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If Alice falls into a black hole wearing a wrist-watch, an outside observer Bob sees Alice freeze at the horizon, her watch seems to have stopped. This also means that all light that originates from Alice is redshifted asymptotically to zero Herz.

Now my question is what Alice sees when she is just a few femtometers above the event horizon and looks back at Bob. Is Bob blueshifted and does his watch seem to spin to the end of time in a wink of her eye? Can Alice see the universe evolve to the end of time and see the black hole below her evaporate? In other words: if Alice is redshifted to zero to Bob, should not Bob be blueshifted to infinity to Alice?

I believe the actual amount of blueshift is not explained by the answers to question: What will the universe look like for anyone falling into a black hole? ?

Deschele Schilder
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anneb
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  • The accepted answer to the other question says "Since you are now moving along a free-fall geodesic, the light from other objects would no longer be blue-shifted because you would be in the same inertial reference frame." Why does this not answer your question? – WillO Oct 02 '20 at 02:18
  • @WillO The question is about Alice hanging just above the horizon. – Deschele Schilder Oct 02 '20 at 04:11
  • @DescheleSchilder : No, the question is quite unambiguously about what Alice sees. – WillO Oct 02 '20 at 04:14
  • @WillO Indeed: Now my question is what Alice sees when she is just a few femtometers above the event horizon and looks back at Bob That is, not about Alice falling freely. – Deschele Schilder Oct 02 '20 at 04:21
  • @Deschele Schilder: The question here says she is falling. You appear to have a totally different question in mind, which is of course fine but not terribly relevant here. – WillO Oct 02 '20 at 04:24
  • @WillO I gave you a quote. Isn't it clear? And the question in the question box? The OP says he thinks the question you (and he/she) refers to doesn't answer this question. – Deschele Schilder Oct 02 '20 at 04:41
  • (I should point out that the highly upvoted but unaccepted answer to the other question is much better and includes relevant links to published literature.) – WillO Oct 02 '20 at 04:46
  • @WillO But that are answers to another question. Are you referring to my answer? If so, what you suggest I should add? Some links? – Deschele Schilder Oct 02 '20 at 04:53
  • @WillO why is it relevant if Alice is free falling or just falling using some force to counter her speed? I understand that at whatever speed Alice approaches the horizon, she will always redshift, not because of her falling speed, but because space-time is warped in such a way that time stops relative to Bob at the horizon? The other answer does not claim that Alice always approaches lightspeed at the horizon relative to Bob , whatever the size of the BH and whatever her starting position? – anneb Oct 02 '20 at 05:52
  • Hi Anne, in my answer to the linked question I compute the shift you are asking about. – John Rennie Oct 02 '20 at 15:16
  • @JohnRennie Why is this a duplicate? In the linked question, Alice isn't able to look at Bob (she is inside the BH), while in this question she is able. – Deschele Schilder Oct 03 '20 at 09:09

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You're right. Let's assume Alice is hoovering just above the horizon (let's not consider the fact that an enormous amount of energy is needed for this).
Her time is almost standing still wrt to the most parts of the Universe. She sees the hole evaporate in a flash (though when the hole gets smaller she'll have to maintain the infinitely small distance to the shrinking horizon.

When the hole has evaporated (how long this takes depends on the mass of the hole), she'll be in sync again with most parts a the Universe (where gravity is weak).

So, indeed, it works in both directions, but oppositely.

EDIT
After reading the comment below by @anneb, the situation gets more complicated.
Of course, Alice needs a huge energy supply (dependent on the BH's mass) of energy to hoover above the BH's horizon. In the part above I assumed to ignore this.
If she could fill her rocket with energy extracted from the BH and her rocket will use photons to propel the rocket and counteract the gravity of the hole. The photons will re-enter the hole though, so the mass of the BH will stay the same. So there will be no difference.

The question remains essentially the same if Alice is hoovering at a considerable distance. In that case, she still needs much fuel to counteract the BH's gravity. If this is done again by ejecting photons, the photons will be absorbed by the hole which thus gets heavier.
Alice makes sure that the distance to the horizon stays the same. While her tank (say positrons and electrons) gets emptier it will in time become easier to stay put where she is.
The BH's mass has increased, so the time to evaporate has increased also. In that case, it takes more time for Alice to get back in sync with the clocks in the major part of the Universe, where gravity is absent. The times on her clock and the times on the clocks in empty space will differ though.

It needs a calculation to find out if the rocket's mass, i.e., including the fuel (which we assume to be positrons) will be that high to form a BH. But since I just woke up, I'll do that later.

Deschele Schilder
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    maybe the required amount of energy is physically relevant to this thought experiment because that energy would increase the gravity between Alice and the BH? if Alice could somehow convert the mass of the BH to anti-falling energy, it wouldn't surprise me if she would need to consume the complete BH to make the hovering experiment work? – anneb Oct 02 '20 at 06:46
  • @anneb Good point! I made an edit. in the answer. – Deschele Schilder Oct 03 '20 at 10:11