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Solid Angle of a Sphere 📂Geometry

Solid Angle of a Sphere

Definition1

A solid angle $\Omega$ of a 3-dimensional sector with a radius of $r$ and a surface area of $A$ is defined as follows:

$$ \Omega := \dfrac{A}{r^{2}} $$

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The unit is called steradian and is denoted as $\mathrm{sr}$.

Explanation

Considering how the radian angle in a circle is defined as the ratio of the arc length to the radius, this definition seems natural.

$$ \theta := \dfrac{s}{r} $$

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However, the reason why $r$ is replaced with $r^{2}$ in the denominator is that, while the arc length is proportional to the radius, the surface area is proportional to the square of the radius. Since the surface area of a sphere is $4\pi r^{2}$, the solid angle is $4\pi$.

$$ \Omega = \dfrac{4 \pi r^{2}}{r^{2}} = 4\pi $$

This is equivalent to integrating over all angles, excluding the radius, in a spherical coordinate system volume integral, thus confirming that the solid angle is well-defined.

$$ \begin{align*} \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi}\sin\theta d\theta d\phi = 4\pi \end{align*} $$

Now, consider a unit sphere with a radius of $r=1$ as shown in the image below. Here, a specific direction refers to the $z$ axis.

3.jpg

Then, the solid angle of a 3-dimensional sector with an angle of $\theta$ is as follows.

$$ \Omega (\theta) = \dfrac{A}{r^{2}} = A = \int_{\theta ^{\prime} = 0}^{\theta}\int_{\phi=0}^{2\pi} \sin \theta d\theta d\phi = 2\pi (1-\cos\theta) $$

Therefore, the following holds true.

$$ \dfrac{d\Omega}{d\theta} = 2\pi \sin \theta \implies d\Omega = 2\pi \sin\theta d\theta $$

Hence, in physics, where integration in spherical coordinate systems is frequent, the following notation is commonly used.

$$ \int_{0}^{\pi} 2\pi \sin\theta d\theta = \int_{0}^{4\pi} d\Omega = 4\pi $$


  1. Stephen J. Blundell and Katherine M. Blundell, Concepts in Thermal Physics, translated by Lee Jae-woo (2nd Edition, 2014), p72-73 ↩︎