetime, derived by whatever tedious route you can derive them.

8. May 20, 2016

### Orodruin

Staff Emeritus
While I agree with what you want to say here, it is not really what you are saying.

A vector field is a map from each event $p$ in space-time $M$ to the tangent space at that point $T_pM$ (more technically it is a section of the tangent bundle). The mapping from the vectors to the corresponding event is the projection $\pi$ from $TM$ to $M$ for which every vector field $V$ satisfies $\pi(V(p)) = p$.

9. May 20, 2016

### Elnur Hajiyev

@PeterDonis , I appreciate your effort and time on writing so detail answer. Your explanation were very helpful for me to grasp the concept. As this would be my final question on this topic, I want to ask about surface gravity. Although to my understanding, surface gravity is the "gravitational force" applied on the particular surface of a spherically symmetric object, I have difficulties on relating it to Killing Horizon. I would like to know more than an algebric relation between these terms.

10. May 20, 2016

### Staff: Mentor

Yes, I was being sloppy; your restatement is much more precise.

11. May 20, 2016

### Staff: Mentor

Yes, the concept of "surface gravity" has to be carefully defined for a Killing horizon, since nobody can actually sit at rest on the horizon (because "sitting at rest" implies a timelike worldline and the horizon is a null surface). The definition can be heuristically understood as the "redshifted proper acceleration" at the horizon. In other words, imagine an object being held at rest at some altitude above the horizon by a string that is suspended from infinity. The force that has to be exerted at infinity to hold the object at rest will be the proper acceleration of the object times the "redshift factor" between the object's altitude and infinity. Then we take the limit of this force exerted at infinity as the object's altitude goes to zero, i.e., as the object approaches the horizon. (We have to take a limit because, as above, no object can actually hover at rest at the horizon.) This limit is the surface gravity.

For example, consider an object hovering at rest at radial coordinate $r$ above a Schwarzschild black hole. The object's proper acceleration is (in "geometric units" where $G = c = 1$)

$$a = \frac{M}{r^2 \sqrt{1 - 2M / r}}$$

The "redshift factor" between the object's altitude and infinity is $\sqrt{1 - 2M / r}$, so multiplying the above proper acceleration by this factor gives

$$\bar{a} = \frac{M}{r^2}$$

This is the force that would have to be exerted at infinity to hold the object at radius $r$ by a string. Now take the limit as $r \rightarrow 2M$; clearly this gives

$$\kappa = \frac{1}{4M}$$

which is the surface gravity of the hole.

12. May 21, 2016

### Elnur Hajiyev

Thank you very much!

13. May 22, 2016

### Elnur Hajiyev

While I read again about Killing Horizon in Carrol's book, I have just seen that it says Stationary Limit Surface is timelike surface in Kerr metric, therefore it is not a Killing Horizon(for it to be Killing horizon, it should be a null surface) for spinning black holes. This statement has baffled me. Becuase it means, every Killing Horizon must be an event horizon, which is defined by being a null hypersurface, so once crossed this surface, you should not be able to return.

14. May 22, 2016

### Elnur Hajiyev

@PeterDonis , I have read your answer very carefully just now, and this point has drawed my attention. Isn't everything should have a timelike worldline, whether it is moving or at rest? And does redshift factor come from the expansion of universe?

Last edited: May 22, 2016
15. May 25, 2016

### George Jones

Staff Emeritus
What is an event horizon?

16. May 25, 2016

### Elnur Hajiyev

It means, once crossed this boundary, it is impossible to return. In other words, once crossed event horizon, time and space swap their roles. In Kruskal coordinates, this corresponds to null hypersurface. That is why, I am asking, if Killing horizon needs to be a null surface, is that mean, every Killing horizon should be aj event horizon?

17. May 25, 2016

### George Jones

Staff Emeritus
No.

Are you using Carroll's on-line notes, or his actual textbook? It seems that Carroll's discussion of event horizons and Killing horizons is much better in his book than in his notes.

18. May 25, 2016

### Elnur Hajiyev

I am using textbook. I read Killing Horizon topic there and found some conflicts with the statements posted here.

And if I ask you to explain why no, would you explain it?

19. May 25, 2016

### George Jones

Staff Emeritus
A black hole is a region of spacetime from which it is impossible for a material (test) particle to escape to infinity. An even horizon is the boundary of this region.

Asymptotic flatness is used to define what "escape to infinity" means.

The standard definition of the black hole region of an asymptotically flat spacetime is the region of spacetime from which it is impossible to escape to future null infinity. An event horizon is the boundary of this region. See page 240 in Carroll.

See page 245, starting with "It's important to point out ..." for an an example of a Killing horizon that is not an an event horizon.

20. May 25, 2016

### Elnur Hajiyev

Yes, I agree. But isn't it another form of saying of the same statement(more technically, it is the region where grr component of the metric becomes 0)? Whatever it is, I still cannot find exact answer to my question. If you answer it, I would appreciate.
Btw, thanks for the reference. :)

21. May 25, 2016

### martinbn

That is not a good enough definition of an event horizon. The null cone of an event in Minkowski space-time has the same proerty. Once crossed you cannot cross it again. It is also a null surface. But it is definitely not an event horizon. The part about escaping to infinite in George Jones' answer is important.

22. May 25, 2016

### George Jones

Staff Emeritus
To illustrate what is happening, let's, for now, move away from Kerr spacetime and Killing vectors, and let's consider a specific vector field in 3-dimensional (for visual simplicity) Minkowski spcetime.

Let $\left(t,x,y\right)$ be an inertial coordinate system for this spacetime. Consider the vector field

$$v = \frac{\partial}{\partial t} - x \frac{\partial}{\partial x}$$

When is $v$ spacelike? Null? Timelike?

Again, for simplicity, consider only $x>0$.

23. May 25, 2016

### Elnur Hajiyev

On $x<1$, $v$ will be spacelike, $x=1$ $v$ will be null, $x>1$ $v$ will be spacelike.

I don't know how to write in mathematical script here. Because I am new in physicsforums. Sorry for that.

Last edited: May 25, 2016
24. May 25, 2016

### Elnur Hajiyev

Hmm... It seems reasonable statement. Thanks for the clarification.

25. May 25, 2016

### George Jones

Staff Emeritus
Yes. (I think you have a typo for null, i.e., you meant x = 1.

Can you describe the surface in this spactime on which v is null?