Editing Openai/6897769e-4ee4-800f-aba5-69cca34f701c (section)

=== Let θ(λ) be the physically relevant angle produced by the geometry at wavelength λ. (For a raindrop-style geometry/primary rainbow this is the scattering angle at the stationary point: θ(λ) = 180° − D_min(λ), where D(θᵢ; n(λ)) is the ray-deviation function and D_min is its extremum.) ===

Given a spectral distribution S(λ) (radiance, intensity), the spectrum-weighted mean angle is

θˉ  =  ∫S(λ) θ(λ) dλ∫S(λ) dλ.\bar\theta \;=\; \frac{\int S(\lambda)\,\theta(\lambda)\,d\lambda}{\int S(\lambda)\,d\lambda}.θˉ=∫S(λ)dλ∫S(λ)θ(λ)dλ.
Two special S(λ) cases you should consider:

• Continuous (Planck) weighting for a blackbody at temperature T:

S(λ)=Bλ(T)(Planck law).S(\lambda)=B_\lambda(T)\quad\text{(Planck law).}S(λ)=Bλ(T)(Planck law).
• Line spectrum (a small set of narrow emission lines), modeled as narrow Gaussians or delta functions:

S(λ)=∑iAi δ(λ−λi)(idealized lines),S(\lambda)=\sum_i A_i\,\delta(\lambda-\lambda_i)\quad\text{(idealized lines)},S(λ)=i∑Aiδ(λ−λi)(idealized lines),
or more realistically

S(λ)=∑iAiexp⁡ ⁣[−(λ−λi)2/(2σi2)].S(\lambda)=\sum_i A_i \exp\!\big[-(\lambda-\lambda_i)^2/(2\sigma_i^2)\big].S(λ)=i∑Aiexp[−(λ−λi)2/(2σi2)].
Compute θ(λ) by solving (numerically) for the stationary point of the ray-deviation function D(θᵢ; n(λ)) for a spherical refracting surface (standard rainbow geometry). Then evaluate the weighted mean.

If you want the angle difference relative to the golden angle θ_G (golden angle expressed as degrees, e.g. 137.5077°), compute

Δθ=θˉ−θG.\Delta\theta = \bar\theta - \theta_G.Δθ=θˉ−θG.
Target: can Δθ ≈ −0.47° (i.e. bring 137.508° down to ~137.036°) for plausible S(λ) choices?