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The aberration of light will cause an observer to still see a black hole as "distant" when the event horizon is crossed.

This means that if the observer looks directly toward the center of the black hole, it will not encompass a proportion of their field of view approaching 50%; this limit instead occurs at the singularity.

It is supposedly difficult for an observer to actually determine when they have crossed the horizon, due to the equivalence principle implying that nothing special happens.

My friend and I are thinking that an observer could determine when they have crossed the event horizon based on how much of their field of view is taken up by it, assuming they have equipment with them to accurately measure angles. What would this proportion even be?

I have absolutely insufficient knowledge to determine this using general relativity and the like, so any reasoning below is almost certainly incorrect. All we did was play around with trigonometry for a bit.

An infinitesimally small observer, in a static spacetime, should see half of their field of view taken up at the surface of a sphere (like lying on the ground and seeing half of one's field of view being taken up by the sky, therefore making the ground also take up half). So the proportion of the field of view taken up is $1/2$, i.e., $\Omega=2\pi$. The subtension is therefore $\theta=\pi/2$, which can be proven with a simple drawing.

But, since the observer is not in a static spacetime, they're moving. We know that $\vert\mathrm{d}r/\mathrm{d}\tau\vert$ cannot exceed $c$ outside the event horizon, so it must happen inside it. We assumed it happens arbitarily close to it. Locally, spacetime is still flat, and the speed of light is still $c$ relative to the observer, so we figured that maybe the inscribed angle is equal to the correct angle at this point along the journey. That would be $\pi/4$ at the event horizon, which results in $\Omega=\left(2-\sqrt{2}\right)\pi$. This is just under 15% of the field of view, or more precisely, $\mathrm{hav}\,\pi/4$. Then, when the observer gets to the singularity, or right before rather, they should see half of their field of view taken up, i.e., $\Omega=2\pi$. The subtension is therefore $\theta=\pi/2$. This fits with our previous assumption because $\mathrm{hav}\,\pi/2=1/2$.

Is our guess of 15%, or exactly $\dfrac{2-\sqrt{2}}{4}$ anywhere close to the correct answer? Please also demonstrate the correct way to calculate this.

Patrick O'Brien
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  • This is, if I recall correctly, an exercise in Moore's A General Relativity Workbook. One important point is that it depends on the observer's speed as they fall in, due to relativistic aberration effects. – Michael Seifert Dec 17 '23 at 13:24
  • Correction: the equation for the angle subtended by the black hole, as seen by an observer freely falling from infinity, is derived in Chapter 12 of that book. – Michael Seifert Dec 17 '23 at 13:32
  • @MichaelSeifert “it depends on the observer's speed” - Perhaps during the fall, but intuitively it should not at the horizon where the black hole should take the entire field of view and the rest of the universe should shrink to single dot exactly behind the observer. Is this not the case? (I doubt the OP will spend $50 on this book.) – safesphere Dec 17 '23 at 16:21
  • @safesphere But aberration of light is still happening inside the horizon, isn't it? The rest of the universe can't possibly shrink to a single dot if dΩ/dτ is positive in the past (behind the observer), or whatever quantity it is that describes the distortion of the field of view. I don't know how to calculate it though, but this is what I would think is the case. – Patrick O'Brien Dec 21 '23 at 06:16
  • @MichaelSeifert Yes, I don't plan to buy a $50 book, which is why I wanted to ask for help here. But what you say implies that the equation for the angle subtended by the black hole is different from what I have. Is it only dependent on the observer's speed, or also other factors? – Patrick O'Brien Dec 21 '23 at 06:18
  • As you approach the horizon, we “see” you infinitely flattened while you see us infinitely farther away. To send a light beam to us, you must send it radially. Your light in any other direction hits the black hole. Light is reversible. I can see you only if you can see me. So you can receive only radial light from outside when you are at the horizon. I don’t have the book and cannot post an answer, but see this question for the reversed case: https://physics.stackexchange.com/questions/787403 - The angle is coordinate dependent and your question is in the local coordinates near the horizon. – safesphere Dec 21 '23 at 17:20

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