We know that the area of an spherical triangle (in a unit sphere) is given by $A(\triangle) = \alpha + \beta + \gamma - \pi$, where $\alpha$, $\beta$, and $\gamma$ are the interior angles of the spherical triangle.

I would like to see how plane (Euclidean) geometry works as a limit when the radius of the sphere goes to infinity. Clearly the curvature of the sphere $1/r$ becomes zero and a sphere turns into a plane. What happens to the area of the triangle?

If we say that the area of the triangle is \begin{equation} A(\triangle) = r^2 [(\alpha + \beta + \gamma) - \pi] \end{equation} clearly $\alpha + \beta + \gamma - \pi$ go to zero, but not at the rate that $r^2$ goes to infinity. It seems that this limit is infinity.

There seems to me that we can not find something like $b h/2$ (base times height over two) from spherical geometry. Right?

Of course objects become amplified in area by $r^2$ or length by $r$ so we would need to have something to pull them back.


Update: One way to pull back is to think that the actual arc lengths of the stretched triangle segments are $a=r \alpha$, $b=r \beta$, and $c= r \gamma$, so we can pull one $r$ inside the formula above and have

\begin{equation} A(\triangle) = r [(a+b+c) - \pi r] \end{equation}

where now $a,b$, and $c$ are the actual lengths of the sides. Pulling $r$ inside again shows me the area of a circle and... it seems that we better point toward

Heron's formula

and forget about base x height/2. Heron's formula is fine to me.

  • $\begingroup$ Wow! you are fast! I just sent this. Thanks for the tip. $\endgroup$ – Herman Jaramillo Nov 17 '15 at 23:18
  • $\begingroup$ @HermanJaramillo Normal Human is (mostly) a bot, the comment will be deleted automatically in a few minutes. $\endgroup$ – Travis Nov 17 '15 at 23:19
  • $\begingroup$ The arc lengths on the surface of the sphere are not $r \alpha$, $r \beta$, and $r \gamma$. This would be the case if all three angles were the angles subtended by the arcs, but $\alpha$, $\beta$ and $\gamma$ are the angles between the arcs in the formula you gave. See here, for example. $\endgroup$ – Michael Seifert Nov 18 '15 at 18:10
  • $\begingroup$ @Michael Selfert : The angle $\alpha$ is by definition the dihedral angle which is formed by the two lunes (or better, the two planes) which form the vertex where $\alpha$ is assigned. So even though in the figures shown in books and websites $\alpha$ looks like an angle in the surface, it goes the same all the way to the center of the sphere. $\endgroup$ – Herman Jaramillo Nov 18 '15 at 18:11
  • $\begingroup$ @MichaelSeifert : I believe you are right. The letters $a$, $b$, and $c$ which measure the arc length of each side of the spherical triangle are also the angles of the intersection with the plane having those segments and the sphere, if the sphere is unitary. If the sphere grows as $r$ in the radial direction, then those angles shrink if we want to preserve the arc lengths of $a$, $b$, and $c$. Thanks. $\endgroup$ – Herman Jaramillo Nov 18 '15 at 21:32

I found a connection here. We need to use Cagnoli's Theorem

That is, given excess $E=\alpha+\beta+\gamma-\pi= A(\triangle)$ Cagnoli's Theorem establishes that:

\begin{equation} \sin \frac{E}{2} = \frac{\sqrt{ \sin s \sin (s-a) \sin (s-b) \sin(s-c)}}{ {2 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}}} \end{equation}

Then as we write the trigonometrical functions as Taylor series:

\begin{eqnarray*} \sin x &=& x - \frac{x^3}{3} + H.O.T. \\ \cos x &=& 1 - \frac{x^2}{2} + H. O. T. \end{eqnarray*}

When $r \to \infty$ we get an asymptotic solution by retaining only the leading order terms here. That is

\begin{equation} \lim_{r \to \infty} \sin \frac{E}{2} = \lim_{r \to \infty} \frac{E}{2}. \end{equation} (note that the radius $r$ is implicit on these equations, in addition when $r \to \infty$, $E \to 0$ since the excess will be nothing once you get from a sphere to a plane.). Similarly for the right hand side term, as $r \to \infty$ we find

\begin{equation} \lim_{r \to \infty} \frac{\sqrt{ \sin s \sin (s-a) \sin (s-b) \sin(s-c)}}{ {2 \cos \frac{a}{2} \cos \frac{b}{2} \cos \frac{c}{2}}} = \lim_{r \to \infty} \frac{\sqrt{s(s-a)(s-b)(s-c)}}{2} \end{equation} Please observe that for very large $r$ the interior angles $a$, $b$, and $c$ become really small and so their semi-perimeter $s$. That is, if $r=1$, then $a$, $b$, and $c$ are simultaneously lengths of the triangle sides (arc segments) and central angles in the sphere. If we want to increase $r \gg 1$, then, in order to preserve the size of the segments, we need to shrink the angles by a factor of $1/r$, so that the central angles $a$, $b$, and $c$ shrink to zero, but the length of the segments $a$, $b$, and $c$ remain constant. Hence there is a duality on the meaning of the symbols $a$, $b$, and $c$. As arguments of the sine and cosine functions, they are angles in radians, but as lengths they are the fixed lengths for $r=1$ and they should be preserved as the sphere explodes. What I call "pull back" in my question is this shrinking of $a$, $b$ , and $c$ as central angles of the sphere to keep the arc lengths of $a$, $b$, and $c$ constant.

From the previous two equations we find that in the limit as $r \to \infty$:

\begin{equation} E = \sqrt{s (s-a) (s-b) (s-c)}. \end{equation}

which is Heron's formula. The conversion from here to base x altitude/2 should be a common problem solved elsewhere.


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