Is this answer from here stating that an external observer can see a black hole mass grow, correct?

This question (is an external observer can ever see the mass of a black hole increase--motivated by the standard view that it will take an infinite time for the external observer to see the mass crossing the event horizon) is a question that comes time and again in [different variations][2] but asking basically the same. Many are marked as duplicates and when you go there that one is also marked as a duplicate in an endless loop (well, it ends somewhere). I also see this question is other physics forums. The multiple answers that really try to address the point are always similar: No, we cannot see it grow, or we "see" the mass growing but is actually outside the horizon (but at the same time we can never measure that), and many other variations. Many of the answers actually contradict a little each other. I do not know the actual identity of any of those who answer, and of the few that I know their identity; they are not really experts in the field.

Then I found the only answer different to all of them. It is here and it says that "he new matter to cross a critical surface that is outside the original event horizon of the smaller black hole. This occurs in finite time, even from the external observer's viewpoint, and the black hole simply grows in size" and " The time is infinite in the limit when the mass of the infalling object goes to zero and there's nothing else that makes the black hole grow. So the infalling object is just a "probe".

  • $\begingroup$ Are you asking if you can visually see ingoing matter cross the new (expanded) event horizon? If so no, light sent outward at the event horizon stays on the event horizon forever in GR. Or are you asking if it will cross the event horizon at some finite coordinate time? It can, depending on the choice of coordinate system. Or are you asking if the spacetime curvature in some region outside the black hole of initial mass M will transform into the appropriate curvature for a black hole of mass M + m after some falling mass m passes through that region? In this case I believe it will. $\endgroup$
    – Hypnosifl
    Commented Nov 14, 2014 at 5:15
  • $\begingroup$ possible duplicate of How can anything ever fall into a black hole as seen from an outside observer? $\endgroup$
    – user4552
    Commented Nov 14, 2014 at 6:18

1 Answer 1


The point Luboš Motl makes in the answer you link is that the infinite time to reach the event horizon is calculated using the Schwarzschild metric. However the Schwarzschild metric is a time independant solution i.e. it only describes a black hole that has existed unchanged for an infinite time and will continue to exist unchanged for an infinite time into the future.

Any mass dropped into the black hole will change the black hole because obviously the additional mass increases the mass of the black hole. Because of this the black hole is no longer time independant and the Schwarzschild metric can't be used to describe its spacetime geometry.

However for small masses and big black holes the perturbation created by the infalling mass is so small we can ignore it, and in the limit where the mass of the infalling object is zero the perturbation disappears and the Schwarzschild metric is an exact description. This is what Luboš means by a probe mass - it's a mass so close to zero that it's affect on the black hole can be ignored. So a probe mass will take an infinite time to reach the event horizon of an eternal black hole.

But consider the opposite extreme. Suppose you consider the merger of two equally massive black holes. This clearly can't be described by the Schwarzschild metric because the geometry is completely different. So just because the Schwarzschild metric predicts infalling matter takes an infinite time to reach the horizon does not mean that two merging black holes will take an infinite time to merge.

I have to confess that I do not know whether merging black holes take a finite time as observed from a large distance. The implication of Luboš' answer is that they do not, though he doesn't provide evidence for that claim. After some brief Googling I can find no definitive statement on this issue. I suspect the merger process is not well enough understood for anyone to give a definitive answer.

There is some related discussion in the answers to the question Why does Stephen Hawking say black holes don't exist? and you may be interested to read this.

  • $\begingroup$ "I have to confess that I do not know whether merging black holes take a finite time as observed from a large distance." No matter could ever be observed crossing either horizon (including matter that had been right between the two event horizons before the collision), if that's what you mean--the event horizon is defined in a "teleological" way that takes into account the entire future history of the black hole, so that if a light ray is emitted on the horizon in an outward direction, by definition its entire future worldline lies on the horizon. $\endgroup$
    – Hypnosifl
    Commented Nov 14, 2014 at 17:02
  • $\begingroup$ See for example the discussion "Teleological nature" of the event horizon on pages 173-175 of Black Hole Physics: Basic Concepts and New Developments, viewable on google books here, in particular the comment starting at the bottom of p. 174 saying that the horizon "bounds not so much a region with an especially strong gravitational field ... but rather a region with very specific global properties; namely, no rays escape from this region to infinity." $\endgroup$
    – Hypnosifl
    Commented Nov 14, 2014 at 17:06
  • $\begingroup$ However, there is also an apparent horizon which is defined in a non-teleological way that doesn't depend on the future--apparent horizons can form and "jump" in discontinuous ways, and they also depend on the definition of simultaneity in the coordinate system you're using, not sure whether one could ever see light emitted from behind an apparent horizon. A diagram of apparent horizons in a collision of two black holes can be seen here. $\endgroup$
    – Hypnosifl
    Commented Nov 14, 2014 at 17:36