Why Unruh temperature is the temperature measured by observer with $\xi = 0$?

Question

My coordinate transformation is: \begin{align} t=\frac{1}{a} e^{a \xi} \sinh (a \eta), \quad x=\frac{1}{a} e^{a \xi} \cosh (a \eta). \end{align} The free scalar field is quantized under this coordinate by: \begin{align} \phi=\sum_{k}\left(\hat{b}_{k} g_{k}+\hat{b}_{k}^{\dagger} g_{k}^{*}\right). \end{align} By Bogoliubov transformation, the particle number distribution in Minkowski vacuum is: \begin{align} \left\langle 0_{\mathrm{M}}\left|\hat{b}_{k}^{ \dagger} \hat{b}_{k}^{}\right| 0_{\mathrm{M}}\right\rangle \propto \frac{1}{e^{2 \pi \omega / a}-1}. \end{align}
This is a thermal spectrum of bosons with temperature $T=a/2\pi$. Then many textbooks claim that it is the temperature measured by observer $\xi=0$, but I dont quite understand. As far as I see, $\hat{b}_{k}\hat{b}_{k}^{\dagger}$ is not a local operator, thus the temperature should be the average temperature of the whole Rindler wedge. Due to the red shift, Rindler observer with larger $\xi$ should measure a lower temperature: \begin{align} T(\xi) = e^{-a\xi} \cdot T(\xi=0). \end{align} I think the average temperature should be defined as: \begin{align} \bar{T} = \frac{\int{T(\xi)\sqrt{\gamma}}d\xi}{\int{\sqrt{\gamma}}d\xi}, \end{align} but it does not give the right result. However, if I define in this way: \begin{align} \bar{T} = \frac{\int{T(\xi)\sqrt{\gamma}}d\xi}{\int{d\xi}}, \end{align} it really gives $\bar{T} = T(\xi=0)$. But the denominator in this definition is coordinate-dependent. Do I miss something? Or my average temperature argument is wrong?

I know that Unruh effect can also be obtained by considering the interaction of an accelerating detector with the field. There is no ambiguity in this case. But I still wish to find a physical explanation under the framework of Bogoliubov transformation.

Answer 1

To define creation and annihilation operators in a curved spacetime, one needs to have a timelike Killing field to work as the generator of time-translations. This will then lead to an associated notion of energy and allow to obtain a positive and negative energy decomposition of modes of the field, allowing for the definition of the creation and annihilation operators. For a more detailed discussion of this construction see this answer I wrote a while back, which is based on the treatment given on R. Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics and on a seminal paper by Ashtekar & Magnon on QFT in Stationary Spacetimes.

In the Rindler wedge, one such vector field is given by $$\beta^{\mu} = a\left(x \left(\frac{\partial}{\partial t}\right)^{\mu} + t \left(\frac{\partial}{\partial x}\right)^{\mu}\right),$$ which is the generator of boosts in the $x$ direction. In the Rindler wedge, this field is timelike, and hence it can be taken to be the generator of a different notion of time translations.

It remains to consider which observers experience this notion of time translation. To do so, we can simply impose $\beta^{\mu} \beta_{\mu} = - 1$, which corresponds to requiring that the Killing field corresponds to a four-velocity. This condition translates to $$a^2 (-x^2 + t^2) = -1, \tag{1}$$ which specifies a particular orbit of the Killing field. Hence, for the observers following this particular orbit of the Killing field, the experienced notion of time is precisely that defined by the Killing field, and hence it makes sense to state that the particle content seem by them is the same described by the creation and annihilation operators defined with respect to this Killing field.

Replacing your expressions for $t$ and $x$ in terms of $\xi$ and $\eta$ on Eq. (1) leads to the condition that $\xi = 0$, while not imposing any new restrictions on $\eta$. It is not possible for an observer to be travelling parallel to the Killing field with $\xi \neq 0$: that would break the normalization of the four-velocity.