In his 1928 paper, Nyquist derives the famous Johnson-Nyquist noise relation which gives the voltage fluctuations around a resistor at thermal equilibrium.
To show it, he starts to consider two resistors connected via a waveguide at thermal equilibrium. To know the absorbed and emitted power from those resistors, he fastly removes the resistor and put the two ends of the line to the ground.
It is simply a trick he does to have a situation he can be familiar with: finding the energy of stationary waves in a waveguide.
After some calculations he finds that the forward and backward travelling waves on this line have a noise power spectral density $\frac{dP}{d \nu}=k_b T$.
Extra informations about how I understood he find the power flow:
He has a resistor in $x=0$ and a resistor in $x=L$. They are connected via a waveguide of length $L$. We assume the impedance of the wire is the same as the value of the resistance so that all the incoming wave on a resistor are not reflected.
He now wants to compute the total energy in the line. To go to a familiar situation, he removes the two resistors and fastly connect the waveguide to the ground. This process doesn't dissipate any energy conceptually. Then the modes in the line are stationnary waves. At this point, the only thing to remain is to compute the energy of a waveguide with the two ends grounded. And then relate it to power flows.
There are two polarizations in the waves on the line. As it is a free propagation, the hamiltonian is harmonic oscillator. From equipartition theorem, we can find the energy in the line being in average: $\frac{k_b T}{2}$ per quadratic term.
The stationnary waves are indexed by $k_n = \frac{n \pi}{L}$. There are $2\cdot\frac{L}{\pi}dk$ modes inside $[k, k+dk]$, where the $\cdot 2$ counts for the two polarizations. It gives $\frac{4 L}{c} d \nu$ modes in $[\nu, \nu+d\nu]$ where $c$ is the velocity of the waves in the waveguide. Thus, inside a frequency interval $[\nu, \nu+d\nu]$, there is an energy in the line being:
$$dE=\frac{k_b T}{2}\cdot\frac{4 L}{c}\cdot d\nu$$
The energy for the propagating left to right modes is then the half:
$$dE_{\rightarrow}=k_b T\cdot\frac{L}{c}\cdot d\nu$$
The energy per unit length of the propagating left to right mode inside a frequency interval $[\nu, \nu+d\nu]$ is:
$$\frac{dE_{\rightarrow}}{d L}=\frac{k_b T}{c}\cdot d\nu$$
The energy that is contained in the length $cdt$ corresponds to the energy that will cross any section of the cable. this energy is: $\frac{k_b T}{c}\cdot d\nu\cdot cdt=k_b T\cdot d\nu\cdot dt$. I finally divide by $dt$ and I find the power power flow of the forward propagating waves inside $[\nu, \nu+d \nu]$:
$$d P_{\rightarrow}=k_b T d\nu$$
Thus, here, I never needed any resistor to get to this result. After he uses this result and apply it back on resistor. But for me this derivation shows that there is a noise spectral density in a waveguide even if you don't have any element on it. Simply because some modes can exist and propagates. Thus I am confused when I see that this noise power spectral density is always related to resistor, for me that's two different thing. Using it we can find the voltage fluctuation around a resistor. But we don't need any resistor to have some noise propagating on a line, it is just electromagnetic field thermalized.
Would you agree ?
