# If I square a measure in radians, is the result measured in steradians?

Originally, SI units included radians and steradians as base units (now they're considered derived). The fact that they bothered to include radians and steradians is confusing to me. I usually think of the SI system as trying to be quite minimal. For example, there's no unit for square meters; you just square the meter.

Like... suppose you decided to measure square angles in terms of the surface area on a sphere swept by an arc of $x$ radians. Initially the area swept by $x$ would be close to $\pi x^2$, but as you approached $x \rightarrow 2\pi$ the swept area would not be increasing very fast anymore and instead of sweeping $\pi x^2 = \pi^3$ you end up sweeping $\pi 4$. Clearly this particular measure of angle area fails to simply be the square of an angle, since it ends up only increasing linearly due to wrapping around the sphere.

Another example. Suppose you defined an angle measure as "area divided by radius". This is a bad measure, since doubling the size of your sphere will double your measure. But it also wouldn't be radians squared.

Are steradians just area over radius squared, or is there something more? When can I confidently multiply measures in radians and say they are in steradians? Are they like meters, where multiplying any two values in meters gives a value in square meters (e.g. a perimeter times a diameter gives you a value measured in square meters, though it may not be the area of the shape)?

If I have an equation that multiplies two values measured in radians, is the result measured in steradians?

The radian measures an angle in the plane (2 dimensional space). The steradian measures a vertex in 3 dimensional space. If you had a pyramid, for example, you could use radians to measure each of the angles at the vertex, and steradians to measure the vertex as a whole.

To measure a vertex in steradians, you would imagine a unit sphere with the vertex at the center, and the measure the area of the sphere inside the vertex.

There is an alternative way to measure it. If the intersection of the vertex and the sphere forms a polygon. If you measure the angels and sum them, and compare that to what that number would be for a polygon with the same number of sides on a plane, the difference is the measure of the vertex.

If you had a regular tetrahedron. The measure of each angle at the vertex is $\frac \pi3$ radians, but the vertex is $3\cos^{-1} \frac 13 - \pi$ steradians.

• Right, but the thing I'm worried about is more basic. Suppose I have an equation that multiplies two values measured in radians. Is the result in steradians? Jul 6, 2016 at 17:08
• I wouldn't think so, but would have to look at the example. Radians (and steradians) are dimensionless. Jul 6, 2016 at 17:15
• Note: I realize that there may be a scalar conversion factor, and that it may differ by situation. That's fine. The same sort of thing happens if you try to multiply perimeter by diameter. You get a value in square meters, but if you wanted the computation to return the area of the shape instead of some random area you'd need to apply a shape-specific factor. Jul 6, 2016 at 17:22

This works for a solution (from wikipedia):

The solid angle of a cone with apex angle $2\theta$, is the area of a spherical cap on a unit sphere $$\Omega =2\pi \left(1-\cos {\theta }\right).$$ For small $\theta$ such that $\sin \theta \approx \theta$, this reduces to the area of a circle $2\pi\theta$. The above is found by computing the following double integral using the unit surface element in spherical coordinates: \begin{align} \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) \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) \end{align}

This means if you have an angle $x$, the solid angle created by sweeping out $x$ radians in two dimensions is $2 \pi (1-\cos(x/2))$, not $x^2$.

I was wondering the same thing because cameras give pixel IFOV in terms of radians, but the view of the pixel is actually a solid angle where the IFOV is swept out left to right and up to down.

• I'm not sure this answers the Question as asked. Angles, which are measured in radians, are dimensionless quantities even though they can be related to lengths of circular arcs on a circle of radius one. So while a solid angle might be measured as area cutoff on the surface of a sphere of radius one, it might not be edifying to claim that the latter dimensionless quantity is obtained by "squaring" the former dimensionless quantity. Sep 16, 2017 at 21:01
• Right but my answer shows that if you give an angle in radians, x, then corresponding solid angle created by sweeping by x in two orthogonal dimensions is $2 \pi (1-\cos(\frac{x}{2})$ not $x^2$ which is what I think the OP had in mind. Sep 17, 2017 at 15:02
• I take your point, but note the OP's Comment on the earlier Answer, "...but the thing I'm worried about is more basic. Suppose I have an equation that multiplies two values measured in radians. Is the result in steradians?" So I suspect your interpretation of what "the OP had in mind" is more particular (and sophisticated) than merely an issue of units of measure. Sep 17, 2017 at 15:11

I agree that Ω = 2π[1−cos(θ)] is correct for the solid angle subtended by a cone having half-angle θ. The angle θ can range from 0 (a point having zero solid angle) to π/2 (a hemisphere with Ω = 2π steradians) to π (the full sphere with Ω = 4π sr). In many imaging applications, θ is small -- perhaps π/10 (a 36 degree FOV) or less. Expanding the cos(θ) in a power series = 1 -θ2/2 + .... and keeping just the first term with θ gives Ω ≈ πθ2, which is the "angular area" subtended by the cone (compare this to Area = πr2). So, for sufficiently small angles, the product of two orthogonal angles in radians gives a solid angle in steradians.