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When does a black hole grow in size as measured by a distant observer? A falling body takes infinite amount of time to reach the horizon as measured by a remote observer, and the horizon of the black hole grows in radius only after it falls into the hole - which happenes after 'infinity' as measured by a remote observer. But that doesn't prevent a remote observer from observing a black hole getting bigger, apparent horizon growing larger and its gravity getting stronger. Doesn’t it pose a paradox? If the horizon increases in radius by 10 meters after Alice falls into a black hole, shouldn't a remote observer see Bob (another black hole visitor who jumps later) freeze 10 meters away from the previous horizon?

Response to the 1st answer: By 'previous' I meant the previous state of the same horizon.

Edit: When does the accretion disk grow, as seen by a distant observer? Shouldn’t it grow after the horizon grows?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/21319/2451 , physics.stackexchange.com/q/5031/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Nov 16, 2021 at 8:56
  • $\begingroup$ A number of comments removed. To answer a question, please post an answer, not a comment. $\endgroup$
    – rob
    Commented Nov 17, 2021 at 15:43
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    $\begingroup$ BTW, it takes around 10× the mass of Jupiter to increase $r_s$ by 10 metres. $\endgroup$
    – PM 2Ring
    Commented Nov 17, 2021 at 23:44

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There are two misunderstandings that I would like to help clarify:

  1. We do observe objects in our universe (like the M87 black hole), but in reality these objects are not fully formed black holes (the event horizon never forms in a finite time as seen from Earth). Why are they black? Because of something called infinite redshift. Photons that come from the object towards us, are redshifted to infinity so that we cannot detect them. Hence, the object looks black.

The light from that matter is ludicrously redshifted. The redshift doubles in a time comparable to the light-crossing time of the hole, which for the M87 hole is around 1 day, so after a year it's roughly 1 googol. Therefore, unless the hole ate a star quite recently, you can't actually see whatever the ray hits, and it'll appear perfectly black (plus some orange for the glowing semitransparent matter that the ray also passed through).

If an event horizon never forms for an outside observer, then what do (or don't) we see in the middle area on this real image of an actual black hole?

  1. When an object approaches the black hole, the reason we do not see it anymore is again, infinite redshift. Now what is very important to understand, is that when an object approaches this black hole (which is not fully formed as seen from Earth), it does not need to cross the horizon (or more precisely, we do not need to see it cross) to add to the black hole's mass. The gravitational field of the black hole extends outside the (not even fully formed) horizon. When the object (as seen from our view) asymptotically approaches the horizon, it already adds to the strength of the gravitational field of the black hole. By the no-hair theorem, and the definition (there are a few, but in this case they all agree on this) of the mass of the black hole, the black hole's mass stems from the gravitational field. The gravitational field's strength itself defines the mass of the black hole, now including the object that seems to asymptotically approach the horizon (then disappears because of redshift).

This isn't some accounting trick; it means we will never see an event horizon form. At this point someone will usually pop up and say that means black holes don't really exist. In a sense that's true in our coordinate system, but all that means is that our coordinate system does not provide a complete description of the universe.

How can anything ever fall into a black hole as seen from an outside observer?

So the answer to your question is that yes, we do observe from Earth black holes, and we can observe them growing as they accumulate more matter. But this is not a contradiction because we observe black holes that are not fully formed (in our view), and the objects that seem to asymptotically approach them add to the gravitational field of the black hole thus increasing the (from our view not fully formed) event horizon.

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  • $\begingroup$ What is a "fully formed" event horizon? $\endgroup$ Commented Nov 17, 2021 at 23:12
  • $\begingroup$ Does that mean, even though we can just see a frozen star, not even a black hole - the accretion disk grows facing its growing gravity in finite time, and we see a growing disk around it? $\endgroup$
    – Nayeem1
    Commented Nov 18, 2021 at 2:10
  • $\begingroup$ So a "fully formed event horizon" is an absolute horizon, rather than an apparent horizon? $\endgroup$ Commented Nov 18, 2021 at 8:26
  • $\begingroup$ @safesphere This is confusing me. IF a far away observer does not ever see an event horizon form, then how does the far away observer never see an object pass the event horizon that hasn't formed? How is that logically consistent, when the not fully formed event horizon is the thing that's supposed to cause the infinite redshift? $\endgroup$ Commented Nov 20, 2021 at 9:09
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It's time that confuses you. If a black hole has formed stuff falling in from the outside will still be able to fall in, which can be seen if you place yourself in the frame of a point particle falling in. The same particles seem to head for the horizon seen from faraway. The old horizon seems to grow if the particles appear to reach for it. It seems a paradox, because from our perspective, the particles never cross the horizon. But neither did the particles forming the hole initially. From our perspective, the hole is just a 2d shell, without volume, on which the particles get stuck. This surface grows if more particles fall in. Ater infinite time they actually reach the surface, while particles in the infalling frame actually reach far beyond the horizon (this strange ambiguity gives rise to the correspondence of surface information and bulk information, but that aside). So, seen from us there is nothing to cross. All matter seems to accumulate on the horizon, and it's this accumulation which defines the horizon.

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  • $\begingroup$ "It's this accumulation that defines the horizon' - So, as seen by a remote observer, there seems to be a surface where objects accumulate, which he perceives as the horizon. Very well. But that surface grows only when objects get past is, which is literally never as measured by a distant observer. Does that mean he will never see objects accumulate around a larger surface? $\endgroup$
    – Nayeem1
    Commented Nov 18, 2021 at 2:54
  • $\begingroup$ @Nayeem1 No. You assume here that there is a surface common to all black holes. There is only an imaginary surface. Imaginary and depending on the mass of the black hole. Even if this mass lays outside, as seen by us, then there is this surface. Even our planetary system has such a surface. It's about 4 kilometers in radius, if I remember correctly. It's to this surface that the planetary mass seems to implode. Within the infalling frame, together with mass and sun particles, you see everything tidally stretched to infinity in the blink of an eye. Space stretches FTL inside a developing hole. $\endgroup$ Commented Nov 18, 2021 at 6:13
  • $\begingroup$ @Nayeem1 From the outside it takes an infinite amount of time before all matter has accumulated at that surface, which is there already for all mass distributions you can imagine. Even the observable universe has one. With a radius bigger than the universe. But matter further away prohibits matter from collapse, so all matter is not collapsing to one sigularity, inside, as we are. If matter outside the universe would suddenly dissappear, we would be in a collapsing black hole, but this can't happen off course. $\endgroup$ Commented Nov 18, 2021 at 6:18
  • $\begingroup$ @Nayeem1 You write But that surface grows only when objects get past it. That is your "mistake". The surface is already there before any matter falls to it. Also the Sun has one, corresponding with its mass. $\endgroup$ Commented Nov 18, 2021 at 6:30
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I would say it never stops growing. Even when there’s no visible mass being absorbed, there’s always photons with energy going in.

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  • $\begingroup$ This is not relevant to my question. $\endgroup$
    – Nayeem1
    Commented Nov 18, 2021 at 2:13
  • $\begingroup$ @Nayeem1 Maybe not the body of your question but it’s definitely relevant to your actual question. $\endgroup$ Commented Nov 18, 2021 at 7:09

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