# Infinite Redshift in Schwarzschild Solution: What would you actually “see”?

Given the Schwarzschild solution

$$ds^2 = -\left(1-\frac{2 G M}{r}\right) dt^2 + \left(1-\frac{2 G M}{r}\right)^{-1} dr^2 + r^2d\Omega$$

the slope of a light cone is given by

$$\frac{dt}{dr}=\pm\left(1-\frac{2 G M}{r}\right)^{-1}$$

which is approaching infinity when approaching the Schwarzschildradius $$r_S=2GM$$.

For an outside observer, a light ray would never reach the horizon. This is due to the badly suited Schwarzschild coordinates. It does not appear using other coordinates, e.g. Tortoise or Kruskal.

But what would an outside observer "see"? The experience cannot depend on the metric one uses, as this would imply that there is a superior metric or true metric of nature, which is not what I would expect.

An outside observer can only "see" light rays coming the other way. Because the slope of the light cones reach infinity at the event horizon, then it would take an infinite time for light rays to reach an external observer, so the observer doesn't see them.

This behaviour of outwardly travelling light rays is not dependent on choice of coordinate system. e.g. in Kruskal-Szekeres coordinates the event horizon is also parallel to the geodesic of an outwardly travelling light ray.