Angular momentum per unit energy in the Schwarzschild spacetime I am reading Chapter 9 of Hartle's Gravity book, and I'm completely stumped on the following thing. In equation 9.45 he claims that for circular orbits in a Schwarschild geometry:
$$\frac{l}{e}=\left(Mr\right)^{1/2}\left(1-\frac{2M}{r}\right)^{-1}\tag{9.45}$$
And my question is how the above equation comes about. 
l and e are derived from the killing vectors:
$$e=\left(1-\frac{2M}{r}\right)\frac{dt}{d\tau}$$
$$l=r^2 \sin^2\theta \frac{d\phi}{d\tau}$$
He hints that this equation (9.45) comes from the fact that the energy equals the effective potential:
$$e^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$
and that the radius of the orbit must correspond to the one that minimizes the effective potential:
$$r_{\rm min}=\frac{l^2}{2M}\left[1+\sqrt{1-12\left(\frac{M}{l}\right)^2}\right]$$
I get these last two equations, but I am lost on how they can be used to derive his 9.45. 
 A: Once you have introduce $e$ and $l$ the remaining equation of motion is
$$e^2 = \left(\frac{dr}{d\tau}\right)^2 +V(r) $$
with
$$ V(r) = \left(1-\frac{2M}{r}\right)\left(1+\frac{l^2}{r^2}\right)$$.
Being a circular orbit requires two things. First of all the radius must not change, i.e. $\frac{dr}{d\tau}=0$. Second, this condition must be maintained over time meaning that $\frac{d^2r}{d\tau^2}=0$ as well.
The first condition simply implies,
$$e^2 = V(r)$$.
This is one of the equations you already had.
The second can be obtained from the first equation above by taking the time derivative and applying the chain rule which leads to
$$ 0 = -2 \frac{d^2r}{d\tau^2} = V'(r) = \frac{M}{r^2} - 2 l^2\frac{r-3M}{r^4}$$.
That is, $r$ must be at the minimum of the the potential $V(r)$. Now instead of solving this equation for $r_{\rm min}$, as you did, solve it to find $l$,
$$l^2 = M\frac{r^2}{r-3M}$$.
Plugging this into the equation for $e^2$ gives
$$e^2 = \frac{(r-2M)^2}{r(r-3M)}$$.
The rest should be obvious.
