# Falling into a Black Hole

1. Dec 19, 2006

### Wallace

I have an interesting question for you all to ponder. The answer can be found in a few places, but not a lot, but I have an inkling that the standard answer given is wrong, or at least, not comprehensive.

The question is this:

You and a friend fall into a black hole together (lets make it a non-rotating neutrally charge black hole for simplicity). At the point exactly where you pass the event horizon you fire a handy dandy rocket pack that you earlier strapped onto you back. You point yourself away from the center of the black hole, and hence are accelerating outwards. Your friend does nothing except adopt a zen-like calm about their approaching doom in the central singularity.

Now obviously your jetpack cannot get you back outside the event horizon, but intuitively you might think it could delay your inevitable meeting with the singularity. The question I have is who will experience the most proper time (who's watch will tick the most seconds) before they hit the central singularity.

Apologies for being a little cagey, but I'd like to see what the smart people here think about this, and what references you may be aware of that discuss this problem, before I reveal what I think is the 'right' answer.

2. Dec 19, 2006

### MeJennifer

Well once you pass the event horizon you really cannot point yourself away from the center since all directions are now pointed towards the singularity. Hence, any acceleration is helping you get to the singularity faster. At least that is my understanding.

3. Dec 19, 2006

### Chris Hillman

Another ambush question?

Hi, Wallace,

Your question is not as well formulated as might be desired, but it appears to resemble a discussion we had here only a few weeks ago, comparing elapsed times measured by slowfall and Lemaitre observers. Here, slowfall observers fall in radially "from rest at infinity", but the slowfall observers radially accelerate outward with the obviously "interesting" nonzero magnitude $$m/r^2$$, while the Lemaitre observers are freefalling. Note that static observers accelerate radially outward with acceleration of magnitude
$\frac{m/r^2}{\sqrt{1-2m/r}}$

Your challenge also appears to resembe a problem in MTW, which every serious student knows by heart. See problem 31.4.

What do you believe is the "standard answer" to these questions and why? (Why the belief, that is; why the reason only if your answer really does disagree with MTW--- I am guessing from the way you phrased the question that it does not.)

MeJennifer: think of attaching a tiny spatial hyperplane orthogonal to the world line of the infalling geodesic, at each event on that geodesic. In this hyperplane, one can define the acceleration vector and then it can be said to point radially outward, as happens in the cases of the slowfall and static observers mentioned above. What is this acceleration vector? Well, if the tangents to the world line belong to the timelike unit vector field $$\vec{X}$$, the acceleration is just $\nabla_{\vec{X}} \vec{X}$
(You got the right answer to the "challenge", though.)

Last edited: Dec 19, 2006
4. Dec 19, 2006

### Wallace

Not exactly. To someone crossing the horizon, there is nothing special about it. In fact you would have to do the calculations in advance to know when you've crossed it and it's time to turn the rocket on. Say you eyeball a distant star as you fall in that lies on the straight line radially out from the black hole passing through you. As you fall through the event horizon you wouldn't notice anything strange happening to that star (apart from it blue-shifting by an increasing amount, but this is a smooth function of your distance from the black hole centre with nothing odd happening and the horizon). You can still look around and define an 'out' and 'in' direction while inside the event horizon, and make sure your rocket is boosting you 'out'.

Edit: Chris posted as I typed.

Sorry if this comes across as an 'ambush' question. I'm not trying to trick everyone and make myself look smart, apologies if I gave that impression! I just wanted to avoid slanting the discussion to a narrow focus before stating the way I'm thinking about this problem. Clearly this approach is not appreciated, fair enough.

Anyway, thanks for the input Chris. I don't have MWT (Isn't that the order not MTW?), though as you suggest it's a pretty standard text. There is a problem in Hartle that is essentially the question I'm asking, though I don't have it on hand to give you the exact question number.

So, my thoughts on this. Hartle says that the free-falling observer will have the maximum proper time since they are on a geo-desic, though dosn't calculate in the solutions what an accelerated observer would experience. He sums this up with 'the harder you struggle, the shorter you live'. However, I've always been under the impression that geodesics maximize the proper time between two points $$(t_1,x_1)$$ and $$(t_2,x_2)$$. However in this case, while both observers end up at the same x (the central singularity), nothing prevents them from doing so at different co-ordinate times. The start and end points are not the same and hence a simple appeal to geo-desics cannot answer the problem.

If the MWT problem (or more specifically the solution) goes into this I'd be interested to hear it. I'll try and grab a copy from the library tomorrow and look up that problem though. Thanks for pointing it out.

Last edited: Dec 19, 2006
5. Dec 19, 2006

### Garth

Wallace You'll have to search for MTW!

Garth

6. Dec 19, 2006

### Chris Hillman

Hi, Wallace,

Not really, I just happened to come to this thread after reading the one asking for "ambush questions" for a visit by an unnamed physicist to some classroom. (So, whatever happened to him, anyway? did he survive?)

If you mean that you wanted to... uhm... shape the discussion, that can make sense for pedagogical purposes... oh dear, now I've probably made teaching sound like cold-war style Bulgarian propaganda. Never mind, let's start over.

MTW (alphabetical order, by age I think it would be WMT although I won't say which age is longest... highest... er... weightiest?). Hartle is also a good textbook, I think, but unfortunately the only widely used textbook of which don't yet possess an example! (This is rather like obsessively collecting the highest denomination bills from All the World's Currency!) So I haven't studied it as closely as the others.

That's the standard answer. I guess you know, by the way, that all four of these people were participants in the Golden Age of Relativity? (Very roughly 1960-1975; I seem to give different dates every time I mention this; ideally this Age should start with something like Bondi radiation theory c. 1959 and end with Hawking radiation c. 1974---as the first truly landmark result outside classical gravitation--- or perhaps with the positive mass theorem c. 1979).

In a Lorentzian manifold, yes, modulo some quibbles about small deviations from a geodesic arc (keeping the endpoints fixed) decreasing the length of the curve.

Right, you can't just quote the above mentioned result, you need to toss in another idea or two. (Since we have lotsa symmetry here, it makes sense to try to take advantage of this happy circumstance!)

Sure thing! Let us know what you think after reading it. BTW, if you ever have the chance to purchase a copy of MTW, I would urge you to do so. In my opinion this is one of the great scientific books of all time--- I put it right up there with the Principia, another Great Book which is more often cited than read. Fortunately, MTW is actually very readable for the most part.

7. Dec 19, 2006

### Wallace

Well the MTW problem 31.4 is precisely the problem I'm trying to get my head around, it's essentially the same question as in Hartle, except Hartle gets you to calculate $$\tau$$ for the geodesic path. I must be missing something though, since the hint in the problem suggests that the longest proper time lapsed between $$r=2M$$ and $$r=0$$ is the radial geodesic. My contention is that $$r=x$$ is not a point in space-time but a line, and therefore I don't understand how a simple argument about geodesic paths can solve this problem.

Consider this thought experiment. In Minkowski space-time, two test particles are moving towards some test point in space. They start at the same place and time, one get one of those magical infinite instantaneous accelerations at the start and then travels at a constant speed to the target. The other starts from rest and accelerates at a constant rate to the target. Using the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html?" [Broken] equations, you can work out the proper time experienced by the accelerated particle, and simple Lorentz transformations for the constant velocity particle. Now, the constant velocity particle is on a geodesic, so by the logic in the hint above should maximize the proper time, however it is easy to find values of the acceleration and velocity that do not give this answer. For instance using natural units, if the acceleration is 0.01 and the velocity is 0.1 then the accelerated observer experiences over 50 times the proper time of the geodesic observer over a distance of 1.

In this case the two observers get to the target spatial point at very different co-ordinate times. To me, this is why we can't use the simple geodesic argument, since we are talking about an interval in space, not an interval in space-time.

I don't see what is fundamentally different about a black hole that changes this argument, though I would be interested to hear what it is! (I assume I havn't found an error in MTW :surprised )

Last edited by a moderator: May 2, 2017
8. Dec 19, 2006

### cesiumfrog

If anything, surely it's an interval in time. Naively, this suggests zen state maximises experience. (To have velocity wrt. this observer is to measure fewer seconds, and acceleration voids counter-claim.)

Last edited: Dec 19, 2006
9. Dec 19, 2006

### Wallace

It's an interval in space-time, a line joining two points (defined by the spatial and temporal co-ordinates) that the two paths cross. What I don't understand is how you can get rid of the temporal part and just consider the radial spatial path between two spatial points. Does something force both objects to elapse the same co-ordinate time during the journey to the black hole center, regardless of acceleration? Or is it more subtle than that?

10. Dec 19, 2006

### cesiumfrog

You realise the time-like direction is radial there, right?

11. Dec 19, 2006

### Chris Hillman

Come again?

Good. Since I have MTW but not Hartle let's focus on that statement.

What is this locus $$r=x$$?

Sorry, I couldn't understand what scenario you have in mind. Maybe you can go back to black holes and write enough math to specify particular geodesics whose proper time (from passing the horizon to striking the curvature singularity) you wish to compare?

That is a good working assumption!

12. Dec 20, 2006

### George Jones

Staff Emeritus
Wallace, I think I understand your concerns. Let's see how I do.

First, your SR example. Let coordinates refer to a particular, fixed inertial frame, and consider two observers, not in the inertial frame, that move between the planes x = 0 and x = 1. Suppose both observers experience the same event A on the plane x = 0, and that the worldlines of the observers intersect x = 1 at different events, say B and C. As you say, it can easily be arranged that so an observer who accelerates from A to B has greater elapsed proper time than an observer who follows a geodesic from A to C.

You compare this to the (vacuum) Schwarzschild case, where two observers start at the same event A on the event horizon r = 2M, but end up at the singularity r = 0 at (speaking loosely) two differents "events", B and C. Arguing by analogy from the SR case, you wonder whether an accelerated observer can expend more proper time going from A to B than does a geodesic observer going from A to C.

But the SR and Schwarschild cases really are different.

Back to the SR example. There always exists a geodesic (straight line) from A to B, and an observer that follows this worldline expends more proper time in going from A to B than does the accelerated observer. Similary, for the Schwarzschild case.

Here's the difference. In the SR case, the set of all possible proper times for a worldline that goes from x = 0 to x = 1 is unbounded. You can take as long as you want to go from x = 0 to x = 1.

This isn't true for Schwarzschild. The set of all psossible poper times is bounded above. This bound is found by considering all possible geodesics. The case of an accelerated observer is a bit of a red herring, siince, if an accelerated observer experiences one event on r = 2M and another event "on" r = 0, there always a geodesic between the same two events for which the proper time is longer.

13. Dec 20, 2006

### Wallace

Thanks for the insight George. I appreciate your assistance free of a condescending tone!

Obviously the SR and Black Hole space times are very different, so yes you would expect different behavior, the key is to work out what behavior is different and why. The issue of the set of proper times being bounded in the black hole case is, as you suggest, probably the key to this problem. There is still something I don't get though, apologies for my slowness! I can do the calculation for the free fall proper time, that's fine. What I can't do is calculate analytically the proper time for the outwardly accelerating fall. Even if it is shorter than the free fall, how much shorter is it? When I try and do the calculation (in EF co-ordinates) I end up with hopelessly coupled equations that I can't solve. The solution is clearly going to be hyperbolic, but it's tricky to fine what it is. I haven't seen any textbook even ask this question, let alone give a solution.

Anyway, I'll attempt to explain my remaining dilemma. Imagine a light cone diagram for a black hole in EF co-ordinates. On this you could draw free-fall geodesic paths for particles starting at rest at the event horizon (or arbitrarily close to the horizon for the pedantic) at a series of co-ordinate times. Intersecting each path at the event horizon you could also draw the ingoing and 'outgoing' null paths (which actually point in inside the horizon of course) that define the light cones of the in falling particles.

Now, this is the uncertain part. How would the infall path change on this diagram for the outwardly accelerating observer? This is what I have been unable to solve analytically, so I will use intuition, which of course is always dangerous in GR! In any case, intuitively I would imagine that the acceleration would cause this path to run closer to the outgoing null path than a pure free fall path would. If you imagine a series of the free-fall paths then the accelerated path would cross a number of the free fall paths (i.e. the paths of free fall particles released from the horizon at a later co-ordinate time). The outgoing null path defines the limit of the deviation from the free fall path, clearly the particles cannot cross this, but with a high enough acceleration, should be able to get arbitrarily close to it.

Now, when the accelerated particle does hit the singularity, it will do so at a later co-ordinate time than a free fall particle that left the horizon at the same co-ordinate time. By the same argument, the free fall particle that hits the singularity at the same co-ordinate time as the accelerated particle will have left the horizon at a different co-ordinate time.

This is why I struggle to understand how to apply the statement 'there always a geodesic between the same two events for which the proper time is longer', since I can't see any geodesic that goes between the two events (crossing the horizon and hitting the singularity) that the accelerated particle experiences.

Now, clearly if my assumption about what a EF diagram would look like for an accelerated particle is correct, then the accelerated particle takes a longer co-ordinate time to reach the singularity. Since it travels at a different velocity to the free fall particle their will be time dilation effects that make comparing the proper times experienced less simple. Obviously this would be sorted if I could just solve the accelerated fall path equations!

14. Dec 20, 2006

### George Jones

Staff Emeritus
Now I'm going to argue by analogy!

In the SR case, for convenience, let A be the origin of the inertial frame coordinates, and call x = t the ingoing ("towards" the destination x = 1) null path and x = -t the outgoing null path. Consider the situation where: one observer is accelerated and one observer moves with with constant velocity in the fixed frame; the two observers are coincident at event A on x = 0; at A the observers are comoving; the accelerated observer has "outgoing" (i.e., in the negative x direction) acceleration. Then the worldline of the accelerated bends towards the outgoing null path, and the accelerated observer reaches x = 1 at a later coordinate time than does the constant velocity observer. This is very much like Schwarzschild situation about which you wrote, but I want to stay with SR for a bit more.

In order for the constant velocity observer to reach x = 1 at the same event as the accelerated observer, the constant velocity observer cannot be comoving at A with the accelerated observer, the constant velocity observer must have a worldline that, at A, is closer to outgoing null than is the accelerated observer.

Now consider Schwarzschild and an outgoing accelerated observer that is on the event horizon at event A. Just like in SR, there are an infinite number of radial timelike geodesics that start from A, with one of these comoving at A with the accelerated observer. However, different freely observers will have different initial conditions ("speeds") at A. It seems to me that, just as in SR, one of these freely falling observers will "hit" the singularity at the same event as the accelerated observer.

This is just a plausibiliy argument; I have done no calculations and I have proved nothing.

Last edited: Dec 20, 2006
15. Dec 20, 2006

### Wallace

Ah yes I hadn't considered that. So if you allow the freely falling observers to have an initial velocity, rather than just staring from rest, then there will be some free fall path that does intersect the accelerated observers path. That makes sense, I hadn't considered that.

As you say it remains to be formerly proven that such a path exists for any accelerated path, but it does sound reasonable. At of course if this is the case, then this path, being a geo-desic, will maximize the proper time.

Thanks George!

I'd still love to be able to solve the accelerated case properly, but I'm happy with the qualitative argument now.

16. Dec 20, 2006

### Chris Hillman

Solving for motion of test particles

Hi, Wallace,

Looks like George Jones is giving you some generous and useful assistance--- but I certainly hope you weren't thinking of me when you mentioned "condescending tone"! (If you did mean me, I hope you noticed "Let's start over!" above.)

No doubt others here can help you with that, if you mean what I think you do (an infalling observer with specified radial acceleration as a function of an appropriate coordinate), but I have made many computations with various black hole models, so let me know in the unlikely event that otherwise insuperable technical difficulties arise.

17. Jan 2, 2007

### MeJennifer

But don't we have a negative expansion scalar inside the black hole?
So, doesn't it go inwards even if it points outwards?
Or am I lost now?

Last edited: Jan 2, 2007
18. Jan 2, 2007

### Chris Hillman

Radially outward acceleration of static observers, etc.

Hi, MeJennifer,

Sorry, my comment wasn't very clear.

Recall how I defined the slowfall observers: at each Schwarzchild radius, they accelerate radially outward with magnitude of acceleration $m/r^2$, which would of course be just sufficient to hold them stationary in Newtonian theory, but (in this sense) the gravitational attraction of a test particle to a Schwarzschild object is just a bit stronger, so these slowfall observers fall radially inward. Note that the acceleration of the slowfall observers depends upon their radial coordinate, but is finite as they fall through the horizon. (Note too that in relativistic units, ${\rm length}^{-1}$ is the unit of path curvature or acceleration, so $m/r^2$ makes sense.)

The acceleration points radially outwards (for both the slowfall observers and the static observers--- the latter are only defined in the exterior region, of course!) because these observers must accelerate outwards away from the world line they would follow if they were not accelerating, which is a timelike geodesic and, because the initial radial velocity is nonpositive, must move radially inward.

It might be easiest to consider first the case of static observers; if one of them drops a test particle it falls radially into the hole, and in order to maintain their position, these guys must be firing their thrusters radially inward. Of course that corresponds to accelerating radially outward. Similarly, someone standing on the surface of a spherical planet is accelerating "upward" (i.e. radially outward) because the ground is preventing him/her from falling radially inward.