I first explain my reasoning. I know that a Schwarzschild metric is the right metric outside any spherical object, but I also know that it is correct in the empty space. So I thought: why we don't talk about Hawking's radiation for a star? Because, I thought, the metric inside a star in not the same outside. So I thought that an event horizon was necessary because, since inside of it all the matter tends toward the singularity, all the space time (singularity excepted) is described by the Schwarzschild metric, and so I can quantize a field using that metric. However, in Rindler space it's not so. In Rindler space I still have horizons, but now the Rindler coordinates describe only some part of the whole space time (they cover only the Rindler Wedges), and still I have Unruh effect. So what I thought before cannot be the reason for the importance of the horizon. So I ask, what is the horizon function, both in Schwarzschild and Rindler space?
2 Answers
One should always be wary of argument that rely on coordinates, since they are just a tool to describe the spacetime. Rindler coordinates are somewhat convenient for accelerated observers but you can derive the Unruh effect in Minkowski coordinates just fine (this is done in the Review of Modern Physics on the Unruh effect, by Crispino, Matsas and Higuchi for instance).
To understand the importance of horizons we must consider what is a thermal bath. There are two ways to do so. The first one is to say that you know you are in a thermal bath if the expected value of the number operator is a thermal distribution. The second one is that a thermal bath is a mixed state whose density matrix is given by the Boltzmann form $e^{-\beta H}$. The second way implies the first, but the first does not imply the second, because you might have thermal distribution for the number operator but still have significant correlations between degrees of freedom such that the state is not strictly thermal.
In the first definition the horizons are not necessary, only that the trajectories of observers behave in such a way as to produce the thermal distribution in the number operator.
In the second definition the horizon does play a role. The Minkowski vacuum is a pure state but there is no transformation in quantum mechanics that send a pure state to a mixed state, in this case the thermal one. What does happen in the Unruh effect is that accelerated observers have a horizon separating them from the rest of spacetime. In other words there are degrees of freedom of the field inside the horizon which are not accessible to the observer. And in quantum mechanics whenever you do not have access to certain degrees of freedom you must trace over them in you current state. So given a Minkowski vacuum, if you trace over all degrees of freedom inside the horizon you get a thermal state of Boltzmann form.
In short, the horizon is relevant because it means there are degrees of freedom in the field which the observer cannot measure. The ignorance regarding this part of the field is what converts a pure state to a mixed state (in this case a thermal one)
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$\begingroup$ Hello, I understood the fact that the uniform acceleration of the observer and the consequent formation of the horizon provoke an increase of the uncertainty, thus trasforming a pure state in a mixed one. However I cannot still figure out how increasing acceleration increases the Unruh temperature, thus decreasing the purity of the state. The horizon seems to me always the same, regardless of the acceleration, so I do not understand how the information can decreases with the augment of the acceleration. $\endgroup$ Nov 27, 2019 at 22:13
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$\begingroup$ @silviozzo, I'm not very experienced with quantum information, but I can say that the state is a global object, so regardless of acceleration all Rindler observers witness the same state and take the trace with respect to the same degrees of freedom. Therefore it seems strange to suggest that different observers have different information regarding the state. Why the inertial intuition of purity as information about mixing not really applying here is something I cannot explain. There is a lot of literature on Unruh effect and quantum information though, I'll try to find something about that $\endgroup$ Nov 29, 2019 at 1:52
Strictly speaking, event or observer horizons are not necessary for the Hawking & Unruh effects.
For example, one can consider a particle detector moving in Minkowski spacetime with time-dependent acceleration. If during a time interval acceleration is approximately constant, the detector would be registering quanta corresponding to Unruh effect. But such detector may not have an observer horizon at all, if acceleration is subsequently switched off.
One can also consider “black hole-like objects” without horizon, such as a collapsing star perpetually approaching would be horizon yet never quite forming it. (Such a geometry may require modifications to GR or alternatively it may be realized as analogue gravity model). Such object would be producing a flux of “Hawking-like” radiation despite not having event horizon.
But Schwarzschild and Rindler spacetimes offer considerable computational advantages for the derivation of Hawking & Unruh effects since horizons there are Killing horizons allowing us to consider static situation only.
For more discussion of the necessary and sufficient conditions for the Hawking effect see e.g. paper 1 and paper 2.
