# Falling into an evaporating black hole

If

• It takes an infinite amount of coordinate time to fall into the horizon of a black hole
• It takes a finite amount of coordinate time for a black hole to evaporate due to Hawking radiation
• All observers will agree on physical events even though they may assign different spacetime labels

Then wouldn't it be the case that:

• A black hole will evaporate before an observer falls into it
• An observer will never cross the event horizon
• Did you search for similar questions before posting this one? Apr 5, 2021 at 22:57
• @DvijD.C. There are questions about falling into black holes, but I can't find any questions which are asking the same thing. If there is then I would be happy to see them Apr 5, 2021 at 23:09

No. The coordinates one uses are completely independent of the physics involved.

This is the Penrose diagram for the evaporating black hole. Someone who falls into the black hole while it still exists crosses the $$r = 2M$$ line, the horizon, and then hits the singularity, the $$r = 0$$ line. The black hole will not have evaporated yet.

While an external observer will never see a falling object cross the event horizon, an observer riding on the falling object will cross the event horizon and reach the singularity in finite proper time.

An amusing calculation to do is to assume your falling object is emitting some continuous electromagnetic signal, with any wavelength and any intensity you like. Since the total energy emitted before the object crosses the horizon is finite, an external observer must see the signal redshifted away to nothing in the long-time limit. Your complaint is that, in the very-long-time limit, the black hole evaporates. The calculation is how long it takes until the Hawking radiation from the event horizon is brighter than your inflating object's signal. This always happens in finite time: your signal gets cooler and dimmer as it redshifts away, while the event horizon (on a much longer timescale) gets hotter and brighter as the black hole evaporates.

(Asking where that radiation comes from may lead you to think about the hypothetical black hole firewall, though the motivating logic there seems to be different.)

• I think the complaint is maybe better put as "While an external observer will never see a ... reach the singularity in finite proper time" is true when analyzing a static black hole in Schwarzchild coordinates HOWEVER if the black hole evaporates then an external observer will see the black hole completely evaporated WHILE the falling object is still outside the event horizon; so the falling observer will never actually cross the horizon in proper time Apr 6, 2021 at 4:27

I believe that Urb is correct. user1379857's Penrrose diagram is not the whole story. Rather than considering the situation form the viewpoint of the distant observer, we can consider it from the viewpoint of observer2 that is falling into the black hole. But we do not merely consider proper time, but events that observer2 can (at least theoretically) detect. The events that I am referring to are clock pulses transmitted from outside the capture range of the black hole. Observer2 is still seeing these coming in as the hole evaporates - and he sees all of these before he penetrates the 'event horizon'. Obviously therefore, his proper time has not yet progressed to the point of penetrating the 'event horizon' when the hole evaporates. A corollary of this is that Schwarzschild event horizons never actually form. (As Einstein famously said "such entities do not exist in the real world"; on the other hand he did not say anything against the existence of entities that were externally very similar).

If a black hole evaporates before an observer falls through the its event horizon then it is tautologous that an observor will never cross the event horizon.

Moreover, if I recall what I read from Hawkings paper correctly, the energy released in the final moments of such an evaporating black hole is huge. In fact, Hawking writes in his paper, Particle Creation by Black Holes published in 1975:

When the temperature gets upto $$10^{12} K$$ or when the mass got down to $$10^{14} g$$, the number of different species of particles might be so great that the black hole radiated away all its remaining rest mass on the strong interaction scale of $$10^{-23}$$. This would produce an explosion of $$10^{35}$$ ergs.

An erg is an old-fashioned unit of energy, equivalent to $$10^{-7}J$$, so the resulting explosion would generate $$10^{28}J$$. Now the energy released by the sun prr second is around $$4 \times 10^{26}J$$. So we're looking at the energy released by around twenty suns in a second.

In other words, the observer won't cross the event horizon since they'd be blown to smithereens.

• "In other words, the observer won't cross the event horizon since they'd be blown to smithereens." This is incorrect. The Hawking radiation that a freefalling observer experiences the horizon is extremely modest, and they can easily enter before the end stages of the evaporation process. Apr 6, 2021 at 4:10
• @user1379857: It's not. This is because the question assumes that the black hole is in yhe final phases of evaporatimg away and as Hawking points thos means in the final phase it explodes with enormous energy. Apr 6, 2021 at 11:17
• @safesphere: I think the explosion will be appatent from whatever frame you're in. Apr 8, 2021 at 11:10
• @safesphere: Explosions are explosions whatever frame of reference you are in. I don't think you can frame away Hawking rafiation as easily as you're making out. QFT on curved spaces is not as simple as you're making out. Apr 8, 2021 at 14:34
• It's not me, I'm just repeating what they say. I have deleted my comments. Apr 8, 2021 at 21:20

I had this same question when I first encountered relativity, believing that there should be a "shell" of matter surrounding a black hole from everything that's ever tried to fall into it.

My, limited, understanding (I'm really not into relativity) of it now is: An object will cross the event horizon for some finite coordinate time, the photons it emits at the event horizon just wont be observed in finite coordinate time.

This might be totally wrong Please take it with a pinch of salt.

• It takes an infinite amount of coordinate time to fall into the horizon of a black hole
• It takes a finite amount of coordinate time for a black hole to evaporate due to Hawking radiation

These statements are meaningless unless you specify what time coordinate you're using. There is no spacetime manifold and causal time coordinate defined on the manifold for which they're both true.

The reason it appears to take an infinite amount of coordinate time to cross the event horizon in Schwarzschild coordinates is that Schwarzschild coordinates don't cover the event horizon at all. In coordinate systems that do cover the horizon, like Eddington-Finkelstein or Kruskal-Szekeres, it takes a finite amount of coordinate time to reach it.

The geometry of evaporating black holes isn't understood, but if (as suggested in Hawking's original paper) the horizon-crossing event is in the causal past of the evaporation event, then the crossing will precede the evaporation in coordinate time, if the coordinate time respects causality.