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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.

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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.

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    $\begingroup$ 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. $\endgroup$ – Kevin Murray 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. $\endgroup$ – Jerry Schirmer Oct 2 '14 at 18:39
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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.

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

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