Unruh temperature in gravitational field

[jcap]

According to the Unruh effect an observer who is has an acceleration $g$ will observe the temperature of the vacuum to be
$$T=\frac{\hbar g}{2 \pi c k_B}.$$
According to the equivalence principle the observer should measure the same Unruh temperature if he is sitting on a planet whose surface gravitational field has a strength of $g$.

Is this correct?

As the Unruh and Hawking temperature are very similar does this mean that all gravitating bodies have a Hawking/Unruh temperature that could in principle be detected by a distant observer?

 

[PeterDonis]

According to the equivalence principle the observer should measure the same Unruh temperature if he is sitting on a planet whose surface gravitational field has a strength of $g$.

Is this correct?

If the planet is surrounded by vacuum and is in thermal equilibrium with it, yes. But of course this never actually happens. See below.

As the Unruh and Hawking temperature are very similar does this mean that all gravitating bodies have a Hawking/Unruh temperature that could in principle be detected by a distant observer?

No. The temperature in question is the temperature of a vacuum--the Hawking temperature of a black hole assumes that the hole is vacuum, with no matter or energy present (just spacetime curvature). As soon as you add matter and energy, you change the scenario. As above, the only way to recover a scenario similar to the vacuum one would be if all of the matter and energy were in thermal equilibrium with the vacuum. But in the real universe that never happens. (Why? If nothing else, because the CMBR is present in the real universe and its temperature is way, way above the Unruh/Hawking temperature for any achievable acceleration and any achievable mass for a black hole).

 

[Demystifier]

According to the Unruh effect an observer who is has an acceleration $g$ will observe the temperature of the vacuum to be
$$T=\frac{\hbar g}{2 \pi c k_B}.$$
According to the equivalence principle the observer should measure the same Unruh temperature if he is sitting on a planet whose surface gravitational field has a strength of $g$.

Is this correct?

Yes.

As the Unruh and Hawking temperature are very similar does this mean that all gravitating bodies have a Hawking/Unruh temperature that could in principle be detected by a distant observer?

If we interpret the surface of the Earth as one big particle detector, then the Earth's surface should get heated by Unruh effect. This heat must create standard thermal radiation, which is not Hawking radiation. This thermal radiation can, in principle, be seen by a distant observer. Of course, the whole effect is too small to be detectable.

 

[jcap]

Yes.If we interpret the surface of the Earth as one big particle detector, then the Earth's surface should get heated by Unruh effect. This heat must create standard thermal radiation, which is not Hawking radiation. This thermal radiation can, in principle, be seen by a distant observer. Of course, the whole effect is too small to be detectable.

So where does the Unruh effect energy come from? If the matter in the Earth is stable then it must come from the vacuum itself when it is subjected to a gravitational field. In that case is the law of conservation of energy violated?

 

[Demystifier]

So where does the Unruh effect energy come from? If the matter in the Earth is stable then it must come from the vacuum itself. Is the law of conservation of energy violated?

Why do you think that Earth is stable? The gravitational collapse of the Earth is prevented by some non-gravitational forces between atoms. These non-gravitational forces are a source of potential energy, which might act as a source of energy for Unruh radiation.