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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?

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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.

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  • $\begingroup$ I do not think you are answering my question, e.g., there is no reason why the power density flow be $k_b(T_2-T_1)$ at frequencies that the transmission line does not pass because of the filter. In fact, I wish to understand the dynamics of what is happening to the noise waves that are reflected back to their respective sources by the filter and how that affects the resistors temperature in time. By ignoring such dynamics is just ignoring the question. $\endgroup$
    – hyportnex
    Nov 19, 2020 at 12:33
  • $\begingroup$ @ hyportnex Sorry if I missed the point. Noise power reflected back to to the source resistor at a specific frequency will be thermalized almost immediately by the source resistor and will become part of the 'white' thermal (Johnson) noise. This will occur on very short time-scales as electrons interact. At frequencies where the two resistors are effectively isolated the measured voltages will be sqrt(4kT1R$\Delta$B) and sqrt(4kT2R$\Delta$B). At frequencies where the two resistors are effectively in parallel the measured voltage will be sqrt(4k((T1+T2)/2)(R/2)$\Delta$B) $\endgroup$
    – Roger Wood
    Nov 19, 2020 at 21:42
  • $\begingroup$ OK, place in the middle of a 1m long coaxial line a 9-pole ideal lossless elliptic filter centered at 1MHz with bandwidth $B=1Hz$, $T_2 = 500K$ and $T_1=100K$, in-band return loss (ripple) = -0.01dB, out-band return loss (ripple) = -100dB; how long will it take for the resistors to reach the common $300K$ within $\pm 1K$? $R_1=R_2=Z_0=50\Omega$, $C_1=C_2=1J/K$ $\endgroup$
    – hyportnex
    Nov 19, 2020 at 22:45

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