Direction of Unruh radiation

Question

In case of a black hole, the direction of the Hawking radiation is from the horizon to the observer. The corresponding effect in the Rindler spacetime is the Unruh radiation.

Intuitively, a rapidly accelerating observer should face the Unruh radiation coming from the direction of motion, like a wind felt by a biker. The faster I accelerate forward, the stronger "wind" hits me in the face.

Yet, if the Unruh radiation comes from the horizon like the Hawking radiation does, it would hit the observer in the back. The faster I accelerate forward, the stronger "wind" hits me in the back. This seems counter-intuitive.

From which direction does the Unruh radiation hit an accelerating observer? In the face from the direction of motion or in the back from the Rindler horizon?

Answer 1

An interesting feature of the Unruh effect at any one point is that it is isotropic. So the intuition that the radiation comes from the horizon and "hits you in the back" is wrong. One way to interpret this is to say that the radiation comes from the horizon, rises up high, and then falls back down again, in such a way that the net result at any one point is isotropic. I note that Valter Moretti's good answer prefers to avoid the word "radiation" for technical reasons, but I think it remains an acceptable word in this context, as a way of discussing energy-momentum transfers between an observer and the field. The point here is that flux at the observer is isotropic, and I believe a detector which absorbs or reflects the radiation will undergo Brownian motion consistent with isotropic fluctuating illumination.

The temperature of the radiation is not homogeneous; it gets smaller as you move away from the horizon. The radiation arriving at any given height $x_1$ from other heights $x_2$ gets just the right Doppler shift to make it all arrive at $x_1$ with the same temperature and flux independent of what height $x_2$ it came from.

This feature of the Unruh effect is different from Hawking radiation. In the case of Hawking radiation, once you are far from the black hole, the radiation approaches you from the black hole and not the other way. For observers near the black hole horizon (within a Schwarzschild radius or two) the situation is more complicated.

Answer 2

To answer the question, a comparison is convenient with the phenomenology of the Hawking radiation to make evident some crucial differences.

First of all, what actually happens for an accelerated observer moving in the Minkowski invariant state is that this state appears as a thermal bath with respect to the Rindler Killing time. A thermal bath is different from a radiation state: the former is an equilibrium state, the latter is not.

Strictly speaking the phenomenology is different from the one of an observer in the spacetime of a large black hole observing the Hawking radiation exiting from the horizon. There a net flux of particles exiting the horizon exists (with thermal properties with respect to the Schwarzshild Killing time) and nothing enters it. That is quite different from an equilibrium state, in fact the black hole eventually evaporates!

This situation, in the Kruskal extension of the Schwarzshild spacetime, is described by the so-called Unruh state of the background quantum field.

Conversely, what describes a thermal bath at the Hawking temperature is the so-called Hartle-Hawking state. Here the flux of particles entering the horizon is equal to the one of particles exiting the horizon.

What happens in the Rindler wedge for an accelerated observer in the Minkowski vacuum is an approximation of the phenomenology of Hartle-Hawking state (in accordance with the equivalence principle) and not of the Unruh state.

A crucial difference, distinguishing black hole phenomenology from Rindler phenomenology, is however that the particles of HH and U state around a black hole are standard particles. In the sense that, far from the black hole where the spacetime becomes flat, they are described by modes of standard QFT in flat spacetime.

Conversely, particles used to describe the thermal bath for the accelerated observer are Rindler particles without physical direct meaning. Their existence is furthermore confined to the Rindler wedge, so that their physical relevance is disputable. This does not automatically means that the abovementioned thermal properties do not exist since different theoretical descriptions of extended thermal states of a quantum field are at our disposal, in particular, the one relying on the KMS identity.

In summary, there is no experienced radiation of Rindler particles for an accelerating observer but a thermal equilibrium state of those particles takes place. A quantitative description of the effect of the action of this bath on physical devices is more difficult. In particular, the geometrical setup is a delicate matter here: the rest space of the Rindler observer appears to be homogeneous, but the norm of the Killing time used to define the thermal equilibrium depends on the spatial non-Cartesian coordinate usually denoted by $\rho$. Therefore to quantitatively answer your question one should fix suitably adapted definitions.