F ⃗ = 1 4 π ϵ 0 q Q r 2 r ^ ⃗ \vec{F} = {1 \over {4\pi\epsilon_0}}{qQ\over{r^2}}\vec{\hat{r}} F=4πϵ01r2qQr^
其中 r ^ ⃗ \vec{\hat{r}} r^为单位矢量。其中 ϵ 0 \epsilon_0 ϵ0为真空电容率,其大小为 8.8542 × 1 0 − 12 F ⋅ m − 1 8.8542 \times 10^{-12} F \cdot m^{-1} 8.8542×10−12F⋅m−1
∯ E ⃗ ⋅ d S ⃗ = 1 ϵ 0 ∑ q i = 1 ϵ 0 ∭ ρ d V \oiint\vec{E}\cdot \rm d\vec{S} = {1\over{\epsilon_0}}\sum q_i = {1\over{\epsilon_0}}\iiint\rho \rm dV ∬E⋅dS=ϵ01∑qi=ϵ01∭ρdV
其中, E ⃗ : = F ⃗ q \vec{E} := {\vec{F} \over {q}} E:=qF
∮ E ⃗ ⋅ d l ⃗ = 0 \oint\vec{E}\cdot \rm d\vec{l} = 0 ∮E⋅dl=0
∯ B ⃗ ⋅ d S ⃗ = 0 \oiint\vec{B}\cdot \rm d\vec{S} = 0 ∬B⋅dS=0
d B ⃗ = μ 0 4 π I d l ⃗ × r ^ ⃗ r 2 \rm d\vec{B} = {\mu_0\over{4\pi}}{I\rm d\vec{l}\times \vec{\hat{r}}\over r^2} dB=4πμ0r2Idl×r^
其中, μ 0 \mu_0 μ0为真空磁导率,其值为 4 π × 1 0 − 7 H ⋅ m − 1 4\pi\times10^{-7}H\cdot m^{-1} 4π×10−7H⋅m−1
∮ B ⃗ ⋅ d l ⃗ = μ 0 ∑ I i \oint\vec{B}\cdot \rm d\vec{l} = \mu_0\sum I_i ∮B⋅dl=μ0∑Ii
∯ j ⃗ ⋅ d S ⃗ = − d Q d t \oiint\vec{j}\cdot \rm d\vec{S} = -{\rm dQ\over \rm dt} ∬j⋅dS=−dtdQ
其中 j ⃗ \vec{j} j为电流密度, ∯ j ⃗ ⋅ d S ⃗ = 0 \oiint\vec{j}\cdot \rm d\vec{S} = 0 ∬j⋅dS=0称为恒定电流条件。其微分形式为
∇ ⃗ ⋅ j ⃗ = − ∂ ρ ∂ t \vec\nabla\cdot\vec j = -\frac{\partial \rho}{\partial t} ∇⋅j=−∂t∂ρ
∮ B ⃗ ⋅ d l ⃗ = μ 0 ∬ ( j ⃗ + ϵ 0 ∂ E ⃗ ∂ t ) ⋅ d S ⃗ \oint\vec{B}\cdot \rm d\vec{l} = \mu_0\iint(\vec{j} + \epsilon_0{\partial\vec{E} \over {\partial t}})\cdot \rm d\vec{S} ∮B⋅dl=μ0∬(j+ϵ0∂t∂E)⋅dS
∮ E ⃗ ⋅ d l ⃗ = − d d t ∬ B ⃗ ⋅ d S ⃗ \oint\vec{E}\cdot \rm d\vec{l} = - {\rm d\over \rm dt}\iint\vec{B}\cdot \rm d\vec{S} ∮E⋅dl=−dtd∬B⋅dS
F ⃗ = q ( E ⃗ + v ⃗ × B ⃗ ) \vec{F} = q(\vec{E} + \vec{v}\times\vec{B}) F=q(E+v×B)
F ⃗ = I l ⃗ × B ⃗ \vec{F} = I\vec{l}\times\vec{B} F=Il×B
安培定律是洛伦兹1磁场力的宏观体现,实际上, F ⃗ = I l ⃗ × B ⃗ = I t l ⃗ t × B ⃗ = q v ⃗ × B ⃗ \vec{F} = I\vec{l}\times\vec{B} = It{\vec{l}\over t}\times\vec{B} = q\vec{v}\times\vec{B} F=Il×B=Ittl×B=qv×B
∭ ∇ ⃗ ⋅ A ⃗ d V = ∯ A ⃗ ⋅ d S ⃗ \iiint\vec\nabla\cdot\vec{A} \rm dV = \oiint\vec{A}\cdot \rm d\vec{S} ∭∇⋅AdV=∬A⋅dS
∬ ∇ ⃗ × A ⃗ ⋅ d S ⃗ = ∮ A ⃗ ⋅ d l ⃗ \iint\vec\nabla\times\vec{A}\cdot\rm d\vec{S} = \oint\vec{A}\cdot\rm d\vec{l} ∬∇×A⋅dS=∮A⋅dl
高斯电场定理 ∯ E ⃗ ⋅ d S ⃗ = 1 ϵ 0 ∭ ρ d V \oiint\vec{E}\cdot \rm d\vec{S} = {1\over{\epsilon_0}}\iiint\rho \rm dV ∬E⋅dS=ϵ01∭ρdV
高斯磁场定理 ∯ B ⃗ ⋅ d S ⃗ = 0 \oiint\vec{B}\cdot \rm d\vec{S} = 0 ∬B⋅dS=0
法拉第定律 ∮ E ⃗ ⋅ d l ⃗ = − ∬ ∂ B ⃗ ∂ t ⋅ d S ⃗ \oint\vec{E}\cdot \rm d\vec{l} = -\iint{\partial\vec{B}\over\partial t}\cdot \rm d\vec{S} ∮E⋅dl=−∬∂t∂B⋅dS
安培-麦克斯韦定律 ∮ B ⃗ ⋅ d l ⃗ = μ 0 ∬ ( j ⃗ + ϵ 0 ∂ E ⃗ ∂ t ) ⋅ d S ⃗ \oint\vec{B}\cdot \rm d\vec{l} = \mu_0\iint(\vec{j} + \epsilon_0{\partial\vec{E} \over {\partial t}})\cdot \rm d \vec{S} ∮B⋅dl=μ0∬(j+ϵ0∂t∂E)⋅dS
高斯电场定理 ∇ ⃗ ⋅ E ⃗ = ρ ϵ 0 \vec\nabla\cdot\vec{E} = {\rho \over \epsilon_0} ∇⋅E=ϵ0ρ
高斯磁场定理 ∇ ⃗ ⋅ B ⃗ = 0 \vec\nabla\cdot\vec{B} = 0 ∇⋅B=0
法拉第定律 ∇ ⃗ × E ⃗ = − ∂ B ⃗ ∂ t \vec\nabla\times\vec{E} = -{\partial\vec{B}\over\partial t} ∇×E=−∂t∂B
安培-麦克斯韦定律 ∇ ⃗ × B ⃗ = μ 0 ( j ⃗ + ϵ 0 ∂ E ⃗ ∂ t ) \vec\nabla\times\vec{B} = \mu_0(\vec{j} + \epsilon_0{\partial\vec{E} \over {\partial t}}) ∇×B=μ0(j+ϵ0∂t∂E)
a ⃗ × ( b ⃗ × c ⃗ ) = b ⃗ ( a ⃗ ⋅ c ⃗ ) − c ⃗ ( a ⃗ ⋅ b ⃗ ) \vec a\times(\vec b\times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b) a×(b×c)=b(a⋅c)−c(a⋅b)
由高斯磁场定律,数学上可以知道存在一个 A ⃗ \vec A A满足
B ⃗ = ∇ ⃗ × A ⃗ \vec B = \vec \nabla \times \vec A B=∇×A
物理上将 A ⃗ \vec A A称为电磁场的矢势。由此,法拉第定律可以写为
∇ ⃗ × E ⃗ = − ∂ ∂ t ( ∇ ⃗ × A ⃗ ) = − ∇ ⃗ × ∂ A ⃗ ∂ t \vec\nabla\times\vec{E} = -{\partial\over\partial t}(\vec \nabla \times \vec A) = -\vec \nabla \times{\partial\vec A\over\partial t} ∇×E=−∂t∂(∇×A)=−∇×∂t∂A即 ∇ ⃗ × ( E ⃗ + ∂ A ⃗ ∂ t ) = 0 \vec \nabla \times(\vec E + {\partial\vec A\over\partial t}) = 0 ∇×(E+∂t∂A)=0,数学上可以知道存在一个 ϕ \phi ϕ满足 E ⃗ + ∂ A ⃗ ∂ t = − ∇ ⃗ ϕ \vec E + {\partial\vec A\over\partial t} = -\vec\nabla\phi E+∂t∂A=−∇ϕ,因此
E ⃗ = − ∇ ⃗ ϕ − ∂ A ⃗ ∂ t \vec E = -\vec\nabla\phi - {\partial\vec A\over\partial t} E=−∇ϕ−∂t∂A
物理上将 ϕ \phi ϕ称为电磁场的标势。这样,安培-麦克斯韦定律就可以写为
∇ ⃗ × ( ∇ ⃗ × A ⃗ ) = ∇ ⃗ ( ∇ ⃗ ⋅ A ⃗ ) − ∇ ⃗ 2 A ⃗ = μ 0 j ⃗ − μ 0 ϵ 0 ∇ ⃗ ∂ ϕ ∂ t − μ 0 ϵ 0 ∂ 2 A ⃗ ∂ t 2 \vec\nabla\times(\vec\nabla\times\vec A) = \vec\nabla(\vec\nabla\cdot\vec A) - \vec\nabla^2\vec A = \mu_0\vec j - \mu_0\epsilon_0\vec\nabla{\partial\phi \over \partial t} - \mu_0\epsilon_0{\partial^2\vec A \over \partial t^2} ∇×(∇×A)=∇(∇⋅A)−∇2A=μ0j−μ0ϵ0∇∂t∂ϕ−μ0ϵ0∂t2∂2A
整理可得
∇ 2 A ⃗ − μ 0 ϵ 0 ∂ 2 A ⃗ ∂ t 2 = − μ 0 j ⃗ + ∇ ⃗ ( ∇ ⃗ ⋅ A ⃗ + μ 0 ϵ 0 ∂ ϕ ∂ t ) \nabla^2\vec A - \mu_0\epsilon_0{\partial^2\vec A \over \partial t^2} = -\mu_0\vec j + \vec\nabla(\vec\nabla\cdot\vec A + \mu_0\epsilon_0{\partial\phi \over \partial t}) ∇2A−μ0ϵ0∂t2∂2A=−μ0j+∇(∇⋅A+μ0ϵ0∂t∂ϕ)
作规范变换
A ⃗ ′ = A ⃗ + ∇ ⃗ χ ϕ ⃗ ′ = ϕ ⃗ − ∂ χ ∂ t \begin{array}{ll} \vec A' &= \vec A + \vec\nabla\chi \\ \vec \phi' &= \vec \phi - {\partial\chi \over \partial t} \end{array} A′ϕ′=A+∇χ=ϕ−∂t∂χ
使得
∇ ⃗ ⋅ A ⃗ ′ + μ 0 ϵ 0 ∂ ϕ ′ ∂ t = 0 \vec\nabla\cdot\vec A' + \mu_0\epsilon_0{\partial\phi' \over \partial t} = 0 ∇⋅A′+μ0ϵ0∂t∂ϕ′=0
这被称为Lorenz2规范。据此,有
∇ 2 A ⃗ − μ 0 ϵ 0 ∂ 2 A ⃗ ∂ t 2 = − μ 0 j ⃗ \nabla^2\vec A - \mu_0\epsilon_0{\partial^2\vec A \over \partial t^2} = -\mu_0\vec j ∇2A−μ0ϵ0∂t2∂2A=−μ0j
同样地,高斯电场定理可以写为
− ∇ ⃗ ⋅ ( ∇ ⃗ ϕ + ∂ A ⃗ ∂ t ) = − ∇ ⃗ 2 ϕ − ∂ ∂ t ( ∇ ⃗ ⋅ A ⃗ ) = ρ ϵ 0 -\vec\nabla\cdot(\vec\nabla\phi + {\partial\vec A\over\partial t}) = -\vec\nabla^2\phi - {\partial \over \partial t}(\vec\nabla\cdot\vec A)={\rho \over \epsilon_0} −∇⋅(∇ϕ+∂t∂A)=−∇2ϕ−∂t∂(∇⋅A)=ϵ0ρ
根据Lorenz规范,就有
∇ ⃗ 2 ϕ − μ 0 ϵ 0 ∂ 2 ϕ ∂ t 2 = − ρ ϵ 0 \vec\nabla^2\phi - \mu_0\epsilon_0{\partial^2\phi \over \partial t^2} = -{\rho \over \epsilon_0} ∇2ϕ−μ0ϵ0∂t2∂2ϕ=−ϵ0ρ
引入达朗贝尔算子 □ 2 = − 1 c 2 ∂ 2 ∂ t 2 + ∇ ⃗ 2 \Box^2 = -\frac{1}{c^2}\frac{\partial^2}{\partial t^2} + \vec\nabla^2 □2=−c21∂t2∂2+∇2,麦克斯韦方程组的势场形式可以表示为
B ⃗ = ∇ ⃗ × A ⃗ E ⃗ = − ∇ ⃗ ϕ − ∂ A ⃗ ∂ t □ 2 A ⃗ = − μ 0 j ⃗ □ 2 ϕ = − ρ ϵ 0 \begin{array}{ll} \vec B &= \vec \nabla \times \vec A \\ \vec E &= -\vec\nabla\phi - {\partial\vec A\over\partial t} \\ \Box^2\vec A &= -\mu_0\vec j \\ \Box^2\phi &= -{\rho \over \epsilon_0} \end{array} BE□2A□2ϕ=∇×A=−∇ϕ−∂t∂A=−μ0j=−ϵ0ρ
定义电磁4-势为
A a = ( − ϕ c , A ⃗ ) A_a = (-\frac{\phi}{c}, \vec A) Aa=(−cϕ,A)
则洛伦兹规范可以表示为
∂ a A a = A a , a = 0 \partial_a A^a = A^a{}_{,a} = 0 ∂aAa=Aa,a=0
其中
∂ a = ( 1 c ∂ ∂ t , ∇ ⃗ ) \partial_a = (\frac{1}{c}\frac{\partial}{\partial t}, \vec\nabla) ∂a=(c1∂t∂,∇)
定义4-电流密度为
J a = ( c ρ , j ⃗ ) J^a = (c\rho, \vec j) Ja=(cρ,j)
借此,电流的连续性方程可以简洁地表示为
∂ a J a = J a , a = 0 \partial_a J^a = J^a{}_{,a} = 0 ∂aJa=Ja,a=0
A a A_a Aa和 J a J^a Ja两者之间的关系构成麦克斯韦方程组的四维势场形式,即
∂ a ∂ a A b = □ 2 A b = − μ 0 J b = − μ 0 η a b J a \partial^a\partial_aA_b = \Box^2A_b = -\mu_0J_b = -\mu_0\eta_{ab}J^a ∂a∂aAb=□2Ab=−μ0Jb=−μ0ηabJa
只要注意到闵可夫斯基度规的符号,就可以验证这与前一节的势场方程是等价的。
再考察一下 A a A_a Aa和 E ⃗ \vec E E、 B ⃗ \vec B B的关系,不难得出如下的结论
− E x c = ∂ t A x − ∂ x A t − E y c = ∂ t A y − ∂ y A t − E z c = ∂ t A z − ∂ z A t B x = ∂ y A z − ∂ z A y − B y = ∂ x A z − ∂ z A x B z = ∂ x A y − ∂ y A x \begin{array}{ll} -\frac{E_x}{c} &= \partial_tA_x - \partial_xA_t \\ -\frac{E_y}{c} &= \partial_tA_y - \partial_yA_t \\ -\frac{E_z}{c} &= \partial_tA_z - \partial_zA_t \\ B_x &= \partial_yA_z - \partial_zA_y \\ -B_y &= \partial_xA_z - \partial_zA_x \\ B_z &= \partial_xA_y - \partial_yA_x \end{array} −cEx−cEy−cEzBx−ByBz=∂tAx−∂xAt=∂tAy−∂yAt=∂tAz−∂zAt=∂yAz−∂zAy=∂xAz−∂zAx=∂xAy−∂yAx
这启发我们定义如下的二阶、反称、协变张量
F a b = ∂ a A b − ∂ b A a F_{ab} = \partial_aA_b - \partial_bA_a Fab=∂aAb−∂bAa
称为电磁张量,其分量形式为
F μ ν = [ 0 − E x c − E y c − E z c E x c 0 B z − B y E y c − B z 0 B x E z c B y − B x 0 ] F_{\mu\nu} = \begin{bmatrix} 0 &-\frac{E_x}{c} &-\frac{E_y}{c} &-\frac{E_z}{c} \\ \frac{E_x}{c} &0 &B_z &-B_y \\ \frac{E_y}{c} &-B_z &0 &B_x \\ \frac{E_z}{c} &B_y &-B_x &0 \end{bmatrix} Fμν=⎣⎢⎢⎡0cExcEycEz−cEx0−BzBy−cEyBz0−Bx−cEz−ByBx0⎦⎥⎥⎤
使用电磁张量,麦克斯韦方程组可以简洁地表述为
∂ b F a b = F a b , b = μ 0 J a ∂ [ c F a b ] = F [ a b , c ] = 0 \begin{array}{lll} \partial_bF^{ab} &= F^{ab}{}_{,b} &= \mu_0J^a \\ \partial_{[c}F_{ab]} &= F_{[ab,c]} &= 0 \end{array} ∂bFab∂[cFab]=Fab,b=F[ab,c]=μ0Ja=0
我们对上式两边求导可得
□ 2 F a b = ∂ c ∂ c F a b = − ∂ c ∂ b F c a − ∂ c ∂ a F b c \Box^2F_{ab} = \partial^c\partial_cF_{ab} = -\partial^c\partial_bF_{ca} -\partial^c\partial_aF_{bc} □2Fab=∂c∂cFab=−∂c∂bFca−∂c∂aFbc
注意到 ∂ c F c a = − ∂ c F a c = − u 0 J a \partial^cF_{ca} = -\partial^cF_{ac} = -u_0J_a ∂cFca=−∂cFac=−u0Ja,就有
□ 2 F a b = − μ 0 ( ∂ a J b − ∂ b J a ) = 2 μ 0 J [ a , b ] \Box^2F_{ab} = -\mu_0(\partial_aJ_b - \partial_bJ_a) = 2\mu_0J_{[a,b]} □2Fab=−μ0(∂aJb−∂bJa)=2μ0J[a,b]
在无源( J a = 0 J^a = 0 Ja=0)的情况下,上述方程就是波动方程
□ 2 F a b = 0 \Box^2F_{ab} = 0 □2Fab=0
电磁张量的Hodge对偶为
∗ F μ ν = [ 0 B x B y B z − B x 0 E z c − E y c − B y − E z c 0 E x c − B z E y c − E x c 0 ] ^*F_{\mu\nu} = \begin{bmatrix} 0 & B_x & B_y & B_z \\ -B_x & 0 & \frac{E_z}{c} & -\frac{E_y}{c} \\ -B_y & -\frac{E_z}{c} & 0 & \frac{E_x}{c} \\ -B_z & \frac{E_y}{c} & -\frac{E_x}{c} & 0 \end{bmatrix} ∗Fμν=⎣⎢⎢⎡0−Bx−By−BzBx0−cEzcEyBycEz0−cExBz−cEycEx0⎦⎥⎥⎤
借此,麦克斯韦方程可表示为
d F = 0 ∗ d ∗ F = μ 0 J \begin{array}{ll} dF &= 0\\ ^*d^*F &= \mu_0J \end{array} dF∗d∗F=0=μ0J
1 v 2 ∂ 2 f ∂ t 2 = ∇ ⃗ 2 f {1\over v^2}{\partial^2 f \over \partial t^2} = \vec\nabla^2f v21∂t2∂2f=∇2f
对法拉第定律两边求旋度
∇ ⃗ × ( ∇ ⃗ × E ⃗ ) = ∇ ⃗ ( ∇ ⃗ ⋅ E ⃗ ) − ∇ ⃗ 2 E ⃗ = − ∂ ∂ t ( ∇ ⃗ × B ⃗ ) \vec\nabla\times(\vec\nabla\times\vec{E}) = \vec\nabla(\vec\nabla\cdot\vec{E}) - \vec\nabla^2 \vec E = -{\partial\over\partial t}(\vec\nabla\times\vec{B}) ∇×(∇×E)=∇(∇⋅E)−∇2E=−∂t∂(∇×B)
而对于无源场, ∇ ⃗ ⋅ E ⃗ = 0 \vec\nabla\cdot\vec{E} = 0 ∇⋅E=0, ∇ ⃗ × B ⃗ = μ 0 ϵ 0 ∂ E ⃗ ∂ t \vec\nabla\times\vec{B} = \mu_0\epsilon_0{\partial\vec{E} \over {\partial t}} ∇×B=μ0ϵ0∂t∂E所以
∇ ⃗ 2 E ⃗ = μ 0 ϵ 0 ∂ 2 E ⃗ ∂ t 2 \vec\nabla^2 \vec E = \mu_0\epsilon_0{\partial^2\vec{E} \over {\partial t^2}} ∇2E=μ0ϵ0∂t2∂2E
同理可得
∇ ⃗ 2 B ⃗ = μ 0 ϵ 0 ∂ 2 B ⃗ ∂ t 2 \vec\nabla^2 \vec B = \mu_0\epsilon_0{\partial^2\vec{B} \over {\partial t^2}} ∇2B=μ0ϵ0∂t2∂2B
与波动方程对照,有 1 v 2 = μ 0 ϵ 0 {1 \over v^2} = \mu_0\epsilon_0 v21=μ0ϵ0,因此
v = 1 μ 0 ϵ 0 = 2.9979 × 1 0 8 m / s = c v = {1 \over \sqrt{\mu_0\epsilon_0}} = 2.9979 \times 10^8 m / s = c v=μ0ϵ01=2.9979×108m/s=c
这就是真空中的光速。
∇ ⃗ ⋅ ( A ⃗ × B ⃗ ) = B ⃗ ⋅ ( ∇ ⃗ × A ⃗ ) − A ⃗ × ( ∇ ⃗ × B ⃗ ) \vec\nabla\cdot(\vec A \times \vec B) = \vec B\cdot(\vec\nabla\times \vec A) - \vec A \times(\vec\nabla\times\vec B) ∇⋅(A×B)=B⋅(∇×A)−A×(∇×B)
∇ ⃗ ( A ⃗ ⋅ B ⃗ ) = A ⃗ × ( ∇ ⃗ × B ⃗ ) + B ⃗ × ( ∇ ⃗ × A ⃗ ) + ( A ⃗ ⋅ ∇ ⃗ ) B ⃗ + ( B ⃗ ⋅ ∇ ⃗ ) A ⃗ \vec\nabla(\vec A\cdot \vec B) = \vec A\times(\vec \nabla \times \vec B) + \vec B\times(\vec \nabla \times \vec A) + (\vec A \cdot \vec\nabla)\vec B + (\vec B \cdot \vec \nabla)\vec A ∇(A⋅B)=A×(∇×B)+B×(∇×A)+(A⋅∇)B+(B⋅∇)A
当 A ⃗ = B ⃗ \vec A = \vec B A=B时就有
A ⃗ × ( ∇ ⃗ × A ⃗ ) = 1 2 ∇ ⃗ A ⃗ 2 − ( A ⃗ ⋅ ∇ ⃗ ) A ⃗ \vec A \times(\vec \nabla\times \vec A) = {1\over 2}\vec\nabla\vec A^2 - (\vec A\cdot\vec \nabla)\vec A A×(∇×A)=21∇A2−(A⋅∇)A
参考Nabla算符的运算律以及常用公式
做功为系统增加能量,能量的变化率就是做功的功率。在电磁场中,仅电场力对带电体做功,因此功率为
∭ E ⃗ ρ d V ⋅ u ⃗ = ∭ E ⃗ ⋅ j ⃗ d V \iiint\vec E\rho \rm dV\cdot\vec u = \iiint\vec E\cdot\vec j\rm dV ∭EρdV⋅u=∭E⋅jdV
而 E ⃗ ⋅ j ⃗ = E ⃗ ⋅ ( 1 μ 0 ∇ ⃗ × B ⃗ ) − E ⃗ ⋅ ϵ 0 ∂ E ⃗ ∂ t = 1 μ 0 ∇ ⃗ ⋅ ( B ⃗ × E ⃗ ) + 1 μ 0 B ⃗ × ( ∇ ⃗ × E ⃗ ) − E ⃗ ⋅ ϵ 0 ∂ E ⃗ ∂ t = − 1 μ 0 ∇ ⃗ ⋅ ( E ⃗ × B ⃗ ) − 1 μ 0 B ⃗ ⋅ ∂ B ⃗ ∂ t − E ⃗ ⋅ ϵ 0 ∂ E ⃗ ∂ t = − ∂ ∂ t ( ϵ 0 E ⃗ 2 + 1 μ 0 B ⃗ 2 2 ) − 1 μ 0 ∇ ⃗ ⋅ ( E ⃗ × B ⃗ ) \vec E\cdot \vec j = \vec E \cdot ({1 \over \mu_0}\vec\nabla\times\vec B) - \vec E\cdot\epsilon_0{\partial \vec E \over \partial t} = {1 \over \mu_0}\vec\nabla\cdot(\vec B \times \vec E) + {1 \over \mu_0}\vec B \times(\vec \nabla\times\vec E) - \vec E\cdot\epsilon_0{\partial \vec E \over \partial t} = -{1 \over \mu_0}\vec\nabla\cdot(\vec E \times \vec B) - {1 \over \mu_0}\vec B\cdot{\partial \vec B \over \partial t} - \vec E\cdot\epsilon_0{\partial \vec E \over \partial t} = -{\partial \over \partial t}({\epsilon_0\vec E^2 + {1 \over \mu_0}\vec B^2 \over 2}) -{1 \over \mu_0}\vec\nabla\cdot(\vec E \times \vec B) E⋅j=E⋅(μ01∇×B)−E⋅ϵ0∂t∂E=μ01∇⋅(B×E)+μ01B×(∇×E)−E⋅ϵ0∂t∂E=−μ01∇⋅(E×B)−μ01B⋅∂t∂B−E⋅ϵ0∂t∂E=−∂t∂(2ϵ0E2+μ01B2)−μ01∇⋅(E×B),带入上式并使用高斯定理,有
∭ E ⃗ ρ d V ⋅ u ⃗ = − d d t ∭ ( μ 0 ϵ 0 E ⃗ 2 + B ⃗ 2 2 μ 0 ) d V − ∯ 1 μ 0 ( E ⃗ × B ⃗ ) ⋅ d S ⃗ \iiint\vec E\rho \rm dV\cdot\vec u = -{\rm d \over \rm dt}\iiint({\mu_0\epsilon_0\vec E^2 + \vec B^2 \over 2\mu_0})\rm dV - \oiint{1 \over \mu_0}(\vec E \times \vec B)\cdot\rm d\vec S ∭EρdV⋅u=−dtd∭(2μ0μ0ϵ0E2+B2)dV−∬μ01(E×B)⋅dS
物理上称 μ 0 ϵ 0 E ⃗ 2 + B ⃗ 2 2 μ 0 {\mu_0\epsilon_0\vec E^2 + \vec B^2 \over 2\mu_0} 2μ0μ0ϵ0E2+B2为电磁场的能量密度, S ⃗ = 1 μ 0 ( E ⃗ × B ⃗ ) \vec S = {1 \over \mu_0}(\vec E \times \vec B) S=μ01(E×B)为电磁场的能流密度矢量,也叫坡印廷矢量。
根据牛顿第二定律,物体受到的力等于其动量的变化率。在电磁场中,物体受到的力为
∭ ( ρ d V ) ( E ⃗ + u ⃗ × B ⃗ ) = ∭ ( E ⃗ ρ + j ⃗ × B ⃗ ) d V \iiint(\rho\rm dV)(\vec E + \vec u \times \vec B) = \iiint(\vec E\rho + \vec j \times \vec B)\rm dV ∭(ρdV)(E+u×B)=∭(Eρ+j×B)dV
而 E ⃗ ρ + j ⃗ × B ⃗ = ϵ 0 E ⃗ ( ∇ ⃗ ⋅ E ⃗ ) − 1 μ 0 B ⃗ × ( ∇ ⃗ × B ⃗ ) + ϵ 0 B ⃗ × ∂ E ⃗ ∂ t \vec E\rho + \vec j \times \vec B = \epsilon_0\vec E(\vec\nabla\cdot\vec E) - {1\over \mu_0}\vec B\times(\vec\nabla\times \vec B) + \epsilon_0\vec B\times{\partial \vec E \over \partial t} Eρ+j×B=ϵ0E(∇⋅E)−μ01B×(∇×B)+ϵ0B×∂t∂E
考虑到 ∂ ∂ t ( E ⃗ × B ⃗ ) = E ⃗ × ∂ B ⃗ ∂ t − B ⃗ × ∂ E ⃗ ∂ t {\partial \over \partial t}(\vec E \times \vec B) = \vec E \times {\partial \vec B \over \partial t} - \vec B \times {\partial \vec E \over \partial t} ∂t∂(E×B)=E×∂t∂B−B×∂t∂E,代入前面就有 E ⃗ ρ + j ⃗ × B ⃗ = ϵ 0 E ⃗ ( ∇ ⃗ ⋅ E ⃗ ) − 1 μ 0 B ⃗ × ( ∇ ⃗ × B ⃗ ) − ϵ 0 E ⃗ × ( ∇ ⃗ × E ⃗ ) − ϵ 0 ∂ ∂ t ( E ⃗ × B ⃗ ) \vec E\rho + \vec j \times \vec B = \epsilon_0\vec E(\vec\nabla\cdot\vec E) - {1\over \mu_0}\vec B\times(\vec\nabla\times \vec B) - \epsilon_0\vec E\times(\vec \nabla\times\vec E) - \epsilon_0{\partial \over \partial t}(\vec E \times \vec B) Eρ+j×B=ϵ0E(∇⋅E)−μ01B×(∇×B)−ϵ0E×(∇×E)−ϵ0∂t∂(E×B),整理得
E ⃗ ρ + j ⃗ × B ⃗ = ϵ 0 [ E ⃗ ( ∇ ⃗ ⋅ E ⃗ ) − E ⃗ × ( ∇ ⃗ × E ⃗ ) ] + 1 μ 0 [ B ⃗ ( ∇ ⃗ ⋅ B ⃗ ) − B ⃗ × ( ∇ ⃗ × B ⃗ ) ] − ϵ 0 ∂ ∂ t ( E ⃗ × B ⃗ ) \vec E\rho + \vec j \times \vec B = \epsilon_0[\vec E(\vec\nabla\cdot\vec E) - \vec E\times(\vec \nabla\times\vec E)] + {1 \over \mu_0}[\vec B(\vec\nabla\cdot\vec B) - \vec B\times(\vec \nabla\times\vec B)] - \epsilon_0{\partial \over \partial t}(\vec E \times \vec B) Eρ+j×B=ϵ0[E(∇⋅E)−E×(∇×E)]+μ01[B(∇⋅B)−B×(∇×B)]−ϵ0∂t∂(E×B)
利用前面的公式,将上式变形为
E ⃗ ρ + j ⃗ × B ⃗ = ϵ 0 [ E ⃗ ( ∇ ⃗ ⋅ E ⃗ ) + ( E ⃗ ⋅ ∇ ⃗ ) E ⃗ − 1 2 ∇ ⃗ E ⃗ 2 ] + 1 μ 0 [ B ⃗ ( ∇ ⃗ ⋅ B ⃗ ) + ( B ⃗ ⋅ ∇ ⃗ ) B ⃗ − 1 2 ∇ ⃗ B ⃗ 2 ] − ϵ 0 ∂ ∂ t ( E ⃗ × B ⃗ ) \vec E\rho + \vec j \times \vec B = \epsilon_0[\vec E(\vec\nabla\cdot\vec E) + (\vec E\cdot\vec \nabla)\vec E - {1\over 2}\vec\nabla\vec E^2] + {1 \over \mu_0}[\vec B(\vec\nabla\cdot\vec B) + (\vec B\cdot\vec \nabla)\vec B - {1\over 2}\vec\nabla\vec B^2] - \epsilon_0{\partial \over \partial t}(\vec E \times \vec B) Eρ+j×B=ϵ0[E(∇⋅E)+(E⋅∇)E−21∇E2]+μ01[B(∇⋅B)+(B⋅∇)B−21∇B2]−ϵ0∂t∂(E×B)
如果使用张量记号,上式可化简为
E ⃗ ρ + j ⃗ × B ⃗ = ∇ ⃗ ⋅ [ ϵ 0 ( E a E b − 1 2 δ a b E 2 ) + 1 μ 0 ( B a B b − 1 2 δ a b B 2 ) ] − ϵ 0 ∂ ∂ t ( E ⃗ × B ⃗ ) \vec E\rho + \vec j \times \vec B = \vec\nabla\cdot[\epsilon_0(E_aE_b - {1\over 2}\delta_{ab}E^2) + {1\over \mu_0}(B_aB_b - {1\over 2}\delta_{ab}B^2)]- \epsilon_0{\partial \over \partial t}(\vec E \times \vec B) Eρ+j×B=∇⋅[ϵ0(EaEb−21δabE2)+μ01(BaBb−21δabB2)]−ϵ0∂t∂(E×B)
引入麦克斯韦应力张量
σ a b : = ϵ 0 ( E a E b − 1 2 δ a b E 2 ) + 1 μ 0 ( B a B b − 1 2 δ a b B 2 ) \sigma_{ab} := \epsilon_0(E_aE_b - {1\over 2}\delta_{ab}E^2) + {1\over \mu_0}(B_aB_b - {1\over 2}\delta_{ab}B^2) σab:=ϵ0(EaEb−21δabE2)+μ01(BaBb−21δabB2)
就有
∭ ( ρ d V ) ( E ⃗ + u ⃗ × B ⃗ ) = − d d t ∭ ϵ 0 ( E ⃗ × B ⃗ ) d V + ∯ σ a b ⋅ d S ⃗ \iiint(\rho\rm dV)(\vec E + \vec u \times \vec B) = -{\rm d \over \rm dt}\iiint\epsilon_0(\vec E \times \vec B)\rm dV + \oiint \sigma_{ab}\cdot\rm d\vec S ∭(ρdV)(E+u×B)=−dtd∭ϵ0(E×B)dV+∬σab⋅dS
物理上称 ϵ 0 ( E ⃗ × B ⃗ ) \epsilon_0(\vec E \times \vec B) ϵ0(E×B)为电磁场的动量密度, σ a b \sigma_{ab} σab为电磁场的动量流密度张量。
能动张量是一个二阶张量 T a b T_{ab} Tab,用来描述4-动量的 a a a分量通过 b b b坐标平面的通量。其中 T 00 T_{00} T00代表能量密度, T 0 i T_{0i} T0i代表能量通过 i i i坐标平面的通量,它等于 T i 0 T_{i0} Ti0即动量的第 i i i分量之密度。而 T i j T_{ij} Tij就是动量的第 i i i分量通过 j j j坐标平面的通量,也就是动量流密度。具体到电磁场,其分量如下
T μ ν = [ 1 2 ( ϵ 0 E 2 + 1 μ 0 B 2 ) S x c S y c S z c S x c − σ x x − σ x y − σ x z S y c − σ y x − σ y y − σ y z S z c − σ z x − σ z y − σ z z ] = − 1 2 μ 0 ( F μ σ F σ ν + ∗ F μ σ ∗ F σ ν ) T_{\mu\nu} = \begin{bmatrix} \frac{1}{2}(\epsilon_0E^2 + \frac{1}{\mu_0}B^2) & \frac{S_x}{c} & \frac{S_y}{c} & \frac{S_z}{c} \\ \frac{S_x}{c} & -\sigma_{xx} & -\sigma_{xy} & -\sigma_{xz} \\ \frac{S_y}{c} & -\sigma_{yx} & -\sigma_{yy} & -\sigma_{yz} \\ \frac{S_z}{c} & -\sigma_{zx} & -\sigma_{zy} & -\sigma_{zz} \end{bmatrix} = -\frac{1}{2\mu_0}(F_{\mu\sigma}F^\sigma{}_\nu + {}^*F_{\mu\sigma}{}^*F^\sigma{}_\nu) Tμν=⎣⎢⎢⎢⎡21(ϵ0E2+μ01B2)cSxcSycSzcSx−σxx−σyx−σzxcSy−σxy−σ