1
$\begingroup$

From my understanding of black hole evaporation, a pair of virtual particles manages to avoid annihilating itself when one of the particles in the pair passes the event horizon and the other escapes the horizon. Wouldn't the same behavior take place at the hubble horizon? Would the hubble horizon radiation look any different from hawking radiation? The only difference between the two is that black holes shrink due to this radiation whereas the hubble horizon grows because of this radiation; both horizons "leave" us, just in different directions. Black holes emit more hawking radiation per volume as they shrink, which would look a lot like the big bang radiation from when our universe was small. Does this also imply that if you created a sphere of black hole spheres (surround a space with event horizons (or a rindler horizon)), you would create a universe?

$\endgroup$
  • $\begingroup$ You ask whether it applies to all other event horizons. This gets tricky because an accelerating observer in flat spacetime sees an event horizon. If you believe in the standard arguments used in semiclassical gravity (which have never been confirmed empirically), then this suggests that an accelerated observer also sees radiation. You then have to invoke the fact that different observers need not agree on the number of quanta. $\endgroup$ – Ben Crowell Apr 17 at 17:27
  • $\begingroup$ @BenCrowell, it's called the OBSERVABLE universe after all. Now that I think about it, wouldn't the forwardly accelerating observer see more radiation (and thus pressure) from the front than from the back? This radiation pressure would act as a force pushing back against acceleration. $\endgroup$ – Hierarchist Apr 17 at 17:37
2
$\begingroup$

There is thermal radiation from cosmological horizons. This is known as Gibbons–Hawking effect for their discoverers:

  • Gibbons, G. W., & Hawking, S. W. (1977). Cosmological event horizons, thermodynamics, and particle creation. Physical Review D, 15(10), 2738, doi:10.1103/PhysRevD.15.2738.

Does this also imply that if you created a sphere of black hole spheres …

This would not work. For black holes $R \sim M $, so if you gather several black holes to cover a sphere they would merge and there would be simply one large black hole.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.