[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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[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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