Uncertainty in the temperature measurement using an oscillator coupled to a bath

Question

I'm considering a harmonic oscillator coupled to a bath (e.g. a gas). The oscillator is subject to a damping and a fluctuating Langevin force

$$ m\ddot{x} + m\gamma\dot{x} + kx =F_\mathrm{fluct}, $$

where $x(t)$ is the position, $m$ the mass, and $k$ the spring constant of the oscillator. $\gamma$ is the damping constant and $F_\mathrm{fluct}$ the fluctuating force that fulfills $\langle F_\mathrm{fluct}(t)F_\mathrm{fluct}(t') \rangle=2m\gamma k_B T\delta(t-t')$. Here $T$ is the bath temperature and $k_B$ the Boltzmann constant. The oscillator shall be in thermal equilibrium with the bath and due to the flucatuations the amplitude of the oscillation will fluctuate over time at a time scale of the order of $1/\gamma$.

The goal is to measure the bath temperature $T$, given a time trace of the oscillator position $x(t)$ of measurement duration $\tau$. I know that the bath temperature can be calculated from the variance of the position

$$ T = \frac{k\langle x^2(t)\rangle_{\tau\to\infty}}{k_B}. $$

However, if the measurement duration $\tau$ is finite, I will have an error on the temperature measurement, because the oscillation amplitude is fluctuating.

What is the measurement uncertainty $\Delta T$ as a function of measurement duration $\tau$ and damping rate $\gamma$.

Note: Measurements suggest that the standard deviation of the temperature measurement is 1% for $\tau\approx \frac{2000}{(\gamma/2\pi)}$ and 5% for $\tau\approx \frac{100}{(\gamma/2\pi)}$. However, these results are purely empirical and I would like to understand if this makes sense from a statistics point-of-view.

Answer 1

It turns out to be as follows:

Assume that you have a long time trace of length $\tau$ which is chopped into $N$ pieces of duration $\Delta \tau=\tau/N$. The energy estimate for every section is $E_i$. The mean of all estimates is $\bar{E}=\sum_i E_i/N$. The energy is Boltzmann distributed, because we have a thermal state. Therefore, the variance for each measurement is $\sigma_{E_i}^2=(k_BT)^2\approx\bar{E}^2$, where $T$ is the bath temperature. The oscillation amplitude changes on a time scale $\Delta t=1/\gamma$ and therefore $n=\Delta t/\Delta \tau$ short traces are correlated. Then we get for the variance of the averaged energy:

$$ \sigma_{\bar{E}}^2 = \frac{2n}{N}\sigma_{E}^2 = \frac{2 \bar{E}^2}{\gamma \tau}. $$

The key in this derivation is that there are correlations between the short sections of the time trace, which don't provide new information on the energy.