15 Jul 2025

Linear, Circular, and Elliptical Polarization

Polarization ellipse derived from orthogonal field components, with linear and circular limits.

bsc semester-v electromagnetic-theory mj-8 unit-iii polarization

For a monochromatic wave travelling along $+z$, the transverse electric field can have two orthogonal components:

\[E_x=A\cos\psi, \qquad E_y=B\cos(\psi+\delta), \qquad \psi=kz-\omega t.\]

$A$ and $B$ are nonnegative peak amplitudes in $\mathrm{V\,m^{-1}}$, and $\delta$ is their phase difference. At a fixed point in space, the tip of $\mathbf E$ traces the polarization curve as time advances.

Derivation of the polarization ellipse

Expand the second component:

\[\frac{E_y}{B}=\cos\psi\cos\delta-\sin\psi\sin\delta.\]

Since $E_x/A=\cos\psi$,

\[\sin\psi\sin\delta =\frac{E_x}{A}\cos\delta-\frac{E_y}{B}.\]

Square this equation and use $\sin^2\psi=1-\cos^2\psi=1-(E_x/A)^2$. After collecting terms,

\[\boxed{\left(\frac{E_x}{A}\right)^2 +\left(\frac{E_y}{B}\right)^2 -2\frac{E_xE_y}{AB}\cos\delta =\sin^2\delta}.\]

This is the equation of an ellipse in the transverse plane. Linear and circular polarization are special limiting forms.

Three polarization states generated directly from the two-component field equations. Editable TikZ source.

Linear polarization

If $\delta=0$ or $\pi$, the right-hand side of the ellipse equation vanishes and the two components are proportional:

\[\delta=0:\quad E_y=\frac BA E_x,\] \[\delta=\pi:\quad E_y=-\frac BA E_x.\]

The field oscillates along a fixed straight line. Linear polarization also results if either $A=0$ or $B=0$, regardless of phase.

Circular polarization

If the components have equal amplitude and differ in phase by a quarter cycle,

\[A=B=E_0, \qquad \delta=\pm\frac\pi2,\]

then

\[E_x^2+E_y^2=E_0^2.\]

The field magnitude is constant and its direction rotates uniformly. The signs $+\pi/2$ and $-\pi/2$ give opposite handedness. A handedness label must always be accompanied by the viewing or propagation convention; here the sign of $\delta$ unambiguously records the state.

Elliptical polarization

All other fully polarized combinations with both components present produce an ellipse. The orientation angle $\theta$ of its principal axis relative to $x$ follows by eliminating the mixed term through a rotation of coordinates:

\[\boxed{\tan2\theta=\frac{2AB\cos\delta}{A^2-B^2}}.\]

Define the ellipticity angle $\chi$ by $\tan\chi=b/a$, where $a\ge b$ are the semi-major and semi-minor axes and the sign carries the rotation sense. Then

\[\boxed{\sin2\chi=\frac{2AB\sin\delta}{A^2+B^2}}, \qquad -45^\circ\le\chi\le45^\circ.\]

$\chi=0$ is linear, while $\lvert\chi\rvert=45^\circ$ is circular. Elliptical polarization is therefore the general state, with linear and circular states obtained by particular amplitude and phase relations.

The eliminated-phase ellipse identity is checked in the Unit III Maxima worksheet; its printed residual is zero.