solid angle

A conical surface may contain a certain portion $\mathrm{\Omega }$ of the space ${ℝ}^3$.  This portion is called solid angle or space angle.  If the conical surface contains a portion $A$ of a spherical surface with radius $R$ and with centre (http://planetmath.org/Sphere) $P$ in the of the solid angle, then the magnitude of the solid angle is given by

$$\mathrm{\Omega }=\frac{A}{R^2}$$

which is on the radius $R$. The spherical surface can be replaced by any surface $a$, through which all the half-lines originating from $P$ and being contained in the solid angle go.  Then the solid angle may be computed from the

$\mathrm{\Omega }=-{\displaystyle {\int }_a}\overrightarrow{da}\cdot \nabla {\displaystyle \frac{1}{r}},$(1)

where $r$ is the length of the position vector $\overrightarrow{r}$ for the points on the surface $a$.The full solid angle, consisting of all points of ${ℝ}^3$, has the magnitude $4\pi$.

The SI of solid angle, analogous to the angle radian, is the steradian ($=1\text{sr}$).  The steradian takes a proportion $\frac{1}{4\pi }$, or approximately 7.957747 %, of the surface area of a sphere.

If the solid angle is bounded by three planes having exactly one common point, it may be called a trihedral angle; cf. the example 2!

Example 1.  The solid angle determined by a right circular cone with the angle $\alpha$ between its axis and is equal to $2\pi (1-\cos \alpha )$, i.e. $4\pi {\sin }^2{\displaystyle \frac{\alpha }{2}}$.

Example 2.  Let  ${\overrightarrow{r}}_1,{\overrightarrow{r}}_2,{\overrightarrow{r}}_3$ be the position vectors of three points in ${ℝ}^3$ and $r_1,r_2,r_3$ their lengths.  Then the solid angle $\mathrm{\Omega }$ of the tetrahedron by the vectors $\overrightarrow{r_i}$ is obtained from the equation

$\tan {\displaystyle \frac{\mathrm{\Omega }}{2}}={\displaystyle \frac{{\overrightarrow{r}}_1\times {\overrightarrow{r}}_2\cdot {\overrightarrow{r}}_3}{({\overrightarrow{r}}_1\cdot {\overrightarrow{r}}_2)r_3+({\overrightarrow{r}}_2\cdot {\overrightarrow{r}}_3)r_1+({\overrightarrow{r}}_3\cdot {\overrightarrow{r}}_1)r_2+r_1{r}_2{r}_3}},$(2)

where the numerator of the is the triple scalar product of the vectors.  This equation is expressed simplier using the unit vectors ${\overrightarrow{u}}_i$ corresponding ${\overrightarrow{r}}_i$:

$$\tan \frac{\mathrm{\Omega }}{2}=\frac{{\overrightarrow{u}}_1\times {\overrightarrow{u}}_2\cdot {\overrightarrow{u}}_3}{1+{\overrightarrow{u}}_2\cdot {\overrightarrow{u}}_3+{\overrightarrow{u}}_3\cdot {\overrightarrow{u}}_1+{\overrightarrow{u}}_1\cdot {\overrightarrow{u}}_2}$$

The result (2) is due to van Oosterom and Strackee 1983.

Example 3.  Using (2), one can easily get the apical (http://planetmath.org/ConeInMathbbR3) solid angle of a right pyramid (http://planetmath.org/ConeInMathbbR3) with square base:

$$\mathrm{\Omega }=\mathrm{\hspace{0.33em}4}\arctan \frac{a^2}{2h\sqrt{2a^2+4h^2}}=\mathrm{\hspace{0.33em}4}\arcsin \frac{a^2}{a^2+4h^2}$$

Here $a$ is the side of the base square and $h$ is the height (http://planetmath.org/ConeInMathbbR3) of the pyramid.  Cf. the solid angle of rectangular pyramid.