레이블이 plasma physics인 게시물을 표시합니다. 모든 게시물 표시
레이블이 plasma physics인 게시물을 표시합니다. 모든 게시물 표시

2018년 4월 15일 일요일

[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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2018년 4월 14일 토요일

[플라즈마 물리][Plasma Physics]CH1 - Distribution Function

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Distribution function

\(\bullet\) The most detailed description of a plasma gives the location and velocity of each plasma particle as a function of time.
\(\bullet\) It is impossible to obtain such a description of a real plasma. Rather than require an exact knowledge of a system with many particles, the behavior of such a particle system can be studied statistically.
\(\bullet\) It is customary to use the distribution function to describe a plasma. The distribution function is the number of particles per unit volume in phase space. \[f(\vec{r},\vec{v},t)d\vec{r}d\vec{v}\] represents the expected number of particles at time \(t\) in \((\vec{r},\vec{v})\) (6D phase) space with coordinates \(\vec{r}\) and \(\vec{r}+d\vec{r}\) and velocity \(\vec{v}\) and \(\vec{v}+d\vec{v}\)
\(\bullet\) A gas in thermal equilibrium has particles of all velocities, and the most probable distribution of these velocities is known as the Maxwellian distribution.

Maxwellian distribution

\(\bullet\) Maxwellian distribution is nothing but a Gaussian distribution. Recall, 1D Gaussian equation is given as

\[\begin{aligned} \label{eq_2} f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(v-v_{\mu})^2}{2\sigma^2}}\end{aligned}\]

where \(v_{\mu}\) is mean or expectation of the distribution (and also its median and mode), \(\sigma\) is the standard deviation, and \(\sigma^2\) is the variance.
\(\bullet\) Let us consider the mean velocity is \(0\); \(v_{\mu}=0\), and thermal speed(standard deviation) is \(\sigma=\sqrt{\frac{K_BT}{m}}\). Then, by simple substitution, we get Maxwellian distribution for particles.

\[\begin{aligned} \label{eq_3} f(u)= \sqrt{\frac{m}{2\pi K_BT}} e^{-\frac{\frac{1}{2}mv^2}{K_BT}}\end{aligned}\]

What is f(u)?

\(f\) would be a fractional distribution. Suppose a teacher gave a 100-point quiz to a large number N of students. \(n_i\) of students got \(s_i\) scores. The fractional distribution would be

\[\begin{aligned} f_i = \frac{n_i}{N} \end{aligned}\]

Notice that \(\sum_i f_i =1\)


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2018년 4월 12일 목요일

[플라즈마 물리][Plasma Physics]사하 공식 Saha Equation

Saha Equation

\(\bullet\) We live in a small part of the universe where plasmas do not occur naturally; otherwise we would not be alive. The reason for this can be seen from the Saha equation, which tells us the amount of ionization to be expected in a gas in thermal equilibrium. \[\begin{aligned} \label{eq_1} \frac{n_i}{n_n} \simeq 2.4 \times 10^{21} \frac{T^{\frac{3}{2}}}{n_i}e^{\frac{-U_i}{KT}} \end{aligned}\]

where \(n_i\) and \(n_n\) are density(number per \(m^3\)) of ionized atoms and of neutral atoms, respectively. \(T\) is the gas temperature in \(^{\circ}K\), \(K\) is Boltzmann’s constant; \(1.38 \times 10^{-23} [\frac{J}{^{\circ}K}]\), and \(U_i\) is the ionization energy of the gas - that is, the number of joules required to remove the outermost electron from an atom.
In room temperature, \(n_n \simeq 3 \times 10^{25}[m^{-3}]\), \(T \simeq 300 ^{\circ}K\), and \(U_i = 14.5eV\) for nitrogen, where \(1eV = 1.6 \times 10^{-10}[J]\). The fractional ionization \(\frac{n_i}{n_n+n_i}\simeq \frac{n_i}{n_n}\) is rediculously low:

\[\begin{aligned} \frac{n_i}{n_n}\simeq 10^{-122}\end{aligned}\]

Let us define degree of ionization \[\alpha = \frac{n_e}{n_e+n_n}\]. Where \(n_e=n_i\) is the number of electrons or ions per volume [\(cm^3\)], and \(n_n\) is the number of neutrals per volume [\(cm^3\)].

  • \(\alpha << 1\): weakly ionized situation (low temp)

  • \(\alpha = 1\): fully ionized (high temp)

Trend of Saha equation tells us how fractional ionization changes depending on the temperature. As the temperature increases, the ionization increases significantly.

  • \(\frac{n_e}{n_n}\rightarrow 0\) at \(K_BT << U_i\)

  • \(\frac{n_e}{n_n}>> 0\): as \(K_BT >> U_i\)

where again, \(U_i\) is the ionization energy and \(K_BT: T_e \simeq T_i\)

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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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2018년 4월 11일 수요일

[플라즈마 물리][Plasma Physics]Debye Shielding 디바이 차폐

Example Problem: Debye Shielding
Consider a positive point charge immersed in a plasma as we discussed in the class. Show that the net charge in the Debye shielding cloud exactly cancels the test charge. Assume that the ions are fixed and that \(e\phi << KT_e\). Note that we assumed that ion distribution is similar to that of the electrons in the class.

Answer:

While deriving, we assume potential to be

\[\begin{aligned} \frac{e\phi}{KT_e} << 1 \quad \quad \quad n_i \simeq n_e\\ \lambda_D = \sqrt{\frac{\epsilon KT}{ne^2}}\\ \phi(r) = \frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}}\end{aligned}\]

Assume \(\frac{m_i}{m_e} \rightarrow \infty\) so that ions are fixed; \(n_i(r)=n\). Further assume electrons obey the Boltzmann distribution.

\[\begin{aligned} f(u) &= A e^{\frac{\frac{-1}{2}mv^2 + q\phi}{KT_e}}\\ n_e(r) &= ne^{\frac{e\phi(r)}{K_B T}}\end{aligned}\]

\[\begin{aligned} \rho &= -e(n_i - n_e) \\ &=-e \left(n - ne^{\frac{e\phi(r)}{K_BT}} \right)\\ &= ne \left(e^{\frac{e\phi(r)}{K_BT}} - 1 \right)\\\end{aligned}\]

for \(\frac{e\phi}{KT_e} << 1\),

\[\begin{aligned} e^{\frac{e\phi(r)}{K_B T}} = 1 + \frac{1}{2}\frac{e\phi(r)}{K_B T} + ...\end{aligned}\]

\[\begin{aligned} \rho &= ne(1+\frac{e\phi(r)}{K_B T} - 1) \\ &=\frac{ne^2 \phi(r)}{K_B T} \\ &= \frac{\epsilon_0}{\lambda_D^2}\phi(r)\end{aligned}\]

\[\begin{aligned} \int \rho dv &= Q \\ &= \int_V \frac{\epsilon_0}{\lambda_D^2}\frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}} r^2 \sin\theta dr d\theta d\phi\\ &= \int \frac{4\pi \epsilon_0}{\lambda_D^2}\frac{e}{4\pi \epsilon_0} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} \int_{0}^{\infty} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} (\lambda_D^2 e^{-\frac{r}{\lambda_D}}-\lambda_D r e^{-\frac{r}{\lambda_D}}) =0\end{aligned}\]

On the last step, when \(r = \lambda_D\), charge becomes zero. Hence, the Debye sheilding cloud exactly cancels the test charge.

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2018년 4월 10일 화요일

[플라즈마 물리][Plasma Physics]CH2 - Single Particle Motion - Nonuniform B-field Curvature Drift

Consider a magnetic field that has curvature.

\(\bullet\) Such \(B\)-field will have gradient which result in the grad-B drift.

\(\bullet\) Once particle has a velocity along the magnetic field, it experiences a centrifugal force due to the field curvature and this force gives a drift.

The centrifugal force on a charged particle with \(\vec{v_{\parallel}}\) along \(\vec{B}\) is

\[\begin{aligned} \vec{F_{cf}}= m\frac{v_{\parallel}^2}{R_c}\hat{r} = \frac{ m v_{\parallel}^2}{R_c}\frac{\vec{R_c}}{R_c}\end{aligned}\]

where \(\vec{R_c}\) is the radius of curvature. The drift velocity due to \(\vec{F_{cf}}\) (the curvature drift) is then

\[\begin{aligned} \boxed{\vec{v_R} = \frac{mv_{\parallel}^2}{qB^2}\frac{\vec{R_c}\times \vec{B}}{R_c^2}}\end{aligned}\]

Curvature drift is \(\vec{v_R}\). Now, total drift in a curved magnetic field is the combination of the grad-B drift and the drift due to the centrifugal force. From Ampere’s Law, \(\nabla \times \vec{B} = \mu_0 \vec{J}\), we have \(\nabla \times \vec{B}=0\) in a vacuum. In the curvature, B-field is defined as \(\vec{B} = B(r) \hat{\phi}\)

\[\begin{aligned} \nabla \times \vec{B} &= \left[ \frac{1}{r} \frac{\partial B_z}{\partial \phi} - \frac{\partial B_{\phi}}{\partial z } \right]\hat{r} + \left[\frac{\partial B_r}{\partial z} - \frac{\partial B_z}{\partial r} \right] \hat{\phi} + \frac{1}{r}\left[ \frac{\partial}{\partial r}(r B_{\phi}) - \frac{\partial B_r}{\partial \phi} \right]\hat{z} \\ &= \frac{1}{r}\frac{d}{dr}(rB) =0\end{aligned}\]

Hence,

\[\begin{aligned} B(r) \propto \frac{1}{r} = \frac{1}{R_c}\end{aligned}\]

Thus,

\[\begin{aligned} \frac{\nabla B}{B} &= \frac{1}{B}\frac{\partial B}{\partial r} \hat{R_c} = -\frac{\vec{R_c}}{R_c^2}\end{aligned}\]

Then the grad-B drift velocity may be written as

\[\begin{aligned} \vec{\nabla B} = \frac{\frac{1}{2}mv_{\perp}^2}{qB} \frac{\vec{B} \times \nabla B}{B^2} = \frac{\frac{1}{2}mv_{\perp}^2}{qB}\frac{\vec{B}}{B} \times \left( \frac{-\vec{R_c}}{R_c^2} \right) = \frac{\frac{1}{2}mv_{\perp}^2}{qB^2}\frac{\vec{R_c}\times \vec{B}}{R_c^2}\end{aligned}\]

Finally, the total drift in a curved magnetic field is sum of grad-B drift and centrifugal force driven drift.

\[\begin{aligned} \boxed{\vec{v_R} + \vec{v_{\nabla B}} = \frac{m}{q}\frac{\vec{R_c}\times \vec{B}}{R_c^2 B^2} \left( v_{\parallel}^2 + \frac{1}{2}v_{\perp}^2 \right)}\end{aligned}\]

As you see, since the equation depends on the charge, we will have charge separation and leads to the rise of current. Notice that \(\frac{1}{2}v_{\perp}^2\) comes from the grad-B drift and \(v_{\parallel}^2\) comes from the centrifugal driven drift; the curvature drift.

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[플라즈마물리][Plasma Physics]CH2 - Single Particle Motion - Nonuniform B-field Grad-B drift

Case Five: Non Uniform Magnetic Field

Let us assume gradient B-field exist in the z-direction with gradient along the y-direction.

\[\begin{aligned} \vec{B}(r) = B(y)\hat{z}\end{aligned}\]

Assume \(B\) is slightly in-homogeneous.

\[r_L\left| \frac{\nabla B}{B} \right|<< 1\]

\(\vec{B}\) can be expressed by the Taylor series.

\[\begin{aligned} \vec{B} = \vec{B_0} + (\vec{r} \cdot \nabla)\vec{B} + ...\end{aligned}\]

so that we have

\[\begin{aligned} \vec{B} = \vec{B_0} + (\vec{r} \cdot \nabla)\vec{B} = \vec{B_0} + y\frac{\partial \vec{B}}{\partial y}\end{aligned}\]

From the Lorentz force, we will compute the average force and determine the guiding center drift.

\[\begin{aligned} \frac{dv_x}{dt} &=\frac{q}{m}v_y B(y) \hat{x}\\ \frac{dv_y}{dt} &=-\frac{q}{m}v_x B(y) \hat{y}\end{aligned}\]

\[\begin{aligned} v_x &= v_{\perp}\cos\omega_c t \\ v_y &= \mp v_{\perp}\sin\omega_c t\end{aligned}\]

\[\begin{aligned} x &= r_L \sin(\omega_c t) \\ y &=\pm r_L \cos \omega_c t\end{aligned}\]

Hence, using above equation, we can approximate the Lorentz force in inhomogeneous B-field.

\[\begin{aligned} F_x &= qv_y B =\mp q v_{\perp}\sin(\omega_c t) \left( B_0 \pm r_L \cos(\omega_c t)\frac{\partial B}{\partial y} \right)\\ F_y &=-qv_x B = -qv_{\perp}\cos\omega_c t \left(B_0 \pm r_L \cos(\omega_c t) \frac{\partial B}{\partial y} \right)\\\end{aligned}\]

Since

\[\begin{aligned} <\sin(\omega t)> &= \frac{1}{T} \int_{0}^{T}\sin \omega t dt = 0\\ <\cos(\omega t)> &= \frac{1}{T} \int_{0}^{T} \cos \omega t dt = 0\\ <\sin^2(\omega t)> &= \frac{1}{T} \int_{0}^{T}\sin^2 \omega t dt = \frac{1}{T} \int_{0}^{T} \frac{1-\cos(2\omega t)}{2}dt = \frac{1}{2}\\ <\cos^2(\omega t)> &= \frac{1}{T} \int_{0}^{T} \cos^2 \omega t dt = \frac{1}{T} \int_{0}^{T} \frac{1+\cos(2\omega t)}{2}dt = \frac{1}{2}\\ <\sin \omega t \cos \omega t> &=\frac{1}{T} \int_{0}^{T} \cos(\omega t) \sin(\omega t)dt = \frac{1}{T} \int_{0}^{T} \frac{1}{2}\sin(2\omega t)dt =0\end{aligned}\]

we get

\[\begin{aligned} <F_x> &=\frac{1}{T}\int_{0}^{T}F_x dt =0\\ <F_y> &=\frac{1}{T}\int_{0}^{T}F_y dt \\ &=\frac{1}{T}\int_{0}^{T}F_y dt \\ &=\mp qv_{\perp}r_L \frac{\partial B}{\partial y}\frac{1}{2}\\ &=-\mu \frac{\partial B}{\partial y}\end{aligned}\]

Notice that \(\mu = \pm \frac{1}{2}q v_{\perp} r_L\). By plugging this into the guiding center drift equation for general case,

\[\begin{aligned} \vec{v}_{\nabla B} &=\frac{\vec{F}\times \vec{B}}{qB^2} \\ &= \frac{F_y \hat{y} \times B \hat{z}}{qB^2}\\ &=\frac{F_y}{qB}\hat{x} \\ &= -\frac{\mu}{qB}\frac{\partial B}{\partial y}\hat{x}\end{aligned}\]

The expression can be generalized as

\[\begin{aligned} \vec{v}_{\nabla B} = \frac{\mu}{q}\frac{\vec{B} \times \nabla B}{B^2} \end{aligned}\]

or

\[\begin{aligned} \boxed{\vec{v}_{\nabla_B} = \frac{\frac{1}{2}mv_{\perp}^2}{qB}\frac{\vec{B} \times \nabla B}{B^2}} \end{aligned}\]

The grad-B drift depends of the charge \(q\), it can cause the plasma currents and charge separation.

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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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