Maximizing survival time when falling into a black hole

[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).

 

[timmdeeg]

It is really interesting that there is one very certain line of constant axial coordinate which maximizes the proper time, the line with a constant acceleration "outwards" of a = 0,5 as mentioned in said article.

 

[PeterDonis]

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.

Yes, agreed. By "make your worldline as close to the horizon as possible", I really meant "make your worldline the constant t/z line that is as close to the horizon as possible".

 

[timmdeeg]

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.

I think this helps intuition. Ever closer to the Null geodesic means closer to a geodesic with zero proper time and thus is consistent with the notion that by this the proper time is decreasing..

 

[PeterDonis]

It is really interesting that there is one very certain line of constant axial coordinate which maximizes the proper time, the line with a constant acceleration "outwards" of a = 0,5 as mentioned in said article

Lines of constant axial coordinate inside the horizon are timelike geodesics; they have no acceleration. Which part of the article are you referring to?

Ever closer to the Null geodesic means closer to a geodesic with zero proper time

No, that's not correct. Consider the analogous case in Minkowski spacetime: timelike geodesics radiating outward from the origin and ending on the hyperbola $t^2 - x^2 = 1$ (in units where $c = 1$). All of these geodesics have the same proper time from origin to hyperbola; the ones "closer" to the null lines that are the asymptotes to the hyperbola do not have shorter proper times.

I haven't tried to prove it, but I think an analogous statement is true for the set of timelike geodesics we have been discussing for a black hole inside the horizon: all of the timelike geodesics from the origin of the Kruskal diagram to the singularity have the same proper time between those two events.

 

[PAllen]

I haven't tried to prove it, but I think an analogous statement is true for the set of timelike geodesics we have been discussing for a black hole inside the horizon: all of the timelike geodesics from the origin of the Kruskal diagram to the singularity have the same proper time between those two events.

Yes, this is true, you just need to write down the integral to verify this.

 

[PAllen]

No, that's not correct. Consider the analogous case in Minkowski spacetime: timelike geodesics radiating outward from the origin and ending on the hyperbola $t^2 - x^2 = 1$ (in units where $c = 1$). All of these geodesics have the same proper time from origin to hyperbola; the ones "closer" to the null lines that are the asymptotes to the hyperbola do not have shorter proper times.

I think @timmdeeg is referring to my point, expanded as follows:

Choose any interim event whatever. There is a constant axial coordinates geodesic through it, that maximizes proper time from this event to the singularity. There is also a null geodesic in the + axial direction from this event to the singularity (this is a 45 degree line to the right in a kruskal diagram). Proper time along this is obviously zero. Paths in between these lines will have intermediate proper time to the singularity, thus less than maximal. You can understand this as being too close to a light like path.

 

[PeterDonis]

I think @timmdeeg is referring to my point, expanded as follows

Ah, ok. Yes, I agree with this.

 

[timmdeeg]

You can understand this as being too close to a light like path.

Actually I was thinking of the twin paradox. The closer the worldline of the traveling twin comes to the lightlike geodesic the more the proper time of the twin slows down.

 

[PeterDonis]

Actually I was thinking of the twin paradox.

In the twin paradox, both twins start and end at the same event. That's not the case in the scenario we are discussing in this thread. The worldlines being compared end on the singularity, but they don't all end at the same event on the singularity.

 

[timmdeeg]

In the twin paradox, both twins start and end at the same event. That's not the case in the scenario we are discussing in this thread. The worldlines being compared end on the singularity, but they don't all end at the same event on the singularity.

Yes, the scenario is different. The reasoning was that 'worldline closer to the Null-geodesic' should reduce its the proper time.

 

[PeterDonis]

The reasoning was that 'worldline closer to the Null-geodesic' should reduce its the proper time.

Yes, and that reasoning is not correct in general. It is only correct with appropriate qualifications.

 

[Ibix]

I think I'm drifting off topic, so if I'm opening a can of worms I'm happy to start a separate thread.

"Space" is ambiguous here.

The context of the quote was that I was talking about space inside the event horizon. Is it really ambiguous?

Outside the horizon, at any point you can pick a basis whose time axis is aligned to the time-like Killing vector at that point. Then you can "make nearby axes point to one another" and construct a set of three volume that you can reasonably call "space" indexed by time. It's just a particular set of three-volumes picked out by symmetry.

Inside the event horizon you can't do that because there's no time-like Killing vector field. But there are four space-like ones, three of which lie in the 2-sphere orthogonal to the Kruskal plane and one of which lies in the Kruskal plane. Can you not reasonably define "time" to be the remaining direction, then apply the same methodology from outside to stitch together points into "space"? Isn't that just a different way to use symmetry to pick out an "obvious" definition of space?

 

[PeterDonis]

I was talking about space inside the event horizon. Is it really ambiguous?

Yes. In a stationary spacetime, or stationary region of spacetime, such as outside the horizon, you can at least somewhat justify a preferred notion of "space" based on the spacelike surfaces that are orthogonal to the timelike Killing vector field. (In Schwarzschild spacetime outside the horizon, these would be the spacelike surfaces of constant Schwarzschild coordinate time.) However, the region inside the horizon is not stationary, so there is no way to pick any notion of "space" that is not arbitrary.

Can you not reasonably define "time" to be the remaining direction, then apply the same methodology from outside to stitch together points into "space"?

No, because the timelike curves orthogonal to the "space" defined this way intersect (these timelike curves are just the curves of constant axial coordinate that @PAllen described, and they all intersect at the origin of the Kruskal diagram) and are not at rest relative to each other, so treating them as distinct "points in space", which is what the construction you describe requires, does not work. (Outside the horizon you don't have this problem, because things are switched around: the "axial" KVF is timelike and the curves of constant "axial" coordinate are spacelike, so as long as we stay outside the horizon, the fact that the "spaces" all intersect at the origin of the Kruskal diagram is OK, since that origin is on the horizon; and all of the integral curves of the timelike KVF are at rest relative to each other, so each one can be considered as a distinct "point in space".)

 

[PeterDonis]

Isn't that just a different way to use symmetry to pick out an "obvious" definition of space?

Oh, and also, the "spaces" defined this way inside the horizon are not spanned by spacelike geodesics: in other words, if we leave out the two angular coordinates, the curves that pick out each "space" (which are hyperbolas on the Kruskal diagram, orthogonal to the curves of constant axial coordinate) are not geodesics. Outside the horizon, the spacelike curves (curves of constant Schwarzschild coordinate time) are geodesics; it is the timelike curves marking out the "points in space" which are not geodesics, but that's OK since it just means observers whose worldlines are those curves have nonzero proper acceleration.

 

[PAllen]

I still think the killing vector based foliation of the interior is the least arbitrary and most informative. Obviously, no physics is affected by this choice versus another. The fact that the timelike geodesics intersect at the kruskal origin is of no import, because this is a horizon point, not an interior point. The whole interior is coverered without intersection by a foliation of killing hyper cylinders. Among other things, this clarifies that the centers of 2 spheres of symmetry are not part of the manifold in a way easy to see by analogy to ordinary cylinders. Note that the standard cosmology foliation has a timelike congruence of geodesics that approach intersection at the origin, and spaces are not static, because they can’t be. Also the timelike congruence geodesics in cosmology are not at mutual rest by any normal criteria - they show Doppler, no reasonable comparison via parallel transport has them at rest, and in the flat limit, they are not at rest. There are thus many similarities between the killing based interior foliation and the standard cosmology foliation. One has an expansion of spatial 3 spheres or hyperbolic 3 surfaces, the other has a contraction of 3 cylinders.

[edit: Also, few of the common GR foliations of exact solutions are geodesic foliations:

1) The exterior is not because 2 spheres are non spacelike geodesics.
2) The standard cosmological foliation is not built from spacelike geodesics.

In fact, a foliation built from spacelike geodesics is almost uniquely a Fermi-Normal foliation, which is informative for many purposes, but is not useful as a global or large scale foliation for most useful solutions. It can cover only pieces of the exterior or interior in kruskal geometry, and it (if I remember right) can only cover to the square root of cosmological horizon distance in standard cosmologies. ]

 

[Ibix]

Thanks @PeterDonis.

I think @PAllen makes good points. I would only make one additional point (with a degree of sophistry, it must be said) which is that the crossing point of the time-like curves doesn't exist in a realistic black hole.

I think I see the point about there needing to be spacelike geodesics in my spacelike volume to be able to call it "space" rather than a space-like foliation. If the natural concept of "going straight ahead not advancing in time" takes you out of the volume then it's a bit difficult to argue that the volume represents a "slice of spacetime at a single time".

Would you argue that the volumes in Minkowski spacetime defined by equal Rindler time aren't "space" on this basis?

 

[PAllen]

In cosmology, pick any comoving timelike geodesic. Then pick any space like geodesic 4 orthogonal to it. This geodesic will not remain in a slice of constant cosmological time. No one pretends this makes the standard cosmological foliation useless or uninformative.

 

[PeterDonis]

I still think the killing vector based foliation of the interior is the least arbitrary and most informative

Ultimately this is a matter of preference, since as you note no physics is affected by a coordinate choice. I agree that the points you bring up are valid ones to consider when making such a choice.

 

[PeterDonis]

crossing point of the time-like curves doesn't exist in a realistic black hole.

Yes, that's correct; it only exists in the maximally extended manifold, which is not physically realistic.

 

[PeterDonis]

Would you argue that the volumes in Minkowski spacetime defined by equal Rindler time aren't "space" on this basis?

These correspond to "space" outside the horizon, not inside it. And this particular foliation is a geodesic one (in flat spacetime, not the curved spacetime outside a black hole).

 

[PAllen]

Just another thought on the importance of spatial 3-cylinders in the SC BH interior. In flat spacetime, and in the asymptotically flat BH exterior it is impossible construct/embed a spacelike 3-cylinder at all. That this is possible in the interior is telling us something significant about the difference in geometry between the interior and exterior.

 

[PeterDonis]

in the asymptotically flat BH exterior it is impossible construct/embed a spacelike 3-cylinder at all

If we don't include the origin of the Kruskal diagram in the "exterior", this is true. But spacelike surfaces passing through the origin of the Kruskal diagram, and extending to infinity in both directions (the right-hand wedge and the left-hand wedge of the Kruskal diagram) are spacelike 3-cylinders. The origin is on the horizon, so it's not strictly speaking "exterior", but it's not "interior" either.

In the more realistic geometry of an actual black hole formed by gravitational collapse, the point corresponding to the origin of the Kruskal diagram is not there, nor is any of the left-hand wedge (the region occupied by the collapsing matter is there instead), so in that geometry, yes, the only spacelike 3-cylinders are in the BH interior. (These 3-cylinders "end" in one direction inside the collapsing matter, but they extend infinitely in the other direction, at least for the classical case where Hawking radiation is excluded.)

 

[PAllen]

If we don't include the origin of the Kruskal diagram in the "exterior", this is true. But spacelike surfaces passing through the origin of the Kruskal diagram, and extending to infinity in both directions (the right-hand wedge and the left-hand wedge of the Kruskal diagram) are spacelike 3-cylinders. The origin is on the horizon, so it's not strictly speaking "exterior", but it's not "interior" either.

By geometric 3-cylinder, I am including the fact that the area of 2-sheres is constant. So, no, the best you can do including the whole horizon is to have a 3-cylinder whose axis is lightlike. In fact, what you are describing, as you get further and further from the origin, becomes geometrically more and more just concentric spheres about a common origin (as the spacetime becomes asymptotically flat).

 

[PeterDonis]

By geometric 3-cylinder, I am including the fact that the area of 2-sheres is constant.

Ah, ok.