Time paradox inside a black hole

At the event horizon of a black hole, time and the spatial direction toward the center exchange places. The direction inside the black hole from the event horizon to the the singularity in the center is the direction in time.

Assume a symmetrical non-rotating black hole and also assume that things can actually fall to the black hole. Consider that over 10 billion years a number of relatively small objects fall in. Finally, at the end, a number of larger objects also fall in that are big enough to increase the size of the event horizon.

Now let's look what's happening inside. The distance from the event horizon to the center is a time coordinate. Therefore, inside a symmetrical black hole, does the sphere of the event horizon represent the same moment of time? If so, then all things that have been falling in over 10 billion years appear all at once on the inside at the same moment that the sphere of the event horizon represents. Is this correct?

Finally, when the larger objects fall in and increase the size of the event horizon, on the inside, a larger radius would represent an earlier time than a smaller radius. Correct? If so, the larger things that fell in at the very end appear on the inside before everything else that's been falling in over 10 billion years. Is this the proper understanding?

In fact, if a black hole has been growing over time gradually increasing in size by sucking the external matter in, on the inside this process would appear happening in reverse, because time inside goes from a larger to smaller radius of the event horizon. Is this correct?

And if it is, then would this not violate causality? For example, consider that a large asteroid was supposed to fall in at the very end, but got hit by another asteroid and both avoided being sucked into the black hole. However, on the inside, this asteroid falls in "before" everything else and therefore is already there when everything else falls in.

• All blackhole are symmetric: spherical or cylindrical for $J=0$ or $J\ne 0$--with the possible exception of course, of those still ringing down from a merger. – JEB Feb 10 '18 at 3:23

Time and space don't swap places inside a black hole.

In order to describe what happens in spacetime we have to attach labels to spacetime points, and these labels are our coordinate system. That is, we choose some coordinates $t$, $r$, $\theta$ and $\phi$ then we can label points in spacetime by their coordinates $(t, r, \theta, \phi)$.

But there are lots of ways to form a coordinate system. The usual one for describing the exterior of black holes are called the Schwarzschild coordinates and these correspond to the physical measurements made by an observer an infinite distance from the black hole. So the Schwarzschild $t$ coordinate is the time measured on the Schwarzschild observer's clock. However there are lots of other ways to make a coordinate system. Gullstrand-Painlevé coordinates are superficially similar but now $t$ is the time measured on a clock held by an observer falling freely into the black hole. Or Kruskal-Szekeres coordinates use abstract coordinates that don't correspond to anything measured by an observer.

The point of all this is that the coordinates are not spacetime - they are just labels we attach to spacetime. What happens inside an event horizon is not that time and space swap places but rather that the labels we call Schwarzschild coordinates behave oddly inside a black hole. This is an important distinction. If we use the Kruskal-Szekeres coordinates instead then we have a spacelike coordinate $u$ and a timelike coordinate $v$ and these do not swap places inside the black hole.

It is fairly easy to see why the Schwarzschild coordinates are not a good description for the interior of a black hole. If we drop something into a black hole and start timing its fall we discover that the object takes an infinite time to even reach the event horizon. That is, no matter how long we time for the falling object never reaches the horizon. In fact for Schwarzschild observers black holes don't exist because any black hole will take an infinite time to form. So by using Schwarzschild coordinates to label the interior of a black hole we are labelling something that doesn't exist. Is it then any wonder that the behaviour of our labels, our coordinate system, inside the black hole is bizarre?

Where things get confusing is that even though in the Schwarzschild coordinates black holes can never form this does not mean they don't exist. For example if you're standing on the surface of a collapsing star then in your rest frame not only does the black hole form in a finite time, but you will fall through the event horizon and to your fatal encounter with the singularity in a finite time. The Schwarzschild $t$ coordinate only covers your trajectory up to the formation of the horizon, but your trajectory carries on past the point where the Schwarzschild $t$ coordinate reaches infinity. But once again let me emphasise that this is just a failing of the Schwarzschild coordinate system. Time carries on for you as normal - it's just just that the Schwarzschild coordinates are not a good way to label time in those circumstances.

• Are you not overlooking the fact that in the frame of a falling observer the universe dies in the infinite future before he crosses the event horizon? For example, hypothetically, imagine the Big Crash scenario, in which the universe crashes into itself before the falling observer crosses the horizon. And yes, it happens very fast for him indeed. Calling coordinate systems "labels" in order to invalidate them is not helpful, because coordinate systems are consistent and valid and also related mathematical concepts. If black holes don't exist in one, they don't exist in any, as described above. – safesphere Aug 30 '17 at 17:22
• It is well known that the time coordinate inside the black hole runs from the horizon to the singularity. I am not sure what you mean by "swapping" or by "time and space don't swap places", but this is just a terminology irrelevant to my question. My understanding is that the spacetime is bent at the horizon such that the time coordinate turns toward the singularity, so it is not "swapped", but bent. Also, the Schwarzschild coordinates are problematic only at the horizon where they become singular. They are fine inside the horizon. I appreciate your insight, but my question isn't answered. – safesphere Aug 30 '17 at 17:40
• "... in the frame of a falling observer the universe dies in the infinite future before he crosses the event horizon?" - That's not true according to the answers here. – Hal Hollis Aug 30 '17 at 21:22
• @Hall Hollis: It is according to the answer of asperanz. – safesphere Aug 31 '17 at 3:18
• +1 for "Time and space don't swap places inside a black hole." :) On Schwarzschild coordinates, I would say they don't behave well near the horizon, however they are fine for the interior in general. I do think Gullstrand-Painleve coordinates are more intuitive in the interior, and have the added bonus of being regular across the horizon. But Schwarzschild still have their uses, for example they conveniently describe the very unique case of $t=\textrm{const}$ observers (for $r<2M$). – Colin MacLaurin Feb 7 '18 at 14:52

I think you are making some confusions on the terms. Time is time, space is space inside and outside a black hole. There is no such thing as time is space. What happens specifically inside the Schwarzschild black hole is that the coordinate "r" can be now understood as time and the "t" coordinate as distance, given the tangent space for both vectors $\partial_t$ and $\partial_r$, i. e. $<\partial_t, \partial_t>$ is negative outside the horizon and positive inside it, the contrary being true for r.

Not sure about other assurances, but I think that all your points are based on the assumption of interchange of time and space, which is not correct.

• Your answer describes what I stated in the first sentence. Then your conclusion contradicts the rest of your answer. Perhaps it is you who makes some confusion on the terms, because your description is the same as mine, but you don't seem to see it. – safesphere Aug 30 '17 at 4:42

There is a difference between the coordinate directions and the falling observer's directions. Before realising this, I was also confused by the poetic rhetoric "swapping of time and space". The Schwarzschild coordinate basis vectors are $\partial_t$, $\partial_r$, $\partial_\theta$, $\partial_\phi$ which have components $(1,0,0,0)$, $(0,1,0,0)$, etc. So indeed the coordinate vector $\partial_t$ becomes spacelike inside the horizon and $\partial_r$ becomes timelike, so we can say the same for the coordinates $t$ and $r$. (Actually this only follows in the case of a diagonal metric, but I'm trying to keep things simpler.)

However the "time" direction for a falling observer is its $4$-velocity: $$\Big(\frac{1}{1-2M/r},-\sqrt{2M/r},0,0\Big)$$ And similarly its spatial radial direction is $$\Big(-\frac{\sqrt{2M/r}}{1-2M/r},1,0,0\Big)$$ These remain timelike and spacelike respectively! These do not swap natures at the horizon. So in summary, the Schwarzschild $t$ and $r$ coordinates are not the pure space and time as determined by the observer.

No, the event horizon does not represent a single moment of time. It is a null surface, meaning an outward directed photon could remain at constant $r=2M$. But this sets up causality: for (many) events on the horizon, it gives an unambiguous before and after, based on passing the photon. Time does not reverse inside the horizon, only the direction of the Schwarzschild $t$-coordinate does (what I mean is, the $t$-coordinate decreases for a typical observer -- but not for all observers, incidentally). The picture is clearer in say Gullstrand-Painleve coordinates, see my answer elsewhere, which includes a diagram from Taylor & Wheeler's excellent book Exploring Black Holes.

• Thanks for your answer. I will carefully review it. Since I asked this question, I have realized that on the inside of the black hole, the time that the event horizon represents on the time axis from the event horizon to the singularity is the past infinity. This throws a wrench into the argument in my question, because things crossing the event horizon at different outside times would show up on the inside in the infinite past. On the other hand, a falling observer never crosses the event jorizon shrinking in front of him as he falls into the singularity. Enough to ponder :) Thanks again! – safesphere Feb 10 '18 at 2:47
• I hope it's helpful! Are your comments based on the spacetime diagram of a falling object in Schwarzschild coordinates? While correct, such diagrams are misleading, and even fooled the early experts. Don't over-interpret the Schwarzschild $t$-coordinate as "time". Time is relative, and there are an infinite number of choices of "time". Much more intuitive and natural is to use the proper time of the observer. Hence I strongly advocate Gullstrand-Painleve aka "raindrop" coordinates for this context. See diagram linked above. Note the horizon occurs at finite time under that choice. – Colin MacLaurin Feb 10 '18 at 3:05