I know that the Schwarzschild line-element is:
$$ds^2=c^2\left(1-\frac{2\mu}{r}\right)dt^2-\left(1-\frac{2\mu}{r}\right)^{-1}dr^2-r^2d\theta^2-r^2\sin^2\theta d\phi^2$$
Dividing by $d\lambda^2$, where $\lambda$ is an affine parameter, I get:
$$\left(\frac{ds}{d\lambda}\right)^2=c^2\left(1-\frac{2\mu}{r}\right)\dot{t}^2-\left(1-\frac{2\mu}{r}\right)^{-1}\dot{r}^2-r^2\dot{\theta}^2-r^2\sin^2\theta \dot{\phi}^2$$
where $\dot{X}$ is $\frac{dX}{d\lambda}$.
To my understanding, this is the Lagrangian, so Euler-Lagrange equations hold.
So I must have:
$$\frac{d}{d\lambda} \left( \frac{\partial L}{\partial \dot{\phi}} \right)=\frac{\partial L}{\partial \phi}$$
Evaluating LHS: $-2r^2\sin^2\theta\ddot{\phi}$.
RHS: $\frac{\partial L}{\partial \phi} = 0$.
Ie: $\ddot{\phi}=0$
Geodesic equation for $\phi$:
$$\frac{d\phi^2}{d\lambda^2}+\Gamma^{\phi}_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^nu}{d\lambda}=0$$
So I conclude that $\Gamma^{\phi}_{\mu\nu}$ for any $\mu, \nu$.
This, however, is in conflict with what I find on the internet. Equation 5.4 from here, Wikipedia, disagrees with this result.
What am I doing wrong?
