What if acceleration changes in the Unruh effect?

Question

Based on the Unruh effect, when a observer accelerates then he will see a thermal bath. mathematically the vacuum state for a non-inertial observer is $$|0\rangle=\text{cos}^{-1}(r)\sum_{n=0}^{\infty}\text{tanh}^n(r)|n\rangle_\mathrm{I}|n\rangle_{\mathrm{II}},$$ where $\mathrm{I}$ and $\mathrm{II}$ indicate the first and second Rindler regions that are causally disconnected. Also we have $$\text{sinh}(r)=(e^{2 \pi c \omega /a}-1)^{-1/2},$$ where $\omega$, $c$ and $a$ are the frequency of the particle, speed of light and the acceleration of the particle, respectively.

My question is what if the acceleration is not constant? Does the above relations hold when the acceleration changes?

Answer 1

No, they don't.

Thermality is a very special property of the uniformly accelerated observers. This is intimately related to the fact that the accelerated observers are following boost symmetries, and with respect to this specific trajectories the spectrum is thermal.

To give a quick intuition, consider the trajectory that stays at rest for proper time $\tau < \tau_1$, accelerates with acceleration $a$ for $\tau_1 < \tau < \tau_2$, and accelerates with acceleration $2a$ for $\tau > \tau_2$ ($\tau_1$ and $\tau_2$ are constants). At each point of this trajectory, the temperature is different. Hence, overall, you do not have a thermal spectrum.

There is an important caveat in my example: the Unruh effect does not assume you are accelerated for a little while. It assumes you have always been and will always be accelerating at a constant acceleration. It can, of course, be taken to be an approximation for sufficiently long accelerations, but we don't have any idea about what happens in between these "epochs". Hence, the spectrum in my example is not merely three thermal spectra patched together: it is definitely more complicated.