Editing Solid angle (section)

== Solid angles for common objects ==

=== Cone, spherical cap, hemisphere  ===
[[Image:Steradian cone and cap.svg|thumb|right|250px|Diagram showing a section through the centre of a cone (1) subtending a solid angle of 1 steradian in a sphere of radius r, along with the spherical "cap" (2). The external surface area A of the cap equals <math>r^2</math> only if solid angle of the cone is exactly 1 steradian. Hence, in this figure {{math|''θ'' {{=}} ''A''/2}} and {{math|''r'' {{=}} 1}}.]]
The solid angle of a [[cone (geometry)|cone]] with its apex at the apex of the solid angle, and with [[Apex (geometry)|apex]] angle 2{{math|''θ''}}, is the area of a [[spherical cap]] on a [[unit sphere]]

<math display=block>\Omega = 2\pi \left (1 - \cos\theta \right)\ = 4\pi \sin^2 \frac{\theta}{2}.</math>

For small {{math|''θ''}} such that {{math|cos ''θ'' ≈ 1 − {{sfrac|''θ''<sup>2</sup>|2}}}} this reduces to {{math|π''θ''<sup>2</sup>  ≈  π''r''<sup>2</sup>}}, the area of a circle. (As {{math|h → 0, ''θ'' → r}}.)

The above is found by computing the following [[double integral]] using the unit [[Spherical coordinate system#Integration and differentiation in spherical coordinates|surface element in spherical coordinates]]:

<math display=block>\begin{align}
\int_0^{2\pi} \int_0^\theta \sin\theta' \, d \theta' \, d \phi
&= \int_0^{2\pi} d \phi\int_0^\theta \sin\theta' \, d \theta' \\
&= 2\pi\int_0^\theta \sin\theta' \, d \theta' \\
&= 2\pi\left[ -\cos\theta' \right]_0^{\theta} \\
&= 2\pi\left(1 - \cos\theta \right).
\end{align}</math>

This formula can also be derived without the use of [[calculus]]. 

Over 2200 years ago [[Archimedes]] proved that the surface area of a spherical cap is always equal to the area of a circle whose radius equals the distance from the rim of the spherical cap to the point where the cap's axis of symmetry intersects the cap.<ref>{{cite journal |year = 2015 |title = Archimedes on Spheres and Cylinders |journal = Math Pages |url = http://www.mathpages.com/home/kmath343/kmath343.htm}}</ref> [[File:Archimedes-spherical-cap.png|thumb|Archimedes' theorem that surface area of the region of sphere below horizontal plane H in given diagram is equal to area of a circle of radius t.]]In the above coloured diagram this radius is given as
<math display=block> 2r \sin \frac{\theta}{2}. </math>
In the adjacent black & white diagram this radius is given as "t".

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

<math display=block> \Omega = 4\pi \sin^2 \frac{\theta}{2} = 2\pi \left (1 - \cos\theta \right). </math>

{{anchor|hemisphere}}
When {{math|''θ''}} = {{sfrac|{{pi}}|2}}, the spherical cap becomes a [[Sphere|hemisphere]] having a solid angle 2{{pi}}.

The solid angle of the complement of the cone is
 
<math display=block>4\pi - \Omega = 2\pi \left(1 + \cos\theta \right) = 4\pi\cos^2 \frac{\theta}{2}.</math>

This is also the solid angle of the part of the [[celestial sphere]] that an astronomical observer positioned at latitude {{math|''θ''}} can see as the Earth rotates. At the equator all of the celestial sphere is visible; at either pole, only one half.

The solid angle subtended by a segment of a spherical cap cut by a plane at angle {{mvar|''γ''}} from the cone's axis and passing through the cone's apex can be calculated by the formula<ref name = Mazonka>{{cite arXiv |last =  Mazonka |first = Oleg |year = 2012 |title = Solid Angle of Conical Surfaces, Polyhedral Cones, and Intersecting Spherical Caps |eprint=1205.1396 |class = math.MG}}</ref>

<math display=block> \Omega = 2 \left[ \arccos \left(\frac{\sin\gamma}{\sin\theta}\right) - \cos\theta \arccos\left(\frac{\tan\gamma}{\tan\theta}\right) \right]. </math>

For example, if {{math|1=''γ'' = −''θ''}}, then the formula reduces to the spherical cap formula above: the first term becomes {{pi}}, and the second {{math|{{pi}} cos ''θ''}}.

===Tetrahedron===
Let OABC be the vertices of a [[tetrahedron]] with an origin at O subtended by the triangular face ABC where <math>\vec a\ ,\, \vec b\ ,\, \vec c </math> are the vector positions of the vertices A, B and C. Define the [[vertex angle]] {{mvar|θ<sub>a</sub>}} to be the angle BOC and define {{mvar|θ<sub>b</sub>}}, {{mvar|θ<sub>c</sub>}} correspondingly. Let <math>\phi_{ab}</math> be the [[dihedral angle]] between the planes that contain the tetrahedral faces OAC and OBC and define <math>\phi_{ac}</math>, <math>\phi_{bc}</math> correspondingly. The solid angle {{math|Ω}} subtended by the triangular surface ABC is given by

<math display=block> \Omega = \left(\phi_{ab} + \phi_{bc} + \phi_{ac}\right)\ - \pi.</math>

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:

<math display=block> \sum_{i=1}^4 \Omega_i = 2 \sum_{i=1}^6 \phi_i\ - 4 \pi,</math>

where <math>\phi_i</math> ranges over all six of the dihedral angles between any two planes that contain the tetrahedral faces OAB, OAC, OBC and ABC.<ref>{{cite journal |last1=Hopf |first1=Heinz |title=Selected Chapters of Geometry |journal=ETH Zurich |date=1940 |pages=1–2 |url=http://pi.math.cornell.edu/~hatcher/Other/hopf-samelson.pdf |archive-url=https://web.archive.org/web/20180921122755/http://pi.math.cornell.edu/~hatcher/Other/hopf-samelson.pdf |archive-date=2018-09-21 |url-status=live}}</ref>

A useful formula for calculating the solid angle of the tetrahedron at the origin O that is purely a function of the vertex angles {{mvar|θ<sub>a</sub>}}, {{mvar|θ<sub>b</sub>}}, {{mvar|θ<sub>c</sub>}} is given by [[Simon Antoine Jean L'Huilier|L'Huilier]]'s theorem<ref>{{cite web|url=http://mathworld.wolfram.com/LHuiliersTheorem.html|title=L'Huilier's Theorem – from Wolfram MathWorld |publisher=Mathworld.wolfram.com |date=2015-10-19|access-date=2015-10-19}}</ref><ref>{{cite web|url=http://mathworld.wolfram.com/SphericalExcess.html|title=Spherical Excess – from Wolfram MathWorld |publisher=Mathworld.wolfram.com |date=2015-10-19|access-date=2015-10-19}}</ref> as

<math display=block> \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)}, </math>

where
<math display=block> \theta_s = \frac {\theta_a + \theta_b + \theta_c}{2}. </math>

Another interesting formula involves expressing the vertices as vectors in 3 dimensional space. Let <math>\vec a\ ,\, \vec b\ ,\, \vec c </math> be the vector positions of the vertices A, B and C, and let {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} be the magnitude of each vector (the origin-point distance). The solid angle {{math|Ω}} subtended by the triangular surface ABC is:<ref>{{cite journal| first=Folke| last=Eriksson| title= On the measure of solid angles| journal= Math. Mag.| volume=63|issue=3|pages=184–187|year=1990| doi=10.2307/2691141| jstor=2691141}}</ref><ref>{{cite journal| last = Van Oosterom| first = A|author2=Strackee, J | year = 1983| title = The Solid Angle of a Plane Triangle| journal = IEEE Trans. Biomed. Eng.| volume = BME-30| issue = 2| pages = 125–126| doi = 10.1109/TBME.1983.325207| pmid = 6832789| s2cid = 22669644}}</ref>

<math display=block>\tan \left( \frac{1}{2} \Omega \right) =
  \frac{\left|\vec a\ \vec b\ \vec c\right|}{abc + \left(\vec a \cdot \vec b\right)c + \left(\vec a \cdot \vec c\right)b + \left(\vec b \cdot \vec c\right)a},
</math>

where
<math display=block>\left|\vec a\ \vec b\ \vec c\right|=\vec a \cdot (\vec b \times \vec c)</math>

denotes the [[triple product|scalar triple product]] of the three vectors and <math>\vec a \cdot \vec b</math> denotes the [[scalar product]].

Care must be taken here to avoid negative or incorrect solid angles. One source of potential errors is that the scalar triple product can be negative if {{mvar|a}}, {{mvar|b}}, {{mvar|c}} have the wrong [[determinant|winding]]. Computing the absolute value is a sufficient solution since no other portion of the equation depends on the winding. The other pitfall arises when the scalar triple product is positive but the divisor is negative. In this case returns a negative value that must be increased by {{pi}}.

===Pyramid===
The solid angle of a four-sided right rectangular [[Pyramid (geometry)|pyramid]] with [[apex (geometry)|apex]] angles {{mvar|a}} and {{mvar|b}} ([[dihedral angle]]s measured to the opposite side faces of the pyramid) is
<math display=block>\Omega = 4 \arcsin \left( \sin \left({a \over 2}\right) \sin \left({b \over 2}\right) \right). </math>

If both the side lengths ({{math|''α''}} and {{math|''β''}}) of the base of the pyramid and the distance ({{math|''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

<math display=block>\Omega = 4 \arctan \frac {\alpha\beta} {2d\sqrt{4d^2 + \alpha^2 + \beta^2}}. </math>

The solid angle of a right {{mvar|n}}-gonal pyramid, where the pyramid base is a regular {{mvar|n}}-sided polygon of circumradius {{mvar|r}}, with a 
pyramid height {{mvar|h}} is

<math display=block>\Omega = 2\pi - 2n \arctan\left(\frac {\tan \left({\pi\over n}\right)}{\sqrt{1 + {r^2 \over h^2}}} \right). </math>

The solid angle of an arbitrary pyramid with an {{math|''n''}}-sided base defined by the sequence of unit vectors representing edges {{math|{''s''<sub>1</sub>, ''s''<sub>2</sub>}, ... ''s''<sub>''n''</sub>}} can be efficiently computed by:<ref name ="Mazonka"/>

<math display=block> \Omega = 2\pi - \arg \prod_{j=1}^{n} \left(
    \left( s_{j-1} s_j \right)\left( s_{j} s_{j+1} \right) -
    \left( s_{j-1} s_{j+1} \right) +
    i\left[ s_{j-1} s_j s_{j+1} \right]
  \right).
</math>

where parentheses (* *) is a [[scalar product]] and square brackets [* * *] is a [[scalar triple product]], and {{mvar|i}} is an [[imaginary unit]]. Indices are cycled: {{math|''s''<sub>0</sub> {{=}} ''s''<sub>''n''</sub>}} and {{math|''s''<sub>1</sub> {{=}} ''s''<sub>''n'' + 1</sub>}}.  The complex products add the phase associated with each vertex angle of the polygon.  However, a multiple of
<math>2\pi</math> is lost in the branch cut of <math>\arg</math> and must be kept track of separately.  Also, the running product of complex phases must scaled occasionally to avoid underflow in the limit of nearly parallel segments.

=== Latitude-longitude rectangle ===
The solid angle of a latitude-longitude rectangle on a [[globe]] is
<math display=block>\left ( \sin \phi_\mathrm{N} - \sin \phi_\mathrm{S} \right ) \left ( \theta_\mathrm{E} - \theta_\mathrm{W} \,\! \right)\;\mathrm{sr},</math>
where {{math|''φ''<sub>N</sub>}} and {{math|''φ''<sub>S</sub>}} are north and south lines of [[latitude]] (measured from the [[equator]] in [[radian]]s with angle increasing northward), and {{math|''θ''<sub>E</sub>}} and {{math|''θ''<sub>W</sub>}} are east and west lines of [[longitude]] (where the angle in radians increases eastward).<ref>{{cite journal| year = 2003| title = Area of a Latitude-Longitude Rectangle| journal = The Math Forum @ Drexel| url = http://mathforum.org/library/drmath/view/63767.html}}</ref> Mathematically, this represents an arc of angle {{math|''ϕ''<sub>N</sub> − ''ϕ''<sub>S</sub>}} swept around a sphere by {{math|''θ''<sub>E</sub> − ''θ''<sub>W</sub>}} radians. When longitude spans 2{{pi}} radians and latitude spans {{pi}} 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.

=== Celestial objects ===
By using the definition of [[angular diameter]], the formula for the solid angle of a celestial object can be defined in terms of the radius of the object, <math display="inline">R</math>, and the distance from the observer to the object, <math>d</math>:

<math display=block>\Omega = 2 \pi \left (1 - \frac{\sqrt{d^2 - R^2}}{d} \right ) : d \geq R.</math>  

By inputting the appropriate average values for the [[Sun]] and the [[Moon]] (in relation to Earth), the average solid angle of the Sun is {{val|6.794|e=-5}} steradians and the average solid angle of the [[Moon]] is {{val|6.418|e=-5}} steradians. In terms of the total celestial sphere, the [[Sun]] and the [[Moon]] subtend average ''fractional areas'' of {{val|0.0005406}}% ({{val|5.406|u=[[part per million|ppm]]}}) and {{val|0.0005107}}% ({{val|5.107|u=ppm}}), respectively. As these solid angles are about the same size, the Moon can cause both total and annular solar [[Solar eclipse|eclipses]] depending on the distance between the Earth and the Moon during the eclipse.