# Spaceship trapped inside event horizon - serious flaw

1. Jun 4, 2010

### tom.stoer

Perhaps you know the question what the captain of a spaceship trapped inside the black hole event horizon shall do in order to maximize the time left to being sucked into the singularity.

I know the following idea (and I used to believe it over years :-)

The geodesic equation of "free-fall" motion and the proper time calculation are both based on the same integral, namely

$$S[C] = \int_C ds$$

As geodesics maxime S and as the question is to maxime proper time, the answer is simply: "free-fall"! So the caption should not start the engine and try to accelerate outwards, try to reach an orbit around the singularity or something like that.

This reasoning is well-know from the twin paradox, too. But thinking about it in more detail shows that the two scenarios are not equivalent.

For the twin paradox the situation is as follows: both twins start from one spacetime point

$$C_1(x^0=0) = C_2(x^0=0)$$

and they will meet again after some time (coordinate time T) in the same spacetime point

$$C_1(x^0=T) = C_2(x^0=T)$$

$$x^0$$ is the coordinate time in one reference frame.

For the spaceship the situation is different: one compares curves (labelled by an index a) all starting start from one spacetime point

$$C_a(x^0=0)$$

but the end of the curves is not always in the same spacetime point! Instead one has

$$C_a(x^0=T_a) \to x^\mu = (x^0=T_a, 0, \theta, \phi)$$

where I have used polar coordinates; the singularity is located at radial coordinate 0, angles are unspecified. Any other coordinate system will do as well.

Now the problem is that we have to check this larger class of curves, namely all curves which eventually meet the singularity. But this applies to all physically allowed curves. That means that we can classify the set of all those curves by the coordinate time when they will meet the singularity. Of course we can calculate the proper time and the coordinate time when a geodesic (free-fall curve) will meet the singularity; let's call this coordinate time $$T^0$$. And we know that for all other curves meeting the singularity at the same coordinate time $$T_a = T^0$$, the geodesic will be the one which maximizes the proper time for this specific subset of curves.

But how can we prove that all other curves (resulting from some acceleration of the spaceship) meeting the singularity at a different coordinate time $$T_a \neq T^0$$ have smaller proper time?

Last edited: Jun 4, 2010
2. Jun 4, 2010

### Passionflower

Good topic.

I assume you are familiar with http://arxiv.org/abs/0705.1029

But, as an interesting question, are you sure that different coordinate times for a singularity excludes the idea that everything will enter the singularity at the same time? Perhaps it is just an effect of the chosen chart? (think spacelike vs. timelike singularity)

Last edited: Jun 4, 2010
3. Jun 4, 2010

### tom.stoer

No, I just found this paper a couple of minutes ago :-)

I was not thinking about different charts, but only about two different world lines, one for free-fall, one for accelerated motion. Both world lines will eventually hit the singularity, but this will happen at two different spacetime points (regardless of the chart). Transforming between different charts this may change the way we have to talk about space and time; perhaps what I called "time" is rather a "spacelike coordinate" insde the black hole; using Kruskal coordinates or something like that means that we should forget about "space" and "time" at all. Nevertheless the singularity is a one-dim. object in a Kruskal- or Penrose diagram. So starting from one spacetime point inside the Horizon one can hit the singularity at different points along a one-dim line.

No I am wondering
a) if all these curves have the same proper time (I guess not)
b) how one can show which curves maximizes its proper time

4. Jun 4, 2010

### tom.stoer

Look at the diagrams in the reference you cited. The dashed lines show exactly what I was thinking about: the curves always meet the singularity at r=0 but at different t (here in Eddington-Funkelstein coordinates which allow us still to call it "time").

Interesting paper!

5. Jun 4, 2010

### George Jones

Staff Emeritus
6. Jun 4, 2010

### tom.stoer

Yes, it is essentially the same discussion.

The problem seems to be that mostly one does not specify the initial conditions carefully. For an observer at rest at the event horizon free-fall maximizes his proper time. For an observer fallng inwards = crossing the event horizon it is preferrable to accelerate outwards and turn off the engine once he has "compensated his initial condition" and returning to free-fall.

The situation is really very complicated as can be seen from the paper cited above.

My point was not so much to understand all the details but I was hit by the fact that I misunderstood the problem for years! I was always thinking that it's like the twin paradox - it's not! In the twin paradox both twins meet eventually at the same event, whereas two spaceships need not meet again necessarily because they may arrive at the singularity at two different coordinate times.

So the setup of the two problems is different which I overlooked completely.