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If we would hypothetically be exactly on the event horizon, we should see our own back, because of the circular motion of photons on the event horizon, right?

But what would be the image size, or $-$ asking differently $-$ how far away, would our back seem to be? Would it be magnified or minified, when compared to the image of a person $2 \pi R_{Schwarzschild}$ away?

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up vote 8 down vote accepted

The distance where light has a circular orbit is actually $1.5r_s$ not the event horizon. This distance is known as the photon sphere. In principle a shell observer hovering at this distance could indeed see their own back.

The proper distance is indeed just $2\pi r$, however the object would look bigger than expected because the curvature of spacetime has a focussing effect. Light slightly outside the photon sphere will spiral outwards away from the black hole while light slightly inside the photon sphere will spiral inwards towards the black hole. Therefore the light will not travel in a straight line. If we unroll the circular object to make it easier to draw the light rays we get something like:

Ray diagram

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+1 for a great graphical explanation. – Hennes Jul 5 '14 at 13:14

$r=1.5r_s$ for the Schwarzschild solution corresponds to the unstable maximum of the effective potential for a photon, therefore you won't be able to see much in practice, since practically every photon on this orbit will either fall in the black hole or escape to infinity.

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