Whenever two massive bodies interact via gravity it is useful to characterize the resulting motion. For example, is the situation more akin to scattering? Will an orbit result? If so, what shape? For both Newtonian gravity and general relativity circular orbits can result. In this section we explore the conditions for this to happen in both classical mechanics and general relativity.
For a classical particle subject to Newtonian gravity, one begins with the force equation
Fm = −GmM r/r3. (6.1)
This is a relatively straightforward equation to solve. What about for general relativity? In general relativity the force equation is even easier:
Fm = 0, (6.2)
since in general relativity gravity does not create a force on objects. Instead gravity warps (curves) space-time, and space-time's curvature is responsible for particle motion. If we are to look for something comparable to Eq. (1) which gives us an equation of motion, we must look at the geodesic equation. The geodesic is the shortest path in space-time for a particle to follow.
For general relativity, one can also determine the energy per mass, E/m, for an orbiter. It can be shown that:
(E/m)2 = (dr/dt)2 + (1 − 2M/r)[1 + (L/m)2/r2] , (6.3 --- EBH 30, p 4-15)
where we can assign U/m = [1 − 2M/r]1/2 [1 + (L/m)/r2]1/2 as the effective potential. This is the U shown in our simulations.
For circular motion to occur, dr/dt = 0, and hence
(E/m)2 = (1 − 2M/r)[1 + (L/m)2/r2]. (6.4)
In this formalism, for what values of r, L/m, and E/m do we get circular motion? For circular motion we must have a minimum in the effective potential energy diagram and the particle must be situated at that radius with that energy. If the particle's radial position and energy correspond to the minimum of the effective potential energy function, it cannot deviate in r and hence is in circular motion. To determine the minimum of the potential energy function one takes the derivative with respect to r and solves for the r that minimizes the derivative of this function:
d(U/m)/dr = [2U/m]-1 [− (L/m)2/r3 + M/r2 + 3M(L/m)2/r4 ]= 0, (6.5)
which yields:
r = (L/m)2/2M [1 +/− {1 − 12[Mm/L)]2}1/2], (6.6)
which gives a minimum value (the minus sign) and a maximum (the plus sign) for the orbit. For circular motion we only have the minimum radius.
ro = (L/m)2/2M [1 − {1 − 12[Mm/L]2}1/2]. (6.7)
The smallest possible value occurs when L/m = (12)1/2 M. which yields ro min = 6M.
This simulation shows the motion of objects in orbit about a central mass. You can drag particles to a new position. You can also double-click the particle or click on the U(r) check-box and see the effective potential that it experiences. Along with the plot of the effective potential, you can see the values of the parameters describing the motion: position, energy, angular momentum, radial and angular speeds.
Run the simulation by double-clicking the green arrow, do not press the Run button yet. This simulation shows an orbiter in the innermost stable circular orbit (ISCO), r = 6M, around a black hole of mass M = 1. E/m and L/m are displayed on the simulation, and by clicking in the checkbox the effective potential energy diagram, U/m vs. r, is shown as well. Recall that t is far-away time and that r and φ are the r- and φ-coordinates, respectively.