# How can matter falling into a black hole make it grow?

None of the answers to this question or this question answer my question. Once an object falls past the event horizon, it can no longer act as a source of gravity. Also, if one tiny object falls into the black hole then later a larger object falls into the black hole, we can make a simplifying approximation that neglects the increase in size caused by the tiny object falling into the black hole and treat it is thought the tiny object doesn't change space-time at all. Then the tiny object will pass the event horizon in finite time long before the larger object falls into the black hole and makes it bigger. Even long after the larger object passes the event horizon, we will still see the tiny object as not yet having passed the event horizon. It seems like we can derive the contradiction that we see the tiny object near the event horizon of a larger black hole but I think I have the solution to it. Could this be the explanation of how an object can make a black hole bigger after it falls in?

The reason is similar to the reason why gravity can escape a black hole. The gravity outside a black hole is not caused by the matter in it. Instead, just like one Quora answer said, the gravitational field outside a black hole is a self-sustaining gravitational field.

Similarly after an object falls into a black hole, it is not the source of the stronger gravitational field. Instead, because of its own gravitational field, it changes space outside the black hole before it passes the event horizon then the stronger gravitational field sustains itself.

In the case where a tiny object falls into the black hole then a larger object falls into the black hole, after the larger object falls into the black hole, we're really seeing the tiny object from an earlier time before the black hole got bigger. That might make it's appearance look very weird because we're looking through the space outside a larger black hole into the space closer to the event horizon the way it was a long time ago before the black hole got bigger.

According to this answer, an object actually takes infinite time to reach the event horizon and after the black hole gets bigger, the object moves further away from the singularity with the event horizon but I believe that information is wrong. I think that's only what happens to a photon hovering at the event horizon from the way the event horizon is defined. The apparent horzion is where a photon can hover at a constant distance from the singularity but the event horizon is defined to be moving away from the apparent horizon exponentially because a photon following the apparent horizon will be unable to escape because the black hole will get bigger moving the apparent horizon further out and preventing the photon from escaping.

If you have a shell of mass ℳ₁ which is collapsing to a black hole, in the system of an external observer its radius will, because of time dilation, only converge to, but always stay a little bit larger than r=2Gℳ₁/c².

If you now decide to let a second shell with the mass ℳ₂ collapse into the first shell, the radius of the second shell will, in the system of the external observer, also always stay an infinitesimal bit larger than the event horizon of the combined masses r=2G(ℳ₁+ℳ₂)/c², while the first shell will still stay larger than r=2Gℳ₁/c² after it is surrounded by the second shell (and it shrinks even slower, since the second shell maked the time dilation relative to an external observer even larger).

So from the perspective of an external observer, no mass or mass combination gets smaller than its own event horizon (because of the shell theorem you can only count the mass inside a shell, and this, in the system of the external observer, always stays larger than its horizon), but mass ℳ₁ can be inside the future event horizon of ℳ₁+ℳ₂.

Let's say ℳ₁ is the 1 gramm in the center of the earth and the total mass of the earth is ℳ₁+ℳ₂. The center of the earth is also inside the Schwarzschildradius of the earth, but since the earth is larger than its Schwarzschildradius this does not mean that ℳ₁ is actually inside the event horizon of the earth. It is inside the Schwarzschildradius of ℳ₁+ℳ₂, but like the radius of ℳ₁ is larger than 2Gℳ₁/c², and the radius of the earth larger than 2G(ℳ₁+ℳ₂)/c², nothing is behind a true horizon.

The same goes for black holes, with the difference that their mass eventually does become smaller than their event horizon in a finite proper time of an infalling observer, but not in a finite coordinate time of an external observer.

In the above animation you have a red mass M which is always an infinitesimal bit larger than its own event horizon 2GM/c², and a black Kugelblitz shell of mass equivalent m which is collapsing onto the red mass. The black shell always stays larger than 2G(M+m)/c² (see the numeric display on the bottom of the animation, where the units of the combined mass M+m and G=c=1), so no event horizon is ever crossed, it can only be engulfed by the future horizon of additional masses.

For astrophysical black holes you integrate over a bunch of infinitesimal shells, and every shell is always a bit larger than the event horizon of the total mass inside the shell.

• no mass or mass combination gets smaller than its own event horizon that is wrong (especially its own part). For a given black hole there is only one event horizon. That is a global entity, and contrary to what shell theorem makes you believe event horizon position can depend on the motion of an outside shell (for example on whether this shell in the future will collapse around you). – A.V.S. Aug 13 '18 at 5:30
• we are talking about the reference frame of an external observer, and in that frame no mass can shrink below its horizon radius because of the time dilation, it can only converge to that radius, and the black hole needs an infinite coordinate time to form. – Yukterez Aug 13 '18 at 6:02
• You are mixing the notion of apparent horizon (for example the first shell 'feels' an apparent horizon at $r=2M_1$) and event horizon. In your two shells example that would be surface of rotation of null geodesics that approach from the inside $r=2(M_1+M_2)$ as $t\to \infty$. And this null geodesic crosses the worldsurface of the 1st shell at finite Schw. $t$ and outside the first shell's apparent horizon. – A.V.S. Aug 13 '18 at 7:11
• If you're rather being smart than reading what I wrote, why don't you write your own answear instead? – Yukterez Aug 14 '18 at 2:25