# 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.

• This is for exam prep – grE Jun 7 '17 at 6:57
• The curve through spacetime is parametrized as $x^\mu=x^\mu(\lambda)$. The parameter $\lambda$ can be thought of as time for time like paths. In that case the quantity with parentheses is nothing more than the square of the four velocity. – Damian Sowinski Jun 7 '17 at 7:06
• Related: physics.stackexchange.com/q/149082/2451 and links therein. – Qmechanic Jun 7 '17 at 7:42

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.

• Downvoted because this is wrong: Applying the Euler-Lagrange equations, one obtains the geodesic equations. You don't obtain the geodesic equation in its usual form. This has to do with reparametrization invariance of the action. You have to explicitly apply additional gauge fixing conditions to recover the usual geodesic equation. – Prof. Legolasov Jun 7 '17 at 13:38
• @SolenodonParadoxus Yes, it's the same subtlety for the Nambu-Goto and Polyakov actions, but what I said is not outright wrong. Maybe I should say one applies the principle of stationary action. – JamalS Jun 7 '17 at 14:42
• Could you add this so I can upvote your answer? :) – Prof. Legolasov Jun 8 '17 at 3:59
• @SolenodonParadoxus The modification has been made and is sufficient for you to at least remove the downvote, as you cannot say anything is wrong at this point. Whether you upvote or not is not my concern. – JamalS Jun 8 '17 at 5:32