Forces along equidistant particle paths (Q1876640)

Let \(M\) be a surface of constant nonzero curvature \(K\). An {inertial path} is a path along a geodesic such that the speed is constant (and then equal to the initial speed). A {constant speed path} is a path without tangential acceleration component. Let two particles, each of mass \(m\), travel along a constant speed path such that they move with constant speed \(v\) at a distance \(\frac{d}{2}\) from a central inertial path. The force required to keep each on its path is called {coupling force} and denoted \(F_{c}\). The main result of the paper is that \(F_{c}=2mv^{2}k\) where \(k=\root{2}\of{K}\tan \left( \frac{\root{2}\of{K}d}{2}\right) \) in the case of positive curvature and \(k=-\root{2}\of{-K}\tanh \left( \frac{\root{2}\of{-K}d}{2}\right) \) in\ the case of negative curvature.