Give a formula for the stereographic projection \(z\in\mathbb C\) of a point on \(S^2\) with latitude and longitude \((\alpha,\beta)\).
Find the distance between Boston \((42^\circ\mathrm N,71^\circ\mathrm W)\) and Rio \((23^\circ\mathrm S,43^\circ\mathrm W)\), assuming the circumference of the Earth is \(40{,}000\) km.
Solution
1. A point of latitude \(\alpha\) and longitude \(\beta\) on the unit spherehas Cartesian coordinates\[ P=(\cos\alpha\cos\beta,\,\cos\alpha\sin\beta,\,\sin\alpha). \]
Latitude \(\alpha\) gives height \(\sin\alpha\) above the equatorial plane and equatorial radius \(\cos\alpha\). Resolving that equatorial component at longitude \(\beta\) gives the stated coordinates.
2. With stereographic projection from the north pole to the equatorialplane, the complex coordinate of \(P\) is\[ \boxed{ z=e^{i\beta}\frac{\cos\alpha}{1-\sin\alpha} =e^{i\beta}\tan\left(\frac\pi4+\frac\alpha2\right).} \]
Let \(N=(0,0,1)\). The line through \(N\) and \[ P=(X,Y,Z) \] meets the plane \(Z=0\) at \[ \left(\frac{X}{1-Z},\frac{Y}{1-Z},0\right). \] Substituting \[ X+iY=\cos\alpha\,e^{i\beta}, \qquad Z=\sin\alpha \] gives \[ z=e^{i\beta}\frac{\cos\alpha}{1-\sin\alpha}. \] Multiplying numerator and denominator by \(1+\sin\alpha\) yields \[ \frac{\cos\alpha}{1-\sin\alpha} =\frac{1+\sin\alpha}{\cos\alpha} =\tan\left(\frac\pi4+\frac\alpha2\right), \] which is the desired formula.
3. If two points have latitudes \(\alpha_1,\alpha_2\) and longitudes\(\beta_1,\beta_2\), their central angle \(\gamma\) satisfies\[ \cos\gamma =\sin\alpha_1\sin\alpha_2 +\cos\alpha_1\cos\alpha_2\cos(\beta_1-\beta_2). \]
The points lie on the unit sphere, so the cosine of the angle between their position vectors is their Euclidean dot product. Using the coordinates in 1, \[ \begin{aligned} P_1\cdot P_2 &=\cos\alpha_1\cos\alpha_2 (\cos\beta_1\cos\beta_2+\sin\beta_1\sin\beta_2) +\sin\alpha_1\sin\alpha_2\\ &=\cos\alpha_1\cos\alpha_2\cos(\beta_1-\beta_2) +\sin\alpha_1\sin\alpha_2. \end{aligned} \] This equals \(\cos\gamma\).
4. For Boston and Rio,\[ \alpha_B=42^\circ, \quad \beta_B=-71^\circ, \quad \alpha_R=-23^\circ, \quad \beta_R=-43^\circ, \]so\[ \cos\gamma =\sin42^\circ\sin(-23^\circ) +\cos42^\circ\cos23^\circ\cos28^\circ \approx0.3425464. \]Thus\[ \gamma\approx1.2211704\text{ radians} \approx69.9679^\circ. \]
West longitudes are negative and south latitudes are negative, so the longitude difference is \[ |-71^\circ-(-43^\circ)|=28^\circ. \] Substitution into 3 gives the displayed value and hence the stated angle.
5. The great-circle distance between Boston and Rio is therefore\[ \boxed{\frac{40{,}000}{2\pi}\gamma\approx7{,}774\text{ km}.} \]
An arc subtending angle \(\gamma\) occupies the fraction \(\gamma/(2\pi)\) of a full great circle. Multiplying that fraction by the given circumference \(40{,}000\) km gives \[ d=40{,}000\frac{\gamma}{2\pi} \approx7{,}774.2\text{ km}. \]