2

I read in an answer to a question if Hawking radiation can be explained without too much mathematics that this is impossible insofar the vacuum energy is not boost invariant in a curved spacetime. Different observers in a curved spacetime see a different vacuum energy density. Hawking made in his 1975 paper a rigorous mathematical derivation of this fact.

In popular science, this is made clear without mathematics, by the popping in and out of existence in the vacuum of pairs of particles, one with positive and one with negative energy. On the event horizon (where the temperature is very high) these two are separated. The one with the negative energy disappears into the black hole, the other one comes out as Hawking radiation.

Isn't it possible to say that the vacuum energy depends on the metric of spacetime, and therefore on the varying curvature, which makes energy flow?

Community
  • 1
Deschele Schilder
  • 1
  • 5
  • 45
  • 105

1 Answers1

1

I don't really know anything about Hawking radiation but one thing bothers me in this question. In flat spacetime, generators of boosts are Killing vectors (generators of isometries: the symmetries of spacetime itself). In curved spacetime there are no such Killing fields. You cannot boost your spacetime and get the same thing. It is not true that all frames are equivalent, due to gravitational tidal forces. There might be a privileged direction in a curved spacetime. Therefore regardless of the precise dynamics of the theory, it is not required that it is "the same in every frame", whatever that would mean. I don't think it is possible for a theory to be Lorentz invariant when the spacetime itself isn't.

Blazej
  • 2,267