0

I'm having difficulty proving the LHS equals the RHS. Both the limit and integral is giving me problems.

$$ 4\pi\,\mathrm{rad}^2 ≟ \lim_{n\to\infty}n\int_{-\pi/2}^{\pi/2} \cos^{-1}(\sin(φ)² + \cos(φ)² \cdot \cos(\frac{2\pi\,\mathrm{rad}}{n}))\,dφ \qquad ①$$

The RHS is formed by trying to show when angular displacement in one plane (θ longitude) is integrated with angular displacement in an orthogonal plane (φ latitude) that the maximum 3d angle is 4π rad².

If δ is the angular distance (great circle angle) for two points at same latitude φ given a rotation of Δθ, then a 3d longitudinal strip of angles can be approximated by integrating longitude angular distance δ across the full range of latitude angles for small Δθ.

$$ δ = cos^{-1}(\sin(φ)² + \cos(φ)²\cos(Δθ)) $$

$$ \text{longitudinal slice} ≈\int_{-\pi/2}^{\pi/2} δ\,dφ $$

This can then be multiplied by the number of strips $n$ required for a full rotation, thus $n=\frac{2π\,\mathrm{rad}}{Δθ}$

$$ \text{full spherical angle} = \lim_{n\to\infty}\int_{-\pi/2}^{\pi/2} δ\,dφ $$

The expected value is 4π rad²= 4π sr, and from numerical integration it seems to converge on 4π. However, I am having a hard time proving ①.

FxT
  • 11

1 Answers1

1

OK I think I finally figured this out. The trick was bounding the arccos function with the inequality

$$ \sqrt{1-t^2} ≤ cos^{-1}(t) ≤ {\sqrt{1-t^2} \over t} $$

(See Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$)

and using the Maclaurin expansion of cos(τ/n) where τ= 2π [rad] is one full turn. The limit can then be evaluated first which turns out to be

$$ \lim_{n\to\infty}n \cos^{-1}(\sin(φ)² + \cos(φ)² \cdot \cos(\frac{τ}{n})) \;=\; {τ^2 \over 2π}\cos(φ) $$

This is true since the limit of lower bound and upper bound are equal when we let $ t = \sin(φ)² + \cos(φ)² \cdot \cos(\frac{τ}{n}) $

$$ \lim_{n\to\infty}n \sqrt{1-t^2} = \lim_{n\to\infty}n {\sqrt{1-t^2} \over t} = {τ^2 \over 2π}\cos(φ)$$

Thus $$ \text{full spherical angle = } \int_{-\pi/2}^{\pi/2} {τ^2 \over 2π}\cos(φ) \;dφ \;=\; {τ^2 \over π} \;=\; 4π \;\mathrm{rad}^2 $$

FxT
  • 11