Driven harmonic oscillator with thermal Langevin force. How to extract temperature from $x(t)$?

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

• Look perhaps at the particle diffusivity $D$, which should be i) related to the position variance as $\bar{x^2} = 2Dt$, and ii) proportional to temperature, $D \sim k_bT$. See for instance web2.clarkson.edu/projects/crcd/me537/downloads/2_Brownian.pdf – udrv Sep 15 '16 at 3:22
• On a 2nd thought $\bar{x^2} = 2Dt$ at large $t$, but you could still look at $\bar{x^2}(t)$ and its derivatives even at shorter time scales. The proportionality to temperature via the Langevin force intensity should still be there. – udrv Sep 15 '16 at 3:35

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.

• Thanks for your answer. Could you explain the last step? It seems Var diverges with Deltat. Is it normal? – pierebean Oct 4 '16 at 14:59
• Can't this be done the Langevin way, i.e. write the equation of motion in the frequency domain and solve for $\langle x^2 \rangle$? – DanielSank Oct 5 '16 at 4:47
• @alarge Yes, according to my simulations your variances are really sensitive to noise. It seems there are no free lunch, if you want to know the energy of oscillator (thermal energy), you need to wait.... Indeed, if you look at the spectrum of <x²>, you will see a lorentzian of width $$\gamma$$ at frequency $$\omega_{0}$$ . If you want to resolve this lorentzian, you need to wait at least a few \gamma. The integral below your lorentzian will give you for sure the temperature. The essence of my question was "could you afford not to wait a few $$\gamma$$. It seems not, but why... – pierebean Oct 5 '16 at 11:15
• @pierebean Can you add some specific values of the variables and I'll see if I have any better luck inferring the temperature? – alarge Oct 5 '16 at 15:08
• omega0=2*pi*1e9; gammaM=2*pi*1e9/1e3; (Q factor 1000) meff=9,025e-15; k=omega0^2*meff; Deltat=0.0001*gammaM; – pierebean Oct 6 '16 at 8:13