Answer_for_HW4

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1. 带电半球受力

带电球电场:

\vec{E} (r) = {\begin{array}{r} \frac{Q r}{4 π ϵ_{0} R^{3}} {\vec{e}}_{r}, r \leq R \\ \frac{Q}{4 π ϵ_{0} r^{2}} {\vec{e}}_{r}, r > R \end{array}

得到:

T_{i j} = ϵ_{0} E_{i} E_{j} - \frac{1}{2} ϵ_{0} E^{2} δ_{i j} \to T_{x z} = T_{y z} = 0, T_{z z} = - \frac{1}{2} ϵ_{0} E^{2}, a t z = 0

$r \to \infty$ 时 $\int T \cdot d \vec{S} \to 0$ 忽略, 则:

\begin{aligned} \vec{F} = {\vec{e}}_{z} \int T_{z z} d S_{z} & = {\vec{e}}_{z} \int_{0}^{\infty} (- \frac{1}{2} ϵ_{0} E^{2}) (- 2 π r d r) \\ = {\vec{e}}_{z} \int_{0}^{R} ϵ_{0} {(\frac{Q r}{4 π ϵ_{0} R^{3}})}^{2} (π r d r) + {\vec{e}}_{z} \int_{R}^{\infty} ϵ_{0} {(\frac{Q}{4 π ϵ_{0} r^{2}})}^{2} (π r d r) \\ = \frac{3 Q^{2}}{64 π ϵ_{0} R^{2}} {\vec{e}}_{z} \end{aligned}

2. 磁矢势的洛伦茨规范

已知 $\vec{A} = \frac{μ_{0}}{4 π} \int \frac{\vec{j} ({\vec{x}}^{'})}{| \vec{r} |} d V^{'}$ , 其中 $\vec{r} = \vec{x} - {\vec{x}}^{'}$ . 记 $\nabla^{'} = \frac{\partial}{\partial {\vec{x}}^{'}}$ , 有:

\begin{aligned} \nabla \cdot \vec{A} & \propto \nabla \cdot \int \frac{\vec{j} ({\vec{x}}^{'})}{| \vec{r} |} d V^{'} \\ = \int \vec{j} ({\vec{x}}^{'}) \nabla \frac{1}{r} d V^{'} \\ \nabla \frac{1}{r} = - \nabla^{'} \frac{1}{r} \to & = - \int \vec{j} ({\vec{x}}^{'}) \nabla^{'} \frac{1}{r} d V^{'} \\ = \underset{边 界 无 电 流 = 0}{\underset{⏟}{- \int \nabla^{'} \cdot \frac{\vec{j} ({\vec{x}}^{'})}{r} d V^{'}}} + \int \underset{稳 恒 磁 场 \nabla \cdot \vec{j} = \frac{\partial ρ}{\partial t} = 0}{\underset{⏟}{\frac{\nabla^{'} \cdot \vec{j} ({\vec{x}}^{'})}{r} d V^{'}}} \\ = 0 \end{aligned}

因此对于静磁场, $\partial_{t} ϕ = 0$ , 得到洛伦茨规范成立:

◻ \cdot \vec{A} = 0

3. 静电场多极展开

(1) 零阶和一阶场

已知离散形式的电场: $\vec{E} (\vec{x} - {\vec{x}}^{'}) = \frac{1}{4 π ϵ_{0}} \sum q (x^{'}) \frac{\vec{x} - {\vec{x}}^{'}}{| \vec{x} - {\vec{x}}^{'} |^{3}} = {\vec{E}}^{0} + {\vec{E}}^{1} + \dots$ , 记 $r = | \vec{x} | ≫ | {\vec{x}}^{'} |$ , $\hat{r} = \frac{\vec{x}}{r}$ , 偶极矩 $\vec{p} = \sum q {\vec{x}}^{'}$ , 则根据泰勒展开有:

{\vec{E}}^{0} = \vec{E} (\vec{x}) = \frac{1}{4 π ϵ_{0}} \sum q \frac{\hat{r}}{r^{2}}

以及:

\begin{aligned} {\vec{E}}^{1} & = - {\vec{x}}^{'} \cdot \nabla \vec{E} (\vec{x}) \\ = - {\vec{x}}^{'} \cdot \nabla \frac{1}{4 π ϵ_{0}} \sum q \frac{\hat{r}}{r^{2}} \\ = \frac{1}{4 π ϵ_{0}} \sum q [- \frac{\hat{r}}{r^{2}} + \frac{(3 {\vec{x}}^{'} \cdot \hat{r}) \hat{r}}{r^{3}}] \\ = \frac{1}{4 π ϵ_{0}} \frac{- \vec{p} + 3 \vec{p} \cdot \hat{r} \hat{r}}{r^{3}} \end{aligned}

(2) 小电荷体系受力矩

小电荷体系位于 ${\vec{r}}^{'}$ , 所受力矩可写作:

\vec{N} ({\vec{r}}^{'}) = \sum {\vec{r}}^{'} \times q \vec{E} ({\vec{r}}^{'}) = \sum \vec{p} \times \vec{E} ({\vec{r}}^{'}) = {\vec{N}}^{0} + {\vec{N}}^{1} + \dots

于是根据麦克劳林展开有:

{\vec{N}}^{0} = \sum q_{i} \vec{0} \times \vec{E} (0) = 0

\begin{aligned} {\vec{N}}^{1} & = \sum {\vec{r}}^{'} \cdot \nabla q \vec{r} \times E (\vec{r}) |_{\vec{r} = 0} \\ = \sum \nabla ({\vec{r}}^{'} \cdot q \vec{r} \times E (\vec{r})) |_{\vec{r} = 0} \\ = \sum \nabla (q \vec{E} (\vec{r}) \cdot {\vec{r}}^{'} \times \vec{r}) |_{\vec{r} = 0} \\ = \sum q {\vec{E} \times ({\vec{r}}^{'} \times \underset{= 0}{\underset{⏟}{(\nabla \times \vec{r})}}) + ({\vec{r}}^{'} \times \vec{r}) \times \underset{= 0}{\underset{⏟}{(\nabla \times \vec{E})}} + {\vec{r}}^{'} \times \underset{= \vec{E}}{\underset{⏟}{(\vec{E} \cdot \nabla) \vec{r}}} + (({\vec{r}}^{'} \times \underset{= 0}{\underset{⏟}{\vec{r}}}) \cdot \nabla) \vec{E}} \\ = \sum q {\vec{r}}^{'} \times \vec{E} \\ = \vec{p} \times \vec{E} (0) \end{aligned}

4. 均匀带电球的全空间能量

电荷分布

ρ (r) = {\begin{aligned} \frac{3 Q}{4 π R^{3}} & r \leq R \\ 0 & r > R \end{aligned}

于是由电场散度得到电场的分布:

\vec{E} (r) = {\begin{aligned} \frac{Q r}{4 π ϵ_{0} R^{3}} & r \leq R \\ \frac{Q}{4 π ϵ_{0} r^{2}} & r > R \end{aligned}

电势分布

ϕ (r) = {\begin{aligned} \int_{r}^{R} \frac{Q r}{4 π ϵ_{0} R^{3}} d r + \int_{R}^{+ \infty} \frac{Q}{4 π ϵ_{0} r^{2}} d r = \frac{Q}{4 π ϵ_{0} R} (\frac{3}{2} - \frac{r^{2}}{2 R^{2}}) & r \leq R \\ - & r > R \end{aligned}

于是得到能量:

W = \frac{1}{2} \int_{0}^{R} ρ (r) ϕ (r) r^{2} d r \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} d φ = \frac{3 Q^{2}}{20 π ϵ_{0} R}

同理

W = \frac{1}{2} ϵ_{0} \int_{0}^{+ \infty} E^{2} r^{2} d r \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} d φ = \frac{3 Q^{2}}{20 π ϵ_{0} R}

5. 线性介质的静电场能量密度

已知:

\begin{aligned} \frac{\partial w}{\partial t} + \nabla \cdot (\vec{E} \times \vec{H}) = - \vec{E} \cdot \vec{j} & (1) \\ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} & (2) \\ \nabla \times \vec{H} = \vec{j} + \frac{\partial \vec{D}}{\partial t} & (3) \end{aligned}

(3)带入(1)右侧得到:

\begin{aligned} \vec{E} \cdot \vec{j} & = \vec{E} \cdot (\nabla \times \vec{H} - \frac{\partial \vec{D}}{\partial t}) \\ = (\vec{E} \cdot \nabla \times \vec{H} - \vec{H} \cdot \nabla \times \vec{E}) + \vec{H} \cdot \nabla \times \vec{E} - \vec{E} \cdot \frac{\partial \vec{D}}{\partial t} \\ \overset{(2)}{=} - \nabla \cdot (\vec{E} \times \vec{H}) - \vec{H} \cdot \frac{\partial \vec{B}}{\partial t} - \vec{E} \cdot \frac{\partial \vec{D}}{\partial t} \\ = - \nabla \cdot (\vec{E} \times \vec{H}) - \frac{1}{2} \frac{\partial \vec{H} \cdot \vec{B}}{\partial t} - \frac{1}{2} \frac{\partial \vec{E} \cdot \vec{D}}{\partial t} \end{aligned}

所以在没有磁场时:

w = \frac{1}{2} \vec{E} \cdot \vec{D}

6. 课本题

1.7

(1)
易知:

\vec{E} (r) = {\begin{cases} 0 & r < r_{1} \\ \frac{ρ_{f}}{4 π ϵ} \frac{\frac{4}{3} π (r^{3} - r_{1}^{3})}{r^{2}} = \frac{ρ_{f} (r^{3} - r_{1}^{3})}{3 ε r^{2}} {\vec{e}}_{r} & r < r < r_{2} \\ \frac{ρ_{f}}{4 π ε_{0}} \frac{\frac{4}{3} (r_{2}^{3} - r_{1}^{3})}{r^{2}} = \frac{ρ_{f} (r_{2}^{3} - r_{3}^{3})}{3 ε_{0} r^{2}} {\vec{e}}_{r} & r > r_{2} \end{cases}

(2)
由 $\vec{p} = (ε - ε_{0}) \vec{E}$ 可得:

σ_{p} = {\begin{array}{lr} [\vec{p} (r < r < r_{2}) - \vec{p} (r < r_{1})] \cdot (- {\vec{e}}_{r}) = - (ε - ε_{0}) \frac{ρ_{f} (r^{3} - r_{r}^{3})}{3 ε r^{2}} = 0 & r = r_{1} \\ [\vec{p} (r_{1} < r < r_{2}) - \vec{p} (r > r_{2})] \cdot {\vec{e}}_{r} = (ε - ε_{0}) \frac{ρ_{f} (r_{2}^{3} - r_{1}^{3})}{3 ε r^{2}} & r = r_{2} \end{array}

ρ_{p} = - \nabla \cdot \vec{P} (r_{1} < r < r_{2}) = - (1 - \frac{ε_{0}}{ε}) \frac{ρ_{f}}{3} [\nabla \cdot \vec{r} - r_{1}^{3} \nabla \cdot \frac{\vec{r}}{r^{3}}] = - (1 - \frac{ε_{0}}{ε}) ρ_{f} r_{1} < r < r_{2}

1.9

\begin{aligned} ρ_{f} & = - \nabla \cdot \vec{P} \\ = - \nabla \cdot (ϵ - ϵ_{0}) \vec{E} \\ \nabla \cdot \vec{E} = \frac{ρ_{f}}{ϵ} \to & = - (1 - \frac{ε_{0}}{ε}) ρ_{f} \end{aligned}

2.1

ρ_{p} (r < R) = - \nabla \cdot \vec{p} = - k \nabla \cdot \frac{\vec{r}}{r^{2}} = - k [\frac{3 r^{2} - 2 (x^{2} + y^{2} + z^{2})}{r^{4}}] = - k / r^{2}

σ_{p} (r = R) = \vec{p} \cdot {\vec{e}}_{r} |_{r = R} = k / R

ρ_{f} = ρ_{p} / (\frac{ε_{p}}{ε} - 1) = \frac{k}{(1 - \frac{ε_{0}}{ε}) r^{2}} (r < R)

E (r) = {\begin{cases} \int_{0}^{R} \int_{0}^{π} \int_{0}^{2 π} ρ_{f} r^{2} \sin θ d r d θ d φ \cdot \frac{1}{4 π ε_{0} r^{2}} = \frac{k R}{(1 - \frac{ε_{0}}{ε}) ε_{0} r^{2}} {\vec{e}}_{r} r \geq R \\ \int_{0}^{r} \int_{0}^{π} \int_{0}^{2 π} ρ_{f} r^{2} \sin θ d r d θ d φ \cdot \frac{1}{4 π ε r^{2}} = \frac{k}{(1 - \frac{ε_{0}}{ε}) ε r} {\vec{e}}_{r} 0 < r < R \end{cases}

∴ φ (r) = {\begin{cases} \int_{r}^{+ \infty} E (r \geq R) d r = \frac{k R}{(1 - \frac{ε_{0}}{ε}) ε_{0} r} & r \geq R \\ \int_{R}^{+ \infty} E (r \geq R) d r + \int_{r}^{R} E (r < R) d r = \frac{k}{ε - ε_{0}} (\frac{ε}{ε_{0}} + \ln \frac{R}{r}) & r < R \end{cases}

\begin{aligned} w = \frac{1}{2} \int ρ_{f} φ & = \frac{1}{2} \int_{0}^{R} ρ_{f} (r < R) φ (r < R) r^{2} d r \int_{0}^{π} \sin θ d θ \int_{0}^{2 π} d φ \\ = \frac{2 π ε k^{2}}{{(ε - ε_{0})}^{2}} \cdot \int_{0}^{R} (\frac{ε}{ε_{0}} + \ln \frac{R}{r}) d r \\ = \frac{2 π ε (ε + ε_{0}) k^{2}}{ε_{0} {(ε - ε_{0})}^{2}} R \end{aligned}

7. 场源不分离的极化场的电势

(a)

记源点 $\vec{r}$ , 场点 ${\vec{r}}^{'}$ , 相对位移 $\vec{R} = \vec{r} - {\vec{r}}^{'}$ , 于是单个电荷 $q, \vec{r}$ 在 $R_{0}$ 球内产生的平均电场可以写作

\begin{aligned} {\vec{E}}_{a v e} (\vec{r}) & = \frac{1}{V^{'}} \int \frac{q (- {\vec{e}}_{R})}{4 π ϵ_{0} R^{2}} d V^{'} \\ ρ = - \frac{q}{V^{'}} \to & = \frac{1}{4 π ϵ_{0}} \int \frac{ρ \vec{e_{R}}}{R^{2}} d V^{'} \\ = {\vec{E}}_{ρ} (\vec{r}) \end{aligned}

等于均匀带电球在球内 $\vec{r}$ 产生的平均电场 ${\vec{E}}_{ρ}$ , 将 q 的数量增加到球内电荷的数目, 记每个电荷为 $q_{i}, {\vec{r}}_{i}$ , 产生的等效量是 ${\vec{E}}_{ρ i}, ρ_{i}$ , 于是由高斯定理和叠加原理得到:

E_{ρ_{i}} \cdot 4 π r_{i}^{2} = \frac{ρ_{i}}{ϵ_{0}} \cdot \frac{4}{3} π r_{i}^{3} \to {\vec{E}}_{a v e} = \sum {\vec{E}}_{ρ i} = \sum \frac{ρ_{i} \cdot \frac{4}{3} π r_{i}^{3} {\vec{e}}_{r i}}{ϵ_{0} 4 π r_{i}^{2}} = - \frac{\sum q_{i} \vec{r_{i}}}{4 π ϵ_{0} R_{0}^{3}} = - \frac{\vec{p}}{4 π ϵ_{0} R_{0}^{3}}

记极化强度 $\vec{P} = \frac{\vec{p}}{V^{'}}$ , 得到

{\vec{E}}_{a v e} = - \frac{\vec{P}}{3 ϵ_{0}}

(b)

设 $\vec{P} = P {\vec{e}}_{z}$ , $\vec{p} = \vec{P} V$ 则 $σ_{p} = \vec{P} \cdot \vec{n} = P \cos θ$ , 求解:

\nabla^{2} φ = 0

已知边界条件:

{\begin{aligned} φ (r = 0) = 0 \\ φ (R_{0}^{-}) = φ (R_{0}^{+}) \\ \frac{\partial}{\partial r} φ (R_{0}^{-}) - \frac{\partial}{\partial r} φ (R_{0}^{+}) = \frac{σ_{p}}{ϵ_{0}} \\ φ (r ≫ R_{0}) = \frac{\vec{p} \cdot \vec{r}}{4 π ϵ_{0} r^{3}} \end{aligned}

得到:

\begin{aligned} φ (r < R_{0}) & = \frac{1}{3 ϵ_{0}} P z \\ \vec{E} (r < R_{0}) & = - \nabla φ (r < R_{0}) = - \frac{\vec{P}}{3 ϵ_{0}} \end{aligned}

(c)

由于 ${\vec{r}}^{'} ≫ \vec{r}$ , 球外电荷 ${\vec{r}}^{'}, q$ 在球内(遍历 $\vec{r}$ )的平均电场, 与场点 $\vec{r}$ 处的电场相同, 即球内电场近似均匀, 同时

\begin{aligned} {\vec{E}}_{a v e} & = \frac{1}{V} \int \frac{q (- {\vec{e}}_{R})}{4 π ϵ_{0} R^{2}} d V \\ {\vec{r}}^{'} ≫ \vec{r}, \vec{R} = {\vec{r}}^{'} - \vec{r} \to & = \frac{q (- {\vec{e}}_{r}^{'})}{4 π ϵ_{0} r^{' 2}} \end{aligned}

第二个等号只保留了零阶项, 即等于外部电荷在球心处的电场.

8. poison 方程猜想解

L = \sum \frac{1}{2} (\partial_{i} φ)^{2} - \frac{ρ φ}{ϵ_{0}} = \frac{1}{2} (\nabla φ)^{2} + φ \nabla^{2} φ = - \frac{1}{2} (\nabla φ)^{2} + \frac{1}{2} \nabla^{2} φ^{2}

\nabla^{2} φ + \frac{ρ}{ϵ_{0}} = 0

9. 静电势的球谐展开

数理方程中提到, 勒让德函数的母函数表示是:

(1 - 2 x t + t^{2})^{- \frac{1}{2}} = \sum_{0}^{+ \infty} P_{n} (x) t^{n} | t | < 1

将 $t = \frac{r^{'}}{r}, x = \cos α$ 代入上式就可以得到 (14). 于是可以求解球谐系数:

\begin{aligned} ϕ (\vec{x}) & = \sum \sum \frac{4 π}{2 l + 1} q_{l m} r^{- (l + 1)} Y_{l m} \\ = \frac{1}{4 π ϵ_{0}} \int \frac{ρ ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d V^{'} \\ (14) \to & = \frac{1}{4 π ϵ_{0}} \int ρ ({\vec{x}}^{'}) \sum r^{' l} r^{- (l + 1)} P_{l} (\cos α) d V^{'} \\ (15) \to & = \sum \sum \frac{1}{4 π ϵ_{0}} \int ρ ({\vec{x}}^{'}) r^{' l} r^{- (l + 1)} \frac{4 π}{2 l + 1} Y_{l m}^{*} (θ^{'}, φ^{'}) Y_{l m} (θ, φ) d V^{'} \\ = \sum \sum \frac{4 π}{2 l + 1} (\frac{1}{4 π ϵ_{0}} \int ρ ({\vec{x}}^{'}) r^{' l} Y_{l m}^{*} (θ^{'}, φ^{'}) d V^{'}) r^{- (l + 1)} Y_{l m} \end{aligned}

于是 $q_{l m} = \frac{1}{4 π ϵ_{0}} \int ρ ({\vec{x}}^{'}) r^{' l} Y_{l m}^{*} (θ^{'}, φ^{'}) d V^{'}$ .

~~# 补充 1. 电磁场的拉氏量是猜出来的吗?~~

补充 2. 球谐函数

第九题的泊松方程课上已经用 green 函数求解讲过. 来看更简单的 Laplace 方程. 三维拉普拉斯方程方程在球坐标下:

\begin{aligned} Δ u & = \nabla^{2} u = \nabla \cdot \nabla u = 0 \\ = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial u}{\partial r}) + \frac{1}{r^{2} \sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial u}{\partial θ}) + \frac{1}{r^{2} \sin^{2} θ} \frac{\partial^{2} u}{\partial^{2} φ} \end{aligned}

分离变量 $u (r, θ, φ) = R (r) Y (θ, φ)$ , 得到

\frac{1}{R} \frac{d}{d r} (r^{2} \frac{d R}{d r}) = - \frac{1}{Y} (\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) - \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial ϕ^{2}}) = λ

1. 径向衰减

常常选取 $λ = l (l + 1)$ , 于是

\frac{d}{d r} (r^{2} \frac{d R}{d r}) - l (l + 1) R = 0

这是欧拉方程, 根据换元 $r = e^{s}$ 得到它的通解是

R (r) = A r^{l} + B r^{- (l + 1)}, l \geq 0

通常情况下, 场不会传播到无穷远处, 也就是说, 在解 $R (r)$ 的定义域中包含 $r = \infty$ 时, 因为 $r^{l} \to \infty$ 不合适, 所以通常取 $A = 0$ . 同理, 在解的定义域包含 $r = 0$ 时, 通常取 $B = 0$ .

2. 球面谐波

由上述已知

\frac{1}{\sin θ} \frac{\partial}{\partial θ} (\sin θ \frac{\partial Y}{\partial θ}) - \frac{1}{\sin^{2} θ} \frac{\partial^{2} Y}{\partial ϕ^{2}} + l (l + 1) Y = 0

进一步分离变量 $Y (θ, φ) = Θ (θ) Φ (φ)$ , 有

\begin{array}{r} {\begin{cases} \frac{d^{2} Φ}{d^{2} φ} + m^{2} Φ = 0 \\ Φ (φ) = Φ (φ + 2 π) \end{cases} a n d \begin{array}{l} \sin θ \frac{d}{d θ} (\sin θ \frac{d Θ}{d θ}) + [(l + 1) l \sin^{2} θ - m^{2}] Θ = 0 \end{array} \end{array}

第一个方程没什么好说的, 解是 $Φ_{m} (φ) = e^{i m φ}$ , 对第二个方程做和数理课上一样的换元 $θ = \arccos x$ , 得到

(1 - x^{2}) \frac{d^{2} Θ}{d x^{2}} - 2 x \frac{d Θ}{d x} + [l (l + 1) - \frac{m^{2}}{1 - x^{2}}] Θ = 0 \dots \dots (1)

被称为连带勒让德方程, 特别的当 $m = 0$ 时变成勒让德方程, 他的解就是勒让德函数 $P_{l} (x)$ .

做换元 $y (x) = Θ (x) (1 - x^{2})^{- \frac{m}{2}}$ , 可得到

\begin{matrix} Θ = (1 - x^{2})^{\frac{m}{2}} y \\ Θ^{'} = {(1 - x^{2})}^{\frac{m}{2}} y^{'} - m {(1 - x^{2})}^{\frac{m}{2} - 1} x y \\ Θ^{″} = {(1 - x^{2})}^{\frac{m}{2}} y^{''} - 2 m {(1 - x^{2})}^{\frac{m}{2} - 1} x y^{'} - m {(1 - x^{2})}^{\frac{m}{2} - 1} y + m (m - 2) {(1 - x^{2})}^{\frac{m}{2} - 2} x^{2} y \\ ⇓ \\ (1 - x^{2}) y^{''} - 2 (m + 1) x y^{'} + [l (l + 1) - m (m + 1)] y = 0 \dots \dots (2) \end{matrix}

将上式与与下面的勒让德方程(式 $(1)$ )做比较, 注意下面的方程还没有做 $y (x) = Θ (x) (1 - x^{2})^{- \frac{m}{2}}$ 的换元.

(1 - x^{2}) \frac{d^{2} P_{l}}{d x^{2}} - 2 x \frac{d P_{l}}{d x} + l (l + 1) P_{l} = 0

对它求导 $m$ 次, 记 $P^{(m)}$ 表示对 $P$ 求 $m$ 次导, 注意到 $P_{l} (x)$ 是 $l$ 阶多项式, 可以得到

\begin{aligned} 0 & = [(1 - x^{2}) P_{l}^{(m + 2)} - 2 m x P_{l}^{(m + 1)} - m (m - 1) P_{l}^{(m)}] - 2 [x P_{l}^{(m + 1)} + m P_{l}^{(m)}] + l (l + 1) P_{l}^{(m)} \\ = (1 - x^{2}) P_{l}^{(m + 2)} - 2 (m + 1) x P_{l}^{(m + 1)} + [l (l + 1) - m (m + 1)] P_{l}^{(m)} \end{aligned}

恰好和 $(2)$ 式形式一致, 也就是说如果将方程 $(1 - x^{2}) \frac{d^{2} Θ}{d x^{2}} - 2 x \frac{d Θ}{d x} + [l (l + 1) - \frac{m^{2}}{1 - x^{2}}] Θ = 0$ 的解记为连带勒让德函数 $P_{l}^{m} (x)$ , 那么

P_{l}^{m} (x) = (1 - x^{2})^{\frac{m}{2}} \frac{d^{m}}{d x^{m}} P_{l} (x), | m | \leq l

最后的球谐函数就是

Y_{l}^{m} (θ, φ) = P_{l}^{m} (\cos θ) Φ_{m} (φ)

如果要归一化 $Y$ , 也就是令 $\int_{0}^{2 π} d φ \int_{0}^{π} Y_{l}^{m} (θ, φ) \sin θ d θ = 1$ , 可以得到归一化的球谐函数:

Y_{l}^{m} (θ, φ) = \sqrt{\frac{1}{2 π} \frac{2 l + 1}{2} \frac{(l - m)!}{(l + m)!}} P_{l}^{m} (\cos θ) e^{i m φ}

于是通解表示为

u (r, θ, φ) = \sum_{l = 0}^{\infty} \sum_{m = - l}^{l} (A r^{l} + B r^{- (l + 1)}) \sqrt{\frac{1}{2 π} \frac{2 l + 1}{2} \frac{(l - m)!}{(l + m)!}} P_{l}^{m} (\cos θ) e^{i m φ}

另外, 连带勒让德函数的前几项为:

\begin{aligned} P_{0}^{0} (x) & = 1, \\ P_{1}^{0} (x) & = x, P_{1}^{1} (x) = - \sqrt{1 - x^{2}}, \\ P_{2}^{0} (x) & = \frac{1}{2} (3 x^{2} - 1), P_{2}^{1} (x) = - 3 x \sqrt{1 - x^{2}}, P_{2}^{2} (x) = 3 (1 - x^{2}), \\ P_{3}^{0} (x) & = \frac{1}{2} (5 x^{3} - 3 x), P_{3}^{1} (x) = - \frac{3}{2} (5 x^{2} - 1) \sqrt{1 - x^{2}}, P_{3}^{2} (x) = 15 x (1 - x^{2}), P_{3}^{3} (x) = - 15 (1 - x^{2})^{3 / 2} . \end{aligned}

2.1 $Y_{1}^{m}$ 和偶极场

在静磁学中, 如果场点不存在电流, 那么磁感应强度可以表示成磁标势的梯度 $\vec{B} = - \nabla Φ_{m}$ , 并且 $\nabla^{2} Φ_{m} = 0$ .

归一化球谐函数中, $Y_{1}^{0} (θ) = \sqrt{\frac{3}{4 π}} \cos θ \propto \cos θ = \frac{z}{r}$ , $R_{1} (r) = A r + \frac{B}{r^{2}} = \frac{B}{r^{2}}$ (取 $A = 0$ ). 于是 $Φ_{m}$ 的一个解可以写作

Φ_{p o l e} (r, θ, φ) = \frac{C}{r^{2}} \cos θ

求梯度得到

\vec{B} = - (\partial_{r} Φ_{p o l e}, \frac{1}{r} \partial_{θ} Φ_{p o l e}, 0) = C \cdot (\frac{2 \cos θ}{r^{3}}, \frac{\sin θ}{r^{3}}, 0)

回忆课上所学到的磁偶极子在远处的磁场表达式:

\vec{B} = \frac{μ_{0}}{4 π} \frac{3 (\vec{n} \cdot \vec{m}) \vec{n} - \vec{m}}{r^{3}} = \frac{μ_{0}}{4 π} \frac{3 \cos θ {\vec{e}}_{r} - {\vec{e}}_{z}}{r^{3}} = \frac{μ_{0}}{4 π} \frac{3 \cos θ {\vec{e}}_{r} - (\cos θ {\vec{e}}_{r} - \sin θ {\vec{e}}_{θ})}{r^{3}} = \frac{μ_{0}}{4 π} \frac{2 \cos θ {\vec{e}}_{r} + \sin θ {\vec{e}}_{θ}}{r^{3}}

发现两式形式一致, 只要取 $C = \frac{μ_{0}}{4 π}$ , 就可以用球谐函数描述偶极磁场.

另外, 由于 $Y_{1}^{\pm 1}$ 是虚数, 所以选取它们的实数线性组合 $Y_{1}^{x} = \frac{Y_{1}^{1} + Y_{1}^{- 1}}{\sqrt{2}}$ 和 $Y_{1}^{y} = \frac{Y_{1}^{1} - Y_{1}^{- 1}}{i \sqrt{2}}$ , 显然 $Y_{1}^{x} \propto \sin θ \cos φ = \frac{x}{r}$ , $Y_{1}^{y} \propto \sin θ \sin φ = \frac{y}{r}$ . 对于三维旋转, 考察球坐标的变换, 比如绕 $y$ 轴旋转矩阵, 直角坐标系中的变换矩阵是

R_{y} (β) = (\begin{matrix} \cos β & 0 & \sin β \\ 0 & 1 & 0 \\ - \sin β & 0 & \cos β \end{matrix})

即:

\begin{aligned} θ^{'} & = \arccos (\frac{z^{'}}{r}) = \arccos (\frac{- x \sin β + z \cos β}{r}) \\ = \arccos (- \sin θ \cos φ \sin β + \cos θ \cos β) \end{aligned}

\begin{aligned} φ^{'} & = \arctan 2 (y^{'}, x^{'}) = \arctan 2 (y, x \cos β + z \sin β) \\ = \arctan 2 (\sin θ \sin φ, \sin θ \cos φ \cos β + \cos θ \sin β) \end{aligned}

如果取 $β = \frac{π}{2}$ , 就有

\begin{array}{r} θ^{'} = \arccos (- \sin θ \cos φ) \end{array}

\begin{array}{r} φ^{'} = \arctan (\frac{\sin θ \sin φ}{\cos θ}) \end{array}

也就是说

\begin{aligned} Y_{1}^{x} (θ^{'}, φ^{'}) \propto \sin θ^{'} \cos φ^{'} & = \sin (\arccos (- \sin θ \cos φ)) \cdot \cos (\arctan (\frac{\sin θ \sin φ}{\cos θ})) \\ = \sqrt{1 - (\sin θ \cos φ)^{2}} \cdot \frac{1}{\sqrt{1 + {(\frac{\sin θ \sin φ}{\cos θ})}^{2}}} \\ = \sqrt{1 - (\sin θ \cos φ)^{2}} \cdot \frac{\cos θ}{\sqrt{\cos^{2} θ + {(\sin θ \sin φ)}^{2}}} \\ = \cos θ \\ \propto Y_{1}^{0} (θ) \end{aligned}

因此, $Y_{1}^{x}$ 描述沿 $x$ 轴方向的磁偶极子, 类似的知道 $Y_{1}^{y}$ 描述沿 $y$ 轴方向的磁偶极子. 在量子力学里面(量C不讲这个东西), 上面的旋转操作可以用 $W i n g e r - D$ 矩阵作用于球谐函数得到, 这个矩阵的数值形状很丑, 可以在在 $M M A$ ( $M a t h e m a t i c a$ )中直接调用 $W i n g e r - D$ 函数快速完成.

2.2 $Y_{2}^{m}$ 和四极场

同理, 对于归一化球谐函数 $Y_{2}^{0} (θ) = \sqrt{\frac{5}{16 π}} (3 \cos^{2} θ - 1)$ , 磁标势 $Φ_{quad} (r, θ) = R \cdot g_{2}^{0} {(\frac{R}{r})}^{3} \sqrt{\frac{5}{16 π}} (3 \cos^{2} θ - 1)$ , $g_{2}^{0}$ 为常数, 计算得到

\begin{matrix} B_{r} = - \frac{\partial Φ}{\partial r} = 3 g_{2}^{0} (\frac{R^{4}}{r^{4}}) \sqrt{\frac{5}{16 π}} (3 \cos^{2} θ - 1) \\ B_{θ} = - \frac{1}{r} \frac{\partial Φ}{\partial θ} = 3 g_{2}^{0} (\frac{R^{4}}{r^{4}}) \sqrt{\frac{5}{16 π}} \sin 2 θ \\ B_{φ} = 0 \end{matrix}

关于球谐系数 $g_{l}^{m}$ 的具体拟合值, 可在第 $14$ 版国际参考地磁场(IGRF)的官方文件中查阅. 对于 $m \neq 0$ 的球谐函数, 同样选取他们的实数线性组合, 可以得到他们表示的四极场只是 $Y_{2}^{0} (θ)$ 的旋转, 但是形式更加复杂.

3. 球谐系数

现在需要确定系数 $B_{l m}$ , 假设 $A = 0$ , 通解写成:

u (r, θ, φ) = \sum_{l = 0}^{\infty} \sum_{m = - l}^{l} \frac{B_{l m}}{r^{l + 1}} Y_{l}^{m} (θ, φ)

其中 $Y_{l}^{m} (θ, φ) = \sqrt{\frac{(2 l + 1)}{4 π} \frac{(l - m)!}{(l + m)!}} P_{l}^{m} (\cos θ) e^{i m φ}$ , 它是正交归一化的:

\int_{0}^{2 π} \int_{0}^{π} Y_{l}^{m} (θ, φ) Y_{l^{'}}^{m^{'} *} (θ, φ) \sin θ d θ d φ = δ_{l l^{'}} δ_{m m^{'}}

将通解两边乘以 $Y_{l^{'}}^{m^{'} *} (θ, φ)$ , 然后积分:

\int_{0}^{2 π} \int_{0}^{π} u (r, θ, φ) Y_{l^{'}}^{m^{'} *} (θ, φ) \sin θ d θ d φ = \sum_{l = 0}^{\infty} \sum_{m = - l}^{l} \frac{B_{l m}}{r^{l + 1}} \int_{0}^{2 π} \int_{0}^{π} Y_{l}^{m} (θ, φ) Y_{l^{'}}^{m^{'} *} (θ, φ) \sin θ d θ d φ

利用正交归一关系, 右边只剩下 $l = l^{'}$ 且 $m = m^{'}$ 时的项：

\int_{0}^{2 π} \int_{0}^{π} u (r, θ, φ) Y_{l^{'}}^{m^{'} *} (θ, φ) \sin θ d θ d φ = \frac{B_{l^{'} m^{'}}}{r^{l^{'} + 1}}

倘若已知边界条件 $u (r, θ, φ) |_{r = R} = u_{R} (θ, φ)$ , 就可解出:

B_{l^{'} m^{'}} = R^{l^{'} + 1} \int_{0}^{2 π} \int_{0}^{π} u (R, θ, φ) Y_{l^{'}}^{m^{'} *} (θ, φ) \sin θ d θ d φ

这就是拉普拉斯方程通解对应的球谐系数. 不同的边界条件会得到不同的系数表达式, 上述只是较为简单的一种.