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I haven't seen this explained clearly anywhere. Solid angles are described usually as a fraction of the surface area of a unit sphere, similar to how angles are the fraction of the circumference of a unit circle. However, I don't know how solid angles are actually quantified.

Are solid angles just a single number, the describes this fraction of the area? It's confusing to me since often times, I've seen integrals that integrate over a sphere using solid angles, which seems to imply that solid angles are multi-dimensional quantities (e.g. when integrating using spherical coordinates, the solid angle would have to consist of the azimuthal and polar angles covered by the differential solid angle).

Following from this, how would you write down a solid angle that covers the entire surface of a unit sphere?

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Steradians, like radians, are a dimensionless quantity. See en.wikipedia.org/wiki/Solid_angle and en.wikipedia.org/wiki/Steradian. There are $4\pi$ steradians in a sphere. –  Nathaniel Feb 12 at 9:56
    
The reason we carry these dimensionless "labels" is partly to keep track of what we're doing, and partly to make sure nobody confuses, e.g., radians with degrees of arc. You're correct that one could define a polar cap in terms of 2D $ \theta , \phi $ just as well as in terms of steradians. –  Carl Witthoft Feb 12 at 12:47
    
a steradian is a radian squared. that is of course, unitless. –  Dimensio1n0 Feb 13 at 3:07
    
Related: Are angles ever multiplied? –  Dimensio1n0 Feb 13 at 3:08

3 Answers 3

up vote 7 down vote accepted

The solid angle is defined as the area on the unit sphere subtended by the angle divided by one unit area. It's a ratio so it's a single dimensionless number.

I see why you think it should be a 2D quantity, because the surface of a sphere, and any patch on it, is a 2D manifold and you need two quantities (traditionally $\theta$ and $\phi$) to map it. When you calculate an area on the sphere you are basically calculating a definite integral over $\theta$ and $\phi$, and the result is of course just a single number. You do lose information in the process - for example you just know the total area not the shape of the patch on the sphere.

The solid angle that covers the whole sphere is of course $4\pi$/1 or $4\pi$.

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When doing surface integrals like you say, you would always normalise by $R^2$.

So, if you give a ray's direction by the spherical co-ordinates $\theta,\,\phi$, and you want the solid angle subtended by a bundle of such rays (e.g. a light field) converging on some point, then it is by definition the area of the part of the sphere "pierced" by the bundle. The point is, when using a sphere like this, the radius is immaterial in defining directions. The direction defined by the position vector of a point on the unit sphere or on a point on a sphere of any radius is exactly the same direction. So, by convention, the radius is chosen to be one unit. Alternatively, the solid angle is the fraction of the sphere pierced by the ray system, but multiplied by $4\pi$. Hopefully you can see that the $4\pi$ is a convention: we could equally well define solid angle as simply the fraction of the area of the sphere. However, the $4\pi$ does make intuitive sense from the standpoint of units when we think about the Gauss-Bonnet theorem, which is a thoroughly fascinating idea linked to solid angle.

Here we talk about a compact two-dimensional manifold $M$, which the sphere is an example of. It may be boundaryless, like the sphere, or it might have a hole cut through it so that there is a boundary $\partial\, M$, like a soap bubble billowing out of its bounding wire bubble-blower (essential kit for anyone with children of between two and five years old in tow!). The Gauss-Bonnet theorem is:

$$\int_M \,K_G(\vec{r})\,\mathrm{d}\,A + \int_{\partial M} K_{FS}(\vec{r})\,\mathrm{d}\,s = 2\,\pi\,\chi(M)$$

Here $K_G(\vec{r})$ is the Gaussian curvature of the manifold. This is defined by imagining a plane at right-angles to the point $\vec{r}$ on the surface in question. The intersection of the plane and the surface will be a space curve confined to the plane, and it will have some curvature $\kappa = \mathrm{d}_s \theta$ in radians bent per unit distance run along the curve. There are two principal planes through the point in question which are those for which $\kappa$ is extremal. The Gaussian curvature is the product of the two sectional extremal curvatures. If we have to do the second integral over the boundary, $K_{FS}$ is simply the curvature from a Frenet-Serret description of the boundary.

You'll notice that the Gaussian curvature, being the product of two curvatures, has dimensions of inverse length squared. So this dimension cancels the dimension of the area element $\mathrm{d}\,A$. Likewise for the integral over the boundary: the curvature has the dimensions of inverse length which cancel with the dimensions of the line element $\mathrm{d} s$.

For our sphere there is no hole $\partial M = \emptyset$ and the Gaussian Curvature is constant $R^{-2}$. So the solid angle subtended by the whole sphere agrees with the Gauss-Bonnet theorem only if we use $4\pi$ steradians = solid angle subtended by all directions. The Gauss Bonnet theorem has a very deep meaning: $\chi(M)$ is the manifold's Euler characteristic, the topoligical invariant found from a simple "quantum calculation" derived from any triangulation of the surface. We can warp, shrink, squash the sphere into any shape we want: as long as we don't tear or cut it, then $\chi(M) = 2$.

So in the Gauss-Bonnet theorem you can both see the units cancelling out and a generalisation of the concept of solid angle! One could imagine, however, getting rid of the $2 \pi$ in the Gauss-Bonnet theorem by redefining angles, but it would be a bit messy (we'd have to have $2 \pi$ before the first term of the LHS of the GB theorem). So, whilst many conventions can be agreed on, we're stuck with the one that defines a sphere as $4\pi$ steradian.

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John Rennie's answer seems fine to me (+1). I'll only add the relevant pieces of the BIPM brochure (PDF, p. 118). BIPM rules.

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