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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?

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  • $\begingroup$ Is the statement in your first sentence correct? $\endgroup$ May 23 '20 at 12:38
  • $\begingroup$ This is not true. For any infalling observer the proper-time to the singularity is finite. $\endgroup$
    – user107153
    May 23 '20 at 12:44
  • $\begingroup$ 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. $\endgroup$ May 23 '20 at 12:50
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    $\begingroup$ Proper distance is not ds, but an integral of ds. Do the math and you’ll see the integral is not infinite. $\endgroup$
    – safesphere
    May 23 '20 at 13:37
  • $\begingroup$ Right. If at one point ds = Infinite, how is it possible that the total S is finite? $\endgroup$ May 23 '20 at 22:07
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It is only infinite for an infinitesimaly short coordinate length, so the integral is of course finite:

$\int_{r_{\rm s}}^{r_{0}} \sqrt{|g_{rr}|} \, dr =\int_{r_{\rm s}}^{r_{0}} \frac{1}{\sqrt{1-\frac{r_{\rm s}}{r}}} \, dr = \sqrt{(r_{0}-r_{\rm s}) r_{0}}+\ln \left(r_{0}+\sqrt{(r_{0}-r_{\rm s}) r_{0}}-\frac{r_{\rm s}}{2}\right) = {\rm finite}$

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.

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