Why is stressed in Kruskal metric that the part in $u$ and $v$ is conformally flat?

Question

Many books show how Kruskal metric, derived putting together both Eddington-Finkelstein results, is conformal flat (if just the first 2 coordinates are considered): $ds^2 = \left(1-\frac{2GM}{r} \right) dv \, du$. Indeed the final result is : $$ds^2 = \frac{32G^3M^3}{rc^6} \operatorname{exp}\left(-\frac{rc^2}{2GM}\right) \left(dv^2-du^2\right),$$ that is conformally flat. To be conformal means that the physic is unchanged so makes sense to stress that solutions maintain the physics. But for other coordinates changes, like Eddington-Finkelstein, this is not used. I'm missing something. For example some books neither talk about conformally flatness, just do the calculations.

Answer 1

First lets clear up some confusion. You say:

To be conformal means that the physic is unchanged

This is not what "conformal" means. Conformal means that the angles are unchanged. In the context of GR, this in particular means that the causal structure remains unchanged. The phyisics in the meantime can be quite different.

In particular, the $uv$-block of the Kruskal metric being conformal to Minkowski space means that (radial) null rays follow diagonals. This makes it relatively easy to further understand the causal structure of the extended Schwarzschild solution.

One of the reasons that books often stress this property is that writing a metric in this "conformally flat" form is a key first step towards constructing the Penrose diagram for a Schwarzschild black hole, which can now be compactified using the same conformal transformations as used for Minkowski space.