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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 $2*\frac{L}{\pi}dk$ modes inside $[k, k+dk]$, where the $*2$ 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:

$$dE=\frac{k_b T}{2}*\frac{4 L}{c}*d\nu$$

The energy for the propagating left to right modes is then the half:

$$dE_{\rightarrow}=k_b T*\frac{L}{c}*d\nu$$

The energy per unit length of the propagating left to right mode inside a frequency interval $[\nu, \nu+d\nu]$ is:

$$\frac{dE_{\rightarrow}}{d L}=\frac{k_b T}{c}*d\nu$$

The energy that is contained in the length $cdt$ corresponds to the energy that will cross any section of the cable. this energy is: $\frac{k_b T}{c}*d\nu*cdt=k_b T*d\nu*dt$. 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 ?

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  • $\begingroup$ Can you expand more on how you got from "a waveguide at temperature $T$ where the two ends are at the ground," to "I simply apply the equipartition theorem..., and I find $\frac{dP}{d \nu}=k_b T$"? $\endgroup$ – The Photon Apr 17 '20 at 15:31
  • $\begingroup$ @ThePhoton I edited ! $\endgroup$ – StarBucK Apr 17 '20 at 16:18
  • $\begingroup$ First, you determine the number of modes. Then you say, "Thus, [in a given frequency interval] there is an energy in the line..." I don't see how the existence of a certain number of modes means there must be energy present in those modes. $\endgroup$ – The Photon Apr 17 '20 at 16:35
  • $\begingroup$ @ThePhoton because your field is at thermal equilibrium. The energy comes from the bath $\endgroup$ – StarBucK Apr 17 '20 at 16:40
  • $\begingroup$ What bath? You took away the resistors at the end. Weren't they the radiators (or bath) that were exciting the waveguide in the original formulation? $\endgroup$ – The Photon Apr 17 '20 at 16:43
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The Johnson-Nyquist noise is a special case of the Fluctuation-dissipation Theorem. Any system with some sort of dissipation mechanism built into it will be noisy.

If in your system there is no is a process that dissipates energy, turning it into heat, conversely, there will no be a fluctuation in some sort of physical quantity due to the finite temperature of your system.

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  • $\begingroup$ I edited, look at my post. Inside a waveguide some modes can exist. The temperature will excite them. I don't see why we would need any dissipation for those mode to be excited. This is how I understood the derivation made by Nyquist and I would like to check my understanding. $\endgroup$ – StarBucK Apr 17 '20 at 16:18
  • $\begingroup$ @StarBucK Maybe you have some power in the waveguide, but this power is sterile without something to dissipate to. Hence no noise. $\endgroup$ – Davide Dal Bosco Apr 17 '20 at 23:29
  • $\begingroup$ I dont ser what you mean. Why would we need to dissipate this power to have noise ? $\endgroup$ – StarBucK Apr 17 '20 at 23:37
  • $\begingroup$ @StarBucK noise is always related to a signal. Which is the signal that you are measuring in your system? $\endgroup$ – Davide Dal Bosco Apr 18 '20 at 6:57
  • $\begingroup$ I see what you mean. To in practice see this noise I must use something that will convert this power flow in some "visible" signal. A resistor does the job. But conceptually this resistor is just here to make me able to see that there is power flow density in the waveguide that exists even if I didn't put a resistor. This power flow exists just because of thermal equilibrium between the bath and the waveguide $\endgroup$ – StarBucK Apr 18 '20 at 10:17
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In Nyquist's analysis the resistors represent black body cavities with a small hole from which we extract energy in a single mode and polarization via a transmission line and transfer it to an identical black body cavity. This way the two cavities and the connecting transmission line establish a single thermodynamic system in equilibrium, a single composite cavity without opening whose internal thermal equilibrium is under investigation. Each cavity with its opening in either end represents electromagnetically a terminating resistor for the propagating mode of the transmission.

For Nyquist's argument to work he assumes that the each resistor on either end is matched to the transmission line's wave impedance, this is how the noise waves get launched on the line without reflection and get absorbed at the other end again without reflection. While you might say that the wave impedance of a finite length transmission line is a fictitious one in the sense of not being resistive in nature and not having a temperature the terminating impedances are resistive electromagnetically and have temperature, the cavity temperature.

If the terminating impedances were different, i.e, unmatched, then the argument would not work for then the energy content on the line would be fluctuating in time and location instead of being stationary. When matched the energy distribution depends only on the temperature and not on the where and when. Recall what happens to a pulse when it is launched on or reflected by a mismatched line.

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  • $\begingroup$ Hello. Thank you for your answer. I agree that he needs in the derivation to have impedance matching between the lines and the resistor so that all the wave arriving on resistor are fully absorbed (and not reflected). What confuses me is that the power flow that is propagating in the line is independant from the fact we put resistor at the end or not. Indeed the waveguide in itself is a physical system that has a sum of harmonic oscillator hamiltonian. Some modes can thus be excited because of temperature. Then if we want we can put resistors on the two ends but it is an "extra". $\endgroup$ – StarBucK Apr 18 '20 at 16:27
  • $\begingroup$ Assume that I considered a cable grounded in $x=0$ and $x=L$, I never talk about resistor at all. Then I just count the stationnary mode that can exist. They can be excited by temperature (equipartition theorem if we deal with the system classically). From it I find that my waveguide has forward and backward propagating waves and power flow associated. Of course the net power flow will be $0$. What is wrong in what I am saying ? We do not need any resistor to have this. Then if we want we plug resistor and we see that it will induce on them voltage fluctuation. But the underline physics is $\endgroup$ – StarBucK Apr 18 '20 at 16:29
  • $\begingroup$ linked to the waveguide hamiltonian, not to the resistor (that are just here to dissipate those excitations). $\endgroup$ – StarBucK Apr 18 '20 at 16:30
  • $\begingroup$ ask yourself: how do I get the forward and propagating waves on the line? $\endgroup$ – hyportnex Apr 18 '20 at 16:46
  • $\begingroup$ What do you mean by "how do I get" ? How do I measure them ? Or how do I create them ? $\endgroup$ – StarBucK Apr 18 '20 at 16:47

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