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In electromagnetism, all possible monochromatic electric field polarization states can be described in terms of the shape and orientation of the polarization ellipse, which maps the locus of points traced out by the electric field over a 2D plane.

There are several ways to derive the equation of the polarization ellipse. The derivation described in R.M.A. Azzam's seminal 'Ellipsometry and Polarized Light' first expresses the field in terms of two components (parallel and perpendicular to established basis vectors):

$$E_\parallel = a_1\cos(\omega t)-a_2\cos(\gamma)\sin(\omega t)$$

$$E_\perp=-a_2sin(\gamma)sin(\omega t)$$

Here $\omega t$ is the time varying phase, $a_1$ and $a_2$ are constant multipliers, and $\gamma$ is the angle between previously established unit vectors.

Azzam proceeds to deduce the equation of the polarization ellipse by "elimination of $t$":

$$\frac{E_\parallel^2}{a_1^2}+\frac{E_\perp^2}{a_2^2\sin(\gamma)^2}-\frac{2E_\parallel E_\perp\cot(\gamma)}{a_1^2}=1$$

However, it is unclear how $t$ is eliminated. The use of obvious trigonometric identities does not yield the equation of the ellipse. How can we eliminate $t$ to derive the equation of the ellipse shown above?

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  • $\begingroup$ Try squaring both equations and adding them together as a first step. Then remember (cos(x)^2 + (sin(x)^2) = 1. $\endgroup$
    – S. McGrew
    Jul 5, 2020 at 19:03

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Eliminate $t$ from equations $(1)$ and $(2)$ to get equation $(3)$. \begin{align*} E_\parallel &= a_1\cos(\omega t)-a_2\cos(\gamma)\sin(\omega t)\tag{1}\\ E_\perp&=-a_2\sin(\gamma)\sin(\omega t)\tag{2}\\ \frac{E_\parallel^2}{a_1^2}&+\frac{E_\perp^2}{a_2^2\sin(\gamma)^2}-\frac{2E_\parallel E_\perp\cot(\gamma)}{a_1^2}=1\tag{3} \end{align*}

Let's write each term of the equation $(3)$ of the equation separately. \begin{align*} E_\parallel^2&=a_1^2\cos^2(\omega t)+a_2^2\cos^2(\gamma)\sin^2(\omega t)-2a_1a_2\cos(\gamma)\sin(\omega t)\cos(\omega t)\tag{4}\\ E_\perp^2&=a_2^2sin^2(\gamma)\sin^2(\omega t)\tag{5}\\ E_\parallel E_\perp&=a_2^2\sin(\gamma)\cos(\gamma)\sin^2(\omega t)-a_1a_2\sin(\gamma)\sin(\omega t)\cos(\omega t)\tag{6} \end{align*} Eventually, one should seek to obtain $\sin^2(\omega t)+\cos^2(\omega t)=1$ somewhere, that removes the dependence on $t$. In situations like \begin{align*} E_\parallel &= a_1\cos(\omega t)+a_2\sin(\omega t)\\ E_\perp&=a_1\sin(\omega t)+a_2\cos(\omega t)\\ \end{align*} , squaring and adding alone suffices. So, in the natural course of action, one notices the first term of equations $(1)$ and $(2)$ (or $(4)$ and $(5)$) and compensates for the extra factors as follows \begin{align*} \frac{E_\parallel^2}{a_1^2}+\frac{E_\perp^2}{a_2^2\sin(\gamma)^2}&=\cos^2(\omega t)+\sin^2(\omega t)+\frac{a_2^2}{a_1^2}\cos^2(\gamma)\sin^2(\omega t)-2\frac{a_2}{a_1}\cos(\gamma)\sin(\omega t)\cos(\omega t)\\ &=1+\frac{a_2^2}{a_1^2}\cos^2(\gamma)\sin^2(\omega t)-2\frac{a_2}{a_1}\cos(\gamma)\sin(\omega t)\cos(\omega t)\\ \end{align*} , but only by getting two nasty extra terms. Thanks to the single term $E_\perp$, we can take it right outside as \begin{align*} &\Rightarrow\frac{a_2^2}{a_1^2}\cos^2(\gamma)\sin^2(\omega t)-2\frac{a_2}{a_1}\cos(\gamma)\sin(\omega t)\cos(\omega t)\\ &=\frac{a_2\sin(\gamma)\sin(\omega t)}{a_1^2}\left[\frac{a_2\cos^2(\gamma)\sin(\omega t)}{\sin(\gamma)}-2\frac{a_1\cos(\gamma)\cos(\omega t)}{\sin(\gamma)}\right]\\ &=-\frac{a_2\sin(\gamma)\sin(\omega t)}{a_1^2}\cot(\gamma)\left[2a_1\cos(\omega t)-a_2\cos(\gamma)\sin(\omega t)\right]\\ &=\frac{E_\perp}{a_1^2}\cot(\gamma)\left[2a_1\cos(\omega t)-a_2\cos(\gamma)\sin(\omega t)\right]\\ &\neq\frac{2E_\parallel E_\perp\cot(\gamma)}{a_1^2} \end{align*}


So, either the equation you have written is incorrect or I am mistaken while trying to break the nasty terms into $E_\parallel$ and $E_\perp$.

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  • $\begingroup$ I believe that there is a typo in the book. $\endgroup$ Aug 12, 2020 at 17:17

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