PHALIN THAKKAR

Fitting a direction without searching every angle

How does the double angle identity turn the fit into a linear algebra problem?

Note 9 · Research extension

Analytical result · Hypothetical examples are not physical measurements.

Write:

I(a)=b+I0cos⁡2(ϕ−a)=A+Ccos⁡2a+Ssin⁡2a,I(a)=b+I_0\cos^2(\phi-a) =A+C\cos2a+S\sin2a,

where:

A=b+I0/2,C=(I0/2)cos⁡2ϕ,S=(I0/2)sin⁡2ϕ.A=b+I_0/2,\qquad C=(I_0/2)\cos2\phi,\qquad S=(I_0/2)\sin2\phi.

For known analyzer angles, form one row of a design matrix as [1, cos2a, sin2a]. Estimate A, C and S with linear least squares. With known unequal independent uncertainties, use weighted least squares. Use a stable QR or SVD implementation.

Recover:

ϕ=12atan2⁡(S,C),R=C2+S2.\phi=\frac12\operatorname{atan2}(S,C),\qquad R=\sqrt{C^2+S^2}.

Illustrative readings at 0°, 45°, 90° and 135° are 83, 63, 23 and 43 units. They give A = 53, C = 30, S = 10, R ≈ 31.6228 and φ ≈ 9.2175°.

Under the assumed fully linear input plus constant background model, I0 = 2R ≈ 63.2456 and b = A − R ≈ 21.3772.

That last interpretation depends on the model. Unpolarized light and detector background both contribute to the constant term. The fit alone cannot uniquely assign that term to one cause. A dark calibration adds independent information.

If R = 0, no linear axis is identified. Do not use the numerical behaviour of atan2 at the origin as an inference.

Continue the reasoning

Finding the hidden direction

All twelve notes

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