greg@harvard.UUCP (Greg) (12/14/85)
Somone posted a long time ago asking about geodesics on a circular ellipsoid, and whether or not they had anything to do with elliptic integral. The answer is yes. Here is how you do it: I will use the coordinates phi and theta, where: x = a*cos(phi)*sin(theta) y = a*sin(phi)*sin(theta) z = b*cos(theta) where a does not have to equal b. Let c^2 = b^2 - a^2. The metric is given by ds^2 = dx^2 + dy^2 + dz^2 = (a^2+c^2*cos(theta)^2)*dtheta^2 + a^2* sin(theta)^2*dphi^2, where s is the distance along the geodesic. We know that because rotation about the z-axis is a symmetry, L = a^2*sin(theta)^2 * dphi/ds is a conserved quantity along the geodesic. We have: 1 = (a^2+c^2*cos(theta)^2)*(dtheta/ds)^2 + a^2*sin(theta)^2*(dphi/ds)^2 = (a^2+c^2*cos(theta)^2)*(dtheta/ds)^2 + (L/a)^2/sin(theta)^2 If u = cos(theta), du = -dtheta * sin(theta), so the above simplifies to: 1 = (a^2 + c^2*u^2)*(du/ds)^2/sin(theta)^2 + (L/a)^2/sin(theta)^2, or 1-u^2 = (a^2 + c^2*u^2)*(du/ds)^2 + (L/a)^2. With a little more algebra, one gets: du*(a^2+c^2*u^2)/sqrt((1-(L/a)^2-u^2)*(a^2+c^2*u^2)) = ds For the solution, then, we must integrate both sides. I think that the right side can be reduced to an elliptic integral, but since elliptic integrals are really messy, and since I don't know them very well, I don't care to work out the solution any farther. -- gregregreg