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.


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.

  • 1
    $\begingroup$ Thanks so much, that helped a lot. I got confused by the integration variable (mistook if for time, and then got confused that I didn't have time as a coordinate) and didn't realize I could associate it with the affine parameter in the geodesic equation. $\endgroup$ Oct 2 '14 at 17:28
  • $\begingroup$ @KevinMurray: no problem! Also, as a bonus that I intended to include above and forgot, notice that all of the variation with respect to $\theta$ is due to variation with respect to $\dot \theta$. Therefore, when you integrate with parts, it should be clear that $\frac{d}{ds}\left(r^{2}{\dot \theta}\right) = 0$, which means that $r^{2}{\dot \theta} = C$ for some constant value $C$ on your geodesic. This is related to the fact that $\partial_{\theta}$ is a killing vector of the 2-plane. That trick can make the task of actually solving for geodesics go a lot more quickly. $\endgroup$ Oct 2 '14 at 18:39

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.

  • $\begingroup$ Love the concision :D $\endgroup$ Dec 27 '15 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.