[*integer*]{} multiple of $2 \pi$, for otherwise the operator appearing in the exponential function on the right–hand side would not be skew-adjoint. By using the proof of Theorem 6.2 in [@BB] with $\As(\Os)$ replaced by $\As(W_0)$, one sees that its only possible value is $\beta = 2 \pi$. Proceeding to the proper time scale of the observer, we conclude that he is exposed to the Unruh temperature $(1/2\pi)((\dot{\lambda}_{0 1}(0)\, x_O)^2){}^{-1/2}$, in accordance with the value found in computations for some particular models [@DeLe; @Jac] and also by more general considerations [@BEM].

For geodesic observers, Lemma \[nogroundstate\] is not applicable. In fact, $\omega$ cannot be a KMS-state for $e^{itM_{0n}}$ on $\As$, the laboratory observable algebra for geodesic observers. Indeed, since the covariance assumption (iii) implies $U(\lambda)\As U(\lambda)^{-1} = \As$, for all $\lambda \in {\mbox{SO$_0(2,n\!-\!1)$}}$, if $e^{itM_{0n}}$ were the modular group for $\Omega$ on $\As$, then modular theory would necessitate $U(\lambda)e^{itM_{0n}} = e^{itM_{0n}}U(\lambda)$, for all $\lambda \in {\mbox{SO$_0(2,n\!-\!1)$}}$ (cf. Thm. 3.2.18 in [@BR]). But this would only be possible if the representation $U({\mbox{SO$_0(2,n\!-\!1)$}})$ were trivial, which is excluded by assumption (i). So $\omega$ must be a ground state for $e^{itM_{0n}}$.

We have therefore established the following general facts:

\[temperature\] Let Standing Assumptions 