= Solution
Set $v=\dot x$, $\zeta=6\pi\mu a$, and $\tau=m/\zeta$. To keep dimensions and the thermal amplitude explicit, write the <force> covariance as $\langle F(t)F(s)\rangle=A\delta(t-s)$. The printed unit coefficient is a choice of noise normalization, not a general dimensional thermal <force>. Assume the initial <velocity> is independent of the future <white noise>. The <Underdamped Langevin dynamics> has integrating-factor solution
$$
v(t)=v_0e^{-t/\tau}+\frac1m\int_0^t e^{-(t-s)/\tau}F(s)\,ds.
$$
Taking its <expectation> and using the delta covariance in the double integral gives
$$
\boxed{\langle v(t)\rangle=\langle v_0\rangle e^{-t/\tau},\qquad \langle v(t)^2\rangle=\langle v_0^2\rangle e^{-2t/\tau}+\frac{A}{2m\zeta}(1-e^{-2t/\tau}).}
$$
Here the noise contribution is $A m^{-2}\int_0^t e^{-2(t-s)/\tau}ds$. The <equipartition theorem> requires the long-time value $\langle mv^2/2\rangle=k_BT/2$. Consequently the <fluctuation-dissipation relation for a Langevin particle> is
$$
\boxed{A=2\zeta k_BT,\qquad \langle F(t)F(s)\rangle=2\zeta k_BT\delta(t-s).}
$$
With an initially thermal <velocity>, $\langle v_0\rangle=0$ and $\langle v_0^2\rangle=k_BT/m$, so the <velocity> variance is stationary. If one insists on the printed coefficient $A=1$ in dimensional units, the model has <temperature> fixed by $k_BT=1/(2\zeta)$ in those units; it cannot describe arbitrary <temperature> while retaining that coefficient.
Back to article page
