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Let's say we have a particle free-falling directly towards the center of a black hole. Initially it's at rest at distance r from the center, that is equal to double the Schwarzschild radius of this black-hole. It is followed by a photon that is initially at a distance d behind the particle.

This photon was initially at such distance that it will catch up with this particle exactly when the particle passes the event horizon of this black hole.

How to calculate the initial distance d?

safesphere
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Kamil Szot
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    Here is the coordinate time of an object falling from $R$ to $r$: https://physics.stackexchange.com/questions/450032/524855#524855 - You can find the same formula for light from the radial Schwarzschild metric. Then set the difference to zero in the limit of $r\to 2M$. – safesphere Sep 06 '21 at 16:20
  • @safesphere Will the formula for t(r) for the light ray that's behind the particle be independent of the (changing) velocity of the particle? It's inertial frame of reference but as the particle accelerates the distance behind it from it's point of view gets shorter. Where or how to find that formula? – Kamil Szot Sep 07 '21 at 08:59
  • You can take the radial null Schwarzschild metric and solve it for $r(t)$ or for $t(r)$. For radial, drop the angular part. For null, set $ds\equiv d\tau=0$. You get the coordinate speed of light. Just integrate it. – safesphere Sep 07 '21 at 15:38
  • @safesphere I found really good answer to almost exactly my question here https://physics.stackexchange.com/questions/82678/does-someone-falling-into-a-black-hole-see-the-end-of-the-universe/82710#answer-396157 – Kamil Szot Sep 08 '21 at 11:04

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