# A Maximizing survival time when falling into a black hole

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1. Nov 28, 2017

### PAllen

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.

Last edited: Nov 28, 2017
2. Nov 28, 2017

### Staff: Mentor

Ah, ok. Yes, I agree with this.

3. Nov 28, 2017

### timmdeeg

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

4. Nov 28, 2017

### Staff: Mentor

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.

5. Nov 28, 2017

### timmdeeg

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

6. Nov 28, 2017

### Staff: Mentor

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

7. Nov 29, 2017

### 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.
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?

8. Nov 29, 2017

### Staff: Mentor

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.

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".)

9. Nov 29, 2017

### Staff: Mentor

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.

10. Nov 29, 2017

### 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. ]

Last edited: Nov 29, 2017
11. Nov 29, 2017

### 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?

12. Nov 29, 2017

### 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.

13. Nov 29, 2017

### Staff: Mentor

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.

14. Nov 29, 2017

### Staff: Mentor

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

15. Nov 29, 2017

### Staff: Mentor

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).

16. Nov 29, 2017

### 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.

17. Nov 29, 2017

### Staff: Mentor

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.)

18. Nov 30, 2017

### PAllen

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).

Last edited: Nov 30, 2017
19. Nov 30, 2017

Ah, ok.