According to the General Theory of Relativity, the coordinate time distance per spacetime distance traveled by a particle freely falling into a black hole gets closer and closer to $0$ as the particle approaches the event horizon

$$\mathrm{d}s^2 ~=~ -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left( 1-\frac{2M}{r} \right)^{-1}\mathrm{d}r^2 + \dots.$$

Equivalently, an observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily long time. However, Hawking radiation implies that black holes don't last forever, but instead shrink and fade away? What happens to everything that is "currently" falling into the black hole during that time?

Edit: What this seems to imply is that as a particle gets close to the event horizon, the rate and intensity of Hawking radiation will increase and increase until most of the mass of the black hole is ejected, in the form of Hawking radiation, out towards the falling particle.


If, as is canonically accepted, the black hole evaporates within a finite time, then the freely falling particle will be released from its gravitational attraction. By the premise of the question, in the observer frame the particle can not have passed the event horizon during the intervening period. This would also solve the quantum information paradox, as the information represented by the particle would never have been lost.


The short answer is that noone knows, since your question is basically a statement of the black hole information paradox. Here's another way of phrasing the question:

Suppose we have a big pile of toasters that we plan to squish together really hard to form a black hole. We also have a big pile of houseplants of the same mass, which we can also form a black hole out of. We now have two equal mass nonspinning black holes: one made of toasters and one made of houseplants.

While the holes exist, the toasters/houseplants are in principle "painted" on the black hole horizons, so we can always in principle tell which hole is which (in practice, of course, this will be basically impossible).

However, real black holes emit Hawking radiation and will eventually evaporate. That radiation is exactly thermal with a distribution that depends only on the mass. So the toaster and houseplant holes will emit identical radiation. Once completely evaporated, they will therefore end as two identical radiation fields. We have therefore constructed a map from two different states (toasters and houseplants) to the same radiation state (the post-evaporated hole). This map is not one-to-one and therefore non-unitary, which is impossible quantum mechanically.

Basically, something has to be wrong with the above reasoning, but I think it's safe to say that noone knows quite what it is. There is now fairly strong evidence that information is not lost, and whatever happens, at the end of the day it is possible to recover the toasters from the radiation. For example, for black holes in anti-de Sitter space it is possible to show that the map is in fact unitary. Perhaps something goes very wrong in asymptotic flatness, but this seems unlikely.

One tempting argument is to ascribe all the information conservation to Planck-scale quantum gravity effects: the Hawking radiation proceeds as normal until the hole fills about a Planck volume, then suddenly quantum gravity kicks in and saves the information. The problem with this reasoning is that it essentially implies that a Planck volume can hold an arbitrary amount of information, since the original hole can be of arbitrary size. So this is almost certainly not the answer.


We know there is only one reality, measured from different reference points. So the pertinent question is what is the reality in this case? Does the particle cross the event horizon or does it not?

I don’t believe it does and here is why: The equation you set out is the Schwarzschild metric which describes the space/time around a point mass. This is also used to model a stationary black hole with an event horizon. As a particle falls radially towards the event horizon, time progresses differently depending upon the reference frame from which it is measured, because of the effect that gravity and motion has on the passage of time. This is all reflected in the Schwarzschild metric

A first integral equation that is derived from the Schwarzschild metric allows calculation of the passage of coordinate time, as experienced by the distant observer. A second integral equation, also derived from the Schwarzschild metric allows calculation of the passage of local time, as experienced by the particle. As long as the particle is outside the event horizon of the black hole, both the integrands are defined and the integrals are perfectly well-behaved. So the journey of the particle can be tracked with certainty all the way until the event horizon.

As the falling particle gets closer to the event horizon, time goes quicker when measured in local time than the journey measured in coordinate time, because of the relativistic effects of motion and gravity.

Since velocity equals distance divided by time, there is a different perception of speed depending upon whether the distant observer or the particle is making measurements. For each location reached, the particle thinks it got there quickly, so it thinks it is going fast. The distant observer thinks the particle got there slower, so it thinks the particle is slowing down. The particle arrives at the same location, there is just a different perception of the amount of time it took to get to the location.

This perception of the speed slowing down from the perspective of the distant observer continues until the forward progress of the particle seems to be very slow. So slow, in fact, that at 10^60 years or so, when the black hole has completely evaporated, the particle is still outside the event horizon. The particle’s journey stops at an end location that is near but outside the location of the event horizon of the no longer existing black hole.

Now, from the particle’s perception, it was going pretty fast when it reached the end location. It was trucking along at a fine speed, when all of the sudden, in the local time of the particle, the black hole rapidly evaporated.

This is the scenario from the calculations of the integrals derived from the Schwarzschild metric. As long as the particle is outside the event horizon, the integrals used to calculate the journey are perfectly behaved, so there is no need to doubt the results. Further, according to Einstein, all coordinate systems (reference frames) all will observe the same reality (e.g., logical sequencing of events), although the time to complete the journey will vary depending upon from which reference point the calculations are made.

Given the above scenario, it seems that regardless of the start time, nothing is able to cross the event horizon.


Doesn't the Schwarzschild metric combined with Hawking radiation imply that nothing ever gets past the event horizon of a black hole?

No. First of all, let's set Hawking radiation aside, we don't need it. Let's look at this:

An observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily long time.

This is accepted as being true, because at the event horizon the "coordinate" speed of light is zero, and nothing can go faster than light. So the obvious question is how can the black hole grow? One can find various opinions on this, such as The Formation and Growth of Black Holes on Mathspages, see http://mathpages.com/rr/s7-02/7-02.htm. This favours the view that the infalling body goes to the end of time and back in finite proper time, wherein like Susskind's elephant, it's in two places at once. I'm confident that this is incorrect, and that the alternative frozen-star interpretation is correct. In consequence, the black hole grows like a hailstone. Imagine you're a water molecule. You alight upon the surface of the hailstone, but you can't pass through the surface. However other water molecules surround you and bury you. So whilst you didn't pass through the surface, the surface passed through you. In similar vein nothing ever gets past the event horizon, but the event horizon gets past it.


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