Confusion about Frame Dragging

[tzimie]

Wiki here: https://en.wikipedia.org/wiki/Frame-dragging
Claims that:

Another interesting consequence is that, for an object constrained in an equatorial orbit, but not in freefall, it weighs more if orbiting anti-spinward, and less if orbiting spinward. For example, in a suspended equatorial bowling alley, a bowling ball rolled anti-spinward would weigh more than the same ball rolled in a spinward direction. Note, frame dragging will neither accelerate or slow down the bowling ball in either direction. It is not a "viscosity". Similarly, a stationary plumb-bob suspended over the rotating object will not list. It will hang vertically. If it starts to fall, induction will push it in the spinward direction.

It sounds strange. What happens if an observer is hovering in equatorial plane of the rotating black hole, few meters above the ergosphere? Now an observer starts to release a chain, very slowly. Wiki claims that a chain will be vertical. Obviously, it can't be vertical any longer when it touches the ergosphere.

 

[PeterDonis]

Now an observer starts to release a chain, very slowly. Wiki claims that a chain will be vertical.

Yes.

Obviously, it can't be vertical any longer when it touches the ergosphere.

More precisely, the portion of the chain within the ergosphere cannot be vertical. The reason is that, to remain vertical, i.e., not move in the spinward direction, it would have to be moving faster than light, and it can't. So if we assume that the observer himself remains static, then the chain will break at some point above the boundary of the ergosphere: the portion remaining above the ergosphere will stay static, and the portion that broke off will start moving spinward.

 

[tzimie]

The problem with it as I see that the solution is not continuous. The chain is under vertical stress, as it attracts to the black hole, but that stress is not infinite as we are still far above the event horizon (I am aware that the concept of 'force' is not accurate in GR). So when we slowly release the chain, the horizontal component of the force is 0 until the bottom link of a chain touches the ergosphere - then the horizontal component suddenly becomes infinite, and it happens instantly without any warning!

 

[PeterDonis]

The chain is under vertical stress, as it attracts to the black hole, but that stress is not infinite as we are still far above the event horizon

This is not correct. The chain is under stress not just because of its height above the horizon, but because of its being static--exactly vertical--close to the boundary of the ergosphere. If a piece of the chain drops into the ergosphere, the stress on that piece if it "tries" to stay static increases without bound. See further comments below.

when we slowly release the chain, the horizontal component of the force is 0 until the bottom link of a chain touches the ergosphere - then the horizontal component suddenly becomes infinite

You have to be careful thinking of "force" here. Spacetime doesn't exert any "force" on the chain; the force is exerted by whatever is holding it static. All spacetime does is determine the light cones: which curves are timelike, null, and spacelike. Above the ergosphere, curves with constant spatial coordinates (static) are timelike; on the ergosphere boundary, a curve with constant spatial coordinates becomes null. And below the ergosphere boundary, a curve with constant spatial coordinates is spacelike. This is perfectly continuous: just above the ergosphere boundary, a curve with constant spatial coordinates is "almost null".

Also, the statement in the Wikipedia article about the chain being vertical without requiring any sideways force applies to a very short chain; but I'm not sure it continues to apply as the chain is lowered more and more.

 

[Ibix]

How would you go about confirming this? For the bowling ball presumably you calculate the four acceleration associated with a four velocity parallel to $(1,0,0,\omega)$, sitting on the equator and working in Boyer-Lindquist $(t,r,\theta,\phi)$ coordinates. Then you show that the resulting proper acceleration is not equal for $\omega$s with opposite signs?

For the chain you'd presumably get the four acceleration from a four velocity parallel to $(1,0,0,0)$, then show that non-radial components are zero (arbitrary chain hangs vertically) or small (short chain hangs vertically, or close enough). Right?

 

[jartsa]

Wiki here: https://en.wikipedia.org/wiki/Frame-dragging
Claims that:
It sounds strange. What happens if an observer is hovering in equatorial plane of the rotating black hole, few meters above the ergosphere? Now an observer starts to release a chain, very slowly. Wiki claims that a chain will be vertical. Obviously, it can't be vertical any longer when it touches the ergosphere.

I interpret what the Wikipedia says like this:

Induction tries to push the chain in the spinward direction. At the ergosurface the force that is supposed to keep the chain stationary with respect to an outside observer at a great distance should make the chain to move at the speed of light with respect to the local spacetime. So it should be a force that approaches infinity as the chain approaches the ergosurface.

At the ergosurface the induction force becomes infinite, if you resist it with an equal opposite force.

 

[tzimie]

So it should be a force that approaches infinity as the chain approaches the ergosurface.

in that case continuity is restored, but the chain is no longer vertical as it approaches the ergosphere.
(and yes, I am aware that the notion of "force" is not valid in GR)

 

[tzimie]

Also, the statement in the Wikipedia article about the chain being vertical without requiring any sideways force applies to a very short chain; but I'm not sure it continues to apply as the chain is lowered more and more.

obviously, if a SHORT chain is EXACTLY vertical than a chain of ANY length is vertical.

 

[jartsa]

in that case continuity is restored, but the chain is no longer vertical as it approaches the ergosphere.

The chain is never vertical if there is an induction force that is not canceled by a counter force. And there is an induction force whenever we move the chain. If we move it slowly, the force is small, but it also lasts a long time.

If we do not move the chain there is no induction force. So using force we can set the chain to hang in any orientation at almost any altitude, and it stays in that orientation. Except when one end is in the ergosphere, then we can not do that.

 

[PeterDonis]

Induction tries to push the chain in the spinward direction.

More precisely, induction tries to push a radially moving chain in the spinward direction. As you say in your follow-up post.

 

[PeterDonis]

obviously, if a SHORT chain is EXACTLY vertical than a chain of ANY length is vertical.

No, that's not obvious, because as @jartsa points out, the chain has to be moved in order to extend downward. And the closer the chain is to the ergosphere, the more the spinward frame dragging will "pull" it as soon as it starts to move downward.

Even if you set up a very long chain that is initially static (so the frame dragging doesn't affect it--but note that in order to get the chain into this state, you will have to exert force on it to line it up vertically), with the lower end just above the ergosphere boundary, as soon as you try to lower the lower end, that end will be "grabbed" very hard by frame dragging. And there is no way to set up a chain that starts out vertical and static with one end already at or below the ergosphere boundary, because there are no static timelike worldlines at or below the boundary.

 

[Ibix]

So I had a crack at the maths I laid out in #5. The result is something of a mess, but agrees with what Peter said. Each element of a hanging chain has a four-velocity parallel to (1,0,0,0), and the proper acceleration associated with this has only radial components. And the proper acceleration is divergent at the ergosphere. So the chain will snap above the ergosphere and start to fall.

If the four-velocity of a chain element includes a radial component - i.e. is parallel to (1,$v_r$,0,0) - then some force in the $d\phi$ direction is needed to keep the chain vertical.

So a hanging chain hangs vertically. As the chain is winched out it will curve spinward, but it will settle when the winching stops. It must break before crossing the ergosphere, no matter how strong it is.

Maxima code:

load(ctensor);
ct_coordsys(kerr_newman);
lg:substitute(0,e,lg); /* Uncharged */
cmetric(false);
ratchristof:true;
christof(false);

v0:[1,vr,0,0]; /* Unnormalised 4-v */
v0.substitute(%pi/2,theta,lg).transpose(v0);
v:v0/sqrt(%); /* Normalised 4-v */

/* Four acceleration: V^{i}nabla_{i}V^{j} */
covdif(v):=block(
  [z:zeromatrix(4,4),mu,nu,lamda],
  for mu: 1 thru dim do
    for nu: 1 thru dim do (
      z[mu,nu]:diff(v[nu],ct_coords[mu]),
      for lamda: 1 thru dim do
        z[mu,nu]:z[mu,nu]+v[lamda]*mcs[lamda,mu,nu]
    ),
  z
);
calcA(v):=block(
  [a:[0,0,0,0],cdv,mu,nu],
  cdv:covdif(v),
  for mu: 1 thru dim do (
    for nu: 1 thru dim do
      a[nu]:a[nu]+v[mu]*cdv[mu,nu]
  ),
  a
);

/* Calculate the four-acceleration and extract the r-component */
ar:substitute(%pi/2,theta,calcA(v)[2]);

/* Plot for a BH with mass mv and angular momentum */
/* parameter av, with a chain element with four-velocity */
/* parallel to (1,vrv,0,0) over (2-rng)*mv<r<(2+rng)*mv */
plotar(mv,av,vrv,rng):=block ([],
  plot2d(substitute(mv,m,
          substitute(av,v,
            substitute(vrv,vr,ar))),
         [r,(2-rng)*mv,(2+rng)*mv])
);
plotar(1,0.1,0,0.0002);

 

[tzimie]

No, that's not obvious, because as @jartsa points out, the chain has to be moved in order to extend downward. And the closer the chain is to the ergosphere, the more the spinward frame dragging will "pull" it as soon as it starts to move downward.

You can move it INFINITELY SLOWLY - it is just a thought experiment anyway.

But I have another question, what's about a realistic static chain in a frame-dragging environment? Even if it is not moving, metal contains atoms moving in all directions. I suspect that a rigid body would try to rotate, but I am not sure. What do you think?

 

[phinds]

You can move it INFINITELY SLOWLY

Sure. Then it will take an infinite amount of time and you'll never know WHAT happens

 

[tzimie]

Sure. Then it will take an infinite amount of time and you'll never know WHAT happens

You can just define a pre-existing chain around the eternal black hole
For example, schwarzschild solution itself is an idealization about the "eternally existing" black hole.

 

[PeterDonis]

You can move it INFINITELY SLOWLY

This amounts to treating the chain as static. You can treat a chain as static, yes, but then you obviously can't have a chain whose lower end is at or below the boundary of the ergosphere, since it is impossible for anything to be static there. So you can't even formulate your thought experiment if you treat the chain this way.

You can just define a pre-existing chain around the eternal black hole

Not if it's static and one end is at or below the ergosphere boundary. I already addressed this in post #11, which you apparently didn't read all of.

 

[jartsa]

But I have another question, what's about a realistic static chain in a frame-dragging environment? Even if it is not moving, metal contains atoms moving in all directions. I suspect that a rigid body would try to rotate, but I am not sure. What do you think?

Well it seems quite simple:

Atom moves down - induced force points to the left.
Atom moves up - induced force points to the right.
Net induced force is zero.But how about this:

Atom moves infinitely slowly 1 m down - induced infinitely small force points to the left.
The same atom moves infinitely slowly 1 m up, but not along the same path that it moved down - induced infinitely small force points to the right.

What is the net torque multiplied by time that is induced to the atom? (Torque multiplied by time is angular momentum)

 

[jartsa]

So I had a crack at the maths I laid out in #5. The result is something of a mess, but agrees with what Peter said. Each element of a hanging chain has a four-velocity parallel to (1,0,0,0), and the proper acceleration associated with this has only radial components. And the proper acceleration is divergent at the ergosphere. So the chain will snap above the ergosphere and start to fall.

If the four-velocity of a chain element includes a radial component - i.e. is parallel to (1,$v_r$,0,0) - then some force in the $d\phi$ direction is needed to keep the chain vertical.

So a hanging chain hangs vertically. As the chain is winched out it will curve spinward, but it will settle when the winching stops. It must break before crossing the ergosphere, no matter how strong it is.

So the 'drag force' does not become infinite at the ergosphere, if winching time is finite?

But the force that we call weight becomes infinite?

Or maybe they both become infinite?

 

[tzimie]

This amounts to treating the chain as static. You can treat a chain as static, yes, but then you obviously can't have a chain whose lower end is at or below the boundary of the ergosphere, since it is impossible for anything to be static there. So you can't even formulate your thought experiment if you treat the chain this way.

Not if it's static and one end is at or below the ergosphere boundary. I already addressed this in post #11, which you apparently didn't read all of.

Peter,
I read all your posts very carefully, and of course, I meant the chain completely ABOVE the ergosphere.

 

[PeterDonis]

of course, I meant the chain completely ABOVE the ergosphere

But a static chain entirely above the ergosphere makes the question in your OP meaningless, since in the OP you specifically talked about what happens if the chain touches the ergosphere. So are you now saying your question in the OP can be ignored? If so, we can just close this thread.

 

[tzimie]

But a static chain entirely above the ergosphere makes the question in your OP meaningless, since in the OP you specifically talked about what happens if the chain touches the ergosphere. So are you now saying your question in the OP can be ignored? If so, we can just close this thread.

I am interested in a case when a chain approaches infinitely the ergosphere, but is not touching it.

Lets looks at a sequence of hovering points above and below the ergosphere. Below the ergosphere their path is spacelike and is physically impossible. Above the ergosophere it is timelike. Exactly on the ergosphere it is NULL.

Then if we look at an infinite sequence of hovering observers, approaching but not touching the ergosphere, then the factor of time dilation for them will diverge to infinity (as it is infinite for a NULL path). A little bit weird but looks like it is true.

 

[tzimie]

Well it seems quite simple:

Atom moves down - induced force points to the left.
Atom moves up - induced force points to the right.
Net induced force is zero.

I believe if would start to disintegrate - atom on a surface moves down - is pulled to the left very hard - escapes from surface. This is just one of many effects.

 

[PeterDonis]

if we look at an infinite sequence of hovering observers, approaching but not touching the ergosphere, then the factor of time dilation for them will diverge to infinity (as it is infinite for a NULL path).

Yes, that is correct.

 

[nomadreid]

I am not sure whether I should start a new thread with my question, because whereas it does not directly deal with the OP's question, my question deals with a related question. I leave the fate of the question to the moderators. (Feel free to move it or to omit it and have me re-post as a new thread.)
So, the question: please point out whether the following reasoning is correct, wrong in certain steps, or simply off-the-wall.
If one decides, instead of lowering a chain, to commit suicide by entering the event horizon of black hole that is spinning enough to have jets, one would indeed commit suicide in the attempt without however being able to enter, even when one makes the assumptions:
[1] the black hole is a super-massive one, so that it is not the tidal forces outside the event horizon that would kill the suicidal astronaut) and
[2] supposing there is no firewall at the event horizon,
as the frame dragging would tear the astronaut apart before reaching the event horizon, unless she tried to enter when one does not experience any frame-dragging, but that would be at the poles, where the jets would mean a quick death.

 

[PeterDonis]

the frame dragging would tear the astronaut apart before reaching the event horizon

Why do you think that? If tidal gravity is negligible at the horizon, then any difference in frame dragging between the astronaut's head and feet will also be negligible. So his whole body will be frame dragged by the same amount and he will feel no force due to it, so he won't be affected at all.

 

[nomadreid]

Ah, thank you, PeterDonis. My fault for reading popular science sites.

 

[nomadreid]

Ah, coming back to that answer... if there were no significant frame dragging around a supermassive black hole, then where do the jets of those who have them (OK, only a small portion, but nonetheless) come from? I thought the consensus was that the frame dragging around them was one of the causes. ?

 

[Ibix]

Peter isn't saying that frame dragging isn't significant. He's saying that the difference in the frame dragging effect between your head and your toes isn't significant (with the stipulation you made that this is a super massive black hole). You'll go round the black hole, and fast, but you won't be torn apart because your toes won't be trying to go much faster than your head.

 

[tzimie]

Yesterday I learned that there may be ergospeheres around neutron stars too. Wow.

 

[nomadreid]

Thanks, Ibix. so

You'll go round the black hole, and fast,

I would end up spiraling into the black hole?

 

[Ibix]

Thanks, Ibix. so

I would end up spiraling into the black hole?

I wouldn't like to take a position on whether you would complete a full circle without doing some maths, but basically yes. If I drop you from a great distance from a non-rotating hole you will always be between me and the hole as you fall in. But with a rotating hole you would be dragged spinwards to some extent.

In practice the accretion disc around the hole would have a straightforward frictional drag effect as well (not to mention cooking you extra-crispy). I suspect that would tend to make you orbit more than a vacuum GR analysis would suggest, but I haven't done the maths for that either.

 

[nomadreid]

Many thanks for that, Ibix. That makes sense, even without the calculations. (The various things that can, as you put it, cook an astronaut extra-crispy, seem to do away with all science-fiction scenarios about an astronaut entering the event horizon alive.)

 

[PeterDonis]

Yesterday I learned that there may be ergospeheres around neutron stars too.

From where? Please give a reference.

 

[PeterDonis]

if there were no significant frame dragging around a supermassive black hole, then where do the jets of those who have them (OK, only a small portion, but nonetheless) come from?

The jets are believed to be coming out along the rotation axis of the hole; as I understand it, frame dragging basically collimates them to shoot out that way. But that is frame dragging considered globally, on the scale of the entire hole: as @Ibix said, that is very different from the difference in frame dragging between the feet and the head of an astronaut in a free-fall trajectory near the hole.

 

[tzimie]

From where? Please give a reference.

https://arxiv.org/pdf/astro-ph/0212429.pdf