Variational principle - how to solve for geodesic? I am given the variational principle:
$$\delta \int_{\lambda 1}^{\lambda_2} \left(\frac{ds}{d\lambda}\right)^2d\lambda=0$$
where $ds^2=-dt^2+e^{2\Psi}dr^2+r^2(d\theta^2+\sin^2(\theta)d\phi^2)$ with $\Psi$ only depending on $r$.
I want to use this to write the geodesic equation for $r$ with respect to $\lambda$.
My first problem is interpreting $$\left(\frac{ds}{d\lambda}\right)^2$$
So $ds^2$ is the line element, where $ds^2=g_{\mu\nu}dx^\mu dx^\nu$ and since we are, I imagine, working with timelike separated paths, we then have: $$ds = \sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$$
and hence I have:
$$\left(\frac{\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}}{d\lambda}\right)^2,$$
but what does this expression even mean?
Nothing seem to depend on $\lambda$? Furthermore I don't know what this has to do with the geodesic equation.
 A: The idea is to consider the distance between two time-like separate points and extremise,
$$S=\int ds $$
where $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$. We can choose an affine parameter $\lambda$ to parametrise the path, from which we can define an action,
$$S= \int d\lambda \, \sqrt{g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}.$$
Thus, $(ds/d\lambda)^2$ can be interpreted as the square of the four-velocity essentially. Applying the principle of stationary action, modulo some subtleties, one obtains the geodesic equations,
$$\frac{d^2 x^\mu}{ds^2} = -\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}.$$
For the metric,
$$ds^2 = dt^2 - e^{2\Psi(r)} dr^2-r^2\left(d\theta^2 + \sin^2 \theta \, d\phi^2\right)$$
the non-vanishing Christoffel symbols are $\Gamma^r_{rr} = \Psi'(r)$, $\Gamma^r_{\theta\theta} = -e^{-2\Psi}r$, $\Gamma^r_{\phi\phi} = -e^{-2\Psi}r\sin^2 \theta$, $\Gamma^\theta_{r\theta}= r^{-1}$, $\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$, $\Gamma^\phi_{r\phi} = r^{-1}$ and $\Gamma^\phi_{\theta\phi} = \cot\theta$ remembering they are symmetric in the two lower indices. You should be able to take it from here to find the geodesics.
