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
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. Therefore your feet and your head cross at the same proper time of your head. 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.
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?

Edit: The comments on that question have been moved to the discussion here. Please help me understand.
 A: 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.
A: 
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.
A: 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)
A: 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.
A: 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:

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*(asymptotically) infinite time dilation


*(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).
Now your question is even trickier worded. You are specifically asking whether the object (human in your example) crosses the horizon all at once or gradually. At once you ask? Are you asking in the temporal dimension or the spatial? Again, the spatial and temporal dimensions start behaving oddly at the horizon. What is correct to say, is that your foot will "happen" first, and your head will "happen" later. In your own frame. This is very important, as this is what creates the (apparent) contradiction between the two answers/comments.
You have to decide whether your object is relatively small enough so that it fits into a local frame, and then you have no problem, the foot "happens" first, and the head "happens" later, and you can easily observe your foot while you are falling. Take an object too long (like the stick in the other question), and you have to consider the far away observer to be in a different frame and you have to deal with time dilation and redshift, and the observer will not be able to see the other end of the stick cross the horizon. From an external observer's view, the black hole is kind of "frozen", because of these effects, and as the observer approaches the horizon, when that observer can be considered to be part of the same/single frame (as the other parts of the extended object), so you can disregard time dilation and redshift, that is your choice.
