Cone is one scenario to think about, but what it really comes down to is the fraction of the full sphere that your solid angle includes.
To make a 2D analogy, you can think of 2D angles as a portion of a circle represented as a fraction of 2pi radians. Cut it in half and you have pi radians, cut that in half (so a quarter of the circle) and you have pi/2 radians, or 90 degrees.
That 90 degrees "quarter of a circle" example is looking at it as the "point of a pie slice", but it's the same 90 degree 2D angle as you have in the corner of a square.
You can look at the corner of a cube the same way. The full sphere of solid angle is 4pi, a hemisphere is 2pi. Now take that hemisphere and cut it into quarters (1/8ths of a sphere). Each of those quarters is pi/2 steradians, and the solid angle at the center is the same solid angle represented by 1/8th of the sphere is the same solid angle you have at the corner of a cube.
Or to put it another way, you could pack the corners of 8 cubes around a point and it would leave no empty gaps, so the corner of each cube is occupying 1/8th of a "full" 4pi steradians.
Now that I've thought about it, I don't think we're on solid ground measuring pointiness in n-dimensional radians. A perfectly non-pointy angle in 2D is half the circle, or π radians. A perfectly non-pointy angle in 3D is half the sphere, or 2π steradians. π is less than 2π, but the straight-line 2D angle is not sharper than the flat-plane 3D angle. What's relevant to the pointiness is the proportion of the sphere, not the numeric value of the surface area.
It corresponds to area if and only if you're talking about a unit sphere.
In the same way that an angle of pi radians corresponds to pi length around the perimeter of a unit circle, but if you're measuring on the perimeter of a larger circle you'll have a correspondingly larger length.
The "4pi steradians in a sphere" comes from the area of a unit sphere (4×pi×r^2 = 4×pi). I've never had cause to use 4+ dimensional solid angle for anything, but you'd determine full n-dimensional radians the same way, I guess with a 4-dimensional sphere the "surface area" is a 3D volume on the perimeter of the 4D unit sphere's space?
> In the same way that an angle of pi radians corresponds to pi length around the perimeter of a unit circle, but if you're measuring on the perimeter of a larger circle you'll have a correspondingly larger length.
I understand this. It's not what I'm talking about.
There was a claim upthread that the corner of a square and the corner of a cube are equally sharp because (1) one of them is π/2 radians; (2) the other is π/2 steradians; and (3) π/2 is equal to π/2, hence equally sharp.
I'm saying I don't think this can be right, because when we consider two angles that we know are equally sharp, the 2D flat line and the 3D flat plane, one of them is π and the other is 2π, so unit-free radian measurements cannot be a valid measure of sharpness. The radian measurement needs to be normed against the full sphere in an equal number of dimensions, so that the flat line equals 1/2 (of a circle) and the flat plane equals 1/2 (of a sphere). I want to conclude that these two angles are equally sharp because 1/2 is equal to 1/2. But even if I can't do that, I need to avoid concluding that a flat line is twice as sharp as a flat plane, because I know that's untrue.
But my reasoning would tell us that the corner of a cube is twice as sharp as the corner of a square, 1/8 to 1/4.
Ah gotcha. I'm not even sure it's meaningful to compare the "relative pointiness" of angles between different dimensions. It's like trying to compare whether the area of a circle is bigger or smaller than the volume of a sphere. And then what if I only have 70% of a circle, but 80% of the sphere?
You could make up some metric like "the unit sphere occupies a greater volume fraction of the unit cube compared to the area fraction of a unit circle in a unit square", and compare like that if you only have some fraction of each circle/sphere instead of the whole thing, but is there any value in making that comparison?
> I'm not even sure it's meaningful to compare the "relative pointiness" of angles between different dimensions.
I think this is actually something we can't avoid. The problem with dodging the question this way is that it's really, really easy to translate a low-dimensional angle up into a higher-dimensional space.
Imagine I have two 2D vectors, U and V, with an angle between them. I can think of them as representing an infinite 1D surface consisting of the points (m·u_1 + n·v_1, m·u_2 + n·v_2) for all nonnegative real m and n.
It's trivial to define a 2D surface in 3D space representing exactly the same angle: it is the points (m·u_1 + n·v_1, m·u_2 + n·v_2, a) for all nonnegative real m and n and all real a. When the angle between U and V is θ radians, this surface will always express an angle of 2θ steradians. I don't think it's a stretch to say that the two angles, θ radians and 2θ steradians, must be equally sharp. (And indeed, the flat-line / flat-plane example is a special case of this one.)
I need to correct myself: I've defined my surfaces as linear combinations of the vectors forming the angle, which is wrong. Rather, the 1D surface expressing the 2D angle consists of the points in the set (mU ∪ nV) for all nonnegative m and n, and the 2D surface extending the angle into 3D space consists of the set [(m·u_1, m·u_2, a) ∪ (n·v_1, n·v_2, a)] for all nonnegative m, n and all a.
To make a 2D analogy, you can think of 2D angles as a portion of a circle represented as a fraction of 2pi radians. Cut it in half and you have pi radians, cut that in half (so a quarter of the circle) and you have pi/2 radians, or 90 degrees.
That 90 degrees "quarter of a circle" example is looking at it as the "point of a pie slice", but it's the same 90 degree 2D angle as you have in the corner of a square.
You can look at the corner of a cube the same way. The full sphere of solid angle is 4pi, a hemisphere is 2pi. Now take that hemisphere and cut it into quarters (1/8ths of a sphere). Each of those quarters is pi/2 steradians, and the solid angle at the center is the same solid angle represented by 1/8th of the sphere is the same solid angle you have at the corner of a cube.
Or to put it another way, you could pack the corners of 8 cubes around a point and it would leave no empty gaps, so the corner of each cube is occupying 1/8th of a "full" 4pi steradians.