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A recent spacetime video about Kerr's objection to the existence of singularities has made want to clear up something about geodesics towards the origin in the Schwarzschild solution.

It is said that null geodesics of this kind always terminate at the singularity - but what about the worldlines of free-falling objects? If the (non-null) geodesic of such an object was parameterized on its local time or distance, would that terminate too? In other words, would an observer actually arrive at the singularity in finite time and after travelling a finite distance?

I have a background in differential geometry from a long time ago, but it is way too rusty to do the calculation myself.

This question is not about what would actually happen physically, I'm only interested in what the Schwarzschild metric implies.

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    $\begingroup$ Yes, I would have to look up the calculation to remind myself how to do it, but it is immediately obvious if you use the Kruskal-Szekeres coordinates. See Does someone falling into a black hole see the end of the universe? for some discussion of this. $\endgroup$
    – John Rennie
    Commented Mar 14 at 8:30
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    $\begingroup$ Related: Why does Roy Kerr claim that the Kerr black hole does not contain a singularity? $\endgroup$
    – PM 2Ring
    Commented Mar 14 at 9:02
  • $\begingroup$ @JohnRennie How is that immediately obvious from the Kruskal-Szekeres coordinates? In their discussions I only ever see the questions about the event horizon and whether a falling observer reaches that in proper time are discussed, and nothing about the same issue regarding the singularity (the parabolas in the diagram). $\endgroup$
    – John
    Commented Mar 14 at 9:36
  • $\begingroup$ @John there are null geodesics that hit the singularity after the massive body. $\endgroup$
    – John Rennie
    Commented Mar 14 at 11:13

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Null geodesics are of interest because their collection can give you the set of null-cones on the spacetime, which, in turn show the constraints on massive particle worldlines. I think what John Rennie was saying in the comments is that once you see the alignment of null cones in Kruskal-Szekeres coordinates, it is clear that a time-like worldline, after crossing the event horizon, will end at the singularity. And, because there is no compactification for KS coordinates (like on a Penrose diagram), "it is obvious" that this end must occur in finite proper time.

To calculate the infall time you need to integrate the geodesic equation. You could, e.g., follow these lecture notes by Joel Franklin (Reed College) on radial freefall. They show that the proper time in going from some initial coordinate $r_0$ to some final value $r$ is: $$ \tau(r) =\frac{ \pm 2}{3 \sqrt{2 M}} \left( r^{3/2} - r_0^{3/2} \right) $$ where the sign is chosen to be "$-$" for infall (and the initial velocity, $dr/d\tau$, has been chosen specially to get this simple form; see comments of Cham below). Thus the infall time to $r=0$ is $$ \tau_\text{infall} = \frac{2 \, r_0^{3/2}}{3 \sqrt{2M}} \quad \rightarrow \quad c \, \tau_\text{infall} = \frac{2 r_0^{3/2}}{3 \sqrt{ 2 M \left( \frac{G}{c^2}\right)}} \quad \rightarrow \quad \tau_\text{infall} = \frac{2 r_0^{3/2}}{3 \sqrt{2 G M}} $$ where I've followed Wald's prescription for returning to physical units. Alternatively, you could write this as: $$ c \, \tau_\text{infall} = \frac{2}{3} \sqrt{\frac{r_0^3}{r_s}} $$ where the Schwarzschild radius is $r_s = \frac{2 GM}{c^2}$. Taking the initial point to be the Schwarzschild radius, $r_0 = r_s$, the time inside the black hole is: $$ \tau_\text{inside BH} = \frac{2}{3} \frac{r_s}{c} = \left\{\begin{array}{ll} \frac{2}{3} \frac{2950\,\text{m}}{c} \approx 6.6\, \mu\text{s} & M = 1\, M_\odot\\ \frac{2}{3} \frac{1.2\times10^{10}\,\text{m}}{c} \approx 40\, \text{s} & M = 4\times10^6\, M_\odot\\ \end{array} \right. $$

Figure 31.1 in those lecture notes nicely shows the finite (proper) time of infall for the object, as compared to the infinite infall time as seen by an observer at infinity.

And there is no need for the infalling worldline to traverse any "distance". Once you cross the event horizon, the future singularity will arrive to you in that finite proper time. You don't need to go anywhere to find it.

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    $\begingroup$ I suspect there's a wrong numerical factor in your $\tau_{\text{infall}}$. I get $$\tau_{\text{infall}} = \frac{\pi}{2} \sqrt{\frac{r_0^3}{r_s}}.$$ $\endgroup$
    – Cham
    Commented Mar 14 at 13:49
  • $\begingroup$ Huh, I did not do the integration myself here, I just took Eq. (31.4) from the linked notes. Do you see an error there? $\endgroup$
    – Ben H
    Commented Mar 14 at 13:51
  • $\begingroup$ I just checked the linked lecture. The authors doesn't seem to tell clearly what was the initial velocity. Apparently ("(losing a constant of integration)" on page 2), he sets $v_0 = 0$ at infinity (I'm not sure). My own calculation sets $v_0 = 0$ at $r = r_0$ (and $\tau = 0$). My own value gives the maximal infall proper time. $\endgroup$
    – Cham
    Commented Mar 14 at 13:58
  • $\begingroup$ Yes, I see what you mean. Thanks for pointing it out. In addition to the "lost" integration constant, I don't immediately see the result in the second integration. I'll take look later today and update. $\endgroup$
    – Ben H
    Commented Mar 14 at 14:01
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    $\begingroup$ The result from the lecture is still nice (but he should clearly state what was the initial velocity at start: $r = r_0$). As you noted, dropping the integration constant means that $$\frac{dr}{d\tau} = \sqrt{\frac{2M}{r}}$$ which implies that $v_0 \ne 0$ at $r = r_0$. Then crossing the event horizon with a velocity implies that the infall proper time would obviously be shorter than dropping from rest. $\endgroup$
    – Cham
    Commented Mar 14 at 14:18

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