Suppose you have driven harmonic oscillator (parameters: mass,gamma,omega0) by a deterministic force Fdrive (a sine wave say). Now suppose that you add stochastic Langevin force FL which is related to the bath temperature T.
The question is how to extract the information about the temperature T by looking at the time trace of x(t) by looking at it for a time MUCH SMALLER THAN 1/gamma.
So you can only look at x(t) a fraction of 1/gamma and you want to know the temperature of the bath. You already know omega0, gamma and mass.
I think it is possible but I cannot prove it.
NB: omega0 is the resonant frequency of the oscillator gamma is the damping rate FL is defined as =2gammakBTdeltadirac(t2-t1) and =0
Answer
Taking $$m\frac{d^2x}{dt^2} = - kx - \gamma v + F(t) + \eta$$ and writing this as $$\mathrm{d}\mathbf{x}_t= A\mathbf{x}_t\mathrm{d}t + \mathbf{F}_t\mathrm{d}t + \sigma\mathrm{d}W_t$$ where $\mathbf{x}_t = (x, v)^\mathrm{T}$, $A = \begin{pmatrix}0 & 1 \\ -\frac{k}{m} & -\frac{\gamma}{m}\end{pmatrix}$, $\mathbf{F}_t = (0, F(t))^\mathrm{T}$, $\sigma = (0, \sqrt{2 \gamma k_BT}/m)^\mathrm{T}$.
Solving this, as usual, $$\mathbf{x}_t = e^{tA}\mathbf{x}_0 + \int_0^t e^{-(s-t)A}\mathbf{F}_s\mathrm{d}s + \int_0^t e^{-(s-t)A}\sigma\mathrm{d}W_s$$
The general solution here is a bit messy thanks to the matrix exponential, but if you set $k = 0$ it all simplifies a great deal and you recover the Ornstein-Uhlenbeck process.
Now I don't have proof for this (I'm guessing that at least under typical conditions the integrated process $\int_0^t \int_0^{t'} f(s,t') \mathrm{d}W_s\mathrm{d}t'$ has a lower variance than $\int_0^t f(s,t) \mathrm{d}W_s$, which I think is equivalent to the statement $\left(f(s,t)\right)^2 > \left(\int_s^tf(s,t')\mathrm{d}t'\right)^2$), but testing with simulations it seemed to be quite difficult to recover the temperature from the variance of $x_t$: I computed $x_{t+\Delta t}$ given $x_t$ using the formula above, then took the variance of the difference of the thus predicted $x_{t+\Delta t}$ vs. the actual $x_{t+\Delta t}$. This still left a residual term due to the external force, perhaps because of numerical noise (in the sense that Euler-Maruyama, the method I used, does not numerically speaking match the way I computed the integrals accurately enough). This is all to say that this approach is quite sensitive to noise. It however worked much better for the velocity (again, as its variance is larger),
$$\operatorname{Var}(v_{t+\Delta t} - v_t) = \int_0^{\Delta t} \left((0, 1)e^{-(s-\Delta t)A}\sigma\right)^2\mathrm{d}s$$
which as you can see depends linearly on $T$.
If you don't need a very automated process of doing this, you can probably get rid of the residuals in a more manual fashion.
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