Polarization of Plane Waves
We have followed the wave's amplitude, its speed, and its power — but never asked which way the electric field actually points as time ticks by. That orientation, and how it changes from instant to instant, is polarization. A field can hold its direction, sweep out a circle, or trace an ellipse, and the difference is the whole basis of polarized sunglasses, 3D cinema, satellite links that ignore a tumbling antenna, and radar that sees through rain. It all comes from one idea: a transverse field is two perpendicular waves, and their phase difference decides everything.
- How a transverse field splits into two orthogonal components with a phase difference \(\delta\).
- Linear polarization when the components are in phase (\(\delta=0\)) or anti-phase (\(\delta=\pi\)).
- Circular polarization when amplitudes are equal and \(\delta=\pm90^\circ\).
- Elliptical polarization as the general case, and the polarization ellipse.
- The axial ratio and tilt angle that pin down any ellipse.
- The sense of rotation (right/left-handed) and where polarization matters in practice.
Describing Polarization
For a wave travelling along \(z\), Chapter 18 fixed the electric field to the transverse plane. Any such field is the sum of two perpendicular pieces — one along \(\hat{a}_x\), one along \(\hat{a}_y\) — which may differ in amplitude and, crucially, in phase by an angle \(\delta\):
Polarization is the figure traced by the tip of \(\vec{E}\) at a fixed point as time runs forward. Whether that figure is a line, a circle, or an ellipse depends entirely on the amplitude ratio \(E_2/E_1\) and the phase difference \(\delta\) — nothing else. The three possibilities are the whole story:
Linear Polarization
When the two components rise and fall together — \(\delta=0\) (in phase) or \(\delta=\pi\) (anti-phase) — their ratio is fixed at every instant. The tip moves back and forth along a single straight line, so the field is linearly polarized:
A single component (\(E_2=0\)) is the simplest case — vertical or horizontal polarization, the kind a dipole antenna radiates and a TV aerial is rotated to match. Two equal in-phase components give a line at \(45^\circ\). Anti-phase (\(\delta=\pi\)) tilts the line the other way. The defining mark of linear polarization is that the field direction never changes — only its magnitude does.
Circular Polarization
Now make the amplitudes equal (\(E_1=E_2=E_0\)) and put the components a quarter-cycle out of step (\(\delta=\pm90^\circ\)). One component peaks exactly as the other crosses zero, so the tip keeps a constant length while its direction sweeps steadily around — a circle:
Frozen in time, a circularly polarized wave is a perfect corkscrew in space; frozen in space, its field hand-cranks around a circle. The sign of \(\delta\) picks which way it turns — clockwise or counter-clockwise — and that choice is the wave's handedness, taken up in Section 21-6.
Elliptical Polarization
Linear and circular are the tidy extremes; the generic wave is elliptically polarized. With unequal amplitudes, or a phase difference that is neither \(0\), \(\pi\), nor \(\pm90^\circ\), the tip traces an ellipse. Eliminating time between the two components gives the equation of that ellipse directly:
Every polarization is a special case of this one curve. Set \(\delta=0\) and the right side vanishes, collapsing the ellipse to a straight line (linear). Set \(E_1=E_2\) and \(\delta=\pm90^\circ\) and it rounds out to a circle. Everything in between is a genuine ellipse, generally tilted with respect to the axes.
The Polarization Ellipse
Two numbers fully describe an ellipse's shape and orientation: the axial ratio \(AR\), the ratio of major to minor axis, and the tilt angle \(\tau\), the lean of the major axis from \(\hat{a}_x\):
The axial ratio is the single most useful number in practice and is usually quoted in decibels, \(AR_{\text{dB}}=20\log_{10}AR\). It runs from \(AR=1\) (\(0\ \text{dB}\), perfectly circular) to \(AR\to\infty\) (linear). A "circularly polarized" antenna is really one whose axial ratio stays near \(0\ \text{dB}\) over its band. The table gathers the conditions for each state:
| Polarization | Amplitudes | Phase δ | Axial ratio |
|---|---|---|---|
| Linear | any | \(0\) or \(\pi\) | \(\infty\) |
| Circular | \(E_1=E_2\) | \(\pm90^\circ\) | \(1\) (0 dB) |
| Elliptical | general | any other | \(1 |
Rotation Sense and Uses
For circular and elliptical waves the tip turns one of two ways, giving a right-handed or left-handed wave. The practical rule: point the thumb of the matching hand along the propagation direction; if your fingers curl the way \(\vec{E}\) rotates in time, that is the handedness. The sign of \(\delta\) flips it. (Beware: the IEEE convention used in antennas and the optics convention name the two senses oppositely, so always state which you mean.)
A receiver fully captures a co-polarized wave and, ideally, nothing from the orthogonal one. That single fact powers a remarkable range of technology.
Satellite and GPS links use circular polarization so a spinning or mis-aligned antenna still receives, and to sidestep the Faraday rotation the ionosphere inflicts on linear waves. Frequency reuse doubles a link's capacity by sending independent signals on horizontal and vertical polarization at once. Polarized sunglasses block the horizontally polarized glare off roads and water; 3D cinema feeds each eye an oppositely circularly polarized image; and weather radar compares horizontal and vertical returns to tell rain from hail. Same wave equation, one extra degree of freedom, enormous practical payoff.
Worked Examples
Problem. \(\vec{E}=3\cos(\omega t-\beta z)\,\hat{a}_x+4\cos(\omega t-\beta z)\,\hat{a}_y\). Classify it.
Solution. The components share a phase (\(\delta=0\)), so it is linear; the line tilts at \(\tau=\tan^{-1}(4/3)\):
Problem. \(\vec{E}=E_0\cos(\omega t-\beta z)\,\hat{a}_x+E_0\cos(\omega t-\beta z+90^\circ)\,\hat{a}_y\). Classify it.
Solution. Equal amplitudes and \(\delta=+90^\circ\) give circular; \(E_y=-E_0\sin(\omega t-\beta z)\), so \(|\vec{E}|=E_0\):
Problem. \(\vec{E}=2\cos(\omega t-\beta z)\,\hat{a}_x+5\cos(\omega t-\beta z-90^\circ)\,\hat{a}_y\). Find the axial ratio.
Solution. With \(\delta=\pm90^\circ\) the axes align with \(x,y\), so the semi-axes are the amplitudes \(5\) and \(2\):
Problem. Write the phasor of a circularly polarized wave of amplitude \(E_0\) travelling in \(+z\).
Solution. Put the \(\hat{a}_y\) component in quadrature with a factor \(\mp j\):
Problem. An antenna has an axial ratio of \(3\). Express it in dB, and say how close to circular it is.
Solution. Apply \(AR_{\text{dB}}=20\log_{10}AR\):
Problem. For \(E_1=3\), \(E_2=4\), \(\delta=60^\circ\), find the tilt angle of the polarization ellipse.
Solution. Use \(\tan 2\tau=2E_1E_2\cos\delta/(E_1^2-E_2^2)\):
Chapter Summary
\(\vec{E}=E_1\cos(\cdot)\hat{a}_x+E_2\cos(\cdot+\delta)\hat{a}_y\); \(\delta\) decides the shape.
\(\delta=0\) or \(\pi\); fixed direction, \(\tau=\tan^{-1}(E_2/E_1)\); \(AR=\infty\).
\(E_1=E_2\), \(\delta=\pm90^\circ\); \(|\vec{E}|\) constant; \(AR=1\).
General case; tip traces the polarization ellipse; \(1
\(AR=E_{\max}/E_{\min}\); \(\tan2\tau=2E_1E_2\cos\delta/(E_1^2-E_2^2)\).
Sign of \(\delta\) sets right/left sense; match to capture, cross to reject.
Problems
For each item, read off the amplitude ratio and the phase \(\delta\) first — they decide the state — then compute the tilt or axial ratio as asked. Difficulty rises down the list.
- Classify \(\vec{E}=5\cos(\omega t-\beta z)\,\hat{a}_x\). What is its tilt angle?
- Classify \(\vec{E}=2\cos(\omega t-\beta z)\,\hat{a}_x-2\cos(\omega t-\beta z)\,\hat{a}_y\) and give its tilt.
- Is \(\vec{E}=E_0\cos(\omega t-\beta z)\,\hat{a}_x+E_0\sin(\omega t-\beta z)\,\hat{a}_y\) circular? Find \(\delta\) and \(AR\).
- Find the axial ratio of \(\vec{E}=4\cos(\omega t-\beta z)\,\hat{a}_x+1\cos(\omega t-\beta z-90^\circ)\,\hat{a}_y\).
- Convert axial ratios of \(1\), \(2\), and \(6\) to decibels.
- An antenna spec demands \(AR\le 3\ \text{dB}\). What is the largest linear \(AR\) allowed?
- Write the time-domain field of the phasor \(\vec{E}_s=E_0(\hat{a}_x+j\hat{a}_y)e^{-j\beta z}\) and state its handedness sense.
- Show that \(\delta=0\) reduces the polarization-ellipse equation to a straight line.
- For \(E_1=E_2\) and \(\delta=90^\circ\), reduce the ellipse equation to a circle.
- For \(E_1=2\), \(E_2=3\), \(\delta=90^\circ\), find \(AR\) and the tilt angle.
- Explain why a circularly polarized link is robust to a rotating receive antenna.
- A linearly polarized receiver is rotated \(60^\circ\) from an incoming linear wave. By Malus-type reasoning, what fraction of the power is captured?