Area of a triangle on a torus

Question

On a sphere the area of a trinagle is $A=\dfrac{\pi r^2}{180^{\circ}}\cdot(\alpha+\beta+\gamma-180^{\circ})$ where $\alpha, \beta, \gamma$ are the angles of the triangle.

Is there a similar easy formula for the torus?

Answer 1

There is an analogous formula, but it doesn't give the formula for the area of a (geodesic) triangle.

The formula for the sphere is a special cases of the celebrated Gauss–Bonnet Theorem, which relates a local property (curvature) with a global property (topology). A version of the theorem specialized to geodesic triangles $\Delta$---that is, surfaces with boundary whose boundary consists of three geodesics---is the following:

$$\alpha + \beta + \gamma = \pi + \int_\Delta K \,dA .$$

Here, $K$ is the Gaussian curvature of the triangle $\Delta$. In particular, if $K$ is constant on $\Delta$, the integral is $K \int_\Delta dA = K \cdot \textrm{area}(\Delta)$. So, if $K$ is constant:

• If $K \neq 0$, the area is given by $$\textrm{area}(\Delta) = \frac{\alpha + \beta + \gamma - \pi}{K} .$$ For a sphere of radius $R$, $K = R^{-2}$, which recovers the formula in the question statement.
• If $K = 0$, as is the case for the Euclidean plane, the equation specializes to the familiar statement that the sum of the measures of the angles of a triangle in the Euclidean plane is $\pi$. (Clearly we cannot recover $\textrm{area}(\Delta)$ from $\alpha, \beta, \gamma$ in this case, as similar triangles have identical angles but not necessarily the same areas.)

Another version of the Gauss–Bonnet Theorem implies that for any metric on a torus $T$, the integral of Gaussian curvature is $\int_T K \,dA = 0$, so if $K$ is constant, necessarily $K = 0$ (such a torus is called a flat torus). So, as for the Euclidean plane, the sum of the measures of the angles of any triangle in a flat torus is $\pi$, but we cannot determine the area of the triangle from the measures of its angles.

If $K$ is not constant (as is the case for any torus embedded in $\Bbb R^3$ carrying the induced metric) we cannot pull $K$ out of the integral, and consequently we cannot express the area of a triangle $\Delta$ in terms of $\alpha, \beta, \gamma$. Since $\int_T K \,dA = 0$, a torus with nonconstant curvature will have $K > 0$ at some points and $K < 0$ at others. If we pick a triangle on which $K$ is negative everywhere, our formula tells us that $\alpha + \beta + \gamma < \pi$, and if we pick a triangle on which $K$ is positive everywhere, $\alpha + \beta + \gamma > \pi$.

Answer 2

HINT:

It would be much more involved because there are two parameters ( meridian shift $c$ along radius and radius $a$).

The Gauss-Bonnet theorem provides a basis to find the area $A$ in the manner outlined.. hope there are no errors.

The area $A$ has two triangular regions/parts of a curvilinear triangle enclosed by three geodesics and a meridian of a circular torus; $(k_g , \chi )$ vanish due to geodesics and topology.

$$ \int \int K dA + [3 \pi -(\alpha+\beta+\gamma)] =0 $$

The square brackets above indicate sum of external angles at triangle vertices

$$ \int \int K dA = (\alpha+\beta+\gamma) - 3 \pi $$ $$ r=(c+a\cos \phi); ~ dA = a ~rdr ~d\theta/2 ~$$

$$ \frac{1}{2a}\int\int { \cot \phi ~dr ~d \theta} = (\alpha+\beta+\gamma) - 3 \pi $$

further integrating two triangular regions/parts of a curvilinear triangle enclosed by three geodesics on a circular torus between three $\theta,\phi $ limits gives further guidance to the problem when $r=f(\theta)$ geodesic relation is also derived as per Clairaut's Law ....