这是一份liverpool利物浦大学PHYS103的成功案例

$$
\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \epsilon-\frac{p}{\rho^{2}}\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \rho=0
$$
In the presence of energy sources or heat transport, the right-hand side of this equation would not be zero. For an ideal gas, with $\epsilon=p /[\rho(\gamma-1)]$, this equation reduces to a particularly useful form:
$$
\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) p-c_{s}^{2}\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \rho=0 .
$$

PHYS103 COURSE NOTES ：

$$
\nabla \cdot \boldsymbol{E}=4 \pi k_{1} \rho_{c}
$$
the absence of magnetic monopoles,
$$
\nabla \cdot B=0
$$
Faraday’s law
$$
\nabla \times \boldsymbol{E}=-k_{3} \frac{\partial \boldsymbol{B}}{\partial t},
$$
and Maxwell’s generalization of Ampere’s Law
$$
\nabla \times \boldsymbol{B}=\frac{k_{2}}{k_{1} k_{3}} \frac{\partial \boldsymbol{E}}{\partial t}+4 \pi \frac{k_{2}}{k_{3}} \boldsymbol{J} .
$$