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I'm trying to find the angle between two sides drawn from center for a tetrahedron.

I assumed that each face makes solid angle $ \pi$ at the center and angle between three sides is equal. So I did the following $$\int_0^x\int_0^x \sin \theta \; d\theta \;d\phi = \pi $$ I am assuming that the surface is equal to solid angle in this figure if radius is 1 enter image description here

I am getting $x = 2.094 \approx 120 $, degrees. I should be getting close to 109.

Furthermore I'm trying to generalize it to any solid angle made by three sides having equal angle between them. Can you suggest what assumption did I make wrong? Or taking tetrahedron as particular example was incorrect?

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I don't understand the "So". How is making the integral go up to $x$ in both angles related to the (correct) statement preceding the integral? –  joriki Oct 5 '12 at 20:36
    
@joriki I assumed both $\theta$ and $\phi$ should have the same value. –  Monkey D. Luffy Oct 5 '12 at 20:38
    
I see that you assumed that, but I don't see why. –  joriki Oct 5 '12 at 20:44
    
@joriki shouldn't $\iint \sin \theta \; d\theta \;d\phi $ give the surface area when radius equal 1? –  Monkey D. Luffy Oct 5 '12 at 20:48
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Of course I have an idea for it; I thought the point of the question was to understand why your idea gives the wrong answer. If you're looking for an approach that gives the right answer, put the vertices of the tetrahedron at $(4,0,0,0)$, $(0,4,0,0)$, $(0,0,4,0)$, $(0,0,0,4)$ and the centre at $(1,1,1,1)$ and calculate the cosine from the dot products of the differences. –  joriki Oct 5 '12 at 21:05

1 Answer 1

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Here are two flaws: if one vertex of the tetrahedron is at the north pole ($\theta=0$), and another vertex is located at $(\theta,\phi)= (x,0)$, then it is not the case that a third vertex is at $(x,x)$. The difference in "longitude" between two points on the sphere is not the angle between them, even if they are on the same latitude; the difference in longitude is instead the angle their projections onto the $xy$-plane make.

In addition, lines of latitude are not great circles: the region given by $0\leq \phi\leq 2\pi/3$ and $0\leq \theta \leq x$ will never be the projection of one face of the tetrahedron onto the sphere.

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