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I am studying quantization in Schwarzschild spacetime. In class the Boulware vacuum $\left| B \right>$ has been defined using the o.n. modes $u_I(x) = \frac{1}{4\pi \sqrt{\omega}}e^{-i\omega v}$, $u_R(x) = \frac{1}{4\pi \sqrt{\omega}}e^{-i\omega u} \Theta(r - 2m)$ and $u_L(x) = \frac{1}{4\pi \sqrt{\omega}} e^{i\omega u}\Theta(2m-r)$ (the $u_R$ are modes outside the horizon with positive norm and frequency, while the $u_L$ are modes inside the horizon with positive norm and negative frequency corresponding to partner states). Then one clearly sees the modes are singular on the horizon since there $u = \infty$ and thus the Boulware vacuum is singular on the horizon, however it can still be used to describe the polarized vacuum outside of a star since then the horizon does not really exists as the interior metric will not be Schwarzschild.

The Unruh vacuum $\left|U\right>$ has then been introduced, the o.n. modes are the same for the ingoing sector while they are $\frac{1}{4\pi\sqrt{\omega_K}}e^{-i\omega_K U}$ for the outgoing one, $U = \pm \frac{1}{k}e^{-ku}$ being the Kruscal coordinate and $k$ the surface gravity.

Now I know the outgoing mode at late time in a gravitational collapse behave in general as the outgoing modes of the Unruh vacuum. Now to derive Hawking's radiation the following is computed (in the s-wave and no backscattering approximation) $\left<U\right|N_{\omega}^R\left|U\right> = \frac{1}{e^{8\pi m \omega} - 1}$ having $N^R_\omega$ be the number operator associated to the modes $u_R$.

There are multiple reasons why this is not clear to me (please validate the italic statements):

  • I understand the outgoing modes of the Unruh vacuum and of a late time gravitational collapse are the same, thus it makes sense that the Unruh vacuum is the vacuum at a late time in a gravitational collapse, here what is not clear to me is that respect to which observer this is the vacuum: I suppose the vacuum of an inertial observer at the horizon since $U$ is the local inertial coordinate on the future horizon.
  • why do we use $N_\omega^R$, here I suppose because the Boulware vacuum is the vacuum of an inertial observer at infinity since the modes become Minkowski modes, however the modes are singular on the horizon, which does exist in the case of gravitational collapse (unlike a static star), I feel like I can neglect this because the observer is at infinity even though it seems odd.
  • if my above statements are correct here the vacuum depends on the position of the observer (different on the horizon and at infinity).
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Your definition of the word "vacuum" seems to be "the state in which a given observer sees no particles". This is not a common definition within QFTCS (in fact, we typically use "vacuum" to mean any pure Gaussian state, regardless of observers). With this in mind, let us move on to your questions.

I suppose the vacuum of an inertial observer at the horizon since $U$ is the local inertial coordinate on the future horizon.

Any static observer on Schwarzschild spacetime will observe particles in the Unruh vacuum at late times. Even if the observer is at the horizon. In a gravitational collapse spacetime, observers in the far past (before the formation of the black hole) see no particles. After the star collapses, they see particles appearing. Hence, in Schwarzschild spacetime, static observers in the far past will see no particles.

I suppose because the Boulware vacuum is the vacuum of an inertial observer at infinity since the modes become Minkowski modes

In the Boulware vacuum, no static observer sees any particles. It does not matter where they are in spacetime, as long as they are not on the horizon (which is nonphysical). The Boulware vacuum does not make sense in a gravitational collapse spacetime because it is a nonphysical state. It only makes sense in a star or planetary spacetime. This is also the reason why we have no Hawking effect on star or planetary spacetimes, as I discussed in this answer.

if my above statements are correct here the vacuum depends on the position of the observer (different on the horizon and at infinity).

The notion of particle does depend on the observer. The physical state of the field does not. The correct state to work with depends on the physical situation at hand, just like in a laboratory you should work with the state in which the system is prepared. In a planetary spacetime, for example, the Unruh vacuum is nonphysical because it predicts modes falling into a horizon that does not exist in the spacetime you are considering. Hence, it cannot be taken seriously as a good model for the quantum field. Similarly, in a gravitational collapse spacetime, the Boulware vacuum is not a physically acceptable model for the state of the quantum field because it is singular at the horizon. I discussed this in more length at the answer I previously mentioned.

The state does not depend on the observer. The particle interpretation does.

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  • $\begingroup$ "Any static observer on Schwarzschild spacetime will observe particles in the Unruh vacuum at late times. Even if the observer is at the horizon" however an inertial observer at the horizon is not static, no? "The notion of particle does depend on the observer." I guess this is what I meant since in my mind a vacuum for an observer is the quantum state where such observer sees no particles, as you were saying this is not the correct definition in QFTCS, though. $\endgroup$
    – ekardnam_
    Feb 26, 2023 at 15:20
  • $\begingroup$ I already read your answer before posting this question. It was enlightening in many ways however I was still doubtful and I posted this $\endgroup$
    – ekardnam_
    Feb 26, 2023 at 15:22
  • $\begingroup$ @ekardnam_ Oops, I misread your previous statement. Yes, an inertial observer at the horizon is not static and they will not see any particles $\endgroup$ Feb 26, 2023 at 16:17

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