Introduction

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

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

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

$$l^2 = x^2 + y^2$$ (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.

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

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

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

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

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

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

The length of the axes is then given by

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

Inclination $\tau$

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

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

Ellipticity $\varepsilon$

The ellipticity $\varepsilon$ results to

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

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

what we directly use in the equations for the new amplitudes

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

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

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

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

which can be simplified to the coordinate transformation

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

The backward transformation

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

Component maxima

In the rotated coordinate system the ellipse is given by

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

In the base coordinate system this equals to

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

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

Using that result the field strength components are given by

$$\begin{array}{rcl} E_x &=& X\,\cos(\alpha)\cos(\tau) - Y\,\si... ...,\sin\,(\beta)\cos(\tau) + X\,\cos(\beta)\sin(\tau) \end{array}$$ (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. ()$_2$)

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

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

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

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

The new axes lengths thus become

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

Footnotes

... 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.

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²

we simply use the absolute values instead of the vectors here