Borehole Oscillators

Summary

Borehole Oscillators by Greg Egan There is a simple thought experiment in Newtonian gravity: drill a thin radial borehole all the way through a solid ball of uniform density, and drop a test particle into the hole, starting from rest at the very top. What happens? The result is that the particle (blue, in the image on the left) falls all the way through the borehole, comes to a halt at the opposite end, then falls back, undergoing simple harmonic motion (the same kind of motion as an idealised version of a weight bouncing on the end of a spring) with exactly the same period as another test particle (red) orbiting the ball in a circular orbit that grazes the surface. On this page, we will start by proving that result, but then we will also look at radial motion in the vacuum around the ball in Newtonian gravity, and examine how these systems work in General Relativity — featuring the famous Schwarzschild solution, but starring its much less famous cousin, the second Schwarzschild solution! Newtonian Radial Oscillators A solid ball of radius r and uniform density ρ has mass: m(r) = (4/3) π ρ r3 In Newtonian physics, the acceleration due to gravity from any spherically symmetrical arrangement of matter is directed towards the centre of symmetry, and it is the same as if all the matter that lies closer to the centre than the point where we are computing the acceleration was concentrated at the centre. (This result is known as the shell theorem, and it was proved by Newton himself.) So, anywhere inside our solid ball, the radial acceleration of a test particle is given by: r''(t) = –G m(r(t)) / r(t)2 = –(4/3) G π ρ r(t)3 / r(t)2 = –(4/3) G π ρ r(t) This is the same kind of equation as the one that governs a weight on a spring, where an object experiences a “restoring force” proportional to its displacement from some point. If the particle starts from rest at r = R at time t = 0, the solution is: rsolid(t) = R cos(ω t) where ω2 = (4/3) G π ρ This is easily checked by taki...