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If an observer can measure the Hawking radiation or surface gravity in his co-moving/rest frame, what happens when we make a coordinate transformation to a moving frame of reference? Can we make such coordinate transformation that the Hawking radiation and surface gravity are equal to zero in the moving frame (and vice versa)?

Edit: I had in mind inertial frames when I asked the question.

Dee
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  • We can not transform the radiation away because that would violate the third law of thermodynamics (the local speed of light being a constant guarantees that, too... no observer can outrun photons). The free falling observer, on the other hand, does, to first order, not feel the gravity. – CuriousOne May 08 '15 at 07:53
  • Does that mean that the Hawking radiation can't be measured/seen by the free falling observer? – Dee May 08 '15 at 09:21
  • See Hawking radiation from point of view of a falling observer. – John Rennie May 08 '15 at 10:00
  • No, it means quite the opposite. While I am not a specialist, I would say you can safely discount any analysis that violates the third law. An observer in orbit around a black hole is freely falling, by the way, so that is not a criterion to eliminate Hawking radiation by any means. – CuriousOne May 08 '15 at 10:27

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