[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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프로젝트 팀: 논리설계 카드뉴스

산동일크무크 프로젝트 팀: 논리설계 카드뉴스

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[대학원-전자기학]Vector Calculus Problem 2

Problem 2: Prove \(-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\)

Let \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\]

\[\nabla \frac{1}{|\vec{r}-\vec{r}\prime|} = \frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\frac{1}{|\vec{r}-\vec{r}\prime|}\]

Let us consider only \(\hat{x}\) components. \[\begin{aligned} \frac{\partial}{\partial x}\hat{x}\frac{1}{\zeta^{\frac{1}{2}}} &= \frac{1}{2}\zeta^{-\frac{3}{2}}\zeta\prime \nonumber \\ &=\frac{x_1 - x_2}{\zeta^{\frac{3}{2}}}\hat{x} \nonumber \\ &=\frac{x_1 - x_2}{|\vec{r}-\vec{r}\prime|^3}\hat{x} \nonumber\end{aligned}\]

Same procedure occur for the other two component. When we add all, final result become \[-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\]

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[대학원-전자기학]Vector Calculus Problem 1

Problem 1: Prove \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

Answer:
\[\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x_1-x_2}{|\vec{\eta}^3|} & \frac{y_1-y_2}{|\vec{\eta}^3|} & \frac{z_1-z_2}{|\vec{\eta}^3|} \end{vmatrix}\] Where \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\vec{\eta} = \vec{r}-\vec{r}\prime\], \[|\vec{\eta}|=\sqrt{(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\] \[\begin{aligned} \nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3} &= (\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\eta}^3|}))\vec{x} \nonumber \\ & -(\frac{\partial}{\partial x}(\frac{z_1-z_2}{|\vec{\eta}^3|}) - \frac{\partial}{\partial z}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{y} \nonumber \\ & +(\frac{\partial}{\partial x}(\frac{y_1-y_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial y}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{z} \nonumber\end{aligned}\]

Since \[\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(z_1 - z_2)(y_1 - y_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

\[\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(y_1 - y_2)(z_1 - z_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

They cancel out each others. Same procedure occur for the other two components. Hence, \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

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[플라즈마 물리]Normalization Constant & Average Kinetic Energy

Problem The 3D Maxwellian distribution is given by
\[f(\vec{v}) = A_3 e^{\left( -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} \right)}\]
with \[n = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z\]
(a) Show that the normalization constant is given by \[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
(b) Show that the average kinetic energy is \[E_{av} = \frac{3}{2}KT\]
Answer (a)

Notice that
\[\begin{aligned} \label{eq_gauss} \int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\end{aligned}\]
\[\begin{aligned} n &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z \nonumber \\ &= A_3 \int_{-\infty}^{\infty} e^{ -\frac{m(v_x^2)}{2KT}}dv_x \int_{-\infty}^{\infty} e^{ -\frac{m(v_y^2)}{2KT}}dv_y \int_{-\infty}^{\infty} e^{ -\frac{m(v_z^2)}{2KT}}dv_z \nonumber \\ &= A_3 \left( \sqrt{\frac{ 2 K T \pi}{m}} \right)^{3}\end{aligned}\]
Hence,
\[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
Answer (b)

Average energy can be calculated as
\[\begin{aligned} E_{av} &= \frac{\int_{-\infty}^{\infty} \frac{1}{2}mv^2 f(\vec{v})dv}{\int_{-\infty}^{\infty}f(\vec{v})dv} \nonumber \\ &=\frac{m}{2n}\int_{-\infty}^{\infty} (v_x^2 + v_y^2 + v_z^2) A_3 e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv\end{aligned}\]
The denominator is \(n\) as we calculated in previous problem. Notice that, \[\int x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]
If we calculate the first term of equation,
\[\begin{aligned} E_{av_x} &= \frac{m}{2n} \int_{-\infty}^{\infty} A_3(v_x^2 ) e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv_x \nonumber \\ &=\frac{m}{2n}n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}} \frac{1}{2} \left( \frac{2 KT}{m} \right)^{\frac{3}{2}}\sqrt{\pi}\frac{2KT \pi}{m} \nonumber \\ &=\frac{1}{2}KT\end{aligned}\]
Hence, if we calculate \(v_y\) and \(v_z\) components, final result becomes \[E_{av} = \frac{3}{2}KT\]
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프로젝트 팀: 논리설계 카드뉴스

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[대학원-전자기학]문제풀이 델타함수와 발산

Problem Prove that \[\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})\] where \(r=|\vec{x}|\)
Answer

\[\begin{aligned} \nabla^2 \frac{1}{r} = \nabla \cdot (\nabla \frac{1}{r}) = \nabla \cdot (-\frac{1}{r^2}\hat{r}) = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \left( \frac{-1}{r^2}\right) \right) \nonumber\end{aligned}\]
Hence, the intergral of \(\nabla^2 \frac{1}{r} = 0\)
\[\int_v \nabla^2 \frac{1}{r} dv = 0\]
However, when we use the divergence theorem, we get different result.
\[\begin{aligned} \int_v \nabla^2 \frac{1}{r} dv &= \oint_s -\frac{1}{r^2}\hat{r} d\vec{a} \nonumber \\ &=\oint_s - \frac{1}{r^2} r^2\sin\theta d\theta d\phi \nonumber \\ &= -4\pi \nonumber\end{aligned}\]
This problem arise in the origin. Hence, we introduce delta function to solve this problem.
\[\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})\]
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[대학원-전자기학]문제풀이

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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