Solid angle

The solid angle, Ω, is the two-dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large that object appears to an observer looking from that point. A small object nearby may subtend the same solid angle as a larger object farther away (for example, the small/near Moon can totally eclipse the large/remote Sun because, as observed from a point on the Earth, both objects fill almost exactly the same amount of sky). An object's solid angle is equal to the area of the segment of unit sphere (centered at the vertex of the angle) restricted by the object (this definition works in any dimension, including 1D and 2D). A solid angle equals the area of a segment of unit sphere in the same way a planar angle equals the length of an arc of unit circle.

The area on a sphere, equal to the square of its radius and observed from its center, subtends precisely one steradian.

The units of solid angle can be called steradian (abbreviated "sr") according to SI. From the point of view of mathematics and physics solid angle is dimensionless and has no units, thus "sr" might be skipped in scientific texts. 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 faces is one-sixth of that, or 2π/3 sr. Solid angles can also be measured in square degrees (1 sr = (180/π)² square degree) or in fractions of the sphere (i.e., fractional area), 1 sr = 1/4π fractional area.

In spherical coordinates, there is a simple formula as

$d\Omega =\sin \theta \,d\theta \,d\varphi .$

The solid angle for an arbitrary oriented surface S subtended at a point P is equal to the solid angle of the projection of the surface S to the unit sphere with center P, which can be calculated as the surface integral:

$\Omega =\iint _{S}{\frac {{\vec {r}}\cdot {\hat {n}}\,dS}{r^{3}}}=\iint _{S}\sin \theta \,d\theta \,d\varphi .$

where ${\vec {r}}$ is the vector position of an infinitesimal area of surface $\,dS$ with respect to point P and where ${\hat {n}}$ represents the unit vector normal to $\,dS$. Even if the projection on the unit sphere to the surface S is not isomorphic, the multiple folds are correctly considered according to the surface orientation described by the sign of the scalar product ${\vec {r}}\cdot {\hat {n}}$.

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 to the fact that there is an analogous theorem to the theorem that "The sum of internal angles of a planar triangle is 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 subtended by 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:

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

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.

When implementing the above equation care must be taken with the atan function to avoid negative or incorrect solid angles. One source of potential errors is that the determinant can be negative if a,b,c have the wrong winding. Computing abs(det) is a sufficient solution since no other portion of the equation depends on the winding. The other pitfall arises when the determinant is positive but the divisor is negative. In this case atan returns a negative value that must be biased by $\pi$.

float tri_projection(float3 a, float3 b, float3 c)
{
  float det = abs(dot(a,cross(b,c)));

  float al = length(a);
  float bl = length(b);
  float cl = length(c);

  float div = al*bl*cl + dot(a,b)*cl + dot(a,c)*bl + dot(b,c)*al;
  float at = atan2(det, div);
  if(at < 0) at += M_PIf; // If det>0 && div<0 atan2 returns < 0, so add pi.
  float omega = 2.0f * at;

  return omega;
}

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

$\theta _{s}={\frac {\theta _{a}+\theta _{b}+\theta _{c}}{2}}.$

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.
• Deriving Gauss's Law.
• Calculating emissive power and irradiation in heat transfer.
• Calculating cross sections in Rutherford scattering.

Solid angles for common objects

Cone, spherical cap, hemisphere

Section of cone (1) and spherical cap (2) inside a sphere. In this figure θ = a/2 and r = 1.

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 coordinates):

$\int _{0}^{2\pi }\int _{0}^{\theta }\sin \theta '\,d\theta '\,d\phi =2\pi \int _{0}^{\theta }\sin \theta '\,d\theta '=2\pi \left[-\cos \theta '\right]_{0}^{\theta }\ =2\pi \left(1-\cos \theta \right).$

Over 2200 years ago Archimedes proved, without the use of calculus, that the surface area of a spherical cap was always equal to the area of a circle whose radius was equal to the distance from the rim of the spherical cap to the point where the cap's axis of symmetry intersects the cap. 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.

The solid angle of the complement of the cone (picture a melon with the cone cut out) is clearly:

$4\pi -\Omega =2\pi \left(1+\cos {\theta }\right).\,\!$

A terran astronomical observer positioned at latitude $\theta \,\!$ can see this much of the celestial sphere as the earth rotates, that is, a proportion

${\frac {1}{2}}\left(1+\cos {\theta }\right).\,\$

At the equator you see all of the celestial sphere, at either pole only one half.

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

$\Omega =4\arcsin \left(\sin {a \over 2}\sin {b \over 2}\right).\,\!$

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 sphere) are known, then the above equation can be manipulated to give

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

The solid angle of a right n-gonal pyramid, where the pyramid base is a regular n-sided polygon of radius (r), with a pyramid height (h) is

$\Omega =2\pi \ -\ 2n\arctan \left({\frac {\tan {\pi \over n}}{\sqrt {1+{\ r^{2} \over h^{2}}}}}\right).\,\!$

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 an apparent diameter of about 0.5°, thus they each cover a solid angle of about 0.20 deg² or square degrees, thus they each cover a fractional area of approximately 0.00048% of the total celestial sphere which is about 6×10⁻⁵ 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

$\Omega _{d}={\frac {d\pi ^{d/2}}{\Gamma \left({\frac {d}{2}}+1\right)}}$

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

$\Omega _{d}={\begin{cases}{\frac {d\pi ^{d/2}}{\left({\frac {d}{2}}\right)!}}&d={\dot {2}}\\{\frac {2^{d}\left({\frac {d-1}{2}}\right)!}{(d-1)!}}\pi ^{(d-1)/2}&d\neq {\dot {2}}\end{cases}}$

This gives the expected results of 2π rad for the 2D circumference and 4π sr 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 set ${-1,1}$, which indeed has a measure of 2.

References

• Arthur P. Norton, A Star Atlas, Gall and Inglis, Edinburgh, 1969
• F. M. Jackson, Polytopes in Euclidean n-Space. Inst. Math. Appl. Bull. (UK) 29, 172-174, Nov./Dec. 1993.
• M. G. Kendall, A Course in the Geometry of N Dimensions, No. 8 of Griffin's Statistical Monographs & Courses, ed. M. G. Kendall, Charles Griffin & Co. Ltd, London, 1961
• Weisstein, Eric W. "Spherical Excess". MathWorld.
• Weisstein, Eric W. "Solid Angle". MathWorld.