Understanding Time Dilation at the Event Horizon

Question

I was recently reading about the event horizon of black holes and came across the fact that, to a "stationary" observer, it takes forever for someone to fall into a black hole. The sources claim that this is because, with Einstein's theory of relativity, the clock of the person who is falling into the black hole will be essentially stopped. However, I find this slightly confusing.

For example, I am picturing a person inside of a rocket ship that is entering a black hole. Although it is not possible, imagine that the rocket ship is entering the black hole at a speed of c. So, although the person's clock may be stopped, isn't the rocket ship still traveling at a speed of c? And, because there is a finite distance between the rocket ship and the black hole, shouldn't the rocket ship reach (and enter) the black hole? I know that my thinking is flawed, but I am not sure how. I am more knowledgeable about special relativity than general relativity (and I know that general relativity is the one that would apply here), so am I just confusing the two?

Answer 1

I think this is a good question. Let me start by saying that the subjective experience of the rocket ship is that it passes the event horizon unscathed. (Well... probably not unscathed due to the spaghetti effect unless it is a super, super, supermassive black hole which has a relatively flat event horizon.) An outside observer would see the rocket ship slowly sinking into the event horizon, edging closer but never passing it.

General Relativity allows non-trivial topology in spacetime. For instance, it's possible that the universe is 'closed' in the sense that there is a finite amount of space in it, like there is a finite surface area on the surface a sphere. Consequently, it is not generally possible to use a single coordinate frame (i.e. a single choice of $(x,y,z,t)$ coordinates in one reference frame) which can map out every point in spacetime.

Edit: You said you're familiar with Special Relativity so here's something extra. The Earth's surface, $S^2$, is locally diffeomorphic to $\mathbb E^2$, and likewise, General Relativity models spacetime as a $4$-dimensional pseudo-Riemannian manifold locally diffeomorphic to Minkowski Space, which is the space of Special Relativity.

In your black hole example, here on Earth with our natural $(x,y,z,t)$ coordinates, anything which passes the event horizon no longer has any sensible $t$ coordinate, because the $t$ coordinate explodes to $\infty$ as an object approaches the event horizon (see below). Another way to look at this situation is that our usual choice of coordinates (Schwarzschild) cannot extend into (or out of) the black hole. The event horizon is a singularity in coordinate choice only, not a physical singularity. Much like how the North and South poles of Earth are coordinate singularities in a lot of the common map projections of Earth. Is there something supernatural happening at the poles? Of course not. It's just because we chose to chart our Earth in a particular way.

The same happens when charting $4D$ spacetime.

Schwarzschild Coordinates (Earthly coordinates)

The coordinates we would generally use are Schwarzschild coordinates. Forget about $(x,y,z)$ and let's just look at the $1$-dimensional line straight from Earth to a distant black hole. The spacial distance along this line is $r$. This is the model we would use for the gravitational field of a single black hole where spacetime is otherwise flat (i.e. no dark energy; no dark matter; no expansion).

Here's what the Schwarzschild coordinates look like as we approach a black hole (taken from George Jaroszkiewicz's lecture notes at the University of Nottingham). The black hole singularity is at $r=0$, and the event horizon at $r=1$. We may consider Earth as being at some $r >> 1$:

The lines are called null geodesics and indicate the path that rays of light would take. As you can see, it takes light an infinite amount of time to approach the event horizon from the outside. In fact it takes everything an infinite amount of time to get to the event horizon. But there's an important semantic distinction here: when I say time, I'm really referring to the time that we here on Earth measure, denoted by the $t$ coordinate. It has nothing to do with the subjective experience of someone who's actually falling into the black hole. Their measure of time is called proper time and denoted $\tau$. The yellow 'light-cones' contain all trajectories that objects with mass could take. Here they are drawn for an infalling mass.

If we were to parametrise the path of the rocket ship with $\tau$ in these coordinates, then there would be a special value $\tau_{critical}$ at which it reaches that vertical asymptote $ct \rightarrow \infty$ and its position in these coordinates becomes completely ill-defined. The rocket ship doesn't spontaneously cease to exist at this point; it's just that the rocket ship hits 'uncharted waters' which our chosen coordinate system cannot comprehend.

Retarded Eddington-Finkelstein Coordinates

But there do exist coordinate systems which can map out the trajectory of a mass falling into a black hole. For example, the retarded Eddington-Finkelstein coordinates. Here's a diagram (again taken from George Jaroszkiewicz's lecture notes at the University of Nottingham) of such coordinates applied to a black hole:

Here's what I want you to take home from this diagram: as the mass falls closer and closer to the event horizon, just before it passes it, follow your eye along the null geodesics leading out from the black hole and toward us at $r \rightarrow \infty$. These lines indicate the path that rays of light follow. No matter how far you advance in our Earthly time coordinate, you are still receiving rays of light from when the rocket ship was falling in-- that moment right before $\tau_{critical}$. Any light emitted by the rocket ship once it passes the event horizon, beyond $\tau_{critical}$, is trapped inside the black hole.

Clasically, nothing can escape a black hole once past the event horizon. Not even information. No matter how much time passes on Earth, we receive information (i.e. light) from the instant just before the rocket passed the event horizon, and none from afterwards.

Here I've plotted our time coordinate $ct$ with respect to the retarded Eddington-Finkelstein coordinates:

Following the contour lines, you can see the path that light rays take to reach us from the black hole. No matter which contour line you look at, (i.e. no matter what $t$ coordinate; no matter how long we sit and wait) the light rays always originate from some point with $r>1$ because the contours do not pass through the critical $r=1$ event horizon. And so we will always see the rocket as it was just before $\tau_{critical}$.

Answer 2

Proper time is different for stationary far-away observer and the person entering the event horizon of a black hole (let's call him the traveler). As you said, for the observer it takes infinite amount of time to see someone crossing the event horizon (although you will eventually stop seeing the light, or a signal, coming from him because of the huge redshift and decrease of the signal frequency), but for the traveler nothing special happens at the horizon, supposing the black hole is big enough to not tear him apart. His proper time is doing just fine. Horizon singularity is just a coordinate singularity, you can choose other coordinates (see Kruskal-Szerekes coordinates) to get rid of it. The real physical singularity, whatever it really is, is the center of a black hole.

Answer 3

I believe that these answers embrace a fallacy: that "proper time" necessarily extends to a time when the rocket passes the event horizon. We can see this by considering what is (theoretically) visible from the rocket.

Considering (for now) only what is observable while the rocket remains outside the event horizon. If we were able to transmit a light pulse towards the rocket, it would stll reach the rocket before it passes the event horizon. In principle, the rocket could send an acknowledgement. Moreover, there does not appear to be a time (in the outside universe) when this ceases to be true: however, the longer we delay sending the pulse, the closer the rocket will be to any event horizon.

A secondary implication of course is that GR does not indicate that Schwarzschild event horisons ever form. This is not merely a case of our external observation, but based also on the rocket continuing to be able to observe our time.
Naturally, we can still call the entity a black hole (especially given that even Hawking Radiation has not reversed the description). However (even if we could somehow create a non-rotating hole), so far as I can see it does not have a central point singularity, nor even an event horizon.
Given that compact massive objects have finite lives due to radiation, we don't even have to worry about the time being truncated as the rocket asymptotically approaches the ever-forming event horizon: proper time simply gets arbitrarilly close to the time when it would theoretically "enter" the hole, but accelerates to approach that of the local universe as the hole evaporates.

(Naturally, there are other reasons that Schwarzchild horizons cannot form, not least the requirement of zero angular momentum on three axes)

Should I be mistaken, I would welcome reference to a proof of horizon crossing prior to hole evaporartion

Answer 4

Absolutely good question. And the answer is not complicated, and it's been written about in many places, but I'll summarise it here in very simple terms.

First, what does infinite time dilation mean?

When we compute time dilation in the usual Schwarschild coordinate system, we imagine a body that is stationary ($dr/dt = 0$) relative to the black hole, and a distant observer (so to speak infinite distant, there is obviously no such thing, but very distant is adequate). The body periodically sends signals to the distant observer at a frequency of say 1 signal/second, or in other words sends a radio burst of 1 Hz to the observer. The observer receives this signal at some frequency, and the quotient of the two is the time dilation. Basically, it is how much slower the observer perceives the body clock to be moving.

Now, this works for all cases as long as $r > r_{EH}$ (event horizon), because it is possible for the body to be stationary relative to the black hole. But as $r$ goes to $r_{EH}$ it becomes harder and harder for the body to do its job. Because the only way to remain stationary is to accelerate, so the body needs a drive to accelerate outwards. Now the event horizon is exactly the distance where infinite acceleration, i.e. infinitely strong propulsion, is needed to keep the body stationary. This is a physical impossibility, expressed by infinite time dilation.

Or, to put it another way, if there were a body with an infinitely powerful engine that stopped on the horizon at infinite acceleration, it would have infinite time dilation for any outside observer. Which is logical, just physically impossible.

But that's one of the important things that tends to cause misunderstandings: gravitational time dilation is only the time dilation of a body in a stationary, continuously super-high acceleration, not the time dilation of a body falling towards a black hole!

The case of a free falling body

More realistic and interesting is the case of a free falling (geodesic) body. Consider a body that simply falls into the black hole in radial free fall (no matter what the initial velocity, as long as it does not escape the black hole)! This body also sends signals to the distant observer and receives signals from it. So now the remote observer is also sending signals. Let the remote observer be Alice, the free-falling be Bob!

The question is, at what frequency will Alice receive Bob's signals and Bob receive Alice's signals?

The answer is surprisingly simple (whether or not it is a rotating black hole): for Bob at the event horizon, the frequency of Alice's signals will be half the frequency at which Alice sends them. Conversely, for Alice, the received frequency of the signals Bob sends will decrease and decrease, and will go to exactly zero as Bob approaches the event horizon.

Crossing the horizon

It also follows from the above that when Bob crosses the event horizon, he will not see that Alice has aged infinitely, and that an infinite amount of time has passed in the whole external world (but this has been discussed here: Does someone falling into a black hole see the end of the universe?). All he will see is that Alice continues to send signals towards him, which he will continue to receive with some frequency. As he falls inwards Bob will continue to receive Alice's signals, he will see Alice. Only Alice isn't receiving Bob's signals, she can't see him.

So Alice will not receive Bob's last signal that Bob sends out exactly when he crosses the event horizon. The ones sent before that she will receive them, albeit with quite a delay and redshift.

But what about this signal that Bob emitted perfectly precisely radially outwards towards Alice on the horizon? Well, this light signal (treated as a point, forgetting quantum physics) will be stuck right there on the horizon. So Alice will never get it, it is in fact infinite time dilation for Alice. (But as we have seen there is no infinite dilation for Bob for any signal from Alice.)

Suppose Charlie jumps into the black hole after Bob! Charlie could send signals to Bob and receive Bob's signals. Charlie can even communicate with Bob, but the signal that Bob sends out over the horizon is only received by Charlie when he reaches the horizon. This is where Charlie will bump into this "stuck" light signal.

What happens next? Charlie and Bob will fall further inwards inevitably, and in the meantime they can continue to communicate, but always in a strange way. Because Bob's outward signals are also falling inwards, it's just that Charlie is falling inwards faster than these signals are falling inwards, so he can still catch them. But here it matters a lot whether we are talking about the unrealistic Schwarschild or the more realistic Kerr black holes... That's not important now.

Speed of horizon crossing

Another interesting question is: what is the speed of a body (Bob) falling freely from infinity when it crosses the horizon? If the escape velocity is exactly $c$ on the horizon (by definition), then wouldn't it be $c$?

Of course not. Or ... well ... the question is meaningless. After all, compared to what is its speed? Compared to Alice? But how can Alice measure Bob's speed if she never receives Bob's signal, which Bob transmits from the horizon to her? So she never sees Bob fall into the black hole.

We have a problem if we want to define velocities globally, because in principle we can only guarantee that the velocity of a body (with mass) is always less than $c$c if it is measured locally. So take an observer exactly on the horizon when Bob crosses the horizon! What will the this observer see?

Okay, but how is the observer moving? If we assume that the observer is on the horizon stationary relative to the black hole, then he/se would measure $c$. But what observer can be stationary on the horizon? There are two:

a) A hypothetical spacecraft with infinitely strong thrusters and infinitely high outgoing acceleration.

b) A beam of light heading exactly outgoing.

Again, a) is unphysical. So b) remains as a possibility. But this is an obvious result: since we already know that the local relative velocity of light with respect to a timelike observer is always $c$.

But suppose we have two bodies that reach the horizon at the same time from the same place, but originally started from different places. They may have relative velocities with respect to each other. For example, one just free-fell in from somewhere, while the other was still pushing forward at a good speed towards the black hole to catch up with the first body. Then, of course, it will be true that the relative velocity of the two bodies will be less than $c$, and also that both will locally measure the velocity of any light ray as $c$. Everything fits.

Waterfall (River) model

To make the whole phenomenon easy to understand, a very useful model is the Waterfall (River) model, presented by its author here in an easy to understand format: https://jila.colorado.edu/~ajsh/insidebh/waterfall.html

It describes the black hole as a space (using a certain coordinate system) flowing towards the black hole, and things in this flowing space move according to their own local velocity. In effect, this model is based on a system of observers radially free-falling from infinity. The model makes Bob and Alice's (and Charlie) situation understandable in a very pictorial way.

I also highly recommend the original article: https://arxiv.org/abs/gr-qc/0411060

Answer 5

The standard answer you get when you ask this question is that an observer can fall through the event horizon of a black hole 'without noticing anything'. This doesn't really pan out in thought experiments however.

One of the basic issues here is that no external observer can ever see something fall across the event horizon. From an external perspective anything falling into the black hole 'freezes' at the horizon, trapped by an asymptotic slope of time dilation.

Here's the problem, when we say 'external observer' that doesn't mean an astronaut or a spaceship - it means ANY particle, at ANY distance above the event horizon that is experiencing a gravitational gradient less steep than the one that is more closely approaching the event horizon below it.

If your astronaut is falling feet first into the black hole, their head is necessarily approaching their feet due to differences in length contraction and time dilation. This is an unavoidable phenomena, and it extends to the relationship between every particle in your body, to the point where individual atoms in your body are falling in radically different time frames, even though they are mere angstroms away from each other above the horizon.

This effect is like gravitational tidal forces stretching you apart - but in reverse. Because no particle in your body can EVER see any particle below it cross the horizon they should all necessarily be crushed down into a plank scale density layer. Even a single proton would be far too large to survive this effect intact as it approaches the event horizon at the plank scale.

I'm honestly not sure how this can be waved off as a 'coordinate horizon' when there is no way to detach these physical relationships from what is observed. The differences in time dilation and length contraction experienced by objects falling towards the EH aren't illusionary.

In short, nothing should ever be able to fall through a black hole event horizon. From the perspective of an infalling object, the black hole's 'interior' will have been compressed so absurdly by length contraction by the time you reach the EH that the entire diameter of the event horizon is no more than a plank length across, regardless of how large it appeared further away.

There is no-where left for the particle to fall, so it will be probabilistically smeared across the entire event horizon as its position becomes largely indeterminate. It should basically occupy every point on the event horizon simultaneously, sharing space with all the other stuff that fell onto it in a plank density shell of rather exotic matter. From the perspective of the infalling object, it has in essence actually fallen into a singularity, but from a distant perspective this singularity takes the form of a 2D surface of plank density.