# When an object crosses a black hole event horizon, does the entire object cross the event horizon "all at once?"

This is a follow-up question based on some discussions in one of my other questions posted here.

Imagine a standard Schwarzschild black hole of sufficiently large size so that tidal forces are negligible at the event horizon. The question is, when a spatially extended object enters the black hole, does the event horizon surface sweep past the object or do all parts of the object fall into the black hole at the same instant?

In one of the answers, Dale says,

The event horizon is a lightlike surface. In a local inertial frame it moves outward at c. So while it is true that there is an antipodal piece going the other way it doesn’t matter. The antipodal piece is going slower than c in the local inertial frame. So the horizon is going faster and the antipodal piece cannot possibly cross back through the horizon.

However, in the comments of my question, safesphere says,

The flaw in your question is assuming that things cross the horizon gradually. For example, if you fall feet forward, you assume your feet cross the horizon before your head. This is incorrect. This thinking follows the intuition based on the flat spacetime. The horizon is not a place, it is not spacelike, but lightlike. This means that your head and your feet cross at the same instant. The entire flywheel, no matter how large, crosses the horizon all at once. There is no “front” or “back” when you cross. There is no direction in space pointing “back” to the horizon from the inside.

[...]

See Kevin Brown: “One common question is whether a man falling (feet first) through an even horizon of a black hole would see his feet pass through the event horizon below him. As should be apparent from the schematics above, this kind of question is based on a misunderstanding. Everything that falls into a black hole falls in at the same local time, although spatially separated, just as everything in our city is going to enter tomorrow at the same time.” - Falling Into and Hovering Near A Black Hole

Perhaps the best description is as follows, because it is both intuitive and coordinate independent. Consider two objects falling along the same radius one after the other. Consider the events of these objects crossing the same given radius. These events are timelike separated outside, spacelike separated inside, and lightlike separated at the horizon. So the spacetime interval between the events of two sides of your flywheel crossing the horizon is zero (or “null” as it is commonly called) in any coordinate system.

These two quotes seem like they contradict each other. Which one is correct?

## Questions

I want to do some analysis of my own involving Eddington-Finkelstein coordinates and Kruskal–Szekeres coordinates, but I am short on time today. I intend to update my post with more info later.

• In order to even make the claim that all parts of an object pass the event horizon at the same instant, we need a notion of local simultaneity. What notion of local simultaneity is safesphere using in their claim?
• User safesphere says, "This means that your head and your feet cross at the same instant. The entire flywheel, no matter how large, crosses the horizon all at once." But I don't understand this claim. If you send an arbitrarily long pole, say extending from Earth to Sagittarius A*, this claim seems to imply that if one end of the stick crosses the Schwarzschild radius, then entire pole would be instantaneously sucked into the black hole at once. This violates the speed of light limit and causality. Doesn't this thought experiment demonstrate that your head and feet don't cross the horizon all at once?
• Is the conflict between the two quotes a matter of using different coordinates? Could both be true? Or is there something deeper?

• I would like to draw your attention to my question from some time ago, physics.stackexchange.com/questions/160276/… -- it discusses a closely-related scenario and the answers were excellent.
– zwol
Jan 17 at 16:42
• Physics doesn't allow anyone to agree on what "all at once" even means.
– J...
Jan 17 at 17:14
• @Holger Is that true for a supermassive black hole? Jan 17 at 18:39
• @Holger Read Misner, Thorne, Wheeler Gravitation pages 820 to 823. They calculate the Riemann tensor for an orthonormal frame of a radially infalling observer. They find that the nonzero components of R are of order ~1/M^2 at the event horizon. Therefore, a sufficiently massive black hole will not tear apart a human. Certainly not by the tidal forces. "There are no special rules for human sized bodies, it’s the same whether you consider a human body, a planet, a galaxy, or the entire universe." Tidal forces on a body are size relative. An ant has relatively less tidal forces vs a human. Jan 18 at 17:56
• @Holger "why should the mass of the black hole affect the fact that space time is continuous and doesn’t have sharp kinks?" The EV doesn't have any sharp kinks, I agree. No one implied that it does have kinks. I think you are misunderstanding the EV. What the EV is, is the part of spacetime beyond which all light is forced to go into the center singularity. No infinite tidal forces or sharp kinks required. Jan 18 at 18:01

These two quotes seem like they contradict each other. Which one is correct?

They do contradict. Please be aware that comments cannot be downvoted so they often serve as a haven for content that an author suspects would be severely downvoted.

One thing that both the answer and the comment share is that the horizon is a lightlike surface. If a flash of light occurs below your feet then it reaches your feet before it reaches your head. It does not reach both at the same time. The event horizon, being also lightlike, follows that same pattern of motion locally.

This is true in every local inertial frame. The temporal ordering of lightlike separated events is frame invariant. So any nearby inertial observer, regardless of their relative velocity, will agree that the horizon reaches your feet first and then your head.

• In addition to that light flash below your feet, if you're falling feet-first into a SMBH and there's a light attached to your feet, you can watch that light as you're crossing the EH. You won't see the light go out just because your feet have crossed the horizon before your head does. Jan 17 at 3:42
• Moderator note: the purpose of comments is to clarify or improve the posts they are attached to. If you happen to notice a comment which is serving as a “haven for content,” but isn’t clarifying or improving its parent post, flag the comment for removal. This includes comments which should have been posted as complete or partial answers (whether the answer is correct or not).
– rob
Jan 17 at 5:34
• @PM2Ring seeing your feet depends on how much tall you are. And plunging into a stellar BH, one becomes vary tall. Jan 17 at 11:15
• @fraxinus Sure, see this comment, which contains my spaghettification calculator. But this question is about BHs with small tidal force at the EH, and my previous comment specifically mentioned a Super Massive Black Hole. Jan 17 at 12:03
• "The horizon is a lightlike surface" means that it appears to an observer to approach with the speed of light: That is obviously only true for an infinitely close observer -- to observers on Earth, thank god, the event horizon stays put. This is relevant to the question: How do observers at different distances -- stick or not -- perceive the event horizon, if at all; and perhaps: How is what's happening to them perceived by third parties farther away? Jan 19 at 10:21

Dale is right. Falling feet-first through an event horizon means the horizon sweeps over you in the feet-to-head direction at the speed of light, which means your feet cross first. The event of your feet crossing is in the causal past of the event of your head crossing.

safesphere writes a lot of comments on questions and answers related to general relativity, despite not seeming to understand the subject well. I don't see any interpretation of the comment that can make it correct. Rather than trying to answer your three questions, I'll respond to parts of the comment.

if you fall feet forward, you assume your feet cross the horizon before your head. This is incorrect. This thinking follows the intuition based on the flat spacetime.

The intuition based on flat spacetime is correct, because the spacetime in the region of interest is close to flat (if we're talking about a human being falling into a stellar-mass black hole, at least).

In coordinates of any external observer, no matter where he is or how he moves, nothing crosses or touches the horizon ever. Thus in the coordinates of your head, your feet don’t cross for as long as your head is outside.

Coordinate systems are just ways of assigning numeric labels to spacetime points, and can only be more or less useful, not more or less correct, for any observer. There is no such thing as the coordinate system of your head.

Judging from this and some other comments, safesphere believes that different observers occupy different "private universes," and it can be true in one observer's coordinate system that an object crosses the horizon and in another's that it never does. That simply isn't true.

Schwarzschild coordinates don't cover the event horizon. $$r=2M$$ is not the horizon, but a coordinate singularity. If you plot a worldline that crosses the horizon on a Schwarzschild chart, it has to leave the chart to reach the horizon, and the actual crossing can't be seen. This has no bearing on whether the worldline crosses the horizon on the physical manifold (which it does by assumption).

Also, everything that ever falls crosses at the same coordinate time of r=rs where r is the coordinate time inside the horizon in the Schwarzschild coordinates. Note that in these coordinates everything becomes infinitely thin in the radial direction near the horizon and crosses all at once.

Nothing crosses the horizon in Schwarzschild coordinates at any $$t$$ or $$r$$. The limit that safesphere is trying to take here doesn't make sense; the coordinates don't behave sensibly in this limit.

Perhaps the best description is as follows [...] So the spacetime interval between the events of two sides of your flywheel crossing the horizon is zero (or “null” as it is commonly called) in any coordinate system.

Most of this paragraph is technically correct. The crossing events are lightlike separated. It doesn't really make sense to talk about the interval between them, except in the flat-space limit, nor to say "in any coordinate system" since this isn't related to coordinates in the first place.

But none of that really matters. Lightlike separated or not, one of the events causally precedes the other. This is no different from the causal relationship of events at $$(x,t)=(0,0)$$ and $$(x,t)=(1,1)$$ in Minkowski space.

• If I draw a 2D spatial diagram of a Schwarzschild BH using $r, \phi$ Schwarzschild coordinates, then the EH is at $r=2M$ in natural units (i.e., $r_s=\frac{2GM}{c^2}$). However, if I add time to this diagram then no worldline ever reaches the EH because the Schwarzschild $t$ coordinate approaches $\infty$ as $r$ approaches $2M$. Jan 17 at 3:36
• @PM2Ring But we know a worldline can cross into the black hole in finite proper time. The $t$ going to $\infty$ is just an inadequacy of the Schwarzschild coordinates. Jan 17 at 4:17
• @MaximalIdeal Certainly! Jan 17 at 4:20
• @MaximalIdeal BTW, if you'd like to experiment with null geodesics near a Schwarzschild BH check out my program in physics.stackexchange.com/a/680961/123208 My trajectories have no trouble crossing the EH because I've totally eliminated the time parameter. ;) Jan 17 at 4:25
• It seems like in order to make sense of safesphere's comments, you need to define a notion of local simultaneity that ends up saying lightlike separated events are simultaneous, which is non-sensical. Am I correct in this assertion? Simultaneity can only be meaningfully established for spacelike separated events, so it makes no sense to make the surface of the event horizon as a surface of simultaneity. Would you agree in my thinking? Jan 17 at 8:28

As you approach the event horizon, from the point of view of an external observer your head approach your feet. You become shorter and shorter. But, this process takes an infinite amount of time. You just get shorter and shorter tending to zero height in the infinite limit, but never being of zero height at any finite time.

In a sense you head and feet both reach the event horizon at time infinity. But, that is only in terms of a limit, and not a value of time for the external observer.

To you this takes only a finite time.

Also, the speed of light - from the point of view of said external observer - drops to zero as one gets near to the event horizon. So, if you measure your height in terms of how long it take light to get from your head to your feet, this is a different issue. From your local point of free falling view - your height does not change. And neither does the speed of light.

From your point of view, there is no horizon. You do not see any surface in space sweeping past you at all. What you see is everything outside getting further and further away and faster and faster. At the point that corresponds to you getting to infinite time for the external observer, you would see everything in the external universe vanish off to infinity.

In that sense, to you, the event horizon is a point in time, not a surface in space. So, in a sense, you cross it all at once. But, not as space, just as time.

• I don't think I understand your last two paragraphs. From an outside observer the event horizon is a well defined surface at $r=r_S$ at all points in time. Are you saying that if you transform this surface to the frame of an infalling observer the observer will cross the event horizon all at once instead of it sweeping through the observer? Jan 17 at 11:15
• Sure, the Schwarzschild time coordinate goes to infinity at the EH, so it's not very useful. There's some good info in the answers to Does someone falling into a black hole see the end of the universe? Jan 17 at 12:49
• Why should a distant external observer choose to use Schwarzschild coordinates? I see no reason why an external observer is obligated to choose those coordinates as opposed to any other coordinates that don't have an event horizon singularity in them. I appreciate your explanation and I apologize if I might come off as aggressive, but I'm not convinced your post makes any sense. At the very least, can you answer the three questions I posted? I don't see how your post has enough info to resolve them. I would be very happy to be proven wrong in my skepticism of your post. Jan 17 at 18:36
• @MaximalIdeal I can (will when I get a moment) expand on some of the details to clarify. But, I have a question for you - do I need to justify or explain safesphere to answer your question? Or is deconstruction also an acceptable answer? My position is that safesphere is not following the logic of general relativity. In particular - the external and inward observers have a rather different view of what the event horizon is. One says it a spatial surface and the other says it is a temporal surface - more or less. Jan 18 at 4:36
• My position is that safesphere is not following the logic of general relativity. In particular - the external and inward observers have a rather different view of what the event horizon is. One says it a spatial surface and the other says it is a temporal surface - more or less.” - Please don’t judge if my logic follows GR based on a brief comment. I don’t disagree with your quoted statement. In fact, your answer is the only correct answer here so far, so +1. Jan 18 at 7:56

There is a very nice tool that might help you better understand such questions:

https://www.mathworks.com/matlabcentral/fileexchange/72254-schwarzschild-black-hole-simulation

you can simulate two observers free falling into a black hole starting from slightly different locations (one being the "head" and the other the "feet"), and you can see how light propagates between them so it tells you what each one can see at any moment.

Note that you would want to run it in the Kruskal coordinates mode, since the Schwarzschild coordinates don't cover the part of spacetime after they enter the event horizon.

(And the short answer to the question is no, the entire object does not enter all at once)

I feel like I need to answer because however strange it may seem, both @dale and @safesphere is right in their own reasoning, and there is a way to ease this seemingly contradictory situation.

As you push the stick towards the black hole, the subjective time of the end of your stick moves more and more slowly relative to your subjective time. The Schwartzschild metric can tell us dτdt, the rate of time passage at a particular radius r compared to the rate of time passage infinitely far from the black hole: Notice, that as Rs approaches rs, that ratio approaches zero! By the time your stick is nearly at the event horizon, the end of your stick is experiencing almost no passage of time compared to you.

the proper time of the back end is not the proper time of the front end. In fact, they will no longer be causally linked when the front end crosses the horizon.

Thought Experiment - Poking a stick across a Black Hole's Event Horizon

Now in this question, basically what we have is a extremely tall human (a extremely long stick), where the front end of the stick (your foot), and the back end of the stick (your head) is so far from each other, that we make a convention.

This convention is what creates this apparent contradiction. We say, for certain (calculational convenience) reasons, that we consider the observer at the back end (that is farther from the black hole) of the stick (the head in your example), a far away observer. Why? Because the observer (head) and the front end of the stick (foor in your example) are so far that they cannot be considered to be in the same local (freely falling) frame. This causes two main effects:

1. (asymptotically) infinite time dilation

2. (asymptotically) infinite redshift

For these reasons, the observer far away (head in your example) will never see the front end of the stick (foot in your example) to reach the horizon in a finite amount of time, it will just seem to the observer that the front end is slowing down (time dilation), fades away (redshift), and disappears from observability.

Now in a normal human scale (the human is relatively small object), we say that (again for calculational convenience), we use a convention. This means, that the human, as a whole object (both foot and head) can be considered to be in the same/single (freely falling) frame of reference. What does this mean? Time dilation between the foot and head? Nope. Redshift? Nope. You (as a normal scale human) will just pass the horizon foot first, and you can easily look at your foot and will not notice anything strange.

As a side note, inside the horizon, spatial and temporal dimensions behave oddly (they do not swap), and as you fall into the black hole foot first, you could say that your foot "happens" first, and your head "happens" later. This part is utterly hard to intuitively understand. Nevertheless, the answer to your question is, if you consider the human to be relatively small enough, and say that the whole human as an object can be considered to be in the same (freely falling) frame of reference, you agree that there will be no time dilation and redshift between the observer (head), and the foot, thus, as Dale says, the horizon will simply just swoosh past you at the speed of light, foot first, and head after. As safesphere says, this becomes problematic when the stick (object) is too long, and the observer at one end is far from the other end. There we have to deal with time dilation and redshift between the observer and the other end of the stick, and you will not see the other end of the stick cross the horizon (fades away from observability).