Time paradox inside a black hole

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

In the most appropriate coordinate system for describing the situation, "time and space switch roles and the direction inwards, towards the origin of spherical coordinates, becomes time. Hence, it is as difficult to get back out of the black hole as it is to go backwards in time."

This quotation is from p.4 of the Nobel Prize Committee's 19-page report on the 2020 award of a Nobel Prize to the mathematical physicist Roger Penrose, after it had had more than half a century to carefully consider that award, which was based on Penrose's singularity theorems. The verbalism I've quoted is illustrated by a simple diagram on that report's p.5, and the entire report can be seen at https://www.nobelprize.org/uploads/2020/10/advanced-physicsprize2020.pdf . Its main point appears to be the fact that Penrose's analysis of the situation, taking account of every factor (including the mass equivalence of the energetic aspect of gravity itself), rendered the gravitational singularity at the heart of huge non-rotating black holes inescapable.

I'd agree with Safesphere that the "larger things" falling through the hole's horizon might, from the viewpoint of hypothetical observers somehow extant within any such colossal black hole, tend to fall through its horizon last, given the universality of the gravitational field and its effects on visible matter over time: That visible matter is generally considered to have previously consisted entirely of isolated particles. However, from Penrose's own imagined perspective, the last motion evident anywhere around the black hole would involve the decoherence of its own particulate contents, as described in the 2010 popularization of his past- and future-eternal cosmological model, a book titled "Cycles of Time".

That book includes lengthy discussions of entropy, which is generally considered to provide the physical basis for the time dimension. What gives Penrose's model its eternality is the "Weyl Curvature Hypothesis", which is the possibility that angles might be preserved during changes in scale: This would allow the thermal equilibrium of each quasi-eternal "aeon", in an endless sequence of them, to become the "Big Bang" of the next, in the dynamic version of GR which Einstein adopted when he added the Cosmological Constant.

Entropy, seen as "the measure of energy which cannot be converted into mechanical work in a thermodynamic system", is, in modern applications of relativistic physics, generally considered to be the basis of time, and is often loosely described as disorder: The situation Safesphere has considered seems to describe "dis-order" pretty exactly.

The sequence of occurrences in the "inside observer's" causality might be the reverse of the sequence of occurrences in the "outside observer's" causality, but, because entropy (the physical basis of time) occurs regardless of the direction of an observer's passage thru time, causality would hold regardless of the way the sequence seen by each would be seen by the other: Although each would consider the directionality of passage of the occurrences in the sequence seen by the other to be "paradoxical" and unobservable because of the horizon, its causal separation would hold.

Answer 5

If so, all things that have been falling in over ten billions years appear all at once in the inside.

This isn't quite right. From the perspective of a far-off observer the debris that has been falling into the black hole form a kind of sedimentary layer of matter around the black hole. Notably, it is a kind of thin shell, and hence the area of the black hole and not it's volume is paramount. This is why Hawkings calculation of the entropy of a black hole is naturally written in terms of area and not volume as one would naively expect.