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.
Best Answer
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.