Solid Angles on Spheres

Solid Angles on Spheres

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1. Introduction

In plane geometry, the idea of an “angle” has long been built into our intuition. The image of two rays meeting at a point sparks a vague notion of how “far apart” the two rays are: this notion is made rigorous with the definition of a \textit{radian}, the angle required to intercept a circular arc with equal length and radius.

But what happens when we leave our sheet of paper and enter the real world? How can we describe the “size” of an opening stemming from a point? For instance, if we’re given a cone of arbitrary shape, is there a good way to capture the “size” of its vertex opening? What about more complicated shapes?

Analogous to the radian, the most natural choice of unit for these “solid-angles” is the \textit{steradian}. Essentially, if a sphere of radius 1 is centered around a point of reference, a solid angle of 1 steradian emanating from that point will capture a surface area of 1 on the sphere. Just like in plane geometry, where the maximum possible angle is 2\pi radians, the maximum possible solid-angle about a point is 4\pi steradians, as the surface area of the entire reference unit-sphere is 4\pi(1^3) = 4\pi. Steradians also provide us valuable insight into perspective: the apparent size of any object from a reference point O is proportional to the steradian solid-angle captured by the object from O.

2. Conical Solid-Angles

Starting with one of the simpler cases, what is the solid-angle emanating from the vertex of a cone? Assume WLOG that the cone is scaled down so it has slant length 1, and its base radius and height are r and h respectively (so that r^2 + h^2 = 1). We center a sphere of radius 1 on the vertex, so that it contains the rim of the cone’s base as shown:

Figure 1 – Cone Solid Angle
Figure 2 – Cone Cross Section

To find the area of the yellow surface of revolution, we integrate using polar coordinates. Consider an infinitesimally small segment of the yellow arc located at polar coordinates (1, \phi), with length 1<em>d\phi = d\phi. When this segment is swept around the x-axis, it moves in a circle of radius 1</em>\sin(\phi) = \sin(\phi), and thus sweeps an area of 2\pi\sin(\phi) d\phi.

This observation is slightly nontrivial, as one would expect the orientation of the segment to play a role in the area: indeed, a finite segment will sweep different areas if it is parallel to the x-axis (forming a cylinder), perpendicular (forming an annulus/ring), or anything in between. However, these differences are second-order in the length of the segment, so they vanish when considering our infinitesimally small segment.

Returning to the problem at hand, we see that the area of the yellow surface of revolution in Figure 1 can be expressed as

    \begin{equation*} $\int_{0}^{\theta} 2\pi\sin(\phi) d\phi = 2\pi[-\cos(\phi)]_{0}^{\theta} = 2\pi\left(1 - \cos(\theta)\right)$ \end{equation*}

Since \cos(\theta) = h, our final result for the yellow area, and consequently the steradian measure of the cone’s vertex opening, is simply 2\pi(1 - h). In a general (unscaled) cone with base radius r and height h, \cos(\theta) would be written as \frac{h}{\sqrt{r^2 + h^2}}, so our answer in the most general form is 2\pi\left(1 - \frac{h}{\sqrt{r^2 + h^2}}\right).

This result comes with a pleasant surprise:

The surface area of any “slice” of a sphere, that is, and portion of a sphere contained between two parallel planes, is proportional to the distance between the planes.

WLOG, consider our previous sphere of radius 1. We showed that given a cone of radius r, height h, and slant length \sqrt{r^2+h^2}=1, the area on the sphere cut off by the base of the cone is 2\pi(1-h). However, observe that 1-h simply represents the horizontal “thickness” of the the region cut off by the cone’s base! In other words, if the cone were standing vertically inside the sphere, and the sphere was filled from the bottom with water until the level reached the cone’s base, 1-h would represent the height of the wetted region (i.e. the height of the water). The surface area of the wetted region is just 2\pi times this height. Thus, we have shown that slicing a sphere resting on the ground at a height of x produces an area under the slice that is proportional to the height x, specifically by a factor of 2\pi. The desired result quickly follows, as the area between any two parallel planes can be described as the difference between two such “areas from the ground.”

There might be a second installment to this post where I discuss solid angles intercepted by a variety of other shapes, but for now, all this cone-talk has reminded me of ice cream. Until next time!

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