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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