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到北极的弦长期望

Distance from North Pole

专题
Probability / 概率
难度
L6

题目详情

地球半径取 1，从单位球面上均匀随机选取一点 PP。

求点 PP 与北极点之间的最短距离的期望。

注意：这里的最短距离是穿过球体内部的直线距离（弦长），不是沿球面的大圆距离。

提示：按极角把球面分成薄环积分。

What is the expected distance of any point on Earth and the north pole? Take Earth radius 1.

Clarification: Shortest distance cuts through the sphere, instead of lying on surface.

Further thinking: Is this question same as choosing two random points on unit sphere and asking their expected distance?

Hint

Imagine a ring of some thickness, whose distance from N is constant from all points on that ring. Get the average of all such rings.

解析

答案是 43\frac{4}{3}。

设点的极角为 θ∈[0,π]\theta\in[0,\pi]，则北极到该点的弦长为 2sin⁡(θ/2)2\sin(\theta/2)。

球面面积元素与 sin⁡θ\sin\theta 成比例，因此

E=∫0π2sin⁡(θ/2)⋅sin⁡θ dθ∫0πsin⁡θ dθ=43.E=\frac{\int_0^{\pi} 2\sin(\theta/2)\cdot \sin\theta\,d\theta}{\int_0^{\pi}\sin\theta\,d\theta}=\frac{4}{3}.

Original Explanation

4/3

Solution

2 * sin(x/2)is the distance of north pole from a point on the ring at angle x from the z axis. So, integrate from 0 to pi, (2 * pi * sin(x) * 2 * sin(x/2) dx) and divide by the total area which is 4 * pi. Answer:4/3

Another approach is to imagine a horizontal ring of dy thickness at distance y from N (north pole).

Area of ring = 2 * pi * dy Probability of choosing point on this ring = dy/2 Distance of N & a point on ring = sqrt(2y)

Exp length = integral (y=0 to 2) of sqrt(2y) dy/2 = 4/3

And yes, taking two random points on surface of sphere and asking their expected distance is same as this very question.