# Ergosphere and angular velocity

1. Nov 19, 2015

### Neutrinos02

Hello,

i have some problems with the ergosphere of a rotating black hole.

I can show that an object must rotate with the black hole and a can determine the intervall of possible velocitys. But is it possible to calculate the exact velocity? How it depends on the initial velocity of the object?
If the object has an velocity which matches to the intervall in the ergosphere, would it be constant or will it also be affected by the ergosphere?
What about the conservation of angular momentum (for example if it rotates in the other direction), would it be taken from the black hole (or given to the black hole)?

If we are intressted in a stable oribit in the ergosphere, how will this depend on the frame-dragging effect?

Neutrino

2. Nov 19, 2015

### Staff: Mentor

"Exact velocity" of what relative to what?

I think getting a clear answer to this question will help to resolve all of the issues you raise (to the extent I can understand what they are--your description is not very clear).

3. Nov 19, 2015

### m4r35n357

It is possible to calculate prograde and retrograde orbits for Kerr Black holes. Try this link for a random example. Also it is theoretically possible to extract energy from a black hole (assuming that is what you are getting at, see this link about the Penrose process)

4. Nov 19, 2015

### Neutrinos02

To an infinite observer at rest.

Ok a try to rewrite the questions.
If there is an object in the ergosphere an inifinite observer see that the object is rotating with a velocity ωmin<ω<ωmax. I like to know how it is possible to calculate the value of ω (not only the range (ωmin, ωmax)).
I think that there should a dependence of the initial angular momentum Linitial (which the object had befor it came in the ergosphere)?

As soon as the object enters the ergosphere its angular momentum is changed from Linitial to Lergo, but what happens with the difference Linitial-Lergo?

What happens if the velocity related to Linital is allready in the intervall ω∈(ωmin, ωmax) would there also be a change of Linitial?

For the stable orbit i have the problem that the effective potential depends on the angular momentum L and since L is changed at each point of the ergosphere how i can find a cirular orbit? (I mean if L is given in the classical case i can calculate the minimum of the effectiv potential and get the radius r of the circular orbit. But if L is not fixed and will be changed at each r, the minimum will also change its position)

I hope it is more clear what i mean.

5. Nov 19, 2015

### Staff: Mentor

This doesn't have any physical meaning. The observer at rest at infinity is spatially separated from the object inside the ergosphere, so there is no way to measure their speed relative to each other. In fact, inside the ergosphere there isn't even a well-defined notion of "at rest relative to infinity" at all; curves with constant spatial coordinates $r, \theta, \phi$ are spacelike, not timelike.

There is a way for the observer at infinity to assign an angular velocity to an object inside the ergosphere (see below), but angular velocity is not the same as velocity.

An angular velocity in this range, yes. More precisely, if the observer at infinity watches the object in the ergosphere through a very powerful telescope, and times how long it takes for the object to complete one orbit (meaning to return to the radial line of sight of the observer at infinity), and converts this to an angular velocity, he will get a result in the given range.

Obviously you would have to know something about the specific object whose angular velocity $\omega$ you want to calculate.

If the object is in free fall, angular momentum is a constant of the motion; it can't change. There is a relationship between it and angular velocity, yes.

What makes you think that? As I said above, angular momentum is a constant of the motion for an object in a free-fall trajectory; it doesn't change.

6. Nov 20, 2015

### Neutrinos02

So there would be a general forumlar for this with the initial states of the object?

But if the object came from infinity to the black hole without any angular momentum relativ to the black hole and enters the ergosphere it will rotate. So its angular momentum gets a value >0? Is ist really conserved? In general relativity there are not always conserved quantities?

Or if we think about an object which has an angular momentum in the opposit direction of the black hole and if it enters the ergosphere it will rotate with the black hole, so i thougth there must be a change in the angular momentum?

7. Nov 20, 2015

### Staff: Mentor

Zero angular momentum in the spacetime around a rotating black hole is not the same as zero angular velocity. Zero angular momentum means having a particular angular velocity that depends on your radius $r$ (and latitude $\theta$) with respect to the hole. As $r \rightarrow \infty$, this angular velocity goes to zero; but it is nonzero at any finite radius. So an object that starts falling towards the hole on a purely radial trajectory, but then gradually acquires angular velocity, has zero angular momentum the whole time; its angular velocity at any radius is just the "zero angular momentum" angular velocity at that radius.

I recommend reading Matt Visser's introduction to Kerr spacetime:

http://arxiv.org/abs/0706.0622

It gives a good overview of the math behind it all, and discusses counterintuitive properties like the one I described above.

8. Nov 20, 2015

### Neutrinos02

But how is it possible to have zero angular momentum but angular velocity >0?

9. Nov 20, 2015

### Staff: Mentor

Heuristically, because the spacetime itself is "rotating", in a sense; and zero angular momentum means "rotating at the same rate as spacetime", rather than "not rotating at all".

It's hard to be more precise than that without considerable mathematical background; again, I suggest reading the Visser paper I linked to.

10. Nov 21, 2015

### Neutrinos02

Now I read the paper but in the chapter of the ergosphere I didn't really found somthing about the angular momentum? Maybe I overlooked it? Can you give me the page of it?

11. Nov 21, 2015

### Staff: Mentor

Hm, you're right. I hadn't looked at this paper in a while, and I forgot that, while it discusses the angular momentum of the hole, it doesn't really discuss the angular momentum of a free-falling observer in the spacetime around the hole.

http://relativity.livingreviews.org/Articles/lrr-2013-1/articlese2.html#x5-50002.2 [Broken]

The section I linked to discusses the distinction between zero angular velocity and zero angular momentum.

Last edited by a moderator: May 7, 2017
12. Nov 22, 2015

### Neutrinos02

Thank you. I understood a bit more, but it answered not all questions like:
1, Where can I found the rotating of spacetime in the mathematical expressions?
2, If there is an oberserver which rotates in the opposite direction as the black hole: What happens to his angular momentum, since his rotation will be reversed?
3, According to the situation in 2,: If the spacetime rotates and the observer rotates in the rotating spacetime, why isn't it possible for him to rotate in the opposite φ-direction as the black hole?

13. Nov 22, 2015

### m4r35n357

14. Nov 22, 2015

### Neutrinos02

Perhaps I'm at a loss but i don't see how this answers the questions?

15. Nov 22, 2015

### m4r35n357

Section 6.4 is all about the ergosphere, and 6.5 (the first one!) is all about circular orbits with and against the BH spin. That's the simplest analysis I am aware of.

16. Nov 22, 2015

### Staff: Mentor

This is possible, but only outside the ergosphere. Inside the ergosphere, an observer who tried to rotate in the opposite direction from the hole would have to move faster than light to do so. (In fact, he would have to move faster than light to have zero angular velocity, not rotating at all.)

It will be negative, just as it would be in a non-rotating spacetime. The difference in a rotating spacetime is that an object which is not rotating at all (zero angular velocity), or rotating in the same direction as the hole, but with angular velocity less than the "zero angular momentum" angular velocity, will also have negative angular momentum.

17. Nov 22, 2015

### Neutrinos02

I thougth you said the angular momentum will be conserved? So if he had zero angular momentum shouldn't this be conserved?

Last edited by a moderator: Nov 22, 2015
18. Nov 22, 2015

### Staff: Mentor

You were asking about an object rotating around the hole in the opposite direction to the hole's own rotation. Such an object does not have zero angular momentum. The "zero angular momentum" angular velocity is in the same direction as the hole's rotation.

Last edited by a moderator: Nov 22, 2015
19. Nov 27, 2015

### Neutrinos02

Ok i try to make a short summarization:

The angular momentum along a geodesic is a constant of motion. The angular momentum inside the ergosphere can be negative since the Killinfield is spacelike.
If a particle is falling into the ergosphere with zero (positive) angular momentum at infinite it will have zero (positive) angular momentum at all points of its geodesic?

How is the total angular velocity of a particle is defined (an a geodesic)?

20. Nov 27, 2015

### Staff: Mentor

Yes.

You are confusing angular momentum with energy at infinity. Energy at infinity is the constant of motion corresponding to the Killing field $\partial / \partial t$ (in Boyer-Lindquist coordinates), which is timelike outside the ergosphere but spacelike inside the ergosphere--and the latter fact does make it possible to have geodesic trajectories with negative energy at infinity inside the ergosphere. (Such trajectories can never go outside the ergosphere, though they can fall through the horizon.)

Angular momentum is the constant of motion corresponding to the Killing field $\partial / \partial \phi$ (again, in Boyer-Lindquist coordinates), which is spacelike everywhere. Negative angular momentum just means that the object is "rotating in the opposite sense to the hole"--but here "in the opposite sense" does not necessarily mean negative angular velocity (as it would around a non-rotating hole); it means angular velocity that is either negative, zero, or positive but less than the "zero angular momentum" angular velocity (which is a function of $r$ and $\theta$ in Boyer-Lindquist coordinates).

Yes; its angular velocity at every point of its trajectory will be the "zero angular momentum" angular velocity at that point.

In Boyer-Lindquist coordinates, it is defined as $d \phi / dt$. This is true for any worldline, whether it is a geodesic or not.