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There is a formula of to calculate the angle between two normalized vectors:

$$\alpha=\arccos \frac {\vec{a} \cdot\ \vec{b}} {|\vec {a}||\vec {b}|}.$$

The formula of 3D solid angle between three normalized vectors is (borrowed from wikipedia):

$$\alpha = 2\arctan \frac{|\vec {a}\ \vec {b}\ \vec {c}|}{|\vec{a}| |\vec{b}| |\vec{c}| + (\vec{a} \cdot \ \vec{b})|\vec{c}| + (\vec{a} \cdot \vec {c})|\vec{b}| + (\vec{b} \cdot \vec{c})|\vec{a}|}.$$

How to figure out a formula of solid angle between $n$ normalized vectors in $n$-dimensional space?


OK, I've found out the following formula for the angle between two vectors in 2d space through the $\arctan$:

$$\alpha= 2\arctan \frac {|\vec{a} \wedge \vec {b}|} {|\vec{a}| |\vec{b}| + \vec{a}\ \cdot\ \vec{b}}$$

How can I apply this $\arctan$ formula for the 4D space, for example?

For more detail, I explain my assumptions.
The solid angle of orthogonal basis in 4D space must be $\alpha = \frac {2 \cdot \pi ^ 2 \cdot R^3} {2 ^ 4} = \frac {\pi ^ 2 R^3} {8} $. Here we can see a $\pi^2$ factor. Does this mean that the 4D solid angle formula contains multiplication of two $\arctan$ there is one $\arctan$ in this formula? I think yes, but still have some difficulties with such formula inference.

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Your second formula is quite awkward, because you apply arctan against a vector. – shuhalo Nov 2 '12 at 21:14
@Martin, you are right. Sorry for my mistake. I meant Exterior product there and I've already fixed my answer with it. – KvanTTT Nov 4 '12 at 9:13
up vote 3 down vote accepted

See the paper by Ribando, "Measuring Solid Angles Beyond Dimension Three", published in Discrete & Computational Geometry 2006. An electronic version may be found here:

It seems there is no closed formula for solid angle in dimemson > 3, but a multi-variable Taylor series is given in Theorem 2.2 there. Later in this paper its radius of convergence is discussed.

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Solid angles are a little more complicated than angles. From wolfram's mathworld:

The solid angle subtended by a surface is defined as the surface area of a unit sphere covered by the surface's projection onto the sphere.

Did you notice the phrase 'surface area'? We're going to need some integrals!

If the surface is called $S$, then the solid angle $\Omega$ is given by:

$$\Omega = \int_S \hat{\mathbf{n}}/\mathbf{r}^2 \cdot d \mathbf a $$

where $\hat{\mathbf{n}}$ is the unit vector from the origin, $d \mathbf{a}$ is the infinitesimal area of a path of a surface from $S$, and $r$ is the distance from the origin to the patch of surface.

See here for examples, and more info.

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I saw this formulas. But i don't know how to implement their to vectors case. – KvanTTT Sep 26 '12 at 11:44
Is the issue that you don't know surface integrals yet? – Maus Sep 26 '12 at 17:58

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