Recent content by Yukterez

Not really:

Correct, I didn't mean the freefaller is outside the external observer's future light cone, I meant that in his future you are always outside of the event horizon. The external observer is outside of the freefaller's future light cone once he passed the horizon, I made a mistake in my last...

I understand all of this, but what you don't seem to understand is that in the future light cone of the external observer, the freefaller is always outside of the horizon, so it would make no sense to say the crossing of the horizon was "possibly now". But whatever, better having you against me...

That is correct, I calculated that a lot. It is by the way also a coordinate independend fact what time the stationary observer reads off his clock when he receives a given timestamp from the freefaller, so that statement goes both ways, with the only difference that all the signals the...

I am well aware that any signal sent right at the horizon can only be received when the receiver crosses the horizon himself, that is because an outgoing lightray at the horizon has a constant radial coordinate. So the photon does not come to you, you have to come to the photon, so to speak...

Again, we are still speaking about the external observer, and the time we are talking about is the time he reads off his clock. The light ray itself does not even have a proper time at all. The fact that the external observer receives the timestamp τ=x from the freefaller when his own clock...

They are not the same, but you can calculate the one if you have the other. For example, you can always say "the signal I am receiving right now was emitted such and such years ago". In the case of a black hole, the stationary observer can do that all eternity long, and he will always receive...

That would be a big loss, since they really help calculating stuff, for example when which observer reveives what time stamp from another. For the external observer there is always a signal sent right now, and a later time when he receives that signal. The light travel times from the freefaller...

Far away from the black hole you will see almost no difference between the freefalling and the stationary clock at all, so I don't see what difference that is supposed to make. Sure, but the choice is not arbitrary (for example, just because we have a little gravity here on Earth you still...

I am not the one who claims that the horizon exists for any of those two observers 1 Gyr away from the BH, even for the 2nd observer who is moving slowly towards the black hole the formation of the horizon is in the future, but for sure not in his present. There is no way that any moving...

Because Lemaitre coordinates don't use the time he can read off his clock (t), but the proper time of the infalling observer (τ), see Wikipedia. We are still talking about the external observer, not the infalling one.

Sure, but in this case we were talking about the external observer, so it should be clear that it is his plane of simultaneity in which nothing falls through. Whenever you ask the external observer "where is the test-particle now?" (and this question has nothing to do with "where do you see the...

Why this, I didn't claim "the black hole doesn't form in any reference frame", I only said "the black hole doesn't form in the reference frame of the stationary observer, but it does so in the frame of the infalling observer".

That's not true, at t=34m41s he explicity says that it takes a finite proper time for the infalling observer to cross the horizon. That is no contradiction to the fact that in the bookeeper's frame of reference he stays outside of the horizon eternally. Susskind is completelly right with...

The coordinate time of the external bookkeeper is also an invariant, since it is the proper time of a field free stationary observer (stationary with respect to the fixed stars). Therefore we can surely say that in his frame of reference the matter never gets compressed beyond it's critical radius.