Black Hole infinite distance finite time paradox

Schwarzschild metric at the event horizon shows that, a small distance as perceived by a distant observer is in fact an infinite distance for a falling observer. Yet the falling observer crosses the event horizon and untimately reaches the centre singularity in a finite time. Doesn’t that create a paradox?

How can someome cross an infinite distance in finite time?

• Is the statement in your first sentence correct? May 23 '20 at 12:38
• This is not true. For any infalling observer the proper-time to the singularity is finite.
– user107153
May 23 '20 at 12:44
• It is true. At event horizon the proper distance dS = dr times infinity. And, I am not talking about proper time, I am referring to proper distance here. May 23 '20 at 12:50
• Proper distance is not ds, but an integral of ds. Do the math and you’ll see the integral is not infinite. May 23 '20 at 13:37
• Right. If at one point ds = Infinite, how is it possible that the total S is finite? May 23 '20 at 22:07

$$\int_{r_{\rm s}}^{r_{0}} \sqrt{|g_{rr}|} \, dr =\int_{r_{\rm s}}^{r_{0}} \frac{1}{\sqrt{1-\frac{r_{\rm s}}{r}}} \, dr = \sqrt{(r_{0}-r_{\rm s}) r_{0}}+\ln \left(r_{0}+\sqrt{(r_{0}-r_{\rm s}) r_{0}}-\frac{r_{\rm s}}{2}\right) = {\rm finite}$$
For the stationary bookkeeper this is still larger than $$r_{0}-r_{\rm s}$$, but smaller than $$\infty$$.
If you are in freefall with the negative escape velocity the gravitational depth expansion also exactly cancels out with the kinematic length contraction since $$v_{\rm esc}=c \sqrt{r_{\rm s}/r}$$, therefore $$|g_{rr}|$$ in raindrop coordinates is $$1$$, and the proper distance becomes exactly the coordinate distance.