Sphere with spherical triangle and normals

Given: a unit-radius sphere: $x^2+y^2+z^2 = 1.$

Three unit vectors $A,B,C$ going out from center $O$ of the sphere. Vectors form three intersecting planes that form a spherical triangle with angles $\alpha,\beta,\gamma$ on the sphere surface.

Determine the vector function $F(\alpha,\beta,\gamma)$, or $F(A,B,C)$, which is a sum of external normals over the surface of a spherical triangle, normalized by the area.

The task is equivalent to find the flow (surface integral) of sphere normals vector field limited by spherical triangle per each axis $i,j,k$. Three integrals per each axis $i,j,k$; therefore $F$ - is a vector $<Ii,Ij,Ik>$, not scalar. Where $Ix$ - flow of ort $x$ per sphere surface, limited by spherical triangle. The main problem is to parametrize the spherical triangle to take the surface integral for arbitrary angles ($\alpha,\beta,\gamma$), or arbitrary coordinates ($A,B,C$).

This is a completely practical problem. If we split the unit sphere by some number of vectors from the center, we'll get several spherical triangles, which completely cover surface of the sphere. Each spherical triangle creates a normal flow by its own area. For the whole sphere this flow is zero because for each normal there is a negative normal on the other side of the sphere. The sum of the flows of all spherical triangles of whole sphere will also equals zero. But for each spherical triangle it would be some vector $F(\alpha,\beta,\gamma)$, or $F(A,B,C)$.


There is no need for any parameterization. We can use following lemma:

Let $\Omega$ be any region with piecewise smooth boundary $\partial \Omega$. Let $\vec{n}$ and $dS$ be the outward pointing normal and area element of $\partial \Omega$. We have: $$\int_{\partial \Omega} \vec{n} dS = 0$$

To prove the lemma, let $dV$ be the volume element for $\Omega$ and $\vec{k}$ be any constant vector. By divergence theorem, we have

$$\vec{k}\cdot \int_{\partial\Omega} \vec{n} dS = \int_{\partial\Omega} \vec{k}\cdot\vec{n} dS = \int_{\Omega} \nabla \cdot \vec{k}\,dV = \int_{\Omega} 0\,dV = 0 $$ Since this is true for all $\vec{k}$, the lemma follows.

Back to original problem. We will assume $A, B, C$ is ordered in a counterclockwise manner on $S^2$.

Set $\Omega$ to the convex hull formed by $O$ and the spherical triangle $ABC$. $\partial \Omega$ now consists of 4 pieces, the spherical triangle $ABC$ and 3 circular sectors $OBA$, $OCB$, $OAC$.

Apply above lemma, we find $$\begin{align} F(A,B,C) \stackrel{def}{=} \int_{ABC} \vec{n} dS &= -\int_{OBA} \vec{n} dS - \int_{OCB} \vec{n} dS - \int_{OAC} \vec{n} dS\\ &= \frac{\theta_{AB}}{2} \frac{\vec{A}\times\vec{B}}{|\vec{A}\times\vec{B}|} + \frac{\theta_{BC}}{2} \frac{\vec{B}\times\vec{C}}{|\vec{B}\times\vec{C}|} + \frac{\theta_{CA}}{2} \frac{\vec{C}\times\vec{A}}{|\vec{C}\times\vec{A}|} \end{align} $$ where $\begin{cases} \theta_{AB} &= \cos^{-1}\vec{A}\cdot\vec{B},\\ \theta_{BC} &= \cos^{-1}\vec{B}\cdot\vec{C},\\ \theta_{CA} &= \cos^{-1}\vec{C}\cdot\vec{A}. \end{cases}$

  • $\begingroup$ Perfect! Thank you very much! $\endgroup$ – Vladislav Volga Nov 5 '16 at 3:15
  • $\begingroup$ I had almost the same question, but didn't see this question as relevant because of the title (why "flow"? why no "normal vector"?). So I asked separately, and I did find an answer, that agrees with yours. math.stackexchange.com/questions/2666587/… $\endgroup$ – mr_e_man May 28 '18 at 5:42

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