[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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