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>As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. You have it backwards. From the POV of an ext

>As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. You have it backwards. From the POV of an external observer you never cross the event horizon, because that's a function time dilation, and light that you emit being more and more red-shifted as you get closer to the event horizon. From the falling observer POV, nothing really changes - or rather, what happens close to the event horizon or just past the event horizon depends on what the theory of quantum gravity says.

simonh 6y ago
I didn't address the POV of an external observer but I'm sure you're right. But what you describe for the POV of someone falling in seems consistent with what I suggested. I don't understand what you think is backwards about it.
- macspoofing 6y ago
  You said that from the perspective of a falling observer, they never see themselves cross the event horizon. That's not right. The falling observer certainly will see themselves cross the event horizon. In their frame of reference, time flows normally. For smaller black holes, they may get spaghettified, but for larger ones, crossing event horizon will be an unremarkable occurrence.
  - simonh 6y ago
    There will always be an event horizon below them beyond which light cannot reach them, and they will always be able to observe the external universe. So how will they see themselves crossing the event horizon?
    - smallnamespace 6y ago
      The usual definition of event horizon is from the perspective of the external observer—the falling astronaut can calculate the threshold at which no amount of thrust can ever move them away from the singularity (in fact, this is what is meant that a time and space coordinate switch places—you can now only move 'down' in space, just like you can only move forward in time).
      - simonh 6y ago
        I suppose they could calculate the theoretical event horizon from infinity, but they won't observe it, however they will observe one below them.
        pdonis 6y ago
        > they will observe one below them No, they won't. This is simply wrong. See my other responses upthread.
      - zrm 6y ago
        > no amount of thrust Are you sure that's right? Thrust implies acceleration, not just velocity. Suppose you're just inside the event horizon, from the perspective of an external observer. You're traveling outward at a velocity of .95c. That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. But if you were close to the event horizon you might cross back over it before you fall back in, right? At which point the external observer could see you again. And once you're outside the event horizon, e.g. to the place where escape velocity is .9c, you might by then have slowed to .8c, but if you still have the capacity to accelerate back above .9c again, you could still get out, couldn't you?
        pdonis 6y ago
        > Are you sure that's right? It's right. > Suppose you're just inside the event horizon, from the perspective of an external observer. Whether or not you are inside the event horizon is not a matter of perspective. All observers will agree on it. > You're traveling outward at a velocity of .95c. Relative to what? Outside the horizon, velocities, such as the escape velocity you mention, are usually meant as velocities relative to static observers--observers who are maintaining the same altitude above the horizon. But below the horizon, there are no static observers. It is impossible to maintain a constant "altitude" (which here just means the surface area of the 2-sphere at your location, centered on the hole) even for an instant. So "velocity" relative to a static observer has no meaning inside the horizon. > That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. No, that's not what happens. Relativity is not Newtonian physics. The event horizon is not a place in space. It is an outgoing null surface: a surface formed by radially outgoing light rays. Spacetime around the black hole is curved in such a way that this particular outgoing null surface has a constant surface area: in other words, even though there is a full 2-sphere's worth of radially outgoing light rays at the horizon, the surface area of the 2-sphere formed by those light rays is constant--the 2-sphere does not expand, as you would expect a radially outgoing spherical wave front to do, because of the curvature of spacetime. If you are just a little bit inside the horizon, you are on a slightly smaller 2-sphere, and on that 2-sphere, even radially outgoing light rays cannot stay in the same place, much less expand; they contract--that spherical wave front, even though it is radially outgoing, decreases in area with time, i.e., it is falling. So you, having to move slower than light, will fall even more. You can't move outward at all; you can't even stop moving inward. (The fact that there are no static observers inside the horizon is a consequence of this.) That is true no matter how hard you accelerate, since no amount of acceleration will let you move faster than light.
    - pdonis 6y ago
      > There will always be an event horizon below them beyond which light cannot reach them Wrong. See my other response upthread. > they will always be able to observe the external universe But the light coming to them from the external universe will be more and more redshifted, and they can calculate from the observed redshift when they have crossed the horizon. Also, they will only see a limited portion of the future of the universe; anything that happens later than a fairly short time after they cross the horizon, they will never see, because they will hit the singularity inside the hole before light from such events can reach them.
pdonis 6y ago
> From the POV of an external observer you never cross the event horizon More precisely, the external observer never sees you cross the horizon. But he can still deduce that you did from other data. So the external observer does not say "you never cross the horizon", period. He just can't see it happen.
- gizmo686 6y ago
  This still isn't quite right. In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time. In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. There is a coordinate singularity around this switch, which is where the infinite coordinate time comes from. An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon, which they would find to be finite.
  - pdonis 6y ago
    > In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time No: the path of the free-falling observer at and beneath the horizon is not covered at all by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make any claims about what happens. > In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. No: there is a separate coordinate patch that covers the region inside the horizon (but not the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense. > There is a coordinate singularity around this switch Yes. > which is where the infinite coordinate time comes from. No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all. > An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will not be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, all four coordinates are spacelike below the horizon.)

simonh 6y ago

I didn't address the POV of an external observer but I'm sure you're right. But what you describe for the POV of someone falling in seems consistent with what I suggested. I don't understand what you think is backwards about it.

macspoofing 6y ago
You said that from the perspective of a falling observer, they never see themselves cross the event horizon. That's not right. The falling observer certainly will see themselves cross the event horizon. In their frame of reference, time flows normally. For smaller black holes, they may get spaghettified, but for larger ones, crossing event horizon will be an unremarkable occurrence.
- simonh 6y ago
  There will always be an event horizon below them beyond which light cannot reach them, and they will always be able to observe the external universe. So how will they see themselves crossing the event horizon?
  - smallnamespace 6y ago
    The usual definition of event horizon is from the perspective of the external observer—the falling astronaut can calculate the threshold at which no amount of thrust can ever move them away from the singularity (in fact, this is what is meant that a time and space coordinate switch places—you can now only move 'down' in space, just like you can only move forward in time).
    - simonh 6y ago
      I suppose they could calculate the theoretical event horizon from infinity, but they won't observe it, however they will observe one below them.
      - pdonis 6y ago
        > they will observe one below them No, they won't. This is simply wrong. See my other responses upthread.
    - zrm 6y ago
      > no amount of thrust Are you sure that's right? Thrust implies acceleration, not just velocity. Suppose you're just inside the event horizon, from the perspective of an external observer. You're traveling outward at a velocity of .95c. That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. But if you were close to the event horizon you might cross back over it before you fall back in, right? At which point the external observer could see you again. And once you're outside the event horizon, e.g. to the place where escape velocity is .9c, you might by then have slowed to .8c, but if you still have the capacity to accelerate back above .9c again, you could still get out, couldn't you?
      - pdonis 6y ago
        > Are you sure that's right? It's right. > Suppose you're just inside the event horizon, from the perspective of an external observer. Whether or not you are inside the event horizon is not a matter of perspective. All observers will agree on it. > You're traveling outward at a velocity of .95c. Relative to what? Outside the horizon, velocities, such as the escape velocity you mention, are usually meant as velocities relative to static observers--observers who are maintaining the same altitude above the horizon. But below the horizon, there are no static observers. It is impossible to maintain a constant "altitude" (which here just means the surface area of the 2-sphere at your location, centered on the hole) even for an instant. So "velocity" relative to a static observer has no meaning inside the horizon. > That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. No, that's not what happens. Relativity is not Newtonian physics. The event horizon is not a place in space. It is an outgoing null surface: a surface formed by radially outgoing light rays. Spacetime around the black hole is curved in such a way that this particular outgoing null surface has a constant surface area: in other words, even though there is a full 2-sphere's worth of radially outgoing light rays at the horizon, the surface area of the 2-sphere formed by those light rays is constant--the 2-sphere does not expand, as you would expect a radially outgoing spherical wave front to do, because of the curvature of spacetime. If you are just a little bit inside the horizon, you are on a slightly smaller 2-sphere, and on that 2-sphere, even radially outgoing light rays cannot stay in the same place, much less expand; they contract--that spherical wave front, even though it is radially outgoing, decreases in area with time, i.e., it is falling. So you, having to move slower than light, will fall even more. You can't move outward at all; you can't even stop moving inward. (The fact that there are no static observers inside the horizon is a consequence of this.) That is true no matter how hard you accelerate, since no amount of acceleration will let you move faster than light.
  - pdonis 6y ago
    > There will always be an event horizon below them beyond which light cannot reach them Wrong. See my other response upthread. > they will always be able to observe the external universe But the light coming to them from the external universe will be more and more redshifted, and they can calculate from the observed redshift when they have crossed the horizon. Also, they will only see a limited portion of the future of the universe; anything that happens later than a fairly short time after they cross the horizon, they will never see, because they will hit the singularity inside the hole before light from such events can reach them.

pdonis 6y ago

> From the POV of an external observer you never cross the event horizon More precisely, the external observer never sees you cross the horizon. But he can still deduce that you did from other data. So the external observer does not say "you never cross the horizon", period. He just can't see it happen.

gizmo686 6y ago
This still isn't quite right. In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time. In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. There is a coordinate singularity around this switch, which is where the infinite coordinate time comes from. An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon, which they would find to be finite.
- pdonis 6y ago
  > In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time No: the path of the free-falling observer at and beneath the horizon is not covered at all by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make any claims about what happens. > In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. No: there is a separate coordinate patch that covers the region inside the horizon (but not the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense. > There is a coordinate singularity around this switch Yes. > which is where the infinite coordinate time comes from. No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all. > An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will not be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, all four coordinates are spacelike below the horizon.)