Solid angle

The solid angle, Ω, is the angle in three-dimensional space that an object subtends at a point. It is a measure of how big that object appears to an observer looking from that point. For instance, a small object nearby could subtend the same solid angle as a large object far away. The solid angle is proportional to the surface area, S, of a projection of that object onto a sphere centered at that point, divided by the square of the sphere's radius, R. (Symbolically, Ω = k S/R2, where k is the proportionality constant.) A solid angle is related to the surface of a sphere in the same way an ordinary angle is related to the circumference of a circle.

If the proportionality constant is chosen to be 1, the units of solid angle will be the SI steradian (abbreviated "sr"). Thus the solid angle of a sphere measured from a point in its interior is 4π sr, and the solid angle subtended at the center of a cube by one of its sides is one-sixth of that, or 2π/3 sr. Solid angles can also be measured (for k = (180/π)2) in square degrees or (for k = 1/4π) in fractions of the sphere (i.e., fractional area).

One way to determine the fractional area subtended by a spherical surface is to divide the area of that surface by the entire surface area of the sphere. The fractional area can then be converted to steradian or square degree measurements by the following formulae:


 * 1) To obtain the solid angle in steradians, multiply the fractional area by 4π.
 * 2) To obtain the solid angle in square degrees, multiply the fractional area by 4π × (180/π)2, which is equal to 129600/π.

More rigorously, the solid angle for a surface S subtended at a point P is given by the surface integral:


 * $$\Omega = \iint_S \frac { \vec{r} \cdot \vec{n} dS }{r^3}.$$

where $$ \vec{r} $$ is the vector position of an infinitesimal area of surface $$ \, dS $$ with respect to point P and where $$ \vec n $$ represents the unit vector normal to $$ \, dS $$.

Practical applications

 * Defining luminous intensity and luminance
 * Calculating spherical excess E of a spherical triangle
 * The calculation of potentials by using the boundary element method (BEM)
 * Evaluating the size of ligands in metal complexes, see ligand cone angle.
 * Calculating the electric field and magnetic field strength around charge distributions.

Tetrahedron
Let OABC be the vertices of a tetrahedron with an origin at O subtended by the triangular face ABC where $$\vec a\ ,\, \vec b\ ,\, \vec c $$ are the vector positions of the vertices A, B and C. Define the vertex angle $$ \theta_a \, $$ to be the angle BOC and define $$ \theta_b ,\, \theta_c $$ correspondingly. Let $$ \phi_{ab} \, $$ be the dihedral angle between the planes that contain the tetrahedral faces OAC and OBC and define $$ \phi_{bc} ,\, \phi_{ac} $$ correspondingly. The solid angle at O subtended by the triangular surface ABC is given by


 * $$ \Omega = \phi_{ab} + \phi_{bc} + \phi_{ac} - \pi. \, $$

This follows from the theory of spherical excess and it leads on to the fact that there is an analogous theorem to the sum of internal angles of a triangle equal to $$ \pi $$, for the sum of the four internal solid angles of a tetrahedron as follows:


 * $$ \sum_{i=1}^4 \Omega_i = 2 \sum_{i=1}^6 \phi_i - 4 \pi $$

where $$ \phi_i \, $$ ranges over all six of the dihedral angles between any two planes that contain the tetrahedral faces OAB, OAC, OBC and ABC.

An efficient algorithm for calculating the solid angle at O that subtends the triangular surface ABC where $$\vec a\ ,\, \vec b\ ,\, \vec c $$ are the vector positions of the vertices A, B and C has been given by Oosterom and Strackee (IEEE Trans. Biom. Eng., Vol BME-30, No 2, 1983):



\tan \left( \frac{1}{2} \Omega \right) = \frac{[\vec a\ \vec b\ \vec c]}{ abc + (\vec a \cdot \vec b)c + (\vec a \cdot \vec c)b + (\vec b \cdot \vec c)a}, $$

where


 * $$[\vec a\ \vec b\ \vec c]$$

denotes the determinant of the matrix that results when writing the vectors together in a row, e.g. $$M_{i1}=\vec a_i$$ and so on--this is also equivalent to the scalar triple product of the three vectors;


 * $$\vec a$$ is the vector representation of point A, while $$ \, a $$ is the magnitude of that vector (the origin-point distance);


 * $$\vec a \cdot \vec b$$ denotes the scalar product.

Another useful formula for calculating the solid angle of the tetrahedron at the origin O that is purely a function of the vertex angles $$\theta_a ,\, \theta_b ,\, \theta_c $$ is given by L' Huilier's theorem as


 * $$ \tan \left( \frac{1}{4} \Omega \right)

= \sqrt{ \tan \left( \frac{\theta_s}{2}\right) \tan \left( \frac{\theta_s - \theta_a}{2}\right) \tan \left( \frac{\theta_s - \theta_b}{2}\right) \tan \left( \frac{\theta_s - \theta_c}{2}\right)} $$

where
 * $$ \theta_s = \frac {\theta_a + \theta_b + \theta_c}{2}. $$

Cone, spherical cap, hemisphere
The solid angle of a cone with apex angle $$2 \theta \,\!$$, is the area of a spherical cap on a unit sphere


 * $$\Omega = 2 \pi \left (1 - \cos {\theta} \right) .\,\!$$

(The above result is found by computing the following double integral using the unit surface element in spherical polars):


 * $$\int_0^{2\pi} \int_0^{\theta} \sin \theta' \, d \theta' \, d \phi = 2\pi\int_0^{\theta} \sin \theta' \, d \theta' = 2\pi\left[ -\cos \theta' \right]_0^{\theta} \ = 2\pi\left(1 -\cos \theta \right).$$

Over 2200 years ago Archimedes proved, without the use of calculus, that the surface area of a spherical cap mapped identically onto the area of a circle whose radius was equal to the distance from the rim of the spherical cap to the lowest point on the surface of the spherical cap below the rim. In the diagram opposite this radius is given as:


 * $$ 2r \sin \left( \frac{ \theta}{2} \right).\,\ $$

Hence for a unit sphere the solid angle of the spherical cap is given as:


 * $$ \Omega = 4 \pi \sin^2 \left( \frac{ \theta}{2} \right) = 2 \pi \left (1 - \cos {\theta} \right) .\,\ $$

When θ = π/2, the spherical cap becomes a hemisphere having a solid angle 2π.

Pyramid
The solid angle of a four-sided right rectangular pyramid with apex angles


 * $$a \,\!$$ and $$b \,\!$$ (dihedral angles measured to the opposite side faces of the pyramid) is $$4 \arcsin \left (\sin {a \over 2} \sin {b \over 2} \right). \,\!$$

If both the side lengths (α and β) of the base of the pyramid and the distance (d) from the center of the base rectangle to the apex of the pyramid (the center of the sfere) are known, then the above equation can be manipulated to give


 * $$\Omega = 4 \arcsin \frac {\alpha\beta} {\sqrt{(4d^2+\alpha^2)(4d^2+\beta^2)}}. \,$$

Latitude-longitude rectangle
The solid angle of a latitude-longitude rectangle on a globe is $$\left ( \sin \phi_N - \sin \phi_S \right ) \left ( \theta_E - \theta_W \,\! \right)$$, where $$\phi_N \,\!$$ and $$\phi_S \,\!$$ are north and south lines of latitude (measured from the equator in radians with angle increasing northward), and $$\theta_E \,\!$$ and $$\theta_W \,\!$$ are east and west lines of longitude (where the angle in radians increases eastward). Mathematically, this represents an arc of angle $$\phi_N - \phi_S \,\!$$ swept around a sphere by $$\theta_E - \theta_W \,\!$$ radians. When longitude spans 2π radians and latitude spans π radians, the solid angle is that of a sphere.

A latitude-longitude rectangle should not be confused with the solid angle of a rectangular pyramid. All four sides of a rectangular pyramid intersect the sphere's surface in great circle arcs. With a latitude-longitude rectangle, only lines of longitude are great circle arcs; lines of latitude are not.

Sun and Moon
The Sun and Moon are both seen from Earth at a fractional area of 0.001% of the celestial hemisphere or about 6 steradian.

Solid angles in arbitrary dimensions
The solid angle subtended by the full surface of the unit n-sphere can be defined in any number of dimensions $$d$$. One often needs this solid angle factor in calculations with spherical symmetry. It is given by the formula



\Omega_{d} = \frac{2\pi^{d/2}}{\Gamma \left (\frac{d}{2} \right )} \qquad \forall d \in \mathbb{N}

$$

where $$\Gamma$$ is the Gamma function. Since $$d$$ is an integer, the Gamma function can be computed explicitly. It follows that



\Omega_{d} = \begin{cases} \frac{d\pi^{d/2}}{ \left (\frac{d}{2} \right )!} & d = \dot{2} \\ \frac{2^d\left (\frac{d-1}{2} \right ) !}{(d-1)!} \pi^{(d-1)/2} & d \ne \dot{2} \end{cases}

$$

This gives the expected results of 2&pi; rad for the 2D circumference and 4&pi; srad for the 3D sphere. It also throws the slightly less obvious 2 for the 1D case, in which the origin-centered unit "sphere" is the interval $$[ -1, 1 ]$$, which indeed has a measure of 2.