Finding 3-Sphere Christoffel connection coefficients using variational calculus, Sean Carrol problem I have A 3-Sphere with coordinates $x^{\mu} = (\psi,\theta,\phi)$ and the following metric:
\begin{equation}
ds^2 = d\psi^2 + \text{sin}^2\psi(d\theta^2 + \text{sin}^2\theta d\phi^2)
\end{equation}
I know how to get the connection coefficients using the metric derivatives etc, but I'm looking for a way to do this through calculus of variations. A problem in Sean Carroll (Exercises 3.11 question 8 a) Introduction to General Relativity suggested varying the following integral to find the connection coefficients:
\begin{equation}
I = \frac{1}{2}\int g_{\mu \nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{v}}{d\tau} d\tau
\end{equation}
So I have a lagrangian:
\begin{equation}
\mathcal{L} = \dot{\psi}^2 + (\text{sin}^2\psi) \dot{\theta}^2 + (\text{sin}^2\psi)(\text{sin}^2\theta)\dot{\phi}^2  
\end{equation}
Which I put into the Euler-Lagrange equation:
\begin{equation}
\frac{\partial}{\partial \tau}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}^\mu}\right) - \frac{\partial \mathcal{L}}{\partial x^\mu} = 0 
\end{equation}
Am I on the right track here? What is the strategy for relating this back to the connection symbols? The literature isn't too clear and I'm struggling to make the connection.
 A: I'll show you how to do this for the 2-plane in polar coordinates.  Once you work this out, it should be doable to work it out in your case.
You start with the metric
$$ds^{2} = dr^{2} + r^{2}d\theta^{2}$$
Since the geodesics of this metric (i.e., straight lines) minimizes distance, we know that the geodesics are an extremum of:
$$I = \frac{1}{2}\int ds \left({\dot r}^{2} + r^{2}{\dot \theta}^{2}\right)$$
We take the variation of this, and get
$$\delta I = \int ds \left({\dot r}\delta {\dot r} + r{\dot \theta}^{2} \delta r + r^{2}{\dot \theta} \delta{\dot \theta}\right)$$
Per our usual procedure, we want to vary with respect to the original variables and not their time derivative.  We also neglect the variation on the boundary, and assume that $\delta {\dot x} = \frac{d}{ds}\delta x$.  So, we integrate by parts, and we get:
$$\delta I = \int ds\left(\left(-{\ddot r} + r{\dot \theta}^{2}\right)\delta r + \left(-{\ddot\theta}r^{2} - 2r{\dot r}{\dot\theta}\right)\delta \theta\right)$$
Since the geodesic must be zero independently of the variations $\delta r$ and $\delta \theta$, we know that the terms inside of the parentheses must be independently zero, and we get:
$$\begin{align}
0 &= {\ddot r} - r{\dot \theta}^{2}\\
0 &= {\ddot \theta} + \frac{1}{r}\left({\dot r}{\dot \theta} + {\dot \theta}{\dot r}\right)
\end{align}$$
Now, we have this as a system of equations, and we remember that the geodesic equation, in terms of Christoffel symbols, is $0={\ddot x}^{a} + \Gamma_{bc}{}^{a}{\dot x}^{b}{\dot x}^{c}$, and we conclude that $\Gamma_{\theta \theta}{}^{r} = -r$, $\Gamma_{r\theta}{}^{\theta} = \Gamma_{\theta r}{}^{\theta} = \frac{1}{r}$, and that all others are zero.
A: The strategy is to recall the geodesic equation,
$$
\frac{d^2x^\lambda}{dt^2}+\Gamma^\lambda_{\,\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}=0\tag{1}
$$
From your Lagrangian, you'll end up with equations of the form
\begin{align}
\ddot{\psi}&=f(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\
\ddot{\theta}&=g(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\
\ddot{\phi}&=h(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\
\end{align}
to which you relate to (1) index-by-index.
