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I study molecular biology; my skills in maths aren't the best – so I'm asking for answers that aren't purely mathematical, if possible. I ask you to elaborate on formulas or diagrams, if included in the answer.

I have a long standing interest in black holes/event horizons (I started reading up on them, every now and then, more than 10 years ago) and many of the concepts/problems that lay-people on this site ask about, and the corresponding answers, are not new to me. So I am not asking for ELI5-type answers either – if possible.

If this question has been asked before I would be thankful for a link. I did not find any answers, neither on this nor on other boards.

To the question: The title basically says it. I want to talk about Schwarzschild black holes/corresponding event horizons (for simplicity). As I "understand" it, to an outside observer an event horizon seems to be unreachable. An infalling object appears to freeze or "get stuck" just above the horizon, and vanish by redshifting. So let's suppose I start falling towards the EH myself: How can I change my outside-observer-POV to an inside-observer-POV? When would that happen (on my clock)?

I heard that there seems to be no experiment that I could conduct, to tell the moment when I cross the horizon. But the transition from outside to inside has to happen anyway, whether I realize it or not – doesn't it?

Closely related question (as I understand it, it is the same problem as above, just framed differently): I throw something toward the BH. I see it approach, get closer, slow down, redshift. I start moving toward the "frozen object". As I get ever closer, it appears to recede ever farther away – it is impossible for me to pick it up, even if I move close to the speed of light. Doesn't that mean that the EH itself appears to recede from me? Don't I have an infinite amount of space (and time) to cross, to get to the EH (or the object, for that matter)?

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So let's suppose I start falling towards the EH myself: How can I change my outside-observer-POV to an inside-observer-POV? When would that happen (on my clock)?

If you are freely falling towards a black hole then you really want to use better coordinates that express your situation. These are called Eddington-Finkelstein coordinates. In these you are going to be able to derive a proper time to cross the event horizon and also a proper time before you get crunched at the singularity.

As I get ever closer, it appears to recede ever farther away – it is impossible for me to pick it up, even if I move close to the speed of light. Doesn't that mean that the EH itself appears to recede from me? Don't I have an infinite amount of space (and time) to cross, to get to the EH (or the object, for that matter)?

Let's say you are a mile away from the event horizon of a huge black hole. In such a situation it could occur that tidal effects are negligible at the event horizon so that you wouldn't notice crossing it. To remain where you are you must be accelerating because the black hole will be pulling you. You let your apple fall from your grasp. The light that you see emanating from the apple is being redshifted due to your acceleration. You never see it cross the event horizon but just fade from sight.

Now you decide to take the plunge, i.e you turn your jets off. You are now freely falling after your apple - switch to E-F coordinates. The acceleration between you both is pretty constant so you are just moving after it and follow it into the event horizon. The red-shift is less. You are correct that you could accelerate to catch up with it to eat it. You don't need to go near the speed of light, just fast enough.

However, the singularity may be far away but crunch-time and spaghettification is coming!

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    $\begingroup$ @flippiefanus I'm not sure I understand you. I think it is true that we could pass the event horizon of a black hole without feeling any change. Are you saying this is incorrect? Or are you saying that if I freely fall into a black hole and my apple falls out of my hand just before entering that it will suddenly zoom away from me at the speed of light? Or have I totally misunderstood? $\endgroup$ – Matta Dec 9 '16 at 13:24
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    $\begingroup$ @flippiefanus Apologies but I still don't understand. Which statement are you referring to? Or better yet, what is your version of events? Let's say I ate my apple long ago so it's just me falling in, feet first. For a supermassive black hole the tidal forces at the event horizon can be negligible. If I am in free-fall and I can't feel the tidal squeeze then how could I tell that I was crossing the event horizon? Would my feet suddenly be ripped off me at the speed of light? $\endgroup$ – Matta Dec 9 '16 at 14:18
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    $\begingroup$ @Turgon Just to add to what you said: If I attach a clock to my apple then whilst I'm watching it fall in from way out I see the clock gradually slow down and the apple-clock never quite reaching the event horizon. So, yep - there is a coordinate singularity according to me, at the event horizon. If I start chasing my apple in then our relative time begins to tick at a similar rate. The banana I left behind would see us both slowing down. If my apple isn't too far away then there will come a time when our times are more or less in sync and I can catch up to it. $\endgroup$ – Matta Dec 10 '16 at 8:02
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    $\begingroup$ @Turgon In answer to your question: Near the event horizon the escape velocity approaches the speed of light, therefore accelerate hard if you want to escape. However the tidal forces may be very small here This means you won't be so squished and your feet will accelerate at the same speed as your head. However, when you get deeper inside, the gravity will increase and spaghettification will occur. This situation is true for a large black hole. A small one will have high tidal forces at the horizon. Therefore the intuition you gave is very true, but just for the little ones. $\endgroup$ – Matta Dec 10 '16 at 8:16
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    $\begingroup$ @flippiefanus Superb point, however... In SR anything travelling at c relative to an inertial observer will travel at c relative to all inertial observers. When we say we cross the event horizon at c we mean relative to that spacetime boundary, which cannot be treated as an inertial observer. You reason that I can Lorentz transform from a 'horizon' inertial frame to a 'head' inertial frame, which is not the case. John Baez has a nice summary $\endgroup$ – Matta Dec 12 '16 at 8:24
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(Disclaimer: purely layman in terms of relativity - specifics are probably way more complicated than how my thinking works- hopefully this makes sense though)

In reality there are no strictly "insider" or "outsider" viewpoints - It is a continuum of different frames of reference depending what's your position or movement with regards to the infalling object.

How I visualize this mentally is having "shells" around the event horizon with different speeds of time. Then how one might observe time for other objects (falling in or not) depends not only on their time dilation but yours as well. So close to event horizon time might run 10000x faster, but if you are observing it from the same shell, passing of time for that object would look perfectly normal for you (while objects away from BH would appear to fast-forward and objects closer to BH would appear slower). Whether you can simply add/divide these factors (probably not) is not really relevant for understanding how it works in general.

So, in theory, the outsider point of view is assumed to be located at infinity to negate any effect of the BH to that observer (especially no time dilation from the BH); in my visualization this shell is infinitely far away and thus has no time dilation factor (ie say time runs at 1x speed); obviously simply being infinitely long away from the infalling object would mean it'll take infinitely long for light to arrive to the observer thus meaning in reality no observations. So it's more a thought experiment than a real experimental setting.

Edit:

So to answer your actual answer, as a summary, if you are falling towards a black hole, you will always be viewing it as from an "insider" point of view. You may see other objects falling into a black hole which are closer to the black hole, and you would have sort-of outsider point of view for those objects (i.e. relative time dilation). As long as you eventually cross the event horizon, in theory you should be able to also observe these objects crossing the event horizon too. However, if you somehow back up or would remain hovering outside the event horizon, you'd remain as an outside observer for those in falling objects (i.e. never see the cross the event horizon; since relative time dilation between you and them grows asymptotically towards infinity)

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