Question

I have some trouble asking this question, so I will try a roundabout approach to explain what I mean.

If (outside of Earth atmosphere) one looks at the sun from different distances, the light coming from any direction intersecting the Sun's surface has a BB spectrum which is always the same, that of the surface temperature of the Sun. Looking in any other direction it is a BB spectrum at 2.7°K. So the spectrum at any place is thermal by parts. Far from the Sun the high temperature spectrum is within a smaller solid angle than closer but however close one gets (while still just outside the Sun) the high temperature thermal spectrum never concerns more than a 2π
solid angle.

Now consider a black hole emitting Hawking's radiation. Let us assume it is small enough its Hawking temperature is higher than 2.7°K. So if I understand correctly, if I am stationary outside the black hole (not free falling into it, which means I am effectively in an accelerated frame) I will see BB radiation at the Hawking temperature, but only in the solid angle where the line of sight intersects the black hole horizon. For other directions, I only get 2.7°K. Or am I already mistaken ? The closer I am the larger this solid angle. There is the added complication that when I get closer the spectrum is blue-shifted, but anyway the solid angle where I see BB temperature will never be more than 2π
. Indeed if I cross the horizon, then I cannot be stationary anymore, and the logic of Hawking's radiation cannot apply in the same way.

So here is my question. The Hawking radiation is often compared to the Unruh effect, because in an accelerated frame there is also a horizon, and the expression of the Unruh temperature and Hawking temperature are the same.

But from what I read, it always seems that the Unruh temperature concerns a BB radiation in all directions, over a full 4π solid angle. Unless I am mistaken.

So why is the Unruh temperature BB in the full 4π solid anglerather than just in a half-space 2π solid angle like the Hawking radiation at the limit where I get very close to the horizon ?

Answer 1

An observer sees blackbody radiation from objects like the Sun and a black hole. While Hawking radiation is seen in a 2π solid angle from a black hole, the Unruh effect predicts thermal radiation in a full 4π solid angle for an accelerating observer, owing to different physical scenarios.

Step-by-step explanation:

Your understanding of the blackbody spectrum as being the same when observed from various distances from the Sun is correct. The radiation emitted by an object, such as the Sun or a black hole, follows a blackbody spectrum, which is closely linked to its temperature. The intensity of blackbody radiation, as governed by the Stefan-Boltzmann law, is proportional to the fourth power of the absolute temperature of the body. As such, a hot object like a star can emit a large amount of energy across the electromagnetic spectrum, from ultraviolet to infrared.

When discussing the nature of Hawking radiation, it's noted that an observer outside of a black hole will see this radiation coming from the surface at the event horizon within a certain solid angle. As you approach the black hole, this solid angle in which the Hawking radiation can be detected increases, but only up to a point. Upon crossing the horizon, you can no longer be stationary, changing the observer's relationship with the radiation.

The comparison of Hawking radiation to the Unruh effect indeed lies in the similar mathematical expressions for the temperatures observed due to acceleration or gravitation. However, the Unruh effect predicts a uniform temperature distribution in all directions (4π solid angle), which differs from the half-space solid angle experienced with Hawking radiation. This discrepancy arises from the different physical scenarios: Hawking radiation considers an observer outside a black hole event horizon, while the Unruh effect considers an accelerating observer in space.