$2$nd law of thermodynamics of an isolated system as it passes into a black hole?

Question

So I'm quite confused by this.

Let's say I have $2$ friends $A$ and $B$ with metrics:

$$ ds_A^2 = -dx_1^2 + dx_2^2$$

and

$$ ds_B^2 = dx_2^2 - dx_1^2$$

Ideally I would imagine a symmetry in their dynamics say:

$$ S_A(x_1,x_2) = S_B(x_2,x_1)$$

where $S$ is the action.

Let's say I have an isolated system in a box. Consider the second law of thermodynamics for an isolated system:

$$ \Delta S_{system}(t) \geq 0$$

Now, if I throw the box in a blackhole. Now, considering this:

The point of all this is that the coordinates are not spacetime - they are just labels we attach to spacetime. What happens inside an event horizon is not that time and space swap places but rather that the labels we call Schwarzschild coordinates behave oddly inside a black hole.

Is the following true for the box after passing the horizon?

$$ \Delta S_{system}(t) \geq 0$$

Or is it?

$$ \Delta S_{system}(r) \geq 0$$

Answer 1

Second one.

From equivalence principle and relativity principle, the (small) box physics is governed by its proper time and proper time under event horizon grows as $r$ shrinks.