Ergoregions and Energy Extraction

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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

 

[PeterDonis]

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

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.

why is the first term positive and the second term negative?

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.

Why is this the case?

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 this observer is comoving then surely they should have the same momentum as the horizon

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 $-v$.

 

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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 $-v$.

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?

 

[PeterDonis]

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

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

How can an observer comove with the string and see the particle at rest at the same time

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 $v$.

 

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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 $v$.

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 traveling at same velocity as horizon - wouldn't this mean the observer (at horizon) sees particle traveling at zero velocity?

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

 

[PeterDonis]

if $v_H$ is velocity of zero linear momentum observer at horizon ie a comoving observer

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 $v_H = -v$.

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 traveling at same velocity as horizon

No. This is not Kerr spacetime.

 

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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 $v_H = -v$.

Thanks I understand this now. one question about your final comment:

No. This is not Kerr spacetime.

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?

 

[PeterDonis]

Surely there must be induced motion along the boost direction?

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, $v_H = -v$.

If I take a particle with $p_z=0$

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.

 

[PeterDonis]

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 $p_z \neq 0$

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.

 

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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.

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 traveling at the same velocity as the horizon (also boosted) and so the co-moving frame should be the boosted one, no?

 

[PeterDonis]

$p_z$ will be constant but $\dot{z}$ will need to increase as $r$ decreases in order to maintain this

$\dot{z}$ 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 $\dot{z} / \dot{t}$.

presumably up to the horizon at which point the velocity of the particle will match that of the horizon?

Instead of presuming, you should look at the math.

I take your point that what I said above will strongly depend on which frame of reference we use, right?

Yes, obviously.

Why do you say the co-moving frame is the same as the non-boosted frame?

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.

To me, if I take a frame and boost it, surely it would then be traveling at the same velocity as the horizon (also boosted)

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

 

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$\dot{z}$ 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.

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.

 

[PeterDonis]

It is called a boosted black string and to me that would mean the string itself is moving, no?

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.

I would guess the string is moving with respect to an observer at infinity, right?

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".

what is the main difference between this linear case and the Kerr situation?

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.

 

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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.

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?

 

[PeterDonis]

shouldn't the location of this ergoregion be invariant under change of frame (just like the location of the event horizon is).

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.

 

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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.

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?

 

[PeterDonis]

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?

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.)

 

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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.)

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?

 

[PeterDonis]

the boost velocity, $v(r)=\frac{dz}{dt}(r)$

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.

 

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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.

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

 

[PeterDonis]

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

No. First of all, you need to be clear about what "v" you are referring to. You talked about a function $v(r) = dz/dt$; but when I talked about $v$ before, I was talking about a constant $v = \tanh \beta$ that appears in the boosted frame metric, and also in the definition of the KVF $\partial_t = \partial_T + v \partial_Z$. That $v$ is not a function of $r$; it's not a function of anything; it's a constant. (The KVF definition makes that clear; a linear combination of two KVFs is only a KVF itself if it has constant coefficients.)

Then I talked about the physical meaning of the constant $v$: it describes the velocity of an object with constant $z$ coordinate in the boosted frame, as it falls through the horizon at $r = 2M$, relative to an observer with constant $Z$ coordinate in the non-boosted frame who is falling through the horizon at the same event (i.e., the two worldlines cross at an event on the horizon, and the velocity $v$ is their relative velocity at that event).

 

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No. First of all, you need to be clear about what "v" you are referring to. You talked about a function $v(r) = dz/dt$; but when I talked about $v$ before, I was talking about a constant $v = \tanh \beta$ that appears in the boosted frame metric, and also in the definition of the KVF $\partial_t = \partial_T + v \partial_Z$. That $v$ is not a function of $r$; it's not a function of anything; it's a constant. (The KVF definition makes that clear; a linear combination of two KVFs is only a KVF itself if it has constant coefficients.)

Then I talked about the physical meaning of the constant $v$: it describes the velocity of an object with constant $z$ coordinate in the boosted frame, as it falls through the horizon at $r = 2M$, relative to an observer with constant $Z$ coordinate in the non-boosted frame who is falling through the horizon at the same event (i.e., the two worldlines cross at an event on the horizon, and the velocity $v$ is their relative velocity at that event).

OK. I understand. Thanks.

And is it possible to understand the ergoregion not being a true ergoregion in terms of the Killing vectors not being global (as I was trying to argue in post 18)?

 

[PeterDonis]

is it possible to understand the ergoregion not being a true ergoregion in terms of the Killing vectors not being global (as I was trying to argue in post 18)?

Killing vector fields are always global. They can change signature (i.e., they can be timelike in some places, null in others, and spacelike in still others), but that doesn't stop them from existing.

 

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Killing vector fields are always global. They can change signature (i.e., they can be timelike in some places, null in others, and spacelike in still others), but that doesn't stop them from existing.

So I guess my concern here is the following: we know in the linear case that there isn't a true ergoregion since we can change to the non-boosted frame where we see a static metric and thus $r=2m \cosh{\beta}$ isn't a priviliged surface. But is there a way to show using geometrical quantities (Killing vectors or something else perhaps) that this isn't a privilged surface? The reason that I ask is that in the case of Kerr, you say that there is no frame in which the black hole isn't rotating but how can we be sure that there is no such frame? Maybe we just weren't able to find a suitable frame?

 

[PeterDonis]

is there a way to show using geometrical quantities (Killing vectors or something else perhaps) that this isn't a privilged surface?

As I said, you can always take linear combinations of KVFs with constant coefficients and form new KVFs. So given that $\partial_T$ and $\partial_Z$ (in the non-boosted frame) are KVFs, it's always possible to pick some number $v$ with $0 < v < 1$ and form the KVF $\partial_T + v \partial_Z$, which will be null at some surface of larger $r$ than the event horizon. The only thing that picks out the actual event horizon is the fact that it's the Killing horizon with the smallest $r$, out of all the Killing horizons (surfaces where a KVF is null) of KVFs that can be formed as linear combinations of $\partial_T$ and $\partial_Z$.

in the case of Kerr, you say that there is no frame in which the black hole isn't rotating but how can we be sure that there is no such frame

Because, unlike the "black string" spacetime, the KVF in Kerr spacetime that corresponds to $\partial_T$ is not hypersurface orthogonal. In the "black string" spacetime, $\partial_T$ is orthogonal to a set of hypersurfaces that foliate the spacetime (more precisely, they foliate the region exterior to the event horizon). In Kerr spacetime, the corresponding KVF is not. That is why it is impossible to find a coordinate chart on Kerr spacetime in which the metric is diagonal everywhere, whereas in the "black string" spacetime it obviously is.

 

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As I said, you can always take linear combinations of KVFs with constant coefficients and form new KVFs. So given that $\partial_T$ and $\partial_Z$ (in the non-boosted frame) are KVFs, it's always possible to pick some number $v$ with $0 < v < 1$ and form the KVF $\partial_T + v \partial_Z$, which will be null at some surface of larger $r$ than the event horizon. The only thing that picks out the actual event horizon is the fact that it's the Killing horizon with the smallest $r$, out of all the Killing horizons (surfaces where a KVF is null) of KVFs that can be formed as linear combinations of $\partial_T$ and $\partial_Z$.

I don't see why this shows the surface isn't privileged? Don't we do exactly the same thing in the non-boosted frame and yet here we have the interpretation that it is an ergosurface, don't we?

So the full argument is the following:
1, use hypersurface orthogonality to motivate the existence of a non-boosted frame in which observer sees a static metric
2, argue that from the perspective of an infalling particle with negative energy (boosted frame with boost velocity $v$) who sees the string moving at velocity $v_H=-v$ that $|E|<|p_z|$ (via the integral we discussed at the very beginning). But, if we use normal coordinates for the infalling particle in the boosted frame, we have $E^2=p_z^2+m^2 \Rightarrow |E| \geq |p_z|$ leading to a contradiction. Thus we conclude that negative energy particles cannot exist.
Is this correct?

In the paper he says that $0<v<1$. I understand that you can't boost with $v=0$ since this just returns the static limit but why can't we boost with $v=1$. Would this be because it would need to be a massless particle e.g. photon traveling at speed of light in transverse z direction and therefore it would not fall into the black hole and if the negative energy particle doesn't fall in then we can't extract energy, right?Finally, the ergoregion is defined by the changing of a timelike Killing vector to a spacelike one. We still have this property so I would say it should still be called an ergoregion, don't you think? The argument here is that we cannot find a suitable trajectory through the ergoregion to allow for negative energy particles (needed for energy extraction etc) to exist, right?

 

[PeterDonis]

Don't we do exactly the same thing in the non-boosted frame and yet here we have the interpretation that it is an ergosurface, don't we?

No. In the non-boosted frame there is no "ergosurface"; there is no KVF that is picked out as "special" by the form of the metric except for $\partial_T$ itself, and that KVF is null on the event horizon. The reason there is an apparent "ergosurface" in the boosted frame is that the metric looks different, formally, in the boosted frame; the KVF $\partial_t$, which looks like the "time translation" KVF in the boosted frame (the KVF describing the worldlines of objects "at rest" in that frame) is null on a surface that is outside the event horizon. But this is only apparent; there is no actual physical property that picks out this "ergosurface", it's just an artifact of the choice of coordinates.

So the full argument is the following:
1, use hypersurface orthogonality to motivate the existence of a non-boosted frame in which observer sees a static metric

Yes. The non-boosted frame is picked out by an actual physical symmetry of the spacetime; the KVF $\partial_T$, which describes the worldlines of observers "at rest" in the non-boosted frame, is hypersurface orthogonal, which is an invariant property that is independent of coordinates (you could show that it holds in the boosted frame as well, using the description $\partial_t - v \partial_z$ of the KVF in that frame and the appropriate equation for the orthogonal hypersurfaces in that frame). The KVF $\partial_t = \partial_T + v \partial_Z$ does not have this property, for any nonzero $v$.

2, argue that from the perspective of an infalling particle with negative energy (boosted frame with boost velocity $v$) who sees the string moving at velocity $v_H=-v$ that $|E|<|pz|$

ZI'm not sure you've described the hypothetical "negative energy" particle correctly; but the fact that there are no possible timelike worldlines with negative energy at infinity ("energy at infinity" is the more precise term for what we have been calling "energy"--it is the constant of geodesic motion derived from the timelike KVF) can be deduced in the non-boosted frame, and since it is a frame-invariant property, it must also hold in the boosted frame. The paper gives a derivation purely in the boosted frame, but that is more for pedagogical purposes than anything else; you don't have to derive it in the boosted frame to prove it.

why can't we boost with v=1v=1.

You could, mathematically speaking; but the coordinate chart you would get would be a different kind of chart, with different properties from the "boosted" chart given in the paper. For one thing, note that if $v = 1$, there is no way to define $\beta$ as the paper does, because there is no $\beta$ for which $\tanh \beta = 1$. So the whole analysis would have to be done differently from the start.

Would this be because it would need to be a massless particle e.g. photon traveling at speed of light in transverse z direction

I'm not sure, because I haven't done the alternate analysis I just described. My intuitive guess is that the KVF $\partial_T + \partial_Z$ (described in the non-boosted frame--this is what would be the hypothetical "at rest" timelike KVF in the boosted frame for $v = 1$) would be null everywhere, which means it could (somewhat loosely) be interpreted as describing a family of massless particles moving in the $Z$ direction; but it's just a guess, and I suspect there would be other complications involved in any rigorous analysis.

the ergoregion is defined by the changing of a timelike Killing vector to a spacelike one.

No, it isn't. An ergoregion is defined by the presence of timelike worldlines with negative energy at infinity. The above criterion might pick out an "apparent" ergoregion, but even that is problematic. For example, the KVF $\partial_T$ in the non-boosted frame is spacelike inside the event horizon; but that doesn't make the region inside the event horizon an ergoregion. (In fact, there are KVFs in flat Minkowski spacetime that are timelike in one region and spacelike in another.)

 

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No, it isn't. An ergoregion is defined by the presence of timelike worldlines with negative energy at infinity. The above criterion might pick out an "apparent" ergoregion, but even that is problematic. For example, the KVF $\partial_T$ in the non-boosted frame is spacelike inside the event horizon; but that doesn't make the region inside the event horizon an ergoregion. (In fact, there are KVFs in flat Minkowski spacetime that are timelike in one region and spacelike in another.)

Do you have a reference for this definition of an ergoregion. The one I had written down in my notes was as I said above but I take your point that this would imply the region inside an event horizon would be an ergoregion. Although I'm not sure that would necessarily be wrong - the particle could well split into a positive and negative energy pair of daughters inside the horizon and since they can't escape, we'd never be able to extract anyway so nothing would be affected?

Can you give me an example of a KVF in Minkowski with such properties? Presumably something to do with polar coordinates so you can get $\sin{\theta}<0$?

Is there some physical intuition as to why no such timelike worldline with negative energy can't exist? I now accept that the maths shows this but I'm just wondering why? In principle, the Penrose process doesn't need a black hole, it only requires an ergoregion. In fact, in the paper he gives an example of energy extraction from a carousel I believe in Section 4.3.2. Why would we not expect something similar in the linear case e.g. skipping a stone along the surface of a flowing river or something like that?

When people talk about ``boosted black strings'' do they always mean what you described above e.g. a static black string and a boosted reference frame corresponding to an infalling observer with motion in a transverse direction? Or is it possible for the boosted black string itself to be moving (we could for example be able to show there is some momentum traveling along the horizon) - this sounds quite similar to my stone example above and I'm just wondering if Penrose process would occur in this case or if our analysis has ruled it out for all non-axisymmetric systems?

 

[PeterDonis]

Do you have a reference for this definition of an ergoregion.

The paper you linked to implicitly uses this definition; otherwise it would be pointless to show that negative energy at infinity particles are impossible in the boosted black string spacetime.

I don't know that the term "ergoregion" has that specific a definition in the literature, because not all sources use it. But every source I'm aware of that discusses Kerr spacetime is clear on the fact that the existence of timelike worldlines with negative energy at infinity is the important physical property.

this would imply the region inside an event horizon would be an ergoregion. Although I'm not sure that would necessarily be wrong - the particle could well split into a positive and negative energy pair of daughters inside the horizon and since they can't escape, we'd never be able to extract anyway so nothing would be affected?

No, a particle can't split into a positive and negative energy pair just anywhere; it can only do so if it is possible to have a timelike worldline with negative energy at infinity. Otherwise any interaction where a particle splits must produce two particles with positive energy.

Can you give me an example of a KVF in Minkowski with such properties?

Sure, the KVF described by a family of observers at rest in Rindler coordinates is timelike above the Rindler horizon, null on the Rindler horizon, and spacelike below the Rindler horizon.

Is there some physical intuition as to why no such timelike worldline with negative energy can't exist?

I'm not sure what the best way to express the key physical property would be. I think it is something along the lines of the spacetime being stationary but not static, i.e., there being no timelike KVF that is hypersurface orthogonal. (As we discussed, Kerr spacetime has this property, but the boosted black string spacetime does not--it has a hypersurface orthogonal timelike KVF.) But I'm not sure.

When people talk about ``boosted black strings'' do they always mean what you described above e.g. a static black string and a boosted reference frame corresponding to an infalling observer with motion in a transverse direction?

I don't know; the only time I've seen the term "boosted black string" or even "black string" is in the paper you linked to.

 

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The paper you linked to implicitly uses this definition; otherwise it would be pointless to show that negative energy at infinity particles are impossible in the boosted black string spacetime.

I don't know that the term "ergoregion" has that specific a definition in the literature, because not all sources use it. But every source I'm aware of that discusses Kerr spacetime is clear on the fact that the existence of timelike worldlines with negative energy at infinity is the important physical property.
No, a particle can't split into a positive and negative energy pair just anywhere; it can only do so if it is possible to have a timelike worldline with negative energy at infinity. Otherwise any interaction where a particle splits must produce two particles with positive energy.
Sure, the KVF described by a family of observers at rest in Rindler coordinates is timelike above the Rindler horizon, null on the Rindler horizon, and spacelike below the Rindler horizon.

I'm a little confused here. The paper linked to calls it in the title of Section 4.7 "ergoregions without superradiance" suggesting there is still an ergoregion but it just doesn't allow for superradiance/Penrose process. At least that's the impression I get from the paper?

 

[PeterDonis]

The paper linked to calls it in the title of Section 4.7 "ergoregions without superradiance" suggesting there is still an ergoregion but it just doesn't allow for superradiance/Penrose process.

That's because, as I said, "ergoregion" is not a standard term, and is not even used by all sources. This paper is using it in a way that, IMO, is going to cause confusion, but since there is no single standard definition of the term, their usage is not wrong, exactly, just confusing.

In any case, the important point is not words, but physics. The physics is clear: there are no timelike worldlines with negative energy at infinity in the boosted black string spacetime, whereas there are in Kerr spacetime. That is why supperradiance and the Penrose process are possible in the latter but not the former.

 

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That's because, as I said, "ergoregion" is not a standard term, and is not even used by all sources. This paper is using it in a way that, IMO, is going to cause confusion, but since there is no single standard definition of the term, their usage is not wrong, exactly, just confusing.

In any case, the important point is not words, but physics. The physics is clear: there are no timelike worldlines with negative energy at infinity in the boosted black string spacetime, whereas there are in Kerr spacetime. That is why supperradiance and the Penrose process are possible in the latter but not the former.

Earlier we said that there was no frame in Kerr in whcih the black hole was not rotating. But we know that an observer at infinity sees no rotation since $g_{t \phi} \rightarrow 0$. Is the point then that this is only true locally and as soon as we move away from infinity, we would observe some rotation whereas for the boosted black string, the comoving frame will see a static string regardless of the radial coordinate? I guess this is tied to the hypersurface orthogonality but I'm not clear what the connection is...

 

[PeterDonis]

we know that an observer at infinity sees no rotation since $g_{t \phi} \rightarrow 0$.

That doesn't mean an observer at infinity sees no rotation. It means an observer at infinity sees no frame dragging locally. But "locally" does not include the black hole; that is, if we consider a local "frame" for the observer at infinity, that frame does not contain the hole; it only contains a small local region of spacetime around the observer. When we said there was no frame in Kerr spacetime in which the hole is not rotating, we meant a frame that includes the hole.

 

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That doesn't mean an observer at infinity sees no rotation. It means an observer at infinity sees no frame dragging locally. But "locally" does not include the black hole; that is, if we consider a local "frame" for the observer at infinity, that frame does not contain the hole; it only contains a small local region of spacetime around the observer. When we said there was no frame in Kerr spacetime in which the hole is not rotating, we meant a frame that includes the hole.

To include the hole would require what? A global coordinate system in which $g_{t \phi}=0 \forall r$?

 

[PeterDonis]

To include the hole would require what? A global coordinate system

Yes--or at least one that covers enough of the spacetime to include the horizon and the region inside it.

in which $g_{t \phi}=0 \forall r$?

That's what is not possible in Kerr spacetime (whereas its analogue is in the black string spacetime).