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电磁场的能量守恒和动量守恒_电磁场能量守恒方程

作者：煮酒与君饮 | 2024-08-09 06:31:24

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电磁场能量守恒方程

麦克斯韦方程组

{\begin{cases} \nabla \cdot E = \frac{ρ}{ε_{0}} \\ \nabla \times E = - \frac{\partial B}{\partial t} \\ \nabla \cdot B = 0 \\ \nabla \times B = μ_{0} j \end{cases}

$\begin{cases} \nabla\cdot\mathbf E=\dfrac{\rho}{\varepsilon_0}\\ \nabla\times\mathbf E=-\dfrac{\partial B}{\partial t}\\ \nabla\cdot\mathbf B=0\\ \nabla\times\mathbf B=\mu_0\mathbf j \end{cases}$

\nabla \cdot (\nabla \times E) = \nabla \cdot (- \frac{\partial B}{\partial t}) = - \frac{\partial (\nabla \cdot B)}{\partial t} = 0

$\nabla\cdot(\nabla\times\mathbf E)=\nabla\cdot(-\dfrac{\partial B}{\partial t})=-\dfrac{\partial (\nabla\cdot \mathbf B)}{\partial t}=0$

这说明前两个方程是合理、自洽的.但是

\nabla \cdot (\nabla \times B) = \nabla \cdot (μ_{0} j) = μ_{0} (- \frac{\partial ρ}{\partial t})

$\nabla\cdot(\nabla\times\mathbf B)=\nabla\cdot(\mu_0\mathbf j)=\mu_0(-\dfrac{\partial\rho}{\partial t})$

结果并不总是等于 $0$ ,为此，Maxwell引入位移电流 $\mathbf j_d$ ，即

\nabla \times B = μ_{0} ({j + j}_{d})

$\nabla\times\mathbf B=\mu_0(\mathbf{ j+j}_d)$
由此

\nabla \cdot (\nabla \times B) = \nabla \cdot [μ_{0} ({j + j}_{d})] = μ_{0} [\nabla \cdot j_{d} - \frac{\partial ρ}{\partial t}] = μ_{0} [\nabla \cdot j_{d} - ε_{0} \frac{\partial (\nabla \cdot E)}{\partial t}] = 0 \Rightarrow j_{d} \equiv ε_{0} \frac{\partial E}{\partial t}

$\nabla\cdot(\nabla\times\mathbf B)=\nabla\cdot[\mu_0(\mathbf{ j+j}_d)]=\mu_0[\nabla\cdot \mathbf j_d-\dfrac{\partial\rho}{\partial t}]=\mu_0[\nabla\cdot \mathbf j_d-\varepsilon_0\dfrac{\partial(\nabla\cdot\mathbf E)}{\partial t}]=0\\\Rightarrow\mathbf j_d\equiv\varepsilon_0\dfrac{\partial \mathbf E}{\partial t}$

所以方程组修改为

{\begin{cases} \nabla \cdot E = \frac{ρ}{ε_{0}} \\ \nabla \times E = - \frac{\partial B}{\partial t} \\ \nabla \cdot B = 0 \\ \nabla \times B = μ_{0} j + ε_{0} μ_{0} \frac{\partial E}{\partial t} \end{cases}

$\begin{cases} \nabla\cdot\mathbf E=\dfrac{\rho}{\varepsilon_0}\\ \nabla\times\mathbf E=-\dfrac{\partial B}{\partial t}\\ \nabla\cdot\mathbf B=0\\ \nabla\times\mathbf B=\mu_0\mathbf j+\varepsilon_0\mu_0\dfrac{\partial \mathbf E}{\partial t} \end{cases}$

洛伦兹力方程

宏观

F = q E + q v \times B

$\mathbf F=q\mathbf E+q\mathbf{v\times B}$
微观

f = ρ E + j \times B

$\mathbf f=\rho\mathbf E+\mathbf j\times \mathbf B$
Maxwell方程组和洛伦兹力方程一起构成了经典电动力学的基础。

能量守恒

能量守恒的目标形式是

∭ W d τ - \frac{d}{d t} ∭ ω d τ - \oint S \cdot d σ = ∭ W d τ - \frac{d}{d t} ∭ ω d τ - ∭ \nabla \cdot S d τ \Rightarrow W = - \frac{d ω}{d t} - \nabla \cdot S

$\iiint W\mathrm d\tau-\dfrac{\mathrm d}{\mathrm dt}\iiint\omega\mathrm d\tau-\oint\mathbf S\cdot\mathrm d\sigma\\=\iiint W\mathrm d\tau-\dfrac{\mathrm d}{\mathrm dt}\iiint\omega\mathrm d\tau-\iiint\nabla\cdot\mathbf S\mathrm d\tau\\\Rightarrow W=-\dfrac{\mathrm d\omega}{\mathrm dt}-\nabla\cdot\mathbf S$

其中 $W$ 是一定空间 $V$ 内在合体能量增加率， $\omega$ 是电磁场能量密度， $\mathbf S$ 是能流密度.

\begin{matrix} W = E \cdot j \\ = E \cdot (\frac{1}{μ_{0}} \nabla \times B - ε_{0} \frac{\partial E}{\partial t}) \\ = \frac{1}{μ_{0}} E \cdot (\nabla \times B) - ε_{0} E \cdot \frac{\partial E}{\partial t} \\ = \frac{1}{μ_{0}} [- \nabla \cdot (E \times B) + B \cdot (\nabla \times E)] - ε_{0} E \cdot \frac{\partial E}{\partial t} \\ = - \frac{1}{μ_{0}} \nabla \cdot (E \times B) + \frac{1}{μ_{0}} B \cdot (\nabla \times E) - ε_{0} E \cdot \frac{\partial E}{\partial t} \\ = - \frac{1}{μ_{0}} \nabla \cdot (E \times B) + \frac{1}{μ_{0}} B \cdot (- \frac{\partial B}{\partial t}) - ε_{0} E \cdot \frac{\partial E}{\partial t} \\ = - \nabla \cdot [\frac{1}{μ_{0}} (E \times B)] - \frac{1}{2} \frac{\partial}{\partial t} [\frac{1}{μ_{0}} B^{2} + ε_{0} E^{2}] \end{matrix}

$\begin{array} &W=\mathbf{E\cdot j} \\=\mathbf E\cdot(\dfrac1{\mu_0}\nabla\times \mathbf B-\varepsilon_0\dfrac{\partial\mathbf E}{\partial t}) \\=\dfrac1{\mu_0}\mathbf E\cdot(\nabla\times \mathbf B)-\varepsilon_0\mathbf E\cdot\dfrac{\partial\mathbf E}{\partial t} \\=\dfrac1{\mu_0}[-\nabla\cdot(\mathbf E\times \mathbf B)+\mathbf B\cdot(\nabla\times \mathbf E)]-\varepsilon_0\mathbf E\cdot\dfrac{\partial\mathbf E}{\partial t} \\=-\dfrac1{\mu_0}\nabla\cdot(\mathbf E\times \mathbf B)+\dfrac1{\mu_0}\mathbf B\cdot(\nabla\times \mathbf E)-\varepsilon_0\mathbf E\cdot\dfrac{\partial\mathbf E}{\partial t} \\=-\dfrac1{\mu_0}\nabla\cdot(\mathbf E\times \mathbf B)+\dfrac1{\mu_0}\mathbf B\cdot(-\dfrac{\partial\mathbf B}{\partial t})-\varepsilon_0\mathbf E\cdot\dfrac{\partial\mathbf E}{\partial t} \\=-\nabla\cdot\left[\dfrac1{\mu_0}(\mathbf E\times \mathbf B)\right]-\dfrac12\dfrac{\partial}{\partial t}\left[\dfrac1{\mu_0} B^2+\varepsilon_0 E^2\right] \end{array}$

比较一下理想形式，可以得到

S = \frac{1}{μ_{0}} (E \times B) ω = \frac{1}{2} (\frac{1}{μ_{0}} B^{2} + ε_{0} E^{2})

$\mathbf S=\dfrac1{\mu_0}(\mathbf E\times \mathbf B)\\\omega=\dfrac12\left(\dfrac1{\mu_0} B^2+\varepsilon_0 E^2\right)$

动量守恒

动量守恒的目标形式是

∭ f d τ - \frac{d}{d t} ∭ g d τ - \oint \vec{T} \cdot d σ = ∭ f d τ - \frac{d}{d t} ∭ g d τ - ∭ \nabla \cdot \vec{T} d τ \Rightarrow f = - \frac{d g}{d t} - \nabla \cdot \vec{T}

$\iiint \mathbf f\mathrm d\tau-\dfrac{\mathrm d}{\mathrm dt}\iiint\mathbf g\mathrm d\tau-\oint\vec{\mathbf T}\cdot\mathrm d\sigma\\=\iiint \mathbf f\mathrm d\tau-\dfrac{\mathrm d}{\mathrm dt}\iiint\mathbf g\mathrm d\tau-\iiint\nabla\cdot\vec{\mathbf T}\mathrm d\tau\\\Rightarrow \mathbf f=-\dfrac{\mathrm d\mathbf g}{\mathrm dt}-\nabla\cdot\vec{\mathbf T}$

其中 $\mathbf f$ 是一定空间 $V$ 内在合体动能增加率， $\mathbf g$ 是电磁场动量密度， $\vec{\mathbf T}$ 是动量流密度.

\begin{matrix} f = ρ E + j \times B \\ = ε_{0} (\nabla \cdot E) E + (\frac{1}{μ_{0}} \nabla \times B - ε_{0} \frac{\partial E}{\partial t}) \times B \\ = ε_{0} (\nabla \cdot E) E + \frac{1}{μ_{0}} (\nabla \times B) \times B - ε_{0} \frac{\partial E}{\partial t} \times B \end{matrix}

$\begin{array} &\mathbf f=\rho\mathbf E+\mathbf j\times \mathbf B \\=\varepsilon_0(\nabla\cdot \mathbf E)\mathbf E+(\dfrac1{\mu_0}\nabla\times \mathbf B-\varepsilon_0\dfrac{\partial\mathbf E}{\partial t})\times \mathbf B \\=\varepsilon_0(\nabla\cdot \mathbf E)\mathbf E+\dfrac1{\mu_0}(\nabla\times \mathbf B)\times \mathbf B-\varepsilon_0\dfrac{\partial\mathbf E}{\partial t}\times \mathbf B \end{array}$
利用Maxwell方程组的另外两个方程

0 = {\frac{1}{μ}}_{0} (\nabla \cdot B) B + ε_{0} (\nabla \times E) \times E + ε_{0} \frac{\partial B}{\partial t} \times E

$0=\dfrac1\mu_0(\nabla\cdot \mathbf B)\mathbf B+\varepsilon_0(\nabla\times \mathbf E)\times \mathbf E+\varepsilon_0\dfrac{\partial\mathbf B}{\partial t}\times \mathbf E$

两式相加，得

\begin{matrix} f = ε_{0} [(\nabla \cdot E) E + (\nabla \times E) \times E] + {\frac{1}{μ}}_{0} [(\nabla \cdot B) B + (\nabla \times B) \times B] - ε_{0} [\frac{\partial E}{\partial t} \times B - \frac{\partial B}{\partial t} \times E] \end{matrix}

$\begin{array} &\mathbf f=\varepsilon_0[(\nabla\cdot \mathbf E)\mathbf E+(\nabla\times \mathbf E)\times \mathbf E]+\dfrac1\mu_0[(\nabla\cdot \mathbf B)\mathbf B+(\nabla\times \mathbf B)\times \mathbf B]-\varepsilon_0\left[\dfrac{\partial\mathbf E}{\partial t}\times \mathbf B-\dfrac{\partial\mathbf B}{\partial t}\times \mathbf E\right] \end{array}$
因为

\nabla \cdot (E E) = (E \cdot \nabla) E + (\nabla \cdot E) E (\nabla \times E) \times E = (E \cdot \nabla) E - \frac{1}{2} \nabla E^{2}

$\nabla\cdot(\mathbf E\mathbf E)=(\mathbf E\cdot \nabla)\mathbf E+(\nabla\cdot \mathbf E)\mathbf E\\(\nabla\times \mathbf E)\times \mathbf E=(\mathbf E\cdot\nabla)\mathbf E-\dfrac12\nabla E^2$
所以

(\nabla \cdot E) E + (\nabla \times E) \times E = \nabla \cdot (E E) - \frac{1}{2} \nabla E^{2} = \nabla \cdot (E E) - \frac{1}{2} \nabla \cdot (E^{2} \vec{I}) = \nabla \cdot (E E - \frac{1}{2} E^{2} \vec{I})

$(\nabla\cdot \mathbf E)\mathbf E+(\nabla\times \mathbf E)\times \mathbf E=\nabla\cdot(\mathbf E\mathbf E)-\dfrac12\nabla E^2\\=\nabla\cdot(\mathbf E\mathbf E)-\dfrac12\nabla\cdot( E^2\vec{\mathbf I})\\=\nabla\cdot(\mathbf E\mathbf E-\dfrac12 E^2\vec{\mathbf I})$
同理

(\nabla \cdot B) B + (\nabla \times B) \times B = \nabla \cdot (B B - \frac{1}{2} B^{2} \vec{I})

$(\nabla\cdot \mathbf B)\mathbf B+(\nabla\times \mathbf B)\times \mathbf B=\nabla\cdot(\mathbf B\mathbf B-\dfrac12 B^2\vec{\mathbf I})$
又有

\frac{\partial}{\partial t} (E \times B) = \frac{\partial E}{\partial t} \times B + E \times \frac{\partial B}{\partial t} = \frac{\partial E}{\partial t} \times B - \frac{\partial B}{\partial t} \times E

$\dfrac{\partial}{\partial t}(\mathbf E\times \mathbf B)=\dfrac{\partial\mathbf E}{\partial t}\times \mathbf B+\mathbf E\times \dfrac{\partial\mathbf B}{\partial t}\\=\dfrac{\partial\mathbf E}{\partial t}\times \mathbf B- \dfrac{\partial\mathbf B}{\partial t}\times\mathbf E$

所以得到

\begin{matrix} f = - ε_{0} \frac{\partial}{\partial t} (E \times B) - \nabla \cdot (\frac{1}{2} (ε_{0} E^{2} \vec{I} + \frac{1}{μ_{0}} B^{2} \vec{I}) - ε_{0} E E - \frac{1}{μ_{0}} B B) \end{matrix}

$\begin{array} &\mathbf f=-\varepsilon_0\dfrac{\partial}{\partial t}(\mathbf E\times \mathbf B)-\nabla\cdot\left(\dfrac12( \varepsilon_0E^2\vec{\mathbf I}+\dfrac1{\mu_0}B^2\vec{\mathbf I})-\varepsilon_0\mathbf E\mathbf E-\dfrac1{\mu_0}\mathbf B\mathbf B\right) \end{array}$
比较一下理想形式，可以得到

g = ε_{0} E \times B \vec{T} = \frac{1}{2} (ε_{0} E^{2} \vec{I} + \frac{1}{μ_{0}} B^{2} \vec{I}) - ε_{0} E E - \frac{1}{μ_{0}} B B

$\mathbf {g}={\varepsilon_0 }\mathbf E\times \mathbf B\\\vec{\mathbf T}=\dfrac12( \varepsilon_0E^2\vec{\mathbf I}+\dfrac1{\mu_0}B^2\vec{\mathbf I})-\varepsilon_0\mathbf E\mathbf E-\dfrac1{\mu_0}\mathbf B\mathbf B$

g = ε_{0} μ_{0} S = \frac{S}{c^{2}}

$\mathbf g=\varepsilon_0\mu_0\mathbf S=\dfrac{\mathbf S}{c^2}$

本篇主要参考俞允强《电动力学简明教程》

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