Can we slightly modify the Schwarzschild metric to describe an evaporating black hole, and what would the geodesic be like?

Question

For example, we can assume that the black hole is enclosed in a hypothetical sphere which slowly emits photons to the outside and “negative photons” to the black hole. In this case will a free falling observer still cross the event horizon and reach the singularity within a finite amount of proper time?

Answer 1

哲煜黄 asked: "will a free falling observer still cross the event horizon and reach the singularity within a finite amount of proper time?"

The observer will cross the horizon in finite proper time and hit the singularity just as normal. After he hits the singularity negative energy Hawking photons hit the singularity as well and therefore reduce its mass.

The outgoing positive radiation has a local wavelength comparable to the horizon and is therefore practically undetectable to the infalling observer if the black hole is much larger than the observer, while the ingoing negative radiation that hits him from behind is redshifted (at the horizon exactly the half frequency of the radiation received at infinity if the observer falls in with the negative escape velocity) and is therefore even harder to detect if you don't hover close above the horizon to get blueshift instead of redshift, for which you'd need to accelerate very hard.

The negative radiation might reduce your body temperature very slighty and therefore also the total energy of the inner system, but it won't annihilate your atoms since the wavelength is too long for any noticable effect. You will be terminated not by the Hawking radiation but at the singularity, and the singularity will be fed with negative energy until the sum of positive and negative energy becomes 0, and the gravity therefore also 0.

哲煜黄 asked: "Can we slightly modify the Schwarzschild metric to describe an evaporating black hole?"

You could, but the origin of the Hawking radiation is not exactly localized, while in your modified metric you'd have to chose a specific r outside of the horizon where it originates, but as a relativistic approximation that's probably the best you can do without full quantum gravity.

In this paper the Vaidya metric is transformed from {u,r,θ,φ} to {t,r,θ,φ} so that you can plug in the dM/dt of the Hawking radiation, but that's also just a nonquantum approximation.