# Ergoregions and Energy Extraction

1. Nov 27, 2015

Hi,

I'm trying to understand the argument between eqns 4.117 and 4.119 in this paper http://arxiv.org/abs/1501.06570 as to why the penrose process is not possible for a boosted black string.

I understand the basic idea is to show using an argument similar to 4.33 that $$|E| < |p_z|$$ (see 4.118) and then from the relation $$E^2=p^2+m^2$$ that $$|E| \geq |p_z|$$ thus obtaining a contradiction and showing that the negative energy particle (required for Penrose process) could never have existed in the first place.

My problem is with the derivation of 4.117 using an argument similar to 4.33. My problems are the following:

1, Why is there a minus sign in front of the integral in 4.32?

2, On the right hand side of 4.32, he has $$\delta E - \Omega_H \delta L$$. I don't understand the signs here either - why is the first term positive and the second term negative?

3, This is then integrated to give $$E_H=E-\omega_H L$$ which is apparently positive since the locally measured energy must be positive. Why is this the case?

4, To me it seems that in the Kerr case (4.33) he arrives at $$E_H=E-\omega_H L$$ whilst in the linear case we should get $$E_H=E-v_H p_z$$. Then what he does is note above 4.117 that $$v_H=-v$$ for an observer on the horizon with no linear momentum i.e. comoving. If this observer is comoving then surely they should have the same momentum as the horizon i.e. $$v_H=v$$?

Can anyone explain this? It doesn't seem like particularly difficult calculation - just some missing intuition....

Thanks

2. Nov 27, 2015

### Staff: Mentor

I think this is because he is using the spacelike convention for the metric signature; in other words, the component $T^0_0$ of the stress-energy tensor is negative, while the other diagonal components are positive. So the minus sign just reflects the fact that, for an observer at rest in a local inertial frame, for which $T^0_0$ is the only component that contributes to the locally measured energy, that energy must be positive, meaning it must be $- T^0_0$, hence the minus sign.

Because of the signs of the stress-energy tensor diagonal components, as above. Those components, in the paper's notation, are $T^0_0 = - \delta E$ and $T^3_3 = \delta L$ (strictly speaking, these are what come from those components when they are integrated over $d\Sigma$), and the components of the vector $n^\mu$ are $n^0 = 1$ and $n^3 = \Omega_H$. So the two resulting terms are of opposite sign because the two diagonal terms of $T$ are of opposite sign; and the $\delta E$ term is positive for the reason I gave above.

If you're asking why the signs of the integrated terms come out as they do, see above.

If you're asking why the locally measured energy must be positive, the simplest answer is that it's certainly true in an local inertial frame (since it's true in SR), and "locally measured energy" must be an invariant, so it must be true in any frame.

If I'm understanding their notation correctly, $v_H$ is the velocity of the "zero linear momentum" observer relative to the particle falling in; i.e., it is the velocity of an observer "comoving" with the black string in the frame in which the infalling particle is at rest. So, since the particle's velocity in the frame in which the black string is at rest is $v$, the velocity of the comoving observer in the frame in which the particle is at rest must be $-v$.

3. Nov 27, 2015

Thanks very much for this very detailed reply. There are a couple of bits I still don't quite get though:

Can you elaborate on why $n^3=\Omega_H$?

And also I'm afraid I still don't get the last part about $v_H=-v$. How can an observer comove with the string and see the particle at rest at the same time unless both particle and string have the same velocity?

4. Nov 27, 2015

### Staff: Mentor

Because that's how the vector $n^\mu$ is defined; the definition is given between equations 4.32 and 4.33 in the paper.

He can't; he can only do one or the other. At least, if I'm reading that part of the paper right, the particle is moving at $v$ in the frame that is comoving with the string. That means an observer comoving with the string will not see the particle at rest; he will see it moving at $v$.

5. Nov 28, 2015

Thanks again. So if $v_H$ is velocity of zero linear momentum observer at horizon ie a comoving observer then he will see the particle moving at $v$. Why then is $v_H=-v$? We are trying to measure energy flux across the horizon and won't the particle be forced to comove with horizon as soon as they enter ergoregion such that by the time they reach the horizon they are travelling at same velocity as horizon - wouldn't this mean the observer (at horizon) sees particle travelling at zero velocity?

Maybe I'm confused about what these velocities are defined with respect to? An observer at infinity?

Last edited: Nov 28, 2015
6. Nov 28, 2015

### Staff: Mentor

Velocity relative to what? That's the point you appear to be missing. As I read the paper, $v_H$ is the velocity of a comoving observer relative to the particle. So, since $v$ is the velocity of the particle relative to a comoving observer, we must have $v_H = -v$.

No. This is not Kerr spacetime.

7. Nov 28, 2015

Surely there must be induced motion along the boost direction? If I take a particle with $p_z=0$ and drop it into the spacetime at infinity, geodesic analysis will tell me that since $g_{tz} \neq 0$ we will have $p_z \neq 0$ just as a test particle is forced to co-rotate with a Kerr black hole.
Why would its velocity not tend to the boost velocity of the horizon as it travels from ergosurface to event horizon?

8. Nov 28, 2015

### Staff: Mentor

Motion in the boosted frame, yes; that motion will be in the opposite direction to the boost, and at the same velocity as the boost. That is, $v_H = -v$.

Again, $p_z = 0$ relative to what?

If you mean $p_z = 0$ in the non-boosted (i.e., comoving) frame, then this same particle will not have $p_z = 0$ in the boosted frame; it will have $p_z$ equal to some negative value.

Conversely, if you mean $p_z = 0$ in the boosted frame, then this same particle will have $p_z$ equal to some positive value in the non-boosted (i.e., comoving) frame.

Further comments after I've looked at the paper some more.

9. Nov 28, 2015

### Staff: Mentor

No, it will not. $p_z$ (linear momentum) is a constant of geodesic motion in this spacetime, since $\partial_z$ is a Killing vector field; just as $p_\phi$ (angular momentum) is a constant of geodesic motion in Kerr spacetime, since $\partial_\phi$ is a KVF.

What changes in Kerr spacetime as a particle falls is angular velocity, not angular momentum; that is, the angular velocity that corresponds to a given angular momentum varies with the coordinates (specifically, with $r$ and $\theta$). This effect is often called "frame dragging", and it is what forces the angular velocity of an infalling particle to be the same as the angular velocity of the horizon, in the limit as the horizon is approached.

10. Nov 28, 2015

Why would this be different in the linear case? $p_z$ will be constant but $\dot{z}$ will need to increase as $r$ decreases in order to maintain this - presumably up to the horizon at which point the velocity of the particle will match that of the horizon?

I take your point that what I said above will strongly depend on which frame of reference we use, right? Why do you say the co-moving frame is the same as the non-boosted frame? To me, if I take a frame and boost it, surely it would then be travelling at the same velocity as the horizon (also boosted) and so the co-moving frame should be the boosted one, no?

11. Nov 28, 2015

### Staff: Mentor

$\dot{z}$ will need to change as $r$ decreases in order to maintain this; but the change is not necessarily an increase everywhere. Equations 4.120 and 4.121 in the paper are the geodesic equations for $\dot{t}$ and $\dot{z}$ in the boosted frame. They can be solved simultaneously for $\dot{t}$ and $\dot{z}$ as functions of $r$. I haven't had time to work through the solution, but that's what would tell us the behavior as $r$ decreases.

Also, note that $\dot{z}$ is the rate of change of $z$ with proper time along the geodesic, not with respect to the coordinates. The rate of change of $z$ with respect to the coordinates, which is what I think the paper is describing with $v_H$ and $v$, is the ratio $\dot{z} / \dot{t}$.

Instead of presuming, you should look at the math.

Yes, obviously.

Because "comoving" means "at rest relative to the black string", which means "non-boosted". At least, that's my understanding from reading the paper. The black string is at rest in some coordinate chart; we call this chart "comoving", and it is "non-boosted" by contrast with the "boosted" chart, in which the black string is not at rest; it is moving.

The horizon is not "boosted". Changing coordinates doesn't change the state of motion of the horizon.

12. Nov 29, 2015

Why is the horizon not boosted? It is called a boosted black string and to me that would mean the string itself is moving, no? Again this is of course a frame dependent statement - I would guess the string is moving with respect to an observer at infinity, right? Why wouldn't this mean the horizon is boosted (i.e. moving)?

Secondly, what is the main difference between this linear case and the Kerr situation? I still don't fully understand why I can't apply this same analysis to show Penrose process isn't possible for Kerr? Looking under eqn 5 in this paper http://arxiv.org/pdf/gr-qc/0703091.pdf, would it appear to be because a comoving observer sees a static metric i.e. no frame dragging and no ergoregion?

Lastly, surely this energy calculation is frame dependent also - we are computing the energy across the horizon as seen by the infalling particle, right?

Thanks again.

Last edited: Nov 29, 2015
13. Nov 29, 2015

### Staff: Mentor

No, it means the frame in which the string is being described is boosted relative to the string. That's what the paper you linked to says.

No. The string is at rest relative to an observer at infinity. The paper even gives an equation for the metric in the frame in which the string and the observer at infinity are at rest--equation 4.113. Then this frame is boosted to give the metric in equation 4.114. But that just boosts the frame, not the string or the observer at infinity. Note the comment after equation 4.114: "this solution is just a non-boosted black string as seen by a boosted observer".

In the linear case, there is a frame in which the string is at rest--not moving linearly. In the Kerr case, there is no frame in which the black hole is not rotating.

14. Nov 30, 2015

I'm confused about this. We can compute the Killing horizon for $\partial_t$ and find that it is different to that of $\partial_t + v_H \partial_z$: in other words, there is an ergoregion. Furthermore, shouldn't the location of this ergoregion be invariant under change of frame (just like the location of the event horizon is). But the statement that the "solution is just a non-boosted black string as seen by a boosted observer" suggests that I can remove the ergoregion just by changing frame which seems wrong?

15. Nov 30, 2015

### Staff: Mentor

The location of the two surfaces is invariant under change of frame. To distinguish the two frames, I'll use capital letters for the non-boosted frame, and lower-case for the boosted frame. Then, in the boosted frame, we have:

Ergosurface: $\partial_t$ is null at $r = 2M \cosh^2 \beta$.

Event horizon: $\partial_t + v_H \partial_z$ is null at $r = 2M$.

In the non-boosted frame, these two surfaces are defined by:

"Ergosurface": $\partial_T + v \partial_Z$ is null at $r = 2M \cosh^2 \beta$.

Event horizon: $\partial_T$ is null at $r = 2M$.

Note that the relationships implied by the above between $\partial_t, \partial_z$ and $\partial_T, \partial_Z$ are easily verified using $v_H = -v$.

The only difference between the two frames is that, in the non-boosted frame, the "ergosurface" obviously has no special properties. But as the paper shows, it actually has no special properties in the boosted frame either; it "looks like" an ergosurface, but it isn't actually one, because the penrose process is not possible there (nor are any other features of a "real" ergoregion present there). That is, the appearance of an "ergosurface" in the boosted frame is an artifact of the coordinates, not an invariant property of the spacetime.

16. Nov 30, 2015

Thanks this has helped a lot. I've verified the calculation but why assign velocity $v$ to the non-boosted case and $v_H$ (which I thought was only defined at the horizon) to the boosted one?

17. Nov 30, 2015

### Staff: Mentor

Let me unpack the definitions a little more. First, for the non-boosted case: the velocity $v$ is just some velocity that we picked arbitrarily. Remember that the actual black string is not "moving"--i.e., it's at rest in the non-boosted frame. For any velocity $v$ with $0 < v < 1$, we can define a "boosted frame" moving at $v$ relative to the non-boosted frame, where "moving at $v$" is defined by the equality

$$\partial_t = \partial_T + v \partial_Z$$

That is, the Killing vector field $\partial_t$, whose orbits are the worldlines of observers "at rest" in the boosted frame, is just a linear combination of the Killing vector fields $\partial_T$ and $\partial_Z$ in the non-boosted frame. In other words, observers at rest in the boosted frame are moving at velocity $v$ in the $Z$ direction in the non-boosted frame.

Now, once we have picked a "boosted" frame in this way, we can imagine two observers falling through the event horizon at $r = 2M$, one with a constant $Z$ coordinate in the non-boosted frame, the other with a constant $z$ coordinate in the boosted frame. If we imagine the two observers passing each other just at the instant that they both pass the horizon, then the first observer will have a velocity $v_H$ in the $z$ direction relative to the second observer. This is what we mean by saying that the horizon has "velocity" $v_H$ in the $z$ direction in the boosted frame. But that means that the Killing vector field that becomes null on the horizon, which is $\partial_T$ in the non-boosted frame, becomes $\partial_t + v_H \partial_z$ in the boosted frame.

So far, we have not assumed any relationship between $v$ and $v_H$; they are defined independently, as given above, and the definitions, by themselves, do not assume any relationship between them. But once we have the two equations above, relating the Killing vector fields in the two frames, we can use the fact that $\partial_Z = \partial_z$, i.e., the KVF in the $z$ (or $Z$) direction is the same in both frames, to show that $v_H = -v$. Once we know that, of course, we could, for example, express the KVF that becomes null on the horizon, in the boosted frame, as $\partial_t - v \partial_z$ instead of $\partial_t + v_H \partial_z$. But conceptually, the two "velocities" are still independent. (For example, as you note, the physical meaning of $v_H$ is only directly apparent at the horizon.)

18. Dec 1, 2015

Thanks again. Can you clarify a few more things please:

1, If I do the calculation, I see that the boost velocity, $v(r)=\frac{dz}{dt}(r)$ varies between zero at infinity and $-\tanh{\beta}$ at the horizon. Furthermore, it decreases monotonically from 0 at infinity to $\tanh{\beta} \in [-1,0]$ at the horizon. What does this mean? It's measuring the velocity of the boosted frame relative to the string, right? So doesn't it mean that on the horizon $v=-v_H$ (since $v_H=\tanh{\beta}$) and then we have the argument about the infalling particle seeing the string moving backwards from perspective of particle etc. But what's happening at infinity? It looks like the particle is comoving with string ("unboosted" frame if you like). Doesn't this suggest the particle (boosted frame) speeds up as it falls towards the brane? I'm a little uncertain about this because the way you were talking above suggested to me we were treating $v$ as a constant but here I quite clearly find it increases...

2, The whole idea behind there argument seems to be that in the linear case we can find a frame in which the string isn't moving i.e. non boosted frame and in this frame we can show Penrose process isn't possible therefore it can't be possible in any of the other frames either. Is this the same as saying that the generator of the ergosurface $\partial_t + v \partial_z$ isn't a global Killing vector? Since in the non-boosted frame, it still becomes null at the same location but is no longer a Killing vector - is that why you wrote ergosurface in inverted commas in this case? I'm basically just thinking that there must be some stricter way of phrasing this rather than just that "the unboosted observer sees a static metric and therefore no ergoregion" - can we phrase it in terms of Killing vectors?

3, If so, the statement that there is no frame in which the ergoregion of Kerr metric disappears would be equivalent to saying that $\partial_t + \Omega_H \partial_\phi$ is a global Killing vector for Kerr? If this is not true, what is the reason that we cannot remove the Kerr ergoregion by a similar change of frame to the linear case? I know it's because no such frame exists but is there a more geometrical way to see this e.g. using Killing vectors?

19. Dec 1, 2015

### Staff: Mentor

This is a coordinate velocity, not a physically measured velocity. It doesn't have any direct physical meaning. To derive a velocity with physical meaning, you need to know the 4-velocity of the object whose velocity is being measured, and the 4-velocity of the observer doing the measuring.

20. Dec 1, 2015

Didn't we say $v$ was the velocity of the boosted frame relative to the unmoving string?