[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]CH4 Waves In Plasma - General ellipse equation angle derivation (tilted ellipse)

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Problem
Derive equation below given from the lecture note.

\[\begin{aligned} \tan(2\theta) = \frac{2E_{0x}E_{0y}}{E_{0x}^2 - E_{oy}^2}\cos \delta\end{aligned}\]

Answer
Figure below shows the equation of an ellipse tilted at an angle of \(\theta\) to the \(E_x\) axis.

Consider the transformation between the primed and unprimed coordinates.

\[\begin{aligned} E_x^{\prime} = E_x \cos \theta - E_y \sin \theta\\ E_y^{\prime} = E_x \sin \theta + E_y \cos \theta\end{aligned}\]

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

\[E_{oy}^2 E_x^{\prime} + E_{0x}^2 E_y^{\prime} - 2 E_{0x}E_{oy} E_{x}^{\prime}E_y^{\prime}\cos \delta = \sin^2 \delta E_{ox}^2 E_{oy}^2\]

By sbustitution and rearranging it,

\[\begin{aligned} &E_x^2 [ E_oy^2 \cos^2 \theta + E_{0x}^2 \sin^2 \theta - 2 \cos \theta \sin \theta E_{0x} E_{0y} \cos \delta ] +\\ &E_y^2 [ E_oy^2 \sin^2 \theta + E_{0x}^2 \cos^2 \theta + 2 \cos \theta \sin \theta E_{0x} E_{0y} \cos \delta ] \\ &E_x E_y [ \boxed{-2 \cos\theta \sin \theta E_{0y}^2 + 2 \cos\theta \sin\theta E_{0x}^2 - 2 (\cos^2 \theta - \sin^2\theta) E_{0x}E_{0y}\cos \delta}] \\ &= E_{0x}^2 E_{0y}^2 \sin^2 \delta\end{aligned}\]

\(E_xE_y\) term must be zero in \(x^{\prime}\) and\(y^{\prime}\)coordinate system.

\[\begin{aligned} \sin(2\theta) (E_{0x}^2 - E_{0y}^2 ) = 2 \cos(2\theta) E_{0x} E_{0y} \cos \delta\end{aligned}\]

\[\begin{aligned} \tan(2\theta) = \frac{2E_{0x}E_{0y}}{E_{0x}^2 - E_{oy}^2}\cos \delta\end{aligned}\]

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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]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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