Circular orbit in Schwarzschild coordinates This was an example in a general relativity textbook which I've been trying to work through myself.
A spaceship uses its rocket engine to maintain a circular orbit around a Schwarzschild black hole of mass $M$. In standard Schwarzschild coordinates, the orbit is characterised by fixed $r$ in the plane $\theta = \frac{\pi}{2}$, and a constant $\phi$ component of 4-velocity $u^\phi$.
The only other nonzero component of $u$ is $$u^t = (1 - \frac{2M}{r})^{-\frac{1}{2}} (1+r^2u_\phi^2)$$ and the only nonzero component of the acceleration 4-vector is $$a^r = \frac{M}{r^2} - (r - 3M)u_\phi^2.$$
I can get the first part of both equations but never the second half so I'm a bit stumped. Any help is much appreciated.
 A: The metric of the Schwarzschild black hole is
$$
ds^2=\left(1-\frac{2M}{r}\right)dt^2-\left(1-\frac{2M}{r}\right)^{-1}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2).
$$
Therefore, the Lagrangian of a particle in this background is (with proper time $\tau$)
$$
L=\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\tau}\right)^2-\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2-r^2\left(\left(\frac{d\theta}{d\tau}\right)^2+\sin^2\theta \left(\frac{d\phi}{d\tau}\right)^2\right)
$$
The first equation you can get from the constraint L=1 using $r(\tau)=const.$, $\theta(\tau)=\frac{\pi}{2}$. However, I get an additional square root compared to the expression you give, i.e.
$$
u^t=t'(\tau)=\sqrt{\frac{1+r^2 \phi''(\tau)^2}{1-\frac{2M}{r}}}.
$$
In order to get all equations you can just solve the Euler-Lagrange equations for $L$, i.e.
$$
\frac{d}{d\tau}\frac{\partial L}{\partial t'}-\frac{\partial L}{\partial t}=0\\
\frac{d}{d\tau}\frac{\partial L}{\partial r'}-\frac{\partial L}{\partial r}=0\\
\frac{d}{d\tau}\frac{\partial L}{\partial \theta'}-\frac{\partial L}{\partial \theta}=0\\
\frac{d}{d\tau}\frac{\partial L}{\partial \phi'}-\frac{\partial L}{\partial \phi}=0
$$
give you
$$
-\frac{4 M r'(\tau) t'(\tau)}{r^2}-2 \left(1-\frac{2 M}{r}\right) t''(\tau)=0\\
-\frac{2 M r'(\tau)^2}{(-2 M+r)^2}+\frac{2 M t'(\tau)^2}{r^2}-2 r (\theta'(\tau)^2+\phi'(\tau)^2)+\frac{2 r''(\tau)}{1-\frac{2 M}{r}}=0\\
2 r (2 (r'(\tau)\theta'(\tau)+r\theta''(\tau))=0\\
2 r (2 (r'(\tau)\phi'(\tau)+r\phi''(\tau))=0
$$
after plugging in $r(\tau)=r=const.$ and $\theta=\pi/2$. You see, that you can consistently set $r'(\tau)=0$, $\theta'(\tau)=0$, $\theta''(\tau)=0$ and then get the second equation you were looking for from the second Euler-Lagrange equation (with the help of the first equation you found). 
