Why to we take the Minkowski vacuum to calculate the thermal Unruh distribution? Looking at a derivation for the Unruh effect, this generally ends with a calculation of the particle distrubution:
$$\langle{0_M}|b^\dagger b|0_M\rangle=\frac{1}{\exp\left(\frac{\hbar\omega}{k_BT_U}\right)-1}$$
where $b$ is the annihilation operator that annihilates the Rindler vacuum and $|0_M\rangle$ is the Minkowski vacuum. I see that this would be zero if one used the Rindler vacuum $|0_R\rangle$ instead.
But what is the physical reason behind this? Shouldn't an observer see the particles in "his own" vacuum, i.e. the Rindler vacuum? Why can the Rindler-mode operators be applied on the Minkowski vacuum?
 A: Choosing to work in the Minkowski vacuum is something like an experimental constraint—the quantum field is in some state, and that state does not depend on the observer. The question the Unruh effect addresses is: given that the field is in the Minkowski vacuum $|0_M\rangle$, what do different observers see? Once we do the computations, we learn that the inertial observers are freezing in a void of particles, while the accelerated observers are burning in a sea of particles.
The main point is that the state does not depend on the observer, it is an actually physical property and all observers must agree on what is the state. Hence, we consider the same state for different observers. Given the experimental conditions (the state of the field), we compute what each observer would measure and conclude the Unruh effect.
We could, of course, do similar procedures with other states, such as a state with one Minkowski particle. The Rindler vacuum wouldn't be suitable for these sorts of computations with an inertial observer, since it is singular on the origin's light-cones.
