Proper time to reach the singularity - rocket acceleration

[timmdeeg]

I was assuming that the proper time lapse between $r=2M$ and $r=0$ increases with increasing acceleration outwards.

According to this paper

https://arxiv.org/pdf/0705.1029v1.pdf Fig.2.

it turns out however that the proper time to reach the singularity is longer with low acceleration but shorter with high acceleration compared to free fall. And identical to free fall choosing a certain acceleration in between. The acceleration is switched on in each case just after having crossed the event horizon and is kept constant then.

It seems there are two effects working in opposite directions. One I could imagine is that with increasing acceleration the wordline comes closer and closer to the null (proper time) geodesic.
I would appreciate your comments.

 

[a-nobody]

The paper is 100% correct.
The singulatity is in your future so any acceleration towards it will bring you at a later time there.
A simple Penrose diagram will enlighten you why this is the case!

 

[timmdeeg]

The singulatity is in your future so any acceleration towards it will bring you at a later time there.
A simple Penrose diagram will enlighten you why this is the case!

If you say "towards it" you mean acceleration inwards. No, in this case the proper time to reach the singularity is less than in free fall, see Fig. 5.

Instead, I was talking about acceleration outwards. Why has the proper time lapse a maximum value depending on the amount of acceleration as shown in Fig. 2?

 

[a-nobody]

No, I mean towards it. There is no "inbound" or "outbound" acceleration inside a black hole, so to speak. It's just a play of words.
Like the paper says, choosing a different worldline shortens or lengthens the time we are reaching the singularity accordingly.

 

[Orodruin]

It should be the same issue as that of finding the shortest path from a point to a line in Euclidean space. There are many geodesics (in the Euclidean case, straight lines) that reach the line, but only one of those minimize the length. By the same argumentation, there are many time-like geodesics that reach the singularity given a point inside the horizon. One of those will maximize proper time, but it is not at all a given that a “faller” would be on that geodesic. The maximization of proper time would require the faller to instantaneously accelerate to be on the maximizing geodesic and then turn the rockets off.

Of course, if infinite acceleration is not available or undesirable, an accelerating world line may very well lead to a larger proper time than a non-maximizing geodesic. Just as a non-straight curve leading to the line can be shorter than a straight line that is not headed straight for the line in Euclidean space.

 

[pervect]

The principle of maximal aging is oversimplified, but it suggests that the curve that maximizes proper time should be a geodesic. That's the simple approach, but it's not totally correct. But it turns out that a geodesic with a specific initial condition (of zero energy at infinity, corresponding to falling from at rest just outside the event horizon) is a global maximum.

So the argument is actually not a good one (being too simple), but it turns out to give the right answer anyway, even though it's flawed argument.

It would certainly be helpful to look at the whole set of geodesics with different values of the E parameter, and demonstrate that E=0 is the one with the maximum proper time, which is something I haven't actually done but I think should be true. One can also compare non-geodesic curves, but it shouldn't be too surprising that a global maximum for time exists, and that since it does exist, it must also be a local maximum, which implies that it must meet the appropriate Euler-Lagrange equations for a stationary action.

"Not too surprising" is obviously not the same as "rigorously proven", but if you want rigor, that's what the paper is for, i.e. deomsntrating rigorously that there is a global maximum, not assuming that there is one.

 

[timmdeeg]

It would certainly be helpful to look at the whole set of geodesics with different values of the E parameter, and demonstrate that E=0 is the one with the maximum proper time, which is something I haven't actually done but I think should be true. One can also compare non-geodesic curves, but it shouldn't be too surprising that a global maximum for time exists, and that since it does exist, it must also be a local maximum, which implies that it must meet the appropriate Euler-Lagrange equations for a stationary action.

If we stay with geodesics then as MTW mention in exercise 31.4 "the geodesic of longest proper time lapse between $r=2M$ and $r=0$ is the radial geodesic." The radial non-geodesic curves shown in Fig. 2 are related to accelerating outwards. The math yields a maximum for the proper time here. But why? Accelerating outwards slightly increases the time lapse compared to free fall but this is the reverse with stronger acceleration..
I would appreciate any hint as to why the time to reach the the singularity isn't just increasing with increasing acceleration.

 

[PAllen]

If we stay with geodesics then as MTW mention in exercise 31.4 "the geodesic of longest proper time lapse between $r=2M$ and $r=0$ is the radial geodesic." The radial non-geodesic curves shown in Fig. 2 are related to accelerating outwards. The math yields a maximum for the proper time here. But why? Accelerating outwards slightly increases the time lapse compared to free fall but this is the reverse with stronger acceleration..
I would appreciate any hint as to why the time to reach the the singularity isn't just increasing with increasing acceleration.

Read the full context in MTW more carefully. There are infinitely many distinct radial geodesics with different initial conditions (e.g. free fall from different heights). The maximizing radial geodesic is one that is not actually available to any free falling horizon crosser in that fully extended it only reaches the horizon, but doesn't cross it. Free falling from stationary just above the horizon gets you close to this, but all horizon crossers benefit from a sharp acceleration so as to converge to a maximizing radial geodesic, followed by free fall. This was all discussed recently in another thread.

I should also point out a nuance - the set of maximizing geodesics form a congruence that fills the interior. Thus there is one near any event on an interior world line that you can adjust your 4 velocity to match (if you are not already following one).

 

[a-nobody]

If we stay with geodesics then as MTW mention in exercise 31.4 "the geodesic of longest proper time lapse between $r=2M$ and $r=0$ is the radial geodesic." The radial non-geodesic curves shown in Fig. 2 are related to accelerating outwards. The math yields a maximum for the proper time here. But why? Accelerating outwards slightly increases the time lapse compared to free fall but this is the reverse with stronger acceleration..
I would appreciate any hint as to why the time to reach the the singularity isn't just increasing with increasing acceleration.

Outwards in what sense you mean?
That you are free falling with a spherical cloud of dust particles and suddenly you are thrusting your rockets and the spherical cloud goes down below your feet?
Even so, your vector still points at the singularity r=0.
You must realize that different geodesics lead to different proper times and the situation is quite unlike the exterior solution because the singularity is always in your future.

 

[timmdeeg]

Free falling from stationary just above the horizon gets you close to this, but all horizon crossers benefit from a sharp acceleration so as to converge to a maximizing radial geodesic, followed by free fall. This was all discussed recently in another thread.

Thanks, I will search for this thread.
Here https://arxiv.org/pdf/0705.1029v1.pdf Fig.1 shows that the proper time from the horizon to the singularity decreases with increasing r-coordinate of the begin of the fall.
I'm still puzzled by said maximum of proper time of the non-geodesic worldlines show in Fi.2.

 

[timmdeeg]

Outwards in what sense you mean?

The article describes it. Look at the worldlines Fig. 2. Here the rocket is accelerating outwards. Accelerating inwards is shown in Fig.5. Of course the rocket moves towards $r=0$ in both cases, there is no other choice as the light cone is dipped inwards below $r=2M$.

 

[PAllen]

Thanks, I will search for this thread.
Here https://arxiv.org/pdf/0705.1029v1.pdf Fig.1 shows that the proper time from the horizon to the singularity decreases with increasing r-coordinate of the begin of the fall.
I'm still puzzled by said maximum of proper time of the non-geodesic worldlines show in Fi.2.

Here is the other thread, based in part on a different paper exploring similar issues.
https://www.physicsforums.com/threa...l-time-when-falling-into-a-black-hole.926074/

 

[timmdeeg]

Here is the other thread, based in part on a different paper exploring similar issues.
https://www.physicsforums.com/threa...l-time-when-falling-into-a-black-hole.926074/

Thank you.