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Faraday rotation — the series

One essay so far on this idea, and more questions about it than one essay answers.

$A slope, not an angle. Above: the polarisation angle of a background source against the square of the observing wavelength, at the five wavelengths a radio survey actually uses. The plane rotates as it passes through magnetised plasma, by an amount proportional to the integral of the electron density times the field along the path — and to λ². The intercept is the angle the source emitted at, which nobody knows, so a measurement at one wavelength contains no information whatever; the slope is the rotation measure, here 42 radians per square metre, and it needs no knowledge of the source at all. Below: the fractional polarisation that survives. A telescope beam covers many lines of sight with slightly different rotation measures, their angles disagree by more at longer wavelengths, and the vector sum collapses — which is why the useful band has a long-wavelength edge that has nothing to do with sensitivity. Divide the rotation measure by the dispersion measure of the same path, 26.8 in the usual units, and the electron density cancels: the mean line-of-sight field is 1.93 µG, obtained without knowing the distance, the density, or where along the path the field was.$
A slope that needs no source

Polarised light passing through magnetised plasma has its plane rotated, by an amount proportional to the square of the wavelength. A single measurement of the angle is worthless, because nobody knows the angle the source emitted at. A measurement of the slope against wavelength squared does not need to know.
part 1 · starlight

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A slope that needs no source