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[parent] solid angle (Definition)

A conical surface may contain a certain portion $ \Omega$ of the space $ \mathbb{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 $ P$ in the vertex of the solid angle, then the magnitude of the solid angle is given by

$\displaystyle \Omega = \frac{A}{R^2}$
which is independent 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 surface integral
$\displaystyle \Omega = -\int_a \vec{da}\cdot\nabla\frac{1}{r},$ (1)

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

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

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

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

$\displaystyle \tan{\frac{\Omega}{2}} = \frac{(\vec{r_1}\vec{r_2}\vec{r_3})} {\v... ...}r_3+\vec{r_2}\!\cdot\!\vec{r_3}r_1+ \vec{r_3}\!\cdot\!\vec{r_1}r_2+r_1r_2r_3},$ (2)

where the numerator of the right hand side is the triple scalar product of the vectors; the result is due to A. van Oosterom and J. Strackee 1983.

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

$\displaystyle \Omega = 4\arctan{\frac{a^2}{2h\sqrt{2a^2+4h^2}}} = 4\arcsin\frac{a^2}{a^2+4h^2}$
Here $ a$ is the side of the base square and $ h$ is the height of the pyramid.



"solid angle" is owned by pahio.
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See Also: convex angle, radian, area of a spherical triangle

Also defines:  space angle, full solid angle, steradian

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calculating the solid angle of disc (Example) by pahio
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Cross-references: pyramid, side, base, square, triple scalar product, numerator, equation, vectors, tetrahedron, axis, right circular cone, sphere, surface area, Proportion, radian, angle, Si, points, position vector, length, contained, radius, surface, contain, conical surface
There are 5 references to this entry.

This is version 16 of solid angle, born on 2005-07-25, modified 2008-08-22.
Object id is 7266, canonical name is SolidAngle.
Accessed 14960 times total.

Classification:
AMS MSC: 51M25 (Geometry :: Real and complex geometry :: Length, area and volume)
 15A72 (Linear and multilinear algebra; matrix theory :: Vector and tensor algebra, theory of invariants)
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An odd solid angle by pahio on 2008-01-10 15:07:24
Question of Yossarian:

"Hello
I just saw your entry of the Solid Angle on PlanetMath and wondered if you could give me a hint as to how to calculate the solid angle of a cylinder elegantly (if there is such a way).
The cylinder has a radius r, length L and it's center is at a distance R from the origin.
If I set the axis of the cylinder in parallel to the z-axis on the x-Axis, then the azimuth angle (theta) limits are from -ArcSin[r/R] to +ArcSin[r/R]. The polar angle (phi) limits of course depend on the azimuth. Here is what I figured out...
ArcTan[0.5*L/(R-(Cos[theta]^2 (RR*Tan[theta]^2 + Sqrt[r^2 + (r^2-R^2)*Tan[theta]^2])))]
(The biggest phi is in the direction of the Axis and then would simply give phi(0)=ArcTan[0.5*L/(R-r)] ).
I then just calculated the integral...
Omega = intint_S sin(phi)dphi dtheta with the above given limits.
Is this correct and/or is there an elegant way to calculate the solid angle of the cylinder?
As far as I see it, there is no closed form of the result. Do You agree?
Best Regards,
Stephan"

Dear Stephan,
I guess that you mean a right circular cylinder with radius r of the base circle and height L and that the center of the cylinder is on the x-axis and then one forms the smallest solid angle with apex in the origin and enclosing the cylinder. Right? Such a solid angle seems to be quite difficult to calculate. Maybe somebody else could do it. I send your question to the public forum.
Jussi
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Apical solid angle of pyramid by pahio on 2007-10-09 11:12:40
There has long been a wrong result in this solid angle (example 3). I have now corrected it and added the simpler form 4arcsin...
Jussi
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