This article describes the deviation of the formulas used in GLWaves for calculations concerning the polarisation ellipse.

We consider a wave superposed of two linearely polarized waves, one in X direction, the other one in Y direction. They have 90$ ^\circ$ in between. These are defined by

\begin{displaymath}\begin{array}{rcl} \vec x &=& E_x\,\cos(\xi + \varphi_x) \vec e_x\\  \vec y &=& E_y\,\cos(\xi + \varphi_y) \vec e_y \end{array}\end{displaymath} (1)

Both waves have their own amplitude and phase offset. The instantaneous angle is written $ \xi$ here for simplicity. A real wave has $ \xi=\omega\,t$. If we superpose both waves we will get an elliptically polarized wave. The E vector describes an ellipse. Silently we assume the x and y coordinates being proportional to the x and y component of the field strength vector, respectively.

Figure: The polarisation ellipse with all its parameters used in this article

This ellipse is centered at the origin. Its major axis has the length $ X$, its minor axis the lenth $ Y$. The major axis' inclination to the $ x$ axis is denoted by $ \tau$. The rectangle described by the amplitudes of the E field components has a hypotenuse angle called $ \gamma$, the rectangle streched by the ellipse axes has an angle known as $ \varepsilon$.

Ellipse parameters

Handedness $ \gamma$

The angle $ \gamma$ is obviously given by

$\displaystyle \gamma = \arctan\left(\frac{E_y}{E_x}\right)$ (2)

Ellipse axes

Given are the amplitudes $ E_x$, $ E_y$ and the phase offsets $ \varphi_x$ and $ \varphi_y$. We search the inclination angle $ \tau$ and the ellipse axes lengths $ X$ and $ Y$. Therefore we define the length of the compound E vector

$\displaystyle l^2 = x^2 + y^2$ (3)

The maxium of the length appears in the the major axis' direction. The derivation of the length1 by the angle $ \xi$ has to be zero.

$\displaystyle \frac{\partial l^2}{\partial\xi} = - E_x^2\,\sin\left[2\,(\xi + \varphi_x)\right] - E_y^2\,\sin\left[2\,(\xi + \varphi_y)\right] \stackrel{!}{=} 0$ (4)

This supplies the angle $ \phi ::= \left.\xi\right\vert _{l \rightarrow \max}$ when the superposed waves point to the major axis

$\displaystyle \xi = - \varphi_y - \arctan\left({ \frac {E_x^2\,\cos(\delta ) + ...
...E_x^2\,E_y^2\, \cos(\delta ) + E_y^4}}{E_x^2\, \sin(\delta )}} \right) =:: \phi$ (5)

where $ \delta = \varphi_y - \varphi_x$, the phase difference.

Equations [*] with $ \xi = \phi$2 give the x and y coordinate of the major axis

\begin{displaymath}\begin{array}{rcl} \overline{x} &=& E_x\,\cos(\phi + \varphi_x)\\  \overline{y} &=& E_y\,\cos(\phi + \varphi_y) \end{array}\end{displaymath} (6)

whereas $ \xi = \phi + \frac{\pi}{2}$ results in the coordinates of the minor axis

\begin{displaymath}\begin{array}{rcl} \underline{x} &=& E_x\,\cos(\phi + {\texts...
...cos(\phi + {\textstyle \frac{\pi}{2}} + \varphi_y). \end{array}\end{displaymath} (7)

The length of the axes is then given by

\begin{displaymath}\begin{array}{rcl} X &=& \sqrt{\overline{x}^2+\overline{y}^2}\\  Y &=& \sqrt{\underline{x}^2+\underline{y}^2}\\  \end{array}\end{displaymath} (8)

Inclination $ \tau$

The angle $ \tau$ specifying the inclination of the ellipse now is easily determined by

$\displaystyle \tau = \arctan\left(\frac{\overline{y}}{\overline{x}}\right)$ (9)

Ellipticity $ \varepsilon$

The ellipticity $ \varepsilon$ results to

$\displaystyle \varepsilon = \arctan\left(\frac{Y}{X}\right)$ (10)

Wave parameters

This section is a treatise on the derivation of the wave parameters $ E_x$, $ E_y$, $ \varphi_x$ and $ \varphi_y$ when the ellipse parameters $ \gamma$, $ \tau$ or $ \varepsilon$ are given.

Given handedness $ \gamma$

$ \gamma$ is determined by the component amplitudes $ E_x$ and $ E_y$ (see eq. ([*])) so we have to reversely calculate both amplitudes from the given $ \gamma$. Therefore we need a second equation. We can take one of these
  1. $ E_x =$   const.
  2. $ E_y =$   const.
  3. $ E_x^2+E_y^2 =$   const.
The most meaningful is variant [*]. First the diagonal of the current rectangle is determined as

$\displaystyle A = \sqrt{E_x^2+E_y^2}$ (11)

what we directly use in the equations for the new amplitudes

\begin{displaymath}\begin{array}{rcl} E_x &=& A\,\cos(\gamma)\\  E_y &=& A\,\sin(\gamma) \end{array}\end{displaymath} (12)

(remember the unit circle and the trigonometric functions). The phase offsets don't depend on $ \gamma$.

Finally we recalculate $ \tau$ and $ \varepsilon$ since they change with $ \gamma$. The calculation is given in subsections [*] and [*], respectively.

Given inclination $ \tau$

Coordinate system rotation

The polarisation ellipse can be described as a horizontal ellipse in a rotated coordinate system (rotated by the inclination angle $ \tau$). The coordinate transformation will be derived now.

Figure: The blue coordinate system is rotated by the angle $ \tau$ relative to the black coordinate system.

The coordinates in the base coordinate system $ x$ and $ y$ are rendered as

\begin{displaymath}\begin{array}{rcl} x &=& A\,\cos(\varphi)\\  y &=& A\,\sin(\varphi) \end{array}\end{displaymath} (13)

In the rotated coordinate system $ (x',y')$ the coordinates are given by

\begin{displaymath}\begin{array}{lllllll} x' &=& A\,\cos(\varphi - \tau) &=& \ov...
...derset{x}{\underbrace{A\,\cos(\varphi)}} \sin(\tau) \end{array}\end{displaymath} (14)

which can be simplified to the coordinate transformation

\begin{displaymath}\begin{array}{rcl} x' &=& x\,\cos(\tau) + y\,\sin(\tau)\\  y' &=& y\,\cos(\tau) - x\,\sin(\tau) \end{array}\end{displaymath} (15)

The backward transformation

\begin{displaymath}\begin{array}{rcl} x &=& x'\,\cos(\tau) - y'\,\sin(\tau)\\  y &=& y'\,\cos(\tau) + x'\,\sin(\tau) \end{array}\end{displaymath} (16)

Component maxima

In the rotated coordinate system the ellipse is given by

\begin{displaymath}\begin{array}{rcl} x' &=& X\,\cos(\varphi)\\  y' &=& Y\,\sin(\varphi) \end{array}\end{displaymath} (17)

In the base coordinate system this equals to

\begin{displaymath}\begin{array}{rcl}x &=& X\,\cos(\varphi)\cos(\tau) - Y\,\sin\...
...n\,(\varphi)\cos(\tau) + X\,\cos(\varphi)\sin(\tau) \end{array}\end{displaymath} (18)

where the angle $ \phi$ refers to the rotated coordinate system.

The components' amplitudes appear when these coordinates have their maximum. The angle $ \alpha$ relative to the major axis points to the amplitude of $ E_x$, $ \beta$ denotes the amplitude of $ E_y$.

\begin{displaymath}\begin{array}{lllll} \alpha &=& \left. \varphi \right\vert _{...
...n\left(+\frac{Y \cos(\tau)}{X \sin(\tau)}\right)\\  \end{array}\end{displaymath} (19)

Using that result the field strength components are given by

\begin{displaymath}\begin{array}{rcl} E_x &=& X\,\cos(\alpha)\cos(\tau) - Y\,\si...
...,\sin\,(\beta)\cos(\tau) + X\,\cos(\beta)\sin(\tau) \end{array}\end{displaymath} (20)

Phase difference

When the x component of the elliptically polarized electromagnetic wave takes its maximum (this is at $ \varphi = \alpha$), the y component's value must be (compare with eq. ([*])$ _2$)

$\displaystyle y = Y\,\sin\,(\alpha)\cos(\tau) + X\,\cos(\alpha)\sin(\tau) = E_y\,cos(\delta)$ (21)

with $ \delta = \varphi_y - \varphi_x$ being the phase difference between $ x$ and $ y$. Solving this equation for $ \delta$ finishes the problem.

$\displaystyle \delta = \arccos\left(\frac{Y\,\sin\,(\alpha)\cos(\tau) + X\,\cos(\alpha)\sin(\tau)}{E_y} \right)$ (22)

Attention: be careful with this $ \arccos(\cdot)$. Little inaccuracies and the granularity of floating point calculations can lead to an argument $ \vert\cdot\vert > 1.0$ raising a floating point exception.

Since changing the field strength components also affects the handedness $ \gamma$, it has to be calculated as described in section [*]. $ \varepsilon$ doesn't change because the ellipse is rotated as is.

Given ellipticity $ \varepsilon$

The ellipticity $ \varepsilon$ is determined by the ellipse axes $ X$ and $ Y$ so we have to reversely calculate both from the given $ \varepsilon$. Therefore we again need a second equation.
  1. $ X =$   const.
  2. $ Y =$   const.
  3. $ X^2+Y^2 =$   const.
Item number [*] gives the best result. The diagonal of the ellipticity rectangle derives to

$\displaystyle A = \sqrt{X^2+Y^2}$ (23)

The new axes lengths thus become

\begin{displaymath}\begin{array}{rcl} Y &=& A\,\cos(\varepsilon)\\  Y &=& A\,\sin(\varepsilon) \end{array}\end{displaymath} (24)

(recall unit circle and trigonometric functions). The values $ E_x$, $ E_y$, $ \varphi_x$, $ \varphi_y$ and $ \gamma$ ($ \tau$ is held constant when $ \varepsilon$ is changed) are given by the equations derived in section [*].


... length1
Here we differentiate the square of the length $ l^2$. This is allowed since the length is surely greater than zero and the square is a monotone transformation on the positive axis. It simplifies the calculation enormously.
we simply use the absolute values instead of the vectors here
Johann Glaser