# I Confusion about Frame Dragging

1. Dec 17, 2017

### tzimie

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.

2. Dec 17, 2017

### Staff: Mentor

Yes.

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.

3. Dec 18, 2017

### 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!

4. Dec 18, 2017

### Staff: Mentor

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.

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.

5. Dec 20, 2017

### 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?

6. Dec 20, 2017

### jartsa

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.

Last edited: Dec 20, 2017
7. Dec 20, 2017

### tzimie

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)

8. Dec 20, 2017

### tzimie

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

9. Dec 20, 2017

### jartsa

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.

Last edited: Dec 20, 2017
10. Dec 20, 2017

### Staff: Mentor

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

11. Dec 20, 2017

### Staff: Mentor

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.

12. Dec 22, 2017

### 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:
Code (Text):

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);

Last edited: Dec 23, 2017
13. Dec 31, 2017

### tzimie

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

But I have another question, whats 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?

14. Dec 31, 2017

### phinds

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

15. Dec 31, 2017

### tzimie

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.

16. Dec 31, 2017

### Staff: Mentor

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.

17. Dec 31, 2017

### jartsa

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.

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)

18. Dec 31, 2017

### jartsa

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?

19. Jan 1, 2018

### tzimie

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

20. Jan 1, 2018

### Staff: Mentor

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.

21. Jan 2, 2018

### tzimie

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.

22. Jan 2, 2018

### tzimie

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.

23. Jan 2, 2018

### Staff: Mentor

Yes, that is correct.

24. Jan 2, 2018

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.

25. Jan 2, 2018

### Staff: Mentor

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.