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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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"배워서 남주자"라는 가치를 가진 산동일크무크를 응원해주세요!
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
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.
\[\begin{aligned} T=\frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=11600\end{aligned}\]
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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\(\bullet\) 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.
\(\bullet\) 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.
\[\begin{aligned} \label{eq_3} f(u)= \sqrt{\frac{m}{2\pi K_BT}} e^{-\frac{\frac{1}{2}mv^2}{K_BT}}\end{aligned}\]
\(f\) 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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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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Problem Prove the symmetry property of the Green’s function satisfying the Dirichlet boundary condition \[G(\vec{x},\vec{x}^{\prime}) = G(\vec{x}^{\prime}, \vec{x})\]
Refer to Jackson Problem (1.14)
Answer
Green function \(G(\vec{x},\vec{x}^{\prime})\) satisfies Dirichlet boundary conditions; bound region \(\Omega\) with boundary \(d\Omega\), \(G(\vec{x},\vec{x}^{\prime})=0 \quad \forall x^{\prime} \in \partial \Omega\)
\[\int_V (\phi \nabla^2 \psi - \psi \nabla^2 \phi)d^3x = \oint_s \left[ \phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi }{\partial n}\right]da\]
Substitute \(\phi = G(\vec{x},\vec{y})\) and \(\psi = G(\vec{x}^{\prime},\vec{y})\).
\[\begin{aligned} \int_{\Omega} \left[ G(\vec{x},\vec{y})\nabla^2 G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \nabla^2 G(\vec{x},\vec{y}) \right]d^3y =\nonumber \\ \int_{\partial \Omega}\left[ G(\vec{x}, \vec{y})\frac{\partial }{\partial n}G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \frac{\partial}{\partial n} G(\vec{x},\vec{y}) \right]da \nonumber\end{aligned}\]
\(G(\vec{x}, \vec{x}^{\prime})\) satisfies Dirichlet B.C. Hence \(RHS=0\); \(\Phi\) is known on the surface and \(F\) can be chosen to make \(G_D(\vec{x},\vec{x}^{\prime}=0)\) Since \[\nabla^2 G(\vec{x},\vec{y}) = -4\pi \delta^3(\vec{x}-\vec{y})\]
\[\begin{aligned} \int_{\Omega} \left[ G(\vec{x},\vec{y})\nabla^2 G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \nabla^2 G(\vec{x},\vec{y}) \right]d^3y = \nonumber \\ \int_{\Omega}\left[ -4\pi G(\vec{x},\vec{y}) \delta^3(\vec{x}^{\prime} - \vec{y})+G(x^{\prime},\vec{y}) 4\pi \delta^3(\vec{x}-\vec{y}) \right]d^3y \nonumber \\\end{aligned}\]
Finally
\[G(\vec{x},\vec{x}^{\prime}) = G(\vec{x}^{\prime}, \vec{x})\]
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