Nyquist noise and thermal equilibration

Question

Let there be given two identical lumped element resistors $R_1=R_2$ whose heat capacities are also equal and given $C_1=C_2$. We assume the resistors are attached to thermostats, one at temperature $T_1$ and the other at temperature $T_2$ but $T_1 \ne T_2$. Now separate the resistors from their respective thermostats and connect the resistors with a transmission line that has a very low loss (ideally lossless), and such that its metal conductor also has very low thermal conductivity (ideally zero). I know this is a contradiction per Wiedemann-Franz but assume it for the sake of argument. I expect that because of the Nyquist noise emitted from the resistors eventually they will come to a common temperature, and since we assume equal heat capacities, $C_1=C_2$, the common temperature will be $(T_1+T_2)/2$.

Now somewhere along the transmission line whose wave impedance is $Z_0=R_1=R_2$ we place an ideal lossless reactive filter and/or ideal impedance transformer ($I_2=I_1/N, V_2=NV_1$). How will the system equilibrate if not all frequencies are allowed to pass by the filter (e.g., the transformer does not work at $f=0$)? What is the equation that describes the temperature development of each resistor as noise waves are exchanged between them?

Answer 1

The thermal electrical power available at each resistor is given by kTB, where k is Boltzmann's constant, T is temperature in Kelvin and B is the bandwidth in Hertz through which the noise is observed. Power will flow from the hot resistor to the cold resistor at a rate k(T1-T2)B. The result will be an exponential convergence on a temperature (T1+T2)/2 with a time-constant C/kB.
The question, of course, is what to assume for the the bandwidth, B. The thermal electrical noise is roughly constant up to a frequency of kT/h where h is Planck's constant. Any electrical network will have a bandwidth much less than this. You can reasonably take B as the integral vs. frequency of the power transmission coefficient between the two resistors through the circuit that connects them. The specifics of which frequencies are passed and which are stopped are immaterial since the energy quickly equilbrates through electron interactions, etc.