# Does event horizon expand instantly or does it take time for large black holes?

When something falls into a black hole, its mass increases and its event horizon expands a little to reflect the new mass.

In case of very large supermassive black holes, with event horizon spanning light years, will event horizon expand instantly upon object entering it, in all directions for the full event horizon "surface", or it will take few years for the perturbation to propagate to the opposite end?

• The radius of supermassive black holes is much smaller than light years. For example, Sagittarius A* in the center of Milky Way has a mass of 400 million Suns, but the radius of only 17 times larger than the Sun and much smaller than the orbit of Mercury. Secondly, time slows down near a black hole. If you observe from afar, you would see things falling in to freeze at the event horizon, so in your frame nothing ever penetrates the event horizon or falls into a black hole. Thus the answer depends on the chosen coordinates. – safesphere May 1 '18 at 13:15
• @safesphere: 4 million $M_\odot$ not 400 million. – A.V.S. May 1 '18 at 20:22

## 2 Answers

While the first half of the answer (v1) by John Rennie provides correct timescales, for the process we are discussing its second half is completely wrong.

# Objects falling into a black hole enter the horizon in finite time by the clocks of outside observers

Let me elaborate: At this wikipedia page we can see the solution for a particle falling toward a black hole. We are interested in the asymptotic behavior of the radial coordinate as the particle approaches the horizon (or rather old horizon, before this infalling particle is incorporated in the black hole): $$r(t)\approx r_s\left(1+\exp\left(\frac{-c(t-t_0)}{r_s}\right) \right),\tag{*}$$ here $t$ is the Schwarzschild time or time by the clock of outside observer, $r_s$ is Schwarzschild radius (without the mass of a falling object), and a constant $t_0$ is determined by when and how the object started falling into the black hole.

Now consider the outgoing null geodesics (trajectories of massless particles such as photons flying away from the black hole) near the horizon of this black hole. If we disregard the effect of the falling object they would satisfy the equation $$r(t)\approx r_s\left(1+\exp\left(\frac{c(t-t_1)}{r_s}\right)\right).$$ But, if we consider the area near the horizon where our object is falling, we cannot disregard its influence on these trajectories. As radial null geodesics cross the worldline of the infalling object they are deflected by the gravitational field of this object (however small) and as a result they remain near the horizon for a longer time. And if this intersection occurs while that object is at a distance of $r_s+\delta r_s$ then this geodesic would no longer be able to escape the black hole, and be on the new horizon. The new value of horizon radius: $r_s+\delta r_s$ is the sum of old radius and addition from the mass/energy ($\delta m$) of falling object $\delta r_s \approx \frac{2 G \delta m}{c^2}$ (minus the losses of energy on radiation etc.) While the details depend on the geometry of the fall, the most important fact is that according to (*), the value of radial coordinate of $r=r_s+\delta r_s$ is achieved in a finite time according to an external observer: $$\Delta t\approx \frac{r_s}{c}\ln\left(\frac{ r_s}{\delta r_s}\right)\approx \frac{r_s}{c}\ln\left(\frac{M}{\delta m}\right).$$ Even when we have very light object falling into a very large black hole the resulting time interval is quite small by human standards. For example if we take a photon of cosmic microwave background with energy $k_\text{B}\cdot 3\,\text{K}$ falling into Sagittarius A*, then logarithm would be about 175 and $\Delta t$ about two hours. So such photon falling from a radius of $3 r_s$ into a supermassive black hole would cross event horizon of a black hole in a couple of hours by the clock of external observer.

To illustrate this process look at the spacetime diagram from the book by Andrew Hamilton p. 166:

This image uses Eddington–Finkelstein coordinates that are more suited for study of the near horizon processes. Purple curve is an apparent horizon before the falling particle $r_s=0.5$, red curve is the worldline of the falling particle and we see that it crosses the true event horizon when $r=1$. Null geodesics (thin black lines) between purple line and horizon are starting as outgoing but fall back into the black hole after being deflected by the falling particle.

Another point to note is that event horizon is a global construct and it depends not only on the past but also on the future of the black hole. So if in the future something else would fall into a black hole, the event horizon right now is expanding to accommodate the future increase in mass. (Of course if the increase in mass is small, and/or far away into the future, the horizon would be almost constant). So there is no instantaneous expansions.

Also relevant question: At what moment will matter falling into a black hole affect its size?

In this answer I completely ignored the effects of Hawking radiation and potential 'shrinking' of horizons due to it. This is well justified at current epoch for both stellar mass and supermassive black holes. At the very least temperature of cosmic microwave background is much greater than Hawking temperature of black holes, so their horizons would always be growing by absorbing CMB quanta.

• +1 You've answered the question I wanted to ask. Thanks! A follow-up question. What happens to an object of a finite size near the horizon? For example, a gold plated ring 1cm in size increases the event horizon radius, say, by 1um. Per your logic, the gold plating will penetrate the horizon, but wouldn't the rest of the ring appear frozen at the horizon to a remote observer? Also what happens to objects inside the horizon for a remote observer? Specifically, would the gold plating be stripped off and fall separately from the ring or be held by the ring (if we can ignore the tide forces)? – safesphere May 2 '18 at 12:16
• @safesphere: You cannot shift gravity from one part of extended body and reassign it to another part, different parts of extended objects would cross/increase the horizon independently as they fall in. Also note, that proper distances near the horizon stretch in radial direction, so 1cm of proper distance would correspond to much smaller $dr$ as it approaches horizon. So crossing of horizon by one end would not leave the other one outside, it would be following right behind. – A.V.S. May 2 '18 at 15:39
• @safespher: cont. While horizon does not introduces additional forces, tidal effects do cause internal stresses in the body, and so the motion of its parts may not be geodesic, but for large black holes and small bodies this effect is minor. MTW has a problem on stresses experienced by a human body as it falls in, and iirc human body disintegrates at several $r_s$ before it approaches a solar mass black hole, but survives well into horizon for supermassive b.h., a small ring would be much sturdier... – A.V.S. May 2 '18 at 15:45
• cont. Also what happens to objects inside the horizon for a remote observer?, As usual for horizon, nothing about inside objects could be observed from outside. – A.V.S. May 2 '18 at 15:53
• Thanks for the information! I appreciate it. This is what I am still struggling with. According to your logic and diagram, the size of the event horizon is gradually increasing while the object is approaching the black hole. At no point this size "abruptly jumps" to "swallow the object. So, when the object is approaching the already increased event horizon (the point 1,0 on the diagram), why doesn't the object "freeze" there for a remote observer. Despite the horizon increasing, at no point the object suddenly appears inside the increased horizon., but still has to cross it. What am I missing? – safesphere May 3 '18 at 2:57

The obvious example of this is the merger of two black holes as observed by LIGO. This has been extensively studied numerically so we have a good understanding of what happens. As the two black holes merge a single horizon is formed that is not spherical. However the non-spherical horizon oscillates and radiates away the non-spherical components (technically the quadrupole and higher components) as gravitational waves. A spherical horizon, or at least one indistinguishable from spherical, is formed in a few milliseconds.

For a small mass falling into a black hole the behaviour would be qualitatively similar though the initial perturbation to the event horizon would be smaller and would be radiated away more rapidly.

You specifically mention supermassive black holes, and these have diameters of around $10^{10}$ to $10^{11}$ metres or about 100 light seconds. That means any oscillations of the surface are going to take hundreds of seconds to propagate around the supermassive black hole. So while we would get behaviour that is basically similar to the smaller black holes it would take a lot longer, probably some thousands of seconds, for the perturbations to decay.

As viewed by an external observer nothing can ever cross the horizon because it takes an infinite time for any object to even reach the horizon let alone cross it. See for example How can anything ever fall into a black hole as seen from an outside observer? However even though strictly speaking objects never cross the horizon the external observer still sees the horizon become close to spherical very quickly. There are related questions that would be worth reading:

• `-1' As viewed by an external observer nothing can ever cross the horizon That is wrong. Event horizon for a black hole absorbing matter is an expanding surface, and objects falling into it would cross it in finite time by the clocks of outside observers. – A.V.S. May 1 '18 at 16:47
• +1 The logic of A.V.S. does not hold. While the horizon is expanding, it is expanding before the falling object approaches it. John's answer is correct, "As viewed by an external observer nothing can ever cross the horizon". The downvote was undeserved. – safesphere May 6 '18 at 0:00