# 波浪现象 Wave Phenomena PHYS2004

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location is proportional to $\exp \left[-q \phi /\left(k_{B} T\right)\right]$. Then the charge density $\rho_{c}$ in the vicinity of an ion at $\boldsymbol{x}=0$ is
$$\rho_{c}=Z e \delta(0)-n_{c} e \exp \left[\frac{e \phi}{k_{B} T}\right]+n_{i} e Z \exp \left[\frac{-e Z \phi}{k_{B} T}\right]$$
If we assume that $|q \phi| \ll k_{B} T$ and that the plasma is quasi-neutral, then this becomes
$$\rho_{c}=Z e \delta(0)-\frac{e^{2} \phi}{k_{B} T}\left(n_{e}+n_{i} Z^{2}\right)=Z e \delta(0)-\frac{\phi}{4 \pi \lambda_{D}^{2}}$$
At this point we can write Poisson’s equation in spherical coordinates, assuming that the charges are distributed with spherical symmetry, as
$$\frac{1}{r^{2}} \frac{\mathrm{d}}{\mathrm{d} r}\left(r^{2} \frac{\mathrm{d} \phi}{\mathrm{d} r}\right)=-4 \pi Z e \delta(0)+\frac{\phi}{\lambda_{D}^{2}}$$
which (in cgs units) has the solution
$$\phi=\frac{Z e}{r} \mathrm{e}^{-r / \lambda_{D}}$$

## PHYS2004 COURSE NOTES ：

This particular equation helps one see the physics of the wave we are finding. It is a purely longitudinal wave like an acoustic wave, in which the fluctuating electric field and compression by the electron pressure both cause the electron density to vary. The first term on the right-hand side can be evaluated from Poisson’s , which gives in this case
$$\nabla \cdot \boldsymbol{E}{1}=4 \pi\left(Z e n{i o}-e n_{e o}-e n_{e 1}\right)$$
in which the first two terms in parentheses cancel because the plasma is quasineutral. Then assuming the electrons behave as a polytropic gas with index $\gamma_{e}$, we obtain a wave equation
$$\left(\frac{\partial^{2}}{\partial t^{2}}+\omega_{p e}^{2}-\frac{\gamma_{e} p_{e o}}{n_{e o} m_{e}} \nabla^{2}\right) n_{e 1}=0,$$
in which we have introduced the electron plasma frequency,
$$\omega_{p e}=\sqrt{4 \pi e^{2} n_{e o} / m_{e}}=5.64 \times 10^{4} \sqrt{n_{e o}} \mathrm{rad} / \mathrm{s},$$

# 波浪现象 Wave Phenomena PHYS103

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$$\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \epsilon-\frac{p}{\rho^{2}}\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \rho=0$$
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:
$$\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) p-c_{s}^{2}\left(\frac{\partial}{\partial t}+\boldsymbol{u} \cdot \nabla\right) \rho=0 .$$

$$\nabla \cdot \boldsymbol{E}=4 \pi k_{1} \rho_{c}$$
$$\nabla \cdot B=0$$
$$\nabla \times \boldsymbol{E}=-k_{3} \frac{\partial \boldsymbol{B}}{\partial t},$$
$$\nabla \times \boldsymbol{B}=\frac{k_{2}}{k_{1} k_{3}} \frac{\partial \boldsymbol{E}}{\partial t}+4 \pi \frac{k_{2}}{k_{3}} \boldsymbol{J} .$$