# B.1 Evaluation of Planck’s Sum

Planck’s sum (Equation 2.83) for the average energy per mode of blackbody radiation is

 $\langle E\rangle=\frac{\displaystyle\sum\limits_{n=0}^{\infty}nh\nu\exp\Bigl{(% }-\frac{nh\nu}{kT}\Bigr{)}}{\displaystyle\sum\limits_{n=0}^{\infty}\exp\Bigl{(% }-\frac{nh\nu}{kT}\Bigr{)}}.$

It is convenient to introduce the variable $\alpha\equiv 1/(kT)$, so

 $\langle E\rangle=\frac{\sum\limits_{n=0}^{\infty}nh\nu\exp(-\alpha nh\nu)}{% \sum\limits_{n=0}^{\infty}\exp(-\alpha nh\nu)}.$

Next consider the quantity

 $-\frac{d}{d\alpha}\biggl{[}\ln\sum_{n=0}^{\infty}\exp(-\alpha nh\nu)\biggr{]}.$

Using the chain rule to take the derivative yields

 $\displaystyle-\frac{d}{d\alpha}\biggl{[}\ln\sum_{n=0}^{\infty}\exp(-\alpha nh% \nu)\biggr{]}$ $\displaystyle=-\biggl{[}\sum_{n=0}^{\infty}\exp(-\alpha nh\nu)\biggr{]}^{-1}% \frac{d}{d\alpha}\biggl{[}\sum_{n=0}^{\infty}\exp(-\alpha nh\nu)\biggr{]}$ $\displaystyle=\frac{\sum\limits_{n=0}^{\infty}nh\nu\exp(-\alpha nh\nu)}{\sum% \limits_{n=0}^{\infty}\exp(-\alpha nh\nu)}.$

Thus

 $\langle E\rangle=-\frac{d}{d\alpha}\biggl{[}\ln\sum_{n=0}^{\infty}\exp(-\alpha nh% \nu)\biggr{]}.$

Then,

 $\sum_{n=0}^{\infty}\exp(-\alpha nh\nu)=1+[\exp(-\alpha h\nu)]^{1}+[\exp(-% \alpha h\nu)]^{2}+\cdots$

has the form $1+x+x^{2}+\cdots=(1-x)^{-1}$, so

 $\sum_{n=0}^{\infty}\exp(-\alpha nh\nu)=[1-\exp(-\alpha h\nu)]^{-1}$

and

 $\displaystyle\langle E\rangle$ $\displaystyle=-\frac{d\,\ln[1-\exp(\alpha h\nu)]^{-1}}{d\alpha}$ $\displaystyle=-[1-\exp(-\alpha h\nu)](-1)[1-\exp(-\alpha h\nu)]^{-2}h\nu\exp(-% \alpha h\nu)$ $\displaystyle=\frac{h\nu\exp(-\alpha h\nu)}{1-\exp(-\alpha h\nu)}=\frac{h\nu}{% \exp(\alpha h\nu)-1}=\frac{h\nu}{\displaystyle\exp\Bigl{(}\frac{h\nu}{kT}\Bigr% {)}-1}.$

# B.2 Derivation of the Stefan–Boltzmann Law

The Stefan–Boltzmann law for the integrated brightness of blackbody radiation at temperature $T$ (Equation 2.89) is

 $B(T)=\int_{0}^{\infty}B_{\nu}(T)d\nu=\frac{\sigma T^{4}}{\pi},$

where

 $B_{\nu}(T)=\frac{2h\nu^{3}}{c^{2}}\,\frac{1}{\displaystyle\exp\Bigl{(}\frac{h% \nu}{kT}\Bigr{)}-1}$

is Planck’s law and $\sigma$ is the Stefan–Boltzmann constant. Although the Stefan–Boltzmann law and constant were first determined experimentally, both can be derived mathematically from Planck’s law. For simplicity, define

 $x\equiv\frac{h\nu}{kT},$

so

 $B(T)=\int_{0}^{\infty}\frac{2h}{c^{2}}\biggl{(}\frac{kTx}{h}\biggr{)}^{3}% \biggl{(}\frac{1}{e^{x}-1}\biggr{)}\biggl{(}\frac{kT}{h}\biggr{)}dx=\frac{2k^{% 4}T^{4}}{c^{2}h^{3}}\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}.$

The quantity

 $\frac{1}{e^{x}-1}=\frac{e^{-x}}{1-e^{-x}}=e^{-x}\biggl{(}\frac{1}{1-e^{-x}}% \biggr{)}$

can be expanded in terms of the infinite series

 $\displaystyle\sum_{m=0}^{\infty}z^{m}=$ $\displaystyle 1+z+z^{2}+z^{3}+\cdots$ $\displaystyle=$ $\displaystyle 1+z(1+z+z^{2}+z^{3}+\cdots)$ $\displaystyle=$ $\displaystyle 1+z\sum_{m=0}^{\infty}z^{m},$ $\displaystyle\sum_{m=0}^{\infty}z^{m}=$ $\displaystyle\frac{1}{1-z}.$

Thus

 $\frac{1}{e^{x}-1}=e^{-x}\sum_{m=0}^{\infty}e^{-mx}=e^{-x}+e^{-2x}+e^{-3x}+\cdots$

and the integral becomes

 $\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}=\int_{0}^{\infty}x^{3}\Biggl{(}\sum_{% m=1}^{\infty}e^{-mx}\Biggr{)}dx.$

Each integral in this series can be integrated by parts three times:

 $\displaystyle\int_{0}^{\infty}x^{3}e^{-mx}dx$ $\displaystyle=\frac{x^{3}e^{-mx}}{-m}\bigg{|}_{0}^{\infty}-\int_{0}^{\infty}% \frac{3x^{2}e^{-mx}}{-m}dx=\frac{3}{m}\int_{0}^{\infty}x^{2}e^{-mx}dx,$ $\displaystyle\int_{0}^{\infty}x^{2}e^{-mx}dx$ $\displaystyle=\frac{x^{2}e^{-mx}}{-m}\bigg{|}_{0}^{\infty}-\int_{0}^{\infty}% \frac{2xe^{-mx}}{-m}dx=\frac{2}{m}\int_{0}^{\infty}xe^{-mx}dx,$ $\displaystyle\int_{0}^{\infty}xe^{-mx}dx$ $\displaystyle=\frac{xe^{-mx}}{-m}\bigg{|}_{0}^{\infty}-\int_{0}^{\infty}\frac{% xe^{-mx}}{-m}dx=\frac{1}{m}\int_{0}^{\infty}e^{-mx}dx=\frac{1}{m^{2}},$

to give

 $\int_{0}^{\infty}x^{3}e^{-mx}dx=\frac{6}{m^{4}}$

and

 $\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}=\int_{0}^{\infty}x^{3}\Biggl{(}\sum_{% m=1}^{\infty}e^{-mx}\Biggr{)}dx=6\sum_{m=1}^{\infty}\frac{1}{m^{4}}.$

The sum

 $\sum_{m=1}^{\infty}\frac{1}{m^{4}}=\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^% {4}}+\frac{1}{4^{4}}+\cdots=1+\frac{1}{16}+\frac{1}{81}+\frac{1}{256}+\cdots% \approx 1.082$

converges quickly and is the value of the Riemann zeta function $\zeta(4)=\pi^{4}/90\approx 1.082$. Thus

 $\boxed{\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}=\frac{\pi^{4}}{15}.}$ (B.1)

Finally, the integrated brightness of blackbody radiation is

 $B(T)=\frac{2k^{4}T^{4}}{c^{2}h^{3}}\int_{0}^{\infty}\frac{x^{3}dx}{e^{x}-1}=% \frac{2k^{4}T^{4}}{c^{2}h^{3}}\biggl{(}\frac{\pi^{4}}{15}\biggr{)}=\frac{2\pi^% {4}k^{4}}{15c^{2}h^{3}}T^{4}=\frac{\sigma T^{4}}{\pi},$

so

 $\sigma=\frac{2\pi^{5}k^{4}}{15c^{2}h^{3}}\approx 5.67\times 10^{-5}\,\frac{% \mathrm{~{}erg}}{\mathrm{~{}cm}^{2}\mathrm{~{}s~{}K}^{4}\mathrm{~{}(sr)}}$

is the value of the Stefan–Boltzmann constant.

Similarly, the integral

 $\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}$

is needed to evaluate the number density $n_{\gamma}$ of blackbody photons:

 $n_{\gamma}=\frac{8\pi}{c^{3}}\int_{0}^{\infty}\frac{\nu^{2}d\nu}{\displaystyle% \exp\Bigl{(}\frac{h\nu}{kT}\Bigr{)}-1}=\frac{8\pi}{c^{3}}\biggl{(}\frac{kT}{h}% \biggr{)}^{3}\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}.$

Following the derivation above,

 $\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}=\int_{0}^{\infty}x^{2}\Biggl{(}\sum_{% m=1}^{\infty}e^{-mx}\Biggr{)}dx$

and

 $\int_{0}^{\infty}x^{2}e^{-mx}dx=\frac{2}{m}\int_{0}^{\infty}xe^{-mx}dx=\frac{2% }{m}\biggl{(}\frac{1}{m^{2}}\biggr{)}=\frac{2}{m^{3}},$

so

 $\boxed{\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}=2\sum_{m=1}^{\infty}\frac{1}{m% ^{3}}=2\biggl{(}\frac{1}{1^{3}}+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots\biggr{)% }\approx 2.404.}$ (B.2)

# B.3 Complex Exponentials

A complex exponential $e^{i\phi}$, where $i^{2}=-1$ and $\phi$ is any dimensionless real variable, is a complex number in which the real and imaginary parts are sines and cosines given by Euler’s formula

 $\boxed{e^{i\phi}=\cos\phi+i\sin\phi.}$ (B.3)

Euler’s formula can be derived from the Taylor series

 $\displaystyle\cos\phi$ $\displaystyle=1-\frac{\phi^{2}}{2!}+\frac{\phi^{4}}{4!}-\frac{\phi^{6}}{6!}+\cdots,$ $\displaystyle\sin\phi$ $\displaystyle=\phi-\frac{\phi^{3}}{3!}+\frac{\phi^{5}}{5!}-\frac{\phi^{7}}{7!}% +\cdots,$ $\displaystyle e^{\phi}$ $\displaystyle=1+\phi+\frac{\phi^{2}}{2!}+\frac{\phi^{3}}{3!}+\frac{\phi^{4}}{4% !}+\cdots.$

Thus

 $\displaystyle e^{i\phi}$ $\displaystyle=1+i\phi-\frac{\phi^{2}}{2!}-\frac{i\phi^{3}}{3!}+\frac{\phi^{4}}% {4!}+\frac{i\phi^{5}}{5!}-\frac{i\phi^{6}}{6!}-\frac{i\phi^{7}}{7!}+\cdots$ $\displaystyle=\biggl{(}1-\frac{\phi^{2}}{2!}+\frac{\phi^{4}}{4!}-\frac{\phi^{6% }}{6!}+\cdots\biggr{)}+i\biggl{(}\phi-\frac{\phi^{3}}{3!}+\frac{\phi^{5}}{5!}-% \frac{\phi^{7}}{7!}+\cdots\biggr{)}$ $\displaystyle=\cos\phi+i\sin\phi.$

Complex exponentials (or sines and cosines) are widely used to represent periodic functions in physics for the following reasons:

1. 1.

They comprise a complete and orthogonal set of periodic functions. This set of functions can be used to approximate any piecewise continuous function, and they are the basis of Fourier transforms (Appendix A.1).

2. 2.

They are eigenfunctions of the differential operator—that is, the derivatives of complex exponentials are themselves complex exponentials:

 $\frac{de^{i\phi}}{d\phi}=ie^{i\phi},\quad\frac{d^{2}e^{i\phi}}{d\phi^{2}}=-e^{% i\phi},\quad\frac{d^{3}e^{i\phi}}{d\phi^{3}}=-ie^{i\phi},\quad\frac{d^{4}e^{i% \phi}}{d\phi^{4}}=e^{i\phi},\dots.$

Most physical systems obey linear differential equations, a low-pass filter consisting of a resistor and a capacitor, for example. A sinusoidal input signal will yield a sinusoidal output signal of the same frequency (but not necessarily with the same amplitude and phase), while a square-wave input will not yield a square-wave output. The response to a square-wave input can be calculated by treating the input square wave as a sum of sinusoidal waves, and the filter output is the sum of these filtered sinusoids. This is the reason why periodic waves or oscillations are almost always treated as combinations of complex exponentials (or sines and cosines).

Real periodic signals can be expressed as the real parts of complex exponentials:

 $\displaystyle\cos\phi$ $\displaystyle=\mathrm{Re}(e^{i\phi}),$ $\displaystyle\sin\phi$ $\displaystyle=\mathrm{Im}(e^{i\phi}).$

 $\displaystyle e^{i\phi}$ $\displaystyle=\cos\phi+i\sin\phi,$ $\displaystyle e^{-i\phi}$ $\displaystyle=\cos\phi-i\sin\phi$

gives the identities

 $\boxed{\cos\phi=\frac{e^{i\phi}+e^{-i\phi}}{2}}$ (B.4)

and

 $\boxed{\sin\phi=\frac{e^{i\phi}-e^{-i\phi}}{2i}.}$ (B.5)

The advantage of complex exponentials over the equivalent sums of sines and cosines is that they are easier to manipulate mathematically. For example, you can use complex exponentials to calculate the output spectrum of a square-law detector (Section 3.6.2) without having to remember trigonometric identities. A square-law detector is a nonlinear device whose output voltage is the square of its input voltage. If the input voltage is $\cos(\omega t)$, the output voltage is

 $\displaystyle\cos^{2}(\omega t)$ $\displaystyle=\biggl{(}\frac{e^{i\omega t}+e^{-i\omega t}}{2}\biggr{)}^{2}$ $\displaystyle=\frac{e^{2i\omega t}+2+e^{-2i\omega t}}{4}$ $\displaystyle=\frac{2\cos(2\omega t)+2}{4}$ $\displaystyle=\frac{1}{2}[\cos(2\omega t)+1].$

The output spectrum has two frequency components: one at twice the input frequency $\omega$ and the other at zero frequency (DC).

# B.4 The Fourier Transform of a Gaussian

The normalized Gaussian function is usually written as

 $f(x)=\frac{1}{\sqrt{2\pi}\sigma}\exp\biggl{(}-\frac{x^{2}}{2\sigma^{2}}\biggr{% )},$ (B.6)

where $\sigma$ is its rms width. To calculate its Fourier transform

 $F(s)\equiv\int_{-\infty}^{\infty}f(x)\exp(-i2\pi sx)dx,$ (B.7)

it is easier to use the form $f(x)=\exp(-\pi x^{2})$, for which $\sigma^{2}=1/(2\pi)$. Then

 $\displaystyle F(s)$ $\displaystyle=\int_{-\infty}^{\infty}\exp(-\pi x^{2})\exp(-i2\pi sx)dx$ (B.8) $\displaystyle=\int_{-\infty}^{\infty}\exp[-\pi(x^{2}+i2sx+s^{2}-s^{2})]dx$ (B.9) $\displaystyle=\exp(-\pi s^{2})\int_{-\infty+is}^{\infty+is}\exp[-\pi(x+is)^{2}% ]d(x+is)$ (B.10) $\displaystyle=\exp(-\pi s^{2})\int_{-\infty}^{\infty}\exp(-\pi x^{2})dx.$ (B.11)

To evaluate this one-dimensional integral, break it into the product of two integrals and change one dummy variable from $x$ to $y$ to suggest Cartesian coordinates in a plane:

 $\displaystyle\int_{-\infty}^{\infty}\exp(-\pi x^{2})dx$ $\displaystyle=\biggl{[}\int_{-\infty}^{\infty}\exp(-\pi x^{2})dx\int_{-\infty}% ^{\infty}\exp(-\pi y^{2})dy\biggl{]}^{1/2}$ (B.12) $\displaystyle=\biggl{[}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp[-\pi% (x^{2}+y^{2})]dx\,dy\biggr{]}^{1/2}.$ (B.13)

Next transform to polar coordinates $r,\theta$ so $r^{2}=x^{2}+y^{2}$ and $dx\,dy=r\,dr\,d\theta$:

 $\int_{-\infty}^{\infty}\exp(-\pi x^{2})dx=\biggl{[}\int_{r=0}^{\infty}\int_{% \theta=0}^{2\pi}\exp(-\pi r^{2})r\,dr\,d\theta\biggr{]}^{1/2}.$ (B.14)

Finally, substitute $u\equiv\pi r^{2}$ and $du=2\pi r\,dr$ to get

 $\int_{-\infty}^{\infty}\exp(-\pi x^{2})dx=\biggl{[}2\pi\int_{u=0}^{\infty}\exp% (-u)\frac{du}{2\pi}\biggr{]}^{1/2}=\biggl{[}-e^{-u}\bigg{|}_{0}^{\infty}\,% \biggr{]}^{1/2}=1.$ (B.15)

Thus

 $F(s)=\exp(-\pi s^{2}).$ (B.16)

The Fourier transform of a Gaussian is a Gaussian.

# B.5 The Gaussian Probability Distribution and Noise Voltage

The voltage $V$ of random noise has a Gaussian probability distribution

 $P(V)=\frac{1}{(2\pi)^{1/2}\sigma}\exp\biggl{(}\frac{-V^{2}}{2\sigma^{2}}\biggr% {)},$ (B.17)

where $P(V)dV$ is the differential probability that the voltage will be within the infinitesimal range $V$ to $V+dV$ and $\sigma$ is the root mean square (rms) voltage. The probability of measuring some voltage must be unity, so

 $\int_{-\infty}^{\infty}P(V)dV=1.$ (B.18)

The normalization of $P(V)$ in Equation B.17 can be confirmed by evaluating the integral

 $\displaystyle\int_{-\infty}^{\infty}\frac{1}{(2\pi)^{1/2}\sigma}\exp\biggl{(}% \frac{-V^{2}}{2\sigma^{2}}\biggr{)}dV$ $\displaystyle=2\int_{0}^{\infty}\frac{1}{(2\pi)^{1/2}\sigma}\exp\biggl{(}\frac% {-V^{2}}{2\sigma^{2}}\biggr{)}dV$ (B.19) $\displaystyle=\biggl{[}\frac{2}{(2\pi)^{1/2}\sigma}\biggr{]}\int_{0}^{\infty}% \exp\biggl{(}\frac{-V^{2}}{2\sigma^{2}}\biggr{)}dV.$ (B.20)

Equation B.15 immediately yields the definite integral

 $\int_{0}^{\infty}\exp(-a^{2}x^{2})dx=\frac{\pi^{1/2}}{2a}.$ (B.21)

Substituting $a^{2}=(2\sigma^{2})^{-1}$ gives the desired result:

 $\int_{-\infty}^{\infty}P(V)dV=\biggl{[}\frac{2}{(2\pi)^{1/2}\sigma}\biggr{]}% \biggl{(}\frac{\pi^{1/2}}{2}\biggr{)}(2\sigma^{2})^{1/2}=1.$ (B.22)

The rms (root mean square) $\Sigma$ of a normalized distribution is defined by

 $\Sigma^{2}\equiv\langle V^{2}\rangle-\langle V\rangle^{2}.$ (B.23)

For the symmetric Gaussian distribution, $\langle V\rangle=0$, so

 $\displaystyle\Sigma^{2}$ $\displaystyle=\langle V^{2}\rangle=\int_{-\infty}^{\infty}V^{2}P(V)dV$ (B.24) $\displaystyle=2\int_{0}^{\infty}V^{2}\frac{1}{(2\pi)^{1/2}\sigma}\exp\biggl{(}% \frac{-V^{2}}{2\sigma^{2}}\biggr{)}dV$ (B.25) $\displaystyle=\biggl{[}\frac{2}{(2\pi)^{1/2}\sigma}\biggr{]}\int_{0}^{\infty}V% ^{2}\exp\biggl{(}\frac{-V^{2}}{2\sigma^{2}}\biggr{)}dV.$ (B.26)

The definite integral

 $\int_{0}^{\infty}x^{2}\exp(-a^{2}x^{2})dx=\frac{\pi^{1/2}}{4a^{3}}$ (B.27)

can be derived by integrating Equation B.21 by parts. Inserting $a^{2}=(2\sigma^{2})^{-1}$ yields

 $\Sigma^{2}=\langle V^{2}\rangle=\biggl{[}\frac{2}{(2\pi)^{1/2}\sigma}\biggr{]}% \biggl{(}\frac{\pi^{1/2}}{4}\biggr{)}(2\sigma^{2})^{3/2}=\sigma^{2},$ (B.28)

confirming that $\sigma$ in Equation B.17 is the rms of the Gaussian distribution.

# B.6 The Probability Distribution of Noise Power

A square-law detector multiplies the input voltage $V$ by itself to yield an output voltage $V_{\mathrm{o}}=V^{2}$ that is proportional to the input power. The input voltage distribution is a Gaussian with rms $\sigma$ (Equation B.17),

 $P(V)=\frac{1}{(2\pi)^{1/2}\sigma}\exp\biggl{(}\frac{-V^{2}}{2\sigma^{2}}\biggr% {)}.$ (B.29)

The same value of $V_{\mathrm{o}}=V^{2}$ is produced by both positive and negative values of $V$ and $P(V)=-P(V)$, so

 $P_{\mathrm{o}}(V_{\mathrm{o}})dV_{\mathrm{o}}=2P(V)dV$ (B.30)

for all $V_{\mathrm{o}}\geq 0$. Because $dV_{\mathrm{o}}=2VdV$,

 $P_{\mathrm{o}}(V_{\mathrm{o}})=\biggl{[}\frac{V_{\mathrm{o}}^{-1/2}}{(2\pi)^{1% /2}\sigma}\biggr{]}\exp\biggl{(}\frac{-V_{\mathrm{o}}}{2\sigma^{2}}\biggr{)}$ (B.31)

for $V_{\mathrm{o}}\geq 0$. The distribution of detector output voltage is sharply peaked near $V_{\mathrm{o}}=0$ and has a long exponentially decaying tail (Figure 3.33).

The mean detector output voltage follows from Equation B.28: $\langle V_{\mathrm{o}}\rangle=\langle V^{2}\rangle=\sigma^{2}$.

The rms $\sigma_{o}$ of the detector output voltage is

 $\sigma_{\mathrm{o}}^{2}=\langle V_{\mathrm{o}}^{2}\rangle-\langle V_{\mathrm{o% }}\rangle^{2},$ (B.32)

where

 $\displaystyle\langle V_{\mathrm{o}}^{2}\rangle$ $\displaystyle=\int_{0}^{\infty}V_{\mathrm{o}}^{2}P_{\mathrm{o}}(V_{\mathrm{o}}% )dV_{\mathrm{o}}=\int_{0}^{\infty}V^{4}2P(V)dV$ (B.33) $\displaystyle=\biggl{[}\frac{2}{(2\pi)^{1/2}\sigma}\biggr{]}\int_{0}^{\infty}V% ^{4}\exp\biggl{(}\frac{-V^{2}}{2\sigma^{2}}\biggr{)}dV.$ (B.34)

Integrating Equation B.27 by parts yields the definite integral

 $\int_{0}^{\infty}x^{4}\exp(-a^{2}x^{2})dx=\frac{3\pi^{1/2}}{8a^{5}},$ (B.35)

and substituting $a^{2}=(2\sigma^{2})^{-1}$ gives

 $\langle V_{\mathrm{o}}^{2}\rangle=\biggl{[}\frac{2}{(2\pi)^{1/2}\sigma}\biggr{% ]}\biggl{(}\frac{3\pi^{1/2}}{8}\biggr{)}(2\sigma^{2})^{5/2}=3\sigma^{4}.$ (B.36)

Thus

 $\sigma_{\mathrm{o}}^{2}=\langle V_{\mathrm{o}}^{2}\rangle-\langle V_{\mathrm{o% }}\rangle^{2}=3\sigma^{4}-(\sigma^{2})^{2}=2\sigma^{4}.$ (B.37)

The rms $\sigma_{\mathrm{o}}=2^{1/2}\sigma^{2}$ of the detector output voltage is $2^{1/2}$ times the mean output voltage $\sigma^{2}$. The rms uncertainty in each independent sample of the measured noise power is $2^{1/2}$ times the mean noise power. If $N\gg 1$ independent samples are averaged, the fractional rms uncertainty of the averaged power is $(2/N)^{1/2}$. This result is the heart of the ideal radiometer equation (Equation 3.154). According to the central limit theorem, the distribution of these averages approaches a Gaussian as $N$ becomes large.

# B.7 Evaluation of the Free–Free Pulse Energy Integral

The integral in Equation 4.22 is

 $\displaystyle\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi$ $\displaystyle=\int_{0}^{\pi/2}\cos^{2}\psi(1-\sin^{2}\psi)\,d\psi$ $\displaystyle=\int_{0}^{\pi/2}\cos^{2}\psi\,d\psi-\int_{0}^{\pi/2}\cos^{2}\psi% \sin^{2}\psi\,d\psi.$ (B.38)

We have already found that $\langle\cos^{2}\psi\rangle=1/2$ so $\int_{0}^{\pi/2}\cos^{2}\psi\,d\psi=\pi/4$. Integrate the remaining integral by parts using

 $u\equiv\cos^{2}\psi\sin\psi\quad\mathrm{~{}and~{}}\quad v\,dv\equiv\sin\psi\,d\psi.$

Therefore

 $du=\cos^{3}\psi-2\sin^{2}\psi\cos\psi\quad\mathrm{~{}and~{}}\quad v=-\cos\psi,$

and

 $\displaystyle\int_{0}^{\pi/2}\cos^{2}\psi\sin^{2}\psi\,d\psi$ $\displaystyle=-\cos^{2}\psi\sin\psi\cos\psi|_{0}^{\pi/2}$ $\displaystyle\quad-\int_{0}^{\pi/2}-\cos\psi(\cos^{3}\psi-2\sin^{2}\psi\cos% \psi)d\psi$ $\displaystyle=\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi-2\int_{0}^{\pi/2}\cos^{2}% \psi\sin^{2}\psi\,d\psi,$ which has the same integral on both sides, so $\displaystyle\int_{0}^{\pi/2}\cos^{2}\psi\sin^{2}\psi\,d\psi$ $\displaystyle=\frac{1}{3}\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi.$

Using Equation B.38 we get

 $\displaystyle\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi$ $\displaystyle=\frac{\pi}{4}-\frac{1}{3}\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi,$ $\displaystyle\frac{4}{3}\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi$ $\displaystyle=\frac{\pi}{4},$

so

 $\displaystyle\boxed{\int_{0}^{\pi/2}\cos^{4}\psi\,d\psi=\frac{3\pi}{16}.}$ (B.39)

# B.8 The Nonrelativistic Maxwellian Speed Distribution

Let $v\equiv|\vec{v}|$ be the speed of a particle (e.g., an electron) of mass $m$ in a gas in LTE at temperature $T$. From thermodynamics, recall that the average kinetic energy is $kT/2$ per degree of freedom (e.g., per spatial coordinate for a single particle), so

 $\frac{m\langle v_{x}^{2}\rangle}{2}=\frac{m\langle v_{y}^{2}\rangle}{2}=\frac{% m\langle v_{z}^{2}\rangle}{2}=\frac{kT}{2},$ (B.40)
 $\langle v^{2}\rangle=\langle v_{x}^{2}\rangle+\langle v_{y}^{2}\rangle+\langle v% _{z}^{2}\rangle=\frac{3kT}{m}.$ (B.41)

Collisions eventually bring the gas into LTE, leading to identical Gaussian distributions (Appendix B.5) for $v_{x}$, $v_{y}$, and $v_{z}$. Writing out only the $x$-coordinate distribution $P(v_{x})$ yields

 $P(v_{x})=\frac{1}{\sqrt{2\pi}\sigma_{x}}\exp\biggl{(}-\frac{v_{x}^{2}}{2\sigma% _{x}^{2}}\biggr{)},$ (B.42)

where $\sigma_{x}$ is the rms (root mean square) value of $v_{x}$. The definition of this rms is

 $\displaystyle\sigma_{x}^{2}\equiv\langle v_{x}^{2}\rangle$ $\displaystyle=\int_{-\infty}^{\infty}v_{x}^{2}P(v_{x})dv_{x}=\int_{-\infty}^{% \infty}\frac{v_{x}^{2}}{\sqrt{2\pi}\sigma_{x}}\exp\biggl{(}-\frac{v_{x}^{2}}{2% \sigma_{x}^{2}}\biggr{)}dv_{x}$ (B.43) $\displaystyle=\frac{1}{\sqrt{2\pi}\sigma_{x}}\frac{1}{2}\sqrt{\pi}\biggl{(}% \frac{1}{2\sigma_{x}^{2}}\biggr{)}^{-3/2}=\frac{kT}{m},$ (B.44)

so

 $P(v_{x})=\frac{1}{\sqrt{2\pi}}\biggl{(}\frac{m}{kT}\biggr{)}^{1/2}\exp\biggl{(% }-\frac{mv_{x}^{2}}{2kT}\biggr{)}.$ (B.45)

In three dimensions, by isotropy,

 $P(v_{x},v_{y},v_{z})dv_{x}\,dv_{y}\,dv_{z}=P(v_{x})P(v_{y})P(v_{z})dv_{x}\,dv_% {y}\,dv_{z},$ (B.46)
 $P(v_{x},v_{y},v_{z})=\biggl{(}\frac{m}{2\pi kT}\biggr{)}^{3/2}\exp\biggl{(}-% \frac{mv^{2}}{2kT}\biggr{)}.$ (B.47)

All velocities in the spherical shell of radius $v=(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})^{1/2}$ correspond to the speed $v$, so

 $f(v)=4\pi v^{2}P(v_{x},v_{y},v_{z}),$ (B.48)
 $\boxed{f(v)=\frac{4v^{2}}{\sqrt{\pi}}\biggl{(}\frac{m}{2kT}\biggr{)}^{3/2}\exp% \biggl{(}-\frac{mv^{2}}{2kT}\biggr{)}.}$ (B.49)

This is the nonrelativistic Maxwellian distribution $f$ of speeds $v\ll c$ for particles of mass $m$ at temperature $T$. If we normalize the speeds by the rms speed $(3kT/m)^{1/2}$, the Maxwellian speed distribution looks like Figure 4.6.