Does the mass of matter falling into a black hole affect the size of an event horizon the moment it passes through it, or when it has been incorporated into the singularity?

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    $\begingroup$ The asymptomatic behavior of approach as seen from afar certainly affects the question; nearly to the point of rendering the question ill-posed. $\endgroup$ Commented Mar 12, 2018 at 20:39
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    $\begingroup$ "The moment it passes through it" is ill-defined for an outside observer, see physics.stackexchange.com/q/21319/50583. $\endgroup$
    – ACuriousMind
    Commented Mar 12, 2018 at 20:44
  • $\begingroup$ ACuriousMind is correct. In technical terms, the mass of a black hole has to be defined using some definition such as the ADM mass. If you use the ADM mass, it's defined for an asymptotically flat spacetime, and it's conserved, so clearly it can't change -- ever. It doesn't change even during the initial formation of the black hole. $\endgroup$
    – user4552
    Commented Mar 12, 2018 at 21:26
  • $\begingroup$ @BenCrowell. Even if the ADM mass is not changed, matter outside and inside black hole is quite different physical situations that could be discerned by external observer in a finite time. $\endgroup$
    – A.V.S.
    Commented Mar 14, 2018 at 21:29
  • $\begingroup$ @dmckee: the asymptotic behavior in question is $\exp(-t/\tau)$, solving for t we would get logarithm which is 'almost non-diverging' function, so the question is quite well posed even for an outside observer. $\endgroup$
    – A.V.S.
    Commented Mar 14, 2018 at 21:33

1 Answer 1


Indeed, as commenters already suggested, for test particles the question of when they cross horizon (and when they are 'incorporated into singularity') is rather difficult to define if we wish to take the point of view of outside observer. However, your question does have an unambiguous answer (at least in terms of order of magnitude calculations). The key is that usually such 'test particle' is assumed to have zero mass, while we want to calculate when does the mass of this particle is 'felt' by the black hole. To do that we need to assume finite mass of the falling matter and consider the backreaction of this mass on the metric. And so we will see that the point when the mass is incorporated into black hole by increasing its horizon radius happens at a finite time as measured by the asymptotic observer.

For simplicity, instead of one point-like observer, let us consider the spherical shell (of dust-like matter with mass $\delta M$) falling into the Schwartzschild black hole (of mass $M$). This situation possesses spherical symmetry, and so Birkhoff's theorem applies: we have Schwarzschild metric with mass $M$ and gravitational radius $r_g=2 M$ inside the falling shell and with mass $M+\delta M$ and gravitational radius $r_g'= r_g + 2 \delta M$ outside of it. We could find the motion of the shell by using appropriate junction conditions, but if we for simplicity assume that $\delta M \ll M$ then the motion of this shell would coincide with radial geodesic for the unperturbed mass. And so, when the shell has the radius $r=r_g'$ it occurs in finite time (let's denote it $t'$) by the clock of outside observer. That would be the time, when the matter of the shell is fully incorporated into the black hole: no observer outside of the black hole could after that detect any trace of the shell. And if there was say transmitter falling with the shell, the latest moment of time when the signal from it could reach outside is $t'$ (of course such signal must also spend some time climbing to finite distance away from black hole)

If the mass $\delta M$ is falling from the distance of several $r_g$ to a distance $r$ that is slightly more than (unperturbed) $r_g$ that fall would last approximately $$ \Delta t = t - t_0 \approx \frac{r_g}{c} \ln \frac{r_g}{r-r_g}, $$ (the necessary equations could be found e.g. here, we are interested in case of zero angular momentum and leading term diverging at $r=r_g$). Replacing with $r$ with $r_g'$ we obtain: $$ \Delta t \approx \frac{r_g}{c} \ln \frac{M}{\delta M}. $$ That is the (order of magnitude) answer. The factor in front of it is Schwarzschild radius light-crossing time. We see that for $\delta M$ comparable with $M$ logarithm would be rather small and time in question would simply be 'several light-crossing times'. But even for large ratios logarithm function makes the time reasonably small. For example, let us consider astronaut $\delta M\approx 70\,\text{kg}$ falling into the Sagittarius A* black hole. Logarithm would be $\approx 81$ and $r_g/c$ is about 40 s and $\Delta t\approx 53\,\text{min}$, which is quite finite from a human perspective.

The answer would not change much if we include into consideration aspherical configurations and nonzero angular momentum. But in that case one needs to remember that such aspherical accretion tend to produce gravitational radiation which perturbes black hole horizon. This perturbations (so-called 'ringing down') exponentially decay with time constants about Schwarzschild radius light-crossing time.

For more technical discussion one can look at:

Frolov, V., & Novikov, I. (2012). Black hole physics: basic concepts and new developments (Vol. 96). Springer Science & Business Media, google books.

  • $\begingroup$ Nice answer, well thought out $\endgroup$
    – Bob Bee
    Commented Mar 19, 2018 at 0:39
  • $\begingroup$ I see the following problem in the answer: 1) First, take a spherically symmetric, big mass material shell, and, referring to Birkhoff's theorem, construct the formulat Δt with r′g. 2) But then we apply the same formula to an incident small test body, a spacecraft, and get 53 minutes for it.But the small test particle does not satisfy Birkhoff's theorem case, it does not make extra spherically symmetric surrounding matter distribution, and it has negligibly small mass. The formula Δt calculated from Birkhoff's theorem does not apply for the case of the test particle or an astronaut. @A.V.S. $\endgroup$ Commented Feb 18 at 16:24
  • $\begingroup$ It's not a “big mass material shell” it's any mass material shell. And the point is you cannot use “test particle”, test particles do not exist it is an approximation that breaks down since any massive object produces gravitational field that would modify the horizon. As for non-spherically symmetric matter, higher multipoles decay fast, so after several $r_g/c$ times the outer metric is indistinguishable from metric of spherical shell, and in particular the “moment of horizon increase” cannot differ significantly from the estimate provided. $\endgroup$
    – A.V.S.
    Commented Feb 18 at 20:02

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