# Hawking radiation and non black hole gravitational sources?

I see that the formula for hawking radiation is related the the formula for unruh radiation. The accelleration experienced by a body yields an unruh temperature equivalent to a black holes hawking temperature with an equivalent value of g. The unruh effect happens at all accelerations, therefore is hawking radiation a property of all gravitation sources(not just black holes?)

Matterwave and PAllen

## Answers and Replies

PeterDonis
Mentor
2020 Award
The accelleration experienced by a body yields an unruh temperature equivalent to a black holes hawking temperature with an equivalent value of g.
This doesn't mean quite what you think it means. The "g" in the Unruh case is the actual proper acceleration of the body. The "g" in the Hawking case is not; it's the "surface gravity" of the black hole, which is the "redshifted proper acceleration" at the hole's horizon. This is not a direct observable the way the proper acceleration of the body in the Unruh case is. So, although the two formulas are related, they are not physically equivalent.

The unruh effect happens at all accelerations, therefore is hawking radiation a property of all gravitation sources(not just black holes?)
No. "All accelerations" in the Unruh case corresponds (with the caveats given above) to "all black hole masses" in the Hawking case. As noted above, the "g" in the Hawking formula is the surface gravity of the horizon, so the formula doesn't apply to gravitating bodies that don't have a horizon.

bcrelling
stevendaryl
Staff Emeritus
No. "All accelerations" in the Unruh case corresponds (with the caveats given above) to "all black hole masses" in the Hawking case. As noted above, the "g" in the Hawking formula is the surface gravity of the horizon, so the formula doesn't apply to gravitating bodies that don't have a horizon.
What does the "surface gravity of the horizon" mean? If it means the proper acceleration of an object hovering at the horizon, then doesn't that diverge?

Okay, according to Wikipedia, there's sort of a "cancelling infinities" effect going on here. The temperature goes to infinity near the horizon, but also photons emerging from near the horizon and escaping to infinity undergo an infinite red shift. The two effects combine to give a finite photon energy, corresponding to a finite temperature, as measured by an observer far from the black hole.