Subfield: Electromagnetism

2. Johnson-Nyquist noise

In this problem we study thermal noise in electrical circuits. The goal is to derive the JohnsonNyquist spectral (per-frequency, $f$ ) density of noise produced by a resistor, $R$ :

$$
\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R
\tag{2}
$$

Here, \langle\rangle denotes an average over time, so $\left\langle V^{2}\right\rangle$ is the mean-square value of the voltage fluctuations due to thermal noise. $f$ is the angular frequency, $k$ is Boltzmann's constant, and $T$ is temperature. It says that every frequency range $[f, f+d f]$ contributes a roughly equal amount of noise to the total noise in the resistor; this is called white noise.

Electromagnetic modes in a resistor

We first establish the properties of thermally excited electromagnetic modes

$$
V_{n}(x)=V_{0} \cos \left(k_{n} x-\omega_{n} t\right)
\tag{3}
$$

in a resistor of length $L$. The speed of light $c^{\prime}=\omega_{n} / k_{n}$ in the resistor is independent of $n$.
Context question:
(a) The electromagnetic modes travel through the ends, $x=0$ and $x=L$, of the resistor. Show that the wavevectors corresponding to periodic waves on the interval $[0, L]$ are $k_{n}=\frac{2 \pi n}{L}$.

Then, show that the number of states per angular frequency is $\frac{d n}{d \omega_{n}}=\frac{L}{2 \pi c^{\prime}}$.
Context answer:
\boxed{证明题}


Context question:
(b) Each mode $n$ in the resistor can be thought of as a species of particle, called a bosonic collective mode. This particle obeys Bose-Einstein statistics: the average number of particles $\left\langle N_{n}\right\rangle$ in the mode $n$ is

$$
\left\langle N_{n}\right\rangle=\frac{1}{\exp \frac{\hbar \omega_{n}}{k T}-1}
\tag{4}
$$

In the low-energy limit $\hbar \omega_{n} \ll k T$, show that

$$
\left\langle N_{n}\right\rangle \approx \frac{k T}{\hbar \omega_{n}}
\tag{5}
$$

You can use the Taylor expansion $e^{x} \approx 1+x$ for small $x$.
Context answer:
\boxed{证明题}


Context question:
(c) By analogy to the photon, explain why the energy of each particle in the mode $n$ is $\hbar \omega_{n}$.
Context answer:
\boxed{证明题}


Context question:
(d) Using parts (a), (b), and (c), show that the average power delivered to the resistor (or produced by the resistor) per frequency interval is

$$
P[f, f+d f] \approx k T d f .
\tag{6}
$$

Here, $f=\omega / 2 \pi$ is the frequency. $P[f, f+d f]$ is known as the available noise power of the resistor. (Hint: Power is delivered to the resistor when particles enter at $x=0$, with speed $c^{\prime}$, and produced by the resistor when they exit at $x=L$.)
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Nyquist equivalent noisy voltage source 

The formula $\frac{d\left\langle V^{2}\right\rangle}{d f}=4 k T R$ is the per-frequency, mean-squared value of an equivalent noisy voltage source, $V$, which would dissipate the available noise power, $\frac{d P}{d f}=k T$, from the resistor $R$ into a second resistor $r$.

Question: (a) Assume that resistors $R$ and $r$ are in series with a voltage $V . R$ and $V$ are fixed, but $r$ can vary. Show the maximum power dissipation across $r$ is

$$
P_{\max }=\frac{V^{2}}{4 R} .
\tag{7}
$$

Give the optimal value of $r$ in terms of $R$ and $V$.

Solution:
The total resistance in this circuit is $R+r$. The power dissipated in $r$ is therefore

$$
P=r I^{2}=r\left(\frac{V}{R+r}\right)^{2}
\tag{7}
$$

The maximization of $\frac{r}{(R+r)^{2}}$ is equivalent to the minimization of $\phi(r)=\frac{(R+r)^{2}}{r}=\frac{R^{2}}{r}+2 R+r$. Setting $\frac{d \phi}{d r}=0$, for example, gives the solution $r=R$. Substituting $r=R$ into 7 yields

$$
P=\frac{V^{2}}{4 R}
\tag{8}
$$