附录 B 数学推导

B.1 玻尔兹曼和的计算

黑体辐射每个模的平均能量的玻尔兹曼和（式 2.83）为

\[{{\langle E\rangle} = \frac{\sum\limits_{n = 0}^{\infty}{nh\nu{\exp\left( {- \frac{nh\nu}{kT}} \right)}}}{\sum\limits_{n = 0}^{\infty}{\exp\left( {- \frac{nh\nu}{kT}} \right)}}}. \]

引入变量 \(\alpha \equiv {1/{({kT})}}\) 是很方便的，因此

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

接下来考虑量

\[{- {\frac{d}{d\alpha}\left\lbrack {\ln{\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}}} \right\rbrack}}. \]

使用链式法则取导数得到

\[- {\frac{d}{d\alpha}\left\lbrack {\ln{\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}}} \right\rbrack} \]

\[= {- {\left\lbrack {\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}} \right\rbrack^{- 1}\frac{d}{d\alpha}\left\lbrack {\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}} \right\rbrack}} \]

\[{= \frac{\sum\limits_{n = 0}^{\infty}{nh\nu{\exp{({- {\alpha nh\nu}})}}}}{\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}}}. \]

因此

\[{{\langle E\rangle} = {- {\frac{d}{d\alpha}\left\lbrack {\ln{\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}}} \right\rbrack}}}. \]

然后，

\[{\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}} = {1 + {\lbrack{\exp{({- {\alpha h\nu}})}}\rbrack}^{1} + {\lbrack{\exp{({- {\alpha h\nu}})}}\rbrack}^{2} + \cdots} \]

具有 \({1 + x + x^{2} + \cdots} = {({1 - x})}^{- 1}\) 的形式，因此

\[{\sum\limits_{n = 0}^{\infty}{\exp{({- {\alpha nh\nu}})}}} = {\lbrack{1 - {\exp{({- {\alpha h\nu}})}}}\rbrack}^{- 1} \]

以及

\[\langle E\rangle \]

\[= {- \frac{d\ln{\lbrack 1 - \exp{(\alpha h\nu)}\rbrack}^{- 1}}{d\alpha}} \]

\[= {- {{\lbrack{1 - {\exp{({- {\alpha h\nu}})}}}\rbrack}{({- 1})}{\lbrack{1 - {\exp{({- {\alpha h\nu}})}}}\rbrack}^{- 2}h\nu{\exp{({- {\alpha h\nu}})}}}} \]

\[{= \frac{h\nu{\exp{({- {\alpha h\nu}})}}}{1 - {\exp{({- {\alpha h\nu}})}}} = \frac{h\nu}{{\exp{({\alpha h\nu})}} - 1} = \frac{h\nu}{{\exp\left( \frac{h\nu}{kT} \right)} - 1}}. \]

B.2 斯特藩–玻尔兹曼定律的推导

温度为 \(T\) 的黑体辐射总亮度的斯特藩–玻尔兹曼定律（式 2.89）为

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

其中

\[{B_{\nu}{(T)}} = {\frac{2h\nu^{3}}{c^{2}}\frac{1}{{\exp\left( \frac{h\nu}{kT} \right)} - 1}} \]

是普朗克定律，而\(\sigma\) 是斯特藩–玻尔兹曼常数。虽然斯特藩–玻尔兹曼定律和常数最初是通过实验确定的，但两者都可以从普朗克定律中数学推导出来。为简单起见，定义

\[{x \equiv \frac{h\nu}{kT}}, \]

因此

\[{{B{(T)}} = {\int_{0}^{\infty}{\frac{2h}{c^{2}}\left( \frac{kTx}{h} \right)^{3}\left( \frac{1}{e^{x} - 1} \right)\left( \frac{kT}{h} \right){dx}}} = {\frac{2k^{4}T^{4}}{c^{2}h^{3}}{\int_{0}^{\infty}\frac{x^{3}dx}{e^{x} - 1}}}}. \]

该量

\[\frac{1}{e^{x} - 1} = \frac{e^{- x}}{1 - e^{- x}} = {e^{- x}\left( \frac{1}{1 - e^{- x}} \right)} \]

可以用无限级数展开为

\[{\sum\limits_{m = 0}^{\infty}z^{m}} = \]

\[1 + z + z^{2} + z^{3} + \cdots \]

\[= \]

\[1 + {z{({1 + z + z^{2} + z^{3} + \cdots})}} \]

\[= \]

\[{1 + {z{\sum\limits_{m = 0}^{\infty}z^{m}}}}, \]

\[{\sum\limits_{m = 0}^{\infty}z^{m}} = \]

\[\frac{1}{1 - z}. \]

因此

\[\frac{1}{e^{x} - 1} = {e^{- x}{\sum\limits_{m = 0}^{\infty}e^{- {mx}}}} = {e^{- x} + e^{- {2x}} + e^{- {3x}} + \cdots} \]

积分变为

\[{{\int_{0}^{\infty}\frac{x^{3}dx}{e^{x} - 1}} = {\int_{0}^{\infty}{x^{3}\left( {\sum\limits_{m = 1}^{\infty}e^{- {mx}}} \right){dx}}}}. \]

该级数中的每个积分可以分部积分三次：

\[\int_{0}^{\infty}{x^{3}e^{- {mx}}{dx}} \]

\[{= {\left. \frac{x^{3}e^{- {mx}}}{- m} \right|_{0}^{\infty} - {\int_{0}^{\infty}{\frac{3x^{2}e^{- {mx}}}{- m}{dx}}}} = {\frac{3}{m}{\int_{0}^{\infty}{x^{2}e^{- {mx}}{dx}}}}}, \]

\[\int_{0}^{\infty}{x^{2}e^{- {mx}}{dx}} \]

\[{= {\left. \frac{x^{2}e^{- {mx}}}{- m} \right|_{0}^{\infty} - {\int_{0}^{\infty}{\frac{2xe^{- {mx}}}{- m}{dx}}}} = {\frac{2}{m}{\int_{0}^{\infty}{xe^{- {mx}}{dx}}}}}, \]

\[\int_{0}^{\infty}{xe^{- {mx}}{dx}} \]

\[{= {\left. \frac{xe^{- {mx}}}{- m} \right|_{0}^{\infty} - {\int_{0}^{\infty}{\frac{xe^{- {mx}}}{- m}{dx}}}} = {\frac{1}{m}{\int_{0}^{\infty}{e^{- {mx}}{dx}}}} = \frac{1}{m^{2}}}, \]

得到

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

以及

\[{{\int_{0}^{\infty}\frac{x^{3}dx}{e^{x} - 1}} = {\int_{0}^{\infty}{x^{3}\left( {\sum\limits_{m = 1}^{\infty}e^{- {mx}}} \right){dx}}} = {6{\sum\limits_{m = 1}^{\infty}\frac{1}{m^{4}}}}}. \]

求和

\[{\sum\limits_{m = 1}^{\infty}\frac{1}{m^{4}}} = {\frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \frac{1}{4^{4}} + \cdots} = {1 + \frac{1}{16} + \frac{1}{81} + \frac{1}{256} + \cdots} \approx 1.082 \]

快速收敛，是黎曼 ζ 函数\({\zeta{(4)}} = {\pi^{4}/90} \approx 1.082\) 的值。因此

\[\boxed{\int_{0}^{\infty}\frac{x^{3}dx}{e^{x} - 1} = \frac{\pi^{4}}{15}.} \]

(B.1)

最后，黑体辐射的总亮度为

\[{{B{(T)}} = {\frac{2k^{4}T^{4}}{c^{2}h^{3}}{\int_{0}^{\infty}\frac{x^{3}dx}{e^{x} - 1}}} = {\frac{2k^{4}T^{4}}{c^{2}h^{3}}\left( \frac{\pi^{4}}{15} \right)} = {\frac{2\pi^{4}k^{4}}{15c^{2}h^{3}}T^{4}} = \frac{\sigma T^{4}}{\pi}}, \]

因此

\[\sigma = \frac{2\pi^{5}k^{4}}{15c^{2}h^{3}} \approx {{5.67 \times 10^{- 5}}\frac{erg}{{cm}^{2}sK^{4}{({sr})}}} \]

是斯特藩–玻尔兹曼常数的值。

类似地，该积分

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

需要评估黑体光子的数密度 \(n_{\gamma}\)：

\[{n_{\gamma} = {\frac{8\pi}{c^{3}}{\int_{0}^{\infty}\frac{\nu^{2}d\nu}{{\exp\left( \frac{h\nu}{kT} \right)} - 1}}} = {\frac{8\pi}{c^{3}}\left( \frac{kT}{h} \right)^{3}{\int_{0}^{\infty}\frac{x^{2}dx}{e^{x} - 1}}}}. \]

根据上述推导，

\[{\int_{0}^{\infty}\frac{x^{2}dx}{e^{x} - 1}} = {\int_{0}^{\infty}{x^{2}\left( {\sum\limits_{m = 1}^{\infty}e^{- {mx}}} \right){dx}}} \]

以及

\[{{\int_{0}^{\infty}{x^{2}e^{- {mx}}{dx}}} = {\frac{2}{m}{\int_{0}^{\infty}{xe^{- {mx}}{dx}}}} = {\frac{2}{m}\left( \frac{1}{m^{2}} \right)} = \frac{2}{m^{3}}}, \]

因此

\[\boxed{\int_{0}^{\infty}\frac{x^{2}dx}{e^{x} - 1} = 2\sum\limits_{m = 1}^{\infty}\frac{1}{m^{3}} = 2\left( \frac{1}{1^{3}} + \frac{1}{2^{3}} + \frac{1}{3^{3}} + \cdots \right) \approx 2.404.} \]

(B.2)

B.3 复指数

一个复指数 \(e^{i\phi}\)，其中 \(i^{2} = {- 1}\) 和 \(\phi\) 是任意 无量纲 的实变量，是一个复数，其实部和虚部是由欧拉公式给出的正弦和余弦

\[\boxed{e^{i\phi} = \cos\phi + i\sin\phi.} \]

(B.3)

欧拉公式可以从泰勒级数推导出来

\[\cos\phi \]

\[{= {{{{1 - \frac{\phi^{2}}{2!}} + \frac{\phi^{4}}{4!}} - \frac{\phi^{6}}{6!}} + \cdots}}, \]

\[\sin\phi \]

\[{= {{{{\phi - \frac{\phi^{3}}{3!}} + \frac{\phi^{5}}{5!}} - \frac{\phi^{7}}{7!}} + \cdots}}, \]

\[e^{\phi} \]

\[{= {1 + \phi + \frac{\phi^{2}}{2!} + \frac{\phi^{3}}{3!} + \frac{\phi^{4}}{4!} + \cdots}}. \]

因此

\[e^{i\phi} \]

\[= {{{{{1 + {i\phi}} - \frac{\phi^{2}}{2!} - \frac{i\phi^{3}}{3!}} + \frac{\phi^{4}}{4!} + \frac{i\phi^{5}}{5!}} - \frac{i\phi^{6}}{6!} - \frac{i\phi^{7}}{7!}} + \cdots} \]

\[= {\left( {{{{1 - \frac{\phi^{2}}{2!}} + \frac{\phi^{4}}{4!}} - \frac{\phi^{6}}{6!}} + \cdots} \right) + {i\left( {{{{\phi - \frac{\phi^{3}}{3!}} + \frac{\phi^{5}}{5!}} - \frac{\phi^{7}}{7!}} + \cdots} \right)}} \]

\[{= {{\cos\phi} + {i{\sin\phi}}}}. \]

复指数（或正弦和余弦）在物理中被广泛用于表示周期函数，原因如下：

1. 它们构成了一组完整且正交的周期函数。这组函数可以用来近似任何分段连续函数，并且它们是傅里叶变换的基础（附录 A.1）。

2. 它们是微分算子的本征函数——也就是说，复指数函数的导数仍然是复指数函数：

\[{{\frac{de^{i\phi}}{d\phi} = {ie^{i\phi}}},{{\frac{d^{2}e^{i\phi}}{d\phi^{2}} = {- e^{i\phi}}},{{\frac{d^{3}e^{i\phi}}{d\phi^{3}} = {- {ie^{i\phi}}}},{\frac{d^{4}e^{i\phi}}{d\phi^{4}} = {e^{i\phi},\ldots}}}}}. \]

大多数物理系统遵守线性微分方程，例如由电阻和电容组成的低通滤波器。正弦输入信号会产生相同频率的正弦输出信号（但不一定具有相同的幅度和相位），而方波输入则不会产生方波输出。对方波输入的响应可以通过将输入方波视为正弦波的总和来计算，而滤波器输出就是这些滤波后的正弦波的总和。这也是为什么周期波或振荡几乎总是被视为复指数（或正弦和余弦）的组合的原因。

实际周期信号可以表示为复指数的实部：

\[\cos\phi \]

\[{= {{Re}{(e^{i\phi})}}}, \]

\[\sin\phi \]

\[{= {{Im}{(e^{i\phi})}}}. \]

对方程进行加法和减法

\[e^{i\phi} \]

\[{= {{\cos\phi} + {i{\sin\phi}}}}, \]

\[e^{- {i\phi}} \]

\[= {{\cos\phi} - {i{\sin\phi}}} \]

得出恒等式

\[\boxed{{\cos\phi} = \frac{e^{i\phi} + e^{- {i\phi}}}{2}} \]

(B.4)

以及

\[\boxed{\sin\phi = \frac{e^{i\phi} - e^{- {i\phi}}}{2i}.} \]

(B.5)

复指数相对于等效的正弦和余弦的优点在于它们在数学上更容易处理。例如，你可以使用复指数来计算平方律探测器（第 3.6.2 节）的输出频谱，而不必记住三角恒等式。平方律探测器是一种非线性器件，其输出电压是其输入电压的平方。如果输入电压是 \(\cos{({\omega t})}\)，则输出电压为

\[\cos^{2}{({\omega t})} \]

\[= \left( \frac{e^{i\omega t} + e^{- {i\omega t}}}{2} \right)^{2} \]

\[= \frac{e^{2i\omega t} + 2 + e^{- {2i\omega t}}}{4} \]

\[= \frac{{2{\cos{({2\omega t})}}} + 2}{4} \]

\[{= {\frac{1}{2}{\lbrack{{\cos{({2\omega t})}} + 1}\rbrack}}}. \]

输出频谱具有两个频率分量：一个是输入频率 \(\omega\) 的两倍，另一个是零频率（直流）。

B.4 高斯函数的傅里叶变换

归一化的高斯函数通常写作

\[{{f{(x)}} = {\frac{1}{\sqrt{2\pi}\sigma}{\exp\left( {- \frac{x^{2}}{2\sigma^{2}}} \right)}}}, \]

(B.6)

其中 \(\sigma\) 是其 rms 宽度。要计算其傅里叶变换

\[{{F{(s)}} \equiv {\int_{- \infty}^{\infty}{f{(x)}{\exp{({- {i2\pi sx}})}}{dx}}}}, \]

(B.7)

使用表格 \({f{(x)}} = {\exp{({- {\pi x^{2}}})}}\) 更容易，对于 \(\sigma^{2} = {1/{({2\pi})}}\)。然后

\[F{(s)} \]

\[= {\int_{- \infty}^{\infty}{{\exp{({- {\pi x^{2}}})}}{\exp{({- {i2\pi sx}})}}{dx}}} \]

\[= {\int_{- \infty}^{\infty}{{\exp{\lbrack{- {\pi{({{x^{2} + {i2sx} + s^{2}} - s^{2}})}}}\rbrack}}{dx}}} \]

\[= {{\exp{({- {\pi s^{2}}})}}{\int_{{- \infty} + {is}}^{\infty + {is}}{{\exp{\lbrack{- {\pi{({x + {is}})}^{2}}}\rbrack}}d{({x + {is}})}}}} \]

\[{= {{\exp{({- {\pi s^{2}}})}}{\int_{- \infty}^{\infty}{{\exp{({- {\pi x^{2}}})}}{dx}}}}}. \]

(B.8) (B.9) (B.10) (B.11)

为了计算这个一维积分，将其分解为两个积分的乘积，并将一个虚拟变量从 \(x\) 改为 \(y\)，以提示平面上的笛卡尔坐标：

\[\int_{- \infty}^{\infty}{{\exp{({- {\pi x^{2}}})}}{dx}} \]

\[= \left\lbrack \int_{- \infty}^{\infty}\exp{( - \pi x^{2})}dx\int_{- \infty}^{\infty}\exp{( - \pi y^{2})}dy \right\rbrack^{1/2} \]

\[{= \left\lbrack {\int_{- \infty}^{\infty}{\int_{- \infty}^{\infty}{{\exp{\lbrack{- {\pi{({x^{2} + y^{2}})}}}\rbrack}}{dx}{dy}}}} \right\rbrack^{1/2}}. \]

(B.12) (B.13)

接下来将 \(r,\theta\) 转换为极坐标，所以 \(r^{2} = {x^{2} + y^{2}}\) 和 \({dxdy} = {rdrd\theta}\)：

\[{{\int_{- \infty}^{\infty}{{\exp{({- {\pi x^{2}}})}}{dx}}} = \left\lbrack {\int_{r = 0}^{\infty}{\int_{\theta = 0}^{2\pi}{{\exp{({- {\pi r^{2}}})}}r{dr}{d\theta}}}} \right\rbrack^{1/2}}. \]

(B.14)

最后，代入 \(u \equiv {\pi r^{2}}\) 和 \({du} = {2\pi rdr}\) 得到

\[{{\int_{- \infty}^{\infty}{{\exp{({- {\pi x^{2}}})}}{dx}}} = \left\lbrack {2\pi{\int_{u = 0}^{\infty}{{\exp{({- u})}}\frac{du}{2\pi}}}} \right\rbrack^{1/2} = \left\lbrack {- \left. e^{- u} \right|_{0}^{\infty}} \right\rbrack^{1/2} = 1}. \]

(B.15)

因此

\[{{F{(s)}} = {\exp{({- {\pi s^{2}}})}}}. \]

(B.16)

高斯函数的傅里叶变换仍然是高斯函数。

B.5 高斯概率分布与噪声电压

随机噪声的电压 \(V\) 具有高斯概率分布

\[{{P{(V)}} = {\frac{1}{{({2\pi})}^{1/2}\sigma}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}}}, \]

(B.17)

其中 \(P{(V)}dV\) 是电压位于无穷小范围 \(V\) 到 \(V + {dV}\) 内的微分概率，\(\sigma\) 是均方根 (rms) 电压。测量到某个电压的概率必须为一，因此

\[{{\int_{- \infty}^{\infty}{P{(V)}{dV}}} = 1}. \]

(B.18)

可以通过计算积分来确认式 B.17 中 \(P{(V)}\) 的归一化

\[\int_{- \infty}^{\infty}{\frac{1}{{({2\pi})}^{1/2}\sigma}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}{dV}} \]

\[= {2{\int_{0}^{\infty}{\frac{1}{{({2\pi})}^{1/2}\sigma}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}{dV}}}} \]

\[{= {\left\lbrack \frac{2}{{({2\pi})}^{1/2}\sigma} \right\rbrack{\int_{0}^{\infty}{{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}{dV}}}}}. \]

(B.19) (B.20)

式 B.15 立即得到定积分

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

(B.21)

替换 \(a^{2} = {({2\sigma^{2}})}^{- 1}\) 得到所需的结果：

\[{{\int_{- \infty}^{\infty}{P{(V)}{dV}}} = {\left\lbrack \frac{2}{{({2\pi})}^{1/2}\sigma} \right\rbrack\left( \frac{\pi^{1/2}}{2} \right){({2\sigma^{2}})}^{1/2}} = 1}. \]

(B.22)

归一化分布的均方根（rms，root mean square）\(\Sigma\) 定义为

\[{\Sigma^{2} \equiv {{\langle V^{2}\rangle} - {\langle V\rangle}^{2}}}. \]

(B.23)

对于对称高斯分布，\({\langle V\rangle} = 0\)，因此

\[\Sigma^{2} \]

\[= {\langle V^{2}\rangle} = {\int_{- \infty}^{\infty}{V^{2}P{(V)}{dV}}} \]

\[= {2{\int_{0}^{\infty}{V^{2}\frac{1}{{({2\pi})}^{1/2}\sigma}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}{dV}}}} \]

\[{= {\left\lbrack \frac{2}{{({2\pi})}^{1/2}\sigma} \right\rbrack{\int_{0}^{\infty}{V^{2}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}{dV}}}}}. \]

(B.24) (B.25) (B.26)

定积分

\[{\int_{0}^{\infty}{x^{2}{\exp{({- {a^{2}x^{2}}})}}{dx}}} = \frac{\pi^{1/2}}{4a^{3}} \]

(B.27)

可以通过分部积分公式对式 B.21 积分而得到。代入 \(a^{2} = {({2\sigma^{2}})}^{- 1}\) 得到

\[{\Sigma^{2} = {\langle V^{2}\rangle} = {\left\lbrack \frac{2}{{({2\pi})}^{1/2}\sigma} \right\rbrack\left( \frac{\pi^{1/2}}{4} \right){({2\sigma^{2}})}^{3/2}} = \sigma^{2}}, \]

(B.28)

确认 \(\sigma\) 在式 B.17 中是高斯分布的均方根。

B.6 噪声功率的概率分布

平方律检测器将输入电压 \(V\) 自乘，以产生一个与输入功率成正比的输出电压 \(V_{o} = V^{2}\)。输入电压的分布是均方根为 \(\sigma\) 的高斯分布（式 B.17），

\[{{P{(V)}} = {\frac{1}{{({2\pi})}^{1/2}\sigma}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}}}. \]

(B.29)

正值和负值的 \(V\) 和 \({P{(V)}} = {- {P{(V)}}}\) 都会产生相同的 \(V_{o} = V^{2}\)，因此

\[{P_{o}{(V_{o})}dV_{o}} = {2P{(V)}dV} \]

(B.30)

对所有 \(V_{o} \geq 0\) 均适用。由于 \({dV_{o}} = {2VdV}\)，

\[{P_{o}{(V_{o})}} = {\left\lbrack \frac{V_{o}^{- {1/2}}}{{({2\pi})}^{1/2}\sigma} \right\rbrack{\exp\left( \frac{- V_{o}}{2\sigma^{2}} \right)}} \]

(B.31)

适用于 \(V_{o} \geq 0\)。检测器输出电压的分布在 \(V_{o} = 0\) 附近非常集中，并有一个长的指数衰减尾部（图 3.33）。

检测器输出电压的均值由式 B.28 给出：\({\langle V_{o}\rangle} = {\langle V^{2}\rangle} = \sigma^{2}\)。

探测器输出电压的有效值 \(\sigma_{o}\) 为

\[{\sigma_{o}^{2} = {{\langle V_{o}^{2}\rangle} - {\langle V_{o}\rangle}^{2}}}, \]

(B.32)

其中

\[\langle V_{o}^{2}\rangle \]

\[= {\int_{0}^{\infty}{V_{o}^{2}P_{o}{(V_{o})}{dV_{o}}}} = {\int_{0}^{\infty}{V^{4}2P{(V)}{dV}}} \]

\[{= {\left\lbrack \frac{2}{{({2\pi})}^{1/2}\sigma} \right\rbrack{\int_{0}^{\infty}{V^{4}{\exp\left( \frac{- V^{2}}{2\sigma^{2}} \right)}{dV}}}}}. \]

(B.33) (B.34)

对式 B.27 分部积分得定积分

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

(B.35)

并代入 \(a^{2} = {({2\sigma^{2}})}^{- 1}\) 得

\[{{\langle V_{o}^{2}\rangle} = {\left\lbrack \frac{2}{{({2\pi})}^{1/2}\sigma} \right\rbrack\left( \frac{3\pi^{1/2}}{8} \right){({2\sigma^{2}})}^{5/2}} = {3\sigma^{4}}}. \]

(B.36)

因此

\[{\sigma_{o}^{2} = {{\langle V_{o}^{2}\rangle} - {\langle V_{o}\rangle}^{2}} = {{3\sigma^{4}} - {(\sigma^{2})}^{2}} = {2\sigma^{4}}}. \]

(B.37)

探测器输出电压的均方根值 \(\sigma_{o} = {2^{1/2}\sigma^{2}}\) 是平均输出电压 \(\sigma^{2}\) 的 \(2^{1/2}\) 倍。每个独立样本的测量噪声功率的均方根不确定性是平均噪声功率的 \(2^{1/2}\) 倍。如果对 \(N \gg 1\) 个独立样本求平均，平均功率的分数均方根不确定性为 \({({2/N})}^{1/2}\)。这个结果是理想辐射计方程（公式 3.154）的核心。根据中心极限定理，当 \(N\) 变大时，这些平均值的分布趋于高斯分布。

B.7 自由–自由脉冲能量积分的评估

公式 4.22 中的积分为

\[\int_{0}^{\pi/2}{\cos^{4}{\psi d\psi}} \]

\[= {\int_{0}^{\pi/2}{{\cos^{2}\psi}{({1 - {\sin^{2}\psi}})}{d\psi}}} \]

\[{= {{\int_{0}^{\pi/2}{\cos^{2}{\psi d\psi}}} - {\int_{0}^{\pi/2}{{\cos^{2}\psi}{\sin^{2}{\psi d\psi}}}}}}. \]

(B.38)

我们已经发现 \({\langle{\cos^{2}\psi}\rangle} = {1/2}\)，因此 \({\int_{0}^{\pi/2}{\cos^{2}{\psi d\psi}}} = {\pi/4}\)。对剩余的积分使用部分积分法进行积分

\[{{u \equiv {{{\cos^{2}\psi}{\sin\psi}}\operatorname{\quad\:}{and}}}\operatorname{\quad\:}{{vdv} \equiv {\sin{\psi d\psi}}}}. \]

因此

\[{{{du} = {{{\cos^{3}\psi} - {2{\sin^{2}\psi}{\cos\psi}}}\operatorname{\quad\:}{and}}}\operatorname{\quad\:}{v = {- {\cos\psi}}}}, \]

以及

\[\int_{0}^{\pi/2}{{\cos^{2}\psi}{\sin^{2}{\psi d\psi}}} \]

\[= {- {{{\cos^{2}\psi}{\sin\psi}{\cos\psi}}|}_{0}^{\pi/2}} \]

\[\quad - \int_{0}^{\pi/2} - {{\cos\psi}{({{\cos^{3}\psi} - {2{\sin^{2}\psi}{\cos\psi}}})}d\psi} \]

\[{= {{\int_{0}^{\pi/2}{\cos^{4}{\psi d\psi}}} - {2{\int_{0}^{\pi/2}{{\cos^{2}\psi}{\sin^{2}{\psi d\psi}}}}}}}, \]

\[\int_{0}^{\pi/2}{{\cos^{2}\psi}{\sin^{2}{\psi d\psi}}} \]

\[{= {\frac{1}{3}{\int_{0}^{\pi/2}{\cos^{4}{\psi d\psi}}}}}. \]

使用式 B.38 我们得到

\[\int_{0}^{\pi/2}{\cos^{4}{\psi d\psi}} \]

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

\[\frac{4}{3}{\int_{0}^{\pi/2}{\cos^{4}{\psi d\psi}}} \]

\[{= \frac{\pi}{4}}, \]

因此

\[{{\int_{0}^{\pi/2}{\cos^{4}{\psi d\psi}}} = \frac{3\pi}{16}}. \]

(B.39)

B.8 非相对论的麦克斯韦速度分布

设 \(v \equiv {|\overset{\rightarrow}{v}|}\) 为处于温度 \(T\) 的局部热平衡 (LTE) 气体中质量为 \(m\) 的粒子（例如电子）的速度。根据热力学，回忆一下平均动能为 \({kT}/2\) 每个自由度（例如单个粒子的每个空间坐标），因此

\[{\frac{m{\langle v_{x}^{2}\rangle}}{2} = \frac{m{\langle v_{y}^{2}\rangle}}{2} = \frac{m{\langle v_{z}^{2}\rangle}}{2} = \frac{kT}{2}}, \]

(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)

碰撞最终使气体达到局部热平衡，从而使 \(v_{x}\)、\(v_{y}\) 和 \(v_{z}\) 表现出相同的高斯分布（附录 B.5）。仅写出 \(x\) 坐标的分布 \(P{(v_{x})}\) 为

\[{{P{(v_{x})}} = {\frac{1}{\sqrt{2\pi}\sigma_{x}}{\exp\left( {- \frac{v_{x}^{2}}{2\sigma_{x}^{2}}} \right)}}}, \]

(B.42)

其中 \(\sigma_{x}\) 是 \(v_{x}\) 的均方根（rms）值。该均方根的定义为

\[\sigma_{x}^{2} \equiv {\langle v_{x}^{2}\rangle} \]

\[= {\int_{- \infty}^{\infty}{v_{x}^{2}P{(v_{x})}{dv_{x}}}} = {\int_{- \infty}^{\infty}{\frac{v_{x}^{2}}{\sqrt{2\pi}\sigma_{x}}{\exp\left( {- \frac{v_{x}^{2}}{2\sigma_{x}^{2}}} \right)}{dv_{x}}}} \]

\[{= {\frac{1}{\sqrt{2\pi}\sigma_{x}}\frac{1}{2}\sqrt{\pi}\left( \frac{1}{2\sigma_{x}^{2}} \right)^{- {3/2}}} = \frac{kT}{m}}, \]

(B.43) (B.44)

因此

\[{{P{(v_{x})}} = {\frac{1}{\sqrt{2\pi}}\left( \frac{m}{kT} \right)^{1/2}{\exp\left( {- \frac{mv_{x}^{2}}{2kT}} \right)}}}. \]

(B.45)

在三维情况下，由于各向同性，

\[{{P{(v_{x},v_{y},v_{z})}dv_{x}dv_{y}dv_{z}} = {P{(v_{x})}P{(v_{y})}P{(v_{z})}dv_{x}dv_{y}dv_{z}}}, \]

(B.46)

\[{{P{(v_{x},v_{y},v_{z})}} = {\left( \frac{m}{2\pi kT} \right)^{3/2}{\exp\left( {- \frac{mv^{2}}{2kT}} \right)}}}. \]

(B.47)

在半径为 \(v = {({v_{x}^{2} + v_{y}^{2} + v_{z}^{2}})}^{1/2}\) 的球形壳层中，所有速度对应于 速度 \(v\)，因此

\[{{f{(v)}} = {4\pi v^{2}P{(v_{x},v_{y},v_{z})}}}, \]

(B.48)

\[\boxed{f{(v)} = \frac{4v^{2}}{\sqrt{\pi}}\left( \frac{m}{2kT} \right)^{3/2}\exp\left( - \frac{mv^{2}}{2kT} \right).} \]

(B.49)

这是质量为 \(m\)、温度为 \(T\) 的粒子的非相对论麦克斯韦速度分布 \(f\)。在用均方根速度 \({({{3kT}/m})}^{1/2}\) 对速度进行归一化后，麦克斯韦速度分布如图 4.6 所示。