这是一份liverpool利物浦大学MATH427的成功案例

We will also find it convenient to work with the vector potential, $\boldsymbol{A}$, so that
$$
\boldsymbol{B}=\nabla \times \boldsymbol{A}
$$
and
$$
\boldsymbol{E}=-\nabla \Phi-\frac{1}{c} \frac{\partial \boldsymbol{A}}{\partial t},
$$
in which the scalar potential is $\Phi$, using the Coulomb gauge, so $\nabla \cdot \boldsymbol{A}=0$.
We will also need at times to work with the energy density and energy flux of electromagnetic field. In cgs units the energy densities $W_{E}$ and $W_{B}$ of the electric and magnetic fields, respectively, are
$$
W_{E}=\frac{E^{2}}{8 \pi}
$$
and
$$
W_{B}=\frac{B^{2}}{8 \pi} .
$$
Note that when one averages over many cycles of a fluctuating field to obtain an averaged energy density, these quantities are divided by 2 . This is also the case for the Poynting flux,
$$
\boldsymbol{S}=\frac{v_{g}}{4 \pi} \boldsymbol{E} \times \boldsymbol{B}
$$
in which the group velocity of the wave is $v_{g}$.

MATH427 COURSE NOTES ：

$$
\frac{\partial \rho}{\partial t}+\nabla \cdot \rho \boldsymbol{u}=0
$$
momentum,
$$
\rho\left(\frac{\partial \boldsymbol{u}}{\partial t}\right)=-\nabla p+\frac{\boldsymbol{J} \times \boldsymbol{B}}{c}
$$
Ohm’s law,
$$
\boldsymbol{E}+\frac{\boldsymbol{u} \times \boldsymbol{B}}{c}=\eta \boldsymbol{J}
$$
Faraday’s law,
$$
\nabla \times \boldsymbol{E}=-\frac{1}{c} \frac{\partial \boldsymbol{B}}{\partial t}
$$
and Ampere’s law,
$$
c \nabla \times \boldsymbol{B}=4 \pi \boldsymbol{J}
$$