Coupled Brownian Fans Two fans are immersed in different mediums in equilibrium at different temperatures. Given that $\gamma$ and $\gamma^{\prime}$ are the friction coefficients for each fan and $\theta$ and $\theta^{\prime}$ are the angles of the rotating fans that are coupled by a spring constant $K$, one can write:
$$\frac{d \left(\theta - \theta^{\prime} \right)}{dt} = - \left(\frac{1}{\gamma} + \frac{1}{\gamma^{\prime}}\right)K \left(\theta - \theta^{\prime} \right) + \frac{1}{\gamma}\xi(t)  + \frac{1}{\gamma^{\prime}}\xi^{\prime}(t)$$, where $\xi(t)$ is the white Gaussian noise.
Defining the heat as $$\delta Q \equiv \left( -\gamma \dot{\theta} + \xi \right)\delta \theta$$, after some manipulations, we can write:
$$\frac{\delta Q}{dt} + \frac{\delta Q^{\prime}}{dt} = \frac{d}{dt}\left[ \frac{K}{2}\left(\theta - \theta^{\prime} \right)^2\right]$$
After averaging over $\xi$, we notice that $\langle \left(\theta - \theta^{\prime} \right) \rangle$ doesn't depend on time, so we write:
$$\frac{\langle \delta Q \rangle }{dt} = - \frac{\langle \delta Q^{\prime} \rangle}{dt} $$
But, I can't show that:
$$\gamma\frac{\langle \delta Q \rangle }{dt} - \gamma^{\prime}\frac{\langle \delta Q^{\prime} \rangle}{dt} = K k_{B} \left(T - T^{\prime} \right)$$
How can I derive this last result?
Thank you very much!
Edit:
$ \left(\frac{1}{\gamma} - \frac{1}{\gamma^{\prime}}\right)K \rightarrow  \left(\frac{1}{\gamma} + \frac{1}{\gamma^{\prime}}\right)K$
 A: I think that you initially had two equations:
$$
\dot{\theta} = -\frac{K}{\gamma}(\theta-\theta') + \frac1{\gamma}\xi \quad (0)
$$
and
$$
\dot{\theta}' = \frac{K}{\gamma'}(\theta-\theta') + \frac1{\gamma'}\xi' \quad (0')
$$
Then there are misprints in your text. The correct version of the first equation is
$$
\dot{\theta} - \dot{\theta}' = -K\left(\frac{1}{\gamma} + \frac{1}{\gamma'}\right)(\theta-\theta') + \frac1\gamma\xi -\frac1{\gamma'}\xi'\quad (1)
$$
I suppose that $\delta \theta = \dot{\theta} dt$. Then
$$
\frac{\delta Q}{dt} = -\gamma\dot{\theta}^2 + \xi\dot{\theta} = K(\theta-\theta')\dot{\theta},\quad \frac{\delta Q'}{dt} = -K(\theta-\theta')\dot{\theta}' \quad (2)
$$
By the way, the correct version of the heat exchange equation is as follows:
$$
\frac{\delta Q}{dt} + \frac{\delta Q'}{dt} = \frac{d}{dt}\left[\frac{K}2(\theta-\theta')^2 \right].
$$
It is easy to derive the following equality from equalities (0), (0') and (2):
$$
\gamma \frac{\delta Q}{dt} - \gamma' \frac{\delta Q'}{dt} = K(\theta - \theta')(\xi+\xi'). \quad (3)
$$
I'll use the spectral representation of random processes:
$$
\theta(t) - \theta'(t) = \int\limits_{-\infty}^\infty e^{-i\omega t-\varepsilon|\omega|}(\tilde{\theta}_\omega - \tilde{\theta}_\omega') d\omega,\qquad (4)
$$
$$
\xi(t) = \int\limits_{-\infty}^\infty e^{-i\omega t-\varepsilon|\omega|}\tilde{\xi}_\omega d\omega, \quad \xi'(t) = \int\limits_{-\infty}^\infty e^{-i\omega t-\varepsilon|\omega|}\tilde{\xi}_\omega' d\omega.
$$
The random processes $\xi$ and $\xi'$ are independent white noise. According to the theory of the Brownian motion, these processes have the following properties:
$$
\overline{\tilde{\xi}_\omega \tilde{\xi}_{\omega'}} = A\delta(\omega+\omega'), \quad \overline{\tilde{\xi}_\omega' \tilde{\xi}_{\omega'}'} = A'\delta(\omega+\omega'), \quad
\overline{\tilde{\xi}_\omega \tilde{\xi}_{\omega'}'} = 0,\quad (5)
$$
$$
A = \gamma \frac{k_B T}2,\quad A' = \gamma' \frac{k_B T'}2
$$
Substitution of (4) in (1) gives
$$
\tilde{\theta}_\omega - \tilde{\theta}_\omega' = \frac{\gamma' \tilde{\xi}_\omega - \gamma \tilde{\xi}_\omega'}{K(\gamma+\gamma') - i\omega\gamma\gamma'}.\quad (6)
$$
Now we obtained from (3-6):
$$
\gamma \overline{\frac{\delta Q}{dt}} - \gamma' \overline{\frac{\delta Q'}{dt}} = K\overline{(\theta - \theta')(\xi+\xi')} = 
$$
$$
= K\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty e^{-it(\omega+\omega')}e^{-\varepsilon(|\omega| + |\omega'|)}\overline{(\tilde{\theta}_\omega - \tilde{\theta}_\omega')(\tilde{\xi}_{\omega'} + \tilde{\xi}_{\omega'}')} d\omega d\omega' =
$$
$$
= \left.K\int\limits_{-\infty}^\infty e^{-2\varepsilon|\omega|}\frac{\gamma' A - \gamma A'}{K(\gamma+\gamma') - i\omega\gamma\gamma'} d\omega\right|_{\varepsilon\to+0} = Kk_B (T - T').
$$
Voila.
