[플라즈마 물리][Plasma Physics]CH1 Introduction - Debye Shielding 디바이 차폐

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Debye Shielding

Consider a test charge \(e\) and electron cloud around it in a singly charged plasma. Assume that

\(\bullet\) The ions are fixed; \(\frac{m_i}{m_e} \rightarrow \infty\)

\[\begin{aligned} \label{eq_6} n_i(r)=n \end{aligned}\]

\(\bullet\) The electrons obey Boltzman relation. In the presence of a potential energy \(q\phi\), the electron distribution function is

\[f(u)=A e^{-\frac{\frac{1}{2}mu^2+q\phi}{KT_e}}\]

Integrate this over \(du\)

\[\begin{aligned} \label{eq_7} n_e(r)=ne^{\frac{e\phi(r)}{K_BT}} \end{aligned}\]

Poisson’s equation in one dimension is

\[\begin{aligned} \epsilon_0 \nabla^2 \phi = \epsilon_0 \frac{d^2 \phi}{dx^2}=-e(n_i - n_e) \end{aligned}\]

substitution in to the Poisson’s equation,

\[\begin{aligned} \nabla^2 \phi(r) = \frac{e}{\epsilon_0}n(e^{\frac{e\phi(r)}{KT}}-1)\end{aligned}\]

assume that \(|\frac{e\phi}{KT}|<<1\), then by using taylor expansion; \(e^{\frac{e\phi(r)}{KT}}=1+\frac{e\phi(r)}{KT} +...\)

\[\begin{aligned} \nabla^2 \phi(r) &= \frac{e^2n}{\epsilon_0 KT}\phi(r)\\ &=\frac{\phi(r)}{\lambda_D^2}\end{aligned}\]

here we define Debye length \(\lambda_D\) as

\[\lambda_D = \sqrt{\frac{\epsilon_0 K T }{ne^2}}\]

In spherical coordinate

\[\begin{aligned} \frac{1}{r^2}\frac{d}{dr}\left( r^2 \frac{d\phi}{dr}\right) -\frac{\phi}{\lambda_D^2}&=0\\ \phi^{\prime \prime} + \frac{2}{r}\phi^{\prime} - \frac{1}{\lambda_D^2}\phi&=0\\ r\phi^{\prime \prime} + 2\phi^{\prime} - \frac{1}{\lambda_D^2}r\phi&=0\end{aligned}\]

Let \(\psi(r)=r\phi\), then \(\psi^{\prime}=\phi+r\phi^{\prime}\) and \(\psi^{\prime \prime}=2\phi^{\prime}+r\phi^{\prime \prime}\) so that

\[\psi^{\prime \prime}-\frac{1}{\lambda_D^2}\psi =0\]

\[\begin{aligned} \psi(r) &= C_1 e^{-\frac{r}{\lambda_D}}+C_2 e^{\frac{r}{\lambda_D}}\\ \phi(r) &= \frac{C_1}{r} e^{-\frac{r}{\lambda_D}}+\frac{C_2}{r} e^{\frac{r}{\lambda_D}}\end{aligned}\]

Applying the boundary conditions,

\[\begin{aligned} \phi &\rightarrow 0 \quad \quad \quad \quad ,\quad \quad r \rightarrow \infty \\ \phi &\rightarrow \frac{e}{4\pi\epsilon_0 r} \quad \quad , \quad \quad r \rightarrow 0\end{aligned}\]

we can find the constants.

\[C_1 = \frac{e}{4\pi \epsilon_0} \quad \quad \quad C_2 = 0\]

Hence, the solution is given by

\[\begin{aligned} \label{eq_debyelength} \boxed{\phi(r) = \frac{e}{4\pi\epsilon_0 r}e^{-\frac{r}{\lambda_D}}}\end{aligned}\]

The quantity \(\lambda_D\), called the Debye length, is a measure of the shielding distance or thickness of the sheath over which the influence of an individual charged particle is dominant.

\(\bullet\) \(\lambda_D\) = how long the shielding is effective.

\(\bullet\) Notice that electron temperature is used to define Debye length because it is more mobile; most of time this is true.

\(\bullet\) Useful forms of Eq.([eq_debyelength]) are

• \(\lambda_D = 69\sqrt{\frac{T_e}{n}}[m]\) \(T_e\) in \(^{\circ}K\)

• \(\lambda_D = 7430\sqrt{\frac{KT_e}{n}}[m]\) \(KT_e\) in \(eV\)

\(\bullet\) Trend

• effective shielding \(\quad n \uparrow\) \(\lambda_D \downarrow\)

• poor shielding \(\quad T \uparrow\) \(\lambda_D \uparrow\)

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[플라즈마 물리][Plasma Physics]CH1 Introduction - Temperature

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Temperature

The one-dimensional Maxwellian distribution is given by

\[\begin{aligned} f(v)=A e^{-\frac{mv^2}{2K_BT}}\end{aligned}\]

Unlike normal distribution gaussian equation can have a form of

\[\begin{aligned} \label{eq_4} f(x) = \frac{n}{\sqrt{2\pi}\sigma}e^{-\frac{(v-v_{\mu})^2}{2\sigma^2}}\end{aligned}\]

Where \(n\) is the number density. \(fdv\) is the number of particles per [\(m^3\)] with velocity between \(v\) and \(v+dv\), \(\frac{1}{2}mv^2\) is the kinetic energy, and \(K_B\) is the Boltzmann’s constant. The density \(n\), or number of particles per [\(m^3\)], is given by

\[\begin{aligned} n=\int_{-\infty}^{\infty}f(v)dv\end{aligned}\]

so that the constant \(A\) is found to be

\[\begin{aligned} A=n\sqrt{\frac{m}{2\pi K_BT}}\end{aligned}\]

Where \[\begin{aligned} \int_{-\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}} \end{aligned}\] is used.

\(\bullet\) meaning of T = Distribution of the particles

The width of the distribution is characterized by the constant \(T\). By computing the average kinetic energy of particles in the distribution, we can see the exact meaning of \(T\).
\[\begin{aligned} \label{eq_5} E_{av}=\frac{\int_{-\infty}^{\infty}\frac{1}{2}mu^2 f(u) du}{\int_{-\infty}^{\infty} f(u) du}\end{aligned}\]

Defining \(v_{th}=\sqrt{\frac{2K_BT}{m}}\) and \(y=\frac{u}{v_{th}}\), 1-D Maxwellian distribution can be written as

\[\begin{aligned} f(u)=Ae^{-\frac{u^2}{v_{th}^2}}\end{aligned}\]

By substitution average kinetic energy becomes

\[\begin{aligned} E_{av}&=\frac{\frac{1}{2}mAv_{th}^3 \int_{-\infty}^{\infty} e^{-y^2}y^2 dy}{Av_{th}\int_{-\infty}^{\infty} e^{-y^2}dy}\\ &=\frac{\frac{1}{2}mAv_{th}^3 \frac{1}{2}}{A v_{th}}=\frac{1}{4}mv_{th}^2=\frac{1}{2}K_BT\end{aligned}\]

Thus the average kinetic energy is \(\frac{1}{2}K_BT\).
In three dimensions,

\[\begin{aligned} f(u,v,w)=n\left( \frac{m}{2\pi K_BT } \right)^{\frac{3}{2}}e^{-\frac{\frac{1}{2}m\left( u^2 + v^2 + w^2 \right)}{K_BT}}\end{aligned}\]

Using similar calculation we get

\[\begin{aligned} E_{av}=\frac{3}{2}KT\end{aligned}\]

The general result is that \(E_{av}\) equals \(\frac{1}{2}K_BT\) per degree of freedom.

\(\bullet\) Since \(T\) and \(E_{av}\) are so closely related, it is customary in plasma physics to give temperatures in units of energy.

\(\bullet\) To avoid confusion, it is not \(E_{av}\) but the energy corresponding to \(KT\) that is used to denote the temperature.

\(\bullet\) For \(KT=1eV= 1.6\times 10^{-19}[J]\)

\[\begin{aligned} T=\frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=11600\end{aligned}\]

Thus the conversion factor is

\[\begin{aligned} 1eV = 11,600 ^{\circ}K \end{aligned}\]

\(\bullet\) By a \(2eV\) plasma we mean that \(KT=2eV\), or \(E_{av}=3eV\) in three dimensions.

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[플라즈마 물리][Plasma Physics]CH4 Waves In Plasma - General ellipse equation derivation

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Problem
Derive general elliptical equation for an arbitrary phase difference. Consider the case \(E_{ox} \neq E_{oy}\).

\[\begin{aligned} \left( \frac{E_x}{E_{0x}} \right)^2 + \left( \frac{E_y}{E_{0y}} \right)^2 - 2 \left( \frac{E_x}{E_{0x}} \right) \left( \frac{E_y}{E_{0y}} \right) \cos \delta = \sin^2 \delta\end{aligned}\]

Answer
\[\begin{aligned} E_x(z,t) = E_{0x}\cos(\tau + \delta_x ) \\ E_y(z,t) = E_{0y}\cos(\tau + \delta_y ) \end{aligned}\]

\[\begin{aligned} \frac{E_x(z,t)}{E_{0x}} = \cos( \tau) \cos (\delta_x) - \sin(\tau) \sin (\delta x) \\ \frac{E_y(z,t)}{E_{0y}} = \cos( \tau) \cos (\delta_y) - \sin(\tau) \sin (\delta y) \end{aligned}\]

\[\begin{aligned} \frac{E_x}{E_{0x}}\sin(\delta y) - \frac{E_y}{E_{0y}}\sin (\delta_x) = \cos(\tau) \sin(\delta y -\delta x)\\ \frac{E_x}{E_{0x}}\cos(\delta y) - \frac{E_y}{E_{0y}}\cos (\delta_x) = \sin(\tau) \sin(\delta y -\delta x)\end{aligned}\]

square above two equatiion and add togather leads to the general equation of ellipse.

\[\begin{aligned} \left( \frac{E_x}{E_{0x}} \right)^2 + \left( \frac{E_y}{E_{0y}} \right)^2 - 2 \left( \frac{E_x}{E_{0x}} \right) \left( \frac{E_y}{E_{0y}} \right) \cos \delta = \sin^2 \delta\end{aligned}\]

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[플라즈마 물리][Plasma Physics]Introduction

Introduction to Plasma

\(\bullet\) We are surrounded by plasmas starting from the ionosphere a hundred kilometers above us, which is connected to the sun via the plasma of the solar wind.

\(\bullet\) The very tenuous interstellar space is a plasma and so are the largest objects that emit x-rays in the universe.

Plasma State

\(\bullet\) Plasma is a fundamental state of matter: by heating the mater changes from a solid \(\rightarrow\) liquid \(\rightarrow\) gas \(\rightarrow\) plasma.

\(\bullet\) When gas is heated above a certain temperature, or it is subject to strong electromagnetic fields, it gets ionized, and a transition towards the so-called fourth state of matter(coined by W.crookes in 1879), plasma state, is observed.

\(\bullet\) Compare

• Ancient: Universe - Earth, water, air, fire

• Modern: Universe - solid, liquid, gas, plasma

\(\bullet\) Plasma: Introduced by Tonks and Irving Langmuir(Nobel Prize winner) in 1928.

\(\bullet\) Plasma: Greek words - moldable substacne; jelly.

\(\bullet\) Plasma ionized gas with \(n_e = n_i\); macroscopically charged neutral. Much more complicated than many single charged particle becuase of collective effects.

Definition of Plasma

\(\bullet\) The ensemble of freely moving charged particles of both signs, i.e., ionized gas, can be considered as plasma if the Debye length is small compared with dimensions of the volume occupied by the gas. (Langmuir)

\(\bullet\) A plasma is a quasi-neutral gas of charged and neutral particles which exhibits collective behavior (F.F. Chen)

\(\bullet\) A plasma is a gas of charged particles, in which the potential energy of a typical particle due to its nearest neighbor is much smaller than its kinetic energy. (D. R. Nicholson)

\(\bullet\) A plasma may be roughly defined as a system containing mobile charges, in which the electric and magnetic interactions between particles play a dominant role in the dynamics of the systems. (J. M. Dawson)

\(\bullet\) A plasma is collection of charged particles, usually of opposite sign, that tends to be electrically neutral. We often describe a plasma as the fourth state of matter. Adding energy to a solid melts it and it becomes a liquid; adding energy to a liquid boils it and ib becomes a gas; adding energy to a gas ionizes it and it becomes a plasma. (J. L. Shohet)

\(\bullet\)The plasma state is a characteriza

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[플라즈마 물리]F.F. Chen Problem 1.5 Solution Debye Shielding

Problem In a strictly steady state situation both the ions and the electrons will follow the Boltzmann relation

\[n_i = n_0 e^{\frac{-q_i \phi }{KT_i}}\]

For the case of an infinite, transparent grid charged to a potential \(\phi\), show that the shielding distance is then given approximately by

\[\lambda_D^{-2} = \frac{ne^2}{\epsilon_0} \left( \frac{1}{KT_e} + \frac{1}{KT_i} \right)\]

Answer

\[\begin{aligned} \nabla \cdot \vec{E} &=\frac{\rho}{\epsilon_0} = \frac{e}{\epsilon_0}(n_i - n_e) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( e^{-\frac{e\phi}{KT_i}} - e^{\frac{e\phi}{KT_e}} \right) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( 1 -\frac{e\phi}{KT_i} - 1 - \frac{e\phi}{KT_e} \right) \nonumber \\ \frac{d^2 \phi }{dx^2 }&=\frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \phi \nonumber\end{aligned}\]

Hence,

\[\begin{aligned} \frac{1}{\lambda_D^2} = \frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \nonumber\end{aligned}\]

\[\lambda_D = \sqrt{\left( \frac{\epsilon_0 K T_e T_i}{ne^2 (T_i + T_e)} \right)}\]

• \(T_i << T_e \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_i}{ne^2} \right)}\)

• \(T_e << T_i \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_e}{ne^2} \right)}\)

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[플라즈마물리]Single Particle Motion - E cross B drift

Problem Consider a particle of charge \(q\) and mass \(m\), initially at rest at (0,0,0), in the presence of a static magntic field \(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\).

(a) Taking \(E, B > 0\), sketch the orbit of the particle when \(q>0\).

(b) Derive an exact expression for the orbit \([x(t),y(t),z(t)]\) or the particle. Express your answer in terms of \(E\), \(B\), and \(\omega_c\)

(c) Find the drift velocity after averaging the motion in time. If there were many particles of various charges and masses present, would there be any net current?

(d) Suppose the electric field were replaced by a force \(F\) in the \(y\)-direction. What would be the drift velocity?

Answer (a)

Answer (b)

\(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\). From Lorentz equation,

\[m\frac{dv}{dt}=q(\vec{E} + \vec{v}\times \vec{B})\]

\[\begin{aligned} \dot{v_x}&= \quad \quad \quad \quad \frac{q}{m}v_y B \nonumber \\ \dot{v_y}&= \frac{q}{m}\vec{E_y} - \frac{q}{m}v_xB \nonumber \\ \dot{v_z}&= 0\end{aligned}\]

In z-direction \[z(t) = constant = 0\]

\[\begin{aligned} \frac{dv_x}{dt} &= \frac{qB}{m}v_y = \omega_c v_y \nonumber \\ \frac{dv_y}{dt} &= \frac{qE}{m}-\frac{qB}{m}v_x = \frac{qE_y}{m}-\omega_c v_x \nonumber\end{aligned}\]

\[\begin{aligned} \frac{d^2 v_y}{dt^2} &= -\omega_c^2 v_y \nonumber \\ \frac{d^2 v_x}{dt^2} &= -\omega_c^2 \left(v_x-\frac{E}{B} \right) \nonumber\end{aligned}\]

since \(\frac{E}{B}\) is constant, \[\frac{d^2}{dt^2} \left[ v_x-\frac{E}{B} \right] = \frac{d^2 v_x}{dt^2} = -\omega_c^2 \left(v_x-\frac{E}{B} \right)\]

\[\begin{aligned} v_x &= iv_{\perp}e^{i\omega_c t} + \frac{E}{B}\nonumber \\ v_y &= v_{\perp}e^{i\omega_c t} \nonumber \end{aligned}\]

\[\begin{aligned} x(t) &= x(0) + r_L \cos{\omega_c t} +\frac{E}{B}t \nonumber \\ y(t) &= y(0) + r_L \sin{\omega_c t} \nonumber \end{aligned}\]

Answer (c)

\[<v_d> = \frac{\int_{-\infty}^{\infty}v_d f(v) dv}{\int_{-\infty}^{\infty}f(v) dv} = \frac{E}{B}\]

There will be no current because \(\vec{E} \times \vec{B}\) drifts the ions and electrons in the same direction.

Answer (d)

\[\begin{aligned} \vec{v_f}&= \frac{1}{q} \frac{\vec{F} \times \vec{B}}{B^2} \nonumber \\ &=\frac{FE}{qB}\hat{y}\end{aligned}\]

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