This is to provide a bit of elementary but not entirely obvious math to see better what is going on near the event horizon of a black hole. Near the horizon of the black hole things can be described with special relativity much more simply than with general relativity.

Consider a flat space and special relativity. What we will show is that if a ship uniformly accelerates northward, then there is a region behind some plane to the south from which news of events will never reach the ship, as long as it continues to accelerate. This plane is the ship‘s event horizon.

We adopt the Minkowski view of space time but we shall need only one spatial dimension. An event is characterized by a position and time: <x, t>. The interval between two events is (δx² − δt²)^1/2. Time like intervals are imaginary.

We adopt an unprimed coordinate system and then show that the portrayal of the situation is identical in other closely related primed coordinate systems.

There is a photon at <0, 0> traveling in the +x direction (North) with velocity 1. There is a ship at <1, 0> at rest but just beginning to accelerate at rate 1 in the +x direction. There is a clock on the ship currently reading 0. We will call its reading τ.

At this point we could set out differential equations and find <x(τ), t(τ)> as a function of τ. Instead I will write down the answer that I remember and then show that it is correct.

x(τ) = cosh(τ)
t(τ) = sinh(τ).

These equations are a description of the ship’s trajectory in parametric form. These provide the correct initial ship conditions including the rate of acceleration.

The meta physics is that the equations of physics are unaltered if we transform coordinate systems thus:
x' = x•cosh(r) + t•sinh(r)
t' = x•sinh(r) + t•cosh(r).
All of the geometric part of special relativity is expressed in these equations. The parameter r is called the rapidity. The most important thing is that this transformation does not change the interval (Dx² − Dt²)^1/2. If we compose transformations with rapidities r and r', we get a transformation of rapidity r+r'. In particular such a coordinate transformation leaves the ship’s itinerary expressed as:

x'(τ) = cosh(τ+r)
t'(τ) = sinh(τ+r).

from which we can see that the ship is indeed undergoing constant acceleration in its own frame. Now where is that photon? How much headway has it made toward the ship? In the new coordinates, the photon is still at <0, 0>! In the original unprimed coordinate system, when the ship's clock reads τ, the photon is at x = t = sinh(τ) but the ship has reached cosh(τ). The photon is gaining in this frame but sinh(x) never overtakes cosh(x).

In the accelerated frame of the ship the photon is standing still on the event horizon. If the ship should run out of fuel and stop accelerating, the photon will catch the ship.

When we say that an outward headed photon at the event horizon of a black hole does not make progress, we have the same situation as with the photon at <0, 0>.