Hawking Radiation and Dark Energy

Question

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. Would the negative energy input mentioned earlier look (from the perspective of someone inside the black hole horizon) in any way like what we refer to as Dark Energy?

Answer 1

Unfortunately, A complete theory of Hawking radiation does not exist at the moment. Hawking radiation is described by a sort of WKB approximation far from the balck hole horizon. In order to describe the phenomenon in the whole space time, in fact we would need a complete theory of Quantum gravity, or at least, a consistent theory (that means with a generic metric) of quantum field theory in curved spacetime. So I guess no one can answer at your question completely at the moment.

Answer 2

This might not be right, but as I understand it, the particle and anti particle both have mass. The tidal force of the black hole is able to separate the 2 particles, one of them, flying off in to space, the other, flying towards the black hole.

So from inside the event horizon you'd see both particles, well, you'd have to look very close cause particles are hard to see, but in theory, you'd see one flying away from the black hole, losing kinetic energy, gaining potential energy as it escapes, and the other, flying towards the singularity, increasing in kinetic energy as it speeds towards the singularity. The energy lost is done in the work splitting the 2 particles and sending one away. How the energy is transferred, which seems to be what this link implies, I don't know.

Source: http://www.astronomycafe.net/qadir/q2393.html

Answer 3

Here are my musings on the subject...

1. Gravitational time dilation is higher at the center of a massive object than at it's surface.
2. Time dilation reaches infinity at the event horizon of a black hole.
3. A time reversed black hole would effectively be a white hole, which just might look more than a little bit like the big bang.
4. The observable universe's mass has a Schwarzschild radius of approximately 13.7 billion light years.

Personally, I think the arrow of time is effectively reversed "inside" a black hole and that is why the normally positive energy half of the virtual particle pair seems to us to have negative energy.

And what exactly would it look like? I suppose it would look like a photon with negative energy and so would exert a sort of all pervasive negative radiation pressure. I also think that the speed of the evaporation of the black hole our universe is in is speeding up, which we see as an acceleration in the expansion of the universe.

Eventually the black hole will evaporate completely, which the interior residents will interpret as a Big Rip.