3
$\begingroup$

Consider a „Rindler observer“ hovering close to the event horizon of a black hole, whereby flat spacetime is assumed locally.

  • Does this observer see the same Unruh radiation like a Rindler observer accelerating with the same proper acceleration in flat Minkowski spacetime?

  • What follows from the fact that these two scenarios differ due to the locality condition?

  • Does it make sense - and if yes why - to distinguish between Unruh- and Hawking-radiation regarding the radiation our hovering observer sees?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Yes. If you take the metric for the hovering observer and expand it to first order you get a Rindler metric. Take the acceleration that appears in this metric and insert it in Unruh's equation and you get the correct result for the Hawking radiation. In this sense the Hawking and Unruh effects are the same.

The big difference is that while an observer accelerating in flat spacetime sees radiation a non-accelerating observer does not. However with Hawking radiation a non-accelerating observer in (approximately) flat spacetime far from the black holes does observe radiation coming from the black hole. The difference between the two effects is due to the presence of the event horizon in the black hole.

However I'm afraid I don't know an easy way to explain why the presence of the horizon makes the difference.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks for your answer! "However with Hawking radiation a non-accelerating observer in (approximately) flat spacetime far from the black holes does observe radiation coming from the black hole." Is it correct that in contrast a freely falling observer who is inertial doesn't observe Hawking radiation while crossing the horizon? $\endgroup$
    – timm
    Commented May 1, 2018 at 16:45
  • $\begingroup$ do you know a difficult way to explain why the presence of the horizon makes the difference.? If yes, please explain 😊 $\endgroup$
    – magma
    Commented May 4, 2018 at 10:35
  • $\begingroup$ Well Hawking's original calculation considers the scattering of a quantum field from the black hole, and the fact the horizon swallows all the radiation that hits it modifies the scattering calculation and produces the Hawking radiation. Hawking's original paper is around on the web if you want to read it. $\endgroup$
    – John Rennie
    Commented May 4, 2018 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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