电流定义为电荷的变化率
I = d q d t I = \frac{dq}{dt} I=dtdq
电荷密度
ρ = q V \rho = \frac{q}{V} ρ=Vq
电流密度
j = I S j = \frac{I}{S} j=SI
连续方程
∯ j ⋅ d S = − d q d t \oiint j\cdot dS = -\frac{dq}{dt} ∬j⋅dS=−dtdq
其微分形式为
∇ ⋅ j = − ∂ ρ ∂ t \nabla\cdot j = -\frac{\partial\rho}{\partial t} ∇⋅j=−∂t∂ρ
F c = k c q q ′ r 2 F_c = k_c \frac{qq'}{r^2} Fc=kcr2qq′
电场强度
E = F c q ′ = k c q r 2 E = \frac{F_c}{q'} = k_c \frac{q}{r^2} E=q′Fc=kcr2q
考虑 E E E在球壳上的积分
∯ E ⋅ d S = 4 π r 2 k c q r 2 = 4 π k c ∭ ρ d V \oiint E\cdot dS = 4\pi r^2k_c \frac{q}{r^2} = 4\pi k_c\iiint\rho dV ∬E⋅dS=4πr2kcr2q=4πkc∭ρdV
而根据高斯公式
∯ E ⋅ d S = ∭ ∇ ⋅ E d V \oiint E\cdot dS = \iiint\nabla\cdot EdV ∬E⋅dS=∭∇⋅EdV
所以有
∇ ⋅ E = 4 π k c ρ \nabla\cdot E = 4\pi k_c\rho ∇⋅E=4πkcρ
两根距离为 d d d平行导线,其中一根在单位长度上受到的作用力为
d F a d l = 2 k a I I ′ d \frac{dF_a}{dl} = 2k_a\frac{II'}{d} dldFa=2kadII′
这里系数前面的 2 2 2只是为了方便而引入。对比库伦定律和安培定律,容易看出 k c / k a k_c / k_a kc/ka的量纲为速度平方的量纲。通过实验测量可以得到 k c / k a = c 2 k_c / k_a = c^2 kc/ka=c2.
磁感应强度定义为单位电流对上式的贡献(乘一个系数 α \alpha α)
B = 2 k a α I d B = 2k_a\alpha\frac{I}{d} B=2kaαdI
这里也可以算出 E / B E / B E/B的量纲为速度除以 α \alpha α的量纲。考虑 B B B在以其中一根导线为圆心, d d d为半径的圆周上的积分
∮ B d l = 2 k a α I d 2 π d = 4 π k a α ∬ j d S \oint Bdl = 2k_a\alpha\frac{I}{d}2\pi d = 4\pi k_a\alpha\iint jdS ∮Bdl=2kaαdI2πd=4πkaα∬jdS
而根据斯托克斯定理
∮ B d l = ∬ ∇ × B d S \oint Bdl = \iint\nabla\times BdS ∮Bdl=∬∇×BdS
于是
∇ × B = 4 π k a α j \nabla\times B = 4\pi k_a\alpha j ∇×B=4πkaαj
旋度的散度为零,因此 ∇ ⋅ j = 0 \nabla\cdot j = 0 ∇⋅j=0. 而电流密度 j q j_q jq满足连续性方程,一般情况下不满足 ∇ ⋅ j q = 0 \nabla\cdot j_q = 0 ∇⋅jq=0. 于是考虑将 j j j分解为两项之和, j = j q + j d j = j_q + j_d j=jq+jd,结合前述两条件就有
∇ ⋅ j d = d ρ d t \nabla\cdot j_d = \frac{d\rho}{dt} ∇⋅jd=dtdρ
写成积分形式就是
∯ j d ⋅ d S = d q d t \oiint j_d\cdot dS = \frac{dq}{dt} ∬jd⋅dS=dtdq
将 ∂ E ∂ t \frac{\partial E}{\partial t} ∂t∂E代入上式就有
4 π r 2 j d = r 2 k c ∂ E ∂ t 4\pi r^2j_d = \frac{r^2}{k_c}\frac{\partial E}{\partial t} 4πr2jd=kcr2∂t∂E
于是
∇ × B = 4 π k a α j q + k a k c α ∂ E ∂ t \nabla\times B = 4\pi k_a\alpha j_q + \frac{k_a}{k_c}\alpha\frac{\partial E}{\partial t} ∇×B=4πkaαjq+kckaα∂t∂E
∇ × E = − k f ∂ B ∂ t \nabla\times E = -k_f\frac{\partial B}{\partial t} ∇×E=−kf∂t∂B
∇ ⋅ B = 0 \nabla\cdot B = 0 ∇⋅B=0
综上所述,我们现在就可以写出著名的麦克斯韦方程组
{ ∇ ⋅ E = 4 π k c ρ ∇ × B = 4 π k a α j q + k a k c α ∂ E ∂ t ∇ × E = − k f ∂ B ∂ t ∇ ⋅ B = 0 \left\{\begin{aligned} & \nabla\cdot E = 4\pi k_c\rho \\ & \nabla\times B = 4\pi k_a\alpha j_q + \frac{k_a}{k_c}\alpha\frac{\partial E}{\partial t} \\ & \nabla\times E = -k_f\frac{\partial B}{\partial t} \\ & \nabla\cdot B = 0 \end{aligned}\right . ⎩ ⎨ ⎧∇⋅E=4πkcρ∇×B=4πkaαjq+kckaα∂t∂E∇×E=−kf∂t∂B∇⋅B=0
因为 ∇ × ( ∇ × E ) = ∇ ( ∇ ⋅ E ) − ∇ 2 E = − k f ∂ ∂ t ( ∇ × B ) \nabla\times(\nabla\times E) = \nabla(\nabla\cdot E) - \nabla^2E = -k_f\frac{\partial}{\partial t}(\nabla\times B) ∇×(∇×E)=∇(∇⋅E)−∇2E=−kf∂t∂(∇×B),对于无源场有 ∇ ⋅ E = 0 \nabla\cdot E = 0 ∇⋅E=0以及 ∇ × B = k a k c α ∂ E ∂ t \nabla\times B = \frac{k_a}{k_c}\alpha\frac{\partial E}{\partial t} ∇×B=kckaα∂t∂E,代入前式就有
∇ 2 E = α k a k f k c ∂ 2 E ∂ t 2 \nabla^2E = \frac{\alpha k_ak_f}{k_c}\frac{\partial^2E}{\partial t^2} ∇2E=kcαkakf∂t2∂2E
这是波动方程,而 α k a k f k c \frac{\alpha k_ak_f}{k_c} kcαkakf就是波速平方的倒数。实验测得该数值为 1 c 2 \frac{1}{c^2} c21,结合前面的实验数据就有 α k f = 1 \alpha k_f = 1 αkf=1. 于是麦克斯韦方程就可以简化为
{ ∇ ⋅ E = 4 π c 2 k a ρ ∇ × E = − k f ∂ B ∂ t ∇ × B = 4 π k a k f j q + 1 c 2 k f ∂ E ∂ t ∇ ⋅ B = 0 \left\{\begin{aligned} & \nabla\cdot E = 4\pi c^2k_a\rho \\ & \nabla\times E = -k_f\frac{\partial B}{\partial t} \\ & \nabla\times B = 4\pi \frac{k_a}{k_f}j_q + \frac{1}{c^2k_f}\frac{\partial E}{\partial t} \\ & \nabla\cdot B = 0 \end{aligned}\right . ⎩ ⎨ ⎧∇⋅E=4πc2kaρ∇×E=−kf∂t∂B∇×B=4πkfkajq+c2kf1∂t∂E∇⋅B=0
我们定义介质中的场强如下
D = ϵ 0 E + a P H = 1 μ 0 B − b M D = \epsilon_0E + aP \\ H = \frac{1}{\mu_0}B - bM D=ϵ0E+aPH=μ01B−bM
其中 P P P为电偶极矩贡献的电极化强度,对电场有增强作用; M M M为磁矩贡献的磁化强度,对磁场有削弱作用。要求 a , b a, b a,b为无量纲值,这样可确保 D D D与 P P P, H H H与 M M M量纲一致。 ϵ 0 \epsilon_0 ϵ0称为真空电容率, μ 0 \mu_0 μ0称为真空导磁率。于是我们得到介质中的麦克斯韦方程组
{ ∇ ⋅ D = 4 π c 2 k a ϵ 0 ρ ∇ × E = − k f ∂ B ∂ t ∇ × H = 4 π k a μ 0 k f j q + 1 μ 0 ϵ 0 c 2 k f ∂ D ∂ t ∇ ⋅ B = 0 \left\{\begin{aligned} & \nabla\cdot D = 4\pi c^2k_a\epsilon_0\rho \\ & \nabla\times E = -k_f\frac{\partial B}{\partial t} \\ & \nabla\times H = 4\pi \frac{k_a}{\mu_0k_f}j_q + \frac{1}{\mu_0\epsilon_0c^2k_f}\frac{\partial D}{\partial t} \\ & \nabla\cdot B = 0 \end{aligned}\right . ⎩ ⎨ ⎧∇⋅D=4πc2kaϵ0ρ∇×E=−kf∂t∂B∇×H=4πμ0kfkajq+μ0ϵ0c2kf1∂t∂D∇⋅B=0
系数 | 值 | 量纲 |
---|---|---|
k a k_a ka | 1 0 − 7 10^{-7} 10−7 | [ m ] [ l ] [ t ] − 2 [ I ] − 2 [m][l][t]^{-2}[I]^{-2} [m][l][t]−2[I]−2 |
k f k_f kf | 1 1 1 | 1 1 1 |
ϵ 0 \epsilon_0 ϵ0 | 1 0 7 4 π c 2 \frac{10^7}{4\pi c^2} 4πc2107 | [ m ] − 1 [ l ] − 3 [ t ] 4 [ I ] 2 [m]^{-1}[l]^{-3}[t]^4[I]^2 [m]−1[l]−3[t]4[I]2 |
μ 0 \mu_0 μ0 | 4 π × 1 0 − 7 4\pi\times 10^{-7} 4π×10−7 | [ m ] [ l ] [ t ] − 2 [ I ] − 2 [m][l][t]^{-2}[I]^{-2} [m][l][t]−2[I]−2 |
a a a | 1 1 1 | 1 1 1 |
b b b | 1 1 1 | 1 1 1 |
此时
D = ϵ 0 E + P H = 1 μ 0 B − M D = \epsilon_0E + P \\ H = \frac{1}{\mu_0}B - M D=ϵ0E+PH=μ01B−M
E / B E / B E/B的量纲为速度的量纲。而麦克斯韦方程组表现为
{ ∇ ⋅ E = 1 ϵ 0 ρ ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 ( j q + ϵ 0 ∂ E ∂ t ) ∇ ⋅ B = 0 \left\{\begin{aligned} & \nabla\cdot E = \frac{1}{\epsilon_0}\rho \\ & \nabla\times E = -\frac{\partial B}{\partial t} \\ & \nabla\times B = \mu_0(j_q + \epsilon_0\frac{\partial E}{\partial t}) \\ & \nabla\cdot B = 0 \end{aligned}\right . ⎩ ⎨ ⎧∇⋅E=ϵ01ρ∇×E=−∂t∂B∇×B=μ0(jq+ϵ0∂t∂E)∇⋅B=0
以及
{ ∇ ⋅ D = ρ ∇ × E = − ∂ B ∂ t ∇ × H = j q + ∂ D ∂ t ∇ ⋅ B = 0 \left\{\begin{aligned} & \nabla\cdot D = \rho \\ & \nabla\times E = -\frac{\partial B}{\partial t} \\ & \nabla\times H = j_q + \frac{\partial D}{\partial t} \\ & \nabla\cdot B = 0 \end{aligned}\right . ⎩ ⎨ ⎧∇⋅D=ρ∇×E=−∂t∂B∇×H=jq+∂t∂D∇⋅B=0
系数 | 值 | 量纲 |
---|---|---|
k a k_a ka | c − 2 c^{-2} c−2 | [ c ] − 2 [c]^{-2} [c]−2 |
k f k_f kf | c − 1 c^{-1} c−1 | [ c ] − 1 [c]^{-1} [c]−1 |
ϵ 0 \epsilon_0 ϵ0 | 1 1 1 | 1 1 1 |
μ 0 \mu_0 μ0 | 1 1 1 | 1 1 1 |
a a a | 4 π 4\pi 4π | 1 1 1 |
b b b | 4 π 4\pi 4π | 1 1 1 |
此时
D = E + 4 π P H = B − 4 π M D = E + 4\pi P \\ H = B - 4\pi M D=E+4πPH=B−4πM
E / B E / B E/B的量纲为 1 1 1。而麦克斯韦方程组表现为
{ ∇ ⋅ E = 4 π ρ ∇ × E = − 1 c ∂ B ∂ t ∇ × B = 1 c ( 4 π j q + ∂ E ∂ t ) ∇ ⋅ B = 0 \left\{\begin{aligned} & \nabla\cdot E = 4\pi\rho \\ & \nabla\times E = -\frac{1}{c}\frac{\partial B}{\partial t} \\ & \nabla\times B = \frac{1}{c}(4\pi j_q + \frac{\partial E}{\partial t}) \\ & \nabla\cdot B = 0 \end{aligned}\right . ⎩ ⎨ ⎧∇⋅E=4πρ∇×E=−c1∂t∂B∇×B=c1(4πjq+∂t∂E)∇⋅B=0