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In Unruh effect, the temperature of background appears to be proportional to acceleration. On the other hand, the temperature of a black hole is inversely proportional to its mass.

If the two effects have the same origin as mentioned in

https://physics.stackexchange.com/a/259342/85274

temperature in both must include the same proportionality to acceleration and to gravitational acceleration, respectively.

According to equivalence principle, an observer hovering above the horizon of a black hole is equivalent to an observer who accelerates somewhere far from any massive body. But he can measure the thermal radiation, use the formulas for the two effects and infer in which situation he is.

Does it imply that there is distinction between the situations, from an observer's point of view, or is something wrong with my argument?

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  • $\begingroup$ "But he can measure the thermal radiation, use the formulas for the two effects and infer in which situation he is." How? $\endgroup$ – DanielSank Jul 23 '18 at 4:01
  • $\begingroup$ @DanielSank: How to measure temperature or how to infer? $\endgroup$ – N.S. Jul 23 '18 at 19:08
  • $\begingroup$ You must have a very good thermometer if you can measure the temperature of the Hawking radiation of a stellar black hole. $\endgroup$ – PM 2Ring Jul 24 '18 at 8:04
  • $\begingroup$ @PM2Ring: of course! everything is supposed to be possible in principle here. $\endgroup$ – N.S. Aug 1 '18 at 10:04
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You are probably assuming an uniformly accelerating observer, then no, because The Unruh temperature has the same form as the Hawking temperature TH = ħg / 2πckB of a black hole.

Assuming an inertial observer makes a difference. An inertial observer doesn't see Unruh Radiation, but far away from a Black Hole will see Hawking radiation.

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