I wonder why the equation $(2)$ for steradians of a probability density function given in Henyey-Greenstein-Phase function is still dependent on the function of radius $p(\theta)$. So that the solid angle in steradians $\Omega$ is:
$$ \Omega_{p} = \frac{A}{r^2} \,\,\, \stackrel{?}{=} \,\,\, \int_{0}^{\pi} 2\pi \: p(\theta) \: sin(\theta) \: d\theta $$
Although I am certain the source is correct, I do not understand where that equation comes from. Please let me show a reason why I expected the equation to have no $p(\theta)$ or at least to have it as $p(\theta)^2$.
Surface of an object of revolution: Let us calculate the surface area of a revolved object first. Imagine you have an upright axis called the polar axis. Around this axis is a revolved or twisted surface. A sphere would be an arc with constant radius revolved around that axis. Like the meridian in the following image:
If we make the radius from the center of the polar axis varying, so that it depends on $\theta$ we get $r \,(\,\theta\,)$ and we can influence that arc. The following figure shows the polar axis as horizontal $X - axis$. But still the distance from center $(0,0)$ is given as a radius and not as an perpendicular axis $Y$.
Notice, that the resulting revolved object is independent of $\phi$ and is therefore symmetrical around the polar axis.
To calculate the surface of such an object given by $r\,(\,\theta\,)$ with $\theta \in [0,\pi$] and $\phi \in [0,2\pi] $ my initial assumption is to sum up (integrate) each ring around the polar axis that the given radius creates at position $\theta$.
$$\begin{align} A &= RingAlongPolarAxis \cdot DifferentialHeightOfRing\\ & = \int_{0}^{\pi} 2\pi \, sin(\theta) \, r(\theta) \cdot \sqrt{\Big(r(\theta)cos(\theta)\Big)^2+\Big(r(\theta)sin(\theta)\Big)^2} d\theta \\ & = \int_{0}^{\pi} 2\pi \, sin(\theta) \, r(\theta) \cdot \sqrt{ r(\theta)^2 \Big(cos(\theta)^2+sin(\theta)^2\Big) } d\theta \\ & = \int_{0}^{\pi} 2\pi \, sin(\theta) \, r(\theta) \cdot \sqrt{ r(\theta)^2 \cdot 1 } \: d\theta \\ & = \int_{0}^{\pi} 2\pi \, sin(\theta) \, r(\theta) \cdot r(\theta) \: d\theta \\ & = \int_{0}^{\pi} 2\pi \, sin(\theta) \, r(\theta)^2 \: d\theta \\ \end{align} $$
This seems to be correct because for a constant $r$ it yields the well known surface area of a sphere: $A=4\pi \, r^2$. But in here the radius function is squared. In the integral in question $p(\theta)$ is not squared.
Surface area in steradian: To get to $\Omega$ which is in steradian we need to calculate the surface area per $radius^2$. Let us calculate $\Omega$ of a sphere by putting the integral equation into the steradian definition:
$$ \begin{align} \Omega &= \frac{A}{r^2} \\ &= \int_{0}^{\pi} \frac{2\pi \, sin(\theta) \, r(\theta)^2}{r(\theta)^2} \: d\theta \\ &= \int_{0}^{\pi} 2\pi \, sin(\theta) \: d\theta \\ &= 2\pi \int_{0}^{\pi} sin(\theta) \: d\theta \\ &= 2\pi \, 2 = 4 \, \pi \\ \end{align} $$
As I can see, there is a good reason why a sphere has $4\,\pi$ steradians, regardless of its size and why steradians loose the unit of area e.g. $m^2 \Rightarrow$ the radius is just cancelled out and with it goes the unit of length (squared).
Questions:
- So why is the function that produces the radius dependent of $\theta$ still in the integral equation (2) for the surface area of the Henyey Greenstein function?
- If $p(\theta)$ is still in the equation how can it produce a unitless steradian value?