# How Does Black Hole Evaporation Look From the Inside?

Imagine you're inside a black hole event horizon. The curvature of spacetime is such that the direction towards the singularity becomes timelike, as reverse movement becomes literally impossible.

How would Hawking radiative evaporation look like from inside the black hole? From the outside perspective, we know that black holes form (from our time dimension perspective) and then evaporate at a rate that is an inverse function of their mass.

Hawking radiation is created when the vacuum state at the horizon looks as if it is transformed into a non-vacuum state for a distant observer, resulting in a negative energy 'input' to the black hole. If the energy loss this way is not compensated by inputs from outside, the black hole event horizon should gradually shrink, resulting in eventual evaporation. How if at all would that be perceptible by an observer inside the black hole horizon?

Similar question, but no accepted answer.

An infalling observer cannot remain stationary inside the event horizon of a black hole.

According to them, there is a short period of proper time before they meet whatever lies at the centre of the black hole (a singularity in GR).

If the observer looks back, they do not see an infinite amount of time pass by in the outside universe. In fact for the type of black hole where they might (a) survive and (b) have time to make any observations (i.e. a supermassive black hole) then they would see the outside universe only for a relatively short time longer than their experienced proper time and certainly for nowhere near as long as the Hawking radiation evaporation timescale. So the possibility that the black hole is (slowly) evaporating is of no consequence to their final moments.

The necessary material and diagrams to understand the previous paragraph have been covered before on both Physics and Astronomy SE.

https://physics.stackexchange.com/questions/82678/does-someone-falling-into-a-black-hole-see-the-end-of-the-universe

Would time go by infinitely fast when crossing the event horizon of a black hole?

• Thanks Rob, I will confess to being a little confused about the short period of proper time since all my other research suggests that the spacetime bending twists the towards singularity direction into being timelike. So for a (very durable) observer descending towards the singularity, it's unclear to me that the period would have to be short. Commented Jul 12, 2016 at 15:57

The space itself is falling into the black hole. Below the event horizon - in the timelike zone - an observer is falling towards the singularity, together with the space.

Seen from this falling observer, space looks like vacuum outside the black hole. Close to the singularity tidal forces become an increasing problem. But for a sufficiently large and "benign" black hole, an observation can't locally distinguish between spacetime inside and spacetime outside the black hole. Otherwise it would violate the principle of relativity.

So - seen from the falling observer - we get short-lived virtual particles, as in usual vacuum. Due to space-time curvature, the segment of vacuum in the black hole isn't perfectly symmetric, including quantum fluctuations. This tiny asymmetry results in (few) real particles, when seen from outside the black hole, when applied to the (light-like) zone near the event horizon. It should be fundamentally possible to probe the tiny asymmetry of the resonances with particle experiments.

An observer hovering just below the event horizon would be tachyonic in the in-falling spacetime.

• You cannot "hover just below the event horizon". Commented Jul 8, 2016 at 23:32
• "tachyonic" means you would need to be faster than the speed of light. Commented Jul 13, 2016 at 21:05