# Circular orbits in general relativity(GR)

Reading about Schwarzschild geodesics, I found that circular orbits are possible when the effective force

$$F=-\frac{dV}{dr}=-\frac{\mu{c^{2}}}{2r^{4}}\left(r_{s}r^{2}-2a^{2}r+3r_{s}a^{2}\right)=0$$

However, it was later suggest to me by a friend that in standard GR, the force only has the term proportional to $1/r^{2}$ yet circular orbits still exist for $r>0$.

My question is: Does circular orbits only happen when $F=0$, if yes then, how can they exist in $F=1/r^{2}$?

Also in standard general relativity($F=1/r^{2}$) are the orbits circular(i.e. not elliptical).

• Circular orbits happen not when $F=0$, but when $F=r \omega^2$, the centripetal force needed to keep the orbiting object turning in a circle. The first $F$ (GR) probably includes a fictitious force, while the second $F$ (classical) does not. – 2012rcampion Jun 15 '15 at 13:59
• When you say "standard GR," do you really mean classical/Newtonian mechanics? – 2012rcampion Jun 15 '15 at 14:03
• @ 2012rcampion, by standard I mean not Schwarzschild gravity – MrDi Jun 15 '15 at 14:05
• There is no "not Schwarzschild" gravity in GR... $F_G=G\frac{m_1 m_2}{r^2}$ is classical, non-relativistic gravity. – 2012rcampion Jun 15 '15 at 14:08
• @2012rcampion, you're right, but how does circular orbits happen only when $F=0$ not when as you say, $F=r\omega^{2}$. Also does circular also include elliptical orbits too. – MrDi Jun 15 '15 at 15:09

The $F$ you've found above is not the "real" force in GR, but rather an "effective force" derived from an "effective potential". Basically, we can use the fact that there are multiple constants of the motion to reduce the full three-dimensional problem down to an equivalent one-dimensional problem. This procedure works in Newtonian gravity1 as well, as follows:
Let's imagine a particle moving under the influence of an attractive inverse square force ($F_r = -\gamma/r^2$, with $\gamma > 0$) around a fixed center. If we write down Newton's Laws in polar coordinates, they become $$F_r = m \ddot{r} - m r \dot{\phi}^2 \quad \Rightarrow \quad m \ddot{r} = -\frac{\gamma}{r^2} + m r \dot{\phi}^2 \tag{1}$$ $$F_\phi = m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} \quad \Rightarrow \quad m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} = 0 \tag{2}$$ Now, Eq. (2) is equivalent to $$\frac{d}{dt} \left( m r^2 \dot{\phi} \right) = 0,$$ which implies that $m r^2 \dot{\phi} = \ell$ is a constant; in fact, it is the angular momentum of the particle. We can then use this fact to eliminate all references to $\phi$ from equation (1): $$m \ddot{r} = -\frac{\gamma}{r^2} + \frac{\ell^2}{m r^3}. \tag{3}$$ This means that we have reduced our original system of equations (1) and (2) down to a single ODE. The tactic would then be to solve this equation for $r(t)$, and then (if we want) solve for $\phi(t)$ using the equation $\dot{\phi} = \ell/m r^2(t)$.
Unfortunately, this is pretty tricky to do exactly, but we can get some intuition by defining the effective force $F_\text{eff}$ as the right-hand side of (3), and the effective potential $U_\text{eff}$ as $$U_\text{eff}(r) = -\frac{\gamma}{r} + \frac{\ell^2}{2 m r^2}$$
Notice that $F_\text{eff} = dU_\text{eff}/dr$, so the radial motion of this particle is exactly the same as that of a particle moving in a one-dimensional potential $U_\text{eff}(r)$. This means we can get a sense of what the particle will be doing (radially) by sketching $U_\text{eff}$. You can see some of these graphs at the bottom of this page. The effective potential always has a minimum; this corresponds to $r$ being a constant with respect to time, i.e., circular motion. There will also be motions of the "effective particle" that oscillate between a minimum and a maximum value; these correspond to elliptical orbits. Finally, there will be motions of the "effective particle" where the particle comes in from $r = \infty$, gets to some minimum value, and then goes back out to $r = \infty$; these would be the unbounded orbits (parabolae and hyperbolae.)