Maximizing survival time when falling into a black hole

[tom.stoer]

Unfortunately I didn't find a thread discussing this issue.

First I will sketch the standard argument that one should not use the rocket engine and try to accelerate away from the singularity. Then I will try to identify the problematic part of this argument and ask for your comments.

1) For any two points P and Q in spacetime the worldline C_PQ connecting these two points and maximizing proper time τ_PQ = τ[C_PQ] is a geodesic C⁰_PQ.

2) The problem is that once one uses the engine this affects the point at which one will hit the singularity. In an appropriate coordinate system we may have Q = (t,r=0) but Q' = (t', r=0) with t' ≠ t. Therefore we must not refer to a single geodesic connecting P and Q. Now one could try to find a different geodesic C⁰C_PQ' connecting P and Q'. This geodesic would of course maximize the proper time τ_PQ', but a new geodesic is equivalent to a new initial condition at P. So in the very end with given P and an initial condition at P we cannot say anything about maximizing survival time based on geodesics.

Any comments?

Here's paper that addresses this topic numerically and shows that one can find curves with acceleration with longer survival time larger than for geodesaics.

https://arxiv.org/PS_cache/arxiv/pdf/0705/0705.1029v1.pdf
No Way Back: Maximizing survival time below the Schwarzschild event horizon
Authors: Geraint F. Lewis, Juliana Kwan
Abstract: It has long been known that once you cross the event horizon of a black hole, your destiny lies at the central singularity, irrespective of what you do. Furthermore, your demise will occur in a finite amount of proper time. In this paper, the use of rockets in extending the amount of time before the collision with the central singularity is examined. In general, the use of such rockets can increase your remaining time, but only up to a maximum value; this is at odds with the ``more you struggle, the less time you have'' statement that is sometimes discussed in relation to black holes. The derived equations are simple to solve numerically and the framework can be employed as a teaching tool for general relativity.

published in Publications of the Astronomical Society of Australia, Volume 24 Number 2 2007, pp. 46-52.

 

[martinbn]

Just a quick comment, I haven't looked at the paper yet.

Concerning the end point, you could possibly rephrase the question as follows. Finding a curve that maximizes proper time between a starting point an end submanifold (space-like). Then the result is that it will be a geodesic orthogonal to the end manifold. Of course it is problematic to view the singularity as a submanifold. Are you discussing the general case or a specific one, say Schwartzschild black hole. In any case what is true is that a curve that is a maximum will be a geodesic if the maximum is attained. But there is no guaranty unless you impose extra restrictions. (Of course I might be wrong on any of these, so take it with a lot of salt)

 

[tom.stoer]

In any case what is true is that a curve that is a maximum will be a geodesic if the maximum is attained.

The problem is that a geodesic maximizes the proper time between a fixed start- and a fixed endpoint.

1) Starting with fixed initial conditions (spacetime point and 4-momentum) and using the engine later on changes the endpoint. So there is no reason why the original geodesic with original start- and endpoint should have any meaning for this problem with new worldline and new endpoint.

2) Changing the initial condition (same spacetime point but different 4-momentum) such that the new endpoint from (1) is now reached via a geodesic shows that we can find a worldline with even longer proper time than in (1), namely the new geodesic. But changing the initial conditions means that we are comparing apples with oranges.

Therefore from pure logical reasons it seems that geodesics do not help at all.

Let me explain what the problem is all about: some astronauts cross the event horizon with identical initial conditions = at the same spacetime point, with the same 4-velocity and with the same 4-acceleration (zero). Having crossed the horizon, what is the best strategy for an astronaut to maximize his survival time = to maximize his proper time from the horizon to the singularity?

A) Using the engine to accelerate away from the geodesic and later on to return to it, i.e. to cross the horizon at the original spacetime point and to hit the singularity at the original spacetime point, too, results in a shorter proper time than to stay in free fall.

B) Using the engine to accelerate away from the geodesic but not to return to it, i.e. to cross the horizon at the original spacetime point but to hit the singularity at a different spacetime point means that we are no longer allowed to use the reasoning that "geodesics maximize proper time" b/c this is only true for same start- and endpoint.

Therefore (B) could be a better strategy than (A).

My problem is that the "well-known" reasoning seems to be totally wrong. So I wonder if it's not my reasoning which goes astray.

 

[PAllen]

Just a quick comment, I haven't looked at the paper yet.

Concerning the end point, you could possibly rephrase the question as follows. Finding a curve that maximizes proper time between a starting point an end submanifold (space-like). Then the result is that it will be a geodesic orthogonal to the end manifold. Of course it is problematic to view the singularity as a submanifold. Are you discussing the general case or a specific one, say Schwartzschild black hole. In any case what is true is that a curve that is a maximum will be a geodesic if the maximum is attained. But there is no guaranty unless you impose extra restrictions. (Of course I might be wrong on any of these, so take it with a lot of salt)

The set of geodesics orthogonal to a submanifold are likely to reach a given event with only one 4 velocity. If the desired initial condition is different, this argument tells you nothing.

I agree with Tom's questioning of the common argument; as given, it is invalid.

 

[PeterDonis]

Therefore (B) could be a better strategy than (A).

Yes, I agree. And your argument is basically the argument the paper linked to in the OP makes, and backs up with actual calculations.

 

[PAllen]

The gist of the paper's conclusion is that of all the radial free fall paths, there is one with maximal proper time from horizon crossing to reaching the singularity. Not surprisingly it is the one corresponding to being dropped from just above the horizon. Then, for any initial condition for a chosen event inside the horizon, the longest life is achieved by accelerating as quickly as possible to match a horizon droppped trajectory (unless you are already tangent to such a trajectory, in which case you should just free fall), then free fall from there.

 

[martinbn]

My comment was different. If you have a starting event and afour velocity, consider a spacelike hypersurface as near the singularity as you wish. Then the question is look at all timelike curves that start at the given event and end at the hypersurface, which one has the maximum proper time. I agree that the geodesic from that point with the given four velocity need not be the maximum and you can find non-geodesics with longer proper time. But the one that will have the maximum will be a geodesic from the point (with different four velocity) that hits the hypersurface orthogonally. That is if there are no obstructions i.e. focal point along the way.

 

[PeterDonis]

the one that will have the maximum will be a geodesic from the point (with different four velocity) that hits the hypersurface orthogonally

This doesn't make sense. We are specifying a starting event and a 4-velocity as initial conditions. If you go back and pick a geodesic from the same starting event with a different 4-velocity, you are violating the specified initial conditions.

 

[martinbn]

This doesn't make sense. We are specifying a starting event and a 4-velocity as initial conditions. If you go back and pick a geodesic from the same starting event with a different 4-velocity, you are violating the specified initial conditions.

Yes, i keep agreeing with that. It was just a quick comment, perhaps too much of a side comment. My thinking was if you allow acceleration at later times i.e. you allow to change the four velocity at later times, why not allow to change it at the beginning. Of course, I agree that ,this changes the problem.

 

[PAllen]

Yes, i keep agreeing with that. It was just a quick comment, perhaps too much of a side comment. My thinking was if you allow acceleration at later times i.e. you allow to change the four velocity at later times, why not allow to change it at the beginning. Of course, I agree that ,this changes the problem.

Actually, I think your point and the paper's are consistent. Ideally, you instantly change your 4 velocity to the optimal one. In practice, you aim to be tangent with the congruence of optimal geodesics as quickly as possible.

 

[tom.stoer]

Ok.

Have a look at MTW, exercise 31.4: what are we missing?

 

[PAllen]

Ok.

Have a look at MTW, exercise 31.4: what are we missing?

That is consistent with discussion here and the cited paper. Carefullly compare the exercise wording with what I wrote in #6. If you still see a discrepancy, I will explain more, but there is actuallly no discrepancy.

 

[nomadreid]

To inject a question from the lower echelons of understanding of GR: may I conclude that the optimal time is obtained when one abandons the initial conditions of the original article about the motion being radial?

 

[PAllen]

To inject a question from the lower echelons of understanding of GR: may I conclude that the optimal time is obtained when one abandons the initial conditions of the original article about the motion being radial?

No, radial motion is always best to maximize survival, but for most initial conditions of horizon crossing, you benefit from just the right sharp initial radial proper acceleration, followed by free fall. The initial acceleration creates the conditions of an optimal radial free fall path.

 

[nomadreid]

Thank you very much, PAllen. (Rather counter-intuitive, but that's relativity for you :-).) (Counter-intuitive, because one imagines being able to spiral in, perhaps just under a full orbit, for much longer rather than go straight for the jugular.)

 

[PAllen]

Thank you very much, PAllen. (Rather counter-intuitive, but that's relativity for you :-).) (Counter-intuitive, because one imagines being able to spiral in, perhaps just under a full orbit, for much longer rather than go straight for the jugular.)

There are no orbits inside a Schwarzschild BH. In fact, it is a complete misnomer to speak of a radial spatial direction inside the horizon.

[For the more technical, this geometrical description is in terms of the spherical symmetry killing vectors, plus the additional spatial killing direction, that exist inside horizon; I refer to the extra spatial killing direction as axial].

Then, the interior can be considered as time evolutions of 2-sphere X R, a generalization of a cylinder. As with an ordinary cylinder, there is a radius for a circular slice, but the radial direction does not exist in the surface, at all. The radius is just a curvature parameter, or a parameter defined in terms of circumference. Similaraly, for the 3 cylinder, the radius is just a parameter derived from surface area of 2 spheres, and there is no radial spatial direction within the spacetime manifold, at all. The time evolution of the 3 cylinder invovles shrinking the 2-sphere area and stretching the axial direction.

Relating these geometric notions to standard Schwarzschild coordinates, implies that the what is normally called 't' should be relabeled 'z', as it represents axial position along a given 3-cylinder slice. The 'r' coordinate should be relabeled 't', as it describes the time evolution from one killing 3-surface to another. What is all too commonly called radial motion in the interior is really absence of tangential or axial motion in standard coordinates (or geometrically absence of motion relative to a timelike spacetime direction 4-orthogonal to a killing 3-surface). Thus, radial motion really means staying put in a particular geometric sense. Meanwhile, remember that that the SC 'r' coordinate is not a spatial direction at all, and you cannot possibly fire a thrust in this direction.

A realizable radial free fall trajectory ends up with nonzero axial motion the sense described above, on horizon crossing. What such an observer should do to maximize survival time is accelerate in the axial direction, opposite their current axial motion relative to 3-cylinder orthogonal spacetime direction, until they are stationary relative to such a reference trajectory.

One can see, right from the interior SC metric relabeled as described above, that any motion (tangential or axial) relative to such a 'motionless' reference trajectory decreases the progress of proper time as function of t (former r). This means, that you reach a given smaller 2-sphere (closer to singular death) in less proper time. That is, your death comes sooner.

[edit: this post corrects language I as well as others used earlier in the thread, conflating axial acceleration with radial acceleration]

 

[timmdeeg]

To me another counter-intuitive result is shown in Fig. 2 of the article mentioned in the OP. The rockets are dropped at $r=3m$ and start accelerating outwards at $r=2m$. First, if the chosen acceleration is small enough, the survival time starts to exceed that of the free fall case and reaches a maximum. Then however chosing further increased acceleration reduces the survival time, means one can't "buy" additional survival time by this.
Is there an explanation for this without going into technical terms, like killing vectors etc?

 

[PAllen]

To me another counter-intuitive result is shown in Fig. 2 of the article mentioned in the OP. The rockets are dropped at $r=3m$ and start accelerating outwards at $r=2m$. First, if the chosen acceleration is small enough, the survival time starts to exceed that of the free fall case and reaches a maximum. Then however chosing further increased acceleration reduces the survival time, means one can't "buy" additional survival time by this.
Is there an explanation for this without going into technical terms, like killing vectors etc?

Well, without promising to do so (it would be a fair effort to write up), I could justify (but not strictly prove) all key aspects of survival maximization in the following relatively elementary terms (but won't think about doing this if it would not be accessible to you):

1) By arguments from a Kruskal diagram, establish why any realizable infall trajectory ends up with axial motion inside the horizon.
2) By the form of the metric relabeled as described in my prior post, plus algebra and elementary calculus argue that:
a) from any interior event, the proper time maximizing path to the singularity must be a line of constant z (axial coordinate), theta and phi of the relabeled coordinates
b) then it follows that acceleration sufficient to eliminate your axial motion helps survival time, but any further acceleration simple adds axial motion in the other direction, which reduces survival time. Similarly, adding any tangential speed reduces survival time.

If you confirm such a derivation would be accessible to you, I would consider writing it up.

[Note: I have this thread and the linked papers to thank for motivating my own analysis of how to justify all major features of this problem in fairly elementary terms.]

 

[timmdeeg]

Well, without promising to do so (it would be a fair effort to write up), I could justify (but not strictly prove) all key aspects of survival maximization in the following relatively elementary terms (but won't think about doing this if it would not be accessible to you):

1) By arguments from a Kruskal diagram, establish why any realizable infall trajectory ends up with axial motion inside the horizon.
2) By the form of the metric relabeled as described in my prior post, plus algebra and elementary calculus argue that:
a) from any interior event, the proper time maximizing path to the singularity must be a line of constant z (axial coordinate), theta and phi of the relabeled coordinates
b) then it follows that acceleration sufficient to eliminate your axial motion helps survival time, but any further acceleration simple adds axial motion in the other direction, which reduces survival time. Similarly, adding any tangential speed reduces survival time.

If you confirm such a derivation would be accessible to you, I would consider writing it up.

Thanks for your answer. Perhaps others around here would appreciate the derivation, but it wouldn't be of help for me. I'm still struggling with the meaning of axial position and axial motion and failed to find references in the context of the interior of a black hole. The Kruskal diagram shows straight lines of constant 't'. From your explanation in post #16 ('t' should be relabeled 'z') these lines should be labeled 'z' then. But this seems wrong because nothing can move along such a straight line. Would it be possible to describe how a line of constant z looks like in the Kruskal diagram?

 

[PAllen]

To me another counter-intuitive result is shown in Fig. 2 of the article mentioned in the OP. The rockets are dropped at $r=3m$ and start accelerating outwards at $r=2m$. First, if the chosen acceleration is small enough, the survival time starts to exceed that of the free fall case and reaches a maximum. Then however chosing further increased acceleration reduces the survival time, means one can't "buy" additional survival time by this.
Is there an explanation for this without going into technical terms, like killing vectors etc?

I want to make a comment on “outwards”. Once inside the horizon, the spatial direction orthogonal to the tangential plane (defined by theta and phi) is really an axial direction, not a radial direction; the area of 2 spheres is constant along this direction. However, if you fall through the horizon feet first, with no tangential motion, then your feet will be axially displaced from your head, inside the horizon. It is then true that you want to accelerate in the direction away from your feet, for just the right amount, for maximum survival time. Given that you are accelerating in the direction of feet to head, following feet first infall, it is not really wrong to call this outwards; just don’t call it radial.

 

[PAllen]

Thanks for your answer. Perhaps others around here would appreciate the derivation, but it wouldn't be of help for me. I'm still struggling with the meaning of axial position and axial motion and failed to find references in the context of the interior of a black hole. The Kruskal diagram shows straight lines of constant 't'. From your explanation in post #16 ('t' should be relabeled 'z') these lines should be labeled 'z' then. But this seems wrong because nothing can move along such a straight line. Would it be possible to describe how a line of constant z looks like in the Kruskal diagram?

In the region of kruskal exterior to the horizon, lines of constant t are spacelike, and no body can move on such a path. Inside the horizon, these lines are timelike, and they describe the trajectories that maximize proper time from a given interior event to the singularity. These lines, that I relabel z for the interior, are timelike geodesics.

 

[Ibix]

I want to make a comment on “outwards”. Once inside the horizon, the spatial direction orthogonal to the tangential plane (defined by theta and phi) is really an axial direction, not a radial direction; along this direction, areas of 2 spheres remain constant.

Moment of insight - because the coordinate labels from outside no longer make sense. Inside the horizon it's the time-like coordinate that appears in the Schwarzschild-coordinate version of the metric, not the radial coordinate. So 2-spheres inside the horizon change with respect to the "radial" coordinate the same way 2-spheres outside change with respect to the time-like coordinate - they don't. Which is why the space is hyper-cylindrical. Instead they shrink as your time coordinate changes so you must inevitably be crushed.

 

[PeterDonis]

Once inside the horizon, the spatial direction orthogonal to the tangential plane (defined by theta and phi) is really an axial direction, not a radial direction; the area of 2 spheres is constant along this direction.

This is not correct as you state it, because there is no single "spatial direction orthogonal to the tangential plane". There is a 2-manifold orthogonal to the $\theta - \phi$ 2-manifold, and this 2-manifold is what is described by the Kruskal diagram, so I'll call it the Kruskal 2-manifold. At any event, there are a continuous infinity of spacelike directions in the Kruskal 2-manifold that are all orthogonal to the $\theta - \phi$ 2-manifold, since all directions in the Kruskal diagram are orthogonal to that 2-manifold.

What is true is that, at any event, there are only two directions in the Kruskal 2-manifold along which the areas of 2-spheres are constant; these are the "positive" and "negative" directions along the 4th Killing vector field (more precisely, along the integral curve of that KVF that passes through the chosen event). Outside the horizon, these directions are timelike (and "positive" corresponds to "future" while "negative" corresponds to "past"). At the horizon, they are null; inside the horizon, they are spacelike and "axial", to use your term. But these two directions inside the horizon are not the only spacelike directions that are orthogonal to the $\theta - \phi$ 2-manifold.

if you fall through the horizon feet first, with no tangential motion, then your feet will be axially displaced from your head, inside the horizon

To rephrase this, given any integral curve of the 4th KVF, your feet will cross that curve at a more positive Killing parameter than your head does. Outside the horizon, that means your feet pass through the 2-sphere with a given area at a later time than your head does; inside the horizon, it means your feet pass through the 2-sphere with a given area at a more positive axial coordinate than your head does. (In a local inertial frame in which you are momentarily at rest, your feet also cross the given 2-sphere at a later time than your head does, or, to put it another way, your feet cross at a later proper time, according to you, than your head does; but the correspondence between your local inertial frame or your proper time and the global spacetime geometry is much more counterintuitive inside the horizon.)

Given that you are accelerating in the direction of feet to head, following feet first infall, it is not really wrong to call this outwards; just don’t call it radial.

Again, to rephrase this somewhat: outside the horizon, the intuitive meaning of "radial", given that we are already restricting ourselves to the Kruskal 2-manifold (i.e., to the 2-manifold that is everywhere orthogonal to the $\theta - \phi$ 2-manifold), is "orthogonal to the 4th KVF". Outside the horizon, this direction is spacelike; but inside the horizon, it is timelike, and "radially outward" in this sense is past-directed timelike.

However, there is still a "radial" spacelike direction inside the horizon in another sense: if I have some star that is directly overhead, and I free-fall into the hole in such a way that that star remains directly overhead while I am outside the horizon, so that the direction from which I see that star's light coming is "radially outward", then I can still call the direction from which I see that star's light coming "radially outward" inside the horizon. But defining the spacelike vector that corresponds to this direction is more complicated than just finding a vector orthogonal to the 4th KVF.

 

[PeterDonis]

Inside the horizon it's the time-like coordinate that appears in the Schwarzschild-coordinate version of the metric, not the radial coordinate.

No, in Schwarzschild coordinates inside the horizon, the timelike coordinate and the radial coordinate are the same coordinate; the 4th coordinate, instead of being a time coordinate, is a spacelike axial coordinate, as @PAllen says.

So 2-spheres inside the horizon change with respect to the "radial" coordinate the same way 2-spheres outside change with respect to the time-like coordinate - they don't.

No, 2-spheres inside the horizon change with respect to the axial coordinate the same way 2-spheres outside change with respect to the timelike coordinate.

 

[Ibix]

No, in Schwarzschild coordinates inside the horizon, the timelike coordinate and the radial coordinate are the same coordinate; the 4th coordinate, instead of being a time coordinate, is a spacelike axial coordinate, as @PAllen says.

No, 2-spheres inside the horizon change with respect to the axial coordinate the same way 2-spheres outside change with respect to the timelike coordinate.

Ah - got it this time. I hadn't worked all the way through my insight, apparently. So both inside and outside the event horizon we have spheres nested "inside" each other times an axial direction. Outside the event horizon the axial direction is time-like and the radial direction space-like. Inside it's vice-versa. Which is a consequence of (not sure that's quite the right term - maybe "aspect of"?) the interior being non-stationary - a change in time means your "space" uses a 2-sphere of different radius.

 

[PAllen]

This is not correct as you state it, because there is no single "spatial direction orthogonal to the tangential plane". There is a 2-manifold orthogonal to the $\theta - \phi$ 2-manifold, and this 2-manifold is what is described by the Kruskal diagram, so I'll call it the Kruskal 2-manifold. At any event, there are a continuous infinity of spacelike directions in the Kruskal 2-manifold that are all orthogonal to the $\theta - \phi$ 2-manifold, since all directions in the Kruskal diagram are orthogonal to that 2-manifold.

Yes, I was implicitly meaning killing direction orthogonal to the tangential plane (earlier I had been explicit about this, but left too much out here). I agree that any spacelike vector in the (t,z) plane (per my coordinate labeling) is 4-orthogonal to the tangential plane, and can form spacelike direction of an orthonormal tetrad at that point, with tangential directions being the other spacelike directions. Lemaitre coordinates are an example of building orthonormal interior coordinates with a different choice of non-tangential spatial direction at each point. The axial killing direction is no longer manifest, of course.

What is true is that, at any event, there are only two directions in the Kruskal 2-manifold along which the areas of 2-spheres are constant; these are the "positive" and "negative" directions along the 4th Killing vector field (more precisely, along the integral curve of that KVF that passes through the chosen event). Outside the horizon, these directions are timelike (and "positive" corresponds to "future" while "negative" corresponds to "past"). At the horizon, they are null; inside the horizon, they are spacelike and "axial", to use your term. But these two directions inside the horizon are not the only spacelike directions that are orthogonal to the $\theta - \phi$ 2-manifold.

agreed.

To rephrase this, given any integral curve of the 4th KVF, your feet will cross that curve at a more positive Killing parameter than your head does. Outside the horizon, that means your feet pass through the 2-sphere with a given area at a later time than your head does; inside the horizon, it means your feet pass through the 2-sphere with a given area at a more positive axial coordinate than your head does. (In a local inertial frame in which you are momentarily at rest, your feet also cross the given 2-sphere at a later time than your head does, or, to put it another way, your feet cross at a later proper time, according to you, than your head does; but the correspondence between your local inertial frame or your proper time and the global spacetime geometry is much more counterintuitive inside the horizon.)

don't you meant 3d kvf? you have partial theta, partial phi, and axial. As long as we're being precise ...

Also, it seems you are describing headfirst infall, while I was discussing feet first.

Again, to rephrase this somewhat: outside the horizon, the intuitive meaning of "radial", given that we are already restricting ourselves to the Kruskal 2-manifold (i.e., to the 2-manifold that is everywhere orthogonal to the $\theta - \phi$ 2-manifold), is "orthogonal to the 4th KVF". Outside the horizon, this direction is spacelike; but inside the horizon, it is timelike, and "radially outward" in this sense is past-directed timelike.

However, there is still a "radial" spacelike direction inside the horizon in another sense: if I have some star that is directly overhead, and I free-fall into the hole in such a way that that star remains directly overhead while I am outside the horizon, so that the direction from which I see that star's light coming is "radially outward", then I can still call the direction from which I see that star's light coming "radially outward" inside the horizon. But defining the spacelike vector that corresponds to this direction is more complicated than just finding a vector orthogonal to the 4th KVF.

Agreed.

 

[PeterDonis]

So both inside and outside the event horizon we have spheres nested "inside" each other times an axial direction.

No, both outside and inside the horizon we have 2-spheres orthogonal to a 2-manifold, which I called the Kruskal 2-manifold in my response to @PAllen . The "axial" direction is the direction of a Killing vector field on this 2-manifold. Outside the horizon this KVF is timelike; inside the horizon it's spacelike. (On the horizon it's null.)

Outside the event horizon the axial direction is time-like and the radial direction space-like. Inside it's vice-versa.

I'm not sure what you mean by the "radial" direction. Here's how I would say it: both outside and inside the horizon there are an infinite family of curves (in the Kruskal 2-manifold) that are orthogonal to the "axial" KVF. Outside the horizon these curves are spacelike. Inside the horizon they are timelike. (On the Kruskal diagram these curves are straight lines that pass through the origin of the diagram.) The horizon itself is the pair of null curves in this family which, since they are null, are both orthogonal to and parallel to the KVF at the horizon--i.e., the KVF at the horizon, since it is null, is orthogonal to itself.

Which is a consequence of (not sure that's quite the right term - maybe "aspect of"?) the interior being non-stationary

"Non-stationary" means there is no timelike KVF. That's true inside the horizon, but I wouldn't say the other stuff above is a "consequence" of it; it's just the same thing stated in different words.

a change in time means your "space" uses a 2-sphere of different radius

"Space" is ambiguous here. I would say that inside the horizon, there are no timelike curves along which the radius of the 2-spheres remains constant. That's just another way of stating that there is no timelike KVF.

 

[PeterDonis]

don't you meant 3d kvf? you have partial theta, partial phi, and axial.

No, there are 4 KVFs; 3 for spherical symmetry, plus one more. The third spherical symmetry KVF looks more complicated in terms of the $\theta, \phi$ coordinates, but it's there. (A coordinate-free way to see that it's there is to note that the group corresponding to spherical symmetry, SO(3), has three generators--heuristically, corresponding to rotation about any of three orthogonal axes.)

 

[PeterDonis]

it seems you are describing headfirst infall

Yes, you're right, I should have switched "head" and "feet" around to be consistent with your description.

 

[Ibix]

@PeterDonis - thanks. I think I follow, and I think I'm just using confusing terminology because I'm still sorting through the wreckage of my earlier (mis)understanding. I'll go away and digest.

 

[timmdeeg]

In the region of kruskal exterior to the horizon, lines of constant t are spacelike, and no body can move on such a path. Inside the horizon, these lines are timelike, and they describe the trajectories that maximize proper time from a given interior event to the singularity. These lines, that I relabel z for the interior, are timelike geodesics.

If I understood your post #18 correctly a curve with maximally proper time inside the horizon requires a constant axial coordinate (t-coordinate in the Kruskal diagram) which means that there is no axial motion then. These straight lines described by constant t are passing through the origin. Thus being outside the light cone they can't be timelike. What am I missing?

I still have no notion how the red timelike curve in Fig. 2 (maximally proper time) of the article mentioned in the OP would look like in a Kruskal diagram.

 

[PAllen]

If I understood your post #18 correctly a curve with maximally proper time inside the horizon requires a constant axial coordinate (t-coordinate in the Kruskal diagram) which means that there is no axial motion then. These straight lines described by constant t are passing through the origin. Thus being outside the light cone they can't be timelike. What am I missing?

I still have no notion how the red timelike curve in Fig. 2 (maximally proper time) of the article mentioned in the OP would look like in a Kruskal diagram.

They have an angle greater than 45 degrees to the horizontal axis in the kruskal diagram. This defines what is timelike, by construction in a kruskal diagram. I have no idea what you mean about being outside the light cone. It is true that these all go through the origin, and this is why no free fall crossing trajectory would be following them.

Consider some arbitrary horizon crossing trajectory. Then, once inside the horizon by a tiny amount, there is such a line intersecting this trajectory just inside the horizon. If instant accelerations were allowed, the optimal survival strategy would be instantly change your 4 velocity to match this constant t/z line.

 

[timmdeeg]

Consider some arbitrary horizon crossing trajectory. Then, once inside the horizon by a tiny amount, there is such a line intersecting this trajectory just inside the horizon. If instant accelerations were allowed, the optimal survival strategy would be instantly change your 4 velocity to match this constant t/z line.

Got it. Yes this line is inside the light cone. I've been misinterpreting the interior, sorry for the confusion and thanks for clarifying.

 

[PeterDonis]

once inside the horizon by a tiny amount, there is such a line intersecting this trajectory just inside the horizon

Actually, there are an infinite number of such lines; which one you can hit will depend on how hard you can accelerate. If you look at the curves of constant axial coordinate inside the horizon, they are lines radiating from the origin up towards the singularity, at all possible angles in the open set of plus 45 degrees to minus 45 degrees. But the horizon itself is the plus 45 degree line that forms the limit point (on that "side"--the other side is limited by the antihorizon) of this family of curves, so basically the strategy you are describing is: as soon as you are inside the horizon, accelerate to make your worldline as close to the horizon line as possible. How close you can make it will depend on how hard you can accelerate, and how close to the origin of the Kruskal diagram you are when you cross the horizon (the closer you are, the harder you would have to accelerate to hit a given curve of constant axial coordinate). If we allow instant acceleration, then you could make your worldline only infinitesimally different from the horizon line.

 

[PAllen]

Actually, there are an infinite number of such lines; which one you can hit will depend on how hard you can accelerate. If you look at the curves of constant axial coordinate inside the horizon, they are lines radiating from the origin up towards the singularity, at all possible angles in the open set of plus 45 degrees to minus 45 degrees. But the horizon itself is the plus 45 degree line that forms the limit point (on that "side"--the other side is limited by the antihorizon) of this family of curves, so basically the strategy you are describing is: as soon as you are inside the horizon, accelerate to make your worldline as close to the horizon line as possible. How close you can make it will depend on how hard you can accelerate, and how close to the origin of the Kruskal diagram you are when you cross the horizon (the closer you are, the harder you would have to accelerate to hit a given curve of constant axial coordinate). If we allow instant acceleration, then you could make your worldline only infinitesimally different from the horizon line.

I thought I said exactly that in the ideal case. I said once inside by any tiny amount, if you can instantly accelerate to the constant z/t world line intersecting this event on your world line, that is the best you can do from there. The 'sooner' you do this the better, so there is an unreachable LUB of proper time (from crossing to singularity) for a crossing trajectory. But you never want to overshoot. That is, suppose you wait 1 microsecond after crossing, and you have a choice of instantly following the constant t/z line intersecting that event, versus following a closer to lightlike geodesic from that event (that would also exist). The latter would be an inferior choice. It is true that the LUB of such choices has you following ever closer to the horizon, but there is no path realizing the LUB. For any chosen thrust, it should be applied just long enough that your 4-velocity matches an intersecting constant z/t line, then it should be cut off. This must be true because the constant z/t line maximizes proper time from that event to the horizon (over all possible initial conditions at that event).